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10200 | https://www.youtube.com/watch?v=bjJoZCZJP3Q | Q2 Count two digit numbers where digits cannot repeat
Anil Kumar
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Posted: 18 Sep 2017
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a monokuma sharing with you questions where we will understand organized counting factorials and permutations question number two how many two-digit numbers can be formed with following restrictions where digits cannot be repeated right so we have a restriction here where digits cannot be repeated okay yes using digits 1 2 9 and B is using digits 0 to 9 so use for such questions we can actually since the digits are not being repeated we can apply permutations also right let me begin by simple counting principles the fundamental counting principle so what we will do here is two digits let's say these are the two places once and tens we have to form two digit numbers from numbers 1 2 3 4 5 6 7 8 9 so we got 9 numbers in all now this place tens place can be filled by nine numbers since these numbers not being repeated so whatever you felt here cannot be filled in the ones place so we're left with 8 right now using the product rule the combinations could be 9 times 8 which is 72 you see that so that is how we can get the answer by counting principles right so this is your fundamental counting principle so if there are nine ways to do one thing followed by eight ways of the same thing right in that case the number of combinations will be their product that's the fundamental counting principle using multiplication counting principle is it ok now with permutations also you can do it we have nine numbers Suvir nine numbers nine P and we have to select two nine P 2 is the solution right 9 P 2 means what 9 P 2 means 9 factorial divided by 9 minus 2 factorial which is same as 9 factorial divided by 7 factorial which is same as 9 times 8 is 72 is it okay so that is how you can do it correct now let's do Part B you can always pause the video answer the question and then look into mice ratios so we have to form two-digit numbers using numbers 0 to 9 so 0 to 9 means we have 10 digits is it ok now in the first place you cannot place 0 because then it becomes one digit number so still only nine numbers can be placed here but once you place a digit here maybe five you cannot place the same digit in ones place so you're left with nine digits 0 to 9 excluding what you are placed in tens places are okay so total numbers are 9 times 9 which is 81 okay so so you can form 81 numbers using digits 0 to 9 which are all two-digit numbers do you see that ok using permutation also you could have done correct so now in permutation the the idea here is that you are selecting 10 from ten digits right you are selling Hollywood scene using function will do like this we have ten numbers and we are to select two how will take away those we start with zero since if you start with zero you get a one digit number is set okay so B minus nine p1 do you see that this is because you take away numbers starting with zero is okay you do so if I play a zero here if I play zero here that is to say what we do here is that we place ten digits here right ten and nine here and take away those where we start with zero right way we start with zero is that okay so there's zero is fixed so here we have nine options so we start with zero I should have written start with zero nine options right so you just fix this so that becomes 10 P 2 minus 9 P 1 correct so 10 P 2 is 10 factorial over 10 minus 2 factorial minus 9 P 1 is 9 factorial divided by 9 minus 8 vector incorrect and which is 10 times 9 minus 9 is it okay and that gives you the same answer over just 90 minus 9 as 81 correct this a bit tricky using permutations counting principles give you a very straight answer and in general you will see that when working with numbers counting principles is a good way of going about Amal Kumar and I hope that helps you can always share and subscribe my videos feel free to post questions and if you like play some likes thank you and all the best |
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How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
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$\begingroup$
How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.
This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.
calculus
limits
trigonometry
limits-without-lhopital
edited Jun 17, 2020 at 8:34
José Carlos Santos
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asked Oct 23, 2011 at 16:21
FUZxxlFUZxxl
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$\begingroup$ l'Hôpital's rule is easiest: $\lim\limits_{x\to0}\sin x = 0$ and $\lim\limits_{x\to0}x = 0$, so $\lim\limits_{x\to 0}\frac{\sin x}x = \lim\limits_{x\to 0}\frac{\cos x}1 = 1 $ $\endgroup$
– Joren
2011-10-23 20:41:13 +00:00
Commented Oct 23, 2011 at 20:41
$\begingroup$ @Joren: I'm extremely curious how will you prove then that $\sin ' x = \cos x$ $\endgroup$
– SBF
2011-10-24 09:10:44 +00:00
Commented Oct 24, 2011 at 9:10
$\begingroup$ @FUZx44xl: sure, but to be fare you first prove that $\sin x\sim x$ with $x\to 0$. Geometrically $\endgroup$
SBF
– SBF
2011-10-24 15:42:04 +00:00
Commented Oct 24, 2011 at 15:42
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$\begingroup$ @FUZxxl:Exactly what was your definition by "geometrical means"? $\endgroup$
– Platonix
2013-12-22 18:45:43 +00:00
Commented Dec 22, 2013 at 18:45
4
$\begingroup$ Sandwich theorem may be applied to prove it. $\endgroup$
Sufaid Saleel
– Sufaid Saleel
2017-07-04 15:28:05 +00:00
Commented Jul 4, 2017 at 15:28
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31 Answers 31
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The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$ \frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1} $$ Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get $$ \cos(x)\le\frac{\sin(x)}{x}\le1\tag{2} $$ Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3} $$ Also, dividing $(2)$ by $\cos(x)$, we get that $$ 1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4} $$ Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5} $$
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edited Jun 19, 2013 at 14:37
answered Oct 23, 2011 at 17:26
robjohn♦robjohn
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$\begingroup$ Do my homework! Note that $(1)$ says that for $0\le x\le\frac{\pi}{2}$ we have $0\le\sin(x)\le x$; therefore, $\lim\limits_{x\to0^+}\sin(x)=0$. Then $\cos(x)=1-2\sin^2(x/2)$ should finish the job. $\endgroup$
– robjohn ♦
2011-10-23 18:37:33 +00:00
Commented Oct 23, 2011 at 18:37
$\begingroup$ From your comment, I wasn't expecting that you could find it on your own, but "Read the question!" seemed a bit rough around the edges. $\endgroup$
robjohn
– robjohn ♦
2011-10-23 19:04:48 +00:00
Commented Oct 23, 2011 at 19:04
$\begingroup$ Sorry for that. $\endgroup$
– FUZxxl
2011-10-23 19:35:30 +00:00
Commented Oct 23, 2011 at 19:35
$\begingroup$ I think that your justification is slightly circular :). In an introductory calculus course, $(\sin x,\cos x)$ is probably defined to be the point we reach after traveling $x$ units counterclockwise along the unit circle from $(1,0)$. $\endgroup$
Mike F
– Mike F
2011-10-23 20:54:02 +00:00
Commented Oct 23, 2011 at 20:54
$\begingroup$ @robjohn: OK fair enough, I wasn't really trying to define sine and cosine so much as I was trying to point out that radians are usually introduced via the arclength of a wedge, not the area of a wedge (divided by 2). Also, I didn't have a problem with using that arclength and area are proportional to the angle. Rather I think the gap is where we use that the ratio (area of wedge)/(arc length of wedge) is exactly 1/2. $\endgroup$
Mike F
– Mike F
2011-10-23 22:43:06 +00:00
Commented Oct 23, 2011 at 22:43
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You should first prove that for $x > 0$ small that $\sin x < x < \tan x$. Then, dividing by $x$ you get $$ { \sin x \over x} < 1 $$ and rearranging $1 < {\tan x \over x} = {\sin x \over x \cos x }$ $$ \cos x < {\sin x \over x}. $$ Taking $x \rightarrow 0^+$ you apply the squeeze theorem. For $x < 0$ and small use that $\sin(-x) = -\sin x$ so that $${\sin(-x) \over -x} = {\sin x \over x}.$$ As far as why the first inequality I said is true, you can do this completely from triangles but I don't know how to draw the pictures here.
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edited Nov 20, 2017 at 10:10
Domates
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answered Oct 23, 2011 at 16:28
tkrtkr
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$\begingroup$ But how to prove that $\sin x $\endgroup$
FUZxxl
– FUZxxl
2011-10-23 16:31:58 +00:00
Commented Oct 23, 2011 at 16:31
3
$\begingroup$ It is in the picture. The definition of radians makes the picture above true. Maybe that is worth mentioning: this limit explicitly depends on "$x$" being measured in radians. $\endgroup$
tkr
– tkr
2011-10-23 16:33:58 +00:00
Commented Oct 23, 2011 at 16:33
3
$\begingroup$ This is a strange picture! Normally you want the $tan(\theta)$ side to be parallel to the $sin(\theta)$ side. $\endgroup$
user641
– user641
2011-10-23 17:36:51 +00:00
Commented Oct 23, 2011 at 17:36
19
$\begingroup$ If you make $\tan(\theta)$ parallel then you need to make the points $S$ and $Q$ the same. For whatever reason, this is the proof I like the most because it relates the tangent line at the point on the circle to the value we call "tangent". To each his own... $\endgroup$
tkr
– tkr
2011-10-23 18:22:45 +00:00
Commented Oct 23, 2011 at 18:22
6
$\begingroup$ Even with this picture, you could use the area inclusion principle and argue that area of triangle OPQ < Area of sector OPQ < Area of triangle OPS. This translates into $\sin \theta < \theta < \tan \theta$ which is the first line of this answer and hence, nothing strange about it. $\endgroup$
Deepak Gupta
– Deepak Gupta
2015-09-02 21:05:00 +00:00
Commented Sep 2, 2015 at 21:05
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115
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Usually calculus textbooks do this using geometric arguments followed by squeezing.
Here's an Euler-esque way of looking at it---not a "proof" as that term is usually understood today, but still worth knowing about.
Let $\theta$ be the length of an arc along the circle of unit radius centered at $(0,0)$, from the point $(1,0)$ in a counterclockwise direction to some point $(\cos\theta,\sin\theta)$ on the circle. Then of course $\sin\theta$ is the height of the latter point above the $x$-axis. Now imagine what happens if $\theta$ is an infinitely small positive number. Then the arc is just an infinitely short vertical line, and the height of the endpoint above the $x$-axis is just the length of the arc. I.e. when $\theta$ is an infinitely small number, then $\sin\theta$ is the same as $\theta$. It follows that when $\theta$ is an infinitely small nonzero number, then $\dfrac{\sin\theta}{\theta}=1$.
That is how Euler viewed the matter. See his book on differential calculus.
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edited Jan 21, 2016 at 19:11
answered Oct 23, 2011 at 17:21
Michael HardyMichael Hardy
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$\begingroup$ I disagree with the recent edit to my answer and I have reverted to the previous version. "Infinitesmal" means "infinitely small". $\qquad$ $\endgroup$
Michael Hardy
– Michael Hardy
2016-01-21 19:12:29 +00:00
Commented Jan 21, 2016 at 19:12
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$\begingroup$ You read Euler's book ? Was it very difficult to read because of the notation and language of the time ? $\endgroup$
Saikat
– Saikat
2016-02-27 04:03:03 +00:00
Commented Feb 27, 2016 at 4:03
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$\begingroup$ @user230452 : Just some parts of it. I wouldn't say the differences in language and notation were the challenging part. $\qquad$ $\endgroup$
Michael Hardy
– Michael Hardy
2016-02-27 18:18:27 +00:00
Commented Feb 27, 2016 at 18:18
2
$\begingroup$ Wouldn't one rather say that $\frac{sin \theta}{\theta}$ is infinitely close to 1 if $\theta$ is an infinitely small nonzero number? $\endgroup$
Sven
– Sven
2017-04-14 13:05:08 +00:00
Commented Apr 14, 2017 at 13:05
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$\begingroup$ @Sven : That is indeed how it's done in Robinson's "nonstandard analysis". There is another approach to rigorous infinitesimals in which it would be done the way Euler did it, saying that if $\theta$ is infinitely small then $\dfrac{\sin\theta}\theta = 1.$ It's called "smooth infinitesimal analysis". In that approach, the square of an infinitesimal is $0$, so we have $$ \frac{\sin\theta} \theta = \frac{\theta - \dfrac{\theta^3} 6 + \dfrac{\theta^5}{120} - \cdots \cdots} \theta = 1. $$ en.wikipedia.org/wiki/Smooth_infinitesimal_analysis $\qquad$ $\endgroup$
Michael Hardy
– Michael Hardy
2017-04-14 18:48:16 +00:00
Commented Apr 14, 2017 at 18:48
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Look at this link:
Here is the picture I copied from that blog:
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edited Feb 17, 2018 at 17:17
Anshul Laikar
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answered Oct 23, 2011 at 21:15
Paulo SérgioPaulo Sérgio
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$\begingroup$ +1, nice site - What is the length BC? $\endgroup$
NoChance
– NoChance
2011-10-24 05:34:27 +00:00
Commented Oct 24, 2011 at 5:34
31
$\begingroup$ What is the argument for showing that $\theta \cos \theta\le \sin \theta$? The picture isn't immediately convincing. $\endgroup$
Mark Viola
– Mark Viola
2015-04-15 05:31:55 +00:00
Commented Apr 15, 2015 at 5:31
2
$\begingroup$ @Dr. MV: if you rescale the circle in $\cos \theta$ units, then the original $\sin \theta$ is a scaled $\tan \theta$ which is always longer than its arc. $\endgroup$
Matteo Vitturi
– Matteo Vitturi
2016-05-30 20:38:02 +00:00
Commented May 30, 2016 at 20:38
$\begingroup$ @Mattsteel Why is that? Why is it longer than its arc? $\endgroup$
Anshuman Agrawal
– Anshuman Agrawal
2022-07-19 14:12:57 +00:00
Commented Jul 19, 2022 at 14:12
$\begingroup$ This proof is so elegant ð. $\endgroup$
Broken Dreams
– Broken Dreams
2022-08-20 01:48:00 +00:00
Commented Aug 20, 2022 at 1:48
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61
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I am not sure if it counts as proof, but I have seen this done by a High Schooler.
In the given picture above, $\displaystyle 2n \text{ EJ} = 2nR \sin\left( \frac{\pi}{n } \right ) = \text{ perimeter of polygon }$.
$\displaystyle \lim_{n\to \infty }2nR \sin\left( \frac{\pi}{n } \right ) = \lim_{n\to \infty } (\text{ perimeter of polygon }) = 2 \pi R \implies \lim_{n\to \infty}\frac{\sin\left( \frac{\pi}{n } \right )}{\left( \frac{\pi}{n } \right )} = 1$ and let $\frac{\pi}{n} = x$.
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answered May 16, 2013 at 18:01
S LS L
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$\begingroup$ This method is usually used to prove that the perimeter of a circle is $2\pi R$ using the fact $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$ $\endgroup$
user5402
– user5402
2013-07-20 19:15:01 +00:00
Commented Jul 20, 2013 at 19:15
4
$\begingroup$ Santosh, how does one prove that the perimeter of the polygon converges to the $ 2\pi$ ? $\endgroup$
Imago
– Imago
2016-02-05 19:45:38 +00:00
Commented Feb 5, 2016 at 19:45
$\begingroup$ @Imago math.stackexchange.com/questions/720935/… $\endgroup$
user301988
– user301988
2016-06-06 03:32:48 +00:00
Commented Jun 6, 2016 at 3:32
Add a comment |
59
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I claim that for $0Archimedes, On the Sphere and Cylinder Book I). But both the sector and the quadrilateral both have sides $OC$ and $OA$, so we have $$CA=xCD$ because it is the hypotenuse in $\triangle BCD$ we have $$\tan x = BA = BD+DA\gt CD+DA \gt CA=x \gt \sin x$$
So we have $$\sin x \lt x \lt \tan x$$ $$\frac{\sin x}{x} \lt 1 \lt \frac{\tan x}{x}=\frac{\sin x}{x}\cdot\sec x$$ From this we can extract $$\frac{\sin x}{x} \lt 1$$ and $$1 \lt \frac{\sin x}{x}\cdot\sec x$$ $$\cos x \lt \frac{\sin x}{x}$$ Putting these inequalities back together we have $$\cos x \lt \frac{\sin x}{x} \lt 1$$
Because $\displaystyle\lim_{x\to 0}\cos x = 1$, by the squeeze theorem we have $$\lim_{x\to 0}\frac{\sin x}{x}=1$$
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answered Sep 7, 2014 at 21:44
John JoyJohn Joy
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$\begingroup$ The reason that I chose an arc length proof is because most derivations that I've seen of the area of a circle (see mathopenref.com/circleareaderive.html) assume that $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$. Using an area based proof, to me, seems like putting the cart before the horse. $\endgroup$
John Joy
– John Joy
2014-11-09 15:14:59 +00:00
Commented Nov 9, 2014 at 15:14
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$\begingroup$ Actually, what Archimedes assumed was that if 2 curves (both concave in the same direction) have the same endpoints and one of the curves is strictly between a line segment with the same end points and the other curve, then the length of the enclosed curve is shorter than the length of enclosing curve. See axiom #2 archive.org/stream/worksofarchimede00arch#page/4/mode/2up $\endgroup$
John Joy
– John Joy
2015-02-05 14:17:59 +00:00
Commented Feb 5, 2015 at 14:17
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$\begingroup$ OK, so he essentially axiomatizes the idea that among all curves connecting endpoints $A$ and $B$ satisfying some kind of regularity condition - concavity in the same direction, we have this inequality based on enclosure. This makes $arc(CA) $\endgroup$
String
– String
2015-02-05 14:52:54 +00:00
Commented Feb 5, 2015 at 14:52
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$\begingroup$ If you don't have a copy of Archimedes's book, you can approximate sector OCA with a polygon inside OCDA and use cut-the-knot.org/m/Geometry/PerimetersOfTwoConvexPolygons.shtml $\endgroup$
CopyPasteIt
– CopyPasteIt
2017-06-06 16:06:23 +00:00
Commented Jun 6, 2017 at 16:06
2
$\begingroup$ I love this proof! $\endgroup$
emandret
– emandret
2019-12-04 00:43:57 +00:00
Commented Dec 4, 2019 at 0:43
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Here you may see an elementary approach that starts from a very interesting result, see this problem. All you need is a bit of imagination. When we take $\lim_{n\rightarrow\infty} \frac{n\sin(\frac{\pi}{n})}{1+\sin(\frac{\pi}{n})}$ we may notice that we have infinitely many circles surrounding the unit circle with infinitely small diameters that lastly perfectly approximate the length of the unit circle when having it there infinity times . Therefore when multiplying n by the radius under the limit to infinity we get Ï. Let's denote $\frac{\pi}{n}$ by x.
$$\lim_{x\rightarrow0}\frac{\pi\sin(x)}{x(1+\sin(x))}={\pi}\Rightarrow\lim_{x\rightarrow0}\frac{\sin(x)}{x(1+\sin(x))}=1\Rightarrow\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$$
The proof is complete.
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edited Jun 12, 2020 at 10:38
CommunityBot
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answered Jun 7, 2012 at 7:03
user 1591719user 1591719
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$\begingroup$ this assumes the a priori knowledge of the existence of the limit $\lim_{x\to 0}\frac{\sin(x)}{x}$, how would you go about proving it before hand? (+1) by the way for the original and alternative proof $\endgroup$
user153330
– user153330
2016-02-24 14:07:06 +00:00
Commented Feb 24, 2016 at 14:07
Add a comment |
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Don't you feel strange about why most of the proofs are done with a figure? I've had this problem in the beginning, and realized after that this is due to the definition we use for the function $\sin x$. Because the usual definition of $\sin x$ we all study first in high schools depends on classical geometry and usually with a figure, you should depict out the figure and to make it clear.
However, if you use other definitions of $\sin x$ that are equivalent to the former, you'll find it more simple. For example,
$$\sin x = \frac{x^1}{1!} - \frac{x^3}{3!}+ \frac{x^5}{5!} - \cdots + \cdots - \cdots$$
and hence
$$\frac{\sin x}x = \frac{x^0}{1!} - \frac{x^2}{3!}+ \frac{x^4}{5!} - \cdots$$
which obviously tends to $1$ as $x$ approaches 0.
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edited Mar 13, 2015 at 17:57
FUZxxl
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answered Mar 13, 2015 at 17:06
user223261user223261
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$\begingroup$ Indeed, it's easy to see that this holds if one uses a series, but this question starts on the prerequisite that one does not use a series. $\endgroup$
FUZxxl
– FUZxxl
2015-03-13 17:54:54 +00:00
Commented Mar 13, 2015 at 17:54
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$\begingroup$ And how do you know that the derivative of an infinte series is equal to the sum of the derivatives of each term? $\endgroup$
Steven Alexis Gregory
– Steven Alexis Gregory
2016-04-16 04:32:52 +00:00
Commented Apr 16, 2016 at 4:32
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$\begingroup$ @StevenGregory: Where does he use that? He only divides by $x$. What he does use is that $\frac{1}{x}\lim_{n\to\infty} \sum_{k=0}^n a_k = \lim_{n\to\infty}\sum_{k=0}^n \frac{a_k}{x}$ $\endgroup$
celtschk
– celtschk
2016-09-24 09:36:45 +00:00
Commented Sep 24, 2016 at 9:36
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$\begingroup$ @celschk starting with the infinite series as a 'definition' causes more problems than it solves. Now we have to prove that this sine behaves like the sine we learned in high school. This proof is compelling but it is not really a proof. $\endgroup$
Steven Alexis Gregory
– Steven Alexis Gregory
2016-09-24 14:44:59 +00:00
Commented Sep 24, 2016 at 14:44
$\begingroup$ @StevenAlexisGregory Eh that is debatable. I would instead say that the series definition is one of the fundamental definitions of the sine function, and you can very well begin with it $\endgroup$
Max0815
– Max0815
2023-09-26 14:21:50 +00:00
Commented Sep 26, 2023 at 14:21
Add a comment |
35
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Here's one more: $$ \lim_{x \to 0} \frac{\sin x}{x}=\lim_{x \to 0} \lim_{v \to 0}\frac{\sin (x+v)-\sin v}{x}\ =\lim_{v \to 0} \lim_{x \to 0}\frac{\sin (x+v)-\sin v}{x}=\lim_{v \to 0}\sin'v=\lim_{v\ \to 0} \cos v=1 $$
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answered Jul 20, 2013 at 18:37
AlexAlex
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$\begingroup$ Usually, this limit is used to compute the derivative of $\sin(x)$. $\endgroup$
robjohn
– robjohn ♦
2013-07-20 18:43:09 +00:00
Commented Jul 20, 2013 at 18:43
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$\begingroup$ This is an interesting alternative, but it does raise the question of why the limiting operations can be swapped. $\endgroup$
Allawonder
– Allawonder
2019-06-24 03:27:24 +00:00
Commented Jun 24, 2019 at 3:27
Add a comment |
34
$\begingroup$
Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest without appeal to geometry or differential calculus.
Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.
We define the sine function, $\sin(x)$, as the inverse function of the function $f(x)$ given by
$$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt }\tag 1$$
for $|x|< 1$.
NOTE: It can be shown that the sine function defined as the inverse of $f(x)$ given in $(1)$ has all of the familiar properties that characterize the circular function $\sin(x)$.
It is straightforward to show that since $\frac{1}{\sqrt{1-t^2}}$ is positive and continuous for $t\in (-1,1)$, $f(x)$ is continuous and strictly increasing for $x\in (-1,1)$ with $\displaystyle\lim_{x\to 0}f(x)=f(0)=0$.
Therefore, since $f$ is continuous and strictly increasing, its inverse function, $\sin(x)$, exists and is also continuous and strictly increasing with $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.
From $(1)$, we have the bounds (SEE HERE)
$$\bbox[5px,border:2px solid #C0A000]{1 \le \frac{f(x)}x\le \frac{1}{\sqrt{1-x^2}}} \tag 2$$
for $x\in (-1,1)$, whence applying the squeeze theorem to $(2)$ yields
$$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$$
Finally, let $y=f(x)$ so that $x=\sin(y)$. As $x\to 0$, $y\to 0$ and we can write $(3)$ as
$$\lim_{y\to 0}\frac{y}{\sin(y)}=1$$
from which we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0}\frac{\sin(y)}{y}=1}$$
as was to be shown!
NOTE:
We can deduce the following set of useful inequalities from $(2)$. We let $x=\sin(\theta)$ and restrict $x$ so that $x\in [0,1)$. In addition, we define new functions, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$ and $\tan(\theta)=\sin(\theta)/\cos(\theta)$.
Then, we have from $(2)$
$$\bbox[5px,border:2px solid #C0A000]{y\cos(y)\le \sin(y)\le y\le \tan(y)} $$
which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.
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edited Jun 12, 2020 at 10:38
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answered Dec 6, 2016 at 18:47
Mark ViolaMark Viola
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$\begingroup$ This is similar to my answer. You're formalizing the arc-length definition of $\sin$ in the following three steps. 1) Defining $\arcsin$ geometrically 2) Expressing that informal definition using integration 3) Defining $\sin$ as the inverse of $\arcsin$. Also, where you used the sqeeuze theorem, you could equally have used the Fundamental Theorem of Calculus and the Inverse Function Theorem. Then you'd get my answer $\endgroup$
wlad
– wlad
2018-02-05 10:16:52 +00:00
Commented Feb 5, 2018 at 10:16
1
$\begingroup$ Another difference is you used the arc-length definition and I used the area definition $\endgroup$
wlad
– wlad
2018-02-05 10:19:44 +00:00
Commented Feb 5, 2018 at 10:19
2
$\begingroup$ @ogogmad I don't use any geometrical argument. A function $f$ is defined without reference to the arc length interpretation. Both $f$ and its inverse has properties. It can be shown that the inverse function is the well-known sine function. And while one can apply the FOC instead of the squeeze theorem, there is no advantage in doing so. So, why on earth are you leaving these comments? $\endgroup$
Mark Viola
– Mark Viola
2018-02-05 14:50:22 +00:00
Commented Feb 5, 2018 at 14:50
1
$\begingroup$ Because I'm pointing out that your abstract definitions make geometric sense, whether or not your realize it. I'm also pointing out that your squeeze argument is just a special case of FOC $\endgroup$
wlad
– wlad
2018-02-05 16:43:22 +00:00
Commented Feb 5, 2018 at 16:43
1
$\begingroup$ In fact, all of the arguments above are morally the same, AFAICT. The squeeze arguments are just unpacking the proof of FOC on a special case. The FOC itself is proved using a very similar squeeze argument to the one people are using. The areas of the sectors are simply the integrals that are being differentiated. In tkr's argument, he's instead differentiating an arclength integral, which gives the same result anyway. There's the remaining feature that our two answers uses $\arcsin$ while everyone else's uses $\sin$; they may be implicitly unpacking the proof of the inverse function theorem. $\endgroup$
wlad
– wlad
2018-02-05 17:01:34 +00:00
Commented Feb 5, 2018 at 17:01
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Usual proofs can be circular, but there is a simple way for proving such inequality.
Let $\theta$ be an acute angle and let $O,A,B,C,D,C'$ as in the following diagram:
We may show that:
$$ CD \stackrel{(1)}{ \geq }\;\stackrel{\large\frown}{CB}\; \stackrel{(2)}{\geq } CB\,\stackrel{(3)}{\geq} AB $$
$(1)$: The quadrilateral $OCDC'$ and the circle sector delimited by $O,C,C'$ are two convex sets. Since the circle sector is a subset of the quadrilateral, the perimeter of the circle sector is less than the perimeter of the quadrilateral.
$(2)$: the $CB$ segment is the shortest path between $B$ and $C$.
$(3)$ $CAB$ is a right triangle, hence $CB\geq AB$ by the Pythagorean theorem.
In terms of $\theta$ we get: $$ \tan\theta \geq \theta \geq 2\sin\frac{\theta}{2} \geq \sin\theta $$ for any $\theta\in\left[0,\frac{\pi}{2}\right)$. Since the involved functions are odd functions the reverse inequality holds over $\left(-\frac{\pi}{2},0\right]$, and $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$ follows by squeezing.
A slightly different approach might be the following one: let us assume $\theta\in\left(0,\frac{\pi}{2}\right)$. By $(2)$ and $(3)$ we have $$ \theta \geq 2\sin\frac{\theta}{2}\geq \sin\theta $$ hence the sequence ${a_n}_{n\geq 0}$ defined by $a_n = 2^n \sin\frac{\theta}{2^n}$ is increasing and bounded by $\theta$. Any increasing and bounded sequence is convergent, and we actually have $\lim_{n\to +\infty}a_n=\theta$ since $\stackrel{\large\frown}{BC}$ is a rectifiable curve and for every $n\geq 1$ the $a_n$ term is the length of a polygonal approximation of $\stackrel{\large\frown}{BC}$ through $2^{n-1}$ equal segments. In particular
$$ \forall \theta\in\left(0,\frac{\pi}{2}\right), \qquad \lim_{n\to +\infty}\frac{\sin\left(\frac{\theta}{2^n}\right)}{\frac{\theta}{2^n}} = 1 $$ and this grants that if the limit $\lim_{x\to 0}\frac{\sin x}{x}$ exists, it is $1$. By $\sin x\leq x$ we get $\limsup_{x\to 0}\frac{\sin x}{x}\leq 1$, hence it is enough to show that $\liminf_{x\to 0}\frac{\sin x}{x}\geq 1$. We already know that for any $x$ close enough to the origin the sequence $\frac{\sin x}{x},\frac{\sin(x/2)}{x/2},\frac{\sin(x/4)}{x/4},\ldots$ is convergent to $1$, hence we are done.
Long story short: $\lim_{x\to 0}\frac{\sin x}{x}=1$ follows from the fact that a circle is a rectifiable curve, and a circle is a rectifiable curve because it is the boundary of a convex, bounded subset of $\mathbb{R}^2$. The convexity of the disk follows from the triangle inequality: a disk is a closed ball for the euclidean distance.
$(1)$ relies on this powerful Lemma:
Lemma. If $A,B$ are convex bounded sets in $\mathbb{R}^2$ and $A\subsetneq B$, the perimeter of $A$ is less than the perimeter of $B$.
Proof: by boundedness and convexity, $\partial A$ and $\partial B$ are rectifiable, with lengths $L(A)=\mu(\partial A),\,L(B)=\mu(\partial B)$. Always by convexity, there is some chord in $B$ that does not meet the interior of $A$ (a tangent to $\partial A$ at a smooth point does the job, for instance). Assume that such chord has endpoints $B_1, B_2 \in \partial B$ and perform a cut along $B_1 B_2$: both the area and the perimeter of $B$ decrease, but $B$ remains a bounded convex set enclosing $A$. Since $A$ can be approximated through a sequence of consecutive cuts, $L(A) follows.
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edited Jun 12, 2020 at 10:38
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answered Jun 13, 2017 at 0:03
Jack D'AurizioJack D'Aurizio
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$\begingroup$ Late to the party, but I think this is the non-circular (no puns intended) way of thinking that has been missing so far. Nice job! $\endgroup$
String
– String
2017-06-13 09:05:36 +00:00
Commented Jun 13, 2017 at 9:05
1
$\begingroup$ Up voting! I think Archimedes used a weaker form of your Lemma to get the circumference of a circle. $\endgroup$
CopyPasteIt
– CopyPasteIt
2017-06-21 23:49:05 +00:00
Commented Jun 21, 2017 at 23:49
$\begingroup$ Great answer (+1) $\endgroup$
MathLover
– MathLover
2018-01-25 23:07:46 +00:00
Commented Jan 25, 2018 at 23:07
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It depends on your definition of the sine function. I would suggest checking out the geometric proof in ProofWiki.
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answered Oct 23, 2011 at 16:30
Yuval FilmusYuval Filmus
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The strategy is to find $\frac{d\arcsin y}{dy}$ first. This can easily be done using the picture below.
From the above picture, $\arcsin y$ is twice the area of the orange bit. The area of the red bit is ${1 \over 2}y\sqrt{1-y^2}$. The area of the red bit plus the orange bit is $\int_{0}^y \sqrt{1-Y^2} dY$. So $$\arcsin y = 2\int_{0}^y \sqrt{1-Y^2} dY - y\sqrt{1-y^2}.$$ Differentiating with respect to $y$ gives $\frac{d\arcsin y}{dy} = \frac{1}{\sqrt{1-y^2}}$. Using the theorem for the derivative of inverse functions yields $\sin' \theta = \sqrt{1 - \sin^2 \theta} = \cos \theta$.
(A similar thing can be done with the arc length definition of $\arcsin$.)
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edited Mar 7, 2018 at 16:46
answered Apr 11, 2015 at 17:26
wladwlad
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$\begingroup$ While this is indeed an interesting approach, integrals haven't been taught at the point where this limit is proved. Thank you for your answer though. $\endgroup$
FUZxxl
– FUZxxl
2015-04-11 17:38:27 +00:00
Commented Apr 11, 2015 at 17:38
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Let $f:{y\in\mathbb{R}:y\neq 0}\to\mathbb{R}$ be a function defined by $f(x):=\dfrac{\sin x}{x}$ for all $x\in {y\in\mathbb{R}:y\neq 0}$.
We have $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x}=1$ if and only if for every $\varepsilon>0$, there exists a $\delta>0$ such that $|f(x)-1|<\varepsilon$ whenever $0<|x-0|<\delta$.
Let $\varepsilon>0$ be an arbitrary real number.
Note that $\sin x=\displaystyle \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$.
If $x \neq 0$, we have $\dfrac{\sin x}{x}=$$\displaystyle \sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n+1)!}=1+$$\displaystyle \sum_{n=1}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n+1)!}$.
We thus have
$|f(x)-1|=\left|\dfrac{\sin x}{x}-1\right|=\left|\displaystyle \sum_{n=1}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n+1)!}\right|\leq \left|\displaystyle\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n+1)!}\right|\leq \displaystyle\sum_{n=1}^{\infty} \left|\dfrac{x^{2n}}{(2n+1)!}\right|$
Therefore we have
$|f(x)-1|\leq \displaystyle\sum_{n=1}^{\infty} \left|\dfrac{x^{2n}}{(2n+1)!}\right|\leq \displaystyle \sum_{n=1}^{\infty} |x^{2n}|=\sum_{n=1}^{\infty}|x^2|^n$
If $0<|x|<1$, then $0<|x^2|<1$, and the infinite series $\displaystyle\sum_{n=1}^{\infty}|x^2|^n$ converges to $\dfrac{x^2}{1-x^2}$.
Choose $\delta:=\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}$. Then $0<|x-0|<\delta$ implies that $0<|x|<$$\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}<1$, and hence $x^2<\varepsilon-\varepsilon x^2$. But $x^2<\varepsilon-\varepsilon x^2$ implies that $\dfrac{x^2}{1-x^2}<\varepsilon$.
We therefore have $\sum_{n=1}^{\infty}|x^2|^n<\varepsilon$ whenever $0<|x-0|<\delta$. But since $|f(x)-1|\leq\displaystyle\sum_{n=1}^{\infty}|x^2|^n$, we have $|f(x)-1|<\varepsilon$ whenever $0<|x-0|<\delta$.
Since $\varepsilon$ was arbitrary, we have $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x}=1$.
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answered Oct 10, 2016 at 14:55
Supreeth NarasimhaswamySupreeth Narasimhaswamy
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$\begingroup$ The premise of the question was not to use power series. $\endgroup$
FUZxxl
– FUZxxl
2018-12-04 12:35:03 +00:00
Commented Dec 4, 2018 at 12:35
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Let $\sin(x)$ is defined as solution of $\frac{d^2}{dx^2}\textrm{f}(x)=-\textrm{f}(x)$ with $\mathrm f(0)=0,\,\frac{d}{dx}\mathrm f(0)=C$ initial conditions, so exact solution is $\mathrm f(x)=C\cdot\sin(x)$. Define second derivative as $$ \begin{align} \frac{d^2}{dx^2}\textrm{f}(x)=\lim_{\Delta x\to 0}{\frac{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}-\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}{\Delta x}}&=\=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-2\cdot \mathrm f(x-\Delta x)+\mathrm f(x-2\cdot\Delta x)}{\Delta x^2}} \end{align} $$ we can easy check this limit for any (?) functions. Similarly, we can define the first derivative for right, middle and left points: $$ \frac{d}{dx}\textrm{f}(x)\left{ \begin{aligned} &=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}} \ &=\lim_{\Delta x\to 0}{\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}\ &=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-2\cdot\Delta x)}{2\cdot\Delta x}} \end{aligned} \right. $$ Let's use the finite elements method assuming $Td=\Delta x,\,y_n=\mathrm f(x),\,y_{n-1}=\mathrm f(x-\Delta x),\,y_{n-2}=\mathrm f(x-2\cdot \Delta x)$ Override differential equation as $$ \frac{y_n-2\cdot y_{n-1}+y_{n-2}}{Td^2}=-y_n $$ Now solve this implicit equation for $y_n$ to obtain explicit recurrence relation: $$ y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2} $$ Using arbitrarily small but non-zero quantity Td we can plot exponentially decaying sampled sine function (because the poles are inside the unit circle of the transfer function corresponding to the given recurrence relation). Similarly we write three systems for the initial conditions:
$$ \left{ \begin{aligned} &y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2} \ &C=\frac{y_n-y_{n-1}}{Td} \end{aligned}\right. $$ $$ \left{ \begin{aligned} &y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2} \ &C=\frac{y_{n-1}-y_{n-2}}{Td} \end{aligned}\right. $$ $$ \left{ \begin{aligned} &y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2} \ &C=\frac{y_n-y_{n-2}}{2\cdot Td} \end{aligned}\right. $$ Solve this sequence of equations for $y_{n-1}$ and $y_{n-2}$: $$ \left{ \begin{aligned} &y_{n-1} = -C\cdot Td + y_{n}\ &y_{n-2}=-2\cdot C\cdot Td + y_{n}\cdot\left(1-Td^2\right)\ \end{aligned}\right. $$ $$ \left{ \begin{aligned} &y_{n-1} = -C\cdot Td + y_{n}\cdot\left(1+Td^2\right)\ &y_{n-2}=-2\cdot C\cdot Td + y_{n}\cdot\left(1+Td^2\right)\ \end{aligned}\right. $$ $$ \left{ \begin{aligned} &y_{n-1} = -C\cdot Td + y_{n}\cdot\left(1+\frac{Td^2}{2}\right)\ &y_{n-2}=-2\cdot C\cdot Td + y_{n}\ \end{aligned}\right. $$ At zero point $y_n=\mathrm f(0)=0$ and we can see linear dependence: $$ \begin{aligned} &y_{n-1} = -C\cdot Td\ &y_{n-2}=-2\cdot C\cdot Td \end{aligned} $$ for all three solutions. Replace back: $$ \begin{array}{l} \mathrm f(0)&=0\ \mathrm f(0-\Delta x) &= -C\cdot \Delta x\ \mathrm f(0-2\cdot \Delta x) &= -2\cdot C\cdot \Delta x \end{array} $$ So all three $\frac{d}{dx}\mathrm f(0)$ limits is equal to $C$ at $x=0$ and in accordance with $\mathrm f(x)=C\cdot\sin(x)$ by definition we can write $$ \lim_{\Delta x\to 0}{\frac{\mathrm f(0)-\mathrm f(0-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{0-(-C \cdot \Delta x)}{\Delta x}}=C $$ Thus $$ \lim_{\Delta x\to 0}{\frac{\sin(0)-C\cdot\sin(0-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{C\cdot\sin(\Delta x)}{\Delta x}}=C\cdot\lim_{\Delta x\to 0}{\frac{\sin(\Delta x)}{\Delta x}}=C $$ and $\lim_{\Delta x\to 0}{\frac{\sin(\Delta x)}{\Delta x}}=1$
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answered Aug 19, 2016 at 20:14
Timur ZhoraevTimur Zhoraev
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$\begingroup$ While this is an exceptionally neat answer, it is not really appropriate for the level (high school) on which I asked the question, given that computing this limit is an introductory exercise into differentials. Plus we defined $\sin$ geometrically (i.e. though the relationship between angles and sides in a right triangle), so your prerequisites must be established first. $\endgroup$
FUZxxl
– FUZxxl
2016-08-19 20:40:04 +00:00
Commented Aug 19, 2016 at 20:40
4
$\begingroup$ @FUZxxl do you realize that users with different levels and mathematical background may check this question? The answer given above might be helpful for them. $\endgroup$
Xam
– Xam
2017-09-23 03:38:01 +00:00
Commented Sep 23, 2017 at 3:38
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Simple one is using sandwich theorem Which demonstrated earlier.In this method you have to show that $\frac{\sin x}{x} $ lies between other two functions. As $x \longrightarrow 0$ both of them will tends to ONE.
Then as in the case of sandwich (if both the bread part go to one stomach the middle portion will also go to the same stomach) $\frac{\sin x}{x}$ will go to ONE.
You can use geogebra to see the visualization of this phenomena using geogebra.First you input $\sin x$ and $x$ and observe that near to $0$ values of $\sin x$ and $x$ are same.
Secondly input $\frac{\sin x}{x}$ then observe function is approaching to $1$ as $x$ tends to $0$
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edited Dec 13, 2014 at 10:00
Davide Giraudo
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answered Sep 15, 2014 at 2:56
MadhuMadhu
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$\begingroup$ This describes the Sandwich Theorem, but does not answer the question. At best, this should be a comment to the question, $\endgroup$
robjohn
– robjohn ♦
2014-12-13 13:31:27 +00:00
Commented Dec 13, 2014 at 13:31
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Originally posted on the proofs without words post, here is a simple image that explains the derivative of $\sin(x)$, which as we all know, is directly related to the limit at hand.
If one is not so convinced, take a look at the above picture and notice that if $u\pm h$ is in the first quadrant, then
$$\frac{\sin(x+h)-\sin(x)}h<\cos(x)<\frac{\sin(x-h)-\sin(x)}{-h}$$
Notice that
$$ \begin{align}\frac{d}{dx}\sin(x)&=\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}h\\text{picture above}&=\lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}h\\cos(x)&=\lim_{h\to0}\sin(x)\frac{\cos(h)-1}h+\cos(x)\frac{\sin(h)}h\\cos(0)&=\lim_{h\to0}\frac{\sin(h)}h\end{align} $$
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edited Apr 13, 2017 at 12:58
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answered Nov 15, 2016 at 22:52
Simply Beautiful ArtSimply Beautiful Art
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$\begingroup$ Note that your reason is circular: In order to prove the derivative of $\sin$, we need to know the limit of $\sin h\over h.$ Attempting to prove the limit the other way round is counterproductive. $\endgroup$
FUZxxl
– FUZxxl
2016-11-16 00:47:27 +00:00
Commented Nov 16, 2016 at 0:47
2
$\begingroup$ @FUZxxl no actually, the whole point of this was a geometric proof of the fact. $\endgroup$
Simply Beautiful Art
– Simply Beautiful Art
2016-11-16 17:23:13 +00:00
Commented Nov 16, 2016 at 17:23
$\begingroup$ @FUZxxl If you are still unsatisfied about it, there is a whole post on the topic. $\endgroup$
Simply Beautiful Art
– Simply Beautiful Art
2016-11-17 00:44:47 +00:00
Commented Nov 17, 2016 at 0:44
$\begingroup$ @FUZxxl Since you still seem unhappy, I've added a better "squeeze theorem" type thing for the derivative of $\sin(x)$. $\endgroup$
Simply Beautiful Art
– Simply Beautiful Art
2017-01-05 23:14:24 +00:00
Commented Jan 5, 2017 at 23:14
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This is a new post on an old saw because this is one of those things where that I can see how that, all too sadly, the way in which we've structured the current maths curriculum really doesn't make it possible to do these kinds of things the justice they deserve and I think, ultimately, that is a disservice to many learners.
The truth is, this limit cannot really be given an honest proof without an honest definition of the sine function, first. And that is not as easy as it seems. Even if we consider the simple notion from many trigonometric treatments that the sine is equal to the "length of the opposite side of the right triangle divided by the length of its hypotenuse", that doesn't truly solve the problem because there is actually a subtle missing element and that is that sine is not a function of a "right triangle" (though you could define that if you wanted, and it'd be easy!), but of an angle measure. And actually parsing out what "angle measure" means, it turns out, is essentially equivalent to defining the sine function in the first place, so this approach is circular! (pun observed after writing despite not being originally intended!)
So how do we define sine, or angle measure? Unfortunately, any approach to this is such that it must involve calculus. This is because the angle measure we use is "smooth and steady", meaning that, basically, if we have some angle, we'd like fractionating that angle measure to fractionate the angle in the same manner as cutting up pieces of a pie: if I have an angle with the given angle measure $\theta$, then for the measure system to work I should be able to then produce an angle with measure $\frac{\theta}{n}$, should be an angle that is geometrically the $n$-section of the angle into $n$ congruent smaller angles that add up to the full angle.
Yet already, we can see right there that this is not trivial: consider $n = 3$. Then we have the famous "impossible" problem of "trisection of the angle" which vexed even the ancient Greeks and for which people would keep trying to pound at until Pierre Wantzel finally proved it undoable over two thousand years after. We are asking for a mathematical widget that can not only trisect, but 5-sect, 629-sect, etc. angles and in a systematic manner to boot!
Indeed, not only is the sine function not trivial, we could argue that even the exponential function is considerably easier to treat than sine, though I won't give such a treatment here.
Thus, how do we do it? Well, the key observation is that our "steady" angle measure is one which is, effectively, defined by the arc length of a segment of circle intercepted by the angle when drawn at the circle's center and projected outward. In particular, this should be "obvious" from the (circularly-introduced) geometric formula
$$\mbox{Length of circular arc} = r\theta$$
Since this is only a trivial multiplication, all the nontriviality must be in either defining $\theta$ in terms of geometric angles formed by lines, or in terms of defining the "length of a circular arc" and, moreover, these two problems must be equally hard. Hence, we will begin with the arc question first and one will see that this answer will end up using a fair bit of Calculus II material to answer this Calculus I-level question about a supposedly pre-Calculus mathematical object. Indeed, this is what the whole "radian measure" is: it's a measure of angles in terms of the arc length of the piece they cut from a unit circle (i.e. $r = 1$). "Degrees", are then just a weird multiple unit of actual length, equal to $\frac{2\pi}{360}$ (or better, $\frac{\tau}{360}$) of some other unit length.
If you use a somewhat more honest Trigonometry book, you will see something to the effect that sine and cosine are defined as basically being the coordinates on a unit circle when an angle measure $\theta$ has been emplaced from the $x$-axis:
$$C(\theta) := (\cos(\theta), \sin(\theta))$$
Now as said above, $\theta$ is arc length. Thus, what we have above is something called an arc length parameterization of the circle - and that tells us how we need to proceed. First, we need a separate definition of the arc length of a circle.
How do we get that? Well, we will obviously need a more elementary circle equation, first, than the one we just gave, and that means going to the simple algebraic definition,
$$x^2 + y^2 = 1$$
so that now we can solve using good ole' Algebra for $x$ and $y$ in at least a semicircle:
$$x(y) = \sqrt{1 - y^2}$$ $$y(x) = \sqrt{1 - x^2}$$
And now this is where we then must introduce Calculus II-level concept - namely, integration for arc length. The arc length swept between two values of the $x$-coordinate, for a curve given with $y$ as a function of $x$, is
$$\mbox{Arc Length}(x_1, x_2) := \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$
Hence for the circle, now $\frac{dy}{dx} = \frac{1}{2} (1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}$, so
$$\mbox{Arc Length}(x_1, x_2) = \int_{x_1}^{x_2} \sqrt{1 + \frac{x^2}{1 - x^2}} dx$$
which simplifies to
$$\mbox{Arc Length}(x_1, x_2) = \int_{x_1}^{x_2} \frac{1}{\sqrt{1 - x^2}} dx$$
Now, we use the Fundamental Theorem of Calculus to define the inverse sine as
$$\arcsin(x) := \int_{0}^{x} \frac{1}{\sqrt{1 - \xi^2}} d\xi$$
which is the arc length in terms of coordinate, and now the sine is its inverse, coordinate in terms of arc length:
$$\sin(\theta) := \arcsin^{-1}(\theta)$$
.
Finally, at this point, with a full, airtight definition of $\sin(x)$ now in hand, we are ready to evaluate the limit:
$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x}$$
Since the "real", or base, function here is really the inverse function, i.e. $\arcsin$, we first proceed by making a change of variables: we consider instead the limit in terms of $y$ where $y(x) := \arcsin(x)$. Note that, trivially, $\arcsin(0) = 0$ from the integral definition, thus we get
$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = \lim_{y \rightarrow 0} \frac{y}{\arcsin(y)}$$
Now for the right-hand limit, we need only consider the behavior of $\arcsin(y)$ when $y$ is small. Since the integrand, $\frac{1}{\sqrt{1 - x^2}}$, is differentiable at $x = 0$, it can be approximated with its tangent line (which really, should be also how we define tangent lines in the first place, as a "best approximation", a notion that can be done in an airtight, intuitive fashion through the use of a "zoom-in") and so likewise, the integral over a tiny sliver by integral of that same tangent line. By the power rule and chain rule,
$$\frac{d}{dx} \frac{1}{\sqrt{1 - x^2}} = \frac{d}{dx} (1 - x^2)^{-1/2} = \left(-\frac{1}{2}\right) (1 - x^2)^{-3/2} \cdot (-2x) = x(1 - x^2)^{3/2}$$
so the derivative at $x = 0$ is zero and the tangent line is horizontal: since also $\frac{1}{\sqrt{1 - x^2}}$ evaluated at $x = 0$ is $1$, the tangent is
$$T(x) := 1$$
hence
$$\int_{0}^{y} \frac{1}{\sqrt{1 - \xi^2}}\ d\xi \approx \int_{0}^{y} 1\ d\xi$$
when $y \approx 0$, and then the right-hand integral is approximately $y$, hence $\arcsin(y) \approx y$ when $y \approx 0$ and
$$\lim_{y \rightarrow 0} \frac{y}{\arcsin(y)} = \lim_{y \rightarrow 0} \frac{y}{y} = \lim_{y \rightarrow 0} 1 = 1$$
hence
$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$
QED.
Nonetheless, as I mentioned before, this doesn't solve the requirements of the question which, while I'm sure its original asker has long moved on is, nonetheless, still relevant to calculus student after calculus student up to today: prove the limit using only Calculus I/pre-Calculus methods. What I am saying is that, in fact, that is not truly honestly possible and reveals a weakness of the curriculum in that it doesn't actually follow the proper logical buildup of the mathematical edifice.
What really should be done is to leave trig for later, that is, skip trig and go for Calculus first. When I studied maths on my own, I did just that. In fact, I'd say, as many educators have suggested, that most people don't need either, but really need more statistics instead. Then for those who do pursue higher maths, if we've done algebra and statistics, we already have right there a lot of interesting material we can build on for calculus, including the exponential function. There is no need to add trig functions to "sweeten the mix" when it's already plenty sweet with integrals of algebraic functions like $x \mapsto \sqrt{1 - x^2}$ which is a very nice example of the area-integration relation, and such can, if emphasized more heavily, potentially invite more nuanced thinking about integrals beside just "plugging and chugging integration rules". In particular, with a more limited set of functions, we can think about other ways we might approach them like just that and/or a variety of ways to interpret the integral which can only be good, I'd think, to develop more creative thinking about problems and less drilling in methods with little real understanding gained (and rote crunching integrals is even less relevant now with computer algebra software; more important is really being able to understand a problem and how its parts fit together and lead to a solution. That said, rote crunching is not something I suggest banning either but I suggest that ideas, concepts, and creativity should come first, then you get into those techniques because very often they are also still useful in analysis and being fluent at them can also make you able to solve problems more quickly, e.g. you don't want to be hitting up your calculator for 2+3 all the time in grade school and you don't want to be hitting up your CAS all the time to integrate $x \mapsto x^2$). On top of that, we have effectively now two different functions - logarithm and trigonometry - which we define by integrals, which means also that we can consider that we don't have to stop there, and this exposes the artificiality of the sacredness of so-called "elementary functions" and alows us to also perhaps consider a few more artifices of that sort like $\mathrm{erf}(x)$ which seems not too much more difficult but instead we simply reply that
$$\int e^{-x^2}\ dx$$
"can't be done" which, in light of having seen such things early on, feels like another sore cheat/blemish on the curriculum.
And to finish it all off - if you say calc can't be done before trig, I'd say that too bad Archimedes isn't here, as he would probably not have shared your sentiment since in fact he was one of the earliest to develop even a partial concept of integration and not only that but one of his applications of it was precisely the delineation of the arc length of a circle: that is why $\pi$ is called Archimedes' constant.
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edited Jun 24, 2019 at 5:25
answered May 19, 2019 at 3:57
The_SympathizerThe_Sympathizer
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$\begingroup$ If the (real) exponential function is easier to define than the trigonometric functions, then why not go by way of the imaginary exponential? And BTW, $\theta$ is not arc length, except for the unit circle. It is only proportional to arc length. An intuitive way to think about theta is to think rotations. $\endgroup$
Allawonder
– Allawonder
2019-06-24 03:38:29 +00:00
Commented Jun 24, 2019 at 3:38
$\begingroup$ There are many other little "errors" -- shall we say -- to respond to here, but well, to do some calculus richly well you need some nonalgebraic functions -- the exponential, both real and imaginary, suffices, with its inverse. About why we call these elementary. Well, similar to what you noted -- it's somewhat easy to introduce the exponential function by analogy with addition and multiplication -- at least initially. Thus, one can learn much of these functions without any calculus. But note that even all the algebraic functions are not rigorously understood yet without calculus -- in... $\endgroup$
Allawonder
– Allawonder
2019-06-24 03:51:03 +00:00
Commented Jun 24, 2019 at 3:51
1
$\begingroup$ @Allawonder : I did mention that the radian measure specifically is defined with a unit circle. My point was that measuring angles, in the way we do, is equivalent to measuring the arc length of a circle . That said, I also realized after writing this that a simpler method could be to consider repeated bisection of an angle, but you still need a limit process to fill in at all non-dyadic points, and this can introduce degree measure starting with the definition that a right angle is to be measured as 90 degrees. $\endgroup$
The_Sympathizer
– The_Sympathizer
2019-06-24 05:38:53 +00:00
Commented Jun 24, 2019 at 5:38
1
$\begingroup$ Also, reagrding the real line - yes, it is true that technically even elementary algebra on $\mathbb{R}$ requires "calculus" in this sense because that the real number line itself could be considered to be the essence of calculus, and in fact that is what motivates its definition. However, I'd think the intuitive picture of the line as a "line" is actually fairly sufficient for elementary-algebraic needs without getting into the nitty-gritty of the microstructure. The biggest nit for me there is the use of infinite decimals - which leads many people to the $\endgroup$
The_Sympathizer
– The_Sympathizer
2019-06-24 05:43:28 +00:00
Commented Jun 24, 2019 at 5:43
1
$\begingroup$ persistent misconception of $0.9999... \ne 1$, something that can't be truly disabused honestly without a closer look. My solution to that is just don't bother with infinite decimals. Stick with finite decimals, and say that most reals can only be approximated. $\endgroup$
The_Sympathizer
– The_Sympathizer
2019-06-24 05:43:32 +00:00
Commented Jun 24, 2019 at 5:43
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This is a variant of robjohn's answer. The area of the sector $ADE$ is $\frac{1}{2}x\cos^2(x)$; the area of the triangle $ABC$ is $\frac{1}{2}\sin(x)$; and the area of the sector $ABC$ is $\frac{1}{2}x$. By inclusion, we find that for $0, $$ \frac{1}{2}x\cos^2(x) \le \frac{1}{2}\sin(x) \le \frac{1}{2}x \, . $$ If we multiply each term in this inequality by $\frac{2}{x}$, we obtain the necessary bounds to apply the squeeze theorem: $$ \cos^2(x) \le \frac{\sin(x)}{x} \le 1 \, . $$ This inequality also holds for $-\frac{\pi}{2} < x < 0$, as all three functions in the inequality are even.
Since $\cos$ is continuous at $0$, and $t\mapsto t^2$ is continuous at $\cos(0)=1$, the function $x\mapsto \cos^2(x)$ is continuous at $0$, i.e. $$ \lim_{x \to 0}\cos^2(x)=1 \, . $$ Therefore, by the squeeze theorem, $$\lim_{x \to 0}\frac{\sin(x)}{x}=1 \, .$$
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edited Jan 4, 2024 at 14:49
answered Aug 4, 2021 at 22:25
JoeJoe
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$\begingroup$ I like also this approach. It is similar to $\sin x $\endgroup$
Sebastiano
– Sebastiano
2021-08-04 22:33:21 +00:00
Commented Aug 4, 2021 at 22:33
$\begingroup$ This is a late answer, but it is a good approach. $\endgroup$
Тyma Gaidash
– Тyma Gaidash
2021-08-04 23:58:05 +00:00
Commented Aug 4, 2021 at 23:58
Add a comment |
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The answer ultimately depends on how you define $\sin x$ in the first place.
Here is a more fun one! $\sin x$ is the unique function satisfying $$ y'' = -y;\quad y(0)=0,\; y'(0)=1 $$ By Theory of Ordinary Differential Equations a unique function defined on some interval containing $0$ exists. Now that we called it $\sin x$, we see that $$ \frac{\sin x}{x} = \frac{\sin x - \sin 0}{x - 0} \to y'(0) = 1 \ ,$$ by definition of derivative at $x=0$ and the initial conditions defining $y(x)=\sin x \ .$
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edited Feb 6, 2024 at 0:33
Gary
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answered Nov 20, 2019 at 22:32
Behnam EsmayliBehnam Esmayli
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$\begingroup$ I like this answer, though it would not have been very useful for me back then. $\endgroup$
FUZxxl
– FUZxxl
2019-11-21 00:31:34 +00:00
Commented Nov 21, 2019 at 0:31
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How about this proof?
We can check that function defined as \begin{align} \int_{-m}^m e^{2\pi i k x} \mathrm{d}{k} \end{align} is continuous and have a value $2m$ at $x=0$.
It is same with \begin{align} \int_{-m}^m e^{2\pi i k x} \mathrm{d}{k} = \dfrac{e^{2\pi i m x} - e^{-2\pi i m x}}{2\pi i x} = \dfrac{\sin 2 \pi m x}{\pi x} \end{align}
In conclusion, \begin{align} \lim_{x \to 0}\dfrac{\sin 2 \pi m x}{\pi x} = \lim_{x \to 0}\int_{-m}^m e^{2\pi i k x} \mathrm{d}{k} = 2m \end{align}
You can adjust $m$ as what you want.
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answered Nov 21, 2019 at 14:48
ChoMeditChoMedit
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For completeness answers let me suggest axiomatic approach to $\sin$ and $\cos$. One possible definition is here. I find another one(Ilyin, Poznyak: Fundamentals of Mathematical Analysis, 2005, vol.1, pages 146-155, Russian lang.), which claims, that there is only one pair of continuous functions on $\mathbb{R}$ for which fulfill
$$S(x+y)=S(x)C(y)+S(y)C(x)$$ $$C(x+y)=C(x)C(y)-S(y)C(x)$$ $$S^2(x)+C^2(x)=1$$ $$S(0)=0,S\left(\frac{\pi}{2}\right)=1,C(0)=1,C\left(\frac{\pi}{2}\right)=0$$ From this axioms can be obtained monotonic properties and can be proved, particularly, that for $x \in (0, \frac{\pi}{2})$ holds $0~~. Using last is easy to obtain $$\lim_{x \to 0}\frac{S(x)}{x}=1$$~~
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edited Nov 2, 2021 at 8:15
answered Jun 27, 2020 at 8:20
zkutchzkutch
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$\begingroup$ Where is the definition of pi used? I'm asking because it seems the entirety of the proof is relegated to the last sentence with the inequalities 0 ~~$\endgroup$
latbbltes
– latbbltes
2021-04-25 18:49:47 +00:00
Commented Apr 25, 2021 at 18:49~~
1
$\begingroup$ Added source to answer. From proof can be seen, that in place of $\frac{\pi}{2}$, it can be taken any $d>0$. Connection with circle, then will be obtain from definition of $\pi$ based on circle en.wikipedia.org/wiki/Pi#Definition $\endgroup$
zkutch
– zkutch
2021-11-02 08:24:14 +00:00
Commented Nov 2, 2021 at 8:24
$\begingroup$ I'm not literally asking for the definition of pi. I'm asking how you derive the inequality since it doesn't follow simply from the formal properties stated without further work $\endgroup$
latbbltes
– latbbltes
2021-11-02 20:22:00 +00:00
Commented Nov 2, 2021 at 20:22
$\begingroup$ How do I know if you are asking literally or not? You asked a question, I answered. The more precise your questions are, the more accurate my answers will be. You asked where pi is used - you can see it in the fourth line. You state that "it doesn't follow simply", but the word simple is subjective and depends on the person. In my comment, prior to your question, I included in the answer an indication of the source of the proof where the required properties are obtained on 9 pages. I leave it up to the reader to decide what is easy and what is difficult. $\endgroup$
zkutch
– zkutch
2022-01-24 23:39:58 +00:00
Commented Jan 24, 2022 at 23:39
Add a comment |
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$\begingroup$
Here is another approach.
(1) (2)
In the large triangle, $$\tan(\theta)=\frac{opp}{adg}=\frac{z}{1}=z$$ So the triangle has height $$z=\tan(\theta)$$ and base $1$ so it's area is $$Area(big triangle)=\frac{1}{2}z=\frac{1}{2}\tan(\theta)$$
Next the sector area as a fraction of the entire circle, the sector is (see the right hand side of picture (1))$$\frac{\theta}{2\pi}$$ of the entire circle so its area is
$$Area(sector)=\frac{\theta}{2\pi}(\pi)(1)^2=\frac{\theta}{2}$$ The triangle within the sector has height $y$. But $y=\frac{y}{1}=\frac{opp}{hyp}=\sin(\theta)$ so the small triangle has height $y=\sin(\theta)$ and base $1$ so its area is $$Area(small triangle)=\frac{1}{2}y=\frac{1}{2}\sin(\theta)$$ Now we can use the sandwich theorem as $$Area(big triangle)\geq Area(sector)\geq Area(small triangle)$$
using the equations we worked out above this becomes
$$\frac{\tan(\theta)}{2}\geq\frac{\theta}{2}\geq\frac{\sin(\theta)}{2}$$ now multipliying through by two and using the fact that $$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$$ we get that $$\frac{\sin(\theta)}{\cos(\theta)}\geq\theta\geq\sin(\theta)$$ taking reciprocals changes the inequalities so we have that $$\frac{\cos(\theta)}{\sin(\theta)}\leq\frac{1}{\theta}\leq\frac{1}{\sin(\theta)}$$ now finally multiplying through by $\sin(\theta)$ we get $$\cos(\theta)\leq\frac{\sin(\theta)}{\theta}\leq1$$ Now $$\lim \limits_{\theta \to 0}\cos(\theta)=1$$ and$$\lim \limits_{\theta \to 0}1=1$$
so by the sandwich theorem $$\lim \limits_{\theta \to 0}\frac{\sin(\theta)}{\theta}=1$$ also. QED
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edited Jul 17, 2023 at 16:03
J. W. Tanner
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answered Mar 30, 2017 at 19:30
user395952user395952
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Here is a proof to those familiar with power series.
The definition of $\sin(x)$ is
$$\sin(x) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}h^{2k+1}$$
Therefore we get
$$\begin{align} \lim_{x \to 0} \frac{\sin(x)}{x} &= \lim_{x \to 0} \frac{\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1}}{x} \&= \lim_{x \to 0} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k} \&= 1 + \lim_{x \to 0} \sum_{k=1}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k} \&= 1 \end{align}$$
where we have used the fact that the power series $\sum_{k=1}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k}$ has radius of convergence $R=\infty$ and therefore is continuous on $\mathbb R$. This allows us to take the limit inside and we get
$$\lim_{x \to 0} \sum_{k=1}^\infty \frac{(-1)^k}{(2k+1)!} x^{2k} = \sum_{k=1}^\infty \frac{(-1)^k}{(2k+1)!} 0^{2k} = 0$$
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answered Jan 14, 2018 at 18:08
mdcqmdcq
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$\begingroup$ Compare to this answer and this answer, which both use power series. $\endgroup$
robjohn
– robjohn ♦
2018-01-14 19:48:04 +00:00
Commented Jan 14, 2018 at 19:48
4
$\begingroup$ The premise of the question is not to use power series. $\endgroup$
FUZxxl
– FUZxxl
2018-12-04 12:34:23 +00:00
Commented Dec 4, 2018 at 12:34
$\begingroup$ @FuZxxl That depends how you define "sine". The definition such that sine is the ratio of opposite side to the hypotenuse is very vague while using as a proof. Therefore, you can choose to define sine as what philmcole suggested, with illustration of this meaning that sin x is the ratio, then you can find the limit with power rule. $\endgroup$
Kelvin Chan
– Kelvin Chan
2021-09-20 03:00:42 +00:00
Commented Sep 20, 2021 at 3:00
$\begingroup$ @KelvinChan You can, but as I already explained in a bunch of other comments, the definition based on ratio between opposing cathetus and hypothenuse is what is meant to be used. $\endgroup$
FUZxxl
– FUZxxl
2021-09-20 08:49:48 +00:00
Commented Sep 20, 2021 at 8:49
Add a comment |
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$\begingroup$
We can also use Euler's formula to prove the limit:
$$e^{ix} = \cos x + i\sin x$$
$$\lim_{x\to 0}\dfrac{\sin x}{x} = \implies \lim_{x\to 0} \dfrac{e^{ix}- e^{-ix}}{2i x}$$
$$= \lim_{x\to 0} \dfrac{e^{2ix}-1}{2ix}\times\dfrac 1{ e^{ix} }= 1 \times 1 = 1$$
since:
$\lim_{f(x)\to 0}\dfrac{e^{f(x)}-1}{f(x)} = 1$
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answered Jan 13, 2019 at 12:24
ArcherArcher
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$\begingroup$ If in the end you are going to use some "known" limit, why not doing that in the beginning? $\endgroup$
– PierreCarre
2019-04-29 14:16:10 +00:00
Commented Apr 29, 2019 at 14:16
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$\begingroup$
This is not a rigorous proof, but is instead an intuitive argument. Consider the graph of the sine function and in particular consider the origin $(0,0)$ and some arbitrary point $(x,\sin(x))$ a little bit to the right of it. Connect a secant line between the two points like so: Now consider the average rate of change of the sine function on that interval $[0,x]$, or alternatively the slope $m$ of that secant line, namely: $$m(x)=\frac{\sin(x)-\sin(0)}{x-0}$$ But clearly this is just $$m(x)=\frac{\sin(x)}{x}$$ But then "clearly" (again, this is just for intuitive purposes), as $x\to{0}$, the slope of this secant line approaches the slope of the tangent line to the graph at $x=0$ (i.e. it approaches whatever $\sin'(0)$ should be). Looking at the graph then, it shouldn't feel too unreasonable that this limiting procedure results in a $45$-degree diagonal tangent, hence a slope of $1$.
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answered May 20, 2021 at 22:20
SurfaceIntegralSurfaceIntegral
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$\begingroup$ This has already been said before. $\endgroup$
Toby Mak
– Toby Mak
2021-05-21 00:00:16 +00:00
Commented May 21, 2021 at 0:00
Add a comment |
1
$\begingroup$
Here is a slick trick using elementary integration methods. Note that \begin{align} \int_0^1 \ \cos(xt) \ dt & = \left[ \dfrac{1}{x} \cdot\sin(xt) \right]_0^1 \ & =\dfrac{\sin (x)}{x} - \dfrac{\sin (0)}{x} \ & = \dfrac{\sin (x)}{x}. \end{align} Hence, \begin{align} \lim_{x \rightarrow 0} \dfrac{\sin (x)}{x} & = \lim_{x \rightarrow 0} \int _0^1 \ \cos(xt) \ dt \ & = \int_0^1 \cos (0) \ dt \ & =1. \end{align}
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answered Jan 1, 2022 at 7:32
HelloHello
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$\begingroup$ Is there a way to know the antiderivative of $\cos$ without knowing the limit in question? I mean, the limit in question is usually used to prove that $\sin' =\cos$. $\endgroup$
– Gary
2022-01-01 07:40:03 +00:00
Commented Jan 1, 2022 at 7:40
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$\begingroup$
This geometric solution comes form this question according to the following sketch
we have
$$Area(OBP) \le Area(OAP)\le Area(OBP)+Area(ABPQ)$$
that is
$$\frac 12 \cos x |\sin x|\le \frac12 \cdot 1 \cdot |x|\le \frac 12 \cos x |\sin x|+(1-\cos x)|\sin x|=|\sin x|- \frac12 \cos x |\sin x|$$
and dividing by $\frac12|\sin x|>0$
$$ \cos x \le \frac{|x|}{|\sin x|}\le 2- \cos x $$
and since $\frac{|x|}{|\sin x|} =\frac{x}{\sin x}>0$ for $x\neq 0$ we obtain
$$ \cos x \le \frac{x}{\sin x}\le 2- \cos x $$
finally by squeeze theorem since
$\cos x \to 1$
$2- \cos x \to 2-1=1$
we conclude that
$$\lim_{x\to 0}\frac x{\sin x}=1$$
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edited Sep 3, 2022 at 16:41
answered Sep 3, 2022 at 16:31
useruser
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According to the following sketch:
using the properties for the isosceles triangle, the area of a circular sector is given by the limit:
$$\frac{\theta}2 \cdot 1^2 = \lim_{N\to \infty} N\cdot \sin\left(\frac{\theta}{2N}\right)\cdot\cos\left(\frac{\theta}{2N}\right) \iff \lim_{N\to \infty} \frac{ \sin\left(\frac{\theta}{2N}\right)}{\frac{\theta}{2N}}\cdot\cos\left(\frac{\theta}{2N}\right)=1$$
which implies the limit in hand.
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answered Mar 26, 2023 at 18:56
useruser
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Not at all appropriate for a cal one course (or any course), but for the perverse, egregious silliness of it:
A circle is parametrized by $\gamma(t) = (\cos t, \sin t)$. Now, geometrically, the tangent vector $\gamma'(t)$ is perpendicular to the radius vector, $\gamma(t)$. (Given the nature of the question, analytical differentiation is "not allowed" - hence, "geometrically." But if it was, one could differentiate $\gamma(t)\cdot \gamma(t) =1$ to see $\gamma'(t)\perp \gamma(t)$.) Hence, by rotational symmetry, one may conclude that there is a scalar $a$ such that
$$\gamma'(t) = a \pmatrix{-\sin t\ \cos t}, $$ or, in cal one terms, that $\cos' t = -a \sin t$, and $\sin' t = a \cos t$.
Now, one also has that $$ \int_{t=0}^{2\pi} |\gamma'(t)|\,dt = 2\pi,$$ whereby one concludes that $a=\pm 1$. Finally, looking at a single point, say at $t=0$, at which way $\gamma(t)$ is being traced out, one sees that $a = 1$.
In particular (cough?) $$ 1 = \cos 0 = \sin'0 = \lim_{h\to 0}{\sin h \over h},$$ as desired.
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edited Apr 18, 2024 at 17:23
answered Feb 5, 2024 at 14:57
peter a gpeter a g
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10203 | http://dynamicmathematicslearning.com/octagoncentroids.html | A generalization of Varignon's Theorem: Centroids form Parallelo-polygon
Triangle Centroids of a Hexagon form a Parallelo-Hexagon: A generalization of Varignon's Theorem
General Theorem: The 2 n centroids of the n-gons, A 1 A 2 A 3...A n, A 2 A 3 A 4...A n+1, etc. sub-dividing a 2 n-gon, A 1 A 2 A 3...A 2n (where n ≥ 2), form a 2 n-gon with opposite sides equal and parallel.
The above theorem is a generalization of Varignon's theorem (1731), which states that the midpoints of the sides of any quadrilateral form a parallelogram. The dynamic sketch below illustrates it for a hexagon.
Point A
Point B
Segment
Point C
Segment
Point D
Segment
Point E
Segment
Point F
Segment
Segment
Mid point
Segment
Mid point
Segment
Segment
Inter section G
Mid point
Segment
Mid point
Segment
Segment
Inter section H
Mid point
Segment
Mid point
Segment
Segment
Inter section I
Mid point
Segment
Mid point
Segment
Segment
Inter section J
Mid point
Segment
Mid point
Segment
Segment
Inter section K
Mid point
Segment
Mid point
Segment
Segment
Inter section L
Segment
Segment
Segment
Segment
Segment
Segment
Origin Point
Unit Point
Unit Point
Axis
Axis
Coordinate System
Drag any of the vertices of
ABCDEF
to change its
shape. Also look at concave & crossed examples.
These are the keyboard-only Toolplay instructions
Read my 2007 article in The Montana Mathematics Enthusiast A hexagon result and its generalization via proof.
Also see Nick Lord's note "Maths bite: averaging polygons" in The Mathematical Gazette, Vol. 92, No. 523, March 2008, p. 134 that gives the same generalization to 2 n-gons, and proving it very easily with vectors.
Generalization to 3D: Perhaps surprisingly, the general theorem is also true in 3D space as illustrated in the YouTube video above. Below is a dynamic Cabri 3D sketch shown for a spatial, non-planar hexagon ABCDEF, where a 'spatial' parallelo-hexagon GHIJKL is formed 1. Note that even though the hexagon GHIJKL is non-planar, it's opposite side are still equal and parallel (the length measurements are shown and the configuration can be rotated to align opposite sides and see that they are parallel).
Note: This dynamic 3D applet unfortunately no longer works on Internet Explorer or newer versions of Safari and Firefox, in which case it only gives a static image. The dynamic 3D applet also requires the downloading & installation of the free Cabri 3D Plug In, available at Windows(4 Mb) or Mac OS(13.4 Mb).
Download Cabri 3D Plugin.
Spatial Hexagon Centroids
Basic manipulation: 1) Right click (or Ctrl + click) and drag to rotate the whole figure (glassball).
2) Click to select and hold down the left button to drag any of the vertices A, C or D of the hexagon ABCDEF.
Or click Summary of manipulation to open & resize a separate window with instructions.
1Acknowledgement: I'm indebted to Zalman Usiskin from the University of Chicago who in my ICME-12 paper in Seoul, Korea in July 2012, when I referred to this 2D generalization of Varignon's theorem, raised the question of whether it generalizes to 3D, and even more to Roger Howe from Yale University who afterwards showed me a simple vector proof, similar to that of Nick Lord that I'd seen before. But Roger Howe then pointed out that this proof immediately shows that it is also valid in 3D, since vectors are not dimension specific, something which I knew, but had somehow not thought of before! So this is another excellent example of the 'discovery' function of proof, whereby Polya's 'looking-back' strategy, done in the right way, produces an immediate generalization.
HTML export of second sketch by Cabri 3D. Download a 30 day Demo, or for more information about purchasing this software, go to Cabri 3D.
Copyright © 2016 KCP Technologies, a McGraw-Hill Education Company. All rights reserved.
Release: 2015Q4Update2, Semantic Version: 4.5.1-alpha, Build Number: 1026.7-wsp-widgets, Build Stamp: ip-10-149-70-76/20180827113946
Back to "Dynamic Geometry Sketches"
Back to "Student Explorations"
Revised by Michael de Villiers, 23 July 2012; updated 4 January 2019; 16 June 2022. |
10204 | https://www.teacherspayteachers.com/Product/3rd-Grade-One-Step-Multiplication-and-Division-Word-Problem-Review-Game-3OA3-2698512 | 3rd Grade One Step Multiplication and Division Word Problem Review Game 3.OA.3
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Description
Looking for a fun interactive teaching idea for word problems? Well bingo! Look no further as Multiplication and Division Word Problems Game Puzzles, for CCSS 3.OA.3, will serve as an exciting lesson plan for 3rd grade elementary school classrooms. This is a great resource to incorporate into your unit as a guided math center rotation, review exercise, small group work, morning work, remediation, intervention or rti. It can also be used as a quiz, drill, test, or assessment tool to help determine student mastery of the learning target. Whether a student is homeschooled or given this task as a homework assignment, kids will also love working on these at home to study and improve their skills.
This puzzle set includes 20 colorful jigsaw puzzles, answer key, and an optional station instruction page with an example. These come as pdf printable sheets that can be printed on card stock and laminated for long-term use. As a suggestion, store them in a sealable gallon storage bag and place them in a foldable bin or tub for students to use throughout the year for enrichment when they finish early. Another idea, for a craft project, is students can glue the completed puzzles into a spiral notebook or journal as a model reference sheet. They can also be glued on a poster for displaying on a bulletin board or as a wall anchor chart.
By solving the word problems and finding the matching expression and answer puzzle pieces, students can gain confidence in an important and sometimes challenging skill. Your 3rd graders will love to practice and learn to develop strategies to multiply and divide within 100 using word problems!
I hope you download and enjoy this engaging hands-on manipulative activity with your students! So set those worksheets aside and give our puzzles a try!
This product is included in the 3rd Grade Math Puzzles Bundle HERE
Relevant Grade 3 Common Core Standard 3.OA.A.3
Use multiplication and division within 100 to solve word problems.
Cross Reference: Texas TEKS 3.4K
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10205 | https://pubchem.ncbi.nlm.nih.gov/compound/Acetyl-chloride | Acetyl chloride | CH3COCl | CID 6367 - PubChem
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compound Summary
Acetyl chloride
PubChem CID
6367
Structure
Primary Hazards
Laboratory Chemical Safety Summary (LCSS) Datasheet
Molecular Formula
C 2 H 3 ClO
CH 3 COCl
C 2 ClH 3 O
Synonyms
ACETYL CHLORIDE
75-36-5
Ethanoyl chloride
Acetic chloride
Acetylchloride
View More...
Molecular Weight
78.50 g/mol
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Dates
Create: 2005-03-26
Modify: 2025-09-07
Description
Acetyl chloride appears as a colorless, fuming liquid with a pungent odor. Density 9.2 lb / gal. Flash point 40 °F. Vapor, which is heavier than air, irritates the eyes and mucous membranes. Corrosive to metals and tissue.
CAMEO Chemicals
Acetyl chloride is an acyl chloride. It is functionally related to an acetic acid.
ChEBI
1 Structures
1.1 2D Structure
Structure Search
Get Image
Download Coordinates
Chemical Structure Depiction
Full screen Zoom in Zoom out
PubChem
1.2 3D Conformer
Structure Search
Get Image
Download Coordinates
Interactive Chemical Structure Model
Ball and Stick
Sticks
Wire-Frame
Space-Filling
Show Hydrogens
Animate
Full screen Zoom in Zoom out
PubChem
2 Names and Identifiers
2.1 Computed Descriptors
2.1.1 IUPAC Name
acetyl chloride
Computed by Lexichem TK 2.7.0 (PubChem release 2025.04.14)
PubChem
2.1.2 InChI
InChI=1S/C2H3ClO/c1-2(3)4/h1H3
Computed by InChI 1.07.2 (PubChem release 2025.04.14)
PubChem
2.1.3 InChIKey
WETWJCDKMRHUPV-UHFFFAOYSA-N
Computed by InChI 1.07.2 (PubChem release 2025.04.14)
PubChem
2.1.4 SMILES
CC(=O)Cl
Computed by OEChem 2.3.0 (PubChem release 2025.04.14)
PubChem
2.2 Molecular Formula
C 2 H 3 ClO
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; PubChem
CH 3 COCl
C 2 ClH 3 O
ILO-WHO International Chemical Safety Cards (ICSCs)
2.3 Other Identifiers
2.3.1 CAS
75-36-5
Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; CAS Common Chemistry; ChemIDplus; DHS Chemical Facility Anti-Terrorism Standards (CFATS) Chemicals of Interest; DrugBank; EPA Acute Exposure Guideline Levels (AEGLs); EPA Chemical Data Reporting (CDR); EPA Chemicals under the TSCA; EPA DSSTox; EPA Integrated Risk Information System (IRIS); European Chemicals Agency (ECHA); FDA Global Substance Registration System (GSRS); Hazardous Substances Data Bank (HSDB); ILO-WHO International Chemical Safety Cards (ICSCs); New Zealand Environmental Protection Authority (EPA); NJDOH RTK Hazardous Substance List
2.3.2 European Community (EC) Number
200-865-6
European Chemicals Agency (ECHA)
2.3.3 UNII
QD15RNO45K
FDA Global Substance Registration System (GSRS)
2.3.4 UN Number
1717 (ACETYL CHLORIDE)
CAMEO Chemicals; Emergency Response Guidebook (ERG)
1717
ILO-WHO International Chemical Safety Cards (ICSCs)
2.3.5 ChEBI ID
CHEBI:37580
ChEBI
2.3.6 DrugBank ID
DB14623
DrugBank
2.3.7 DSSTox Substance ID
DTXSID2023852
EPA DSSTox
2.3.8 HMDB ID
HMDB0247922
Human Metabolome Database (HMDB)
2.3.9 ICSC Number
0210
ILO-WHO International Chemical Safety Cards (ICSCs)
2.3.10 Metabolomics Workbench ID
55629
Metabolomics Workbench
2.3.11 Nikkaji Number
J1.453B
Japan Chemical Substance Dictionary (Nikkaji)
2.3.12 Wikidata
Q408038
Wikidata
2.3.13 Wikipedia
Acetyl chloride
Wikipedia
2.4 Synonyms
2.4.1 MeSH Entry Terms
acetyl chloride
acetylchloride
Medical Subject Headings (MeSH)
2.4.2 Depositor-Supplied Synonyms
ACETYL CHLORIDE
75-36-5
Ethanoyl chloride
Acetic chloride
Acetylchloride
Acetic acid chloride
Acetic acid, chloride
QD15RNO45K
DTXSID2023852
CHEBI:37580
RefChem:5510
DTXCID203852
200-865-6
CH3COCl
RCRA waste number U006
MFCD00000719
Acetyl chloride-2-13C
CH3-CO-Cl
C2H3ClO
Acetyl chloride, reagent grade, 98%
CCRIS 4568
HSDB 662
EINECS 200-865-6
UN1717
RCRA waste no. U006
UNII-QD15RNO45K
acetylchlorid
acetylchlorine
BRN 0605303
Acetic Acid Chloride; Acetic Chloride; Ethanoyl Chloride; AcCl
acetyl chlorid
acetyl chlorine
acetyl choride
acetyl cloride
actyl chloride
Acteyl chloride
AcCl
Ac-Cl
methyl carbonyl chloride
Acetic acid monochloride
SCHEMBL519
ACETIC ACID,CHLORIDE
EC 200-865-6
Acetyl Chloride, ? 99%
ACETYL CHLORIDE [MI]
4-02-00-00395 (Beilstein Handbook Reference)
ACETYL CHLORIDE [HSDB]
Ethanoyl chloride;Acetylchloride
SCHEMBL1727937
SCHEMBL3750263
SCHEMBL5180151
SCHEMBL11026359
SCHEMBL31196713
UN1717 (DOT)
BCP31842
PDA18679
STR00115
STL264238
AKOS000121189
DB14623
UN 1717
Acetyl chloride, for synthesis, 98.0%
BP-13326
InChI=1/C2H3ClO/c1-2(3)4/h1H
Acetyl chloride, ReagentPlus(R), >=99%
A0082
NS00002086
EN300-18986
Acetyl chloride [UN1717] [Flammable liquid]
Acetyl chloride, puriss. p.a., >=99.0% (T)
F094104
Q408038
F2190-0010
PubChem
3 Chemical and Physical Properties
3.1 Computed Properties
Property Name
Property Value
Reference
Property Name
Molecular Weight
Property Value
78.50 g/mol
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
XLogP3-AA
Property Value
0.8
Reference
Computed by XLogP3 3.0 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Donor Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Acceptor Count
Property Value
1
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Rotatable Bond Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Exact Mass
Property Value
77.9872424 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Monoisotopic Mass
Property Value
77.9872424 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Topological Polar Surface Area
Property Value
17.1 Ų
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Heavy Atom Count
Property Value
4
Reference
Computed by PubChem
Property Name
Formal Charge
Property Value
0
Reference
Computed by PubChem
Property Name
Complexity
Property Value
33
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Isotope Atom Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Covalently-Bonded Unit Count
Property Value
1
Reference
Computed by PubChem
Property Name
Compound Is Canonicalized
Property Value
Yes
Reference
Computed by PubChem (release 2025.04.14)
PubChem
3.2 Experimental Properties
3.2.1 Physical Description
Acetyl chloride appears as a colorless, fuming liquid with a pungent odor. Density 9.2 lb / gal. Flash point 40 °F. Vapor, which is heavier than air, irritates the eyes and mucous membranes. Corrosive to metals and tissue.
CAMEO Chemicals
Liquid
EPA Chemical Data Reporting (CDR)
Colorless fuming liquid; [HSDB] Pungent odor; [Merck Index]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
COLOURLESS FUMING LIQUID WITH PUNGENT ODOUR.
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.2 Color / Form
Liquid
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
Colorless fuming liquid
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 13
Hazardous Substances Data Bank (HSDB)
3.2.3 Odor
Pungent odor
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
Sharp odor
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
3.2.4 Boiling Point
123.6 °F at 760 mmHg (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
51 °C
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 3-6
Hazardous Substances Data Bank (HSDB)
51Â °C
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.5 Melting Point
-170 °F (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
-112.7 °C
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 3-6
Hazardous Substances Data Bank (HSDB)
-112Â °C
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.6 Flash Point
40 °F (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
4 °C
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
5 °C (41 °F) - closed cup
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
4 °C (40 °F) - closed cup
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 325-10
Hazardous Substances Data Bank (HSDB)
5Â °C c.c.
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.7 Solubility
Reaction (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
Decomposed violently by water or alcohol
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
Miscible with benzene, chloroform, ether, glacial acetic acid, petroleum ether
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
Miscible with ether, acetone, benzene, chloroform; soluble in carbon tetrachloride
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 3-6
Hazardous Substances Data Bank (HSDB)
Soluble in ether, acetone, acetic acid
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 13
Hazardous Substances Data Bank (HSDB)
Solubility in water: reaction
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.8 Density
1.1039 at 69.8 °F (USCG, 1999) - Denser than water; will sink
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
1.1051g/cu cm at 20 °C
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 3-6
Hazardous Substances Data Bank (HSDB)
Saturated liquid density: 68.849 Lb/cu ft at 70 °F
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
Saturated vapor density: 0.06130 Lb/cu ft at 70 °F
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
Relative density (water = 1): 1.11
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.9 Vapor Density
2.7 2.1 at 100 °F (NTP, 1992) - Heavier than air; will sink (Relative to Air)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
2.70 (Air = 1)
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 12th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2012., p. V2: 50
Hazardous Substances Data Bank (HSDB)
Relative vapor density (air = 1): 2.7
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.10 Vapor Pressure
135 mmHg at 45.5 °F (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
287.0 [mmHg]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
287 mm Hg at 25 °C
Daubert, T.E., R.P. Danner. Physical and Thermodynamic Properties of Pure Chemicals Data Compilation. Washington, D.C.: Taylor and Francis, 1989.
Hazardous Substances Data Bank (HSDB)
Vapor pressure, kPa at 20Â °C: 32
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.11 LogP
-0.47
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.12 Stability / Shelf Life
Stable under recommended storage conditions.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
3.2.13 Autoignition Temperature
734 °F (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
734 °F (390 °C)
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 325-10
Hazardous Substances Data Bank (HSDB)
390Â °C
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.14 Decomposition
Hazardous decomposition products formed under fire conditions - Carbon oxides, hydrogen chloride gas.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
May decompose during preparation. ... When heated to decomposition it emits highly toxic fumes of phosgene and /chloride/.
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 42
Hazardous Substances Data Bank (HSDB)
3.2.15 Corrosivity
Corrosive
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
Corrosive to metals and tissue
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
3.2.16 Heat of Combustion
-6,000 btu/lb = -3,000 cal/g = -140X10X5 J/kg
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
3.2.17 Heat of Vaporization
160 btu/Lb = 88 cal/g = 3.7X10X5 J/kg
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
3.2.18 Surface Tension
26.7 dynes/cm at 14.8 °C in contact with vapor
Weast, R.C. (ed.) Handbook of Chemistry and Physics. 67th ed. Boca Raton, FL: CRC Press, Inc., 1986-87., p. F-33
Hazardous Substances Data Bank (HSDB)
3.2.19 Refractive Index
Index of refraction: 1.3886 at 20 °C/D
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 3-6
Hazardous Substances Data Bank (HSDB)
Index of refraction: 1.3898 at 20 °C/D
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
3.2.20 Kovats Retention Index
Standard non-polar
542 , 542 , 542
NIST Mass Spectrometry Data Center
3.2.21 Other Experimental Properties
Decomposed violently by water or alcohol
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
Highly refractive
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 13
Hazardous Substances Data Bank (HSDB)
Ratio of specific heats of vapor (gas): 1.1467
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
Liquid heat capacity: 0.350 Btu/lb-F at 70 °F
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
For more Other Experimental Properties (Complete) data for Acetyl chloride (8 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
3.3 SpringerMaterials Properties
Schoenflies notation
Boiling point
Chemical bond
Chemical shift
Corrosion
Density
Diamagnetic susceptibility
Dielectric constant
Electric dipole moment
Heat of sublimation
Hindering potential
Hydrogen bonding potential
Internuclear distance
Lineshape
Magnetic susceptibility
Molar mass
Molecular structure
Moment of inertia
Nuclear quadrupole coupling
Nuclear quadrupole moment
Nuclear quadrupole resonance spectroscopy
Optical coefficient
Point group
Quadrupole coupling
Refractive index
Rotational excitation cross section
Sound absorption
Sound propagation
Sound velocity
Surface tension
Vapor pressure
Vibrational mode frequency
Viscosity
SpringerMaterials
3.4 Chemical Classes
Toxic Gases & Vapors -> Acid Halides
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Corrosives
Flammable agents - 3rd degree
Reactive agents - 2nd degree
NJDOH RTK Hazardous Substance List
3.4.1 Pesticides
Pesticide -> EPA IRIS
EPA Integrated Risk Information System (IRIS)
4 Spectral Information
4.1 1D NMR Spectra
1 of 2 items
1D NMR Spectra
1H NMR: 167 (Varian Associates NMR Spectra Catalogue)
Hazardous Substances Data Bank (HSDB)
2 of 2 items
1D NMR Spectra
NMRShiftDB Link
NMRShiftDB
4.1.1 1H NMR Spectra
Instrument Name
Varian A-60
Copyright
Copyright © 2009-2025 John Wiley & Sons, Inc. All Rights Reserved.
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SpectraBase
4.1.2 13C NMR Spectra
Source of Sample
MCB Manufacturing Chemists, Norwood, Ohio
Copyright
Copyright © 1980, 1981-2025 John Wiley & Sons, Inc. All Rights Reserved.
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SpectraBase
4.1.3 17O NMR Spectra
Copyright
Copyright © 2016-2025 W. Robien, Inst. of Org. Chem., Univ. of Vienna. All Rights Reserved.
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SpectraBase
4.2 Mass Spectrometry
4.2.1 GC-MS
1 of 4 items
View All
NIST Number
228275
Library
Main library
Total Peaks
41
m/z Top Peak
43
m/z 2nd Highest
15
m/z 3rd Highest
63
Thumbnail
NIST Mass Spectrometry Data Center
2 of 4 items
View All
NIST Number
19240
Library
Replicate library
Total Peaks
58
m/z Top Peak
43
m/z 2nd Highest
15
m/z 3rd Highest
14
Thumbnail
NIST Mass Spectrometry Data Center
4.2.2 Other MS
Other MS
MASS: 404 (NIST/EPA/MSDC Mass Spectral Database, 1990 version)
Hazardous Substances Data Bank (HSDB)
4.3 UV Spectra
Max absorption (petroleum ether): 220 nm (log epsilon = 2.01); Sadtler Ref Number: 6030 (IR, prism) (gas)
Weast, R.C. (ed.). Handbook of Chemistry and Physics. 60th ed. Boca Raton, Florida: CRC Press Inc., 1979., p. C-86
Hazardous Substances Data Bank (HSDB)
UV: 7-3 (Organic Electronic Spectral Data, Phillips et al, John Wiley & Sons, New York)
Lide, D.R., G.W.A. Milne (eds.). Handbook of Data on Organic Compounds. Volume I. 3rd ed. CRC Press, Inc. Boca Raton ,FL. 1994., p. V1: 200
Hazardous Substances Data Bank (HSDB)
4.4 IR Spectra
IR Spectra
IR: 6165 (Coblentz Society Spectral Collection)
Hazardous Substances Data Bank (HSDB)
4.4.1 FTIR Spectra
Instrument Name
Bruker IFS 85
Technique
Cell
Source of Sample
Hoechst AG, Frankfurt
Copyright
Copyright © 1989, 1990-2025 Wiley-VCH GmbH. All Rights Reserved.
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SpectraBase
4.4.2 ATR-IR Spectra
Source of Sample
Sigma-Aldrich
Catalog Number
114189
Copyright
Copyright © 2018-2025 Sigma-Aldrich Co. LLC. - Database Compilation Copyright © 2018-2025 John Wiley & Sons, Inc. All Rights Reserved.
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SpectraBase
4.4.3 Near IR Spectra
Instrument Name
BRUKER IFS 88
Technique
NIR Spectrometer= INSTRUMENT PARAMETERS=INST=BRUKER,RSN=10718,REO=2,CNM=HEI,ZFF=2
Source of Spectrum
Prof. Buback, University of Goettingen, Germany
Copyright
Copyright © 1989, 1990-2025 Wiley-VCH GmbH. All Rights Reserved.
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SpectraBase
4.4.4 Vapor Phase IR Spectra
Instrument Name
DIGILAB FTS-14
Technique
Vapor Phase
Copyright
Copyright © 1980, 1981-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
4.5 Raman Spectra
Catalog Number
114189
Copyright
Copyright © 2017-2025 Sigma-Aldrich Co. LLC. - Database Compilation Copyright © 2017-2025 John Wiley & Sons, Inc. All Rights Reserved.
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SpectraBase
5 Related Records
5.1 Related Compounds with Annotation
Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here.
PubChem
5.2 Related Compounds
Same Connectivity Count
13
Same Parent, Connectivity Count
55
Same Parent, Exact Count
43
Mixtures, Components, and Neutralized Forms Count
3410
Similar Compounds (2D)
View in PubChem Search
Similar Conformers (3D)
View in PubChem Search
PubChem
5.3 Substances
5.3.1 PubChem Reference Collection SID
481106850
PubChem
5.3.2 Related Substances
All Count
3925
Same Count
263
Mixture Count
3662
PubChem
5.3.3 Substances by Category
PubChem
5.4 Entrez Crosslinks
PubMed Count
221
PubChem
5.5 NCBI LinkOut
NCBI
6 Chemical Vendors
PubChem
7 Use and Manufacturing
7.1 Uses
Sources/Uses
Used in chemical production and assays; [HSDB]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Industrial Processes with risk of exposure
Toxic Gas from Spilling Chemical in Water [Category: Other]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Acetylating agent; in testing for cholesterol, determination of water in organic liquids.
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
Organic preparations (acetylating agent); dyestuffs; pharmaceuticals.
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 13
Hazardous Substances Data Bank (HSDB)
Acetyl chloride is an efficient acetylating agent for alcohols and amines to produce esters and amides. It is important in the synthesis of dyes and pharmaceuticals. Acetyl chloride is used in the Friedel-Craft acylation of benzene to yield acetophenone.
Le Berre C et al; Acetic Acid. Ullmann's Encyclopedia of Industrial Chemistry 7th ed. (1999-2018). NY, NY: John Wiley & Sons. Online Posting Date: March 26, 2014
Hazardous Substances Data Bank (HSDB)
Acetyl chloride is a powerful acetylating agent. It is used in the manufacture of aspirin, acetaminophen, acetanilide, and acetophenone. Liquid crystal compositions for optical display and memory devices frequently require acetyl chloride.
Wagner FS; Acetyl Chloride. Kirk-Othmer Encyclopedia of Chemical Technology (1999-2018). John Wiley & Sons, Inc. Online Posting Date: July 19, 2002
Hazardous Substances Data Bank (HSDB)
For more Uses (Complete) data for Acetyl chloride (7 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
7.1.1 Industry Uses
Intermediate
EPA Chemical Data Reporting (CDR)
7.1.2 Consumer Uses
Intermediate
EPA Chemical Data Reporting (CDR)
7.2 Methods of Manufacturing
Prepared from acetic acid and chlorine in the presence of phosphorus; from acetic acid and salts of chlorosulfonic acid; from sodium acetate and sulfuryl chloride. ... Lab preparation from acetic anhydride and calcium chloride.
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
By mixing glacial acetic acid and phosphorus trichloride in the cold and heating a short time to drive off hydrochloric acid. The acetyl chloride is then distilled.
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 13
Hazardous Substances Data Bank (HSDB)
The normal industrial method involves reaction of acetic anhydride with anhydrous hydrogen chloride. On the laboratory scale, it can be produced from acetic acid and reagents such as thionyl chloride, phosphorus trichloride, or phosphorus pentachloride.
Le Berre C et al; Acetic Acid. Ullmann's Encyclopedia of Industrial Chemistry 7th ed. (1999-2018). NY, NY: John Wiley & Sons. Online Posting Date: March 26, 2014
Hazardous Substances Data Bank (HSDB)
Acetyl chloride was formerly manufactured by the action of thionyl chloride, Cl2OS, on gray acetate of lime, but this route has been largely supplanted by the reaction of sodium acetate or acetic acid and phosphorus trichloride.
Wagner FS; Acetyl Chloride. Kirk-Othmer Encyclopedia of Chemical Technology (1999-2018). John Wiley & Sons, Inc. Online Posting Date: July 19, 2002
Hazardous Substances Data Bank (HSDB)
Other acetyl chloride preparations include: the reaction of acetic acid and chlorinated ethylenes in the presence of ferric chloride; a combination of benzyl chloride and acetic acid at 85% yield; conversion of ethylidene dichloride, in 91% yield; and decomposition of ethyl acetate by the action of phosgene ... .
Wagner FS; Acetyl Chloride. Kirk-Othmer Encyclopedia of Chemical Technology (1999-2018). John Wiley & Sons, Inc. Online Posting Date: July 19, 2002
Hazardous Substances Data Bank (HSDB)
7.3 Impurities
Acetyl chloride frequently contains 1-2% by weight of acetic acid or hydrochloric acid. Phosphorus or sulfur-containing acids may also be present in the commercial material.
Wagner FS; Acetyl Chloride. Kirk-Othmer Encyclopedia of Chemical Technology (1999-2018). John Wiley & Sons, Inc. Online Posting Date: July 19, 2002
Hazardous Substances Data Bank (HSDB)
7.4 U.S. Production
Aggregated Product Volume
2019: <1,000,000 lb
2018: <1,000,000 lb
2017: <1,000,000 lb
2016: <1,000,000 lb
EPA Chemical Data Reporting (CDR)
(1972) LESS THAN 4.54X10+5 GRAMS
SRI
Hazardous Substances Data Bank (HSDB)
(1975) PROBABLY LESS THAN 4.54X10+5 GRAMS
SRI
Hazardous Substances Data Bank (HSDB)
The total U.S. market may amount to only 500 metric ton annually.
Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V1 (1991) 157
Hazardous Substances Data Bank (HSDB)
Non-confidential 2016 Chemical Data Reporting (CDR) information on the production and use of chemicals manufactured or imported into the United States. Chemical: Acetyl chloride:
Table: National Aggregate Production Volume (pounds)
2012
2013
2014
2015
2012
100,000 - 500,000
2013
25,000 - 100,000
2014
<25,000
2015
<25,000
USEPA; 2016 Chemical Data Reporting Database. Acetyl Chloride (75-36-5). Available from, as of October 24, 2018:
Hazardous Substances Data Bank (HSDB)
7.5 General Manufacturing Information
Industry Processing Sectors
Not Known or Reasonably Ascertainable
EPA Chemical Data Reporting (CDR)
EPA TSCA Commercial Activity Status
Acetyl chloride: ACTIVE
EPA Chemicals under the TSCA
Acetyl chloride is normally consumed at the site of generation since transportation and storage are difficult.
Le Berre C et al; Acetic Acid. Ullmann's Encyclopedia of Industrial Chemistry 7th ed. (1999-2018). NY, NY: John Wiley & Sons. Online Posting Date: March 26, 2014
Hazardous Substances Data Bank (HSDB)
8 Safety and Hazards
8.1 Hazards Identification
ERG Hazard Classes
Water-reactive material (WR)
Emergency Response Guidebook (ERG)
8.1.1 GHS Classification
1 of 5 items
View All
Pictogram(s)
Signal
Danger
GHS Hazard Statements
H225: Highly Flammable liquid and vapor [Danger Flammable liquids]
H314: Causes severe skin burns and eye damage [Danger Skin corrosion/irritation]
Precautionary Statement Codes
P210, P233, P240, P241, P242, P243, P260, P264, P280, P301+P330+P331, P302+P361+P354, P303+P361+P353, P304+P340, P305+P354+P338, P316, P321, P363, P370+P378, P403+P235, P405, and P501
(The corresponding statement to each P-code can be found at the GHS Classification page.)
Regulation (EC) No 1272/2008 of the European Parliament and of the Council
8.1.2 Hazard Classes and Categories
Flam. Liq. 2 (100%)
Met. Corr. 1 (33.7%)
Acute Tox. 4 (12.6%)
Skin Corr. 1B (100%)
Eye Dam. 1 (52.1%)
European Chemicals Agency (ECHA)
View More...
8.1.3 NFPA Hazard Classification
NFPA 704 Diamond
NFPA Health Rating
3 - Materials that, under emergency conditions, can cause serious or permanent injury.
NFPA Fire Rating
3 - Liquids and solids that can be ignited under almost all ambient temperature conditions. Materials produce hazardous atmospheres with air under almost all ambient temperatures or, though unaffected by ambient temperatures, are readily ignited under almost all conditions.
NFPA Instability Rating
2 - Materials that readily undergo violent chemical changes at elevated temperatures and pressures.
NFPA Specific Notice
W - No water: Materials that react violently or explosively with water.
Hazardous Substances Data Bank (HSDB)
8.1.4 DOT Hazard Classification
Substance (Descriptions/Shipping Name)
Acetyl chloride
DOT ID (UN/NA Number)
UN1717
Hazard Class/Division
Class 3 Flammable and combustible liquid (49 eCFR § 173.120)
Label Codes
Class 3 Flammable and combustible liquid (49 eCFR § 173.120)
Class 8 Corrosive material (49 eCFR § 173.136)
Packing Group
PG II: the degree of danger presented by the material is medium
For more information about the packing group assignment, please visit 49 eCFR § 173
Placard/Label(s)
US Code of Federal Regulations, Hazardous Materials, 49 CFR Part 172
8.1.5 Health Hazards
Vapor irritates mucous membranes. Ingestion of liquid or contact with eyes or skin causes severe irritation. (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
ERG 2024, Guide 155 (Acetyl chloride)
· TOXIC and/or CORROSIVE; inhalation, ingestion or contact (skin, eyes) with vapors, dusts or substance may cause severe injury, burns or death.
· Bromoacetates and chloroacetates are extremely irritating/lachrymators (cause eye irritation and flow of tears).
· Reaction with water or moist air may release toxic, corrosive or flammable gases.
· Reaction with water may generate much heat that will increase the concentration of fumes in the air.
· Fire will produce irritating, corrosive and/or toxic gases.
· Runoff from fire control or dilution water may be corrosive and/or toxic and cause environmental contamination.
Emergency Response Guidebook (ERG)
8.1.6 Fire Hazards
Special Hazards of Combustion Products: When heated to decomposition, hydrogen chloride and phosgene, extremely poisonous gases, are evolved.
Behavior in Fire: Vapor is heavier than air and may travel a considerable distance to a source of ignition and flash back. (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
ERG 2024, Guide 155 (Acetyl chloride)
· HIGHLY FLAMMABLE: Will be easily ignited by heat, sparks or flames.
· Vapors form explosive mixtures with air: indoors, outdoors and sewers explosion hazards.
· Most vapors are heavier than air. They will spread along the ground and collect in low or confined areas (sewers, basements, tanks, etc.).
· Vapors may travel to source of ignition and flash back.
· Those substances designated with a (P) may polymerize explosively when heated or involved in a fire.
· Substance will react with water (some violently) releasing flammable, toxic or corrosive gases and runoff.
· Corrosives in contact with metals may evolve flammable hydrogen gas.
· Containers may explode when heated or if contaminated with water.
Emergency Response Guidebook (ERG)
Highly flammable. Many reactions may cause fire or explosion. Gives off irritating or toxic fumes (or gases) in a fire. Vapour/air mixtures are explosive.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.1.7 Hazards Summary
Corrosive to skin; [Quick CPC] A corrosive substance that can cause pulmonary edema; [ICSC] Reacts violently with water; Can cause severe burns; [Merck Index] Human inhalation of 2 ppm for 1 minute causes other changes in olfaction and lungs, thorax, or respiration; Causes spastic paralysis, excitement, and other changes in lungs, thorax, or respiration in oral lethal-dose studies of rats; [RTECS] See the Process, Toxic Gas from Spilling Chemical in Water.
Quick CPC - Forsberg K, Mansdorf SZ. Quick Selection Guide to Chemical Protective Clothing, 5th Ed. Hoboken, NJ: Wiley-Interscience, 2007.
Merck Index - O'Neil MJ, Heckelman PE, Dobbelaar PH, Roman KJ (eds). The Merck Index, An Encyclopedia of Chemicals, Drugs, and Biologicals, 15th Ed. Cambridge, UK: The Royal Society of Chemistry, 2013.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
8.1.8 Fire Potential
Dangerous fire hazard when exposed to heat or flame.
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 42
Hazardous Substances Data Bank (HSDB)
Flammable.
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
8.1.9 Skin, Eye, and Respiratory Irritations
A human systemic irritant by inhalation.
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 42
Hazardous Substances Data Bank (HSDB)
8.1.10 EPA Hazardous Waste Number
U006; A toxic waste when a discarded commercial chemical product or manufacturing chemical intermediate or an off-specification commercial chemical product or a manufacturing chemical intermediate.
Hazardous Substances Data Bank (HSDB)
8.2 Safety and Hazard Properties
8.2.1 Acute Exposure Guideline Levels (AEGLs)
The compound is currently at the Holding Status AEGLs which have been reviewed by the NAC/AEGL Committee and are on hold due to insufficient data to develop AEGL values.
EPA Acute Exposure Guideline Levels (AEGLs)
8.2.1.1 AEGLs Notes
AEGLs Status: Holding
EPA Acute Exposure Guideline Levels (AEGLs)
8.2.2 Flammable Limits
Lower: 5.0% by volume
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 325-10
Hazardous Substances Data Bank (HSDB)
8.2.3 Lower Explosive Limit (LEL)
5 % (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
8.2.4 Critical Temperature & Pressure
Critical temperature: 475 °F = 246 °C = 519 deg K; Critical pressure: 845 psia = 57.5 atm = 5.83 MN/sq m
NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
8.2.5 Physical Dangers
The vapour is heavier than air and may travel along the ground; distant ignition possible.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.2.6 Explosive Limits and Potential
Explosive limits , vol% in air: 7.3-19
ILO-WHO International Chemical Safety Cards (ICSCs)
8.3 First Aid Measures
Inhalation First Aid
Fresh air, rest. Half-upright position. Artificial respiration may be needed. Refer immediately for medical attention.
ILO-WHO International Chemical Safety Cards (ICSCs)
Skin First Aid
Wear protective gloves when administering first aid. First rinse with plenty of water for at least 15 minutes, then remove contaminated clothes and rinse again. Refer immediately for medical attention.
ILO-WHO International Chemical Safety Cards (ICSCs)
Eye First Aid
Rinse with plenty of water (remove contact lenses if easily possible). Refer immediately for medical attention.
ILO-WHO International Chemical Safety Cards (ICSCs)
Ingestion First Aid
Rinse mouth. Do NOT induce vomiting. Give nothing to drink. Refer immediately for medical attention.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.3.1 First Aid
EYES: First check the victim for contact lenses and remove if present. Flush victim's eyes with water or normal saline solution for 20 to 30 minutes while simultaneously calling a hospital or poison control center. Do not put any ointments, oils, or medication in the victim's eyes without specific instructions from a physician. IMMEDIATELY transport the victim after flushing eyes to a hospital even if no symptoms (such as redness or irritation) develop.
SKIN: IMMEDIATELY flood affected skin with water while removing and isolating all contaminated clothing. Gently wash all affected skin areas thoroughly with soap and water. IMMEDIATELY call a hospital or poison control center even if no symptoms (such as redness or irritation) develop. IMMEDIATELY transport the victim to a hospital for treatment after washing the affected areas.
INHALATION: IMMEDIATELY leave the contaminated area; take deep breaths of fresh air. If symptoms (such as wheezing, coughing, shortness of breath, or burning in the mouth, throat, or chest) develop, call a physician and be prepared to transport the victim to a hospital. Provide proper respiratory protection to rescuers entering an unknown atmosphere. Whenever possible, Self-Contained Breathing Apparatus (SCBA) should be used; if not available, use a level of protection greater than or equal to that advised under Protective Clothing.
INGESTION: DO NOT INDUCE VOMITING. Corrosive chemicals will destroy the membranes of the mouth, throat, and esophagus and volatile chemicals have a high risk of being aspirated into the victim's lungs during vomiting. Thus, the risk of increasing the medical problems by inducing vomiting of a volatile corrosive chemical is very high. If the victim is conscious and not convulsing, give 1 or 2 glasses of water to dilute the chemical and IMMEDIATELY call a hospital or poison control center. IMMEDIATELY transport the victim to a hospital. If the victim is convulsing or unconscious, do not give anything by mouth, ensure that the victim's airway is open and lay the victim on his/her side with the head lower than the body. DO NOT INDUCE VOMITING. IMMEDIATELY transport the victim to a hospital. (NTP, 1992)
National Toxicology Program, Institute of Environmental Health Sciences, National Institutes of Health (NTP). 1992. National Toxicology Program Chemical Repository Database. Research Triangle Park, North Carolina.
CAMEO Chemicals
ERG 2024, Guide 155 (Acetyl chloride)
General First Aid:
· Call 911 or emergency medical service.
· Ensure that medical personnel are aware of the material(s) involved, take precautions to protect themselves and avoid contamination.
· Move victim to fresh air if it can be done safely.
· Administer oxygen if breathing is difficult.
· If victim is not breathing:
-- DO NOT perform mouth-to-mouth resuscitation; the victim may have ingestedor inhaled the substance.
-- If equipped and pulse detected, wash face and mouth, then give artificial respiration using a proper respiratory medical device (bag-valve mask, pocket mask equipped with a one-way valve or other device).
-- If no pulse detected or no respiratory medical device available, provide continuouscompressions. Conduct a pulse check every two minutes or monitor for any signs of spontaneous respirations.
· Remove and isolate contaminated clothing and shoes.
· For minor skin contact, avoid spreading material on unaffected skin.
· In case of contact with substance, remove immediately by flushing skin or eyes with running water for at least 20 minutes.
· For severe burns, immediate medical attention is required.
· Effects of exposure (inhalation, ingestion, or skin contact) to substance may be delayed.
· Keep victim calm and warm.
· Keep victim under observation.
· For further assistance, contact your local Poison Control Center.
· Note: Basic Life Support (BLS) and Advanced Life Support (ALS) should be done by trained professionals.
Specific First Aid:
· For corrosives, in case of contact, immediately flush skin or eyes with running water for at least 30 minutes. Additional flushing may be required.
In Canada, an Emergency Response Assistance Plan (ERAP) may be required for this product. Please consult the shipping paper and/or the "ERAP" section.
Emergency Response Guidebook (ERG)
8.4 Fire Fighting
Excerpt from ERG Guide 155 [Substances - Toxic and/or Corrosive (Flammable / Water-Sensitive)]:
Note: Most foams will react with the material and release corrosive/toxic gases. CAUTION: For Acetyl chloride (UN1717), use CO2 or dry chemical only.
SMALL FIRE: CO2, dry chemical, dry sand, alcohol-resistant foam.
LARGE FIRE: Water spray, fog or alcohol-resistant foam. FOR CHLOROSILANES, DO NOT USE WATER; use alcohol-resistant foam. If it can be done safely, move undamaged containers away from the area around the fire. Avoid aiming straight or solid streams directly onto the product.
FIRE INVOLVING TANKS, RAIL TANK CARS OR HIGHWAY TANKS: Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles. Do not get water inside containers. Cool containers with flooding quantities of water until well after fire is out. Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank. ALWAYS stay away from tanks in direct contact with flames. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Use powder, carbon dioxide. NO hydrous agents, water. In case of fire: keep drums, etc., cool by spraying with water. NO direct contact with water.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.4.1 Fire Fighting Procedures
Suitable extinguishing media: Dry powder, dry sand. Unsuitable extinguishing media: Do NOT use water jet.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Advice for firefighters: Wear self-contained breathing apparatus for firefighting if necessary.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
To fight fire, use /carbon dioxide/ or dry chemical.
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 42
Hazardous Substances Data Bank (HSDB)
If material on fire or involved in fire: Use dry chemical or carbon dioxide. Do not use water on material itself. If large quantities of combustibles are involved, use water in flooding quantities as spray and fog. Use water spray to knock-down vapors. Cool all affected containers with flooding quantities of water. Apply water from as far a distance as possible.
Association of American Railroads; Bureau of Explosives. Emergency Handling of Hazardous Materials in Surface Transportation. Association of American Railroads, Pueblo, CO. 2005, p. 8
Hazardous Substances Data Bank (HSDB)
For more Fire Fighting Procedures (Complete) data for Acetyl chloride (6 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
8.4.2 Firefighting Hazards
Varpors are heavier than air and may travel to a source of ignition and flash back. ... Closed containers may rupture violently when heated.
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-10
Hazardous Substances Data Bank (HSDB)
8.5 Accidental Release Measures
Public Safety: ERG 2024, Guide 155 (Acetyl chloride)
· CALL 911. Then call emergency response telephone number on shipping paper. If shipping paper not available or no answer, refer to appropriate telephone number listed on the inside back cover.
· Keep unauthorized personnel away.
· Stay upwind, uphill and/or upstream.
· Ventilate closed spaces before entering, but only if properly trained and equipped.
Emergency Response Guidebook (ERG)
Spill or Leak: ERG 2024, Guide 155 (Acetyl chloride)
· ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area.
· All equipment used when handling the product must be grounded.
· Do not touch damaged containers or spilled material unless wearing appropriate protective clothing.
· Stop leak if you can do it without risk.
· A vapor-suppressing foam may be used to reduce vapors.
· FOR CHLOROSILANES, use alcohol-resistant foam to reduce vapors.
· DO NOT GET WATER on spilled substance or inside containers.
· Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material.
· Prevent entry into waterways, sewers, basements or confined areas.
Small Spill
· Cover with DRY earth, DRY sand or other non-combustible material followed with plastic sheet to minimize spreading or contact with rain.
· Use clean, non-sparking tools to collect material and place it into loosely covered plastic containers for later disposal.
Emergency Response Guidebook (ERG)
8.5.1 Toxic-by-Inhalation (TIH) Gas
HCl - when spill Acetyl chloride into water.
Emergency Response Guidebook (ERG)
8.5.2 Isolation and Evacuation
Excerpt from ERG Guide 155 [Substances - Toxic and/or Corrosive (Flammable / Water-Sensitive)]:
IMMEDIATE PRECAUTIONARY MEASURE: Isolate spill or leak area in all directions for at least 50 meters (150 feet) for liquids and at least 25 meters (75 feet) for solids.
SPILL: See ERG Table 1 - Initial Isolation and Protective Action Distances on the UN/NA 1717 datasheet.
FIRE: If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 800 meters (1/2 mile) in all directions; also, consider initial evacuation for 800 meters (1/2 mile) in all directions. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Evacuation: ERG 2024, Guide 155 (Acetyl chloride)
Immediate precautionary measure
· Isolate spill or leak area in all directions for at least 50 meters (150 feet) for liquids and at least 25 meters (75 feet) for solids.
Spill
· For highlighted materials: see Table 1 - Initial Isolation and Protective Action Distances.
· For non-highlighted materials: increase the immediate precautionary measure distance, in the downwind direction, as necessary.
Fire
· If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 800 meters (1/2 mile) in all directions; also, consider initial evacuation for 800 meters (1/2 mile) in all directions.
Emergency Response Guidebook (ERG)
Isolation
When spilled in water
Small spill:
ISOLATE in all directions: 30 m (100 ft)
Large spill:
ISOLATE in all directions: 60 m (200 ft)
Emergency Response Guidebook (ERG)
Protection
When spilled in water
Small spill:
PROTECT people from downwind during DAY time: 0.1 km (0.1 mi)
PROTECT people from downwind during NIGHT time: 0.1 km (0.1 mi)
Large spill:
PROTECT people from downwind during DAY time: 0.7 km (0.4 mi)
PROTECT people from downwind during NIGHT time: 2.0 km (1.2 mi)
Emergency Response Guidebook (ERG)
8.5.3 Spillage Disposal
Evacuate danger area! Consult an expert! Personal protection: complete protective clothing including self-contained breathing apparatus. Do NOT let this chemical enter the environment. Do NOT wash away into sewer. Collect leaking and spilled liquid in sealable containers as far as possible. Absorb remaining liquid in sand or inert absorbent. Then store and dispose of according to local regulations.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.5.4 Cleanup Methods
ACCIDENTAL RELEASE MEASURES: Personal precautions, protective equipment and emergency procedures: Use personal protective equipment. Avoid breathing vapors, mist or gas. Ensure adequate ventilation. Remove all sources of ignition. Evacuate personnel to safe areas. Beware of vapors accumulating to form explosive concentrations. Vapors can accumulate in low areas. Environmental precautions: Prevent further leakage or spillage if safe to do so. Do not let product enter drains. Discharge into the environment must be avoided. Methods and materials for containment and cleaning up: Contain spillage, and then collect with non-combustible absorbent material, (e.g. sand, earth, diatomaceous earth, vermiculite) and place in container for disposal according to local/national regulations. Do not flush with water.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Wear special protective clothing and positive pressure self-contained breathing apparatus. Eliminate all ignition sources. Stop or control the leak, if this can be done without undue risk. Approach release from upwind. Control runoff and isolate discharged material for proper disposal.
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-10
Hazardous Substances Data Bank (HSDB)
Environmental considerations - Land spill: Dig a pit, pond, lagoon, holding area to contain liquid or solid material. /SRP: If time permits, pits, ponds, lagoons, soak holes, or holding areas should be sealed with an impermeable flexible membrane liner./ Dike surface flow using soil, sand bags, foamed polyurethane, or foamed concrete. Absorb bulk liquid with fly ash, cement powder, or commercial sorbents. Neutralize with agricultural lime (CaO), crushed limestone (CaCO3) or sodium bicarbonate (NaHCO3). Apply "universal" gelling agent to immobilize spill.
Association of American Railroads; Bureau of Explosives. Emergency Handling of Hazardous Materials in Surface Transportation. Association of American Railroads, Pueblo, CO. 2005, p. 9
Hazardous Substances Data Bank (HSDB)
Environmental considerations - Water spill: Neutralize with agricultural lime (CaO), crushed limestone (CaCO3), or sodium bicarbonate (NaHCO3). Use mechanical dredges or lifts to remove immobilized masses of pollutants and precipitates.
Association of American Railroads; Bureau of Explosives. Emergency Handling of Hazardous Materials in Surface Transportation. Association of American Railroads, Pueblo, CO. 2005, p. 9
Hazardous Substances Data Bank (HSDB)
Environmental considerations - Air spill: Apply water spray or mist to knock down vapors. Vapor knockdown water is corrosive or toxic and should be diked for containment.
Association of American Railroads; Bureau of Explosives. Emergency Handling of Hazardous Materials in Surface Transportation. Association of American Railroads, Pueblo, CO. 2005, p. 9
Hazardous Substances Data Bank (HSDB)
8.5.5 Disposal Methods
Generators of waste (equal to or greater than 100 kg/mo) containing this contaminant, EPA hazardous waste number U006, must conform with USEPA regulations in storage, transportation, treatment and disposal of waste.
40 CFR 240-280, 300-306, 702-799 (USEPA); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of April 2, 2015:
Hazardous Substances Data Bank (HSDB)
Product: Offer surplus and non-recyclable solutions to a licensed disposal company. Burn in a chemical incinerator equipped with an afterburner and scrubber but exert extra care in igniting as this material is highly flammable. Contact a licensed professional waste disposal service to dispose of this material; Contaminated packaging: Dispose of as unused product.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Acetyl chloride is a waste chemical stream constituent which may be subjected to ultimate disposal by controlled incineration.
USEPA; Engineering Handbook for Hazardous Waste Incineration p.2-4 (1981) EPA 68-03-3025
Hazardous Substances Data Bank (HSDB)
A potential candidate for liquid injection incineration at a temperture range of 650 to 1,600 °C and a residence time 0.1 to 2 seconds. A potential candidate for rotary kiln incineration at a temperature range of 820 to 1,600 °C and residence times of seconds for liquids and gases, and hours for solids. A potential candidate for fluidized bed incineration at a temperature range of 450 to 980 °C and residence times of seconds for liquids and gases, and longer for solids.
USEPA; Engineering Handbook for Hazardous Waste Incineration p.3-11 (1981) EPA 68-03-3025
Hazardous Substances Data Bank (HSDB)
8.5.6 Preventive Measures
ACCIDENTAL RELEASE MEASURES: Personal precautions, protective equipment and emergency procedures: Use personal protective equipment. Avoid breathing vapors, mist or gas. Ensure adequate ventilation. Remove all sources of ignition. Evacuate personnel to safe areas. Beware of vapors accumulating to form explosive concentrations. Vapors can accumulate in low areas. Environmental precautions: Prevent further leakage or spillage if safe to do so. Do not let product enter drains. Discharge into the environment must be avoided.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Precautions for safe handling: Avoid inhalation of vapor or mist. Flash back possible over considerable distance. Use explosion-proof equipment. Keep away from sources of ignition - No smoking. Take measures to prevent the build up of electrostatic charge.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Appropriate engineering controls: Handle in accordance with good industrial hygiene and safety practice. Wash hands before breaks and at the end of workday.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Gloves must be inspected prior to use. Use proper glove removal technique (without touching glove's outer surface) to avoid skin contact with this product. Dispose of contaminated gloves after use in accordance with applicable laws and good laboratory practices. Wash and dry hands.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
For more Preventive Measures (Complete) data for Acetyl chloride (8 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
8.6 Handling and Storage
8.6.1 Nonfire Spill Response
Excerpt from ERG Guide 155 [Substances - Toxic and/or Corrosive (Flammable / Water-Sensitive)]:
ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area. All equipment used when handling the product must be grounded. Do not touch damaged containers or spilled material unless wearing appropriate protective clothing. Stop leak if you can do it without risk. A vapor-suppressing foam may be used to reduce vapors. FOR CHLOROSILANES, use alcohol-resistant foam to reduce vapors. DO NOT GET WATER on spilled substance or inside containers. Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material. Prevent entry into waterways, sewers, basements or confined areas.
SMALL SPILL: Cover with DRY earth, DRY sand or other non-combustible material followed with plastic sheet to minimize spreading or contact with rain. Use clean, non-sparking tools to collect material and place it into loosely covered plastic containers for later disposal. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
8.6.2 Safe Storage
Fireproof. Store in an area without drain or sewer access. Separated from incompatible materials. See Chemical Dangers. Dry. Well closed.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.6.3 Storage Conditions
Keep container tightly closed in a dry and well-ventilated place. Containers which are opened must be carefully resealed and kept upright to prevent leakage. Keep away from water. Never allow product to get in contact with water during storage. Hydrolyzes readily. Handle and store under inert gas.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Separate from alcohols, alkalies, amines, and strong oxidizing materials. Store in a cool, dry well-ventilated location. Outside or detached storage is preferred. Inside storage should be in a standard flammable liquids storage warehouse, room, or cabinet.
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-10
Hazardous Substances Data Bank (HSDB)
8.7 Exposure Control and Personal Protection
Protective Clothing: ERG 2024, Guide 155 (Acetyl chloride)
· Wear positive pressure self-contained breathing apparatus (SCBA).
· Wear chemical protective clothing that is specifically recommended by the manufacturer when there is NO RISK OF FIRE.
· Structural firefighters' protective clothing provides thermal protection but only limited chemical protection.
Emergency Response Guidebook (ERG)
Exposure Summary
TIH (Toxic Inhalation Hazard) - Term used to describe gases and volatile liquids that are toxic when inhaled. Some are TIH materials themselves, e.g., chlorine, and some release TIH gases when spilled in water, e.g., chlorosilanes. [ERG 2016].
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
8.7.1 Emergency Response Planning Guidelines
Emergency Response: ERG 2024, Guide 155 (Acetyl chloride)
· Note: Most foams will react with the material and release corrosive/toxic gases.
CAUTION: For Acetyl chloride (UN1717), use CO2 or dry chemical only.
Small Fire
· CO2, dry chemical, dry sand, alcohol-resistant foam.
Large Fire
· Water spray, fog or alcohol-resistant foam.
· FOR CHLOROSILANES, DO NOT USE WATER; use alcohol-resistant foam.
· If it can be done safely, move undamaged containers away from the area around the fire.
· Avoid aiming straight or solid streams directly onto the product.
Fire Involving Tanks, Rail Tank Cars or Highway Tanks
· Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles.
· Do not get water inside containers.
· Cool containers with flooding quantities of water until well after fire is out.
· Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank.
· ALWAYS stay away from tanks in direct contact with flames.
Emergency Response Guidebook (ERG)
8.7.2 Inhalation Risk
No indication can be given about the rate at which a harmful concentration of this substance in the air is reached.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.7.3 Effects of Short Term Exposure
The substance is corrosive to the eyes and skin. The vapour is severely irritating to the eyes and respiratory tract. Corrosive on ingestion. Exposure at high concentrations could cause asphyxiation due to swelling in the throat. Inhalation of high concentrations may cause lung oedema, but only after initial corrosive effects on the eyes and the upper respiratory tract have become manifest. The effects may be delayed. Medical observation is indicated.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.7.4 Effects of Long Term Exposure
Repeated or chronic inhalation of the vapour may cause chronic inflammation of the upper respiratory tract. Repeated or prolonged inhalation of high concentrations may cause effects on the lungs.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.7.5 Personal Protective Equipment (PPE)
Safety goggles; rubber or plastic gloves; self-contained breathing apparatus. (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
Eye/face protection: Tightly fitting safety goggles. Faceshield (8-inch minimum). Use equipment for eye protection tested and approved under appropriate government standards such as NIOSH (US) or EN 166(EU).
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Skin protection: Handle with gloves.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Body Protection: Complete suit protecting against chemicals. Flame retardant antistatic protective clothing. The type of protective equipment must be selected according to the concentration and amount of the dangerous substance at the specific workplace.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Respiratory protection: Where risk assessment shows air-purifying respirators are appropriate use a full-face respirator with multipurpose combination (US) or type AXBEK (EN 14387) respirator cartridges as a backup to engineering controls. If the respirator is the sole means of protection, use a full-face supplied air respirator. Use respirators and components tested and approved under appropriate government standards such as NIOSH (US) or CEN (EU).
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
8.7.6 Fire Prevention
NO open flames, NO sparks and NO smoking. NO contact with hot surfaces. Closed system, ventilation, explosion-proof electrical equipment and lighting. Prevent build-up of electrostatic charges (e.g., by grounding). Do NOT use compressed air for filling, discharging, or handling. Use non-sparking handtools.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.7.7 Exposure Prevention
AVOID ALL CONTACT! IN ALL CASES CONSULT A DOCTOR!
ILO-WHO International Chemical Safety Cards (ICSCs)
8.7.8 Inhalation Prevention
Use breathing protection. Use closed system or ventilation.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.7.9 Skin Prevention
Protective gloves. Protective clothing.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.7.10 Eye Prevention
Wear safety spectacles, face shield or eye protection in combination with breathing protection.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.7.11 Ingestion Prevention
Do not eat, drink, or smoke during work.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.8 Stability and Reactivity
8.8.1 Air and Water Reactions
Highly flammable. Fumes in air. Reacts violently with water producing heat and acidic HCl [Merck 11th ed. 1989].
Acetyl chloride reacts vigorously with water to generate gaseous HCl. Based on a scenario where the chemical is spilled into an excess of water (at least 5 fold excess of water), half of the maximum theoretical yield of Hydrogen Chloride gas will be created in 0.11 minutes. Experimental details are in the following: "Development of the Table of Initial Isolation and Protective Distances for the 2008 Emergency Response Guidebook", ANL/DIS-09-2, D.F. Brown, H.M. Hartmann, W.A. Freeman, and W.D. Haney, Argonne National Laboratory, Argonne, Illinois, June 2009.
CAMEO Chemicals
8.8.2 Reactive Group
Acyl Halides, Sulfonyl Halides, and Chloroformates
CAMEO Chemicals
8.8.3 Reactivity Alerts
Highly Flammable
Water-Reactive
Air-Reactive
CAMEO Chemicals
8.8.4 Reactivity Profile
ACETYL CHLORIDE reacts violently with water, steam, methanol or ethanol to form hydrogen chloride and acetic acid. Reacts vigorously with bases, both organic and inorganic. Incompatible with oxidizing agents and alcohols. Produces highly toxic fumes of phosgene gas and chlorine when heated to decomposition [Sax, 9th ed., 1996, p. 35]. Reaction in a confined space with even a small amount of water may cause a violent eruption of gases [Bretherick, 5th ed., 1995, p. 281]. Vapor forms an explosive mixture with air [Kirk-Othmer, 3rd ed., Vol. 1, 1978, p. 162]. Polymerization reaction with dimethyl sulfoxide is particularly violent [Buckley, A., J. Chem. Ed., 1965, 42, p. 674]. May react vigorously or explosively if mixed with diisopropyl ether or other ethers in the presence of trace amounts of metal salts [J. Haz. Mat., 1981, 4, 291].
CAMEO Chemicals
8.8.5 Hazardous Reactivities and Incompatibilities
Incompatible materials: Water, alcohols, oxidizing agents, strong bases.
Sigma-Aldrich; Safety Data Sheet for Acetyl chloride, Product Number: 00990, Version 3.15 (Revision Date 09/28/2017). Available from, as of September 24, 2018:
Hazardous Substances Data Bank (HSDB)
Dimethyl sulfoxide decomposes violently on contact with ... acetyl choride ... .
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 491-77
Hazardous Substances Data Bank (HSDB)
Acetyl chloride reacts violently with ethyl alcohol or water.
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 491-8
Hazardous Substances Data Bank (HSDB)
Water reactive. Violent exothermic decomposition with water produces corrosive hydrochloric and acetic acids. Reacts violently with alcohols, alkalies, amines, and strong oxidizing materials.
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-10
Hazardous Substances Data Bank (HSDB)
For more Hazardous Reactivities and Incompatibilities (Complete) data for Acetyl chloride (9 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
8.9 Transport Information
8.9.1 DOT Emergency Guidelines
If ... THERE IS NO FIRE, go directly to the Table of Initial Isolation and Protective Action Distances /(see table below)/ ... to obtain initial isolation and protective action distances. IF THERE IS A FIRE, or IF A FIRE IS INVOLVED, go directly to the appropriate guide /(see guide(s) below)/ and use the evacuation information shown under PUBLIC SAFETY.
Table: Table of Initial Isolation and Protective Action Distances for Acetyl chloride (when spilled in water) ID: 1717
Small Spills (from a small package or small leak from a large package)
Small Spills (from a small package or small leak from a large package)
Small Spills (from a small package or small leak from a large package)
Large Spills (from a large package or small leak from a large package)
Large Spills (from a large package or small leak from a large package)
Large Spills (from a large package or small leak from a large package)
Small Spills (from a small package or small leak from a large package)
First ISOLATE in all Directions
Small Spills (from a small package or small leak from a large package)
Then PROTECT persons Downwind during DAY:
Small Spills (from a small package or small leak from a large package)
Then PROTECT persons Downwind during NIGHT:
Large Spills (from a large package or small leak from a large package)
First ISOLATE in all Directions
Large Spills (from a large package or small leak from a large package)
Then PROTECT persons Downwind during DAY:
Large Spills (from a large package or small leak from a large package)
Then PROTECT persons Downwind during NIGHT:
Small Spills (from a small package or small leak from a large package)
30 m (100 ft)
Small Spills (from a small package or small leak from a large package)
0.1 km (0.1 mi)
Small Spills (from a small package or small leak from a large package)
0.3 km (0.2 mi)
Large Spills (from a large package or small leak from a large package)
100 m (300 ft)
Large Spills (from a large package or small leak from a large package)
0.9 km (0.6 mi)
Large Spills (from a large package or small leak from a large package)
2.5 km (1.6 mi)
U.S. Department of Transportation. 2016 Emergency Response Guidebook. Washington, D.C. 2016
Hazardous Substances Data Bank (HSDB)
Table of Water-Reactive Materials Which Produce Toxic Gases.
Table: Materials Which Produce Large Amounts of Toxic-by-Inhalation (TIH) Gas(es) When Spilled in Water
Name of Material
TIH Gas(es) Produced
Name of Material
Acetyl chloride
TIH Gas(es) Produced
Hydrogen chloride (HCl)
U.S. Department of Transportation. 2016 Emergency Response Guidebook. Washington, D.C. 2016
Hazardous Substances Data Bank (HSDB)
/GUIDE 155 SUBSTANCES - TOXIC and/or CORROSIVE (Flammable/Water-Sensitive)/ Fire or Explosion: HIGHLY FLAMMABLE: Will be easily ignited by heat, sparks or flames. Vapors form explosive mixtures with air: indoors, outdoors and sewers explosion hazards. Most vapors are heavier than air. They will spread along ground and collect in low or confined areas (sewers, basements, tanks). Vapors may travel to source of ignition and flash back. Those substances designated with a (P) may polymerize explosively when heated or involved in a fire. Substance will react with water (some violently) releasing flammable, toxic or corrosive gases and runoff. Contact with metals may evolve flammable hydrogen gas. Containers may explode when heated or if contaminated with water.
U.S. Department of Transportation. 2016 Emergency Response Guidebook. Washington, D.C. 2016
Hazardous Substances Data Bank (HSDB)
/GUIDE 155 SUBSTANCES - TOXIC and/or CORROSIVE (Flammable/Water-Sensitive)/ Health: TOXIC; inhalation, ingestion or contact (skin, eyes) with vapors, dusts or substance may cause severe injury, burns or death. Bromoacetates and chloroacetates are extremely irritating/lachrymators. Reaction with water or moist air will release toxic, corrosive or flammable gases. Reaction with water may generate much heat that will increase the concentration of fumes in the air. Fire will produce irritating, corrosive and/or toxic gases. Runoff from fire control or dilution water may be corrosive and/or toxic and cause pollution.
U.S. Department of Transportation. 2016 Emergency Response Guidebook. Washington, D.C. 2016
Hazardous Substances Data Bank (HSDB)
For more DOT Emergency Guidelines (Complete) data for Acetyl chloride (10 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
8.9.2 Shipping Name / Number DOT/UN/NA/IMO
UN 1717; Acetyl chloride
Hazardous Substances Data Bank (HSDB)
IMO 3; Acetyl chloride
Hazardous Substances Data Bank (HSDB)
8.9.3 Standard Transportation Number
49 076 01; Acetyl chloride
Hazardous Substances Data Bank (HSDB)
8.9.4 Shipment Methods and Regulations
No person may /transport,/ offer or accept a hazardous material for transportation in commerce unless that person is registered in conformance ... and the hazardous material is properly classed, described, packaged, marked, labeled, and in condition for shipment as required or authorized by ... /the hazardous materials regulations (49 CFR 171-177)./
49 CFR 171.2 (USDOT); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of October 24, 2018:
Hazardous Substances Data Bank (HSDB)
The International Air Transport Association (IATA) Dangerous Goods Regulations are published by the IATA Dangerous Goods Board pursuant to IATA Resolutions 618 and 619 and constitute a manual of industry carrier regulations to be followed by all IATA Member airlines when transporting hazardous materials. Acetyl chloride is included on the dangerous goods list.
International Air Transport Association. Dangerous Goods Regulations. 59th Edition. Montreal, Quebec Canada. 2018., p. 214
Hazardous Substances Data Bank (HSDB)
The International Maritime Dangerous Goods Code lays down basic principles for transporting hazardous chemicals. Detailed recommendations for individual substances and a number of recommendations for good practice are included in the classes dealing with such substances. A general index of technical names has also been compiled. This index should always be consulted when attempting to locate the appropriate procedures to be used when shipping any substance or article. Acetyl bromide is included on the dangerous goods list.
International Maritime Organization. IMDG Code. International Maritime Dangerous Goods Code Volume 2 2016, p. 81
Hazardous Substances Data Bank (HSDB)
8.9.5 DOT Label
Flammable Liquid Corrosive
CAMEO Chemicals
8.9.6 Packaging and Labelling
Airtight. Unbreakable packaging. Put breakable packaging into closed unbreakable container.
ILO-WHO International Chemical Safety Cards (ICSCs)
8.9.7 UN Classification
UN Hazard Class: 3; UN Subsidiary Risks: 8; UN Pack Group: II
ILO-WHO International Chemical Safety Cards (ICSCs)
8.10 Regulatory Information
The Australian Inventory of Industrial Chemicals
Chemical: Acetyl chloride
Australian Industrial Chemicals Introduction Scheme (AICIS)
REACH Registered Substance
Status: Active Update: 02-03-2021
European Chemicals Agency (ECHA)
New Zealand EPA Inventory of Chemical Status
Acetyl chloride: HSNO Approval: HSR001073 Approved with controls
New Zealand Environmental Protection Authority (EPA)
New Jersey Worker and Community Right to Know Act
The New Jersey Worker and Community Right to Know Act requires public and private employers to provide information about hazardous substances at their workplaces. (N.J.S.A. 34:5A-1 et. seq.)
NJDOH RTK Hazardous Substance List
8.10.1 DHS Chemicals of Interest (COI)
Chemicals of Interest(COI)
Acetyl chloride
Sabotage: Minimum Concentration (%)
A Commercial Grade
Sabotage: Screening Threshold Quantities
A Placarded Amount
Security Issue: Sabotage/Contamination
Chemical or material that can be mixed with readily available materials.
DHS Chemical Facility Anti-Terrorism Standards (CFATS) Chemicals of Interest
8.10.2 Atmospheric Standards
This action promulgates standards of performance for equipment leaks of Volatile Organic Compounds (VOC) in the Synthetic Organic Chemical Manufacturing Industry (SOCMI). The intended effect of these standards is to require all newly constructed, modified, and reconstructed SOCMI process units to use the best demonstrated system of continuous emission reduction for equipment leaks of VOC, considering costs, non air quality health and environmental impact and energy requirements. Acetyl chloride is produced, as an intermediate or a final product, by process units covered under this subpart.
40 CFR 60.489 (USEPA); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of October 23, 2018:
Hazardous Substances Data Bank (HSDB)
8.10.3 Clean Water Act Requirements
Acetyl chloride is designated as a hazardous substance under section 311(b)(2)(A) of the Federal Water Pollution Control Act and further regulated by the Clean Water Act Amendments of 1977 and 1978. These regulations apply to discharges of this substance. This designation includes any isomers and hydrates, as well as any solutions and mixtures containing this substance.
40 CFR 116.4 (USEPA); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of October 23, 2018:
Hazardous Substances Data Bank (HSDB)
8.10.4 CERCLA Reportable Quantities
Persons in charge of vessels or facilities are required to notify the National Response Center (NRC) immediately, when there is a release of this designated hazardous substance, in an amount equal to or greater than its reportable quantity of 5000 lb or 2270 kg. The toll free number of the NRC is (800) 424-8802. The rule for determining when notification is required is stated in 40 CFR 302.4 (section IV. D.3.b).
40 CFR 302.4 (USEPA); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of October 23, 2018:
Hazardous Substances Data Bank (HSDB)
8.10.5 RCRA Requirements
U006; As stipulated in 40 CFR 261.33, when acetyl chloride, as a commercial chemical product or manufacturing chemical intermediate or an off-specification commercial chemical product or a manufacturing chemical intermediate, becomes a waste, it must be managed according to Federal and/or State hazardous waste regulations. Also defined as a hazardous waste is any residue, contaminated soil, water, or other debris resulting from the cleanup of a spill, into water or on dry land, of this waste. Generators of small quantities of this waste may qualify for partial exclusion from hazardous waste regulations (40 CFR 261.5).
40 CFR 261.33 (USEPA); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of October 23, 2018:
Hazardous Substances Data Bank (HSDB)
8.11 Other Safety Information
8.11.1 Toxic Combustion Products
Decomposes at fire temperature with release of hydrogen chloride and phosgene.
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-10
Hazardous Substances Data Bank (HSDB)
8.11.2 Other Hazardous Reactions
May cause severe burns. Avoid contact with skin, eyes, mucous membranes.
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 16
Hazardous Substances Data Bank (HSDB)
9 Toxicity
9.1 Toxicological Information
9.1.1 Toxicity Summary
IDENTIFICATION AND USE: Acetyl chloride is a colorless fuming liquid. It is used as an acetylating agent, reagent in testing for cholesterol, and for determination of water in organic liquids. HUMAN STUDIES: There are no data available. ANIMAL STUDIES: In the Ames test using Salmonella culture strains TA97 TA100 TA1535 TA1537 acetyl chloride was not mutagenic with or without exogenous metabolic activation. Genotoxic activity of acetyl chloride was investigated in both somatic and germ cells of Drosophila melanogaster. Only marginal genotoxic activities were observed in these cells.
Hazardous Substances Data Bank (HSDB)
9.1.2 EPA IRIS Information
Substance
Acetyl chloride
EPA Integrated Risk Information System (IRIS)
9.1.3 Evidence for Carcinogenicity
CLASSIFICATION: D; not classifiable as to human carcinogenicity. BASIS FOR CLASSIFICATION: No human data or animal data. HUMAN CARCINOGENICITY DATA: None. ANIMAL CARCINOGENICITY DATA: None.
U.S. Environmental Protection Agency's Integrated Risk Information System (IRIS). Summary on Acetyl chloride (7446-34-6). Available from, as of February 4, 2019:
Hazardous Substances Data Bank (HSDB)
9.1.4 Exposure Routes
Serious local effects by all routes of exposure.
ILO-WHO International Chemical Safety Cards (ICSCs)
9.1.5 Signs and Symptoms
Inhalation Exposure
Cough. Sore throat. Burning sensation. Shortness of breath.
ILO-WHO International Chemical Safety Cards (ICSCs)
Skin Exposure
Redness. Pain. Burning sensation. Blisters. Serious skin burns.
ILO-WHO International Chemical Safety Cards (ICSCs)
Eye Exposure
Redness. Pain. Severe burns.
ILO-WHO International Chemical Safety Cards (ICSCs)
Ingestion Exposure
Cough. Sore throat. Burning sensation behind the breastbone. Abdominal pain. Shortness of breath.
ILO-WHO International Chemical Safety Cards (ICSCs)
9.1.6 Adverse Effects
Dermatotoxin - Skin burns.
Toxic Pneumonitis - Inflammation of the lungs induced by inhalation of metal fumes or toxic gases and vapors.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
9.1.7 Acute Effects
ChemIDplus
9.1.8 Antidote and Emergency Treatment
Immediate first aid: Ensure that adequate decontamination has been carried out. If patient is not breathing, start artificial respiration, preferably with a demand-valve resuscitator, bag-valve-mask device, or pocket mask, as trained. Perform CPR if necessary. Immediately flush contaminated eyes with gently flowing water. Do not induce vomiting. If vomiting occurs, lean patient forward or place on left side (head-down position, if possible) to maintain an open airway and prevent aspiration. Keep patient quiet and maintain normal body temperature. Obtain medical attention. /Organic acids and related compounds/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3rd revised edition, Elsevier Mosby, St. Louis, MO 2007, p. 176
Hazardous Substances Data Bank (HSDB)
Basic treatment: Establish a patent airway (oropharyngeal or nasopharyngeal airway, if needed). Suction if necessary. Watch for signs of respiratory insufficiency and assist respirations if necessary. Administer oxygen by nonrebreather mask at 10 to 15 L/min. Monitor for pulmonary edema and treat if necessary ... . Monitor for shock and treat if necessary ... . For eye contamination, flush eyes immediately with water. Irrigate each eye continuously with 0.9% saline (NS) during transport ... . Do not use emetics. For ingestion, rinse mouth and administer 5 mL/kg up to 200 mL of water for dilution if the patient can swallow, has a strong gag reflex, and does not drool. Activated charcoal is not effective ... . Do not attempt to neutralize, because of exothermic reaction. Cover skin burns with dry, sterile dressings after decontamination ... . /Organic acids and related compounds/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3rd revised edition, Elsevier Mosby, St. Louis, MO 2007, p. 176-7
Hazardous Substances Data Bank (HSDB)
Advanced treatment: Consider orotracheal or nasotracheal intubation for airway control in the patient who is unconscious, has severe pulmonary edema, or is in severe respiratory distress. Early intubation, at the first sign of upper airway obstruction, may be necessary. Positive-pressure ventilation techniques with a bag-valve-mask device may be beneficial. Consider drug therapy for pulmonary edema ... . Consider administering a beta agonist such as albuterol for severe bronchospasm ... . Monitor cardiac rhythm and treat arrhythmias as necessary ... . Start IV administration of D5W /SRP: "To keep open", minimal flow rate/. Use 0.9% saline (NS) or lactated Ringer's (LR) if signs of hypovolemia are present. For hypotension with signs of hypovolemia, administer fluid cautiously. Consider vasopressors if patient is hypotensive with a normal fluid volume. Watch for signs of fluid overload ... . Use proparacaine hydrochloride to assist eye irrigation ... . /Organic acids and related compounds/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3rd revised edition, Elsevier Mosby, St. Louis, MO 2007, p. 177
Hazardous Substances Data Bank (HSDB)
9.1.9 Non-Human Toxicity Excerpts
/GENOTOXICITY/ The Ames test using the preincubation assay was conducted. The test chemical, Salmonella culture /(strains TA97/TA100/TA1535/TA1537)/, and S-9 mix or buffer were incubated at 37 °C, without shaking, for 20 min. The top agar was added and the contents of the tubes were mixed and poured onto the surface of petri dishes containing Vogel-Bonner medium. The histidine-independent (his+) colonies arising on these plates were counted following two days incubation at 37 °C. Acetyl chloride is not mutagenic in these bacterial test systems either with or without exogenous metabolic activation at the dose level investigated /(0, 33, 100, 333, 1000, 3333 and 6666 ug/plate)/.
European Chemicals Agency (ECHA); Registered Substances, Acetyl chloride (CAS Number: 75-36-5) (EC Number: 200-865-6) (Last updated: April 26, 2018). Available from, as of September 27, 2018:
Hazardous Substances Data Bank (HSDB)
/GENOTOXICITY/ The genetic toxicity profiles of vinyl chloride (VCl), vinyl bromide (VBr), ethyl carbamate (EC), vinyl carbamate (VC) and some structurally related chemicals were investigated in both somatic and germ cells of Drosophila melanogaster. In the white/white+ eye mosaic assay, a screening system measuring predominantly homologous recombination in somatic cells, only marginal genotoxic activities were observed for acetyl chloride (ACl), glycolaldehyde (GCA), 2,2'-dichlorodiethyl ether (DDE) and methyl carbamate (MC), whereas VCl, 2-chloroacetaldehyde (CAA), VBr, 2-bromoacetaldehyde (BAA) and EC were clearly recombinogenic in the assay. Those chemicals proven to be recombinogenic in somatic cells were investigated further in postmeiotic male germ cells, utilizing as descriptors of their genotoxicity I(CL/RL) and M(exr-)/M(exr+) indices. The I(CL/RL) index is the rate of induced chromosome loss (CL), a clastogenic event, divided by the forward mutation rate, measured as recessive lethal (RL) mutations in 700 loci of the X-chromosome. The M(exr-)/M(exr+) mutation enhancement ratio is obtained by determining RL under excision repair deficient versus repair proficient conditions. With I(CL/RL) values (2.7-6.9) similar to those obtained for cross-linking agents, vinyl chloride, vinyl bromide, ethyl carbamate and vinyl carbamate are all efficient clastogenic agents in Drosophila germ cells. In the absence of excision repair, however, neither CEO nor CAA gave a hypermutability response (M(exr-)/M(exr+) approximately 1). By contrast, VCl, VBr, EC and VC showed clearly enhanced M(exr-)/M(exr+) ratios, suggesting that these compounds produce some repairable DNA modification(s) that are not generated by their epoxides. This unexpected finding points to the formation of other, yet unknown, metabolites of vinyl chloride, vinyl bromide, ethyl carbamate and vinyl carbamate. Our results support the concept that the epoxides chloroethylene oxide (CEO), bromoethylene oxide (BEO) and vinyl carbamate epoxide (VCO) are the most essential mutagenic intermediates. Compared to chloroethylene oxide (CEO), 2-chloroacetaldehyde (CAA) was approximately 50 times less effective in the induction of RL, whereas BAA was inactive as a mutagen. These findings are consistent with the general view that CAA and BAA play no major role in the genotoxic action of vinyl halides.
PMID:8640917
Ballering LA et al; Carcinogenesis 17 (5): 1083-92 (1996)
Hazardous Substances Data Bank (HSDB)
9.1.10 Non-Human Toxicity Values
LD50 Rat oral 910 mg/kg
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 42
Hazardous Substances Data Bank (HSDB)
9.1.11 Ongoing Test Status
The following link will take the user to the National Toxicology Program (NTP) Test Status of Agents Search page, which tabulates the results and current status of tests such as "Short-Term Toxicity Studies", "Long-term Carcinogenicity Studies", "Developmental Studies", "Genetic Toxicology Studies", etc., performed with this chemical. Testing status for acetyl chloride is available.[Available from, as of February 1, 2019:
Hazardous Substances Data Bank (HSDB)
9.2 Ecological Information
9.2.1 Ecotoxicity Values
LC50; Species: Pimephales promelas (Fathead minnow); Conditions: freshwater, static; Concentration: 42 mg/L for 96 hr (95% confidence interval: 25.2-70 mg/L)
Curtis MW, CH Ward; J Hydrol 51: 359-367 (1981) as cited in the ECOTOX database. Available from, as of October 14, 2018:
Hazardous Substances Data Bank (HSDB)
9.2.2 ICSC Environmental Data
The substance is harmful to aquatic organisms.
ILO-WHO International Chemical Safety Cards (ICSCs)
9.2.3 Environmental Fate / Exposure Summary
Acetyl chloride's production and use in chemical synthesis (acetylating agent) particularly in the manufacture of pharmaceuticals and dyestuffs may result in its release to the environment through various waste streams. If released to air, a vapor pressure of 287 mm Hg at 25 °C indicates acetyl chloride will exist solely as a vapor in the ambient atmosphere. Vapor-phase acetyl chloride will react with atmospheric moisture and its half-life will depend on the humidity of the air. No half-lives for air with different water contents could be found; however, acetyl chloride is known to fume in moist air. Acetyl chloride contains chromophores that absorb at wavelengths >290 nm and, therefore, may be susceptible to direct photolysis by sunlight. If released to soil, acetyl chloride is expected to have very high mobility based upon an estimated Koc of 1. However, acetyl chloride reacts violently in the presence of water and combined with the reported high reactivity of structurally-similar compounds with active hydrogen groups that occur in soil, it is unlikely that acetyl chloride would persist for long in moist soils. Acetyl chloride may volatilize from dry soil surfaces based upon its vapor pressure. Biodegradation data in soil were not available. If released in water, acetyl chloride is not expected to persist as it reacts violently with water. Adsorption to sediment, volatilization, bioconcentration in aquatic species or biodegradation are not expected because of the very short lifetime of this compound in water. Occupational exposure to acetyl chloride may occur through inhalation and dermal contact with this compound at workplaces where acetyl chloride is produced or used. The general public is not expected to be exposed to acetyl chloride. (SRC)
Hazardous Substances Data Bank (HSDB)
9.2.4 Artificial Pollution Sources
Acetyl chloride's production and use in chemical synthesis (acetylating agent), particularly in the manufacture of pharmaceuticals and dyestuffs(1,2), in testing for cholesterol, and determination of water in organic liquids(2) may result in its release to the environment through various waste streams(SRC).
(1) Larranaga MD et al, eds; Hawley's Condensed Chemical Dictionary 16th ed., Hoboken, NJ: John Wiley & Sons, Inc., p. 13 (2016)
(2) O'Neil MJ, ed; The Merck Index. 15th ed., Cambridge, UK: Royal Society of Chemistry, p. 16 (2013)
Hazardous Substances Data Bank (HSDB)
9.2.5 Environmental Fate
TERRESTRIAL FATE: Based on a classification scheme(1), an estimated Koc value of 1(SRC), determined from a structure estimation method(2), indicates that acetyl chloride is expected to have very high mobility in soil(SRC). In view of its violent decomposition in the presence of water(3) and the high reactivity of structurally-similar compounds with active hydrogen groups that occur in soil(4), it is unlikely that acetyl chloride would persist for long in moist soils(SRC). Acetyl chloride is expected to volatilize from dry soil surfaces(SRC) based upon a vapor pressure of 287 mm Hg at 25 °C(5). Biodegradation data in soil were not available(SRC, 2018).
(1) Swann RL et al; Res Rev 85: 17-28 (1983)
(2) US EPA; Estimation Program Interface (EPI) Suite. Ver. 4.11. Nov, 2012. Available from, as of Sept 25, 2018:
(3) O'Neil MJ, ed; The Merck Index. 15th ed., Cambridge, UK: Royal Society of Chemistry, p. 16 (2013)
(4) Thurman EM, Malcolm RL; pp. 1-23 in Aquatic and Terrestrial Humic Materials. Christman RF, Gjessing ET, eds., Ann Arbor, MI: Ann Arbor Sci (1983)
(5) Daubert TE, Danner RP; Physical and Thermodynamic Properties of Pure Chemicals Data Compilation. Washington, DC: Taylor and Francis (1989)
Hazardous Substances Data Bank (HSDB)
AQUATIC FATE: Acetyl chloride decomposes violently in the presence of water(1) and, therefore, will not persist in the aquatic environment(SRC). Adsorption to sediment, volatilization, bioconcentration in aquatic species or biodegradation are not expected because of the very short lifetime of this compound in water(SRC).
(1) O'Neil MJ, ed; The Merck Index. 15th ed., Cambridge, UK: Royal Society of Chemistry, p. 16 (2013)
Hazardous Substances Data Bank (HSDB)
ATMOSPHERIC FATE: According to a model of gas/particle partitioning of semivolatile organic compounds in the atmosphere(1), acetyl chloride, which has a vapor pressure of 287 mm Hg at 25 °C(2), is expected to exist solely as a vapor in the ambient atmosphere. Vapor-phase acetyl chloride may be degraded in the atmosphere by reaction with photochemically-produced hydroxyl radicals(SRC); the half-life for this reaction in air is estimated to be 5 years(SRC), calculated from its rate constant of 9.1X10-15 cu cm/molecule-sec at 25 °C(3). However, acetyl chloride decomposes violently in the presence of water(4) and fumes in the presence of moist air(5). Acetyl chloride contains chromophores that absorb at wavelengths >290 nm(6) and, therefore, may be susceptible to direct photolysis by sunlight(SRC).
(1) Bidleman TF; Environ Sci Technol 22: 361-7 (1988)
(2) Daubert TE, Danner RP; Physical and Thermodynamic Properties of Pure Chemicals Data Compilation. Washington, DC: Taylor and Francis (1989)
(3) Atkinson R; J Phys Chem Ref Data, Monograph 2 p. 122 (1994)
(4) O'Neil MJ, ed; The Merck Index. 15th ed., Cambridge, UK: Royal Society of Chemistry, p. 16 (2013)
(5) Ashford RD; Ashford's Dictionary of Industrial Chemicals, London, England: Wavelength Publ, Ltd. p. 16 (1994)
(6) Lyman WJ et al; Handbook of Chemical Property Estimation Methods. Washington, DC: Amer Chem Soc pp. 8-12 (1990)
Hazardous Substances Data Bank (HSDB)
9.2.6 Environmental Abiotic Degradation
Acetyl chloride decomposes violently with water and alcohol(1), fuming in moist air(2) and being easily hydrolyzed to acetic acid and hydrochloric acid by body moisture (e.g. mucous membrane, skin)(3). It will, therefore, not persist for any length of time in the environment where water is present. The rate constant for the vapor-phase reaction of acetyl chloride with photochemically-produced hydroxyl radicals is 9.1X10-15 cu cm/molecule-sec at 25 °C(4). This corresponds to an atmospheric half-life of about 5 years at an atmospheric concentration of 5X10+5 hydroxyl radicals per cu cm(SRC). Acetyl chloride contains chromophores that absorb at wavelengths >290 nm(5) and, therefore, may be susceptible to direct photolysis by sunlight(SRC).
(1) O'Neil MJ, ed; The Merck Index. 15th ed., Cambridge, UK: Royal Society of Chemistry, p. 16 (2013)
(2) Ashford RD; Ashford's Dictionary of Industrial Chemicals, London, England: Wavelength Publ, Ltd. p. 16 (1994)
(3) Moretti TA; Kirk-Othmer Encycl Chem Technol. 3rd ed New York, NY: Wiley 1: 162-6 (1978)
(4) Atkinson R; J Phys Chem Ref Data, Monograph 2 p. 122 (1994)
(5) Lyman WJ et al; Handbook of Chemical Property Estimation Methods. Washington, DC: Amer Chem Soc pp. 8-12 (1990)
Hazardous Substances Data Bank (HSDB)
9.2.7 Environmental Bioconcentration
Acetyl chloride will decompose violently in water(1) forming acetic acid and hydrochloric acid(2). Because of its short half-life in water, bioconcentration of acetyl chloride in aquatic organisms is very unlikely(SRC).
(1) O'Neil MJ, ed; The Merck Index. 15th ed., Cambridge, UK: Royal Society of Chemistry, p. 16 (2013)
(2) Moretti TA; Kirk-Othmer Encycl Chem Technol. 3rd ed New York, NY: Wiley 1: 162-6 (1978)
Hazardous Substances Data Bank (HSDB)
9.2.8 Soil Adsorption / Mobility
Using a structure estimation method based on molecular connectivity indices(1), the Koc of acetyl chloride can be estimated to be 1(SRC). According to a classification scheme(2), this estimated Koc value suggests that acetyl chloride is expected to have very high mobility in soil(SRC). However, in view of its violent decomposition in the presence of water(3) and the high reactivity of this compound towards molecules with active hydrogen groups such as natural products containing amine, phenol, and alcohol functional groups that occur in soil(4), it is unlikely that acetyl chloride would persist for long in moist soils(SRC).
(1) US EPA; Estimation Program Interface (EPI) Suite. Ver. 4.11. Nov, 2012. Available from, as of Sept 25, 2018:
(2) Swann RL et al; Res Rev 85: 17-28 (1983)
(3) O'Neil MJ, ed; The Merck Index. 15th ed., Cambridge, UK: Royal Society of Chemistry, p. 16 (2013)
(4) Thurman EM, Malcolm RL; pp. 1-23 in Aquatic and Terrestrial Humic Materials. Christman RF, Gjessing ET, eds., Ann Arbor, MI: Ann Arbor Sci (1983)
Hazardous Substances Data Bank (HSDB)
9.2.9 Volatilization from Water / Soil
Acetyl chloride will decompose violently in water(1) preempting volatilization from water or soil surfaces as a fate process(SRC). For the same reason, volatilization from moist soil surfaces is very unlikely(SRC). The potential for volatilization of acetyl chloride from dry soil surfaces may exist(SRC) based upon a vapor pressure of 287 mm Hg(2).
(1) O'Neil MJ, ed; The Merck Index. 15th ed., Cambridge, UK: Royal Society of Chemistry, p. 16 (2013)
(2) Daubert TE, Danner RP; Physical and Thermodynamic Properties of Pure Chemicals Data Compilation. Washington, DC: Taylor and Francis (1989)
Hazardous Substances Data Bank (HSDB)
9.2.10 Probable Routes of Human Exposure
NIOSH (NOES Survey 1981-1983) has statistically estimated that 2535 workers (574 of these are female) are potentially exposed to acetyl chloride in the US(1). Occupational exposure to acetyl chloride may occur through inhalation and dermal contact with this compound at workplaces where acetyl chloride is produced or used. The general population is not likely to be exposed to acetyl chloride(SRC).
(1) CDC; International Chemical Safety Cards (ICSC) 2012. Atlanta, GA: Centers for Disease Prevention & Control. National Institute for Occupational Safety & Health (NIOSH). Ed Info Div. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
Inhalation and ingestion.
Sittig, M. Handbook of Toxic and Hazardous Chemicals and Carcinogens, 1985. 2nd ed. Park Ridge, NJ: Noyes Data Corporation, 1985., p. 34
Hazardous Substances Data Bank (HSDB)
Workers may be exposed through skin contact(1).
(1) NOAA; CAMEO Chemicals. Database of Hazardous Materials. Acetyl chloride (75-36-5). Natl Ocean Atmos Admin, Off Resp Rest; NOAA Ocean Serv. Available from, as of Sept 25, 2018:
Hazardous Substances Data Bank (HSDB)
10 Associated Disorders and Diseases
Associated Occupational Diseases with Exposure to the Compound
Pneumonitis, toxic [Category: Acute Poisoning]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
11 Literature
11.1 Consolidated References
PubChem
11.2 NLM Curated PubMed Citations
Medical Subject Headings (MeSH)
11.3 Springer Nature References
Springer Nature
11.4 Thieme References
Thieme Chemistry
11.5 Wiley References
Wiley
11.6 Chemical Co-Occurrences in Literature
PubChem
11.7 Chemical-Gene Co-Occurrences in Literature
PubChem
11.8 Chemical-Disease Co-Occurrences in Literature
PubChem
11.9 Chemical-Organism Co-Occurrences in Literature
PubChem
12 Patents
12.1 Depositor-Supplied Patent Identifiers
PubChem
Link to all deposited patent identifiers
PubChem
12.2 WIPO PATENTSCOPE
Patents are available for this chemical structure:
PATENTSCOPE (WIPO)
12.3 Chemical Co-Occurrences in Patents
PubChem
12.4 Chemical-Disease Co-Occurrences in Patents
PubChem
12.5 Chemical-Gene Co-Occurrences in Patents
PubChem
12.6 Chemical-Organism Co-Occurrences in Patents
PubChem
13 Biological Test Results
13.1 BioAssay Results
PubChem
14 Classification
14.1 MeSH Tree
Medical Subject Headings (MeSH)
14.2 ChEBI Ontology
ChEBI
14.3 ChemIDplus
ChemIDplus
14.4 CAMEO Chemicals
CAMEO Chemicals
14.5 UN GHS Classification
GHS Classification (UNECE)
14.6 NORMAN Suspect List Exchange Classification
NORMAN Suspect List Exchange
14.7 EPA DSSTox Classification
EPA DSSTox
14.8 EPA TSCA and CDR Classification
EPA Chemicals under the TSCA
14.9 EPA Substance Registry Services Tree
EPA Substance Registry Services
14.10 MolGenie Organic Chemistry Ontology
MolGenie
14.11 Chemicals in PubChem from Regulatory Sources
PubChem
15 Information Sources
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KnowledgeWell
Published 08 / 2023 Updated 03 / 2025 11 min read
The Truth About Vitamin A for Conception, Pregnancy, Breastfeeding
By Ayla Barmmer, MS, RD, LDN
Founder & CEO, FullWell
TL;DR
Vitamin A is a generic term for fat-soluble compounds found as preformed vitamin A (retinoids)in animal products and as provitamin A carotenoids in fruits and vegetables.
Vitamin A needs vary from person to person based on genetics, gut health, health conditions, and stage of life.
The upper limit of vitamin A is set to 10,000 IU (equivalent to 3,000 mcg) a day, but in the past 30+ years, the number of cases where excessive intake has been associated with any negative affects has been low.
Note before we start: This one is a doozy,but we are going to take care to answer some of your most pressing questions, like:
What is vitamin A?
Am I getting enough vitamin A?
Is my male partner getting enough vitamin A?
How can I/we get the correct forms of vitamin A?
How much vitamin A should I/we be getting?
How much vitamin A is in your Women’s Prenatal?
How much vitamin A is in your Men’s Prenatal?
Why does FullWell push the upper limit of vitamin A?
Where can I look for vitamin A in food?
Incorporating this knowledge into your daily life
Here we go!
As you can see, we get asked very in-depth questions about vitamin A a lot.
“Am I taking too much vitamin A?”
“How many mcg RAE of vitamin A are in FullWell Prenatal?”
“Why do you have more vitamin A than other brands?”
“Can I keep eating liver while I take my prenatal vitamins?”
“I’m taking a medication my dermatologist gave me for my skin. Is it safe to take extra vitamin A?”
We all heard it as kids.
"Carrots improve your eyesight. Eat up!"
Our parents weren't lying: carrots are high invitamin A, which is critical for many of our bodies' systems' proper function, including eye health. Vitamin A deficiency is widespread in children and women of childbearing age and is the leading cause of preventable blindness worldwide.
However, unfortunately for those who were fooled into cleaning our plates (special shoutout to those who thought we might get to give up our eyeglasses!), eating orange fruits and vegetables won't help you magically achieve perfect vision.
In fact, for those who went for it, the over-consumption of carrots may have led to a temporary condition called carotenemia, a rare occurrence of an orange coloring of the feet, hands, and thicker skin on the body. That's right: some who ingest too many carotenoid-containing foods - and have a genetic variation that alters how they convert beta-carotene to vitamin A -turn (a little) orange!(1).
Rare conditions aside, vitamin A deficiency is a growing concern and its role in our overall health (particularly in terms of both fertility and pregnancy) cannot be understated.
Vitamin A comes in multiple forms, each playing a different role in human health.
The various forms of vitamin A are essential for the health of mom AND baby’s (2):
Eyes
Skin
Reproductive system
Immune system
Thyroid glands
So what is vitamin A?
Vitamin A is a generic term forfat-soluble compoundsfound aspreformed vitamin A (retinoids)in animal products and asprovitamin A carotenoidsin fruits and vegetables.
The body's three most active forms of vitamin A areretinol, retinal,andretinoic acid. Dietary supplements usually contain variations on these active forms, such asretinyl acetateorretinyl palmitate(preformed vitamin A),beta-carotene(provitamin A, a less active form), or a combination of both preformed and provitamin A.
While critically low vitamin A levels are not typical in developed countries,subclinically low levels are surprisingly common. Defined as serum retinol, or concentrations lower than 0.70 μmol/L or 20 μg/dL, subclinical vitamin A levels are oftendismissed by traditionally trained doctors and healthcare providersbecause they are higher than the conventional lab ranges for deficiency.
Optimal vitamin A levels are essential for
Healthy vision and eye developmentin developing fetuses and adults alike, including normal color vision, vision in dim lighting, and dark adaptation
Regulating thegrowth and differentiation of virtually every cell in the bodyfrom embryos to adults (meaning, deciding which cells become which organs and putting together body systems as a fetus develops)
Healthy immune responseto infectious diseases from fetal development through adulthood
Normalred blood cell production, including integration of iron and oxygen-carrying capacity within those cells
Normaliron levelsin the blood
Healthythyroid functionand thyroid stimulating hormone (TSH) levels
Robustskin cell turnoverand proper healing
Propersperm maturation(due to high antioxidant capacity)
Are we as a population getting enough vitamin A?
Nope. The robust U.S. survey on nutritional status known as NHANES found that from 2017 to 2018, vitamin A intake for women ages 20 and older was 616 mcg daily, less than the RDA of 770 mcg for pregnant women (3).
For men, the same survey found that intake for men over 20 was around 682 mcg daily, which is less than the RDA of 900 mcg (3). So we know that deficiency is common, but we also know its reasons are multi-layered. Let me explain.
Cultural shifts and lifestyle factors
Vitamin A intake has decreased worldwide - even in developed countries like the US - for many reasons. Some are related to cultural shifts, like more people eating less organ meat, restrictive diets (e.g., low-fat, low-calorie), the increasing popularity of vegan and vegetarian diets, and more people ditching dairy (4). Other factors are tied to the economy, like the rising cost of food (especially when it comes to fresh produce) and the lower cost and convenience of processed and packaged food (4).
Genes
Genetics influence how well you convert beta-carotene into vitamin A in your liver. For example, many people have a genetic variation in the Beta-Carotene Oxygenase 1 (BCO1) gene, which breaks down beta-carotene into the active form of vitamin A (5). The science is conflicting on how much this genetic variant impacts Vitamin A levels but for the roughly 40% of the population estimated to have this variant, it is even more important that they consume foods that contain retinoids as beta carotene may not be enough.
Reduced absorption
In addition to genetic factors, many people live with and experience conditions that reduce the absorption or usability of vitamin A, including digestive disorders and hormonal issues.
Crohn’s
Ulcerative Colitis
Irritable Bowel Syndrome (IBS)
Thyroid Issues
Gallbladder Issues
Fat Malabsorption
As a result, they may have higher needs than what the Recommended Daily Values assert.
Pregnancy
Many women live with vitamin A deficiency, but pregnant women are at an even greater risk. Pregnant women are more likely than healthy non-pregnant women to be deficient in vitamin A even with supplementation, which is cause for alarm. This deficiency directly affects baby's vitamin stores at birth (6, 7).
Through each trimester, demand increases to support baby's rapid growth and help prepare for breastfeeding (8). In particular, deficiency during the third trimester can be associated with an increased risk of preterm delivery and anemia in mom (9).
And, in case you’re not convinced of the importance of vitamin A during pregnancy, one study found that vitamin A deficiency in pregnant women with preeclampsia increased the risk of adverse pregnancy outcomes (10).
Similarly, research has shown that pregnant women with gestational diabetes are more likely to suffer from a vitamin A deficiency and should be closely monitored during pregnancy (11,12).
What about men?
When optimizingmen’s health preconception, consuming adequate vitamin A is crucial. In animal studies, sperm production halted in vitamin A deficiency (13). More research is needed, but we know that the antioxidant capacity of beta-carotene plays a crucial role in capturing and neutralizing reactive oxygen species (ROS), which can damage sperm DNA. Retinyl palmitate, conversely, is essential for the production and maturation of sperm which we know is necessary for a healthy pregnancy.
Provitamin A doesnot equal preformed vitamin (animal-based foods matter)!
The best natural foods for consuming the most active forms of vitamin A (aka preformed vitamin A) are:
Beef liver
Cod liver oil
Butter
Eggs (yolk and all, please!)
Whole milk
According to the US Institute for Medicine (IOM), dietary retinol (one of the preformed “active” versions of vitamin A) is12-24 times more activethan dietary beta-carotene and alpha-carotene.
The bioavailability (or the immediate ability to be absorbed and used in the body) varies greatly from food to food. Different combinations, the amount of fat consumed, and each individual’s enzyme conversion activity can play into this variance.
What about plants?
While carotenoids like beta-carotene are not the most active form of vitamin A, they partially convert to active vitamin A in the body. They are also utilized for non-provitamin A tasks and can be potent antioxidants.
These plant-based foods are rich in carotenoids, which convert in varying amounts to the active forms of vitamin A (14):
How much vitamin A should I be getting?
As we mentioned, vitamin A needs vary from person to person based on genetics, gut health, health conditions, and stage of life. The average person can generalize quantities based on sex and age. For example, needs for women generally look something like this (4):
Adults (19+ years): 700 μg/day (2,331 IU)
Pregnant (19+ years): 770 μg/day (2,564 IU)
Breastfeeding (19+ years): 1300 μg/day (4,330 IU)
Upper limit: 3000 μg/day (10,000 IU)
These are (in my professional opinion) highly conservative guidelines fromThe American Pregnancy Associationand theNational Institutes of Health Office of Dietary Supplements, and we’ll go into more detail about these shortly. But first, let’s clarify some confusing metric terminology and conversions to break these numbers down.
Each of the following isequivalent to 1 microgram (μg) of active vitamin A (retinol), with daily intake goals listed above (2):
1 μg supplemental vitamin A (retinyl acetate or retinyl palmitate)
2 μg of supplemental β-carotene
3.33 IU of retinol
When reading labels, units of measurement can be confusing. But understanding how much active vitamin A you’re consuming is essential. Converting micrograms into a standardized unit that measures the activity level is helpful. Retinol Activity Equivalents measure the vitamin A activity of the type of pre- or pro-vitamin A consumed, and1 IU of retinol is equivalent to 0.3 μg retinol activity equivalents (RAE).
Remembering that animal or supplement-based vitamin A in retinoids is more reliable for absorption, their RAE ratio is 1:1. What goes in can usually all be used by a healthy gut.
Alternatively, beta-carotene in food has an RAE ratio of 24:1. In supplements, it holds an RAE ratio of 2:1 because every microgram of beta-carotene in supplemental form equals 1 RAE.This means it takes 12-24 times as much vitamin A from beta-carotene to have the same activity in the body as active vitamin A consumed from animal-based foods.
What this means in practice
If one large egg has 80 μg of preformed vitamin A (retinol), 270 IU (80 μg RAE) is readily used in the body. You don’t need to worry about your genes or how efficiently you convert to the active, usable form because it’s already primed for use! Easy peasy.
Half of a baked sweet potato contains no ready-to-use retinol but has 11,091 beta-carotene (961 μg RAE). In a perfect world, your body can readily convert and use this form of vitamin A. However, when you account for genetics, gut health/digestibility, and vitamin cofactors needed for conversion, real life is not likely to reflect that perfect scenario.
How much vitamin A is in your Women’s Prenatal?
We aim for 1500mcg RAE total per serving. Half of that is from beta carotene, and half is from retinyl palmitate. So because the ratio is 2:1, it winds up being 750 mcg RAE and 750 mcg retinyl palmitate.
Why is your Women’s Prenatal above the upper limit?
So let’s come back to my thought on the RDA. Based on decades of research and experience, I’m among many health practitioners that have concluded that the RDA for vitamin A is set at a highly conservative level, as is theUL (upper limit). The UL is based on ano-observed-adverse-effect level (NOAEL).Governing bodies are simply looking at the highest intake at which no adverse effects have been reported. If data are inadequate to determine a NOAEL value, thelowest-observed-adverse-effect level (LOAEL)is utilized. The recommended ULs you see are usually several-fold lower than the no-observed level or lowest-observed. These levels are divided by an uncertainty factor. The severity of that uncertainty depends on both adverse effects and uncertainty about the data itself. Sure, a safety margin is provided, but the problem with erring on the side of caution means folks are less likely to focus on vitamin A, leading to suboptimal levels or deficiency
The upper levelshouldbe much higher.
The upper limit of vitamin A is set to 10,000 IU (equivalent to 3,000 mcg) a day, but in the past 30+ years, the number of cases where excessive intake has been associated with congenital disabilities has been low. Ethical and safety issues hinder clinical trials on pregnant women (obviously). Still, animal studies have shown that between 25,000 to 37,000 IU/day of vitamin A was safe during human pregnancy when considering comparable weight adjustments from monkey to human (15). One study, in particular, found that high vitamin A intake later in pregnancy (around 50,000 IU per day) did not increase the risk of congenital disabilities (16).
Supplementing has other benefits for pregnant people, too.
Yes, deficiency is common, so repletion is vital. Enter supplements! The bonus is that supplementing with vitamin A during pregnancy and postpartum has other benefits. It can prevent the depletion of vitamin A stores that occurs toward the end of pregnancy and possibly improve mom’s immune system and ability to see at night. Additional research is needed to determine whether vitamin A can positively impact birth outcomes and cognitive/motor development, but some recent studies are promising (14,17).
Vitamin A also plays a significant role in the function of other vitamins and minerals in your body, like iron. Studies on pregnant and lactating women found that vitamin A supplementation reduced the risk ofanemiaby helping to increase hemoglobin levels (a protein in red blood cells that carries oxygen) andserum ferritin levels(your storage form of iron) (18).
Fascinating, right? Ah, the magic of being a Registered Dietitian: we get to examine and educate on how nutrients work together in your body!
If I’m looking to get more vitamin A in my diet, what foods should I incorporate?
Circling back to our quest to increase our intake through real food, the NOAEL value for preformed vitamin A translates to about 3 ounces of beef liver or 4 ounces of chicken liver every day(not a typo).
For many, a side of liver at every meal isn't exactly ideal (yet another reason why supplementing can save the day). If you want more info on how organ meats can fit into your routine with yourFullWell Prenatal, check out thiscomprehensive blog postwritten by my expert colleague and FullWell supporter,Lily Nichols. A fellow RD, Lily goes into exceptional, science-backed detail about liver’s benefits and provides excellent strategies for incorporating organ meats into your diet, especially if you've historically shied away from them! Check it out(4):
Considering the vitamin A dose in your prenatal, can I take it with what I already eat daily?
We get this question a lot. Let me illustrate it with an example day of meals that factors in the amount of vitamin A we get from food and supplements combined.
As you can see, it's tough to overdo it on vitamin A unless you consistently eat large amounts of liver daily and supplement.
Let's put this information to use!
Nothing can replace a healthy diet. Still, our best efforts may not reach our goals when we individually account for stress andenvironmental factors.
Supplementing with active vitamin A in safe amounts can be a great way to fill nutrient gaps, improve health, and ensure usability in the body.
If you and your partner are preparing for pregnancy or are already pregnant or breastfeeding and trying to deliver proper nutrition to baby, assess your diet and lifestyle. Then, speak to a professional specializing in nutrition, supplements, and women’s health. If you have trouble finding one, you can look to our amazing Clinical Directory (coming soon!) or contact us with your questions.
“Our prenatal Multivitamins for both sexes are expert-formulated and include beta carotene and active retinyl palmitate. No need to worry about genetic factors, food combination, or other nuances in supplementing your diet to achieve adequate vitamin A stores. Place your trust in FullWell.”
WOMEN'S PRENATAL 1500mcg vitamin A SHOP NOW MEN'S MULTIVITAMIN 1125 mcg vitamin A SHOP NOW
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References Click to expand references
Al Nasser Y, Jamal Z, Albugeaey M. Carotenemia. In: StatPearls. Treasure Island (FL): StatPearls Publishing; November 17, 2022.
Vitamin A. Vitamin A. Linus Pauling Institute. Published January 2, 2019.
U.S. Department of Agriculture, Agricultural Research Service. What We Eat in America, 2017-2018 external link disclaimer. 2023.
Vitamin A and Carotenoids. NIH Office of Dietary Supplements. Published June 22, 2022. Accessed March 8, 2023.
Reboul E. Mechanisms of Carotenoid Intestinal Absorption: Where Do We Stand?. Nutrients. 2019;11(4):838. Published 2019 Apr 13. doi:10.3390/nu11040838
Baker H, Frank O, Thomson AD, et al. Vitamin profile of 174 mothers and newborns at parturition. Am J Clin Nutr. 1975;28(1):59-65. doi:10.1093/ajcn/28.1.59
Baker H, DeAngelis B, Holland B, Gittens-Williams L, Barrett T Jr. Vitamin profile of 563 gravidas during trimesters of pregnancy. J Am Coll Nutr. 2002;21(1):33-37. doi:10.1080/07315724.2002.10719191
Reifen R, Ghebremeskel K. Vitamin A during pregnancy. Nutr Health. 2001;15(3-4):237-243. doi:10.1177/026010600101500413
Radhika MS, Bhaskaram P, Balakrishna N, Ramalakshmi BA, Devi S, Kumar BS. Effects of vitamin A deficiency during pregnancy on maternal and child health. BJOG. 2002;109(6):689-693. doi:10.1111/j.1471-0528.2002.01010.x
Duan S, Jiang Y, Mou K, Wang Y, Zhou S, Sun B. Correlation of serum vitamin A and vitamin E levels with the occurrence and severity of preeclampsia. Am J Transl Res. 2021;13(12):14203-14210. Published 2021 Dec 15.
Lira, L. Q. de, & Dimenstein, R. (2010). Vitamina A e diabetes gestacional. Revista Da Associação Médica Brasileira, 56(3), 355–359. doi:10.1590/s0104-42302010000300023
Lu J, Wang D, Ma B, et al. Blood retinol and retinol-binding protein concentrations are associated with diabetes: a systematic review and meta-analysis of observational studies. Eur J Nutr. 2022;61(7):3315-3326. doi:10.1007/s00394-022-02859-2
Li X, Long XY, Xie YJ, Zeng X, Chen X, Mo ZC. The roles of retinoic acid in the differentiation of spermatogonia and spermatogenic disorders. Clin Chim Acta. 2019;497:54-60. doi:10.1016/j.cca.2019.07.013
McCauley ME, van den Broek N, Dou L, Othman M. Vitamin A supplementation during pregnancy for maternal and newborn outcomes. Cochrane Database of Systematic Reviews 2015, Issue 10. Art. No.: CD008666. DOI: 10.1002/14651858.CD008666.pub3. Accessed 08 March 2023.
Hendrickx AG, Peterson P, Hartmann D, Hummler H. Vitamin A teratogenicity and risk assessment in the macaque retinoid model. Reprod Toxicol. 2000;14(4):311-323. doi:10.1016/s0890-6238(00)00091-5
Mastroiacovo P, Mazzone T, Addis A, et al. High vitamin A intake in early pregnancy and major malformations: a multicenter prospective controlled study. Teratology. 1999;59(1):7-11. doi:10.1002/(SICI)1096-9926(199901)59:1<7::AID-TERA4>3.0.CO;2-6
Sabrina Cruz, Suelem Pereira da Cruz & Andréa Ramalho (2018) Impact of Vitamin A Supplementation on Pregnant Women and on Women Who Have Just Given Birth: A Systematic Review, Journal of the American College of Nutrition, 37:3, 243-250, DOI: 10.1080/07315724.2017.1364182
Marcela de Sá Barreto da Cunha, Natália Aboudib Campos Hankins & Sandra Fernandes Arruda (2019) Effect of vitamin A supplementation on iron status in humans: A systematic review and meta-analysis, Critical Reviews in Food Science and Nutrition, 59:11, 1767-1781, DOI: 10.1080/10408398.2018.1427552
The information on this website is provided for educational purposes only and should not be treated as medical advice. FullWell makes no guarantees regarding the information provided or how products may work for any individual. If you suffer from a health condition, you should consult your health care practitioner for medical advice before introducing any new products into your health care regimen. For more information, please read our terms and conditions.
Author
Ayla Barmmer, MS, RD, LDN
Founder & CEO, FullWell
Ayla founded FullWell after decades of treating complex fertility cases in her private clinical practice at Boston Functional Nutrition to provide safe, evidence-based, effective, and high-quality supplements for those trying to conceive. She is also a co-founder of the Women's Health Nutrition Academy, an expert-led continuing education platform for healthcare practitioners.
Ayla founded FullWell after decades of treating complex fertility cases in her private clinical practice at Boston Functional Nutrition to provide safe, evidence-based, effective, and high-quality supplements for those trying to conceive. She is also a co-founder of the Women's Health Nutrition Academy, an expert-led continuing education platform for healthcare practitioners.
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8 / 2023 5 Fertility Diet Mistakes Read more 8 / 2023 6 Antioxidants to Know: Your Guide to Boosting Fertility Preconception Read more 8 / 2023 Understanding the Fertility Spectrum Read more
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10207 | https://press.rebus.community/openstaxbiology/chapter/chapter-5-structure-and-function-of-plasma-membranes/ | Chapter 5. Structure and Function of Plasma Membranes – OpenStax Biology
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Contents
Preface to Biology
Unit 1: The Chemistry of Life
Chapter 1. The Study of Life
1.1 The Science of Biology
1.2 Themes and Concepts of Biology
Chapter 2: The Chemical Foundation of Life
2.1 Atoms, Isotopes, Ions, and Molecules: The Building Blocks
2.2 Water
2.3 Carbon
Chapter 3. Biological Macromolecules
3.1 Synthesis of Biological Macromolecules
3.2 Carbohydrates
3.3 Lipids
3.4 Proteins
3.5 Nucleic Acids
Unit 2: The Cell
Chapter 4. Cell Structure
4.1 Studying Cells
4.2 Prokaryotic Cells
4.3 Eukaryotic Cells
4.4 The Endomembrane System and Proteins
4.5 The Cytoskeleton
4.6 Connections between Cells and Cellular Activities
Chapter 5. Structure and Function of Plasma Membranes
5.1 Components and Structure
5.2 Passive Transport
5.3 Active Transport
5.4 Bulk Transport
Chapter 6. Metabolism
6.1 Energy and Metabolism
6.2 Potential, Kinetic, Free, and Activation Energy
6.3 The Laws of Thermodynamics
6.4 ATP: Adenosine Triphosphate
6.5 Enzymes
Chapter 7. Cellular Respiration
7.1 Energy in Living Systems
7.2 Glycolysis
7.3 Oxidation of Pyruvate and the Citric Acid Cycle
7.4 Oxidative Phosphorylation
7.5 Metabolism without Oxygen
7.6 Connections of Carbohydrate, Protein, and Lipid Metabolic Pathways
7.7 Regulation of Cellular Respiration
Chapter 8. Photosynthesis
8.1 Overview of Photosynthesis
8.2 The Light-Dependent Reactions of Photosynthesis
8.3 Using Light Energy to Make Organic Molecules
Chapter 9. Cell Communication
9.1 Signaling Molecules and Cellular Receptors
9.2 Propagation of the Signal
9.3 Response to the Signal
9.4 Signaling in Single-Celled Organisms
Chapter 10. Cell Reproduction
10.1 Cell Division
10.2 The Cell Cycle
10.3 Control of the Cell Cycle
10.4 Cancer and the Cell Cycle
10.5 Prokaryotic Cell Division
Unit 3: Genetics
Chapter 11. Meiosis and Sexual Reproduction
11.1 The Process of Meiosis
11.2 Sexual Reproduction
Chapter 12. Mendel's Experiments and Heredity
12.1 Mendel’s Experiments and the Laws of Probability
12.2 Characteristics and Traits
12.3 Laws of Inheritance
Chapter 13. Modern Understandings of Inheritance
13.1 Chromosomal Theory and Genetic Linkage
13.2 Chromosomal Basis of Inherited Disorders
Chapter 14. DNA Structure & Functions
14.1 Historical Basis of Modern Understanding
14.2 DNA Structure and Sequencing
14.3 Basics of DNA Replication
14.4 DNA Replication in Prokaryotes
14.5 DNA Replication in Eukaryotes
14.6 DNA Repair
Chapter 15. Genes and Proteins
15.1 The Genetic Code
15.2 Prokaryotic Transcription
15.3 Eukaryotic Transcription
15.4 RNA Processing in Eukaryotes
15.5 Ribosomes and Protein Synthesis
Chapter 16. Gene Expression
16.1 Regulation of Gene Expression
16.2 Prokaryotic Gene Regulation
16.3 Eukaryotic Epigenetic Gene Regulation
16.4 Eukaryotic Transcription Gene Regulation
16.5 Eukaryotic Post-transcriptional Gene Regulation
16.6 Eukaryotic Translational and Post-translational Gene Regulation
16.7 Cancer and Gene Regulation
Chapter 17. Biotechnology and Genomics
17.1 Biotechnology
17.2 Mapping Genomes
17.3 Whole-Genome Sequencing
17.4 Applying Genomics
17.5 Genomics and Proteomics
Unit 4: Evolutionary Processes
18.1 Understanding Evolution
Testing
Reproductive Development and Structure (HUGH TESTING)
MATH: INTEGRALS (HUGH TESTING)
Math out of the box (HUGH TESTING)
OpenStax Biology
Chapter 5. Structure and Function of Plasma Membranes
Introduction
Despite its seeming hustle and bustle, Grand Central Station functions with a high level of organization: People and objects move from one location to another, they cross or are contained within certain boundaries, and they provide a constant flow as part of larger activity. Analogously, a plasma membrane’s functions involve movement within the cell and across boundaries in the process of intracellular and intercellular activities. (credit: modification of work by Randy Le’Moine)
The plasma membrane, which is also called the cell membrane, has many functions, but the most basic one is to define the borders of the cell and keep the cell functional. The plasma membrane is selectively permeable. This means that the membrane allows some materials to freely enter or leave the cell, while other materials cannot move freely, but require the use of a specialized structure, and occasionally, even energy investment for crossing.
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10208 | https://www.youtube.com/watch?v=81O1lXb4Cno&pp=0gcJCfwAo7VqN5tD | Finding Excluded Values for Rational Expressions (MATH 095 Goal Topic 9A)
Matthew Salomone
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Posted: 4 Mar 2016
Goal Topic 9A: Since a rational expression is undefined when its denominator is zero, we must exclude any value(s) of its variable that make that happen. To find these values, (1) Set the denominator equal to zero – you can ignore the numerator! – and (2) Solve the resulting equation using any appropriate method.
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Transcript:
in algebra a rational expression is the result of dividing one polinomial by another polinomial such as in this example where we're dividing the polinomial 3x -1 by the polinomial x^2 + 7x - 18 unlike the other types of Expressions that we've seen up to this point in our course a rational expression may not always have a defined value when certain values are plugged in for its variable for example there might be a value of x such that if we were to plug that value of x into this expression the result might be a numerical expression that does not have a value for the reason that it results in a fraction whose denominator is equal to zero because division by Z is an undefined operation in arithmetic this expression will be undefined for any value of x at which its denominator is equal to zero in this example we will find which values of X that is these are called the excluded values for this rational expression we can do that first by reminding oursel that to exclude a value from a rational expression is done because we want to make sure that the denominator is not permitted to equal zero so the entire question of finding the excluded values of this rational expression is going to turn on us finding where this denominator x^2 + 7 x - 18 becomes zero any value of x for which that happens must must be excluded from the possible values of X at which we can evaluate this rational expression notice also that the question of excluded values has only to do with where the denominator becomes zero and has nothing to do with any properties of the numerator of this rational expression so in the actual work that we do the x^2 + 7x - 18 in the denominator is something that we'll work with but the 3x -1 in the numerator is not relevant at all to our decisionmaking in finding excluded values so we can ignore the numerator and focus completely on the denominator so now that we know what the job is to find those values which make the denominator zero that tells us that we should set up an equation where the denominator is set equal to Zer so x^2 + 7 x - 18 is equal to 0 and any X which solves that equation will be an X that we need to exclude from the domain of this expression that means that all we have to do now is to solve this equation for x since this equation happens to be a quadratic equation with a quadratic trinomial on the left hand side and on the right hand side the most efficient method of solution is probably to factor and then to split each of those factors into its own equation we can Factor x^2 + 7 x - 18 using the sum and product method to find a pair of integers which multiply to give us -8 and which add to give us positive 7 such a pair of integers are positive 9 and -2 so we can Factor the left hand side as the product of x + 9 and x -2 since the right hand side of this equation is zero having factored the left hand side means we can split it apart into two pieces one equation x + 9 = 0 and another equation x - 2 is equal to zero solving each of these simpler equations will give us the value values of X that we're looking for we can solve the first by subtracting 9 from both sides to get x = -9 and we can solve the second by adding two to both sides to get xal pos2 each of these values x = -9 and x = 2 makes the expression x^2 + 7 x - 18 equal to Z and therefore if we were to plug either of these values of X back into our original rational expression the result would be a fraction whose denominator is equal to zero and therefore whose value is undefined therefore these are exactly the values that we must exclude from consideration in the domain of this expression so finding the excluded values of a rational expression means finding those values of X or whatever its variable happens to be for which if we were to plug those values into the expression the result would be an undefined fraction and this will happen exactly when the denominator of that fraction is equal to zero so finding the excluded values is simply a process of setting the denominator of this rational expression equal to zero and then using whatever method is appropriate solving that equation for x X |
10209 | https://www.calculatorsoup.com/calculators/math/percent-to-fraction-calculator.php | skip to calculator
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This online calculator converts a percent to a fraction. If the percent value is greater than 100% it is converted into a mixed number fraction. Enter percents to convert them into fractions. The number you enter can also have decimal places as in 3.5% or 0.625%.
To convert a percent to a fraction you first convert the percent to a decimal then use the same procedure as converting a
decimal to fraction.
How to Convert a Percent to Fraction
Divide the percentage by 100 to get a decimal number.
Use that number as the numerator (top) of a fraction. Put a 1 in the denominator (bottom) of the fraction.
Convert the decimal to a whole number: Count how many places are to the right of the decimal. If you have x decimal places then multiply numerator and denominator by 10x.
Reduce the fraction: Find the Greatest Common Factor (GCF) of the numerator and denominator and reduce the fraction by dividing both numerator and denominator by the GCF.
Simplify the remaining fraction to a mixed number fraction if possible.
Example: Convert 35.5% to a fraction
Divide by 100
35.5100=0.355
Rewrite the decimal number number as a fraction (over 1)
0.3551
Multiply numerator and denominator by by 103 = 1000 to remove 3 decimal places
0.3551×10001000=3551000
4.
Find the Greatest Common Factor (GCF) of 355 and 1000 and reduce the fraction, dividing both numerator and denominator by GCF = 5
355÷51000÷5=71200
Therefore,
35.5%=71200
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10211 | https://math.answers.com/other-math/How_do_you_write_400_billion_in_numbers | How do you write 400 billion in numbers? - Answers
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To write 400 billion in numbers, you would write it as 400,000,000,000. This is because each comma represents a grouping of three digits in the numerical system. So, 400 billion is written as 400 followed by nine zeros.
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### How do you write 1.52 Billion in numbers? 1.52 billion written in numbers is 1,520,000,000
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### How do you write six billion in numbers? Six billion = 6,000,000,000
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What is the difference between hyperbola and ellipse, since they both have the equation, $\frac{y^2}{a^2}+\frac{x^2}{a^2-c^2}=1$?
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$\begingroup$
Noted that the equation of ellipse is given by $\dfrac{y^2}{a^2}+\dfrac{x^2}{b^2}=1$,where $b^2=a^2-c^2$ While the equation of hyperbola is $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$,where $b^2=c^2-a^2$
When substitute the the expression of $b^2$ into each equation
the ellipse equation
$\dfrac{y^2}{a^2}+\dfrac{x^2}{a^2-c^2}=1$
The hyperbola equation $\dfrac{y^2}{a^2}-\dfrac{x^2}{c^2-a^2}=1$ ,which is equivalent to $\dfrac{y^2}{a^2}+\dfrac{x^2}{a^2-c^2}=1$
So my conclusion is that hyperbola and ellipse are of the same formula, is my conclusion correct?
analytic-geometry
conic-sections
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edited Dec 3, 2021 at 4:12
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$\endgroup$
3
4
$\begingroup$ $\frac{x^2}{b^2}$ is always positive since $b$ and $x$ are both real numbers. Your stated manipulations makes the $b$ in one of the equations imaginary. $\endgroup$
Prometheus
– Prometheus
2021-12-03 03:10:24 +00:00
Commented Dec 3, 2021 at 3:10
6
$\begingroup$ The main difference is that for the ellipse, $a^2 - c^2$ is positive whereas for the hyperbola it is negative. This corresponds with the fact that for the ellipse, the foci are between the vertices whereas for the hyperbola, the vertices are between the foci. This difference makes for some rather dramatic differences in how the two figures are plotted in ordinary Cartesian coordinates. $\endgroup$
David K
– David K
2021-12-03 03:36:05 +00:00
Commented Dec 3, 2021 at 3:36
1
$\begingroup$ Both hyperbola and ellipse are cases of conic sections (as is the parabola). In that sense they can all be described by the same general equation, though this doesn't mean they have the same properties. $\endgroup$
dxiv
– dxiv
2021-12-03 04:15:47 +00:00
Commented Dec 3, 2021 at 4:15
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Graphing Ellipses Centered at the Origin
Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form x2a2+y2b2=1, a>b for horizontal ellipses and x2b2+y2a2=1, a>b for vertical ellipses.
HOW TO
Given the standard form of an equation for an ellipse centered at (0,0), sketch the graph.
Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
If the equation is in the form x2a2+y2b2=1, where a>b, then
the major axis is the x-axis
the coordinates of the vertices are (±a,0)
the coordinates of the co-vertices are (0,±b)
the coordinates of the foci are (±c,0)
If the equation is in the form x2b2+y2a2=1, where a>b, then
the major axis is the y-axis
the coordinates of the vertices are (0,±a)
the coordinates of the co-vertices are (±b,0)
the coordinates of the foci are (0,±c)
Solve for c using the equation c2=a2−b2.
Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
EXAMPLE 3
Graphing an Ellipse Centered at the Origin
Graph the ellipse given by the equation, x29+y225=1. Identify and label the center, vertices, co-vertices, and foci.
Solution
First, we determine the position of the major axis. Because 25>9, the major axis is on the y-axis. Therefore, the equation is in the form x2b2+y2a2=1, where b2=9 and a2=25. It follows that:
the center of the ellipse is (0,0)
the coordinates of the vertices are (0,±a)=(0,±√25)=(0,±5)
the coordinates of the co-vertices are (±b,0)=(±√9,0)=(±3,0)
the coordinates of the foci are (0,±c), where c2=a2−b2 Solving for c, we have:
c &= ± \sqrt{a^2 - b^2} \
&= ± \sqrt{25-9} \
&= ± \sqrt{16} \
&= ±4
Therefore, the coordinates of the foci are (0,±4).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure 8.
Figure 8
TRY IT #3
Graph the ellipse given by the equation x236+y24=1. Identify and label the center, vertices, co-vertices, and foci.
EXAMPLE 4
Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form
Graph the ellipse given by the equation 4x2+25y2=100. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Solution
First, use algebra to rewrite the equation in standard form.
(\begin{aligned}
4x^2 &+25y^2 = 100 \
\frac{4x^2}{100} &+25y^2 = \frac{100}{100} \
\frac{x^2}{25} &+ \frac{y^2}{4} = 1
\end{aligned})
Next, we determine the position of the major axis. Because 25>4, the major axis is on the x-axis. Therefore, the equation is in the form x2a2+y2b2=1, where a2=25 and b2=4. It follows that:
the center of the ellipse is (0,0)
the coordinates of the vertices are (±a,0)=(±√25,0)=(±5,0)
the coordinates of the co-vertices are (0,±b)=(0,±√4)=(0,±2)
the coordinates of the foci are (±c,0), where c2=a2−b2 Solving for c, we have:
(\begin{aligned}
c &= ± \sqrt{a^2 - b^2} \
&= ± \sqrt{25 - 4} \
&= ± \sqrt{21}
\end{aligned})
Therefore, the coordinates of the foci are (±√21,0).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Figure 9
TRY IT #4
Graph the ellipse given by the equation 49x^2+16y^2=784. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Graphing Ellipses Not Centered at the Origin
When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, (h, k), we use the standard forms \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1, \quad a > b for horizontal ellipses and \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1, \quad a > b for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.
Given the standard form of an equation for an ellipse centered at (h, k), sketch the graph.
Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
If the equation is in the form \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1, where a > b, then
the center is (h, k)
the major axis is parallel to the x-axis
the coordinates of the vertices are (h \pm a, k)
the coordinates of the co-vertices are (h, k \pm b)
the coordinates of the foci are (h \pm c, k)
If the equation is in the form \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1, where a > b, then
the center is (h, k)
the major axis is parallel to the y-axis
the coordinates of the vertices are (h, k \pm a)
the coordinates of the co-vertices are (h \pm b, k)
the coordinates of the foci are (h, k \pm c)
Solve for c using the equation c^{2}=a^{2}-b^{2}.
Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
EXAMPLE 5
Graphing an Ellipse Centered at (h, k)
Graph the ellipse given by the equation, \frac{(x+2)^2}{4} + \frac{(y−5)^2}{9}=1. Identify and label the center, vertices, co-vertices, and foci.
Solution
First, we determine the position of the major axis. Because 9 > 4, the major axis is parallel to the y axis. Therefore, the equation is in the form \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1, where b^{2}=4 and a^{2}=9. It follows that:
the center of the ellipse is (h, k)=(-2,5)
the coordinates of the vertices are (h, k \pm a)=(-2,5 \pm \sqrt{9})=(-2,5 \pm 3), or (-2,2) and (-2,8)
the coordinates of the co-vertices are (h \pm b, k)=(-2 \pm \sqrt{4}, 5)=(-2 \pm 2,5), or (-4,5) and (0,5)
the coordinates of the foci are (h, k \pm c), where c^{2}=a^{2}-b^{2}. Solving for c, we have:
(\begin{aligned}
&c=\pm \sqrt{a^{2}-b^{2}} \
&=\pm \sqrt{9-4} \
&=\pm \sqrt{5}
\end{aligned})
Therefore, the coordinates of the foci are (-2,5-\sqrt{5}) and (-2,5+\sqrt{5}).
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
Figure 10
TRY IT #5
Graph the ellipse given by the equation \frac{(x−4)^2}{36} + \frac{(y−2)^2}{20}=1. Identify and label the center, vertices, co-vertices, and foci.
HOW TO
Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form.
Recognize that an ellipse described by an equation in the form a x^{2}+b y^{2}+c x+d y+e=0 is in general form.
Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
Factor out the coefficients of the x^{2} and y^{2} terms in preparation for completing the square.
Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, m_{1}(x-h)^{2}+m_{2}(y-k)^{2}=m_{3}, w_{h} where m_{1}, m_{2}, and m_{3} are constants.
Divide both sides of the equation by the constant term to express the equation in standard form.
EXAMPLE 6
Graphing an Ellipse Centered at (h, k) by First Writing It in Standard Form
Graph the ellipse given by the equation 4x^2+9y^2−40x+36y+100=0. Identify and label the center, vertices, co-vertices, and foci.
Solution
We must begin by rewriting the equation in standard form.
Group terms that contain the same variable, and move the constant to the opposite side of the equation.
Factor out the coefficients of the squared terms.
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
Rewrite as perfect squares.
Divide both sides by the constant term to place the equation in standard form.
\frac{(x−5)^2}{9} + \frac{(y+2)^2}{4}=1
Now that the equation is in standard form, we can determine the position of the major axis. Because 9 > 4, the major axis is parallel to the x-axis. Therefore, the equation is in the form \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1, where a^{2}=9 and b^{2}=4. It follows that:
the center of the ellipse is (h, k)=(5,-2)
the coordinates of the vertices are (h \pm a, k)=(5 \pm \sqrt{9},-2)=(5 \pm 3,-2), or (2,-2) and (8,-2)
the coordinates of the co-vertices are (h, k \pm b)=(5,-2 \pm \sqrt{4})=(5,-2 \pm 2), or (5,-4) and (5,0)
the coordinates of the foci are (h \pm c, k), where c^{2}=a^{2}-b^{2}. Solving for c, we have:
(\begin{aligned}
&c=\pm \sqrt{a^{2}-b^{2}} \
&=\pm \sqrt{9-4} \
&=\pm \sqrt{5}
\end{aligned})
Therefore, the coordinates of the foci are (5-\sqrt{5},-2) and (5+\sqrt{5},-2).
Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure 11 .
Figure 11
TRY IT #6
Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.
Source: Rice University,
This work is licensed under a Creative Commons Attribution 4.0 License. |
10214 | https://math.berkeley.edu/~bernd/HW2solution.pdf | Math 170: Homework 1 Liz Ferme February 14, 2017 Exercises 1.1, 1.4, 1.7, 1.8, 1.12, 1.14, 1.19 Problem 1.1 Suppose that a function f : Rn →R is both concave and convex.
Prove that f is an affine function.
Solution. Let f(0) = a and let the function g = f −a. Then g(0) = 0 and g is also both concave and convex. Let λ ∈[0, 1], and let x, y ∈Rn. Since g is convex, we have that that g(λx + (1 −λ)y) ≤λg(x) + (1 −λ)g(y). Since g is concave, we also have that g(λx + (1 −λ)y) ≥λg(x) + (1 −λ)g(y). Hence, we have g(λx + (1 −λ)y) = λg(x) + (1 −λ)g(y). This implies that g(λx) = λg(x) for λ ∈[0, 1] and all x ∈Rn, which implies that g is linear. Hence, f is affine.
Problem 1.4 Consider the problem minimize 2x1 + 3 |x2 −10| subject to |x1 + 2| + |x2| ≤5, and reformulate it as a linear programming problem.
Solution. We are looking to minimize the cost function 2x1 + 3 · max{x2 − 10, −x2 + 10}. We replace the max function with the variable z1 and add in the restrictions z1 ≥x2 −10, z1 ≥−x2 + 10 so that the minimum possible value of z1 is |x2 −10|. Similarly, we add in z2 and z3, adding the restrictions z2 ≥x1 + 2, z2 ≥−x1 −2, z3 ≥x2, z3 ≥−z3 so that the original restriction becomes z2 + z3 ≤5. Our linear programming problem is now minimize 2x1 + 3z1 subject to z1 ≥x2 −10, z1 ≥−x2 + 10, z2 ≥x1 + 2, z2 ≥−x1 −2, z3 ≥ x2, z3 ≥−x2, z2 + z3 ≤5.
Problem 1.7 Suppose that Z is a random variable taking values in the set 0, 1, . . . , K,, with probabilities p0, p1, . . . , pK, respectively. We are given the 1 values of the first two moments E[Z] = PK k=0 kpk and E[Z2] = PK k=0 k2pk of Z and we would like to obtain upper and lower bounds on the value of the fourth moment E[Z4] = PK k=0 k4pk of Z. Show how linear programming can be used to approach this problem.
Solution. We let p0, . . . , pK be the variables in our linear program. We would then have two linear programs: one to minimize E[Z4] = PK k=0 k4pk and the other to maximize E[Z4] = PK k=0 k4pk. These would give us our upper and lower bounds. We use the known moments to give us two restrictions on our variables. We also know that the pi sum to 1, and that each pi ≥0. This gives us K + 4 restrictions to use in our two linear programs.
Problem 1.8 Consider a road divided into n segments that is illuminated by m lamps. Let pj be the power of the jth lamp. The illumination Ii of the ith segment is assumed to be Pm j=1 aijpj, where aij are known coefficients. Let I⋆ I be the desired illumination of road i.
We are interested in choosing the lamp powers pj so that the illuminations are close to the desired illuminations I⋆ i . Provide a reasonable linear programming formulation of this problem.
Solution. We are interested in minimizing the differences between the illumi-nation Ii of each segment and the desired illumination of that segment, I⋆ i . So, our linear program will to be to minimize Pn i=1 |Ii −I⋆ i |, where Ii is given by the summation in the problem statement, and our variables are then the pj, subject to the condition that each pj ≥0.
Problem 1.12 Consider a set P described by linear inequality constraints, that is, P = {x ∈Rn|a′ ix ≤bi, i = 1, . . . , m}. A ball with center y and radius r is defined as the set of all points within the distance r from y. We are interested in finding a ball with the largest possible radius, which is entirely contained within the rest P. Provide a linear programming formulation of this problem.
Solution. We are trying to pick a point y that maximizes the shortest distance from y to the boundary of the set P. The dot product of our potential point y with the normal vector to each hyperplane defining the boundary will help us calculate this distance. Our linear program is thus minimize maxi{−|a′ iy −bi|} subject to a′ iy ≤bi for all i.
Problem 1.14 Solution.
(a) Let a be the number of units of the first product and let b be the number of units of the second product. The desired linear program is then maximize 3a + 2.4b 2 subject to 3a + 4b ≤20, 000, 3a −(6 · 0.45)a + 2b −(5.4 · 0.3)b ≤4000, a ≥0, b ≥0.
(b) Solution. Drawing was done by hand. Net income is only restricted by ma-chine hours in this case, and the net income is maximized by only producing product a, generating $20,000.
(c) Solution. Yes, this will help the company.
Problem 1.19 Suppose that we are given a set of vectors in Rn that form a basis, and let y be an arbitrary vector in Rn. We wish to express y as a linear combination of the basis vectors. How can this be accomplished?
Solution. We let a1, . . . , an denote the coefficients we are trying to find, and let b1, . . . , bn denote the given basis vectors.
We would like to impose the constraints Pn i=1 aibi j = yj for all j, where bi j is the jth component of the ith basis vector. These equations will have a unique solution. Thus, we can simply minimize a constant function (such as 0) subject to these constraints in order to find the desired coefficients.
3 |
10215 | https://www.sciencedirect.com/science/article/pii/S0006349524006659 | Structure and function of skin barrier lipids: Effects of hydration and natural moisturizers in vitro - ScienceDirect
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Volume 123, Issue 22, 19 November 2024, Pages 3951-3963
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Structure and function of skin barrier lipids: Effects of hydration and natural moisturizers in vitro
Author links open overlay panel Irene Sagrafena 1, Maxim Morin 2 3, Georgios Paraskevopoulos 1, Emelie J.Nilsson 2 3, Iva Hrdinová 1, Andrej Kováčik 1, Sebastian Björklund 2 3, Kateřina Vávrová 1
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Abstract
Lipid membranes play a crucial role in regulating the body’s water balance by adjusting their properties in response to hydration. The intercellular lipid matrix of the stratum corneum (SC), the outermost skin layer, serves as the body’s primary defense against environmental factors. Osmolytes, including urocanic acid (UCA) and glycerol, are key components of the natural moisturizing factor that help the SC resist osmotic stress from dry environments. This study examines the effects of UCA and glycerol (each at 5 mol %) on isolated human SC lipids. For this, different techniques were employed, offering complementary information of the system’s multiscale characteristics, including humidity-scanning quartz crystal microbalance with dissipation monitoring, infrared spectroscopy, x-ray diffraction, electrical impedance spectroscopy, and studies of water loss and permeability. Our results show that UCA increases water sorption and makes lipid films more liquid-like at high relative humidity, without significantly altering the lipid lamellar structure, chain order, or orthorhombic chain packing. Lipid films containing UCA exhibited higher water loss and significantly higher model drug permeability compared to lipid films without UCA. Further, incorporation of UCA resulted in kinetically faster changes in electrical properties upon contact with aqueous solution compared with control lipids. These observations suggest that UCA reduces lipid cohesion in regions other than the acyl chain-rich leaflets, which may impact SC desquamation. In contrast, glycerol did not influence the hydration or permeability of the SC lipid matrix. However, it increased the proportion of orthorhombic domains at high humidities and slowed the kinetics of the hydration process, as evidenced by slower changes in the dielectric properties of the lipid film. These findings suggest that glycerol enhances lipid cohesion rather than increasing water uptake, which is typically the expected function of humectants. Consequently, UCA and glycerol appear to have distinct roles in maintaining epidermal homeostasis.
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Significance
This study investigates the effects of two natural compounds, urocanic acid (UCA) and glycerol, on the lipid matrix of the skin’s outermost layer, the stratum corneum (SC), which serves as a barrier against environmental factors. Our experiments show that UCA increases lipid hydration and makes the lipid matrix more fluid, potentially facilitating the natural shedding of dead skin cells. In contrast, glycerol reinforces the lipid matrix’s structure without altering its water sorption capacity. These findings improve our understanding of skin barrier maintenance, which is important for developing skincare treatments.
Introduction
Water balance is a key factor in the adaptation of life, and the lipid membranes that compartmentalize, protect, and regulate the exchange of water and molecules are the central structures of this adaptation (1). In this respect, the extracellular lipid matrix in the uppermost layer of the epidermis, the stratum corneum (SC), is crucial for limiting the water loss from the body, as well as protecting it from adverse effects of the external environment. The lipid matrix has a unique multilamellar lipid structure comprised of a mixture of ceramides, cholesterol, and free fatty acids, which are minimally hydrated (2,3,4). By responding to external conditions, SC lipids effectively regulate transepidermal water loss to limit dehydration and ensure physiological conditions in dry or cold climates (5).
The general mechanism by which lipid membranes regulate the water balance of organisms relies on the fact that water controls the properties and phase behavior of lipid membranes (1,6). Thus, even the molecular properties of SC barrier lipids, although very rigid and poorly hydrated, have been shown to depend on the hydration degree (7,8). Furthermore, hydration regulates the permeability across the skin barrier for both hydrophilic and hydrophobic molecules (9,10) and is also important for the flexibility, softness, and pliability of the SC (11,12). Similarly, several studies have emphasized the importance of sufficient water levels for biochemical and enzymatic reactions to take place in the skin barrier, which is most likely crucial for maintaining a healthy status of the skin (13,14,15). Indeed, reduced SC hydration is associated with skin diseases such as atopic dermatitis (16) and psoriasis (17).
The presence of small water-soluble substances in relatively high concentrations is a common property for organisms exposed to osmotic stress, such as dry conditions (18). The outermost SC is notably exposed to a rather arid external environment and the presence of osmolytes in the SC is therefore expected. In the field of skin cosmetics and dermatology, the osmolytes are referred to as the natural moisturizing factor (NMF), and these components are known to be beneficial for skin suffering from dry conditions (19). The NMF substances contribute to the epidermal homeostasis and this effect is usually associated with their water-binding capacity (20). However, it should be noted that water and NMF can affect the physical state of the SC molecular components in a complex manner, which is why it is important to consider the combined effect of both water and the NMF (21,22). For example, the effects of NMF on water binding in the SC vary by depth, with limited binding at the surface due to folded keratin, maximal binding and swelling in the intermediate layers due to unfolded keratin, and minimal swelling at the deeper layers where water binding sites are saturated (23). In addition, some of these effects are dependent on relative humidity (RH); as Vyumvuhore et al. observed, beyond 60% RH, excess unbound water can disrupt lipid and protein structures, underscoring the critical role of hydration in maintaining the SC barrier function (24).
Most components of NMF are formed by the degradation and processing of the intracellular histidine-rich protein filaggrin in response to the water gradient in the SC (25), for example, pyrrolidonecarboxylic acid, trans-urocanic acid (UCA) (Fig.1a), and free amino acids. Although filaggrin degradation products are produced intracellularly, it may be expected that these small and polar substances can, to some extent, be distributed into the polar headgroup regions of the intercellular lipid matrix and thereby influence its physical state and/or hydration degree. This may be particularly true for UCA, which is somewhat less polar than other NMF components (distribution coefficient, a pH-dependent measure of a molecule’s lipophilicity, logD at pH 5 is approximately −1) and has an affinity for lipid membranes (26). In addition, UCA is distributed systemically in the body and is particularly abundant in the SC (in the cis-UCA form (27)), which necessarily implies permeation through the SC intercellular lipid matrix. A previous study showed that UCA affects not only the molecular mobility of keratin in the SC cells, but also the mobility of molecular segments belonging to ceramides, fatty acids, and cholesterol in porcine SC (21).
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Figure 1. Structures of glycerol and UCA (a), the composition of the isolated human SC lipids by weight (b), FTIR (c–g), and SAXD and WAXD (h–j) results at low RH and 32°C. (c) Is an example FTIR spectrum of the lipids, with insets showing the effects of glycerol and UCA on lipid chain order probed by the methylene symmetric stretching band (d–e) and chain packing deduced from the ratio of intensities of the higher and lower wavenumber components of the methylene rocking vibration (f and g). Data are shown as individual points, n ≥ 4; spectra are representative. Ceramide nomenclature in (b) is a combination of letters defining the sphingoid base (sphingosine, S; dihydrosphingosine, dS; phytosphingosine, P; 6-hydroxysphingosine, H) and the N-acyl chain (unsubstituted, N; α-hydroxylated, A; ω-acyloxy substituted, EO).
Another important NMF component in SC is glycerol (Fig.1a). Glycerol enters the SC via the water/glycerol transporter aquaporin-3 in basal keratinocytes (28) and also from the skin surface, where it is produced by hydrolysis of triglycerides from sebaceous glands. Aquaporin-3-deficient mice have reduced SC water and glycerol levels, leading to lower skin elasticity, and impaired barrier recovery after SC removal. Interestingly, these abnormalities are corrected by glycerol administration (29) and not by SC hydration. Similarly, asebia J1 mice showed epidermal hyperplasia, inflammation, and decreased SC hydration, which can be repaired by topical glycerol but not by urea, another potent endogenous humectant (30). The safety assessment of glycerol was described in a recent review (31).
These studies suggest a more complex role for glycerol than that of a mere moisturizer. In lipid models, glycerol inhibited the phase transition from liquid crystalline to the solid phase in a dry atmosphere (32). Similarly, glycerol (32) stabilized the liquid crystalline phase of dimyristoyl phosphatidylcholine at low humidities, which otherwise would induce a solid gel phase (33). Glycerol further prevented major changes in pig skin permeability caused by dehydration (34) and increased the mobility of both keratin and the lipid chains similarly compared with hydration, but only marginally influenced the SC water content (21). Based on these studies, it is likely that glycerol may substitute water rather than attracting it by providing similar hydrogen bonding to lipid polar headgroups (33).
Given the broad use of glycerol in topical products and the implication of both glycerol and UCA in diseases, it is mandatory to better understand their effect on the skin, including the extracellular lipid barrier. In this study, we investigate the effects of 5 mol % glycerol and 5 mol % UCA on isolated human SC lipids, eliminating the confounding effects due to hydration of the protein components of the SC cells, i.e., the corneocytes. We employ a combination of techniques to probe different aspects of the SC lipid lamellar matrix, offering complementary insights into how glycerol and UCA affect both the macroscopic properties and the molecular structure of the lipid film. Specifically, humidity scanning quartz crystal microbalance with dissipation monitoring (HS QCM-D) provides water sorption isotherms and examines how hydration influences the viscoelastic properties of lipids on a macroscopic level. However, HS QCM-D is less sensitive to macroscopic defects in the lipid films, which is why we also employ electrical impedance spectroscopy (EIS), water loss, and permeability experiments to provide information on the macroscopic barrier properties. Finally, Fourier transform infrared spectroscopy (FTIR) and small- and wide-angle x-ray diffraction (SAXD and WAXD) offer molecular insights into the lipid chain order, lateral packing, and organization of lamellar phases. This complementary approach allows us to conclude that UCA increases water sorption and permeability properties of the lipid film, making it more liquid-like, without significantly altering the molecular properties of the lipid lamellae. Conversely, glycerol enhances lipid cohesion, increases the proportion of orthorhombic domains, and slows the hydration process, without affecting the hydration or permeability characteristics of the lipid film. Thus, this study indicates distinct effects of UCA and glycerol on the SC lipid matrix.
Materials and methods
Materials
Silica gel 60 (230–400 mesh) for column chromatography, thin-layer chromatographic (TLC) plates (silica gel 60 F 254, aluminum back), high-performance TLC (HPTLC) glass plates (silica gel 60; 20×10 cm and 10×10 cm), gentamicin sulfate from Micromonospora purpurea (700 U/mg), trypsin from porcine pancreas (1500 U/mg), glycerol, indomethacin, sodium phosphate dibasic dodecahydrate, propylene glycol, solvents (HPLC grade) and buffer components were purchased from Merck (Darmstadt, Germany). trans-UCA was supplied by TCI (Tokyo, Japan). Nylon membranes (pore size 0.45 μ m, diameter 47 mm) were supplied by Fisher Scientific (Pardubice, Czech Republic). All chemicals were of analytical grade and were used without further purification. All aqueous solutions were prepared using Millipore water (Milli-Q system, Millipore, Burlington, MA). Esco microscope cover glasses 22×22 mm 2 were obtained from Erie Scientific (Portsmouth, NH). Nuclepore track-etched polycarbonate membranes (0.015 μ m pore size) were purchased from Whatman (Maidstone, UK). AT-cut SiO 2 sensors (QSX 303, 5 MHz) were from Biolin Scientific AB (Gothenburg, Sweden). DuPont Kapton with 0.013 mm thickness was obtained from Goodfellow Cambridge (Huntingdon, England).
Human SC lipids
Human skin was obtained from healthy female donors who underwent plastic surgery and beforehand had signed a written informed consent. The process was approved by the Ethics Committee of the Sanus First Private Surgical Centre (Hradec Králové, Czech Republic, no. 03/11/22) and was carried out in accordance with the principles of the Declaration of Helsinki. The SC was isolated using trypsin, the lipids extracted using chloroform/methanol mixtures, and purified by column chromatography using a protocol previously described in (35). HPTLC was used to check both the purity and proportions within the isolated human SC lipids subclasses (35,36).
Preparation of SC lipid models with or without NMF
The concentrations of glycerol and UCA in the SC are approximately 0.8 mmol of glycerol per gram of SC protein (30) and 0.6 mmol of UCA per gram of SC protein (37). Given their largely hydrophilic nature, with logP (log of the partition coefficient of a solute between octanol and water, the basic measure of lipophilicity) around −1.8 for glycerol and −1.0 for UCA at pH 5 (26), it is reasonable to assume that they will primarily be found in the corneocytes, the SC cells. Still, it is likely that these substances will also partition to some extent into the extracellular lipid matrix of the SC. In a pilot study, we first tried to incorporate 5, 10, and 50 mol % of these substances NMF into the lipids of the SC to better see any NMF effects. However, confocal Raman microspectroscopy showed phase separated UCA already at 10 mol % (data not shown). Therefore, we chose 5 mol % NMF in the lipid model for this study, which corresponds to approximately 1 wt % glycerol and 1.5 wt % UCA. These concentrations are also plausible to achieve after topical application of skin care products containing these substances.
Control models contained only isolated human SC lipids, while the NMF-containing models incorporated 5 mol % of either glycerol or UCA into the lipids. The total amount of lipids remained the same for all samples. Lipid solutions were prepared at 1.25 mg/mL in hexane/ethanol 96% 2:1 (v/v); glycerol and UCA were added to lipids at 5 mol % from stock solutions in 96% ethanol. The lipid solutions with or without NMF were sprayed (2×100 μ L per 1 cm 2) using the Linomat V (Camag, Muttenz, Switzerland) (36) under a stream of nitrogen on either glass coverslips (for SAXD, WAXD, and FTIR experiments) or nucleopore polycarbonate filters with 0.015 μ m pore size (for EIS and permeation experiment), resulting in lipid films of about 2.5 μ m in thickness. For the HS QCM-D experiment, lipids were sprayed to AT-cut SiO 2 sensors at a concentration of 0.25 mg/mL (≈500 nm thickness) to match the sample thickness requirements of this technique (38). Under spray conditions, the organic solvents evaporated, leaving a film of lipids. Overnight vacuum was used to remove residual solvents. The models were annealed at 70°C in a water vapor-saturated chamber for 20 min and, after that, allowed to cool over 4 h.
SAXD and WAXD
Samples at 25–90% RH
SAXD and WAXD were measured in a humidity chamber on a Xeuss 3.0 instrument (Xenocs, Grenoble, France), with an x--ray beam generated by a CuKα radiation source (λ= 1.542 Å). The samples were scraped off the support base, mounted on Kapton films, and placed in a chamber with controlled temperature and humidity. The measurements were performed at 32°C and three different RHs: 25, 70, and 90%. The detector was a Pilatus3 R 300K hybrid photon counting detector with a sample-to-detector distance (STDD) of 600 mm (SAXD) and 285 mm (WAXD). The samples were measured for 30 min at each relative humidity and STDD. These two STDDs covered the q-range 0.01≤ q (nm−1) ≤ 18, where q is the scattering vector defined as |q|= q= 4 π/λ sin (θ/2) and θ is the scattering angle. The 1D diffraction curves were obtained by azimuthal integration of the 2D pattern, corrected for background scattering and normalized to the direct beam, using the Xenocs XSACT software (version 2.6). The q-scale was calibrated using silver behenate. Data evaluation was performed using MATLAB R2021a with background subtraction, linear regression, and peak fitting with a single Gaussian curve.
Samples at ∼99% RH
These measurements were performed as above at 32°C but using a multipurpose Peltier gel-holder stage. The lipid models on Kapton film were exposed to a drop of PBS (pH 7.4) with water activity of approximately 0.99 for a duration of 4 h (34). Then, the excess buffer was removed and the Kapton with the hydrated lipids was mounted onto a holder for gels (eight positions) and sealed with another empty Kapton disc using an O-ring spacer. Data evaluation was performed using MATLAB R2021a with background subtraction, linear regression, and peak fitting with a single Gaussian curve.
FTIR
Infrared spectra were collected using a Nicolet 6700 spectrometer (Thermo Fisher Scientific, Waltham, MA) with a single-reflection MIRacle attenuated total reflection ZnSe crystal (PIKE Technologies, Madison, WI,). The spectra were collected in a closed chamber either at room temperature or while gradually increasing the temperature in 2°C steps in 15-min intervals at a resolution of 2 cm−1. The final spectra were generated by the coaddition of 256 scans.
HS QCM-D
Water sorption isotherms of thin films of extracted human SC lipids were determined by HS QCM-D (6,38). In brief, the QCM-D technique works by exciting a piezoelectric quartz sensor into resonance by applying an alternating voltage, while the frequency of the resulting oscillating shear motion is monitored. Any mass adsorbed or desorbed from the sensor results in a change of the frequency and this parameter can therefore be used to accurately determine the mass coupled to the sensor surface. The mass of the adsorbed material can be determined by the Sauerbrey equation (Δ m=−Δ f×C) as long as the adsorbed mass is small compared with the mass of the crystal and the material is rigidly adsorbed and homogenously distributed over the active area of the crystal (39). The Sauerbrey equation describes the relationship between the negative frequency change normalized per overtone, Δf, and the change of the areal mass of the adsorbed material, Δm, where C is a constant that depends only on the intrinsic properties and the thickness of the quartz disc (40). In addition to Δf, the viscoelastic properties of the adsorbed material are probed via the changes of the dissipation data, ΔD, which is related to the change in the decay time of the oscillating resonator when the alternating potential is turned off (40).
A Q-sense E4 instrument, equipped with the humidity module QHM 401 and AT-cut SiO 2 (QSX 303, 5 MHz) sensors, was used in this work (Biolin Scientific AB, Gothenburg, Sweden). The general experimental procedures of the humidity scanning experiment are described in detail elsewhere (6,38). In brief, baselines of uncoated sensors were measured in a dry N 2 atmosphere at 25°C. Next, the lipid films were deposited onto the sensor as described above. Subsequently, the coated sensor was dried in the humidity module under the flow of N 2 gas until a stable baseline of the frequency was obtained. Next, the humidity scanning experiment was initiated by flowing a LiCl solution with defined water activity (a w) through the humidity module. Since only water vapor can pass across the Gore membrane of the humidity module, the relative humidity above the sensor is regulated by continuously adjusting the a w of the LiCl solution, from 11% up to ∼100% RH (a w= RH/100%). The mass of the dry lipid film, as well as the amount of water taken up by the lipid film during the water sorption experiments, was calculated according to the Sauerbrey equation. The film thicknesses (h) of the dry lipid film were estimated from the areal mass of the dry film according to h=m dry/ρ, assuming a lipid density equal to be ρ= 0.87 g/cm 3 (41). The Δf and ΔD were primarily evaluated based on overtones three and five by using MATLAB R2021a.
Water loss and permeability
The model lipid films on Nuclepore filter support were sandwiched between Teflon holders with a 0.5 cm 2 circular opening. Then, the holders were mounted between the donor and the acceptor parts of Franz diffusion cells with an approximate acceptor phase of 7 mL filled with PBS buffer (pH 7.4) containing 50 mg/L gentamicin. The specific volume of the acceptor phase was measured and used for data calculations. Once the Franz cells were fully mounted, they were placed in the water bath at 32°C and left to equilibrate for 12 h. The next day, water loss was measured at least 2 times using an AquaFlux AF 200 instrument (Biox Systems, London, UK), with the condenser-chamber measurement method. The cells were then left to equilibrate overnight. The day after, 100 μ L of 2% indomethacin suspension in PBS was applied to the donor section of the Franz cells, onto the membranes. A total of 300 μ L were withdrawn from the acceptor phase every 2 h over 10 h and at the same time were replaced by the same volume of fresh PBS. The concentration of indomethacin in the acceptor phase was detected using a previously published HPLC method (42).
EIS
Nuclepore-supported lipid films were clamped between the donor and acceptor chambers of conventional Franz cells with a 9 mm orifice. However, the surface area of the lipid film being probed by the EIS was adjusted to 0.30 cm 2 by inserting silicone rings in each chamber. A circulating water bath kept the temperature of the Franz cell stable at 32°C for the duration of the experiment. Both the donor and acceptor phases were filled with PBS buffer (pH 7.4). Measurements were performed for up to 4 h in the following manner: scans 1–11 were performed with 60 s in between each measurement (from 0 to 10 min), every 5 min from scans 11 to–17 (from 10 to 40 min), with 10-min intervals between scans 17 and 20 (from 40 to 70 min), after 20 min was obtained scan 21 (90 min), after half an hour from scan 21, measurement 22 (2 h) was recorded, and finally scans 22–24 with 60-min intervals (from 2 to 4 h), which corresponded to a total time of 4 h. The impedance data were modeled with an electrical circuit consisting of a resistor (solution resistance, R sol) in series with a parallel arrangement of a resistor (membrane resistance, R mem) and a constant phase element (43,44). The constant phase element is an empirical element that accounts for the nonlinear distribution of time constants and can be utilized to determine the effective capacitance of the investigated membrane (45). The method used for analyzing the impedance data is described elsewhere (44). Data evaluation was performed using MATLAB R2021a.
Statistical analysis
One-way ANOVA with Tukey’s multiple comparisons test was used to analyze three or more groups (GraphPad Prism version 8.2.1, GraphPad Software, Boston, MA). All data are presented as the mean and standard deviation (SD); with the specific number of replicates (n) in each figure. A p value <0.05 was considered statistically significant.
Results
At low humidity, neither glycerol nor UCA significantly alters the lamellar and lateral arrangement of human SC lipids
To study the specific interactions of NMF with skin barrier lipids, we used lipids isolated from human SC and chromatographically purified to remove surface lipids or triglyceride contamination that may have occurred during tissue processing. HPTLC confirmed the presence of all subclasses of barrier lipids (Fig.1b). Based on this analysis and the chain length distribution of lipids from the literature (46), we approximated the average molar mass of these lipids to be 475 g/mol.
The models prepared from these isolated SC lipids, without added UCA or glycerol, have predominantly well ordered, tightly packed chains, as inferred by FTIR. In particular, the methylene stretching at 2848.3± 0.3 cm−1 suggests predominantly all -trans chains (Fig.1, c–e) and the rocking band doublet at 730 and 720 cm−1 indicates orthorhombic chain packing (Fig.1, f and g). In addition, these lipids are organized into a long periodicity lamellar phase (LPP) with a repeat distance, d= 13.5± 0.1 nm (Fig.1, h and j). In two out of four of the reference samples, one weak reflection, corresponding to d= 5.7± 0.1 nm, which can be assigned to a short periodicity phase (SPP), was also detected. All the diffractograms also had a very weak reflection at q ∼0.65 nm−1 corresponding to a d= 9.7± 1.8 nm. Separated cholesterol was not visible, but it is possible that its first reflection overlapped with the fourth reflection of the LPP. In the wide-angle region, two reflections at q= 15.2 and 16.8 nm−1 (Fig.1i), corresponding to distances between diffracting planes ∼0.41 and ∼0.37 nm, respectively, and confirmed the presence of orthorhombically packed lipid chains, which is typical for the lipids of healthy human SC (47).
Next, we incorporated glycerol and UCA into the SC lipids. At low humidity, no significant effects of glycerol or UCA on lamellar or lateral lipid organization were observed (Fig.1). Interestingly, the reflection attributable to SPP was only found in one out of four samples containing glycerol and its estimated d was 6.3 nm, which is higher than that in control or UCA samples (although this must be treated with caution as only one reflection was found). FTIR revealed a trend to less orthorhombically packed lipids in the presence of UCA compared with control (but not statistically significant).
At high humidity, UCA, but not glycerol, increases water sorption and alters the viscoelastic properties of the human SC lipids
HS QCM-D is a technique that provides water sorption isotherms of relatively thin films of various samples (38). Still, it is important to underline that HS QCM-D provides water sorption isotherms equivalent to the corresponding isotherms obtained with bulk samples (6,48). Thus, this method probes the hydration process of the SC lipid matrix on a macroscopic scale. The results show that the SC lipids only take up water to a very limited extent (about 7 wt %) and only at high environmental humidity, at about 98% RH (Fig.2a). This value roughly corresponds to 2 water molecules per lipid, which agrees well with the literature (41). This behavior did not change significantly in the presence of 5 mol % glycerol (water content was slightly but not significantly increased compared with lipids alone, p= 0.24). However, the presence of 5% UCA in the lipids increased the water sorption at 98% RH approximately twofold.
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Figure 2. Effects of glycerol and UCA on water sorption (a) and dissipation (b) of thin (≈500 nm) human SC lipid films studied by HS QCM-D. Mean± SD (n= 3–4), p values are given for statistically significant differences between UCA and control samples (values in black) and UCA versus glycerol samples (values in red).
The dissipation data offer insights into the overall viscoelastic behavior of the SC lipid films. The dissipation curve corresponding to the control SC lipid film increases slightly at 98% RH compared with lower RH values (Fig.2b), implying that the lipid film becomes more fluid-like upon hydration. Similar behavior (with slightly higher dissipation values compared with control) was also seen for the lipids with glycerol. In the presence of UCA and 94–98% RH, the dissipation reaches 2–4 times higher values compared with the control.
At high humidity, UCA and glycerol modulate SC lipid phase transitions and chain packing, but not lamellar arrangement
The LPP repeat distance in the SC lipids did not change significantly upon hydration, confirming that LPP does not exhibit noticeable swelling (49). While the reflection attributed to SPP was detected in two out of four of the control samples at 25–90% RH, it was found in all four samples at 99% RH. The intensity of the weak ∼10 nm phase (reflection at q= 0.65 nm−1 normalized to the second LPP reflection) decreased linearly with hydration of the system between 25 and 90% RH but was again slightly higher in the systems prepared at 99% RH. Lipids with 5 mol % glycerol did not differ from control at 25–90% RH but at 99% RH neither the SPP nor the ∼10 nm phase were detected and the LPP reflections slightly shifted. The incorporation of 5 mol % UCA did not affect the SAXD patterns. In all samples at all RH levels, WAXD reflections at q= 15.2 and 16.8 nm−1 with similar relative intensities were found indicating that glycerol or UCA did not alter the relative proportion of the orthorhombic and hexagonal lipid packing detected by WAXD (data not shown).
Lipid models with and without glycerol and UCA hydrated at approximately 99% RH were also studied by FTIR. Fig.3a shows the evolution of the methylene symmetric stretching on heating. Both glycerol and UCA slightly changed the pretransition temperature from about 35 to almost 40°C, but the chain order at 32°C was comparable (Fig.3b). The most pronounced differences were observed at temperatures above the main phase transition (around 65°C), where the lipids with UCA were significantly more disordered, whereas glycerol limited this temperature-induced lipid chain disordering suggesting stronger interactions. In fact, Fig.3a shows that the symmetric stretching of methylene groups remained below 2851 cm−1, which could be consistent with a liquid-ordered phase, even at high temperatures with glycerol. The temperature evolution of the rocking doublet (shown as the relative intensity of its 730 cm−1 component to the 720 cm−1 band in Fig.3c) confirmed the persistence of orthorhombically packed chains at higher temperatures in the presence of NMF compared with control samples, showing that UCA and glycerol have noticeable effects on the physical properties of the SC lipids. At 32°C, lipids with glycerol showed a greater relative abundance of orthorhombically arranged chains compared with the control (p= 0.03; Fig.3d), while this effect was not statistically significant in the case of UCA (p= 0.09; Fig.3d).
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Figure 3. Effects of glycerol and UCA on human SC lipids probed by FTIR at ∼99% RH. Lipid chain order probed by methylene symmetric stretching wavenumber in temperature (a) and at 32°C (b). Relative content of the orthorhombic chain packing deduced from the rocking doublet in temperature (c) and at 32°C (d). Data are shown means with SD (a and c) (n= 2) or as individual points (b and d) (n ≥ 5; p values are indicated).
UCA, but not glycerol, increases permeability of the SC lipid films
The effects of incorporated glycerol or UCA on the functional barrier properties of SC lipids were further investigated. We assessed their effect on water loss and the permeability to a model lipophilic substance, indomethacin, using lipid films mounted in Franz diffusion cells (Fig.4). Water loss values were not significantly different among samples, but there was a trend toward higher water loss in the presence of UCA compared with the lipid film without NMF (p= 0.07; Fig.4a). The indomethacin permeation increased approximately twofold in the presence of UCA compared with the control lipid sample, whereas glycerol did not induce any significant changes (Fig.4a).
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Figure 4. Effects of glycerol and UCA at 5 mol % on the barrier properties of human SC lipid films. Lipid films on porous support were sandwiched in Franz diffusion cells and water loss (a) and permeation of a model lipophilic compound indomethacin (b and c) were measured. Indomethacin permeation is shown as cumulative amounts of indomethacin permeated in time in (b) and calculated indomethacin flux values (from the linear portion of the permeation profile) in (c). Mean± SD (n= 4–6), p values are given for relevant comparisons; in (b), UCA samples are significantly different from control at all time points and from glycerol at 4–10 h at p<0.05.
UCA causes a rapid decrease in the electrical resistance of SC lipids upon hydration
Human SC lipid films with/without UCA or glycerol sandwiched in Franz cells were then examined using EIS over time (Fig.5). Similarly to the water loss and permeability experiments presented above, EIS probes the functional barrier properties on a macroscopic scale via the electrical resistance of the lipid film. The initial resistance values of the control and glycerol samples were around 5 MΩ×cm 2 and decreased rapidly (within 15 min) to tens of kΩ×cm 2 (Fig.5a). No differences in either absolute values or kinetics were found between lipids with and without glycerol. The resistance of the lipid films with UCA started at lower values of about 2 MΩ×cm 2 and dropped abruptly by 96% within 5 min and remained at approximately tens of kΩ×cm 2 for the rest of the experiment (120 min).
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Figure 5. Effects of glycerol and UCA at 5 mol % on the electrical properties of human SC lipid films. Lipid films on porous support were sandwiched in Franz diffusion cells and investigated by EIS. (a) Shows the evolution of membrane resistance and (b) the evolution of effective capacitance in time. Mean± SD (n= 4–6).
The second parameter that we focus on by the EIS data is the effective capacitance, which provides macroscopic scale information on the dielectric properties of the lipid film. However, the capacitive currents are obtained at significantly higher frequencies as compared with the electrical resistance and therefore not sensitive to defects of the lipid barrier. The initial values of the effective capacitance were around 1 nF/cm 2 of the control samples and increased to approximately 1.7 nF/cm 2 within 15 min (Fig.5b). The capacitance of the lipids films with UCA started at similar values and then leveled at around 1.4 nF/cm 2 within 5 min. Finally, the capacitance of the glycerol samples started at 0.7 nF/cm 2, followed by a continuous increase for the entire duration of our measurements with final values around 1.7 nF/cm 2 after 240 min.
In the case of the lipid films, showing close to ideal capacitance properties, we can conceptualize the dielectric properties using a simple plate capacitor model with a dielectric lipid medium between the working and counter electrodes; ε=h C e f f/ε 0. Here, h is the thickness of the lipid (2.5 μ m), ε 0 is the permittivity of vacuum (8.9 × 10−14 F cm−1), and C e f f is obtained from the measurements (Fig.5b). As shown in this study, the lipid matrix does not swell due to hydration to any observable degree; implying that h can be treated as constant. Thus, based on the determined values of C eff, we obtain values of dielectric constants summarized in Table 1. The dielectric constant of the SC lipid matrix is not readily available in standard databases. However, these values are comparable with other organic molecules, which generally have dielectric constants in the range of 2–10.
Table 1. Estimated dielectric constant (relative permittivity, ε) values for the investigated human SC lipid films
| Empty Cell | Dielectric constant, ε |
| Initial (0 min) | Final (240 min) | Increase (%) |
| Control | 2.8 | 4.8 | 70 |
| Glycerol | 2.0 | 4.8 | 143 |
| UCA | 2.8 | 3.9 | 40 |
Discussion
A multiscale approach to understand how glycerol and UCA affects the structure and function of skin barrier lipids
In this study, we employed complementary techniques to explore a range of physicochemical parameters of the SC lipid matrix, with careful consideration of the system’s multiscale characteristics. It is worth noting that the x-ray diffraction and FTIR experiments focus on the molecular organization of the lipid lamellae. In particular, SAXD probes the organization of the lamellar phases at length scales of nm (i.e., SPP and LPP), whereas WAXD examines the lateral acyl packing of the lipids with Ångström resolution (i.e., orthorhombic, hexagonal, or fluid/disordered). Finally, FTIR probes the aliphatic chain bond vibrations and gives insight into the molecular dynamics of the lipids, such as lipid fluidity and lateral acyl packing. The FTIR results therefore provide complementary information to SAXD and WAXD data. Although these techniques provide understandings of the molecular properties of the lipid matrix, it is important to underline that they are insensitive to defects of the macroscopic barrier properties of the lipid films. Therefore, water loss and model drug permeability experiments and electrical resistance measurements were conducted to acquire information on the macroscopic scale, which ultimately determines the functional barrier properties. In addition, HS QCM-D measurements provide water sorption isotherms and information about the viscoelastic properties of the lipid films on a macroscopic scale (i.e., bulk properties) and is not particularly sensitive to defects within the lipid film on the molecular scale. The effective capacitance calculated based on EIS data provides insight into how the dielectric properties are affected by the NMF and can also be considered a bulk parameter, as it is not necessarily sensitive to barrier defects in the lipid film. Considering these aspects, the following discussion highlights our understanding of the structure and function of the SC lipids that emerges from this multiscale approach for studying the effects of incorporating 5 mol % of glycerol or UCA into the SC lipid matrix.
Glycerol and UCA affect SC lipid hydration differently and also behave differently when incorporated into lipids compared with their pure forms
The first conclusion from our results is that glycerol and UCA have a minor effect on the structure and organization of the lipid matrix under dry conditions (Fig.1). However, at high humidities, the data show that glycerol and UCA influence the properties of the extracellular lipid matrix of the human skin barrier differently. Specifically, these substances exhibit distinct behaviors when incorporated into lipids compared with their pure forms. For example, pure glycerol absorbs approximately 65 wt % water at equilibrium with 85% RH (21). However, when added to SC lipids, glycerol did not change the system’s ability to take up water (Fig.2a). This lack of increased hydration suggests that glycerol is incorporated into the lipid matrix via hydrogen bonding with the polar lipid headgroups, thereby limiting its ability to disrupt existing hydrogen bonds and form new ones with water molecules. In contrast, UCA increased the water sorption by the lipids approximately twofold at 98% RH. Notably, pure UCA absorbs significantly less water compared with glycerol, with less than 1 wt % water uptake at 85% RH (21). Therefore, this difference is likely due to UCA’s effect on the physical properties of lipids, leading to increased water uptake by the total system rather than UCA alone.
Does UCA interact differently with the SC lipid matrix as compared with glycerol?
The observed differences in water sorption capacity of the SC lipids in the presence of glycerol and UCA should be related to the effects of these substances on the lipid chain order and lateral packing (i.e., Fig.3). Considering that both glycerol and UCA stabilized the orthorhombic phase by shifting the transition to hexagonal chain packing to higher temperatures, it is unlikely that the water sorption is associated with alterations of the orthorhombically organized acyl chains. An alternative explanation is that glycerol and UCA interacts differently with the SC lipids, ultimately leading to different water sorption capacity of the lipid matrix. Given the distinct chemical structures of glycerol and UCA (see Fig.1a), it is logical to take this into account. For example, the carboxyl group of UCA is likely to be located in the polar head regions of the lipid matrix, while its imidazole ring has a potential to engage in π-π interactions and thus be buried deeper in the hydrophobic regions of the SC lipids. Since the incorporation of the imidazole ring between tightly arranged acyl chains would be very disadvantageous, and the relative abundance of orthorhombic lipid chains does not decrease significantly, it can be assumed that UCA is more likely to be incorporated into cholesterol-rich leaflet of the lipid matrix. Similar selective fluidization of cholesterol-rich regions has recently been described for cholesterol sulfate and ionic liquids in skin lipid models, although these were simplified with fewer lipid species than the human SC lipids used here (50,51). This reasoning might account for the distinctly different effects seen at temperatures above the main lipid phase transition, where UCA promoted heat-induced lipid disordering, while glycerol limited this effect, indicating preserved lipid order and stronger interactions. Further support comes from the dissipation data, which provide insights into the viscoelastic behavior of the SC lipid films. In other words, the selective partitioning of UCA may lead to slipping under shear stress induced by the oscillating quartz crystal, which could potentially explain the significant increase in dissipation at high hydration without any significant structural alterations. Consequently, such UCA-induced regions of a less cohesive lipid matrix could enable higher diffusion rates for water, electrolytes, and the model substance.
On the other hand, the inability of glycerol to increase water sorption by human skin lipids indicates its confinement in lipid layers and saturation of its hydrogen bonds by interactions with polar lipid heads. These hydrogen bonds may be responsible for the higher thermostability of the orthorhombic phase and the lower degree of disorder of the lipid chains after the main phase transition. Thus, glycerol appears to contribute to increased cohesion of the individual leaflets of the lipid matrix and increased stability of the orthorhombic domains without altering the structure of the LPP or the permeability of the lipid matrix, rather than acting as a humectant.
UCA, but not glycerol, influences the functional barrier properties of SC lipids
Although UCA did not affect the overall lipid lamellar organization, its impact on the water sorption isotherm and viscoelastic properties aligns with its effects on the functional barrier properties of SC lipids. The latter was assessed by water loss, probing the inside-outside barrier, and the permeability to indomethacin, as a marker of the outside-inside barrier, indicated increased lipid permeability induced by UCA but not glycerol. Furthermore, these observations are in line with the faster drop of the electrical resistance detected in the case of UCA (Fig.5a) as compared with glycerol or control. Taken together, the increased capacity to take up water and the elevated dissipation values induced by UCA, supports the idea that UCA increases the global fluidity of the lipid system thereby making its barrier properties more permeable.
Interestingly, lipids with glycerol showed more orthorhombically arranged lipid chains than control at skin temperature (i.e., 32°C), as shown in Fig.3d. Given that orthorhombic packing is a key feature of skin lipids and is directly associated with water loss (52), it is somewhat surprising that lipids with glycerol did not show lower water loss values compared with control. In fact, the lipid samples containing glycerol showed similar macroscopic barrier properties as compared with the control samples in terms of water loss, model drug permeability, and electrical resistance.
Insights from EIS measurements on lipid films—Eliminating the confounding effects of corneocytes, the SC cells
The ElS data obtained from the SC lipid films showed some interesting features of the electrical properties compared with what is normally observed for excised skin membranes. The initial resistance values around 2–5 MΩ×cm 2 are significantly higher as compared with typical values obtained with excised skin membranes, ranging between 10 and 200 kΩ×cm 2 (44,53). This may be expected given the hydrophobic character of the SC lipid matrix and its foreseen resistance to the passage of hydrophilic charge carriers. However, the abrupt decline in resistance, manifesting within minutes, is rather unexpected. Given that the structural characteristics of the lipid matrix remain largely intact despite hydration, this observation suggests the presence of global structural defects within the lipid films, facilitating the entry of water and the subsequent transport of electrolytes across the lipid film. The rather drastic drop in the resistance implies that this mechanism of water filling defective voids in the lipid films occurs within minutes. After this initial phase, the resistance values reach stable plateaus with comparable values to what is typically obtained with excised skin membranes, thereby affirming that the lipid films closely resemble the primary barrier constituent of the SC.
When comparing the kinetics of the resistance decay, the results revealed that the control and glycerol samples required about 15 min to reach stable values. For the lipid films with UCA, this process was completed in about 5 min. Hence, it is plausible that the integration of UCA into the lipid films leads to a less homogenous lipid matrix with larger or more accessible void regions, facilitating the rapid ingress of water and electrolytes. As discussed above, this is in line with the observed elevated permeability of the lipid films containing UCA toward water and indomethacin and also supports the elevated dissipation values at high hydration in the presence of UCA.
The high-frequency impedance data showed close to 90° phase shift between the applied voltage and the resulting current. This indicates that the high-frequency impedance response for the lipid films is close to purely capacitive, with negligible contribution from resistive or other nonideal elements. In contrast, typical phase shifts obtained with excised skin membranes vary between 60 and 80°, illustrating the complex and nonideal capacitance components of the SC (44). The capacitance of the SC is intricately linked to its dielectric characteristics, which encompass the low-conductivity lipid matrix, lipid-protein domains, and charged lipid and protein species that contribute to double-layer capacitance (44,54,55). Typical values of the effective capacitance of excised skin membranes fall between 10 and 50 nF/cm 2 (44,56). These values are around one order of magnitude higher compared with the values obtained with the lipid films (0.7–1.7 nF/cm 2). This deviation implies that the capacitive domains of the SC are not only comprising the lipid matrix, which is exclusively probed here, but also other components of the SC. For instance, it has been previously suggested that capacitive currents, originating from double-layer charging, may be generated by lipid-protein domains or charged amino acid residues, for example, within the keratin filament (44,54).
Notably, the estimated dielectric constant increases when comparing the initial and final values, which can be explained by incorporation of water, with a dielectric constant of about 79, in the lipid films. However, both the absolute increase and dynamics of this hydration process differ between the samples (cf. Fig.5b where C eff is proportional to ε). For UCA, the change occurs rapidly and to a lesser extent (40%), while for the glycerol sample, the change occurs more linearly resulting in a final increase of 143%. These observations suggest that the dielectric properties of the capacitive domains of the lipid film containing UCA are readily accessible for water and then remain constant, which is in line with more accessible void regions discussed above. The slower hydration process for the lipid film comprising glycerol indicates a more cohesive lipid matrix, which is supported by the observed higher proportion of orthorhombic chain packing. Thus, slower hydration dynamics is expected assuming that glycerol forms hydrogen bonds with the polar lipid headgroup and that the hydration process requires breaking and making of new hydrogen bonds with water molecules.
The observed effects of glycerol presented here should be compared with the extensive research that has explored how glycerol interacts with various lipid types, such as phospholipids and glycosphingolipids, particularly in its role as a cryoprotectant. When comparing glycerol’s effects on these different lipid classes, including the more nonpolar SC lipids investigated in this study, several differences and similarities emerge. For example, compression isotherms have shown that glycerol causes a slight expansion of monolayers of DPPC, DOPC, POPC, and POPE, indicating that glycerol integrates into the headgroup regions, regardless of the phospholipid species (57). In contrast, compression isotherms for glycosphingolipids monolayers showed the opposite effect (58), suggesting that the structural properties of the lipid headgroup significantly influence glycerol’s impact on the overall molecular organization. Notably, molecular dynamics simulations revealed that glycerol has a strong ability to form hydrogen bonds on facing leaflets of DPPC bilayers, creating a denser and more cohesive solvent layer between the phospholipid bilayers (59). In line with this suggestion, another study concluded that glycerol enhances the intermolecular cohesion of water molecules in the headgroup-solvent interfacial region, based on diffusivity measurements of water at the surface of DPPC bilayers, a property correlated with the hydrated volume of lipid headgroups (60). In summary, these findings indicate that glycerol and water interact with lipid headgroups in a comparable manner, while the effective outcome of these interactions may vary depending on the precise lipid headgroup architecture.
Role of glycerol and UCA in the lipid matrix of the epidermal barrier
The observed effects of UCA can be related to the (patho)physiological processes in the upper layers of the SC, where the intercellular lipids are indeed less ordered than in the middle layers of the SC (61), and the processes necessary for desquamation are activated (62). It is possible that UCA, by contributing to lipid loosening and elevated hydration of the intercellular space, is one of the factors that allows adequate activation of the protease system that degrades corneodesmosomes (63). Consistent with this hypothesis, inadequate desquamation has been described in mice or humans with mutations leading to loss of filaggrin and hence UCA (44,53). Another aspect related to UCA is the fact that this substance undergoes cis-trans isomerization upon exposure to sun light. If we assume that the isomerization of trans-UCA to cis-UCA (27) also takes place to some extent in the lipid matrix, the localization of this molecule in a less rigid environment also makes sense. Consequently, the presence of UCA may result in less cohesive regions or leaflets in the lipid matrix upon hydration while leaving the acyl chain-rich domains essentially unchanged. This selective reorganization may lead to slipping under shear stress induced by the oscillating quartz crystal, which could potentially explain the significant increase in dissipation at high hydration without any significant structural alterations. Consequently, such UCA-induced regions of a less cohesive lipid matrix could enable higher diffusion rates for water, electrolytes, and the model substance.
Glycerol, on the other hand, is expected to be distributed more homogeneously throughout the SC lipid matrix, compared with UCA, as it is transported from below by aquaporin-3 (29) and from above by the degradation of surface lipids (30). Therefore, glycerol’s inability to fluidize and permeabilize the barrier lipids (although so far only valid for our in vitro system with 5 mol % glycerol) seems to make sense, as otherwise the lipid barrier would be compromised. Thus, the overall unchanged molecular structure and functional properties of the SC lipids in the presence of glycerol suggests that it is confined within lipid layers, enhancing the stability and cohesion of the lipid matrix through hydrogen bonds with polar lipid heads, rather than acting as a humectant. This proposed mechanism may be associated with previous observations in aquaporin-3-deficient mice (29) and asebia J1 mice (30), showing that skin barrier abnormalities are corrected by glycerol but not by water or urea. Although the effect of glycerol on skin lipids described in this study was not pronounced, it must be pointed out that 5 mol % glycerol means one glycerol molecule per roughly 20 lipid molecules and its native concentration may be higher; for example, water is present at 1–2 molecules per lipid. Thus, glycerol may be one important piece of the puzzle that leads to a competent epidermal barrier.
Data and code availability
Data are available from the authors upon request.
Acknowledgments
This study was supported by the Czech Science Foundation project no. 22-20839K (to I.S., G.P., A.K., and K.V.), Charles University (SVV 260 661, to I.S. and I.H.), by the project New Technologies for Translational Research in Pharmaceutical Sciences/NETPHARM, ID CZ.02.01.01/00/22_008/0004607, cofunded by the European Union (to K.V., G.P., and A.K.), and by the research network Biobarriers– Health, Disorders and Healing, funded by the Swedish Knowledge Foundation (to M.M., E.J.N., and S.B., grant no. 20190010).
Author contributions
Conceptualization, S.B. and K.V.; data curation, I.S., M.M., E.J.N., S.B., and K.V.; funding acquisition, K.V. and I.S.; investigation, I.S., M.M., G.P., E.J.N., I.H., A.K., S.B., and K.V.; methodology, I.S., M.M., G.P., E.J.N., I.H., A.K., S.B., and K.V.; supervision, S.B. and K.V.; validation, I.S., M.M., G.P., E.J.N., and S.B.; visualization, all authors; writing– original draft, I.S., S.B., and K.V.; writing– review& editing, all authors.
Declaration of interests
The authors declare no competing interests.
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References
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EnglishMatic » Free English Cloze Tests & Exercises Online
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10217 | https://cdnx.uobabylon.edu.iq/lectures/obgu3s7MkWXvppXlVpZGw.pdf | Information security department Logic Design 1st class Dr. Rasim Azeez 1 Boolean Algebra In 1854 George Boole introduced a systematic approach of logic and developed an algebraic system to treat the logic functions, which is now called Boolean algebra. In 1938 C.E. Shannon developed a two-valued Boolean algebra called Switching algebra, and demonstrated that the properties of two-valued or bistable electrical switching circuits can be represented by this algebra. Boolean algebra is a mathematical system for the manipulation of variables that can have one of two values. In digital systems, these values are “on” and “off,” 1 and 0, or “high” and “low.” The following Huntington postulates are satisfied for the definition of Boolean algebra on a set of elements S together with two binary operators (+) and (.). 1. (a) Closer with respect to the operator (+). (b) Closer with respect to the operator (.). 2. (a) An identity element with respect to + is designated by 0 i.e., x + 0 = 0 + x = x. (b) An identity element with respect to . is designated by 1 i.e., x.1 = 1. x= x. Two-Valued Boolean Algebra Two-valued Boolean algebra is defined on a set of only two elements, S = {0,1}, with rules for two binary operators (+) and (.) and inversion or complement as shown in the following operator tables, respectively. Information security department Logic Design 1st class Dr. Rasim Azeez 2 The basic rules of Boolean algebra are:- The proving of theorems can be done by using the Postulates or the truth table as illustrated in the following : Information security department Logic Design 1st class Dr. Rasim Azeez 3 Example: prove that x(y + z)= xy + xz Solution: x y z y + z x(y + z) xy xz xy + xz 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 Hence, it is proved because the left side is similar to the right side Information security department Logic Design 1st class Dr. Rasim Azeez 4 Example: prove that x + yz = (x+y)(x + z) Solution: x y z yz x + yz x + y x + z (x+y)(x+z) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 Hence, it is proved because the left side is similar to the right side Operator Precedence The operator precedence for evaluating Boolean expressions is (1) parentheses, (2) NOT, (3) AND, and (4) OR. In other words, expressions inside parentheses must be evaluated before all other operations. The next operation that holds precedence is the complement, and then follows the AND and, finally, the OR. Boolean Function A Boolean function is a relation between the binary inputs and the binary outputs. The value of a function (output) may be 0 or 1, depending on the values of inputs present in the Boolean function. Boolean Function can be described by: 1- a truth table 2- Boolean equation, 3- a logic diagram 1- Truth table Truth table for a function is a list of all combinations of 1’s and 0’s that can be assigned to the binary variables and a list that shows the value of the function for each binary combination. For n variables, there are rows (states) Information security department Logic Design 1st class Dr. Rasim Azeez 5 For example, when the number of variables (inputs) n=3, then the number of rows (states) = = 8 as shown in this table: A B C F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 0 2- Boolean equation (Boolean function form): Boolean equation consists of a binary variable identifying the function (output) followed by an equal sign and a Boolean expression formed with binary variables, the two binary operators AND and OR, one unary operator NOT, and parentheses. When a Boolean expression is implemented with logic gates, each literal in the function is designated as input to the gate. The literal may be a primed or unprimed variable. For example, the Boolean equation of the truth table above is: ̅ ̅ ̅ Where F is the function (output) A, B, C are the input variables (literals) 3- Logic diagram (circuit diagram): The logic diagram composed of logic gates in which are interconnected by wires that carry logic signals. The figure below shows the logic diagram of the Boolean equation ̅ ̅ ̅ Information security department Logic Design 1st class Dr. Rasim Azeez 6 Example: Write the Boolean expression for the logic diagram shown. Solution: The Boolean expression of this circuit is : ̅ Simplification using Boolean algebra: Minimization of the number of literals and the number of terms leads to less complex circuits as well as less number of gates, which should be a designer’s aim. There are several methods to minimize the Boolean function such as Boolean algebra and Karnaugh map (K-Map). Here, simplification or minimization of complex algebraic expressions will be shown with the help of postulates and theorems of Boolean algebra. Example : Simplify the following Boolean expression. ̅ ̅ Solution: ̅ ̅ ̅ ̅ 𝑥𝑦 𝑧 𝑥𝑦 𝑧 Information security department Logic Design 1st class Dr. Rasim Azeez 7 Example : Simplify the following Boolean expression. ̅ ̅ Solution: ̅ ̅ ̅ ̅ ̅ ̅ Example: Simplify the following Boolean expression. ̅ ̅ ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ Solution: using De Morgan theorem ̅ ̅̅̅̅̅̅̅̅̅̅ ̿ ̿ ̅ ̅ Example: simplify and draw the logic diagram of ̅̅̅̅̅ Solution: ̅̅̅̅̅ ̅ ̅ ̅ ̅ ̅ ̅ Information security department Logic Design 1st class Dr. Rasim Azeez 8 Example : Draw the Boolean expression ̅ a) using basic logic gates. b) using NOR gates only. Solution : a) b) |
10218 | https://www.vedantu.com/maths/sin-60-degrees | Sin 60 Degrees: Value, Formula, and Applications in Maths
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Maths
Sin 60 Degrees – Exact Value, Formula & Uses with Examples
Sin 60 Degrees – Exact Value, Formula & Uses with Examples
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Rama Sharma
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What is the value of sin 60 degrees in fraction, decimal, and radians?
The concept of sin 60 degrees plays a key role in mathematics and is widely applicable to both real-life situations and exam scenarios. From geometry to trigonometry and physics, knowing the exact value of sin 60 degrees helps solve triangles, word problems, and even physics questions with speed and accuracy.
What Is Sin 60 Degrees?
Sin 60 degrees is a trigonometric value representing the ratio of the length of the side opposite a 60° angle to the hypotenuse in a right-angled triangle. You’ll find this value applied in topics such as trigonometric ratios, the unit circle, and geometry word problems.
Key Formula for Sin 60 Degrees
Here’s the standard formula: sin60∘=3 2 or approximately 0.866
| Angle (Degrees) | Angle (Radians) | Sin Value (Fraction) | Sin Value (Decimal) |
--- --- |
| 60° | π/3 | √3/2 | 0.866 |
Cross-Disciplinary Usage
Sin 60 degrees is not only useful in Maths but also plays an important role in Physics, Computer Science, and logical reasoning. Students preparing for JEE, NEET, and board exams often encounter problems that use this value for solving triangles, vector decomposition, and even calculation of heights and distances. It’s also a basic building block for advanced trigonometry, as well as other standard trigonometric angles.
Step-by-Step Illustration
Draw an equilateral triangle with all sides = 2 units. Drop a height from one vertex to the base. This height splits the base into 1 unit + 1 unit and creates two 30-60-90 right triangles.
Calculate the height using Pythagoras’ Theorem: Let height = h.
h 2+1 2=2 2⇒h 2=4−1=3⇒h=3
3. In the right-angled triangle, for angle 60°:
Opposite = h = √3, Hypotenuse = 2
sin60∘=Opposite Hypotenuse=3 2
4. Decimal value: sin60∘≈0.866
Sin 60 Degrees in Trigonometric Table
| Angle | Sin Value | Cos Value | Tan Value |
--- --- |
| 0° | 0 | 1 | 0 |
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
| 90° | 1 | 0 | undefined |
Sin 60 Degrees in Different Forms
| Form | Sin 60° Value |
--- |
| Fraction/Surd | √3/2 |
| Decimal | 0.866 |
| Radian form | sin(π/3) |
| Unit circle (coordinates) | (½, 0.866) |
Applications of Sin 60 Degrees
You’ll use sin 60 degrees in lots of practical problems. For example, calculating the height of a triangle given the hypotenuse, working with trigonometric ratios, or decomposing forces at a 60° angle in physics. It’s also vital for MCQ accuracy and last-minute revision before competitive exams like JEE and CBSE boards.
Solved Example:
Find the perpendicular height of an equilateral triangle of side 4 cm.
Each side = 4 cm. Height splits base into 2 cm.
By Pythagoras’ Theorem,
h 2+2 2=4 2⟹h 2=16−4=12⟹h=2 3 cm
So, height = 2 3 cm = 3.464 cm (using sin 60° value)
Speed Trick or Vedic Shortcut
Here’s a fast recall trick: The value of sin 60° is always larger than sin 30° and sin 45° but less than sin 90°. A memory hack is “increasing order for sin: 30° (½), 45° (1/√2 ≈ 0.707), 60° (√3/2 ≈ 0.866), 90° (1).” Vedic methods and quick tables are shared in Vedantu’s live coaching for rapid revision and MCQ speed.
Try These Yourself
Derive sin 60° using a triangle with sides 2 units and a height.
Find the value of sin 60° + cos 30°.
If sin A = √3/2, what is the value of angle A?
Express sin 60 degrees in terms of tan 60 degrees.
Frequent Errors and Misunderstandings
Mixing up sin 60 degrees (√3/2) with sin 30 degrees (½).
Confusing surd and decimal values during MCQ exams.
Incorrect use of calculators: forgetting to switch to degree/radian mode.
Using the trigonometric table for the wrong quadrant or angle.
Relation to Other Concepts
The idea of sin 60 degrees connects closely to sin 30 degrees, sin 90 degrees, and cos 60 degrees. In fact, sin 60° = cos 30°, and these relations help when applying complementary angle formulas or working with the trigonometric table. Mastering sin 60 also makes right triangle and unit circle concepts clear for future topics.
Classroom Tip
A quick way to remember sin 60 degrees is to think of an equilateral triangle and realize: dropping its height always forms a 30-60-90 triangle, and the exact value pops right out as √3/2. Vedantu’s teachers often share such visual and mnemonic tricks so you can learn and recall faster during online live classes.
We explored sin 60 degrees—from its definition, key formula, derivation, applications, and common mistakes. With practice and the right memory hacks, this value will become second nature for you in exams. Keep reviewing with Vedantu to boost your confidence in trigonometry and related maths topics!
Related Links for Quick Revision:
Sin 30 Degrees
Trigonometric Ratios
Cos 60 Degrees
Trigonometry Table
FAQs on Sin 60 Degrees – Exact Value, Formula & Uses with Examples
What is the value of sin 60 degrees?
The exact value of sin 60 degrees is √3/2, which is approximately equal to 0.866. This value is crucial in various trigonometric calculations and applications.
How do you derive the value of sin 60 degrees?
Consider an equilateral triangle with sides of length 2 units. Dropping an altitude from one vertex bisects the base and creates two 30-60-90 right-angled triangles. The altitude has length √3 units (by the Pythagorean theorem). Therefore, sin 60° (opposite/hypotenuse) = √3/2.
What is sin 60 degrees in decimal form?
Sin 60 degrees in decimal form is approximately 0.8660254. This decimal approximation is often used in calculations where a precise surd form is not required.
How is sin 60 degrees represented on the unit circle?
On the unit circle, a 60° angle from the positive x-axis intersects the circle at the point (1/2, √3/2). The y-coordinate, √3/2, represents the value of sin 60°.
What are some common applications of sin 60 degrees?
Sin 60° is frequently used in solving problems related to:
Geometry: Calculating lengths of sides and heights in triangles.
Physics: Resolving vectors, calculating forces, and analyzing motion.
Engineering: Structural calculations, design, and analysis.
Is sin 60 degrees equal to sin 120 degrees?
While both sin 60° and sin 120° have the same numerical value (√3/2), they represent different angles with different geometric interpretations. Sin 120° is equivalent to sin (180° - 60°).
What is the relationship between sin 60 degrees and cos 30 degrees?
Sin 60° and cos 30° are equal (√3/2). This is a consequence of the complementary angle identity: sin θ = cos (90° - θ).
How can I quickly memorize the value of sin 60 degrees?
Use mnemonics or visualization techniques. Remembering the 30-60-90 triangle derivation can help. Practicing problems frequently reinforces memory.
Are there any common mistakes students make when working with sin 60 degrees?
Common mistakes include confusing sin 60° with sin 30° (1/2) or incorrectly using the decimal approximation. Careful calculation and understanding the concept is key.
How does sin 60 degrees relate to radians?
60 degrees is equivalent to π/3 radians. Therefore, sin (π/3) = √3/2.
Can sin 60 degrees be expressed in surd form?
Yes, the surd form of sin 60 degrees is √3/2. This is the exact value and often preferred in mathematical calculations to avoid rounding errors.
Where can I find more practice problems involving sin 60 degrees?
Vedantu provides numerous practice problems, worksheets, and solved examples related to trigonometric functions like sin 60°. Check our website and app for more resources.
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10219 | https://brainly.ph/question/13730110 | C( 11,5 ) = ?How about this? - Brainly.ph
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jacksteve
22.04.2021
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C( 11,5 ) = ?
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karelandea
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Answer:
Formula of Combination
C( 11,5 )
cross out the common
C(11,5)=462
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10220 | https://stackoverflow.com/questions/65874043/dividing-list-of-integers-into-alternating-groups-of-even-and-odd-sums | Stack Overflow
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Dividing list of integers into alternating groups of even and odd sums
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Ok so lets pretend we have a list [1, 3, 5, 7, 9, 11, 13], I want to make a python program which will divide the list into the maximum number of groups alternating between sums of even or odd. (Even always has to go first) For example the list above can be divided into:
[1,3] , [5,7,9] , [11,13] even , odd, , even
If the list was: [11, 2, 17, 13, 1, 15, 3] we would have:
, , [13,1], , [17,3]
This way, we cannot further divide, and we have the maximum number of groups. I have been thinking about various solutions, for example putting the evens first and then continuing with the groupings. My code so far divides the list of all the integers into a list of even integers and a list of odd integers.
Does anyone have any ideas on how to use the number of even and odd integers to calculate how many even and odd groups there should be in the final list.
Edit: The list can have any numbers less than 100 as long as they don't repeat, they are not necessarily consecutive.
this is the structure I've been trying to use so far (its incomplete but you can look at it)
while True: itera = [] if count % 2 == 0 and len(evens) > 0: x = evens itera.append(x) lines.append(itera) evens.remove(x) if count % 2 == 1 and len(odds) > 0: x = odds itera.append(x) lines.append(itera) odds.remove(x) if len(odds) == 0: if count % 2 == 1: print(count+1) if len(evens) == 0: if count % 2 == 0: if count % 2 == 1 and len(odds) : print(count+1) count+=1 if len(breeds) == 0: break
Really appreciated if someone could help out
python
python-3.x
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edited Jan 24, 2021 at 20:17
JohnJohn
asked Jan 24, 2021 at 17:52
JohnJohn
2755 bronze badges
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how many groups can we make?
Tugay
– Tugay
2021-01-24 18:01:21 +00:00
Commented Jan 24, 2021 at 18:01
we need to find the maximum number of possible groups
John
– John
2021-01-24 18:05:29 +00:00
Commented Jan 24, 2021 at 18:05
Why the second group doesn't consist of only 5? It's an odd number and can make an odd group by its own.
sahinakkaya
– sahinakkaya
2021-01-24 18:08:01 +00:00
Commented Jan 24, 2021 at 18:08
@John , order must be preserved, right?
Tugay
– Tugay
2021-01-24 18:09:16 +00:00
Commented Jan 24, 2021 at 18:09
1
This question was in yesterday's USACO contest, and this question was asked yesterday. Coincidence?
Higigig
– Higigig
2021-01-26 01:12:12 +00:00
Commented Jan 26, 2021 at 1:12
| Show 12 more comments
1 Answer 1
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5
After a few failures, I think I finally understand the problem and I think I have it. There are unsolvable cases for lists shorter than 4 elements, so the code below only tests cases with 4 elements and higher.
The first two test cases are the two presented at the top of the question. The next two test the two trivially degenerate cases...longer lists of all even or all odd values. Then I do some random tests.
For the random tests, I try one list of each length from 4 to 14. This doesn't necessarily test all the possible cases, because each case depends on the mix of even odd numbers, but it has a good chance to. Also, you can run this over and over, and you get different results each time, but the results always seem to be correct.
import random def do_it(input): print(input, ":\n ", end="") r = [] # Split the input into two groups, odds and evens odds = [n for n in input if n % 2] evens = [n for n in input if not n % 2] # First, add all of the trivial pairings of a single even and a single odd while we # still have one of each kind while len(odds) and len(evens): r.append([evens.pop(0)]) r.append([odds.pop(0)]) if len(odds): # If we now only have odd numbers... # Add trivial pairs of even then odd sums while we have 3 or more values while len(odds) > 2: r.append([odds.pop(0), odds.pop(0)]) r.append([odds.pop(0)]) if len(odds) == 2: # If we have two left over values, we're good, just create a final pair that will be even r.append([odds.pop(0), odds.pop(0)]) elif len(odds): # We only have one odd. This is the strange case. Here, we collapse the second and third # to last entries, which we can do because one of them is even and so the result will stay odd r[-3].extend(r.pop(-2)) # Now the last entry is wrong because it is even but should now be odd. But hey, we have one # last odd that we can add to it to change it from even to odd! So we just do that. r[-1].append(odds.pop(0)) elif len(evens): # If we now only have even numbers... # Add one last even value to the end of the list r.append([evens.pop(0)]) # Now add the remaining evens to the last list. We could add them to any of the # lists (or multiple lists), but just the last one is as good a choice as any. while len(evens): r[-1].append(evens.pop(0)) print(r, " (", len(r), ")") return r print("Test cases given in question...") do_it([1, 3, 5, 7, 9, 11, 13]) do_it([11, 2, 17, 13, 1, 15, 3]) print() print("Trivially degenerate cases...") do_it([1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35]) do_it([2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30]) print() print("Random cases of increasing list length...") for i in range(4, 15): do_it(random.sample(range(1, 100), i))
Sample run results:
Test cases given in question... [1, 3, 5, 7, 9, 11, 13] : ( 3 ) [11, 2, 17, 13, 1, 15, 3] : ( 5 ) Trivially degenerate cases... [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35] : ( 12 ) [2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30] : ( 1 ) Random cases of increasing list length... [14, 61, 88, 92] : ( 3 ) [38, 98, 77, 87, 76] : ( 5 ) [71, 51, 45, 86, 56, 22] : ( 6 ) [87, 16, 27, 52, 65, 77, 45] : ( 6 ) [84, 27, 13, 51, 58, 12, 46, 7] : ( 8 ) [1, 65, 22, 41, 56, 18, 38, 69, 75] : ( 7 ) [34, 57, 18, 35, 40, 50, 43, 26, 44, 27] : ( 9 ) [53, 47, 18, 17, 79, 36, 42, 31, 73, 92, 25] : ( 10 ) [19, 95, 35, 76, 60, 57, 58, 98, 46, 77, 26, 85] : ( 12 ) [23, 86, 98, 56, 92, 82, 37, 36, 85, 12, 89, 94, 81] : ( 11 ) [86, 79, 74, 42, 50, 80, 75, 40, 97, 54, 66, 70, 60, 39] : ( 9 )
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edited Jan 24, 2021 at 22:28
answered Jan 24, 2021 at 19:00
CryptoFoolCryptoFool
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CryptoFool
CryptoFool
@John - Am I right? Do the numbers in the incoming list really not matter? That's what I gathered. I can change my code to sort the incoming list and put the actual numbers from the input into the groups, but that has nothing to do with the grouping logic. If I'm right about them being independent, then do you see that? - If you just want to know the number of groups for any input size, this is giving you that without you having to worry about constructing a particular input list (since it doesn't matter).
John
John
wait so how would I use your code with my list of [11, 2, 17, 13, 1, 15, 3]. I mean this makes sense for the same list of consecutive odds but if we use something like the list I have I dont think it would work
CryptoFool
CryptoFool
I'll change my code if you want so that it does that. It's a trivial change. It's just harder to test and to demonstrate. I chose not to do that because it doesn't matter to the logic of the code.
CryptoFool
CryptoFool
I hate to give up. So I didn't. @John, please take a look at the latest version of. my answer and see if it fits the bill.
Shane Springer
Shane Springer
Nice job Steve, I had a slow day at work and was working on it there and while we were moving in similar directions you beat me to it. Emailed it to myself to work on it at home, but it looks like you've done it. It definitely ended up a little tougher than it seems at first glance. I was scared that someone was going to solve it with some itertools one liner lol.
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10221 | https://www.reddit.com/r/askmath/comments/16va9n9/i_realized_that_if_nnm_and_n1n1m1_for_example/ | I realized that if nn=m and (n+1)(n-1)=m-1 for example 77=49 and 86=48 is this a known principle of math? : r/askmath
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Tani_palahanov
I realized that if nn=m and (n+1)(n-1)=m-1 for example 77=49 and 86=48 is this a known principle of math?
Algebra
I was thinking about this and concluded that if nn=m then (n+1)(n-1)=m-1 for example 88=64 and 97=63; 77=49 and 86=48. I tried to think about this and came to the conclusion that this may be due to this formula a^2 - b^2 = (a - b)(a + b). So nn=n^2=m and (n+1)(n-1)=n^2-1. Am I correct and does this principle have a name or is it more a general property of algebraic expressions?
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10222 | https://www.sciencedirect.com/science/article/pii/S1876034125000395 | Skip to article
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Journal of Infection and Public Health
Volume 18, Issue 4, April 2025, 102690
Review Advancing the understanding of Naegleria fowleri: Global epidemiology, phylogenetic analysis, and strategies to combat a deadly pathogen
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Abstract
Naegleria fowleri is a rare but deadly pathogen that has emerged as an important global public health concern. The pathogen induces primary amoebic meningoencephalitis (PAM), a rapidly progressive and almost always fatal life-threatening brain infection. The devastating impact of N. fowleri and the high mortality rate underscores a deeper understanding and the development of innovative strategies to tackle this issue. Despite various studies that have been conducted on N. fowleri, a comprehensive review that integrates recent findings and addresses critical gaps in understanding remains lacking. This review provides a detailed overview of N. fowleri epidemiology, transmission dynamics, phylogenetic diversity, state-of-the-art diagnostic techniques, therapeutic approaches, and preventive measures. We identified 488 PAM cases globally, reported since 1962, with the highest numbers in the US, Pakistan, and Australia. A phylogenetic analysis of 41¯N. fowleri ITS-1, 5.8S, ITS-2 region-based sequences showed genotypic diversity, with genotypes II and III being the most prevalent in Asia, North America, and Europe. Effective approaches to preventing N. fowleri transmission include applying free chlorine to water in storage tanks, taking precautions while swimming, and performing ablution with sterilized water (e.g., boiled or distilled) while avoiding deep inhalation of water, especially in regions with high water contamination. This review highlights the global impact of N. fowleri, future surveillance strategies, prompt diagnosis, potential therapeutic options, and vaccine development to prevent PAM outbreaks. It highlights the importance of joint government and public health efforts to combat this deadly pathogen.
Keywords
Free-living amoeba
Primary amoebic meningoencephalitis
N. fowleri phylogenetics
Safe nasal irrigation
PAM treatment
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2. Last Digits: Determining the Last Digit of Powers 2/3
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This text explains the steps to determine the last digit of a number written as an exponent.
For more, check out the Last Digits pathway.
Uploaded June 6, 2022 | Updated June 6, 2022
We will now go through how to calculate the last digit for powers such as . If you do not know what powers are, you can briefly say that powers are repeated multiplication. , in total a count of eight times each other.
A first simplification is that has the same last digit as . We can understand this with the argument for last digits in multiplication that we learned earlier. If we multiply , it will have the same last digit as if we multiply
After that, we can do the multiplication step by step and delete the numbers that are not last numbers because they will not affect the last number in the answer.
| |
| Example 1 What is the last digit in |
Solution
has the same last digit as . The last digit of we can calculate by adding one a at a time and disregarding digits that are not last digits:
which has the last digit .
which has last digit
which has last digit .
which has the same last digit as which has last digit .
has the same last digit as
has the same last digit as
has the same last digit as
has the same last digit as which has the last digit
has the same last digit as
Maybe you noticed that the last digits in the example are repeated. The last digits were . If we think about it it is only the previous last digit that determines what the next last digit will be, since we take that number times 33. Then we could really as soon as we saw that had the last digit , the same as understand that the last digit would be repeated as over and over again. We can use this to calculate the last digit of even larger powers, such as ends up. The digits are repeated in a cycle of length and we wonder what number is in that cycle. We can find out with a remainder from division. If we divide with we get and remainder . We can see this as going laps and then one more step. Then we end up with the first element in the cycle, . Another way of looking at it is that the last digit for is the same as , just rounds earlier, which has last digit .
The method for finding the last digit for large powers is thus to calculate what the initial last digits will be, until they start to repeat themselves. They will always eventually repeat because there are only different last digits possible and as soon as the same last digit has appeared twice, the digits will repeat as they did in between. When you know what digits are included in the cycle and how long it lasts before it repeats itself, the last step is to find out where in the cycle the power you are looking for ends up. We do this with the help of a remainder from division. For example if a cycle is digits long and the power is after removing the part before the cycle starts (if there is such a part) we just need to divide with and see that the remainder is to know that it is the third digit in the cycle.
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10224 | https://www.fortunebuilders.com/p/gross-rent-multiplier/ | How To Value A Property: The GRM Formula In Real Estate
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What Is The Gross Rent Multiplier (GRM) In Real Estate?
Written by Paul Esajian
Key Takeaways
What Is A Gross Rent Multiplier (GRM)?Why Is The GRM Important In Real Estate?Gross Rent Multiplier FormulaWhat Is A Good Gross Rent Multiplier?Pros & Cons Of The Gross Rent MultiplierGross Rent Multiplier Vs. Capitalization RateSummary
Key Takeaways
What is a gross rent multiplier? |Why is the GRM important? |How to calculate the GRM | What is a good GRM? | Pros & cons of the GRM | Gross rent multiplier vs. cap rate
Have you ever found multiple promising properties at once? Wished for a quick and easy way to compare them without having to run a full analysis? The gross rent multiplier (GRM) might be exactly what you’ve been looking for.
It’s a simple formula that investors use to compare and contrast rental property values, and it can help sort desirable properties from less profitable counterparts. You can use GRM as a preliminary filter to determine which properties are worthy of a deeper financial analysis.
Read on for a breakdown of how to use the gross rent multiplier and why it’s a great tool in real estate.
What Is A Gross Rent Multiplier (GRM)?
The gross rent multiplier (GRM) is a formula used by real estate investors to compare the potential rental income of different properties. This valuation technique is a simplified way to analyze properties without conducting a complete analysis. Real estate investors of all skill levels rely on this formula to quickly compare properties across portfolios and make fast-paced investment decisions.
It is worth noting that the GRM is not to be used in place of thorough property analyses. Instead, it is best used to eliminate properties before performing in-depth analyses of promising candidates.
[ Thinking about investing in real estate?Register to attend a FREE online real estate class and learn how to get started investing in real estate. ]
Why Is The GRM Important In Real Estate?
The GRM is important to real estate investors because of its speed and utility. The formula utilizes two variables: rental property value and gross property income. There are several formulas in real estate investing, but almost none are as simple as the GRM. Investors typically have access to both numbers and can easily perform this calculation. These also happen to be the variables that lenders care about the most when evaluating potential investments (price and potential return). In calculating the GRM, investors get their first look at the factors they may present to lenders when raising financing.
The GRM is also particularly beneficial for commercial real estate investors who may be working in highly competitive environments. Commercial real estate often requires investors to be able to make fast-paced decisions about where to delegate their time and resources. The GRM can be quite an effective tool in doing so, as it allows users to easily compare potential investments.
Gross Rent Multiplier Formula
Calculating the gross rent multiplier is simple. You take the market value of a property and divide it by the property’s gross rental income. How you do this is up to you: you can use the sale price, list price, or property appraisal value. You can even choose between monthly or annual income. When using the gross rent multiplier formula, you’ll want to make sure to keep the factors consistent across all the properties you are considering. Otherwise, any comparisons you make will be invalid.
Gross Rent Multiplier = Rental Property Value / Gross Property Income
It can be helpful to practice with an example. Let’s say you found a rental property with a list price of $500,000, and based on your estimate, the gross annual income is $80,000. In this case, your GRM is 6.25 (500,000 / 80,000). Then, you’ll continue to make similar calculations with other properties that you’ve identified. You’ll run a gross rent multiplier appraisal and look for properties with the lowest possible GRMs. (Obviously, you’ll want properties that produce more income. The larger the denominator, the smaller the resulting number will be.)
Keep in mind that the GRM is best used to compare the potential income between properties. It cannot predict how long a specific loan will take to pay off, which property will have fewer expenses, or the amount of debt associated with purchasing a given property. Each of these factors will need to be considered during a more thorough property analysis.
Using The GRM To Estimate Property Value
GRM can also estimate the property value of an investment you are considering. If you ran the gross rent multiplier formula for a few properties and found an average, you could use that number alongside the annual rental income. Together, these variables would allow you to reverse calculate the property value. This exercise would allow you to compare the market value of a property against its sale price, especially if the purchase price changed. For example, if the GRM is around 7% and the rental income is $75,000 this property value would be $525,000. Suppose that same property is currently listed at $600,000. In that case, you could either choose to walk away from the property or conduct a more thorough analysis to negotiate the purchase price in your favor.
How To Use GRM In Real Estate Investments
Let’s take a look at how a real estate investor would use GRM.
First, gather a list of prospective properties you’re interested in. Be sure you know the rental property value and gross property income for each of them. Then, calculate the GRM for each property using the provided formula.
Let’s say that one property has a GRM of 7, while the other two properties in the same area have a GRM of 9 and 10. The property with the GRM of 7 preliminarily appears to be the most profitable opportunity. You would proceed with a deep analysis of this property to decide if it’s a worthy investment.
More Examples Of GRM In Real Estate
The GRM formula is made up of three variables: Gross Rent Multiplier, Rental Property Value, and Gross Property Income.
You don’t always have to use this formula to calculate GRM. If you know two out of the three variables, you can calculate any of the variables in the formula.
For instance, you can reverse-engineer the formula to calculate gross rental income. Let’s say that the average GRM in a neighborhood is 6, and the asking price of the property is $300,000. You can deduce that the gross rental income is $50,000. This is how the calculation is made:
Gross Rent Multiplier = Rental Property Value / Gross Rental Income
Gross Rental Income = Rental Property Value / Gross Rent Multiplier
Gross Rental Income = $300,000 / 6
Gross Rental Income = $50,000
Manipulating these formulas allows investors to analyze properties quickly before they zero in on one or two promising candidates.
What Is A Good Gross Rent Multiplier?
A good gross rent multiplier in real estate is typically one of the smaller numbers within your range. As I mentioned above, this is because a lower GRM generally suggests more rental income in relation to the purchase price. That being said, there is not a universally “good” GRM; that definition will always have to be established when looking at your specific calculations. What I mean by this is this month, you could look at properties with an average GRM of 4 to 5, but next month that number could be between 7 and 8. The gross rent multiplier is all about comparison.
What Is A Good GRM For A Rental Property?
Typically, a GRM between 4 – 7 is considered “good” for a rental property. Again, it is important to note that a healthy GRM is dependent on your local market and the comparable properties within that market. The lower you can make your GRM, the less time you will need to pay off your rental property’s purchase price. To achieve this, you must generate the most rental income you can with the least cost.
Pros & Cons Of The Gross Rent Multiplier
The gross rent multiplier has several advantages, but there are some drawbacks to consider. Keep reading as we pick apart the GRM and what the great advantages and potential downsides are so that you can be mindful when you add it to your investor toolbox.
Pros Of The GRM
The gross rent multiplier equation is a reliable formula for some of today’s best real estate investors, and there are many reasons why. Read through the following benefits to understand why you should add the GRM to your repertoire today:
Unique to rental properties: The gross rental multiplier is a formula that is unique to rental property valuation.
Quick and easy: The formula is easy to use, and doesn’t require any in-depth calculations or analysis.
Rate of return calculator: The GRM in real estate is a way to calculate the rate of return on rental properties.
Benchmark several properties at once: Comparing and contrasting multiple rental properties at once would be quite an undertaking, but luckily, this powerful formula allows you to compare more than one property so that you can determine which ones are worth further exploring.
Establish a threshold: Once you’ve used the GRM in your analyses several times, you’ll start developing your own threshold for what rate of return you’d find acceptable, and you’ll be able to establish a grading system for rental properties.
Keep a pulse on a market: Finally, the gross rent multiplier can be a nifty tool to keep a pulse on a rental market. You can calculate the GRM for a few properties within a market and keep track of whether the GRMs on those properties increase or decrease over time.
Cons Of The GRM
Like with any real estate evaluation, there are potential downsides to using the GRM. However, most of these cons can be mitigated with a little due diligence. Review the following drawbacks to learn more:
Does not consider all costs: As mentioned above, GRM is calculated using top-line revenue, or gross income. This means that it does not factor in any costs of running a rental property, such as operating expenses, vacancy rates, or the cost of insurance or taxes. Here’s a great breakdown of how to quickly estimate rental property expenses.
Accuracy not guaranteed: Because the GRM does not account for property costs, some will argue that it does not paint an accurate picture of the return on investment. When you think you’re comparing apples to apples in a certain market, what you don’t know is that one property may have more operating expenses than the other. That’s why it’s good to use the GRM only as an initial screening tool. Always be sure to perform a more in-depth analysis once you’ve narrowed down your search to several properties.
May cause you to overlook a good property: If you use a certain tool or calculation as a screening tool, the main danger is the potential to overlook a great property. If you’re quickly filtering through a long list of properties and only looking at certain indicators, you run the risk of skipping over a diamond in the rough just because the initial financials didn’t look good or pass your grading system.
[ Thinking about investing in real estate?Register to attend a FREE online real estate class and learn how to get started investing in real estate. ]
Gross Rent Multiplier Vs. Capitalization Rate
Gross rent multiplier and capitalization rate are two invaluable tools used by today’s real estate investors. While both have proven useful, some find it easy to confuse the two.
Both are used to evaluate income-producing properties based on the amount of rental income they can generate. It’s not hard to see why people would mix the two up.
Relationship Of Cap Rate To The Total Return
The capitalization rate, or cap rate for short, calculates property returns by comparing the net operating income (NOI) with the home’s current market value. More specifically, it can evaluate how profitable an income property will be relative to its purchase price. Doing so takes into account the property’s operational expenses and its vacancy rate. Cap rate indicates whether generated income will cover the mortgage.
Some think that cap rate is a better formula than GRM because it accounts for all costs, but keep in mind that costs can be manipulated. In contrast, GRM measures the ratio between the asset’s gross scheduled income and its fair market value. This tool will tell you how much income a property will generate, but doesn’t account for all expenses. This is why GRM is typically used as a preliminary filtering system, followed by additional formulas such as NOI and cap rate when running a deeper analysis on a promising property.
Example of Calculating Cap Rate Vs. GRM
An example of calculating cap rate can help demonstrate how the formula paints a different story from GRM.
Earlier, we used an example property that had a property value of $300,000 and gross rental income of $50,000 and using the 50% Rule, we estimate that the property’s net operating income (NOI) is $25,000. (Half of the gross rental income used to pay operating expenses, not including the mortgage payment.)
To calculate cap rate, we simply divide the property’s NOI by the property value:
Cap Rate = NOI / Property Value
Cap Rate = $25,000 / $300,000
Cap Rate = 0.0833 or 8.3%
In this example, the Cap Rate is 8.3 percent. Similar to GRM, this data point from a single property doesn’t mean very much. However, you can compare this same data point across multiple properties in the same area to identify which property is a better deal because the potential return is relatively higher.
Note that a potentially profitable property will show a relatively higher cap rate and a relatively lower GRM (in comparison to other properties.) An example comparing a property’s cap rate and GRM before and after a rent hike can help illustrate this point.
The same example property with a property value of $300,000 and NOI of $25,000 will have a rent increase of 5%.
Before the rent increase:
GRM = $300,000 Property Value / $50,000 Gross Rental Income = 6
Cap Rate = $25,000 NOI / $300,000 Property Value = 8.3%
After a rent increase of 5%, gross rental income increases from $50,000 to $53,000 and the NOI increases from $25,000 to $26,500:
GRM = $300,000 Property Value / $53,000 Gross Rental Income = 5.7
Cap Rate = $26,500 NOI / $300,000 Property Value = 8.8%
As you can see from this example, the GRM decreased while the Cap Rate increased after a rent hike. Both indicate an improvement to the financial optics on the same property. Keep this in mind as you incorporate both formulas into your deal analysis toolkit.
Other Ways To Evaluate Real Estate Investments
When evaluating your real estate investments, it’s always a good idea to have several helpful calculations memorized. You should never rely solely on one type of calculator or benchmark. Instead, you should run several sets of calculations to evaluate properties from multiple angles. We already discussed the difference between the GRM and cap rate, which is an investor favorite. There are also indicators such as cash flow, rental yield, and internal rate of return. Be sure to check out our guide that breaks down several rental property calculations you should know.
Summary
The best way to think of the gross rent multiplier is as a grading system. By using consistent variables, investors can quickly compare multiple properties. And although the GRM doesn’t factor in the costs associated with property ownership, it can still be a great initial, large-scale screening tool. This is particularly important to remember for those who plan on breaking into commercial real estate.
If you ever happen to find yourself at the hands of too many deals, this formula can be a great way to eliminate properties that do not hold much promise. This will leave you with the time (and energy) to closely examine properties that could add more value to your portfolio in the long run.
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10225 | https://jhpn.biomedcentral.com/articles/10.1186/s41043-022-00287-w | Advertisement
High burden of hypovitaminosis D among the children and adolescents in South Asia: a systematic review and meta-analysis
Journal of Health, Population and Nutrition
volume 41, Article number: 10 (2022)
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Abstract
Background
Vitamin D is vital for the growth and development of children. While deficiency and/or insufficiency of vitamin D among South Asian children are frequently reported in the literature, the lack of a meta-analysis has left its true extent poorly characterized. In this study, we aimed to conduct a systematic review and perform meta-analyses of the prevalence of hypovitaminosis D among the children of the South Asian countries.
Methods
Two major electronic search engines (PubMed and Scopus) and one database (Google scholar) were used; original studies, conducted among South Asian children and adolescents and published between 1 January 2001 and 31 December 2019. A random-effect meta-analysis was also performed to calculate the pooled prevalence of hypovitaminosis D followed by subgroup analyses for countries and age groups.
Results
After applying inclusion and exclusion criteria, a total of 41 studies with a total population size of 18,233 were finally selected. The overall prevalence of hypovitaminosis D was 61% [95% CI: 46% to 71%] with highly significant heterogeneity (I2 = 99.72%; p < 0.0001). The average level of serum vitamin D ranged from 5 ng/mL to 34 ng/mL, with a weighted mean of 19.15 ng/mL (weighted standard deviation 11.59 ng/mL). Country-wise analysis showed that hypovitaminosis D in Afghanistan was the highest [96.2%; 95% CI: 91% to 99%], followed by Pakistan [94%; 95% CI: 90% to 96%], India [64%; 95% CI: 46% to 79%], Bangladesh [35.48%; 95% CI: 32% to 39%], Nepal [35%; 95% CI: 1% to 83%], and Sri Lanka [25%; 95% CI: 16% to 36%]. Age group analyses revealed that hypovitaminosis D was most prevalent among neonates [85%; 95% CI: 76% to 91%], followed by school-going children [57%; 95% CI: 33% to 80%], and preschool children [55%; 95% CI: 35% to 75%].
Conclusion
This study generates quantitative evidence and specific extent of hypovitaminosis D in the South Asian countries as a public health concern. Being the first systematic review for this region, results from this study will create awareness and will facilitate adopting mitigation strategies by the policymakers and the governments to address this problem.
Background
Vitamin D has a widespread role in the early development of children. The deficiency of this vitamin can often lead to suboptimal bone mass in infants, children, and adolescents. Nutritional rickets is a major bone-related disease that is caused by vitamin D deficiency 13
."),[2:118–26.
."),3:161–70.
.")]. According to the World Health Organisation (WHO), the peak incidence of rickets occurs among children and adolescents aged 2–11 years . Besides rickets, vitamin D deficiency can also cause osteomalacia and other bone-related deformities among children [3:161–70.
."), 4]. It has been suggested that up to 200 genes could be regulated by the active form of vitamin D (1,25 dihydroxy vitamin D), indicating its highly pleiotropic role [2:118–26.
."), 5:266–81.
.")]. Moreover, study reports suggested that the biological plausibility of vitamin D deficiency is also correlated with various kinds of chronic diseases like diabetes, cardiovascular disease, cancer, tuberculosis, etc. [2:118–26.
."), 5:266–81.
."),6:97–107.
."),7] So, a child suffering from vitamin D deficiency early in life becomes more susceptible to other kinds of diseases in the latter part of life.
Vitamin D deficiency affects as many as 1 billion people globally and 50% of the world population suffers from vitamin D insufficiency 26
."), [5:266–81.
.")]. Worldwide, the extent of vitamin D deficiency among children and adolescents in both developed and developing countries is highly variable; prevalence as low as 5% to as high as 95% of the study population has been reported [8,9:1–7.
."),10:53–60.")].
A variety of factors have been highlighted as underlying variables to explain this large variance in serum vitamin D status; besides nutrition, the extent of sunlight exposure is arguably the most important determinant 26
."), [11:1060–7.
.")]. Sunlight exposure eventually depends on some other factors like geographical location, people's skin colour, attitude towards sunlight exposure, clothing practice, etc. [2:118–26.
."), 5:266–81.
."), 12:184.
.")]. As such, understanding comparative variation within a region may reveal crucial clues regarding the potential determining factors of vitamin D deficiency or insufficiency (hypovitaminosis D).
South Asia consists of eight countries—India, Bangladesh, Pakistan, Nepal, Bhutan, Maldives, Sri Lanka, and Afghanistan. Together, these South Asian countries occupy an area of 5.131 million square kilometres and have a population of about 1.8 billion [13, 14].
According to UNICEF, around 627 million children (< 18 years of age) live in South Asian countries and cover up approximately 36% of the total population (1.8 billion) . Reported data indicates a high prevalence of nutritional rickets and other bone-related diseases, cardiovascular problems, diabetes, acute respiratory infections, tuberculosis, and other communicable diseases among South Asian children [4, 16:166–71.
."),17:31–44.
."),18:811–5.
.")]. All of these can be potentially linked to the high prevalence of hypovitaminosis D in this region.
Despite the possibility of large-scale hypovitaminosis D in South Asia, we found only one systematic review (on Indian adolescent girls based on a very limited number of studies) in this region [190
.")]. To address this knowledge gap, this study aims to conduct a systematic review and meta-analysis on the prevalence of hypovitaminosis D among South Asian children and adolescents.
Methods
Preferred Reporting Items for Systematic Reviews and Meta-Analyses (PRISMA-P 2015) have been followed as recommendations and guidelines for conducting this systematic review [201
.")]. Since, we did not register the review protocol anywhere (PROSPERO, Cochrane, etc.), a completed copy of the PRISMA checklist (PRISMA 2020) has been added as an Additional file 2.
Data source and search strategy
PubMed and Scopus were the main databases, and Google Scholar was the main search engine, used in this analysis (to prevent personalized results, the search was conducted, after logging out of all Google accounts). Two researchers (BB and MMR) independently investigated these three datasets (from 26 October 2019 until 26 January 2020) to find out the studies conducted from 2001 until the search date. The details about search strategy, original MeSH terms, and alternative terms are available in Additional file 1: Table S1.
The searches were carried out in English. The corresponding author's personal profiles available online (Google Scholar, ResearchGate, ORCID, and organizational profile), as well as the reference list of our selected studies, were further explored to maximize the search efficiency.
To ensure the inclusion of grey literature, we went through online archives of newspapers that have been published in English among South Asian countries as; The Hindu, The Indian Express, New Age, The Nation, Daily Bhutan, Maldives Times, Himalayan Times, Sunday Observer, etc. We also explored governments’ reports and published abstracts (in electronic media) as relevant sources from the conference held in South Asian countries.
Study selection criteria
Clinical Practice Guideline of the Endocrine Society currently defines vitamin D deficiency, insufficiency, and sufficiency as a serum level of vitamin D < 12 ng/mL (< 30 nmol/L), < 20 ng/mL (< 50 nmol/L) and > 20 ng/mL (> 50 nmol/L), respectively 214
."), [22:449.
.")]. In this study, we defined hypovitaminosis D (vitamin D deficiency or insufficiency) as per current guidelines (cut-off < 20 ng/mL). Studies were selected if serum levels of vitamin D < 20 ng/mL in South Asian children and adolescents were reported. The inclusion criteria were: study conducted in South Asian countries from 1 January 2001 to 31 December 2019; study conducted in the hospital setting or community setting among apparently healthy children and adolescents who were up to 18 years of age, children and adolescents with a minor illness whose physical conditions were not correlated with any chronic diseases or coexisting morbidity [e.g. chronic kidney disease, cardiovascular disease, diabetes, rheumatoid arthritis, cancer, tuberculosis, body aches and pain, proximal muscle weakness, osteoporosis, etc.]. Among different study designs, cross-sectional, longitudinal, case–control (only control group), randomized clinical trial (baseline and placebo data) were included.
Studies were excluded if they had a sample size less than 50 [190
.")]; reported vitamin D levels after some form of intervention or supplementation; conducted on other groups of the population rather than children and adolescents such as pregnant women, adults, and elderly; the reported prevalence of hypovitaminosis D associated to any kind of chronic diseases or disease related to any coexisting morbidity; conducted on a special group of children and adolescents such as physically or mentally challenged; did not mention the prevalence of deficiency and mean level of serum vitamin D; letter to the editor, review article, editorial article; Studies which satisfied selection criteria but were not obtainable from the authors after request were also excluded.
To handle the references and prevent duplications, Mendeley Desktop Program (version 1.19.4) was used. After eliminating duplicates, two researchers (MHS and URH) independently reviewed all papers before final selection for meta-analysis. Any disagreements were resolved through discussion with co-authors.
Data extraction
To extract data from all eligible studies, a standardized form was used. The following information has been collected for each study: publication details [e.g. first author, publication date, journal name, and publisher]; research setting, design and population [e.g. country, study area, study design, method of measurements, and sample size]; participants’ characteristics [e.g. gender, age, and socio-economic status] and major findings [mean level of serum vitamin D and prevalence percentage of hypovitaminosis D]. Data extraction was independently performed by two researchers (BB and MMR) and subsequently, it was cross-checked by two other researchers (MHS and URS). In our selected studies, where mean values of serum vitamin D were given in nanomol per unit litre (nmol/L), we converted to nanogram per millilitre (ng /mL) by dividing with 2.5 (according to the international unit conversion system) to maintain uniformity of data.
We resolved any disputes between us through discussion during data selection and data extraction. Eventually, all four researchers fully agreed to the selected studies before the data extraction began. Therefore, no statistical analysis of inter-rater agreements was performed.
Evaluation of study quality
We used a checklist of 10 parameters validated by Hoy et al. (2012). To assess the risk of bias (as weak, moderate, and high) for the selected papers [234
.")]. As per the checklist, studies with a score of 0–3 are considered as low risk, a score within 4–6 is moderate risk, and studies with a score of 7–9 are considered as high risk of bias.
Statistical analysis
The mean value of the serum vitamin D and the prevalence of hypovitaminosis D in South Asian children and adolescents are regarded as summary measurements. The weighted mean level of serum vitamin D was calculated by using Microsoft Excel (version 2016). A random-effect meta-analysis was used to obtain the weighted pooled prevalence with a confidence interval of 95%. Cochran’s Q test and the I2 statistics were used to assess heterogeneity [241
.")]. Substantial heterogeneity was suggested with an I2 of more than 75% [25:557–60.
.")]. The analyses were performed using the metaprop, metabias, metafunnel commands by Stata version 15 (Stata Corp, College Station, TX).
Results
A total of 1903 articles were retrieved from different databases by using our search strategy. Among these 1903 studies, 1862 articles were excluded because they did not fulfil our inclusion criteria. A total of 41 articles, finally qualified for meta-analyses. Figure 1 shows the selection process used in this study.
PRISMA chart showing the flow of information through the different phases of the systematic review
The study design was cross-sectional for most of the studies (30 out of 41) and the rest were either case–control or randomized control trials. Furthermore, more than half of the selected studies did not mention demographic area (25 out of 41), and socio-economic condition (26 out of 41) for the study population. Table 1 shows the summary outlining the characteristics of selected articles.
Studies selected in this systematic review consisted of 18,233 participants. While most of these studies reported serum vitamin D levels of children up to 18 years of age, two studies included participants of up to 20 years 27x
."), [56:452–7.
.")]. The overall pooled prevalence of hypovitaminosis D was 61% [95% CI: 46% to 71%] with a high degree of heterogeneity (I2 = 99.72%; p < 0.0001). Figure 2 shows the overall forest plot about the prevalence of hypovitaminosis D in South Asia.
Forest plot displays the overall prevalence of hypovitaminosis D among South Asian children and adolescents. Each horizontal line of the forest plot represents an individual study and the box is plotted as prevalence for that study. Diamond at the bottom represents overall polled prevalence when all the individual studies are combined and averaged. The horizontal points of the diamond represent the limit of 95% confidence interval
Prevalence of hypovitaminosis D and the average level of serum vitamin D was mentioned in all studies; prevalence ranged from 8 to 96% and average ranged from 5 ng/mL to 34 ng/mL (for the individual studies). The weighted mean level of serum vitamin D was 15.48 ng/mL and the weighted standard deviation (weighted SD) was 7.49 ng/mL.
Effect of geographical location on the prevalence of hypovitaminosis D
We found studies following our inclusion criteria from 6 out of 8 South Asian countries. No studies were found from Bhutan and Maldives. A summary table shows the country-wise result (Table 2). We found that Afghanistan has the highest and Sri Lanka has the lowest prevalence of hypovitaminosis D in South Asia. The Forest plot shows the country-wise prevalence of hypovitaminosis D (Fig. 3). A bar diagram shows the weighted mean level of vitamin D among South Asian children (Additional file 1: Figure S1).
Forest plot displays the country-wise prevalence of hypovitaminosis D among South Asian children and adolescents. In this forest pot, all the diamonds, except the last one (overall pooled prevalence) represent polled prevalence for the individual country; (1) Nepal, (2) India, (3) Afghanistan, (4) Sri Lanka, (5) Bangladesh, (6) Pakistan. Each horizontal line of the forest plot represents an individual study and the box is plotted as prevalence for that study. The horizontal points of the diamond represent the limit of 95% confidence interval
India
We found thirty-one studies from India with a total of 14,497 participants 267
."),[27:680–4.
."),28:244.
."),29:178–83.
."),30:167–73.
."),31:876–82.
."),32:975–81.
."),33:2335–43.
."),34:48."),35:9–14.
."),36:1611–7.
."),37:81.
."),38:108–15.
."),39 D to Vitamin D and calcium supplementation in school-children from a semi-rural setting in India. J Steroid Biochem Mol Biol. 2018;180:35–40.
."),40:345–50.
."),41. 2014;17:41–6.
."),42:1–7.
."),43 concentration and parathyroid hormone (PTH) concentration. Arch Dis Child. 2016;101(4):316–9.
."),44:289–95.
."),45:SC18–SC21.
."),46:446–50.
."),47:563–7.
."),48,49:154–9.
."),50:307–10.
."),51:1671–4.
."),52:429–35.
."),53,54:1383–9.
."),55:609–16.
."),56:452–7.
.")]. The weighted mean level of hypovitaminosis D for study participants was 13.40 ng/mL (SD 6.61 ng/mL) and random-effect meta-analysis showed that the pooled prevalence of hypovitaminosis D was 64% [95% CI: 46% to 79%] with a high level of heterogeneity (I2 = 99.75%; p < 0.0001).
Nepal
There were three studies from Nepal which consisted of 1746 participants 576
."),[58:825.
."),59:470–6.
.")]. The random-effect meta-analysis pointed that the prevalence of hypovitaminosis D was 35% [95% CI: 1% to 83%.] and the weighted mean level of serum vitamin D for study participants was 25.96 ng/mL (weighted SD 7.29 ng/mL).
Sri Lanka
In Sri Lanka, there were three studies comprised of 693 participants 607
."),[61:259–66.
."),62:335–40.
.")]. The weighted mean level of serum vitamin D for study participants was 27.34 ng/mL (weighted SD 9.54 ng/mL) and random-effect meta-analysis showed that the weighted pooled prevalence of hypovitaminosis D was 25% [95% CI: 16% to 36%].
Pakistan
We found two studies from Pakistan 637
."), [64:59–62.
.")] which together consisted of 277 participants and the random-effect meta-analysis showed that 94% [95% CI: 90% to 96%] of participants were hypovitaminosis D with 11.78 ng/mL weighted mean level of serum vitamin D (weighted SD 7.99 ng/mL).
Bangladesh
There was only one study from Bangladesh [653
.")] which included 913 participants with 21.86 ng/mL mean level of serum vitamin D and 35.48% [95% CI: 32% to 39%] of them had hypovitaminosis D.
Afghanistan
We found a single study from Afghanistan [666
.")] which comprised 107 participants and the study result revealed 96.2% [95% CI: 91% to 99%] of them had hypovitaminosis D with 5 ng/mL mean level of serum vitamin D.
Effect of gender on prevalence of hypovitaminosis D
All of the studies we found were conducted either on both genders or only among female children and adolescents. A summary table shows gender-wise results (Table 3). We categorized this section into two parts. Studies which included participants from both gender and studies which consider only female children as their participants. Overall study result shows high degree of heterogeneity (I2 = > 99%; p < 0.0001). Figure 4 shows the gender-wise forest plot.
Forest plot following gender for the prevalence of hypovitaminosis D among South Asian children and adolescents. In this forest plot, all the diamonds except the last one (overall pooled prevalence) represent polled prevalence following gender. Here are two categories; studies that represent participants from both gender and the only female. Each horizontal line of the forest plot represents an individual study and the box is plotted as prevalence for that study. The horizontal points of the diamond represent the limit of 95% confidence interval
Studies included participants from both gender
We found 36 out of 41 studies which included participants from both genders and among these studies, 26 studies were conducted in India, 2 studies in Pakistan, 3 studies in Nepal, 3 studies in Sri Lanka, and one study from Bangladesh and Afghanistan each 267
."),[27:680–4.
."),28:244.
."),29:178–83.
."),30:167–73.
."), 33:2335–43.
."),34:48."),35:9–14.
."),36:1611–7.
."),37:81.
."), 39 D to Vitamin D and calcium supplementation in school-children from a semi-rural setting in India. J Steroid Biochem Mol Biol. 2018;180:35–40.
."),40:345–50.
."),41. 2014;17:41–6.
."), 43 concentration and parathyroid hormone (PTH) concentration. Arch Dis Child. 2016;101(4):316–9.
."), 45:SC18–SC21.
."),46:446–50.
."),47:563–7.
."),48,49:154–9.
."),50:307–10.
."),51:1671–4.
."),52:429–35.
."),53,54:1383–9.
."),55:609–16.
."),56:452–7.
."),57:979–87.
."),58:825.
."),59:470–6.
."),60:144–51.
."),61:259–66.
."),62:335–40.
."),63:318–23.
."),64:59–62.
."),65:1718–28.
."),66:16–20.
.")]. Together, these studies comprised of 16,434 participants, and a random-effect meta-analysis indicated that 58% [95% CI: 43% to 73%] of study participants were hypovitaminosis D with a high degree of heterogeneity (I2 = 98.72%; p < 0.0001) The mean level of serum vitamin D ranged from 5 ng/mL to 34 ng/mL (for the individual studies) among study participants.
Studies included only female participants
We found 5 out of 41 studies that included only female participants and all of these studies were conducted in India 318
."), [32:975–81.
."), 38:108–15.
."), 42:1–7.
."), 44:289–95.
.")]. These studies together comprised 1799 participants and random-effect meta-analysis demonstrated that 76% [95% CI: 46% to 96%] of study participants had hypovitaminosis D with a high number of heterogeneity (I2 = 99.37%; p < 0.0001). The mean level of serum vitamin D among study participants ranged from 9 to 24 ng/mL.
Prevalence of hypovitaminosis D for different age groups
In this section, we categorized study participants into four groups according to their age and these are 1 month (neonates), 1 month to 5 years (infants and preschool children), 6 to 18 years (school children), and others (< 20 years). A summary table shows the age-wise result (Table 4). We found out that, in South Asia, infants and preschool children have the lowest and neonates have the highest prevalence of hypovitaminosis D. Overall study result shows high degree of heterogeneity (I2 = 99.72%; p < 0.0001). Forest plot with further detail is available in Fig. 5.
Forest plot following age for the prevalence of hypovitaminosis D among South Asian children and adolescents. In this forest plot, all the diamonds except the last one (overall pooled prevalence) represent polled prevalence following age. Here are four categories; studies; (1) 6–18 years of age (school) (2) < 5 years (preschool and infants), (3) up to 1-month (neonates), and (4) others. Each horizontal line of the forest plot represents an individual study and the box is plotted as prevalence for that study. The horizontal points of the diamond represent the limit of 95% confidence interval
School children (6–18 years)
We identified 17 out of 41 studies with individuals aged 6 to 18 years old. Only one of these studies was conducted in Nepal and the rest were conducted in India 267
."), [28:244.
."),29:178–83.
."),30:167–73.
."),31:876–82.
."),32:975–81.
."),33:2335–43.
."),34:48."), 36:1611–7.
."),37:81.
."),38:108–15.
."),39 D to Vitamin D and calcium supplementation in school-children from a semi-rural setting in India. J Steroid Biochem Mol Biol. 2018;180:35–40.
."),40:345–50.
."),41. 2014;17:41–6.
."),42:1–7.
."),43 concentration and parathyroid hormone (PTH) concentration. Arch Dis Child. 2016;101(4):316–9.
."),44:289–95.
.")]. Together, these studies consisted of 12,709 participants and random-effect meta-analysis showed that 57% [95% CI: 33% to 80%] of study participants had hypovitaminosis D with a high degree of heterogeneity (I2 = 99.85%; p < 0.0001). The mean level of serum vitamin D among study participants ranged from 6.3 ng/mL to 26.52 ng/mL (for the individual studies).
Infants and preschool children (1 month–5 years)
There were 14 out of 41 studies which included participants who were 1 month to 5 years of age and among these studies 7 were conducted in India, 3 in Sri Lanka, 2 in Nepal, and one study was conducted in Bangladesh and Afghanistan each 457
."), [47:563–7.
."), 48, 51:1671–4.
."), 53, 54:1383–9.
."), 57:979–87.
."),58:825.
."),59:470–6.
."),60:144–51.
."),61:259–66.
."),62:335–40.
."), 65:1718–28.
."), 66:16–20.
.")]. Together, these studies consisted of 4324 participants with 5 ng/mL to 33.71 ng/mL mean serum level of vitamin D. Random-effect meta-analysis showed that 55% [95% CI: 35% to 75%] of study participants had hypovitaminosis D with a high degree of heterogeneity (I2 = 99.47%; p < 0.0001).
Neonates (1 month)
We found 6 out of 41 studies which included participants who were up to 1 month in age. Among, these studies 4 were conducted in India and 2 in Pakistan 463
."), [49:154–9.
."), 52:429–35.
."), 55:609–16.
."), 63:318–23.
."), 64:59–62.
.")]. Together, these studies consisted of 763 participants and random-effect meta-analysis revealed that 85% [95% CI: 76% to 91%] of study participants had hypovitaminosis D with a high degree of heterogeneity (I2 = 84.82%; p < 0.0001). The mean level of serum vitamin D ranged from 6 ng/mL to 20 ng/mL among study participants.
Others
In this group participants' age range was < 20 years. In total, we found 4 studies in this section, and all of these were conducted in India 27x
."), [35:9–14.
."), 50:307–10.
."), 56:452–7.
.")]. These studies together consisted of 437 participants and random-effect meta-analysis showed that 57% [95% CI: 35% to 77%] of study participants had hypovitaminosis D with a high degree of heterogeneity (I2 = 95.20%; p < 0.0001). The average vitamin D level of study participants ranged from 14 to 30 ng/mL.
Effect of study setting on the prevalence of hypovitaminosis D
Most of the selected studies were community-based (24 out of 41) and the rest were hospital-based (17 out of 41). Moreover, community-based study setting (62%; 95% CI: 43% to 80%) showed high prevalence of hypovitaminosis D in compared with hospital-based setting (58%; 95% CI: 41% to 74%). However, high degree of heterogeneity was observed in both of the study setting (Community: I2 = 99.83; p < 0.0001, Hospital: I2 = 98.40; p < 0.0001). Forest plot with additional information is presented in Fig. 6.
Forest plot in accordance with study setting for the prevalence of hypovitaminosis D. In this forest plot, all the diamonds except the last one (overall pooled prevalence) represent polled prevalence following study setting. Here are two categories; studies; (1) Community Based (2) Hospital Based. Each horizontal line of the forest plot represents an individual study and the box is plotted as prevalence for that study. The horizontal points of the diamond represent the limit of 95% confidence interval
Effect of lab methods on the prevalence of hypovitaminosis D
In our selected studies, the serum level of vitamin D was determined by using a variety of lab methods. Among these, Radioimmunoassay (RIA; 14 out of 41), Chemiluminescent Immunoassay (CLIA; 8 out of 41), and Enzyme-linked Immunosorbent assay (ELISA; 4 out of 41) were mostly used. Only two studies did not mention their procedure of vitamin D estimation 498
."), [64:59–62.
.")].
However, ELISA demonstrated the lowest (22%; 95% CI: 17% to 28%) and CLIA (84%; 95% CI: 70% to 94%) showed the highest prevalence of hypovitaminosis D amid all of the measurement methods used. High degree of heterogeneity was also observed in this section (RIA: I2 = 99.86%; p < 0.0001, CLIA: I2 = 98.93%; p < 0.0001, ELISA: I2 = 55.08%; p = 0.08). Detail is accessible in forest plot (Fig. 7).
Forest plot in accordance with lab methods for the prevalence of hypovitaminosis D. In this forest plot, all the diamonds except the last one (overall pooled prevalence) represent polled prevalence following study setting. Each horizontal line of the forest plot represents an individual study and the box is plotted as prevalence for that study. The horizontal points of the diamond represent the limit of 95% confidence interval. Here are eleven categories; studies; (1) RIA or radio immune assay (2) CLIA or chemiluminescence immunoassay, (3) ELISA or enzyme-linked immune sorbent assay, (4) IRMA or immunoradiometric immunoassay, (5) HPLC or high-performance liquid chromatography, (6) CMIA or chemiluminescent microparticle immune assay, (7) ECLIA or electrochemiluminescence Immunoassay, (8) LCMS or liquid chromatography tandem mass spectroscopy, (9) EIA or enzyme immunoassay, (10) CIA or commercial immunoassay, (11) NM or not mention
Quality assessment
Among these selected studies no study was found with a high risk of bias, 14 studies have a moderate risk of bias and the rest contained a low risk of bias. The risk of Bias for selected studies is available in Additional file 1: Table S2. Furthermore, random-effect meta-analysis illustrated that, studies with moderate risk of bias (66%; 95% CI: 51% to 80%) has a high prevalence of hypovitaminosis D compared to studies with low risk of bias (58%; 95% CI: 41% to 73%). Furthermore, high degree of heterogeneity was also observed in risk of bias assessment (Low: I2 = 99.68%; p < 0.0001, Moderate: I2 = 98.97%; p < 0.0001). Figure 8 shows the forest plot.
Forest plot in accordance with Risk of bias for the prevalence of hypovitaminosis D. In this forest plot, all the diamonds except the last one (overall pooled prevalence) represent polled prevalence following study setting. Here are two categories; studies; (1) Low (2) Moderate. Each horizontal line of the forest plot represents an individual study and the box is plotted as prevalence for that study. The horizontal points of the diamond represent the limit of 95% confidence interval
Publication bias
The presence of asymmetry and publication bias was indicated by the funnel plot. The Eggers test was found to be statistically insignificant, implying that small-study effects were not present (p = 0.74). The funnel plot is available in Additional file 1: Figure S2.
Discussion
This study reveals that approximately 6 out of 10 South Asian children and adolescents (up to 18 years) could be affected with hypovitaminosis D (Fig. 2). Comparison of this hypovitaminosis D to the other parts of the world implies that this problem might be worse in South Asia compared to Africa (around 34%; taking < 20 ng/mL cut-off) [677
.")]. We assume that hypovitaminosis D may be co-related to the high prevalence of some childhood health problems in this area. Indeed, this is supported by several reports suggesting a high burden of diseases that are associated with vitamin D deficiency like tuberculosis, obesity and overweight, type 1 diabetes, etc. among South Asian children and adolescents [4, 16:166–71.
."),17:31–44.
."),18:811–5.
."), 68. Levels and trends in child malnutrition worldwide.
."),69: S34.
."),70:1–8.
."),71. Momentum on child TB: South East Asia (SEA).
.")].
While high hypovitaminosis D was the highlight of this study, we also found significant amounts of heterogeneity in the overall result (I2 = 99.72%) and it can be assumed that geography might be acting as one of the significant variables. Because, vitamin D is synthesized naturally in our body when UVB from sunlight penetrates our skin and undergoes some physiological processes 26
."), [5:266–81.
.")]. This is likely since people living in tropical areas are exposed to more sunlight than those living in subtropical regions [2:118–26.
."), 5:266–81.
.")]. Among the South Asian countries, Pakistan and Afghanistan are situated in a subtropical area, while Sri Lanka is situated in a tropical area . In our study, we also found that Pakistan and Afghanistan had the highest prevalence of hypovitaminosis D in South Asia; 94% (nation-wide average of serum vitamin D 11.78 ng/mL) and 96% (nation-wide average of serum vitamin D 5 ng/mL), respectively. On the contrary, Sri Lanka has the lowest level of hypovitaminosis D in South Asia; 25% (nation-wide average of serum vitamin D 27.34 ng/mL). Our hypothesis is further strengthened by reports from other tropical and subtropical countries. For example, Brazil is located in a tropical region and according to published literature, only 28% of the Brazilian population had hypovitaminosis D, while Qatar, a subtropical country had a high prevalence (90%) of hypovitaminosis D [73:2102–9.
."), 74:229.
.")]. However, it must be noted that besides geographical variation, other variables could be affecting the inter-country variations.
Among the factors influencing the production of vitamin D, age, gender, and diet are notable 26
."), [5:266–81.
.")]. In our study, the age-wise analysis revealed that the prevalence of hypovitaminosis D is more among neonates (85%) than preschool (55%) and school-going children (57%) (Fig. 5). This may be associated with the high prevalence of vitamin D insufficiency (65%) among South Asian pregnant women which we have shown in one of our recent studies .
Most of the studies in South Asia enrolled participants from both genders. We did not find any study in South Asia that was conducted on males only. We found five studies that were conducted among females. Gender-wise comparison suggested that the studies that considered only females as participants showed a higher prevalence of hypovitaminosis D (76%; 95% CI: 46% to 96%) compared to those that considered participants from both genders (58%; 95% CI: 43% to 73%) (Fig. 4). This could be suggestive of the fact that female children in South Asia could be more affected with hypovitaminosis D compared to young males. This may be associated with cultural aspects and clothing practices of South Asian where females practice heavier clothing (traditional and religious full-body covering dresses like burqa, hijab, shari, salwar, kurta, etc.). A recent media report also pointed that these practices have increased dramatically in this region over the past three decades [768
. Accessed 30 Dec 2020.")]. However, we also argue that early marriage (which concerning early pregnancy), poor education, and insufficient decision-making ability may have some combined effects resulting in such a high prevalence of hypovitaminosis D among South Asian girls. In this regard, lack of recommended dietary intake, early marriage, and lack of higher education among girls has been reported in this region [77, 78].
Moreover, South Asians are affected with malnutrition of nearly all forms 791
."), [80:e12739.
.")]. While the cod liver oil, mushrooms, egg yolk, fish; salmon, mackerel, tuna, and fortified foods; milk, yogurt, cheese, orange juice, etc. are commonly referred to as the primary dietary sources of vitamin D, the effects of variation in staple foods (rice and wheat) in the South Asian population is not yet fully understood [2:118–26.
."), 5:266–81.
."), 81:395–427.
."), 82]. Therefore, we recommend further studies to understand the influence of staple foods on population-level serum vitamin D levels.
Furthermore, skin complexion may be another factor for such a high prevalence of hypovitaminosis D in this region. According to the Fitzpatrick scale, South Asians are quite darker in comparison to Europeans [836
.")]. A large observational data suggested that the prevalence of hypovitaminosis D (taking < 20 ng/mL cut-off) is 40% among Europeans [84:1033–44.
.")] which is much lower in comparison to what we have seen in this study for South Asia.
Additionally, our analysis also revealed that studies design with community-based setting (62%; 95% CI: 43% to 80%) has high prevalence of hypovitaminosis compared to hospital-based setting (58%; 95% CI: 41% to 74%). In accordance with iceberg phenomenon [856
.")], these findings indicate that a substantial number of South Asian children are affected with hypovitaminosis D that is either subclinical, unreported, or concealed from view. Therefore, community-based study settings demonstrated such a high prevalence of hypovitaminosis in South Asian children.
The high burden of hypovitaminosis D among South Asian children is a public health concern that should be addressed as an emergency. Some researchers also proposed that deficiency of vitamin D should be treated as a pandemic in progress [847
.")]. In this regard, a more recent study reported that vitamin D deficiency is also related to ‘cytokine storm’ (dramatic immune system overreaction) which causes COVID 19 patients more vulnerable . However, we also proposed that further research is needed to check if our findings can be applied to a wider group of general populations. Moreover, our analysis revealed a high degree of heterogeneity, which we attempted to explain explicitly using various independent variables such as geolocation, gender, age, skin colour, and so on. However, we believe that further studies are required to fully comprehend the reasons behind such significant heterogeneity.
Despite the high prevalence of hypovitaminosis D among South Asian children and adolescents, we did not find any national-level nutritional guidelines or policies for vitamin D except in India [87x
.")]. For the prevention of hypovitaminosis D among children and adolescents, China, Japan, and South Korea have similar guidelines [88:466.
."), 89:S43–7.
.")]. Furthermore, supplementation and food fortification programmes have been proved to be successful in Europe to reduce hypovitaminosis D [90: a systematic review of Economic evaluations. Eur J Public Health. 2017;27(2):292–301.
.")]. Moreover, some other challenges need to be addressed in South Asia. A negative attitude towards sunlight exposure can be a big challenge. It has been reported that Indian and Pakistani students had a lack of knowledge about vitamin D and a negative attitude towards sunlight exposure [91:308."), 92:1.
.")]. This lack of knowledge, along with a negative attitude, could be a key factor behind staying away from sunlight which can lead to hypovitaminosis D. Therefore, keeping the socio-cultural aspects of the individual countries (e.g. clothing practice, skin complexion, and economic status) in consideration, awareness campaigns about the relationship between sunlight exposure as a source of vitamin D can be emphasized. Furthermore, active measures should be taken to expand the number of diagnostic tests for detecting the serum vitamin D level. To achieve this, increasing the number of tests centres, reducing the cost of testing the serum level of vitamin D by offering subsidies can also be considered by the governments in the South Asian region.
To the best of our knowledge, this is the first systematic review and meta-analysis to highlight the prevalence of hypovitaminosis D in South Asian children. But this study also has few limitations. We were unable to explore the effect of the season in our analysis due to data insufficiency—only three of our selected studies 360
."), [48, 65:1718–28.
.")] provided season-specific data. Above 75% of our selected studies (31 out of 41) were conducted among Indian children. In this regard, the population of India is also disproportionately higher compared to the other South Asian countries. We did not find any studies from Maldives and Bhutan. So, we could not calculate the prevalence of hypovitaminosis D and the weighted mean level of serum vitamin D for these countries. Moreover, more than 60% of our selected studies did not mention demographic area (urban vs rural) and socioeconomic status (high income vs low income) for their study populations. Therefore, we were unable to find any correlations between hypovitaminosis D and these factors. Furthermore, multiple different methods were used in different studies (Table 1) to assay the serum level of vitamin D, which might have introduced some degree of assay bias. However, this limitation is inherent for studies like ours and was indeed unavoidable. Another limitation is that, since, the definition of children and adolescents were not uniform [93. Adolescence: a period needing special attention.
. Accessed 2 June 2021.")], and that many of our selected studies did not provide age-wise data, we could not perform the subgroup analysis for children and adolescents as separate age groups. Furthermore, vitamin D assessments are often expensive for South Asian children and adolescents due to their poor economic circumstances. So, it is possible that the children and adolescents who were enrolled in at least some studies or who participated in a study or a trial that included assessment of vitamin D were indeed suspected of deficiency/insufficiency of this vitamin. We assume that this is another inherent yet unavoidable limitation of our study. Therefore, we recommend that the readers should exercise caution before generalizing the extent of hypovitaminosis D in South Asia.
Conclusions
Out study unveiled that around six out of ten South Asian children and adolescents could be suffering from hypovitaminosis D. These findings have generated evidence of the actual population-level data that underscores the urgency of prioritizing the mitigation strategies. The subgroup analyses have resulted in several hypotheses to explain the observed heterogeneity of hypovitaminosis D among different countries, age groups, and genders. While this systematic review focused on South Asian children, the knowledge and insight generated from this study can be applied to other regions and countries with comparable geographical and socio-cultural aspects.
Availability of data and materials
Only aggregated summaries of the data are provided in this manuscript. However, all data generated in this study can be made publicly available on request. Please contact the corresponding author for any kind of data request.
Change history
08 April 2022
The Open Access license was not included in the original publication, this article has been updated.
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Acknowledgements
We are thankful to Mohammed Tanveer Hussain for his efforts in checking the grammatical aspects of the manuscript.
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Mahbubul H. Siddiqee and Badhan Bhattacharjee have contributed equally.
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Department of Mathematics and Natural Sciences, School of Data and Sciences, BRAC University, Dhaka, 1212, Bangladesh
Mahbubul H. Siddiqee & Badhan Bhattacharjee
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Mahbubul H. Siddiqee & Badhan Bhattacharjee
Communicable Disease Control Unit (CDC), Directorate General of Health Services, Dhaka, 1212, Bangladesh
Umme Ruman Siddiqi
Biomedical Research Foundation, Dhaka, 1230, Bangladesh
Mohammad Meshbahur Rahman
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MHS and URS conceptualize the idea and supervise the overall project. MHS and BB designed this study. BB played a major role in searching the literature, data extraction, data compilation, and preparing the first draft of the manuscript. MHS and URS reviewed the selected literature for analysis. MHS and BB prepared the bias table for quality assessment. MMR played a major role in searching, data extraction, and data analysis. MHS, URS, and MMR critically revised the manuscript. All authors read and approved the final manuscript.
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Additional file 1: Figure S1
. Comparative weighted mean of serum D levels among the South Asian countries. The error bars represent weighted standard deviation (except for Afghanistan from where only one study was reported).
Additional file 2: Figure S2
. Funnel plot for weighted mean values of serum vitamin D levels reported by the studies included in this systematic review (where a single dot represents the measurement). High number of observations falling outside the expected range on both sides indicate high heterogenity of reported serum vitamin D levels in South Asia.
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Siddiqee, M.H., Bhattacharjee, B., Siddiqi, U.R. et al. High burden of hypovitaminosis D among the children and adolescents in South Asia: a systematic review and meta-analysis.
J Health Popul Nutr 41, 10 (2022).
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How much does symbolic integration mean to mathematics?
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(Before reading, I apologize for my poor English ability.)
I have enjoyed calculating some symbolic integrals as a hobby, and this has been one of the main source of my interest towards the vast world of mathematics. For instance, the integral below
∫π 2 0 arctan(1−sin 2 x cos 2 x)d x=π(π 4−arctan 2–√−1 2−−−−−−−√).∫0 π 2 arctan(1−sin 2x cos 2x)d x=π(π 4−arctan2−1 2).
is what I succeeded in calculating today.
But recently, as I learn advanced fields, it seems to me that symbolic integration is of no use for most fields in mathematics. For example, in analysis where the integration first stems from, now people seem to be interested only in performing numerical integration. One integrates in order to find an evolution of a compact hypersurface governed by mean curvature flow, to calculate a probabilistic outcome described by Ito integral, or something like that. Then numerical calculation will be quite adequate for those problems. But it seems that few people are interested in finding an exact value for a symbolic integral.
So this is my question: Is it true that problems related to symbolic integration have lost their attraction nowadays? Is there no such field that seriously deals with symbolic calculation (including integration, summation) anymore?
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edited May 12, 2011 at 14:13
RegDwight
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asked May 11, 2011 at 15:25
Sangchul LeeSangchul Lee
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5 Manipulationally, an analytic (closed) form would be terribly convenient, in the sense that we know so much about (special) functions and the identities they satisfy that we would like to be able to exploit that whole body of knowledge for the integral at hand. For numerical work, a closed form may or may not be the best thing to have, depending on the circumstances.J. M. ain't a mathematician –J. M. ain't a mathematician 2011-05-11 15:29:55 +00:00 Commented May 11, 2011 at 15:29
3 If you broaden the question, allowing multiparameter (in)definite integrals and sums, then the question is not about only (transcendental) constants but, more generally, about the utility of closed formfunctions solving differential / difference equations, i.e. special functions. Do you intend to ask only about special constants or, more generally, special functions?Bill Dubuque –Bill Dubuque 2011-05-11 15:52:20 +00:00 Commented May 11, 2011 at 15:52
@J.M. : Thanks for your comment. But if you won't mind, can you be a little more specific?Sangchul Lee –Sangchul Lee 2011-05-11 15:56:34 +00:00 Commented May 11, 2011 at 15:56
@Bill Dubuque : Oh, that's my point. I am also interested in special functions.Sangchul Lee –Sangchul Lee 2011-05-11 15:57:39 +00:00 Commented May 11, 2011 at 15:57
2 Since you find symbolic integration interesting, you may find useful the references I mention in this answer: math.stackexchange.com/questions/37088/integration-doubt/…Andrés E. Caicedo –Andrés E. Caicedo 2011-05-12 16:39:53 +00:00 Commented May 12, 2011 at 16:39
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I think it would be appropriate at this point to quote Forman Acton:
...at a more difficult but less pernicious level we have the inefficiencies engendered by exact analytic integrations where a sensible approximation would give a simpler and more effective algorithm. Thus
∫0.3 0 sin 8 θ d θ=[(−1 8 cos θ)(sin 4 θ+7 6 sin 2 θ+35 24)sin 3 θ+105 384(θ−sin 2 θ)]0.3 0=(−0.119417)(0.007627+0.101887+1.458333)(0.0258085)+0.004341=−0.0048320+0.0048341=0.0000021∫0 0.3 sin 8θ d θ=[(−1 8 cos θ)(sin 4θ+7 6 sin 2θ+35 24)sin 3θ+105 384(θ−sin 2 θ)]0 0.3=(−0.119417)(0.007627+0.101887+1.458333)(0.0258085)+0.004341=−0.0048320+0.0048341=0.0000021
manages to compute a very small result as the difference between two much larger numbers. The crudest approximation for sin θ sin θ will give
∫0.3 0 θ 8 d θ=1 9[θ 9]0.3 0=0.00000219∫0 0.3 θ 8 d θ=1 9[θ 9]0 0.3=0.00000219
with considerably more potential accuracy and much less trouble. If several more figures are needed, a second term of the series may be kept.
In a similar vein, if not too many figures are required, the quadrature
∫0.55 0.45 d x 1+x 2=[tan−1 x]0.55 0.45=0.502843−0.422854=0.079989≈0.0800∫0.45 0.55 d x 1+x 2=[tan−1x]0.45 0.55=0.502843−0.422854=0.079989≈0.0800
causes the computer to spend a lot of time evaluating two arctangents to get a result that would have been more expediently calculated as the product of the range of integration (0.1 0.1) by the value of the integrand at the midpoint (0.8 0.8). The expenditure of times for the two calculations is roughly ten to one. For more accurate quadrature, Simpson's rule would still be more efficient than the arctangent evaluations, nor would it lose a significant figure by subtraction. The student that worships at the altars of Classical Mathematics should really be warned that his rites frequently have quite oblique connections with the external world.
It may very well be that choosing the closed form approach would still end up with you having to (implicitly) perform a quadrature anyway; for instance, one efficient method for numerically evaluating the zeroth-order Bessel function of the first kind J 0(x)J 0(x) uses the trapezoidal rule!
On the other hand, there are also situations where the closed form might be better for computational purposes. The usual examples are the complete elliptic integrals K(m)K(m) and E(m)E(m); both are more efficiently computed via the arithmetic-geometric mean than by using a numerical quadrature method.
But, as I said in the comments, for manipulational work, possessing a closed form for your integral is powerful stuff; there is a wholebodyof results that are now conveniently at your disposal once you have a closed form at hand. Think of it as "standing on the shoulders of giants".
In short, again, "it depends on the situation and the terrain".
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answered May 11, 2011 at 17:15
J. M. ain't a mathematicianJ. M. ain't a mathematician
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1 For the original problem, things are similar. After realizing that 1−sin 2 x cos 2 x=[7+cos(4 x)]/8 1−sin 2x cos 2x=[7+cos(4 x)]/8, i.e., 7/8 7/8 plus something oscillating, one is tempted to replace the integrand by arctan(7/8)arctan(7/8). Then one gets π arctan(7/8)/2≈1.129 π arctan(7/8)/2≈1.129 for the integral which coincides with the exact expression 1.126 1.126 up to 3×10−3 3×10−3. I don't know how much time the OP spend for his answer, but Pareto's principle seems to apply.Fabian –Fabian 2011-05-11 20:36:57 +00:00 Commented May 11, 2011 at 20:36
1 Thanks to everyone, especially J.M., for giving insightful answers. It's really convincing that distinction between numerical one and symbolic one is subject not to a particular classification of research area, but rather to a situation.Sangchul Lee –Sangchul Lee 2011-05-12 10:24:24 +00:00 Commented May 12, 2011 at 10:24
2 @sos440: You're very much welcome. :) Don't let my answer deter you from the fun you seem to have in teasing out analytic solutions; what I'm merely saying is that in real-world applications, one must eventually develop a "feel" for choosing the "right tool for the job".J. M. ain't a mathematician –J. M. ain't a mathematician 2011-05-12 10:28:00 +00:00 Commented May 12, 2011 at 10:28
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I don't think your point of view is the right one. To compute an integral analytically and to compute an integral numerically are different things. A numerical analysis professor of mine once said that, in applications (engineering, physics...) it is often more convenient to directly evaluate integrals by numerical means, even if they are integrable analytically! For example, suppose that you need
∫π 2 0 arctan(1−sin 2 x cos 2 x)d x∫0 π 2 arctan(1−sin 2x cos 2x)d x
meters of conducting wire. You make a phone call to the wire factory and ask for what? For π(π 4−arctan 2√−1 2−−−−√)π(π 4−arctan2−1 2) meters of wire? More realistically you will ask for something like 1.13 1.13 meters of wire.
To obtain this number 1.13 1.13 you performed an approximation over the non-rational quantity π(π 4−arctan 2√−1 2−−−−√)π(π 4−arctan2−1 2). In doing so you wasted information. It would have been more convenient (and, maybe, even more accurate) to perform this approximation on the first integral directly, that is, to evaluate it numerically.
Of course this does not render analytical methods useless. You could have a family of integrals depending on a parameter, for example. Numerical methods tell you nothing here. You could run across an integral in the middle of a proof, and need its exact value for theoretical purposes. The possibilities are countless.
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edited May 12, 2011 at 10:31
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answered May 11, 2011 at 15:55
Giuseppe NegroGiuseppe Negro
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Also, if you don't need that much accuracy (which is often in applications), even the simple-minded methods like trapezoidal or Simpson's might end up being faster to compute with than having to use a special routine for some exotic transcendental.J. M. ain't a mathematician –J. M. ain't a mathematician 2011-05-11 16:04:03 +00:00 Commented May 11, 2011 at 16:04
1 In short, if I may borrow the usual piece of military wisdom: "it depends on the situation and the terrain".J. M. ain't a mathematician –J. M. ain't a mathematician 2011-05-11 16:12:03 +00:00 Commented May 11, 2011 at 16:12
Surely you mean something like 1.126 instead of 1.49... Also I'm curious if it is sure or just very likely that this number is not rational, as you state.Myself –Myself 2011-05-12 02:58:42 +00:00 Commented May 12, 2011 at 2:58
@Myself: Oh yes, sorry about that, I must have done some mistake in typing that number into Maple. Regarding the irrationality of π(π 4−arctan 2√−1 2−−−−√)π(π 4−arctan2−1 2), I haven't checked it. Looks like a safe bet, though... Don't you agree?Giuseppe Negro –Giuseppe Negro 2011-05-12 17:26:52 +00:00 Commented May 12, 2011 at 17:26
1 Yes, I completely agree. But for all I know these things are typically hard to prove, so I thought perhaps I had missed something.Myself –Myself 2011-05-12 17:33:57 +00:00 Commented May 12, 2011 at 17:33
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If you're talking about practical engineering applications, then really only numerical approximations are used (and studied in computer science as 'numerical analysis' or more recently 'scientific computing').
As to an academic mathematical field nowadays that deals with symbolic integration, first some perspective. Newton/Leibniz invented integral calculus in ...hm...late 1600's and was popularized (as much as you can say that) in the 1700's. Some basic symbolic integration even occurred (without that name and system) before then. So let's just say there's been at least 300 years of work there.
Also, there's more to inverting derivatives than just integrals. Solving systems of partial differential equations seem to be the big thing (both numerically and symbolically) for almost as long as simple single variable integrals.
That said, there is a small academic group of people working in 'symbolic computation' (with their own journals), and one subarea is symbolic integration. There are proofs of impossibility (i.e. proving that given certain restrictions there is no 'closed form' for a particular integral), and there are algorithms for computing integrals given other certain restrictions (the Risch algorithm). The latter are often implemented in computer algebra packages (Mathematica, Maple, etc.).
There is surely room for solving particular integrals (in the AMM there don't seem to be many integrals though in the Problems section) and for finding patterns. I'd look at those journals to see what particular interest there is for integrals.
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answered May 11, 2011 at 17:17
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2 Back in the day, SIAM Reviews used to maintain a "problems column" where people submitted various requests for simple proofs, and yes, integrals/sums/whatever to evaluate. With the computing environments now, I suppose there is now less reason to submit those sorts of problems.J. M. ain't a mathematician –J. M. ain't a mathematician 2011-05-11 17:25:53 +00:00 Commented May 11, 2011 at 17:25
3 @J.M.: It's been a longstanding (15 years?) challenge by Doron Zeilberger that he (and his computer) can solve automatically any summation sent in to the AMM Problems section. Surely there are summations (and integrals) that are people-solvable and currently not computer-solvable, but as the technology progresses, these will be harder for people to come by.Mitch –Mitch 2011-05-11 17:44:07 +00:00 Commented May 11, 2011 at 17:44
2 I would also note that it may very well be that the symbolic output of current computing environments may be less than optimal, and some further human massaging may be needed. For instance, Mathematica sucks at producing optimal elliptic integral expressions...J. M. ain't a mathematician –J. M. ain't a mathematician 2011-05-12 01:24:16 +00:00 Commented May 12, 2011 at 1:24
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Symbolic integration becomes less popular indeed and most researchers prefer numerical integration. However, significance of the symbolic integration should not be underestimated. This can be shown by using the integration formula:
∫1 0 f(x)d x=2∑m=1 M∑n=0∞1(2 M)2 n+1(2 n+1)!f(2 n)(x)∣∣x=m−1/2 M,∫0 1 f(x)d x=2∑m=1 M∑n=0∞1(2 M)2 n+1(2 n+1)!f(2 n)(x)|x=m−1/2 M,
where the notation f(x)(2 n)|x=m−1/2 M f(x)(2 n)|x=m−1/2 M implies 2n-th derivative at the points x=m−1/2 M x=m−1/2 M. Once the integral is expanded as a series, we can use it either numerically or analytically (i.e. symbolically). It may be tedious to find by hand 2n-th derivatives. However, with powerful packages supporting symbolic programming like Maple, Mathematica or MATLAB this can be done easily. For example, by taking f(x)=θ 1+θ 2 x 2 f(x)=θ 1+θ 2 x 2 even at smallest M=1 M=1 we can find a rapidly convergent series for the arctangent function:
tan−1(θ)=i∑n=1∞1 2 n−1(1(1+2 i/θ)2 n−1−1(1−2 i/θ)2 n−1),tan−1(θ)=i∑n=1∞1 2 n−1(1(1+2 i/θ)2 n−1−1(1−2 i/θ)2 n−1),
where i=−1−−−√i=−1. This example shows that symbolic integration may be highly efficient in many numerical applications.
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edited Jun 5, 2020 at 21:33
answered Jun 5, 2020 at 21:20
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10227 | https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:foundation-algebra/x2f8bb11595b61c86:division-zero/v/why-dividing-by-zero-is-undefined | Why dividing by zero is undefined (video) | Khan Academy
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Skip to lesson content
Algebra 1
Course: Algebra 1>Unit 1
Lesson 6: Division by zero
Why dividing by zero is undefined
The problem with dividing zero by zero
Undefined & indeterminate expressions
Algebra foundations: FAQ
Math>
Algebra 1>
Algebra foundations>
Division by zero
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Why dividing by zero is undefined
CCSS.Math: 7.NS.A.2, 7.NS.A.2.b
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About About this video Transcript
As much as we would like to have an answer for "what's 1 divided by 0?" it's sadly impossible to have an answer. The reason, in short, is that whatever we may answer, we will then have to agree that that answer times 0 equals to 1, and that cannot be true, because anything times 0 is 0.Created by Sal Khan.
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Justin Wei 13 years ago Posted 13 years ago. Direct link to Justin Wei's post “What is 0/0 then? Because...” more What is 0/0 then? Because aren't there several rules that yield different answers for example, any number divided by itself is one, zero divided by any number is zero, etc? At 1:00 Sal says, "any non-zero number divided by zero is undefined". What if you divide zero by zero?
Thanks!
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Mr. Millette 13 years ago Posted 13 years ago. Direct link to Mr. Millette's post “The way I explained it to...” more The way I explained it to my 7th graders is this:
If 0/0 is equal to a number, then let's rewrite it as a multiplication.
0/0 = 0 ==> 0 0 = 0, so that works...
But wait!
0/0 = 1 ==> 0 1 = 0. Oops, that also works.
and so does 0/0 = 2 and 0/0 = 6 and 0/0 = any real number.
This is where it breaks down.
Because there are so many (in fact, an infinite number) of ways that this division could be converted into a valid multiplication, we can conclude that this division isn't valid or indeterminate, to use the correct terminology). For any operation, there should only be at most one operation that "undoes" it.
Hope this helps.
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maxtn 13 years ago Posted 13 years ago. Direct link to maxtn's post “Sal, is it possible that ...” more Sal, is it possible that 0 is the intersection of positive and negative numbers, since all other division results in some other non zero number and multiplication by zero is zero? Therefore infinity exists as an intersection.
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Lucas K. Sorensen 13 years ago Posted 13 years ago. Direct link to Lucas K. Sorensen's post “Well.....if you ask me, I...” more Well.....if you ask me, I think 0 is not really an intersection, but a bridge between positive and negative numbers and the 2nd question, 0+-∞.
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Ronald Das 13 years ago Posted 13 years ago. Direct link to Ronald Das's post “Guys What is square root ...” more Guys What is square root of Zero??
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saurabhag12 5 years ago Posted 5 years ago. Direct link to saurabhag12's post “if you go and check a cal...” more if you go and check a calculator it will show 0 and if you even think square root means its asking to what number we should put ^2 that the number which is being square rooted will come . so the answer for square root of zero is always=0. hope this helps:)
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jude.gallagher 2 years ago Posted 2 years ago. Direct link to jude.gallagher's post “anything divided by zero ...” more anything divided by zero is itself. dividing by nothing is leaving everything there
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Tanner P 2 years ago Posted 2 years ago. Direct link to Tanner P's post “It doesn’t work that way....” more It doesn’t work that way.
You can think of division as finding the missing part to a multiplication problem. For example, 20/4 can be written as 4 x ? = 20. And ? = 5.
Now, say you have 20/0. This can be written as 0 x ? = 20. According to your method, ? = 20. But we know this isn’t possible because 0 x 20 = 0, not 20.
Hope this helps!
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Dennis Fisher 9 years ago Posted 9 years ago. Direct link to Dennis Fisher's post “Why couldn't it just be d...” more Why couldn't it just be defined like this:
0=nothing
If i divide something by nothing, I'm left with that something. I simply didn't divide it.
Therefore, x/0=x
If I divide by .0000000001 or -.00000000001, I am then dividing by something again and therefore have an answer that is quantifiable. This could still allow us to hold true that dividing something by nothing simply means you did not divide it at all.
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Spacepickle 3 years ago Posted 3 years ago. Direct link to Spacepickle's post “The way I learned it is t...” more The way I learned it is this:
If you do division (for example 4/2) you can then do it in reverse and the answer will still be correct. 4/2=2, 22=4. But if you divide by zero (4/0) it is not possible to do it in reverse. If you take the product of 4/0 answer and multiply by zero, you won't get 4. You completely break division. As a side note, reversing the answer will not always get you the original number, but it will get you the absolute value of the original number.
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CR_DragoUchiha 12 years ago Posted 12 years ago. Direct link to CR_DragoUchiha's post “If infinity is not a real...” more If infinity is not a real number, what exactly is it?
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jmascaro 12 years ago Posted 12 years ago. Direct link to jmascaro's post “Hi, Infinity is a concept...” more Hi,
Infinity is a concept. We cannot put a value on infinity because there will always be something bigger that we can come up with.
We tend to think of infinity as "the biggest number" but it's not, infinity is just an idea. If I give you my biggest number, you will be able to come up with a bigger number. Then I can come up with a number bigger than yours, and you come up with an even bigger number, and so on and so on forever.
By trying to put a value on infinity, we are limiting infinity which, by definition, is limitless. Hope that helps :-)
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Leigh 2 years ago Posted 2 years ago. Direct link to Leigh's post “It could be inferred that...” more It could be inferred that division by 0 = everything. So when Sal says dividing by 0 could = 42, it is a reference to Hitchhikers Guide to the Galaxy, where the answer to life and everything is 42.
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Tutoring Maniac YT 13 days ago Posted 13 days ago. Direct link to Tutoring Maniac YT's post “Yeah Leigh, good observ...” more Yeah Leigh, good observation, and Sal is making an indirect reference to Hitchhiker's Guide to the Galaxy, where the answer to life and everything else is 42. Ummm, anyways, stay sharp and have a good time learning. :)
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Alexander L 6 years ago Posted 6 years ago. Direct link to Alexander L's post “Hi, I'm failing to unders...” more Hi, I'm failing to understand the issue here? It seems to me that 2 different questions are trying to be represented here with the same mathematical formula:
What happens when you have nothing to divide?
If you have 7 and you want to divide by nothing then no division is going to happen, so it seems to me that 7/0 = 7
If you owe 1 and you want to divide by nothing then nothing is going to happen either -1/0 = -1 Again, no division is happening because you have nothing to divide.
If we are asking the question: How many times can I fit a number into another like how many times can we fit 0.1 into 1 then is obvious that the smaller the number the more times we can fit it, that is what happens when we do:
1/0.1 = 10
Obviously here the smaller the numbers we are using to divide the more times we can fit them and since we know that there are an infinite amount of numbers in between 0.1 & 1, then as long as we have something to divide, no matter how small, we get a result. But the assumption that because we are dividing by smaller and smaller numbers then 1/0=∞ seems wrong to me because the moment we are dealing with 0 we really have nothing to divide, therefore the division doesn't happen.
CONCLUSION:
So I'm failing to see the issue here and why there is a problem determining the value? Where is the problem with this line of thinking and what are the other ways of reasoning that contradict this?
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CavCave 4 years ago Posted 4 years ago. Direct link to CavCave's post “See, maths cannot just re...” more See, maths cannot just rely on real life logic. It also needs to be consistent with itself. Defining division by 0 like you said would lead to so many problems and inconsistencies, like these people have said here.
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Web a year ago Posted a year ago. Direct link to Web's post “turns out when he says th...” more turns out when he says that 1/0 could equal infinity at 3:44 , it really isn't possible, because infinity is not a number.
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Tutoring Maniac YT a year ago Posted a year ago. Direct link to Tutoring Maniac YT's post “This is true as infinity ...” more This is true as infinity is not a number, as it goes on forever. So you have the right thinking. Infinity more like an infinite series of numbers.
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MathKid 2 years ago Posted 2 years ago. Direct link to MathKid's post “What does this sign mean?...” more What does this sign mean? : ±
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TheReal3A 2 years ago Posted 2 years ago. Direct link to TheReal3A's post “You may see that ± is jus...” more You may see that ± is just a + sign sitting above a - sign. An expression like a ± b means evaluate it if we made it a + and again if we made it a -, so it would be both determined as "a + b" or "a - b". This can be used to show multiple solutions to a variable. Eg. The roots of quadratic equations are in the form x = a ± b where a and b are algebraic numbers.
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Video transcript
Comedian Steven Wright-- and I guess we can credit him with being a bit of a philosopher-- once commented that "Black holes are where God divided by zero." And I won't get in to the physics of it, and obviously the metaphor breaks down in certain ways But it is strangely appropriate, because black holes are where our current understanding of physics seems to break down and dividing by zero, as simple of idea as that seems to be, is where our mathematics also breaks down. This is "undefined." Sometimes when you see "undefined" in math class it seems like a very strange thing. It seems like a very bizarre idea. But it really means exactly what the word means. Mathematicians have never defined what it must mean to divide by zero. What is that value? And the reason they haven't done it is because they couldn't come up with a good answer. There's no good answer here, no good definition. And because of that, any non-zero number, divided by zero, is left just "undefined." 7 divided by 0. 8 divided by 0. Negative 1 divided by 0. We say all of these things are just "undefined." You might say, well if we can just define it, let's at least try to come up with a definition of what it means to take a non-zero number divided by zero. So let's do that right now. We could just take the simplest of all non-zero numbers. We'll just do it with one. But we could have done this with any non-zero number. Let's take the example of one. Since we don't know what it means-- or we're trying to figure out what it means to divide by zero Lets just try out really, really, small positive numbers. Let's divide by really, really small positive numbers and see what happens as we get close to zero. So lets divide by 0.1 Well, this will get us to 10. If we divide 1 by 0.01 that gets us to 100. If I go really, really close to zero. If I divide 1 by 0.000001 this gets us-- so this is not a tenth, hundredth, thousandth, ten thousandth, hundred thousandth. This is a millionth. 1 divided by a millionth, that's going to give us 1 million. So we see a pattern here. As we divide one by smaller and smaller and smaller positive numbers, we get a larger and larger and larger value. Based on just this you might say, well, hey, I've got somewhat of a definition for 1 divided by 0. Maybe we can say that 1 divided by 0 is positive infinity. As we get smaller and smaller positive numbers here, we get super super large numbers right over here. But then, your friend might say, well, that worked when we divided by positive numbers close to zero but what happens when we divide by negative numbers close to zero? So lets try those out. Well, 1 divided by negative 0.1, that's going to be negative 10. 1 divided by negative 0.01, that's going to be negative 100. And, if we go all the way to 1 divided by negative 0.000001-- yup, I drew the same number of zeros-- that gets us to negative 1 million. So you when we keep dividing 1 by negative numbers that are closer and closer and closer and closer to zero, we get a very different answer. We actually start approaching negative infinity. So over here we said maybe it would be positive infinity, but you can make an equally strong argument that it could be a very different number. Negative infinity is going the exact opposite direction. So you could make an equally strong argument that it should be negative infinity. And this is why mathematicians say there's just no good answer here. Especially one that's consistent with the rest of mathematics. They could have just said it's equal to 42 or something like that. But that would make no sense. It's neither one of these values, and it wouldn't be consistent with everything else we know. So they just left the whole thing "undefined."
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10229 | https://www.youtube.com/watch?v=G1BapSikVbw | Diprotic & Polyprotic Acids
MooreChemistry
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Posted: 10 Nov 2016
During this lecture we will learn to calculate concentrations and pH of polyprotic acids.
AP Chemistry
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Transcript:
hello this is Miss Moore and today during chemistry we're going to discuss diprotic and polyprotic acids today's essential question how does the use of polyprotic acids change acid-base constant calculations and big hint here calculations make sure you have your calculators handy okay let's start with a definition and overview of polyprotic acids so definition a polyprotic acid is an acid that contains 2 which would be diprotic or more which would be polyprotic so let's try that again a polyprotic acid is an acid that contains two or more I am this ion double ionisable protons which are hydrogen ions so what did I just say let's look at an example it'll make a lot more sense so sulfuric acid or h2s o3 has right here two H's that can be ionized or released in water and phosphoric acid h3po4 has three hydrogen ions that are ionizable or can be released in water so we have here a diprotic acid and polyprotic acid alright so there are the definitions let's talk about some behavior of polyprotic acids polyprotic acids ionize in a stepwise fashion with each step having its own ionization constant where ka value so that means that they ionize or break apart step by step by step so example we have here h2 so4 mixed with water gives us H three O and hso4 so you can see that one of the two hydrogen's was ionized and this first step has a KA value of one point six times ten to the negative two from there we have hso4 note hso4 hso4 okay so hso4 with water produces more h3o and now so4 to - so these second hydrogen has now ionized and take note of the new ka value it is now six point four times ten to the negative eight so let's look at that those KA values one more time the first ionization constant is 1.6 times 10 to the negative 2 just because sometimes it's hard to visualize scientific notation that means it's zero point zero one six ii k egg constant is six point four times 10 to the negative eight meaning zero point one two three four five six seven six four you know the pattern here the eye the first ionization constant is way way bigger than the second ionization constant because it gets much more difficult to remove successive hydrogens and that's because after that after the first hydrogen's removed the acid now has a negative charge making it more difficult difficult to remove a positive charge another positive charge from giving it an even larger negative charge okay let's just try this out real quick let's write the equilibrium expression in ka values or or expression for the ionization of the diprotic acid h2co3 all right so to do that we're going to start with h2 cro4 this is gonna be aqueous plus water liquid producing we're gonna ionize only one of the hydrogen's so our products are going to be H cro4 - aqueous plus h3o so let's go ahead and do the second step we're going to be taking that conjugate base is now going to be our acid so we're gonna have h up almost put it to there H cro4 aqueous and this is still in the water please keep in mind we're not picking these things out of the water okay as this is going on this this and this is all still in the same environment okay so where are we here producing we're now gonna ionize our second hydrogen so we're gonna end up with C R oops wait let's go back to here forgot to put a - I hope you were shouting that out as I missed that cro4 - - what a mess - - aqueous plus more h3o okay and there is the step for ionizing this acid h2so4 so now let's write the ionization constant equations so for step one we'll have ka one for step one still going to have the products over the reactants so we're going to have concentration of H cro4 minus H three O over concentration of h2 cro4 and remember we don't involve water because it's not an in aqueous State okay and then for our ka - four equation - we will have the concentration of cro4 2 minus H three O over concentration of cro4 - - so there you go that's how you write the equilibrium expression and ionization constant expression or equations for a polyprotic acid alright let's try some calculations here the good news is it's really not any different than these same sorts of calculations with monoprotic acids the bad news is it's often more than one step alright so we're going to do this with a practice problem so practice problem calculate the pH of a point one molar solution of h2 c6h6 oh six solutions and the equilibrium constant trations of all the species involved and we have our k1 and our ka to guess the first thing we should do is write our equilibrium expression so we know we're going to have we're going to start with the acid H to see 6 H 6 o 6 aqueous plus water which is liquid producing we're going to lose 1 hydrogen's giving us H c 6h 6 o 6 - aqueous plus h3o and unfortunately that's only step one so for step two we're going to start with that guy there and we will have each C 6 H 6 0 6 - acquiesce in water producing C 6 H 6 0 6 - negative aqueous and forgot my plus sign + H 3 o okay so there is our step 1 and 2 equilibrium expressions we can only do our calculations one equilibrium expression at a time and being that I have limited space I'm gonna get rid of the second expression for now and let's fill right out our rice box so rice we've got our reaction so H 2 c6h6 0 6 and we don't we can skip the water each c6h6 so 6 negative and h3o are the initial concentration is zero point 1 0 0 and then 0 and 0 we don't have enough information to know the change so we're gonna have - some change and a plus some change so that end that's our C and then at equilibrium we will have zero point 1 0 0 minus X 0 plus X or just X and 0 plus X or just X ok I moved our rice box to a new page so have a little bit more room to work so remember that K a in this case k a 1 equals the concentration of our products over our reactants reactant okay and so that will give us let's see what was our ka one again eight point zero times 10 to the negative fifth so we'll have eight point zero times 10 to the negative fifth equals x and X right so x squared over zero point one zero zero minus X and remember this is a weak acid so we're going to assume that the X here is extremely small so we're gonna ignore it although we do need to check to see if we're right later all right so then let's get rid of the point one by multiplying naught by erasing by multiplying it on this side giving us eight point zero times 10 to the negative sixth equals x squared so to solve for X we will square root both sides giving us two point eight times 10 to the negative third and that's we got 4x so remember to check to make sure that our X was really small we need to take that number and divide it by from divided by our original concentration and as long as that result end up ends up being less than 5% we're good to go and it looks like it is about two and a half percent so we are our assumption that the X was small enough to ignore was a pro all right so we now have some concentration information let me get rid of these X's here give us a little bit more space all right so we know that our hydronium ion concentration is two point eight times 10 to the negative third as is our conjugate base or H c6h6 zero six two point eight times 10 to the negative third and I guess I should not have deleted this last box because it was zero point one zero zero minus X right so that gives us a concentration of our original acid zero point zero nine seven okay guess what steps one done however we're not we've got to move on to step two okay on to the second step of our ionization so we've got to do the whole rice box all over again so starting with our reaction we have h c 6h 6 0 6 and we have c 6h 6 o 6 and we have h3o so there's our initial reaction or a reaction now for the initial concentrations it's a little bit different this time we don't get the initial concentration from the beginning of the problem we get the concentration because this here is what we're looking for right and that is our concentration the equilibrium concentration we just calculated 2.8 times 10 to the third and negative third and there's another difference we actually have a starting concentration of our hydronium ion again 2.8 times 10 to the third because this is all happening at the same time in the same reaction dish so 2.8 times tend to the negative third what we do not have anything of is that guy there all right so now our change just like last time we don't know how much it's going to change exactly but this one will decrease by some X allowing these guys to increase by the same X which means our equilibrium will be 2.8 times 10 to the negative third minus X to point no no no zero plus X or zero no or X and 2.8 times 10 to the negative third plus X all right so next up is to calculate or ka2 this time and our ka - I gotta go back and check right here 1.6 times 10 to the negative 12 and then that's going to equal the product of these two right so we're going to have ups sorry X x times 2.8 times 10 to the negative third plus x over this one here which will be 2.8 times 10 to the negative third minus X this is starting to look complicated but here's the good deal good news we're just going to assume again that these two x's are so small that when we add or subtract them it's not going to make a difference although we'll check so we get to get rid of these so then this turns out to be really easy we divide and now look our x equals or ka2 so x equals one point six times ten to the negative 12th and again to a test our assumption we should the of a small X we need to divide one point six times ten to the negative twelve by our original original for this part of the problem concentration and that will definitely give us less than five percent so we're good to go all right everybody good so far so next let's let's calculate all these concentrations let me get rid of a few things here to make it so there's some room to right so now let's plug X into each of these equilibrium concentrate to figure out our concentrations so 2.8 times ten to then and get 1/3 minus one point six times ten to the negative twelve gives us drumroll please 2.8 times 10 to the negative third and what about for the next one well that's easy one point six times 10 to the negative 12 and how about our h3o 2.8 times 10 to the negative third oh my gosh let's go back and check something well wait let's not go there yet so we now know the concentration is all the species from the first step in the second step we now need to calculate h3o well h3o the final step was two point eight times ten to the third the did it change look at the first step we still basically have the same concentration of h3o so guess what we could have gotten the pH from the first one I mean now before you get mad at me I want you to remember I want you to know something that's only true of a weak acid okay when there's well let's we'll get there in just a second let's calculate the pH of this solution remember that pH equals negative log of the h3o concentration so when we put that in our calculators we should end up with something like two point five five all right so now you know how do to do to figure out concentrations of all the species of a diprotic acid and it would be the same for a polyprotic acid it would just be more steps and you now know if you only are asked the pH and it's a weak acid you can stop at the first step because they're so little let's go to the next one there's like no change no change in the concentration of h3o so you only have to calculate h3o or the ph from the first step you will find if it's a strong acid that the h3o concentration will change a lot in the first step and it'll still change again in the second step so you're gonna have to go through the multiple steps to calculate the ph okay so folks hopefully you I see that besides us being sort of long it's really no different than what we've been doing with a monoprotic acid and that's it for today have a good one |
10230 | https://www.nature.com/articles/s41598-022-26486-3 | Detection of acute thoracic aortic dissection based on plain chest radiography and a residual neural network (Resnet) | Scientific Reports
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Detection of acute thoracic aortic dissection based on plain chest radiography and a residual neural network (Resnet)
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Published: 19 December 2022
Detection of acute thoracic aortic dissection based on plain chest radiography and a residual neural network (Resnet)
Dong Keon Lee1,2na1,
Jin Hyuk Kim3na1,
Jaehoon Oh4,5na1,
Tae Hyun Kim3,4na1,
Myeong Seong Yoon4,
Dong Jin Im6,
Jae Ho Chung4,7,8&
…
Hayoung Byun4,7
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Scientific Reportsvolume 12, Article number:21884 (2022) Cite this article
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An Author Correction to this article was published on 09 February 2023
This article has been updated
Abstract
Acute thoracic aortic dissection is a life-threatening disease, in which blood leaking from the damaged inner layer of the aorta causes dissection between the intimal and adventitial layers. The diagnosis of this disease is challenging. Chest x-rays are usually performed for initial screening or diagnosis, but the diagnostic accuracy of this method is not high. Recently, deep learning has been successfully applied in multiple medical image analysis tasks. In this paper, we attempt to increase the accuracy of diagnosis of acute thoracic aortic dissection based on chest x-rays by applying deep learning techniques. In aggregate, 3,331 images, comprising 716 positive images and 2615 negative images, were collected from 3,331 patients. Residual neural network 18 was used to detect acute thoracic aortic dissection. The diagnostic accuracy of the ResNet18 was observed to be 90.20% with a precision of 75.00%, recall of 94.44%, and F1-score of 83.61%. Further research is required to improve diagnostic accuracy based on aorta segmentation.
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Introduction
Acute thoracic aortic dissection is a life-threatening disease, in which blood leaking from the damaged inner layer of the aorta causes dissection between the intimal and adventitial layers. The diagnosis of this disease is challenging. Although the true incidence rate is difficult to determine, large series of autopsies have reported that the rate of prevalence of aortic dissection is 0.2–0.8%1. However, it is one of the most catastrophic cardiovascular diseases, with a mortality rate of 50% within 48 h if not diagnosed and treated properly.
The most common symptoms of aortic dissection are severe chest or back pain of abrupt onset, but they are highly variable between patients2."),3."). A substantial number of patients complain of nonspecific symptoms, such as abdominal pain, nausea, discomfort, and syncope, but some do not report discomfort. In addition, physical examination results are relatively normal. These characteristics make the diagnosis of aortic dissection difficult.
The modality of choice for diagnosing aortic dissection is contrast-enhanced computed tomography (CT)2."). Contrast-enhanced CT can reliably identify the dissection flap and false lumen, which are the primary diagnostic features of aortic dissection. However, because of possible atypical presentations and other diagnoses that may mimic aortic dissection, diagnostic modalities, such as chest X-ray scanning, laboratory tests including D-dimer, and bed-side transthoracic echocardiography, are used to screen for aortic dissection. Among them, chest X-ray imaging is most commonly used to differentiate the various causes of chest pain rapidly. It is also used as a screening test for acute thoracic aortic dissection. However, the sensitivity of chest X-ray scanning through a widening of the aortic silhouette is only 70%, as reported in previous studies4."),5.").
With recent developments in the application of deep learning technology to image recognition, significant research has been conducted on automatic interpretation of medical images, including X-ray images. In particular, algorithms have been developed to diagnose diseases such as tuberculosis, pneumonia, and pneumothorax using chest X-ray scanning6."),7."),8."),9.").
Given the need for rapid and accurate diagnosis of acute thoracic aortic dissection, the use of deep learning technology may be helpful. During the conception of this study, we hypothesised that the accuracy of diagnosis could be improved by analysing chest x-rays, which are the fastest available screening modality in clinical practice, using deep learning. Therefore, we sought to investigate the accuracy of a deep learning algorithm for screening acute thoracic aortic dissection based on chest X-ray scanning using a Convolutional Neural Network (CNN) model.
Related works
Contrast CT is the modality of choice used to diagnose aortic dissection. Several studies have been conducted on the detection of thoracic aortic dissection based on CT images using CNN. Recently, Hata et al. investigated the diagnostic performance of a deep learning algorithm for aortic dissection based on CT images. In aggregate, the data of 170 patients were considered, including 85 patients with aortic dissection. Only non-contrast CT images were used, and an area under the curve (AUC) of 0.940 was achieved, alongside an accuracy of 90.0%, sensitivity of 91.8%, and specificity of 88.2%10. Another study attempted a similar task based on contrast CT images. The authors constructed a U-Net-based semantic segmentation architecture to segment the aortic lumen and performed aortic circularity analysis on the segmentation results. Their detection results exhibited 85.00% accuracy, 90.00% sensitivity, and 80.00% specificity11. In another study on aorta lumen segmentation, 260 type B aortic dissection patients were enrolled and a mean Dice coefficient exceeding 90% was recorded12.
The accuracy of diagnosis of acute thoracic aortic dissection is high when CT images and a CNN are used, reaching 85–90%. However, the accuracy is also high when physicians use CT images for diagnosis (the scores obtained are: 94.90% accuracy, 82.60% sensitivity, and 100.00% specificity)13. This is because 3-dimensional images of the aorta and pathognomonic CT findings of type A acute thoracic aortic dissection are greatly helpful in diagnosis.
Moreover, CT images can be obtained hours after an emergency department (ED) visit, as intact kidney function must be confirmed with a blood test. Given that the mortality rates of acute thoracic aortic dissection increase by 2% per hour14,15, this time interval might be critical. Even if physicians decide to perform CT without checking renal function, performing CT becomes a challenge for all patients presenting chest pain because of radiation hazard and cost.
Chest x-rays are currently used as a screening tool in the ED because of the low cost, a 150 to 1350-fold lower radiation hazard16,17, and easy accessibility to the early phase of patients’ presentation in the ED. However, the sensitivity for screening acute thoracic aortic dissection is 70% for chest x-rays, which is not high enough4."),5.").
To our knowledge, there has been no study regarding improving the accuracy of acute thoracic aortic dissection with chest x-rays using CNNs. In this study, efforts are made to do that, and this may further improve patient outcomes.
Methods
Study design
This is a multicentre retrospective study aimed at learning and detecting aortic dissection using CNN to analyse chest X-ray images obtained from three tertiary academic hospitals (Seoul and Gyeonggi-do, Republic of Korea) between October 2021 and March 2022. The study was reported in accordance with the Checklist for Artificial Intelligence in Medical Imaging (CLAIM)18.
Data collection
Data collected between 2003 and 2020 at Seoul National University Bundang Hospital (Hospital A) and between 2005 and 2020 at Hanyang University Hospital (Hospital B) were used to construct the training dataset. Considering the versatility of the trained CNN model, data collected between 2018 and 2020 at Yonsei University (Hospital C) was used as testing data, on which the proposed model had not been trained.
Chest X-ray images were obtained from three tertiary academic hospitals. One trained researcher in each institution who did not participate as an author investigated each patient's age, sex, major symptoms, final diagnosis, chest CT readings, and surgical records based on the electronic health records. Based on this, a list was prepared comprising patients whose final diagnoses were consistent with acute thoracic aortic dissection and who concurrently underwent chest CT. In reference to this list, chest x-rays collected during the initial hospital visit of all patients were collected from picture archiving and communication systems. In addition, the type of thoracic aortic dissection was classified following the readings of radiologists. Acute thoracic aortic dissection was classified based on chest CT readings or surgical records as Stanford type A and B—the former involves any part of the aorta proximal to the origin of the left subclavian artery and the latter involves the aorta distal to the left subclavian artery.
Positive images, i.e., chest X-ray images of patients with acute thoracic aortic dissection, were obtained from patients diagnosed with acute thoracic aortic dissection in the emergency department. Diagnosis of acute thoracic aortic dissection included only those confirmed by contrast-enhanced CT or surgical diagnosis. Chest X-ray images of patients who visited the emergency departments after surgery or endovascular treatment at other hospitals were excluded. Negative images were obtained from patients who visited the emergency department for chest pain but did not receive a specific diagnosis, such as aortic dissection, aortic aneurysm, heart failure, pneumothorax, or ischemic heart disease. These diagnoses were excluded by the emergency physician based on the patient's symptoms, laboratory studies, chest x-rays, and, if necessary, CT reports. For example, if a pneumothorax was suspected in the absence of air in the pleural space on a chest X-ray, a CT scan was performed to confirm it. Finally, chest X-rays of chest pain patients who were excluded from specific diagnoses depicting readings within the normal range were used as negative images. The obtained images exhibited a positive-to-negative ratio between 1:3 and 1:5. All chest X-ray images were extracted in Digital Imaging and Communications in Medicine (DICOM) format and converted to JPEG format via image pre-processing.
Data pre-processing and augmentation
Personal information, such as name, gender, and age, was removed from all chest X-ray images and only the images were used. Since chest X-ray images were collected from three different hospitals over an extended period of 18 years, a pre-processing step was necessary to ensure consistency (Fig.1). First, unnecessary black margins were removed, and images were uniformly resized to a 448 × 448 pixel resolution. Next, the training images were augmented via random transformations, including flip, flop, and rotation—this transformation was needed to improve the performance of the model as the collected data included some images that had been flipped, flopped, and rotated. Further, a small amount of data was collected—thus, augmentation was needed to increase the size of the data and to secure robustness against geometric changes, such as scale, translational, and rotational transformation. Finally, histogram equalization, given by the following expression, was performed to reduce the deviation of the contrast:
$$T\left( {r_{k} } \right) = \frac{L}{n}\mathop \sum \limits_{j}^{k} n_{j}$$
(r_{k}): Brightness of input image pixels, (n_{j}): Number of (r_{k}), L : Value of maximum brightness, n : Total number of pixels.
Histogram equalization yielded improved images by distributing the image brightness values such that uniform brightness values in the range between 0 and 255 were used.
Figure 1
Image pre-processing.
Full size image
Diagnosis of acute thoracic aortic dissection via image classification
The proposed image classification architecture was based on residual neural network (ResNet) 18. ResNet is one of the most popular deep learning models in image classification, which successfully resolves the vanishing gradient problem, which is common during the training of traditional convolutional neural networks (CNNs) using residual mapping190
"). To construct the residual mapping, skip connections between layers were used to implement the network over multiple layers. The overall network architecture is illustrated in Fig.220.").
Figure 2
Model architecture of ResNet 18.
Full size image
The Resnet 18 methodology adopted for the detection and classification of aortic dissection is depicted in Fig.3. The primary objective was to classify chest X-ray images into one of two categories—normal and aortic dissection. Two main stages were involved in the model—the pre-processing stage (which further included data augmentation and normalization) and the classification stage (which involved the use of Resnet18 on pre-trained models and prediction). The images were rescaled to 448 × 448 pixel resolution. Moreover, the images were augmented via: (1) rotation (2) horizontal flipping and vertical flipping.
Figure 3
Overview of main architecture used for the diagnosis of aortic dissection.
Full size image
During the training process of the Resnet18 model, performance degradation induced by data imbalance was mitigated by using a weighted random sampling method and a weighted cross-entropy loss function. In addition, an ensemble voting system was implemented to prevent overfitting.
The Gradient-weighted Class Activation Mapping (Grad-CAM) technique was employed to determine the aortic dissection detection transparency. This technique highlights the regions of the input image where the model pays greater attention during the classification process, implying that the feature maps generated in the final convolution layer contain the spatial information that aids the capture of the visual pattern. This visual pattern contributes to distinction between the assigned classes. The Grad-CAM technique was applied by utilizing the layers and extracted features of the trained model.
Network performance and validation
The X-ray images collected from hospitals A and B were combined and used to construct a training dataset to train and validate the network, and the images from hospital C were used to construct the testing dataset to evaluate the performance of the network. K-fold (k = 5) cross-validation was performed on the final network architecture using the training dataset to achieve satisfactory performance while avoiding overfitting. During fivefold cross-validation, the training dataset was randomly divided into five roughly equal-sized subsets—four of them were used to train the network and the remaining one was used to estimate its performance.
An ensemble voting system was used to further improve the robustness and reduce the risk of overfitting. In the proposed model, five models resulting from fivefold cross-validation were ensembled (Fig.4). After using a soft voting classifier to assign probabilities to target variables, the training data were first shuffled and then divided into five groups, and passed to a fivefold training model. Individual predictions were obtained from each model using a voting aggregator and the soft voting technique. Finally, the majority vote was calculated, yielding the final prediction21,22.
Figure 4
Ensemble voting system using fivefold cross-validation.
Full size image
Data imbalance
In the collected training dataset, 20% of the images were positive images and the remaining 80% were negative images. To prevent performance degradation caused by data imbalance, a weighted random sampling method and a weighted cross-entropy loss function were used. Random sampling is a part of the sampling technique, in which each sample is assigned an equal probability of every mini-batch. In the mini-batch (size = 16) for learning, the ratio of the positive and negative images was set to 1:1. The weighted Cross-Entropy loss function was used to solve the negative effect of overfitting on the training dataset on the accuracy of the deep learning model due to a decrease in the imbalance of the convergence speed of the loss function23."). The following standard weighted binary cross-entropy loss function was used:
$$J_{wbce} = - \frac{1}{M}\mathop \sum \limits_{m = 1}^{M} \left[ {w \times y_{m} \times \log \log \left( {h_{\theta } \left( {x_{m} } \right)} \right) + \left( {1 - y_{m} } \right) \times \log \log 1 - h_{\theta } \left( {x_{m} } \right) } \right)]$$
where (x_{m}) and (y_{m}) denote input and target labels during training, and (h_{\theta }) denotes a model with neural network weights, (\theta). Here, (w) denotes a weight, which is taken to be 0.7 in the following experiment based on experience.
Visual verification via grad-CAM
For visual verification of the diagnostic results of the proposed deep learning network, Grad-CAM was implemented on the final convolutional layer. Grad-CAM uses the gradients of any target classes passing through the final convolutional layer of the CNN to generate a highlighted localization map depicting the essential regions of the image for the prediction of acute thoracic aortic dissection. The class-discriminative localization map is given by:
$$L_{Grad - CAM}^{c} = ReLU\left( {\mathop \sum \limits_{k} \alpha_{k}^{c} A^{k} } \right)$$
where (A^{k}) denotes feature map of the (k^{th}) channel, and (\alpha_{k}^{c}) represents partial linearization of the deep network, which is the primary function of the feature map.
Primary outcomes and performance evaluation
The primary outcome of this study was the detection of acute thoracic aortic dissection based on chest X-ray scanning using a CNN model. To validate the performance of five models trained via fivefold cross-validation, accuracy, sensitivity, specificity, positive predictive value (PPV), negative predictive value (NPV), and F-1 score were calculated. In addition, the performance of the ensemble model, which yields the final prediction via soft voting of the five models created during the cross-validation process, on the testing dataset was estimated.
To evaluate the performance of the model, we calculated its precision, recall, F1-score, and accuracy. The normal case and acute thoracic aortic dissection were considered as negative and positive cases, respectively. True positives (TP), true negatives (TN), false positives (FP), and false negatives (FN) were estimate based on the confusion matrices. These were calculated using the following parameters and equations:
True positive (TP)
An image with acute thoracic aortic dissection is classified in the acute thoracic aortic dissection category.
True negative (TN)
A normal image is classified in the normal category.
False positive (FP)
A normal image is incorrectly classified in the aortic dissection category.
False negative (FN)
An image with acute thoracic aortic dissection is incorrectly classified in the normal category.
Precision denotes the fraction of correct positive detection of acute thoracic aortic dissection.
Recall represents the quality of all the positives, which depends on the percentage of total relevant cases correctly classified by the model.
F1-score denotes the harmonic mean of precision and recall.
$$\begin{aligned} Precision & = \frac{{TP}}{{TP + FP}} \ Recall & = \frac{{TP}}{{FN + TP}} \ Accuracy & = \frac{{TP + TN}}{{FP + TP + TN + FN}} \ F1 - score & = 2 \times \frac{{Precision \times Recall}}{{Precision + Recall}} \ \end{aligned}$$
Statistical analysis
All data processing and statistical analyses were performed using the Pytorch (ver.1.6.0, environment for Resnet construction, training, and evaluation. Kolmogorov–Smirnov tests were performed to demonstrate the normal distribution of all datasets. We generated descriptive statistics and presented them as frequency and percentage for categorical data and as either median and interquartile range (IQR) (non-normal distribution) or mean and standard deviation (SD) (normal distribution) or 95% confidence interval (95% CI) for continuous data. The AUC of the receiver operating characteristic (ROC) was used to measure the performance of the deep learning model. Two-tailed p < 0.05 was considered to be significantly different.
Experimental environment
Weighted binary cross-entropy was adopted as the loss function to fit the binary classifier (weight for type A = 3:7, type B = 1:9), and Adam was used as the optimiser function (learning rate = 0.0001). Training and testing were performed using a GeForce RTX 2080 Ti GPU (NVIDIA, Santa Clara, CA, USA). The network weights were initialised based on a pre-trained model on Resnet18, and the network was trained end-to-end using stochastic gradient descent (SGD). We trained the model in batches of 16, with an initial learning rate of 0.0001, which was decreased by 0.5 gamma every 10 epochs.
Ethics approval and consent to participate
This study was approved by the Institutional Review Board (IRB) of Seoul National University Bundang Hospital (B-2002/597–102), IRB of Hanyang University Hospital (2021–01-005), and IRB of Yonsei University Hospital (4–2022-0770). All methods and procedures were carried out in accordance with the Declaration of Helsinki.
Results
In aggregate, 3,331 images, containing 716 positive images and 2615 negative images, were collected from 3,331 patients. Overall, 1,972 images consisting of 507 positive images (gender: 62.7% male; age [SD]: 61 years) from hospital A, 1,155 images consisting of 155 positive images (gender: 56.1% male; age [SD]: 63 years), and 204 images consisting of 54 positive images (gender: 55.6% male; age [SD]: 61 years) were analysed (Table 1). All patients with negative images visited the emergency department with chest pain and no specific diagnosis. 422 (83.2%), 123 (79.4%) and 31 (57.4%) patients were diagnosed with Stanford type A aortic dissection at hospitals A, B, and C, respectively. The datasets of hospitals A and B were separated into training data (80%) and internal validation data (20%) to verify the performance of the proposed method. The dataset of hospital C was used for testing (Fig.5).
Table 1 Baseline characteristics of participants who provided images for the data sets.
Full size table
Figure 5
Flow chart of data collection and analysis during acute thoracic aortic dissection detection based on deep learning algorithms.
Full size image
Performance of the deep learning model
The diagnostic performance matrix and outcomes are presented in Tables 2 and 3, respectively. The average accuracy on the validation set was 90.20%. The testing data set was obtained from data collected at hospital C, and the deep learning model was not trained on this data set. To evaluate the performance of the final deep learning model, the ROC curve was drawn and its AUC was calculated to be 0.955 on the testing data set (Fig.6). The diagnostic accuracy of the deep learning model was 90.20%, with 75.00% precision, 94.44% recall, and 83.61% F1-score, on acute thoracic aortic dissection images.
Table 2 Diagnostic performance matrix.
Full size table
Table 3 Diagnostic performance matrix on the internal validation using fivefold validation and Test data set.
Full size table
Figure 6
The ROC for the trained classification model. The AUC was 0.955.
Full size image
Regions of interest for aortic dissection
Figure7. depicts the classification results obtained using the deep learning network. The model trained using the results of the true positive category emphasised the aortic region. Conversely, the highlighted parts were scattered in the true negative category. In the false positive and false negative categories, cases focusing on the aorta and scattered regions were mixed.
Figure 7
The regions of interest for aortic dissection diagnosis were visualised as heat maps based on Grad-CAM following the confusion matrix categories: (a) true positive, (b) true negative, (c) false positive, and (d) false negative.
Full size image
Discussion
Several studies have investigated the detection accuracy of aortic dissection by applying CNN on CT images. Since contrast chest CT uses a contrast material to enhance blood vessels, it is a modality of choice for the diagnosis of acute thoracic aortic dissection. It aids physicians to distinguish between an enhanced dissected aorta and a normal one. However, the detection of dissected aorta based on chest x-rays is more challenging due to the lack of enhancement of blood vessels, unlike in CT images. Therefore, in this study, we improved the diagnostic accuracy of aortic dissection based chest x-rays using a CNN. To the best of our knowledge, this is the first attempt to improve the diagnostic accuracy of aortic dissection by learning chest X-rays using a CNN. As chest X-ray is the most basic examination modality for patients visiting the emergency department with chest pain and a common screening test for aortic dissection, the proposed model is expected to facilitate clinical screening for aortic dissection.
To determine the model with the best performance, ResNet (18 and 34), DenseNet, and EfficientNet (b0 and b1) were used. Table 1 presents the number of parameters in each network and the accuracy on the test data set. DenseNet and EfficientNet (b0, b1) exhibited poorer performance in representative cases despite a slight increase in parameters compared to the original ResNet (18 and 34). ResNet 18 required fewer parameters than ResNet 34 and exhibited superior performance than the other models. Therefore, we used ResNet18 as our final model (Supplementary table 1).
In chest X-ray scanning images, the aorta is distinguished based on the contrast between the air-filled lungs and the fluid-filled aorta. The ascending aorta can be identified outside the upper-right cardiac silhouette on a normal chest X-ray image. Moreover, the aortic arch is typically small and distinct in the upper left mediastinum, and the descending aorta can be distinguished as a clean, crisp stripe to the left of the vertebral column on a normal chest X-ray image24.
In chest x-ray images, aortic dissection is characterised by certain features, including mediastinal widening, expansion of aortic diameter, presence of double density due to enlargement of the false lumen, irregular contour due to edema and haemorrhage in the tissues, blurred aortic knob, displacement of intimal calcium, discrepancy in diameters of ascending and descending aorta, displacement of trachea/left main bronchus/oesophagus, and pleural effusion25. Moreover, abnormal findings on the lung field are possible, which are indicative of impending aorta rupture, such as pneumonitis caused by transmural aortic bleeding, nonspecific inflammatory reaction, secondary bronchopneumonia, regional compression atelectasis, and para-aortic hematoma26.").
Although X-ray scanning is a useful screening tool based on the aforementioned guidelines, its diagnostic accuracy is not satisfactory considering the fatality of aortic dissection. According to a meta-analysis, the sensitivity of chest X-ray scanning was 64% when evaluated based on wide mediastinum and 71% based on a combination with abnormal aortic contour4."). Even though a sensitivity of 90% was achieved when all nonspecific abnormal findings, such as pleural effusion, were combined, the development of a more accurate method is required since it is difficult to assume that all patients with nonspecific abnormal findings on X-ray scanning suffer from aortic dissection.
Christoph et al. used transthoracic echocardiography (TTE), transoesophageal echocardiography (TEE), CT, and magnetic resonance image (MRI) to diagnose aortic dissection, achieving sensitivities of 59.3%, 97.7%, 93.8%, and 98.3%, respectively; accuracies of 69.8%, 90.0%, 91.1%, and 98.0%, respectively; and precisions of 81.4%, 87.7%, 91.8%, and 98.3%, respectively27."). In the present study, chest X-ray images were analysed using a CNN to detect acute thoracic aortic dissection, yielding a sensitivity of 94.44% and an accuracy of 90.20%—thus the performance was comparable to those of TEE and CT, and better than that of TTE in terms of accuracy. Although the precision achieved was low, the results were deemed to be relevant considering the ease of obtaining chest X-ray images and using them as a screening test.
The results were also notable compared to manual chest x-ray readings of radiologists. William et al. reported a sensitivity of 86%, an accuracy of 60%, and a precision of 53% during the diagnosis of aortic dissection based on abnormal chest x-ray findings, which is lower than the result obtained using the proposed CNN280
(1998).").
The high aortic dissection detection accuracy of the proposed CNN can be explained based on the heat maps. Since the area highlighted in the heat map of the true positive category surrounded the aortic knob, mediastinal widening, aortic diameter expansion, and double contour of the aorta, which are directly related to aortas, were focussed on by the CNN. Additionally, the aortic diameter tends to taper from the origin to the downstream region gradually29."). If the diameter of ascending aorta or descending aorta is larger than the diameter of the origin, the possibility of aortic dissection must be considered even if the range is normal. There may be a possibility that these points were reflected in the CNN, which is why the achieved accuracy was higher than that of a radiologist.
Considering the aforementioned comparisons with other diagnostic tools and radiologists, the proposed deep learning algorithm for acute thoracic aortic dissection based on chest X-ray scanning can be considered to be a reliable, quick method to detect acute thoracic aortic dissection. Its application as a screening test on patients visiting the emergency department is expected to enable the detection of cases that might otherwise have been missed. Based on the preliminary result, immediate confirmatory tests may be performed.
The accurate identification of aortic contour based on medical images and detection of aortic dissection accordingly are challenging. In a previous report, aortic contours were automatically distinguished from non-contrast CT images30, and Hata et al. classified aortic dissection using ML in non-contrast CTs. They achieved an accuracy of 90.0% and a sensitivity of 91.8%10. Even though the identification of the aorta by applying ML algorithm on chest X-ray scanning was not performed in this study, our results are still compelling because the reported accuracy and sensitivity are as high as those of CT studies, including the accuracy of identification of the aorta.
Regarding aorta identification based on chest X-ray scanning, the attention mechanism, which is known to function well in image classification tasks by increasing the representation power, was utilised to increase the accuracy of the diagnostic network. As depicted in Fig.3, the trained model emphasises the thoracic aortic region in the true positive category, while the highlighted parts are scattered in the true negative category. This indicates that the proposed deep learning model concentrates on the thoracic region during the detection of acute thoracic aortic dissection without distinguishing aortic contours in chest X-ray images. Further, in the false positive and false negative categories, a combination of cases of aortic concentration and vice versa were observed, which corroborates the emphasis on the aortic region without thoracic aortic classification. Given that abnormal X-ray findings in acute thoracic aortic dissection are not limited to the aorta, this result seems reasonable. However, further research is required to classify acute thoracic aortic dissection after segmenting the aortic contour.
Limitations
This study suffers from certain limitations. Firstly, the age and sex of the patients were not completely matched to the presence of aortic dissection because aortic dissection is more prevalent corresponding to a certain age and sex. Secondly, radiographic images of assorted sizes taken in different environments were pre-processed and used as input images. Even though they were cropped around the aorta, except for the outer edges and margins of the image, the trained model could not identify the aorta properly in some images and sometimes misclassified it. Therefore, better results can be expected if classification is performed following aortic segmentation. Third, although equal proportions of the training dataset were allocated for acute thoracic aortic dissection and normal imaging, an imbalanced test dataset may reduce the reliability of the test results. Fourth, in this study, binary classification of normal images and acute thoracic aortic dissection images was performed without classifying the type of acute thoracic aortic dissection, and performance was not evaluated by including images of other aortic syndromes and other specific conditions causing chest pain. Finally, although the performance of the aortic dissection classification model was good, our observations are not sufficient to conclude that chest x-rays can be used to replace CT scans completely in patients with suspected aortic dissection.
Conclusions
The detection accuracy of acute thoracic aortic dissection using Resnet 18 was 90.20%. This model is expected to facilitate the screening of patients with suspected acute thoracic aortic dissection among patients who visit the emergency department with chest pain. Given the high severity and acuity of thoracic aortic dissection, early suspicion based on chest X-rays and CNN could accelerate the diagnosis and improve the prognosis. In future works, the accuracy should be improved based on segmentation, and research directly applicable to clinical practice should be prioritised.
Data availability
The data presented in this study are available on request from the corresponding author.
Change history
09 February 2023
A Correction to this paper has been published:
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Funding
This work was partly supported by Institute of Information & communications Technology Planning & Evaluation (IITP) grant funded by the Korea government (MSIT) (No.2020–0-01373, Artificial Intelligence Graduate School Program (Hanyang University)).
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Author notes
1. These authors contributed equally: Dong Keon Lee, Jin Hyuk Kim, Jaehoon Oh and Tae Hyun Kim.
Authors and Affiliations
Department of Emergency Medicine, Seoul National University Bundang Hospital, Seongnam, Republic of Korea
Dong Keon Lee
Department of Emergency Medicine, Seoul National University College of Medicine, Seoul, Republic of Korea
Dong Keon Lee
Department of Computer Science, Hanyang University, 222 Wangsimni‑ro, Seongdong‑gu, Seoul, 04763, Republic of Korea
Jin Hyuk Kim&Tae Hyun Kim
Machine Learning Research Center for Medical Data, Hanyang University, Seoul, Republic of Korea
Jaehoon Oh,Tae Hyun Kim,Myeong Seong Yoon,Jae Ho Chung&Hayoung Byun
Department of Emergency Medicine, College of Medicine, Hanyang University, 222 Wangsimni‑ro, Seongdong‑gu, Seoul, 04763, Republic of Korea
Jaehoon Oh
Department of Radiology and Research Institute of Radiological Science, Severance Hospital, Yonsei University College of Medicine, Seoul, Republic of Korea
Dong Jin Im
Department of Otolaryngology-Head and Neck Surgery, College of Medicine, Hanyang University, Seoul, Republic of Korea
Jae Ho Chung&Hayoung Byun
Department of HY, College of Medicine, KIST Bio-Convergence, Hanyang University, Seoul, Republic of Korea
Jae Ho Chung
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Contributions
Conceptualization, D.K.L. and J.O.; data curation, D.K.L., M.S.Y., D.J.I., and J.H.K.; formal analysis, J.H.K., T.H.K., D.K.L., and J.O.; methodology, T.H.K., J.O., J.H.C., and H.B.; supervision, J.H.C., H.B. and J.O.; validation, J.O. and T.H.K.; visualization, J.H.K. and D.K.L.; writing—original draft, D.K.L. and J.H.K.; writing—review and editing, J.O. and T.H.K. All authors have read and agreed to the published version of the manuscript.
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Correspondence to Jaehoon Oh or Tae Hyun Kim.
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The original online version of this Article was revised: The original version of this Article contained an error in the spelling of the author Jin Hyuk Kim which was incorrectly given as Kim Jin Hyuk. Consequently, the equal contribution statement was incorrect. Full information regarding the corrections made can be found in the correction for this article.
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Lee, D.K., Kim, J.H., Oh, J. et al. Detection of acute thoracic aortic dissection based on plain chest radiography and a residual neural network (Resnet). Sci Rep12, 21884 (2022).
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Auer, J., Berent, R. & Eber, B. Aortic dissection: Incidence, natural history and impact of surgery. J. Clin. Basic3(3), 151–154 (2000).
Google Scholar
Hiratzka, L. F. et al. ACCF/AHA/AATS/ACR/ASA/SCA/SCAI/SIR/STS/SVM guidelines for the diagnosis and management of patients with thoracic aortic disease. A report of the American college of cardiology foundation/American heart association task force on practice guidelines, American association for thoracic surgery, American college of radiology,American stroke association, society of csardiovascular anesthesiologists, society for cardiovascular angiography and interventions, society of interventional radiology, society of thoracic surgeons,and society for vascular medicine. J. Am. Coll. Cardiol.55, e27–e129. (2010).
ArticlePubMedGoogle Scholar
Fan, K. L. & Leung, L. P. Clinical profile of patients of acute aortic dissection presenting to the ED without chest pain. Am. J. Emerg. Med.35, 599–601. (2017).
ArticlePubMedGoogle Scholar
Klompas, M. Does this patient have an acute thoracic aortic dissection?. JAMA287, 2262–2272. (2002).
ArticlePubMedGoogle Scholar
von Kodolitsch, Y. et al. Chest radiography for the diagnosis of acute aortic syndrome. Am. J. Med.116, 73–77. (2004).
ArticleGoogle Scholar
Lakhani, P. & Sundaram, B. Deep learning at chest radiography: Automated classification of pulmonary tuberculosis by using convolutional neural networks. Radiology284, 574–582. (2017).
ArticlePubMedGoogle Scholar
Wu, J.-X. et al. Multilayer fractional-order machine vision classifier for rapid typical lung diseases screening on digital chest x-ray images. IEEE Access8, 105886–105902. (2020).
ArticleGoogle Scholar
Chaudhary, P. K. & Pachori, R. B. FBSED based automatic diagnosis of COVID-19 using X-ray and CT images. Comput. Biol. Med.134, 104454. (2021).
ArticleCASPubMedPubMed CentralGoogle Scholar
Sze-To, A., Riasatian, A. & Tizhoosh, H. R. Searching for pneumothorax in x-ray images using autoencoded deep features. Sci. Rep.11, 9817. (2021).
ArticleADSCASPubMedPubMed CentralGoogle Scholar
Hata, A. et al. Deep learning algorithm for detection of aortic dissection on non-contrast-enhanced CT. Eur. Radiol.31, 1151–1159 (2021).
ArticlePubMedGoogle Scholar
Cheng, J., Tian, S., Yu, L., Ma, X. & Xing, Y. A deep learning algorithm using contrast-enhanced computed tomography (CT) images for segmentation and rapid automatic detection of aortic dissection. Biomed. Signal Process. Control62, 102145 (2020).
ArticleGoogle Scholar
Cao, L. et al. Fully automatic segmentation of type B aortic dissection from CTA images enabled by deep learning. Eur. J. Radiol.121, 108713 (2019).
ArticlePubMedGoogle Scholar
Nienaber, C. A. et al. The diagnosis of thoracic aortic dissection by noninvasive imaging procedures. New England J. Med.328, 1–9 (1993).
ArticleCASGoogle Scholar
LA Pape et al (2015) Presentation, diagnosis, and outcomes of acute aortic dissection: 17 year trends from the international registry of acute aortic dissection J. Am. Coll. Cardiol. 66 350 358
Hirst JR, A. E., Johns JR, V. J. & Kime JR, S. W. Dissecting aneurysm of the aorta: a review of 505 cases. Medicine37(3), 217 (1958).
ArticleCASPubMedGoogle Scholar
Diederich, S. & Lenzen, H. J. C. Radiation exposure associated with imaging of the chest: Comparison of different radiographic and computed tomography techniques. Cancer89, 2457–2460 (2000).
ArticleCASPubMedGoogle Scholar
Jacob, K., Vivian, G. & Steel, J. R. X-ray dose training: Are we exposed to enough?. Clin. Radiol.59(10), 928–934 (2004).
ArticleCASPubMedGoogle Scholar
Mongan, J., Moy, L. & Kahn, C. E. Checklist for artificial intelligence in medical imaging (CLAIM): A guide for authors and reviewers. Radiol. Artif. Intell.2(2), 200029 (2020).
ArticleGoogle Scholar
He, K., Zhang, X., Ren, S., Sun, J. (2016).Deep Residual Learning for Image Recognition. 2016 IEEE Conference on Computer Vision and Pattern Recognition (CVPR).
Byun, H. et al. An assistive role of a machine learning network in diagnosis of middle ear diseases. J. Clin. Med.10, 3198 (2021).
ArticlePubMedPubMed CentralGoogle Scholar
Raschka, S. (2015) Python Machine Learning.
Kumari, S., Kumar, D. & Mittal, M. An ensemble approach for classification and prediction of diabetes mellitus using soft voting classifier. Int. J. Cognit. Comput. Eng.2, 40–46 (2021).
ArticleGoogle Scholar
Zhou, Z., Huang, H. & Fang, B. Application of weighted cross-entropy loss function in intrusion detection. J. Comput. Commun.09, 1–21. (2021).
ArticleCASGoogle Scholar
Damberg, A., Ziganshin, B. A. & Elefteriades, J. A. (2018) New approaches to aortic diseases from valve to abdominal bifurcation (eds Ion C. Ţintoiu et al.) 129–131 Academic Press
Goldstein, S. A. et al. Multimodality imaging of diseases of the thoracic aorta in adults: From the American society of echocardiography and the European association of cardiovascular imaging: Endorsed by the society of cardiovascular computed tomography and society for cardiovascular magnetic resonance. J. Am. Soc. Echocardiogr.28, 119–182 (2015).
ArticlePubMedGoogle Scholar
Dotter, C. T., Niles, N. R. & Steinberg, I. Impending aortic rupture, pathogenesis of x-ray signs. N. Engl. J. Med.265, 214–221. (1961).
ArticleCASPubMedGoogle Scholar
Nienaber, C. A. et al. The diagnosis of thoracic aortic dissection by noninvasive imaging procedures. N. Engl. J. Med.328, 1–9. (1993).
ArticleCASPubMedGoogle Scholar
Armstrong, W. F. et al. Clinical and echocardiographic findings in patients with suspected acute aortic dissection. Am. Heart J.136, 1051–1060. (1998).
ArticleCASPubMedGoogle Scholar
Hager, A. et al. Diameters of the thoracic aorta throughout life as measured with helical computed tomography. J. Thorac. Cardiovasc. Surg.123, 1060–1066. (2002).
ArticlePubMedGoogle Scholar
Kitasaka, T., Mori, K., Hasegawa, J.-I., Toriwaki, J.-I. & Katada, K. (2002) Medical Imaging 2002: Image Processing Vol. 4684 1496–1507.
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10231 | https://pubs.rsna.org/doi/abs/10.1148/radiol.2016161584 | PreviousNext
Original ResearchFree Access
Congenital Brain Abnormalities and Zika Virus: What the Radiologist Can Expect to See Prenatally and Postnatally
Patricia Soares de Oliveira-Szejnfeld,
Deborah Levine,
Adriana Suely de Oliveira Melo,
Melania Maria Ramos Amorim,
Alba Gean M. Batista,
Leila Chimelli,
Amilcar Tanuri,
Renato Santana Aguiar,
Gustavo Malinger,
Renato Ximenes,
Richard Robertson,
Jacob Szejnfeld,
Fernanda Tovar-Moll
Patricia Soares de Oliveira-Szejnfeld,
Deborah Levine,
Adriana Suely de Oliveira Melo,
Melania Maria Ramos Amorim,
Alba Gean M. Batista,
Leila Chimelli,
Amilcar Tanuri,
Renato Santana Aguiar,
Gustavo Malinger,
Renato Ximenes,
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… See all authors
Author Affiliations
From the Dept of Diagnostic Imaging, Federal Univ of São Paulo, São Paulo, Brazil (P.S.d.O.S., R.X., J.S.); Foundation Inst for Education and Research in Diagnostic Imaging, Dept of Diagnostic Imaging, Federal Univ of São Paulo, São Paulo, Brazil (P.S.d.O.S., J.S.); Dept of Radiology, Beth Israel Deaconess Medical Ctr, Harvard Medical School, Boston, Mass (D.L.); Instituto de Pesquisa Professor Amorim Neto, Campina Grande, PB, Brazil (A.S.d.O.M., M.M.R.A.); Instituto de Saúde Elpidio de Almeida, Campina Grande, PB, Brazil (A.S.d.O.M., M.M.R.A.); Faculdade de Ciências Médicas de Campina Grande, Campina Grande, PB, Brazil (A.S.d.O.M.); Hosp Municipal Pedro I, Serviço Municipal de Atendimento Transdisciplinar a Gestantes e Bebês com Infecção Congênita por Zika Virus, Campina Grande, PB, Brazil (A.S.d.O.M., A.G.M.B.); Universidade Federal de Campina Grande, PB, Brazil (M.M.R.A.); Laboratory of Neuropathology, State Inst of Brain, Rio de Janeiro, Brazil (L.C.); Departamento de Genética, Instituto de Biologia, Universidade Federal do Rio de Janeiro, Rio de Janeiro, Brazil (A.T., R.S.A.); Div of Ultrasound in Obstetrics & Gynecology, Lis Maternity Hosp, Tel Aviv Sourasky Medical Ctr, Sackler Faculty of Medicine, Tel Aviv Univ, Tel Aviv, Israel (G.M.); Fetal Medicine Foundation Latinamerica–FMFLA, Centrus–Fetal Medicine, Campinas, Brazil (R.X.); Boston Children’s Hosp, Boston, Mass (R.R.); Inst of Biomedical Sciences and National Ctr for Structural Biology and Bioimaging, Federal Univ of Rio de Janeiro, Rio de Janeiro, Brazil (F.T.M.); and D’Or Inst for Research and Education, Rua Diniz Cordeiro 30, Botafogo, Rio de Janeiro, RJ, Brazil 22881-100 (F.T.M.).
Address correspondence to F.T.M. (e-mail: Fernanda.tovarmoll@idor.org).
Patricia Soares de Oliveira-Szejnfeld
Deborah Levine
Adriana Suely de Oliveira Melo
Melania Maria Ramos Amorim
Alba Gean M. Batista
Leila Chimelli
Amilcar Tanuri
Renato Santana Aguiar
Gustavo Malinger
Renato Ximenes
Richard Robertson
Jacob Szejnfeld
Fernanda Tovar-Moll
Published Online:Aug 23 2016
Abstract
Purpose
To document the imaging findings associated with congenital Zika virus infection as found in the Instituto de Pesquisa in Campina Grande State Paraiba (IPESQ) in northeastern Brazil, where the congenital infection has been particularly severe.
Materials and Methods
From June 2015 to May 2016, 438 patients were referred to the IPESQ for rash occurring during pregnancy or for suspected fetal central nervous system abnormality. Patients who underwent imaging at IPESQ were included, as well as those with documented Zika virus infection in fluid or tissue (n = 17, confirmed infection cohort) or those with brain findings suspicious for Zika virus infection, with intracranial calcifications (n = 28, presumed infection cohort). Imaging examinations included 12 fetal magnetic resonance (MR) examinations, 42 postnatal brain computed tomographic examinations, and 11 postnatal brain MR examinations. Images were reviewed by four radiologists, with final opinion achieved by means of consensus.
Results
Brain abnormalities seen in confirmed (n = 17) and presumed (n = 28) congenital Zika virus infections were similar, with ventriculomegaly in 16 of 17 (94%) and 27 of 28 (96%) infections, respectively; abnormalities of the corpus callosum in 16 of 17 (94%) and 22 of 28 (78%) infections, respectively; and cortical migrational abnormalities in 16 of 17 (94%) and 28 of 28 (100%) infections, respectively. Although most fetuses underwent at least one examination that showed head circumference below the 5th percentile, head circumference could be normal in the presence of severe ventriculomegaly (seen in three fetuses). Intracranial calcifications were most commonly seen at the gray matter–white matter junction, in 15 of 17 (88%) and 28 of 28 (100%) confirmed and presumed infections, respectively. The basal ganglia and/or thalamus were also commonly involved with calcifications in 11 of 17 (65%) and 18 of 28 (64%) infections, respectively. The skull frequently had a collapsed appearance with overlapping sutures and redundant skin folds and, occasionally, intracranial herniation of orbital fat and clot in the confluence of sinuses.
Conclusion
The spectrum of findings associated with congenital Zika virus infection in the IPESQ in northeastern Brazil is illustrated to aid the radiologist in identifying Zika virus infection at imaging.
© RSNA, 2016
Online supplemental material is available for this article.
Introduction
Much has been written recently regarding Zika virus in pregnancy and the increased risk of microcephaly in fetuses exposed to the virus. The outbreak of infection in Brazil, especially in the northeast part of the country (1), has been of particular concern. The virus has been found in the fluids of pregnant mothers and during autopsy in the brains of neonates with microcephaly (2–5). Much of the concern in the media regarding the teratogenicity of Zika virus infection has focused on brain findings of microcephaly. However, as documented in many case series, there are a variety of brain abnormalities that can be found in fetuses exposed to intrauterine Zika virus infection (2–11). These include abnormalities in ventricular size, gray and white matter volume loss, brainstem abnormalities, and calcifications.
Although the current outbreak has centered in Brazil, according to the U.S. Centers for Disease Control and Prevention, there are currently 51 countries or territories in which active transmission of Zika virus has been reported (12). It is important for radiologists to understand the type of abnormalities associated with congenital Zika virus infection to aid in recognition of disease and appropriate counseling of patients. The purpose of this special report is to document the imaging findings associated with congenital Zika virus infection as found in patients seen at the Instituto de Pesquisa in Campina Grande State Paraiba (IPESQ) in northeastern Brazil.
Materials and Methods
This retrospective review includes imaging and autopsy data from an institutional review board–approved study that allowed for imaging and follow-up of presumed Zika virus infection in pregnant women and their neonates. Written informed consent was obtained from the pregnant women and/or the parents of newborns. From June 2015 to May 2016, 438 patients were referred to the IPESQ for one or more of the following: (a) pregnancy with rash, (b) fetal central nervous system (CNS) abnormalities at prenatal ultrasonography (US), and/or (c) postnatal microcephaly or other CNS malformation that was believed to be characteristic of congenital infection. During this period, 384 pregnant women with a rash or a history of CNS abnormality at US (group 1) and 47 neonates with postnatal physical examination findings suggestive of microcephaly (head circumference < 33 cm) (group 2) were enrolled in this protocol. Group 1 exclusion criteria included not returning for imaging at IPESQ, no CNS abnormality identified at US examination at IPESQ, classic findings of isolated CNS abnormality not characteristic of infection, diagnosis of a genetic syndrome or aneuploidy, and lack of postnatal images for review. Group 2 exclusion criteria included not returning for computed tomography (CT) or magnetic resonance (MR) imaging examination at IPESQ and no calcification on postnatal images. Initially, we used a head circumference criterion of 32.5 cm for microcephaly as an exclusion criterion for the postnatal diagnosis, but after review of the imaging findings, there was a newborn with normal head circumference but severe ventriculomegaly and calcifications similar to those seen in the other confirmed and presumed Zika virus infections; therefore, this neonate was included in our population with presumed Zika virus infection. After exclusions, there were imaging studies of 31 fetuses in 30 pregnant women (which included one set of twins) and 45 neonates (which included the 31 fetuses that underwent prenatal imaging, as well as an additional 14 neonates that were enrolled postnatally; Fig 1).
Results in 10 of these patients have been reported previously (4,5); however, these prior publications were not of sufficiently large sample size to provide estimates of the types of abnormalities seen in infected neonates, and the images were not formally reviewed by multiple fetal and neonatal imaging experts.
Prenatal and/or postnatal Zika virus infection investigations were performed on the patients’ fluids (blood, urine, amniotic fluid, and/or cord blood) or tissues (placenta or brain and other organ tissue at autopsy), as described previously (4). In the group with confirmed infection, viral infection was confirmed by means of serologic findings or reverse transcription polymerase chain reaction (RT-PCR). An assay (Rapid Test DPP Zika IgM/IgG Assay; Chembio, Medford, NY) was used to detect immunoglobulin G and immunoglobulin M against Zika virus in the fluids. Virus genome was identified in fluids and/or tissues with RT-PCR by using specific primers and probes to identify the region of Zika virus (13). Dengue virus and chikungunya virus infection were excluded by using enzyme-linked immunosorbent assay and RT-PCR. The patients tested negative for toxoplasmosis, syphilis, varicella-zoster virus, Parvovirus B19, rubella, Cytomegalovirus, and herpes virus (TORCH) infection, as well as human immunodeficiency virus.
As described previously (4,5), prenatal US was performed by fetal medicine specialists using either a Voluson E8 unit (General Electric, Milwaukee, Wis) with transvaginal probes or a Samsung XG or WS80 unit (Samsung, Seoul, South Korea) with 2–9-MHz probes. MR imaging of the fetus was performed with a 3-T Skyra unit (Siemens Healthcare, Erlangen, Germany) or a 1.5-T Espree unit (Siemens Healthcare) with an eight-channel body coil and standard acquisition protocols (Table E1 [online]). Postnatal head CT was performed with a 16-section CT scanner (Siemens Healthcare). Postnatal MR imaging was performed with a 1.5-T Espree brain MR imaging unit (Siemens Healthcare). Postmortem brain examination was performed in stillborns or neonates who died within 48 hours of birth after obtaining parental consent to perform autopsy. Brain tissue images were acquired with a 64-channel multisection CT scanner (GE Healthcare) and a 3-T MR imaging unit (Achieva; Philips, Best, the Netherlands).
Neurological findings on prenatal MR images and postnatal CT and MR images were subjectively described by four radiologists with experience in fetal and/or neonatal neuroradiology (D.L., an obstetric and fetal MR imager with 20 years of experience; R.R. and F.T.M., pediatric neuroradiologists with 20 and 16 years of experience, respectively; and P.S.d.O.S., a neuroradiologist with 12 years of experience). Findings were agreed on in consensus. In a minority of cases, select images from obstetric US were also available for review.
Results
In the 16 women with confirmed infection (including one set of twins, yielding 17 fetuses and/or neonates), testing for Zika virus by means of RT-PCR and/or serologic analysis generated positive results in amniotic fluid in nine women (with 10 gestational sacs), cord blood in seven women, neonatal brain during autopsy in three neonates, and placenta in one neonate, with multiple positive sites in two women (one of whom had twins; Table E2 [online]). A rash characteristic of Zika virus infection was present in the first trimester in 13 of 16 women (81%) with confirmed infection (one with twins) and in 22 of 28 women (78%) with presumed infection.
The imaging findings are described in Table 1. Figures depicting fetal anatomy (Figs 2–8), neonatal microcephaly (Figs 9–15; Figs E1–E3 [online]), and one neonate with normal head circumference but with characteristic calcifications in association with severe ventriculomegaly (Fig 16) are illustrated. Movie clips of sequences of images are also available in Movies 1–E5 (online).
Movie 1. Axial T2-weighted fetal MR imaging of the brain.
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Movie 2. Postnatal CT.
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Movie 3. Postnatal coronal T2-weighted MR imaging.
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Movie 4. Axial T2-weighted fetal MR imaging.
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Movie 5. Postnatal CT performed at 1 day of age.
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Table 1 Imaging Findings in 17 Confirmed and 28 Presumed Zika Virus Infections
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Note.—Numbers in parentheses are percentages.
In all but one fetus with confirmed Zika virus infection at prenatal imaging, the head circumference percentile was at or below the 5th percentile in at least one US examination performed during the second trimester of pregnancy (Table 2). However, one fetus with severe parenchymal and brainstem malformation had severe ventriculomegaly and normal head circumference at a scan conducted at 17 weeks of gestational age, and the circumference remained within the normal range later in pregnancy. In 23 of 26 fetuses that underwent serial prenatal US, head circumference remained under the 5th percentile until birth, which led to a diagnosis of microcephaly at birth. However, it is notable that the three fetuses with head circumference in the normal range at birth showed severe ventriculomegaly, which we presume was due to the enlarged, obstructed ventricles. For this reason, we included Figure 16, which depicts a neonate referred to IPESQ for potential Zika virus infection, but the head circumference was 38 cm at birth. The CT images showed calcifications in the subcortical region, thalamus, basal ganglia, and brainstem. No sulci were seen; however, the parenchyma was extremely thin. There was pontocerebellar hypoplasia and Dandy-Walker spectrum anomaly, and the corpus callosum was not visualized.
Table 2 Head Circumference Percentiles with Respect to Gestational Age in Fetuses with Prenatal US Images
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Note.—Data are percentages, unless indicated otherwise. If more than one examination was conducted in a given time period, measurements from the first examination were used.
The most remarkable change in the brain parenchyma, present on all neonatal images, was the reduction in parenchymal volume. Abnormalities of cortical development associated with volume changes were observed in 16 of 17 confirmed infections (94%) and 28 of 28 presumed infections (100%). Abnormalities of the corpus callosum were present in 16 of 17 confirmed infections (94%) and 22 of 28 presumed infections (78%).
Lateral ventricles were enlarged in 16 of 17 confirmed infections (94%) and 27 of 28 presumed infections (96%). This was asymmetrical in six of 17 confirmed infections and five of 28 presumed infections. Despite ventriculomegaly, the extra-axial spaces are frequently still prominent because of cortical underdevelopment or atrophy. Other findings associated with the ventricles were septations in the ventricle (typically in the occipital horns), which were frequently difficult to distinguish from subventricular cysts. Subependymal cysts were occasionally visualized.
Abnormalities of cortical development were present in all patients but with a substantial variation regarding the type of abnormality, hemispheric symmetry, and severity (Table 1). The most common finding was irregular areas of sulci and/or gyri not otherwise specified, but focal cortical malformation was also observed. In addition, six patients had the appearance of lissencephaly. In three fetuses, this was diagnosed prenatally. Of the other three, two did not have a fetal sonogram for review, and one only underwent a scan at 10 weeks of gestation, which is too early to assign this diagnosis. The cortical development abnormalities were usually asymmetrical. In general, the sulci were less prominent, and wide sylvian and interhemispheric fissures were identified in most neonates, as well as abnormal myelination.
Calcification regions were predominantly located in the gray matter–white matter junction in our series (88% in the confirmed infection cohort and 100% in the presumed infection cohort). Calcifications were also identified in the thalamus, basal ganglia, cortex, and periventricular regions. It is important to mention that the latter were only present in neonates where there was substantial thinning of the brain parenchyma; thus, the precise location of calcifications was difficult to determine. Although less common, infratentorial calcifications were also identified. However, these were usually present in more severe manifestations of infection, being associated with dysmorphic brainstem, stenosis of the aqueduct, and secondary supratentorial hydrocephaly. Calcification in the brainstem was a common finding at autopsy (in three of three of the neonates with confirmed infection that underwent autopsy).
Abnormalities of the brainstem were identified. The pons was often thin and atrophic. There was frequently a kink seen at the pontomedullary junction. The spinal cord was thinned and at times irregular in its appearance. Other posterior fossa abnormalities included cerebellar hemisphere hypoplasia, vermis hypoplasia, and elevation of the vermis, associated with an enlarged cisterna magna.
Abnormalities of the corpus callosum—usually thin, dysgenetic, and hypoplastic or even absent—were frequently observed. Other changes included under-rotation of the hippocampus and thickened fornices.
In some imaging studies, an enlarged confluence of the dural venous sinuses had heterogeneous material. In a few fetal sonograms that were available for review, this was demonstrated to be blood clot (Fig 3g). In many postnatal CT studies, there was hyperattenuating material in this region, which could be either thrombus or hematocrit effect (due to dehydration with hemoconcentration). In many fetal and neonatal MR studies, there was fluid posterior to the confluence of the sinuses that was either similar to cerebrospinal fluid in attenuation or signal intensity or contained fluid with a higher protein content than that of cerebrospinal fluid. The location of the fluid collection is likely related to the unusual head shape and overlapping sutures. This abnormal head shape was frequently associated with redundant skin folds.
Other findings include orbital abnormalities, such as asymmetrical micropthalmia, cataracts, and herniation of the orbital fat into the cranial vault. Body abnormalities included arthrogryposis.
Discussion
Zika virus is a single-strand RNA Flavivirus (1). It is transmitted by infected female mosquito vectors, such as the Aedes aegypti mosquito. Diagnosis of Zika virus infection is complicated by the fact that it is asymptomatic in up to 80% of infections (14). The common symptoms tend to be mild and nonspecific, including headache, fever, and rash. Other symptoms include conjunctivitis, and, in rare instances, Guillain-Barré syndrome (15).
It is well recognized that Zika virus crosses the fetal-placental barrier. Zika virus has been isolated from the brain and cerebrospinal fluid of neonates born with congenital microcephaly and identified in the amniotic fluid and placental tissue of mothers who had experienced clinical symptoms consistent with Zika virus infection during their pregnancies (2–5,8,15,16). Zika virus has also been shown to lead to neurotoxiticity and to impair human neurosphere growth in experimental models (17).
Microcephaly is a nonspecific term that refers to a head circumference smaller than normal for gestational age. There are many causes of microcephaly, the most common being infections (such as TORCH infections and human immunodeficiency virus), teratogens (including maternal exposure to heavy metals, alcohol, and radiation), genetic abnormalities and syndromes, and growth restriction. In the case of Zika virus, it is clear that there are developmental insults that lead to micrencephaly (small brain) and associated microcephaly (small head) (6,18–21). It is important to recognize that almost all of the infections at our institution occurred in women who had a characteristic rash in the late first trimester. This correlates well with the finding of severe cerebral dysmorphisms associated with infection during a time of rapid brain development. According to the U.S. Centers for Disease Control and Prevention, the risk of microcephaly after maternal infection with Zika virus in the first trimester of pregnancy is 1%–13% (22). However, as the flow of women referred to the IPESQ for assessment demonstrates, it is clear that many pregnant women with a rash in Brazil are never shown to have congenitally infected fetuses, although some certainly could have less severe infection that has thus far been undiagnosed. For women who have neonates with findings suggestive of severe microcephaly but who have a history of rash or for those who had a rash in the third trimester, we can hypothesize that there was an unrecognized or asymptomatic first-trimester exposure and/or infection.
As of July 2, 2016, the Brazilian Ministry of Health had been notified of 8301 cases of microcephaly and confirmed 1656 infections (23). In a recent study, França et al reported on the follow-up of 1501 cases in which Zika virus infection was suspected, of which 602 were deemed to be definitely or probably due to Zika virus (76 definite infections, 54 highly probable infections, 181 moderately probable infections, and 291 somewhat probable infections) (24). The incidence of confirmed infection in our study with respect to referral population is similar to what has been seen in the larger Brazilian population. In our study, of 432 women initially screened, only findings in 44 patients are reported here after applying our exclusion criteria. In the report by França et al, one in five definite or probable Zika virus infections yielded head circumferences in the normal range (more than −2 standard deviations below the median of the International Fetal and Newborn Growth Consortium for the 21st Century, or INTERGROWTH-21st, standard), and for one-third of definite and probable infections, there was no history of a rash during pregnancy (24). This suggests that our series is biased to the more severe infections. However, in the series by França et al (24), the more severe infections were localized in the northeast region of Brazil, since 97% of definite or probable infections were from the northeast region, where 28% of all births in Brazil occur. This suggests that there could be additional unknown factors that exacerbate the fetal infection in this region. Coinfections, in addition to those already excluded, as well other environmental factors, will need to be explored further.
There are many nomograms for head circumference size. Current guidance in Brazil is to use the standards from the INTERGROWTH-21st study for fetuses (25), the INTERGROWTH-21st study for infants (25), and World Health Organization criteria for full-term neonates (26). These charts show that a head circumference of 32 cm is at about −2 standard deviations below the mean for both boys and girls at term. However, this threshold will naturally include some normally developing neonates and also be inaccurate for neonates born prematurely. However, it should be recognized that to cast a broad net to find neonates with congenital Zika virus infection, we must bear in mind that not all neonates will have microcephaly at birth.
The striking imaging features of the severe micrencephaly associated with Zika virus include a markedly abnormal head shape. The unusual appearance of the skull, we hypothesize, is due to a combination of the small brain as it develops and a result of what, at some point, was likely a larger head size (due to ventriculomegaly) that then decompresses. Cerebral atrophy may also contribute, giving the skull the collapsed shape with everted and/or cupped sutures and overriding bones in the occipital region, causing redundant and folded skin. In part, this is also likely due to the head and skin continuing to grow, while the size of the brain regresses. However, in some fetuses and/or neonates, the ventricle and/or brain atrophy has not yet occurred, and in these instances, a normal (or even increased) head circumference may be present.
Another unusual finding that suggests skull collapse is that neonates have orbital fat herniation into the cranial vault. Thus, some of the ocular findings could be secondary to the process of skull deformation itself rather than direct infection of the eye. For example, nerve and blood flow interruption could be due to herniated tissue.
We used brain calcifications as inclusion criteria for the postnatal assessment to exclude microcephaly from causes other than infection, such as unrecognized prematurity or congenital syndromes. While this could have led to exclusion of some infections without intracranial calcifications, it led to a homogeneous group of neonates with strikingly similar parenchymal abnormalities. In our series, the most common location for calcifications was the gray matter–white matter junction (88% in the confirmed infection cohort and 100% in the presumed infection cohort), which is an area not classically or commonly targeted in other congenital infections. The location of the calcifications at the gray matter–white matter interface could suggest a vascular component to the infection, as other processes that preferentially affect the gray matter–white matter junction have been posited to be due to changes in arterial configuration from straight vessels in the cortex to coiled vessels in the subcortical white matter (27).
In the classic TORCH infections, the brain calcifications are periventricular and cortical, although rare cases of basal ganglia and thalamus calcifications have been reported (28,29). Other findings include intraventricular adhesions, callosal abnormalities, periventricular pseudocysts, sulcation, and gyral abnormalities (29–31), similar to what we describe in this report.
However, unlike most patients with congenital Cytomegalovirus, the patients with documented or presumed Zika virus infection described in this report had severe microcephaly. This may be due to the first-trimester nature of most of the infections reported herein. It could also be due to the viral load in the Zika virus infections, which we assume are severe infections. It could be that congenital Cytomegalovirus is diagnosed in a range of infections from mild to severe, whereas we may be focusing our results on the severe Zika virus infections. In our cohort, almost all fetuses and neonates had dramatically abnormal cerebral volume, abnormal cortical folding pattern, and/or regions of lissencephaly, pachygyria, and/or polymicrogyria. We hypothesize that the cortical abnormalities visualized are due at least in part to arrested cortical development at various stages.
There have been many reports of small series of imaging findings in fetuses and neonates with congenital Zika virus infection. In 2016, Mlakar et al described one pregnancy at 29 weeks with microcephaly and intracranial calcifications, with an earlier, second-trimester sonogram that showed no abnormality (6). Also in 2016, Calvet et al described two pregnant women who underwent US at 22 weeks, which showed microcephaly (2). Sarno et al described a stillbirth at 32 weeks, with microcephaly, intracranial calcifications, and fetal hydrops (7). Driggers et al in 2016 described decreased fetal head circumference between 16 and 21 weeks, with brain abnormalities (8). In a study by Brasil et al (3), of 88 pregnant women with rash, 72 tested positive for Zika virus. Fetal US was performed in 42 fetuses, and abnormalities were seen in 12, including intrauterine growth restriction with or without microcephaly and ventricular calcifications (3).
Schuler-Faccini et al described 35 neonates with microcephaly, including brain calcifications, ventriculomgaly, and cortical and/or subcortical atrophy (9). In 2016, Hazin et al (10) and de Fatima Vasco Aragao et al (11) each described 23 neonates with microcephaly who had CT findings that included intracranial calcifications, ventriculomegaly, abnormal gryal pattern, and abnormal white matter attenuation. Guillemette-Artur et al described three neonates with congenital Zika virus infection, with micrencephaly in all three, small cerebellum in two, occipital subependymal pseudocysts in two, polymicrogyria in three, corpus callosum abnormalities in two, and hypoplastic brainstem in one (32). These findings are all similar to what we report.
Our study had limitations. What we present here is a convenience sample of imaging findings for illustrative purposes. Our cohort was obtained from a referral center for high-risk pregnancy. Thus, we have no information on incidence of the Zika virus in the general population or risk estimates for transmission to the fetus. Because of the manner in which we accrued subjects, neonates with congenital infection but with normal head size or brain abnormalities without calcifications could have been missed. In addition, it could be that some of the cohort had disease origin for microcephaly other than Zika virus. For example, we excluded a neonate with microcephaly and confirmed Zika virus infection due to aneuploidy with trisomy 13 syndrome. However, other infections or syndromes could be present but not yet identified in either our confirmed infection cohort or our presumed infection cohort. Findings on MR images can lead to underestimation of the incidence of calcifications, and evaluation of CT images makes characterization of subtle parenchymal abnormalities and corpus callosum abnormalities difficult. Further imaging studies on these neonates as they grow will be helpful in further assessment of areas involved with the infection. Finally, we focused on brain findings in this review. Additional sites of infection and associated pathologic abnormalities will likely be identified in the future.
It is well recognized that transplacental transmission of viruses, even in subclinical maternal infection, can lead to severe congenital abnormalities. As in other infections, serial imaging can demonstrate evolution of findings. Prenatal sonograms may show normal or decreased head circumference and, rarely, increased head circumference. Almost all neonates will show intraparenchymal calcifications more severe than what are typically seen in TORCH infections and frequently occur at the gray matter–white matter junction, which is an unusual location for the calcifications of other congenital infections. We hope the illustrations of these many fetuses and neonates will aid others in the event that the unfortunate epidemic of congenital Zika virus continues.
Disclosures of Conflicts of Interest: P.S.d.O.S. disclosed no relevant relationships. D.L. Activities related to the present article: disclosed no relevant relationships. Activities not related to the present article: author received royalties from Elsevier and UpToDate. Other relationships: disclosed no relevant relationships. A.S.d.O.M. disclosed no relevant relationships. M.M.R.A. disclosed no relevant relationships. A.G.M.B. disclosed no relevant relationships. L.C. disclosed no relevant relationships. A.T. disclosed no relevant relationships. R.S.A. disclosed no relevant relationships. G.M. disclosed no relevant relationships. R.X. disclosed no relevant relationships. R.R. disclosed no relevant relationships. J.S. disclosed no relevant relationships. F.T.M. disclosed no relevant relationships.
Acknowledgments
We gratefully acknowledge the members of the Brazilian Network BRAZIKA (Rede Internacional de Estudos Sobre Zika no Brasil).
Author Contributions
Author contributions: Guarantors of integrity of entire study, P.S.d.O.S., D.L., A.S.d.O.M., A.G.M.B., A.T., R.X., F.T.M.; study concepts/study design or data acquisition or data analysis/interpretation, all authors; manuscript drafting or manuscript revision for important intellectual content, all authors; approval of final version of submitted manuscript, all authors; agrees to ensure any questions related to the work are appropriately resolved, all authors; literature research, P.S.d.O.S., D.L., A.S.d.O.M., M.M.R.A., L.C., R.S.A., R.X., R.R., F.T.M.; clinical studies, P.S.d.O.S., A.S.d.O.M., A.G.M.B., L.C., G.M., R.X., R.R., F.T.M.; experimental studies, A.T., R.S.A., F.T.M.; statistical analysis, P.S.d.O.S., D.L., J.S., F.T.M.; and manuscript editing, P.S.d.O.S., D.L., A.S.d.O.M., L.C., R.S.A., G.M., R.X., R.R., J.S., F.T.M.
Article History
Received July 7, 2016; revision requested July 14; revision received July 22; accepted July 27; final version accepted August 3.
Published online: Aug 23 2016
Published in print: Oct 2016
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LeilaChimelli
2021Jan1
+ Hearing and Zika virus infection
Maria Helena de M.Barbosa, Maria Clarade Magalhães-Barbosa, ArnaldoPrata-Barbosa
2021Jan1
+ Urological sequels in the scope of the Congenital Zika Syndrome
Lucia MariaCosta Monteiro
2021Jan1
+ Congenital Zika syndrome: A systematic review
Danielle A.Freitas, ReinaldoSouza-Santos, Liege M. A.Carvalho, Wagner B.Barros, Luiza M.Neves, PatríciaBrasil, Mayumi D.Wakimoto, KazumichiFujioka
15 December 2020 | PLOS ONE, Vol. 15, No. 12
+ The Latin American network for congenital malformation surveillance:
ReLAMC
Iêda MariaOrioli, HelenDolk, JorgeLopez‐Camelo, BorisGroisman, AdrianaBenavides‐Lara, Lucas GabrielGimenez, Daniel MattosCorrea, MartaAscurra, Elianade Aquino Bonilha, Maria AuroraCanessa‐Tapia, Giovanny Vinícius Araújode França, PaulaHurtado‐Villa, MarisolIbarra‐Ramírez, RosaPardo, Dania MariaPastora, IgnacioZarante, Flávia SchneiderSoares, Flávia Martinezde Carvalho, MarianaPiola,
14 December 2020 | American Journal of Medical Genetics Part C: Seminars in Medical Genetics, Vol. 184, No. 4
+ Obstetric and perinatal outcomes in cases of congenital Zika syndrome
AdrianaMelo, Jousilenede Sales Tavares, Mariannyde Assis Costa, RenatoSantana de Aguiar, GustavoMalinger, Fabianade Oliveira Melo, MarianaBalbino da‐Silva, BrunoLuiz Fonseca Schamber‐Reis, GabrielaGama, AmilcarTanuri, LeilaChimelli, PatriciaOliveira‐Szejnfeld, MelaniaM. Ramos de Amorim
27 October 2020 | Prenatal Diagnosis, Vol. 40, No. 13
+ Contemporary Understanding of Ebola and Zika Virus in Pregnancy
LaurenSayres, Brenna L.Hughes
2020Dec1 | Clinics in Perinatology, Vol. 47, No. 4
+ Multiscale 3D phenotyping of human cerebral organoids
AlexandreAlbanese, Justin M.Swaney, Dae HeeYun, Nicholas B.Evans, Jenna M.Antonucci, SilviaVelasco, Chang HoSohn, PaolaArlotta, LeeGehrke, KwanghunChung
8 December 2020 | Scientific Reports, Vol. 10, No. 1
+ Epilepsy surveillance in normocephalic children with and without prenatal Zika virus exposure
KarenBlackmon, RandallWaechter, BarbaraLandon, TrevorNoël, CalumMacpherson, TyhiesiaDonald, NikitaCudjoe, RobertaEvans, Kemi S.Burgen, PiumiJayatilake, VivianOyegunle, OttoPedraza, SamahAbdel Baki, ThomasThesen, DennisDlugos, GeethaChari, Archana A.Patel, Elysse N.Grossi-Soyster, Amy R.Krystosik, A. DesireeLaBeaud, EugeniaCorrales-Aguilar
30 November 2020 | PLOS Neglected Tropical Diseases, Vol. 14, No. 11
+ Two Sides of a Coin: a Zika Virus Mutation Selected in Pregnant Rhesus Macaques Promotes Fetal Infection in Mice but at a Cost of Reduced Fitness in Nonpregnant Macaques and Diminished Transmissibility by Vectors
DaniloLemos, Jackson B.Stuart, WilliamLouie, AnilSingapuri, Ana L.Ramírez, JenniferWatanabe, JodieUsachenko, Rebekah I.Keesler, ClaudiaSanchez-San Martin, TonyLi, CallaMartyn, GlennOliveira, SharadaSaraf, Nathan D.Grubaugh, Kristian G.Andersen, JamesThissen, JonathanAllen, MonicaBorucki, Konstantin A.Tsetsarkin, Alexander G.Pletnev, Charles Y.Chiu, Koen K. A.Van Rompay, Lark L.Coffey, Colin R.Parrish
2020Nov23 | Journal of Virology, Vol. 94, No. 24
+ Cellular Imprinting Proteomics Assay: A Novel Method for Detection of Neural and Ocular Disorders Applied to Congenital Zika Virus Syndrome
LiviaRosa-Fernandes, Raquel HoraBarbosa, Maria Luiza B.dos Santos, Claudia B.Angeli, Thiago P.Silva, Rossana C. N.Melo, Gilberto Santosde Oliveira, BernardoLemos, Jennifer EVan Eyk, Martin R.Larsen, Claudete AraújoCardoso, GiuseppePalmisano
19 July 2020 | Journal of Proteome Research, Vol. 19, No. 11
+ Microcefalia e alterações do sistema nervoso central relacionadas à infecção congênita pelo vírus Zika e outras etiologias infecciosas no estado do Rio de Janeiro: estudo transversal, 2015 a 2017
Claudia CaminhaEscosteguy, RenataEscosteguy Medronho, Renata CoelhoRodrigues, Lucas Dalsenter Romanoda Silva, Bruna Andradede Oliveira, Fernanda BeatrizMachado, Yuri SousaCosta, Silvia Cristinade Carvalho Cardoso, Antonio José Ledo Alvesda Cunha, Robertode Andrade Medronho
2020Nov2 | Revista Panamericana de Salud Pública, Vol. 44
+ Congenital Zika Virus Infection: a Review with Emphasis on the Spectrum of Brain Abnormalities
LeãoVHP, MMAragão, RSPinho, ANHazin, ARPaciorkowski, ACPenalva de Oliveira, Marcelo RodriguesMasruha
3 September 2020 | Current Neurology and Neuroscience Reports, Vol. 20, No. 11
+ White matter calcifications in infants: not always STORCH
Felipe AlbaSCORTEGAGNA, Renato HoffmannNUNES, Luiz CelsoVILANOVA, Felipe TorresPACHECO, Antonio Jose daROCHA
1 November 2020 | Arquivos de Neuro-Psiquiatria, Vol. 78, No. 11
+ Intracranial calcifications in childhood: Part 1
Fabrício GuimarãesGonçalves, LucaCaschera, Sara ReisTeixeira, Angela NicoleViaene, LorenzoPinelli, KshitijMankad, César Augusto Pinheiro FerreiraAlves, Xilma RosaOrtiz-Gonzalez, SavvasAndronikou, ArastooVossough
30 July 2020 | Pediatric Radiology, Vol. 50, No. 10
+ Association between brain morphology and electrophysiological features in Congenital Zika Virus Syndrome: A cross-sectional, observational study
Eduardo BSequerra, Antonio JRocha, Galtieri O Cde Medeiros, Manuel MNeto, Claudia R SMaia, Nívia M RArrais, MylenaBezerra, Selma M BJeronimo, Allan KardecBarros, Patrícia SSousa, AureaNogueira de Melo, Claudio MQueiroz
2020Sep1 | EClinicalMedicine, Vol. 26
+ ISUOG
Practice Guidelines: role of ultrasound in congenital infection
A.Khalil, A.Sotiriadis, R.Chaoui, F.da Silva Costa, F.D'Antonio, P.T.Heath, C.Jones, G.Malinger, A.Odibo, F.Prefumo, L. J.Salomon, S.Wood, Y.Ville
13 May 2020 | Ultrasound in Obstetrics & Gynecology, Vol. 56, No. 1
+ Analysis of the altmetric top 100 articles with the highest altmetric attention scores in medical imaging journals
Ji YoonMoon, Eun JooYun, Dae YoungYoon, Young LanSeo, Young KwonCho, Kyoung JaLim, Ji HyunHong
9 March 2020 | Japanese Journal of Radiology, Vol. 38, No. 7
+ Ultrasound prediction of Zika virus-associated congenital injury using the profile of fetal growth
Christie L.Walker, NoahEhinger, BrittneyMason, ElizabethOler, Marie-Térèse E.Little, Eric O.Ohuma, Aris T.Papageorghiou, UnzilaNayeri, ChristineCurry, Kristina M.Adams Waldorf, IrinaBurd
13 May 2020 | PLOS ONE, Vol. 15, No. 5
+ Neurological outcomes of congenital Zika syndrome in toddlers and preschoolers: a case series
Heloisa Viscaino Fernandes SouzaPereira, Stella Pintodos Santos, Ana Paula Rodrigues LazzariAmncio, Patricia Soaresde Oliveira-Szejnfeld, Edneusa OliveiraFlor, Jousilenede Sales Tavares, Rayssa Vieira BrandãoFerreira, FernandaTovar-Moll, Melania Maria Ramosde Amorim, AdrianaMelo
2020May1 | The Lancet Child & Adolescent Health, Vol. 4, No. 5
+ Role of Prenatal Ultrasonography and Amniocentesis in the Diagnosis of Congenital Zika Syndrome
Laura J.Viens, ShannonFleck-Derderian, Madelyn A.Baez-Santiago, TitilopeOduyebo, Cheryl S.Broussard, SumaiyaKhan, Abbey M.Jones, DanaMeaney-Delman
2020May1 | Obstetrics & Gynecology, Vol. 135, No. 5
+ ZIKV Diagnostics: Current Scenario and Future Directions
ZiltonVasconcelos, RenataCampos Azevedo, AndreaZin, LuizaNeves, DanielaPrado Cunha
29 April 2020
+ Review of neuroimaging findings in congenital Zika virus syndrome and its relation to the time of infection
GracianeRadaelli, MagdaLahorgue Nunes, RicardoBernardi Soder, Júlia Monteirode Oliveira, FernandaThays Konat Bruzzo, FelipeKalil Neto, EduardoLeal-Conceição, MirnaWetters Portuguez, JadersonCosta da Costa
2 January 2020 | The Neuroradiology Journal, Vol. 33, No. 2
+ Congenital Zika syndrome associated with findings of cerebellar cortical dysplasia - Broadening the spectrum of presentation of the syndrome
Bruno Niemeyer de FreitasRibeiro, EdsonMarchiori
2020Mar1 | Journal of Neuroradiology, Vol. 47, No. 2
+ Longitudinal visual acuity development in ZIKV-exposed children
Diego da SilvaLima, Luiz Claudio PortnoiBaran, Russell DavidHamer, Marcelo Fernandes daCosta, Kallene SummerVidal, Francisco MaxDamico, Mirella Telles SalgueiroBarboni, Valtenice de Cássia Rodrigues de MatosFrança, Cristiane Maria GomesMartins, Heydi SegundoTabares, Sarah LeonardoDias, Leonardo AparecidoSilva, DiegoDecleva, MayanaZatz, Ana Paula Antunes PascalicchioBertozzi, Rosa EstelaGazeta, Saulo DuartePassos, Dora FixVentura
2020Feb1 | Journal of American Association for Pediatric Ophthalmology and Strabismus, Vol. 24, No. 1
+ Early Gross Motor Development Among Brazilian Children with Microcephaly Born Right After Zika Virus Infection Outbreak
PalomaA. Ventura, Maria-LúciaC. Lage, AlessandraL. de Carvalho, AdrianaS. Fernandes, TniaB. Taguchi, Cristiana M.Nascimento-Carvalho
2020Feb1 | Journal of Developmental & Behavioral Pediatrics, Vol. 41, No. 2
+ Motor function in children with congenital Zika syndrome
AdrianaMelo, Gabriela LGama, Renan ADa Silva Júnior, Paula LDe Assunção, Jousilene STavares, Mariana BDa Silva, Kamila N F SCosta, Mell LVnia, Morgana AEvangelista, Melania M RDe Amorim
4 April 2019 | Developmental Medicine & Child Neurology, Vol. 62, No. 2
+ Análise do conceito síndrome congênita pelo Zika vírus
Gracimary AlvesTeixeira, Dndara Nayara AzevêdoDantas, Gleyce Any Freire de LimaCarvalho, Aylla Nauana daSilva, Ana Luísa Brandão de CarvalhoLira, Bertha CruzEnders
1 February 2020 | Ciência & Saúde Coletiva, Vol. 25, No. 2
+ Flaviviruses 3
LeilaChimelli
10 January 2020
+ Congenital Zika syndrome is associated with maternal protein malnutrition
J.Barbeito-Andrés, P.Pezzuto, L. M.Higa, A. A.Dias, J. M.Vasconcelos, T. M. P.Santos, J. C. C. G.Ferreira, R. O.Ferreira, F. F.Dutra, A. D.Rossi, R. V.Barbosa, C. K. N.Amorim, M. P. C.De Souza, L.Chimelli, R. S.Aguiar, P. N.Gonzalez, F. A.Lara, M. C.Castro, Z.Molnár, R. T.Lopes, M. T.Bozza, J. L. S. G.Vianez, C. G.Barbeito, P.Cuervo, M.Bellio, A.Tanuri, P. P.Garcez
2020Jan10 | Science Advances, Vol. 6, No. 2
+ Neurologic infections during pregnancy
Angela M.Curcio, PriyankaShekhawat, Alexandra S.Reynolds, Kiran T.Thakur
2020Jan1
+ Sonography of the Fetal Central Nervous System
LucDe Catte, BartDe Keersmaecker, LucJoyeux, MichaelAertsen
2020Jan1
+ Evaluation of the frequency of neuroimaging findings in congenital infection by Zika virus and differences between computed tomography and magnetic resonance imaging in the detection of alterations
Bruno Niemeyer de FreitasRibeiro, Bernardo CarvalhoMuniz, EdsonMarchiori
1 January 2020 | Revista da Sociedade Brasileira de Medicina Tropical, Vol. 53
+ Congenital Zika syndrome: association between the gestational trimester of maternal infection, severity of brain computed tomography findings and microcephaly at birth
Ana Karolina TorresMendes, Marizélia Rodrigues CostaRibeiro, FernandoLamy-Filho, Gláucio AndradeAmaral, Marcella Costa RibeiroBorges, Luciana CavalcanteCosta, Tamires BarradasCavalcante, Rosngela Fernandes LucenaBatista, Patrícia da SilvaSousa, Antônio Augusto Moura daSilva
1 January 2020 | Revista do Instituto de Medicina Tropical de São Paulo, Vol. 62
+ Viral Infections and the Neonatal Brain
Linda S.de Vries
2019Dec1 | Seminars in Pediatric Neurology, Vol. 32
+ Zika virus infection: A correlation between prenatal ultrasonographic and postmortem neuropathologic changes
Luz A.Gutiérrez Sánchez, Diana K.Sandoval Martínez, Luis A.Díaz‐Martínez, Carlos H.Becerra Mojica
11 November 2019 | Neuropathology, Vol. 39, No. 6
+ The most mentioned neuroimaging articles in online media: a bibliometric analysis of the top 100 articles with the highest Altmetric Attention Scores
Eun SooKim, Dae YoungYoon, Hye JeongKim, KwanseopLee, YerimKim, Jong SeokBae, Ju-HunLee
1 May 2019 | Acta Radiologica, Vol. 60, No. 12
+ Diaphragmatic paralysis: Evaluation in infants with congenital Zika syndrome
Vanessavan der Linden, Otavio GomesLins, Natacha Calheirosde Lima Petribu, Ana Claudia Marques Gouveiade Melo, JazmynMoore, Sonja A.Rasmussen, Cynthia A.Moore
9 October 2019 | Birth Defects Research, Vol. 111, No. 19
+ Health outcomes associated with Zika virus infection in humans: a systematic review of systematic reviews
RaphaelXimenes, Lauren CRamsay, Rafael NevesMiranda, Shaun KMorris, KellieMurphy, BeateSander
3 November 2019 | BMJ Open, Vol. 9, No. 11
+ Maternal infection with Zika virus and prevalence of congenital disorders in infants: systematic review and meta-analysis
Saiee F.Nithiyanantham, AlaaBadawi
10 May 2019 | Canadian Journal of Public Health, Vol. 110, No. 5
+ Using Macaques to Address Critical Questions in Zika Virus Research
Dawn M.Dudley, Matthew T.Aliota, Emma L.Mohr, Christina M.Newman, Thaddeus G.Golos, Thomas C.Friedrich, David H.O'Connor
2019Sep29 | Annual Review of Virology, Vol. 6, No. 1
+ Changing Epidemiology, Treatment, and Vaccine Update on Chikungunya, Dengue, and Zika Viruses
AdekunleSanyaolu, OladapoAyodele, LorenaLikaj, AleksandraMarinkovic, JenniferLocke, MiriamAhmed, OdunayoAkanbi, VernerOrish, ChukuOkorie, OlanrewajuBadaru
8 May 2019 | Current Tropical Medicine Reports, Vol. 6, No. 3
+ Zika virus infection
CandaceMoore, RohitSharma, CiléinKearns
17 August 2019
+ Role of adherens junctions and apical-basal polarity of neural stem/progenitor cells in the pathogenesis of neurodevelopmental disorders: a novel perspective on congenital Zika syndrome
Felipe A.Bustamante, MarÍa PazMiró, Zahady D.VelÁsquez, LuisMolina, PamelaEhrenfeld, Francisco J.Rivera, Luis FedericoBÁtiz
2019Aug1 | Translational Research, Vol. 210
+ Association Between Neonatal Neuroimaging and Clinical Outcomes in Zika-Exposed Infants From Rio de Janeiro, Brazil
Kara-LeePool, KristinaAdachi, StelliosKarnezis, NorikoSalamon, TahminehRomero, KarinNielsen-Saines, SheilaPone, MarciaBoechat, MitsueAibe, TallitaGomes da Silva, Carla Trevisan MartinsRibeiro, M. InesBoechat, PatriciaBrasil, AndreaZin, IrenaTsui, Stephanie L.Gaw, PedroDaltro, Bianca GuedesRibeiro, TatianaFazecas, L. CelsoHygino da Cruz, RenataNogueira, ZiltonVasconcelos, Jose PauloPereira, TaniaSaad Salles, Claudia NevesBarbosa, WeiqiangChen, Suan-SinFoo, JaeJung, Maria ElisabethMoreira, MarcosPone
31 July 2019 | JAMA Network Open, Vol. 2, No. 7
+ Zika virus during pregnancy: From maternal exposure to congenital Zika virus syndrome
LéoPomar, DidierMusso, GustavoMalinger, ManonVouga, AlicePanchaud, DavidBaud
1 April 2019 | Prenatal Diagnosis, Vol. 39, No. 6
+ Neuroinflammation During RNA Viral Infections
Robyn S.Klein, ChariseGarber, Kristen E.Funk, HamidSalimi, AllisonSoung, MarleneKanmogne, SindhuManivasagam, ShannonAgner, MatthewCain
2019Apr26 | Annual Review of Immunology, Vol. 37, No. 1
+ Effects of Zika infection on growth
ArnaldoPrata-Barbosa, Marlos MeloMartins, Andreia BittencourtGuastavino, Antônio José Ledo Alves daCunha
2019Mar1 | Jornal de Pediatria, Vol. 95
+ Effects of Zika infection on growth
ArnaldoPrata‐Barbosa, Marlos MeloMartins, Andreia BittencourtGuastavino, Antônio José Ledo Alves daCunha
2019Mar1 | Jornal de Pediatria (Versão em Português), Vol. 95
+ Can We Better Understand How Zika Leads to Microcephaly? A Systematic Review of the Effects of the Zika Virus on Human Brain Organoids
BayuSutarjono
26 September 2018 | The Journal of Infectious Diseases, Vol. 219, No. 5
+ Neurological Complications of Congenital Zika Virus Infection
Vinícius de MeloMarques, Camilla SousaSantos, Isabella GodinhoSantiago, Solomar MartinsMarques, Maria das GraçasNunes Brasil, Talita ToledoLima, Paulo SucasasCosta
2019Feb1 | Pediatric Neurology, Vol. 91
+ Updated Imaging Findings in Congenital Zika Syndrome
Maria de Fatima Viana VascoAragao, Natacha Calheiros de LimaPetribu, Vanessavan der Linden, Marcelo MoraesValenca, Carlos Alexandre Antunes deBrito, Paul M.Parizel
2019Feb1 | Topics in Magnetic Resonance Imaging, Vol. 28, No. 1
+ Perinatal Infections
Vasileios G.Xydis, Vasiliki C.Mouka, Maria I.Argyropoulou
13 March 2019
+ Perinatal Infections
Vasileios G.Xydis, Vasiliki C.Mouka, Maria I.Argyropoulou
22 August 2019
+ Fetal and neonatal neuroimaging
Serena J.Counsell, TomokiArichi, SophieArulkumaran, Mary A.Rutherford
2019Jan1
+ Zika virus and the nonmicrocephalic fetus: why we should still worry
Christie L.Walker, Marie-Térèse E.Little, Justin A.Roby, BlairArmistead, MichaelGale, LakshmiRajagopal, Branden R.Nelson, NoahEhinger, BrittneyMason, UnzilaNayeri, Christine L.Curry, Kristina M.Adams Waldorf
2019Jan1 | American Journal of Obstetrics and Gynecology, Vol. 220, No. 1
+ Putative Cellular and Molecular Roles of Zika Virus in Fetal and Pediatric Neuropathologies
RajendraGharbaran, LatchmanSomenarain
27 August 2018 | Pediatric and Developmental Pathology, Vol. 22, No. 1
+ Association of Prenatal Ultrasonographic Findings With Adverse Neonatal Outcomes Among Pregnant Women With Zika Virus Infection in Brazil
Jose PauloPereira, KarinNielsen-Saines, JeffreySperling, Melanie M.Maykin, LuanaDamasceno, Renan FonsecaCardozo, Helena AbreuValle, Beatriz Ribeiro TorresDutra, Helder DottaGama, KristinaAdachi, Andrea A.Zin, IrenaTsui, ZiltonVasconcelos, PatriciaBrasil, Maria E.Moreira, Stephanie L.Gaw
28 December 2018 | JAMA Network Open, Vol. 1, No. 8
+ Intraamniotic Zika virus inoculation of pregnant rhesus macaques produces fetal neurologic disease
Lark L.Coffey, Rebekah I.Keesler, Patricia A.Pesavento, KevinWoolard, AnilSingapuri, JenniferWatanabe, ChristinaCruzen, Kari L.Christe, JodieUsachenko, JoAnnYee, Victoria A.Heng, ElizaBliss-Moreau, J. RachelReader, Wilhelmvon Morgenland, Anne M.Gibbons, KennethJackson, AmirArdeshir, HollyHeimsath, SalliePermar, ParanthamanSenthamaraikannan, PietroPresicce, Suhas G.Kallapur, Jeffrey M.Linnen, KuiGao, RobertOrr, TracyMacGill, MichelleMcClure, RichardMcFarland, John H.Morrison, Koen K. A.Van Rompay
20 June 2018 | Nature Communications, Vol. 9, No. 1
+ Second-trimester Ultrasound and Neuropathologic Findings in Congenital Zika Virus Infection
Cheng-YingHo, NicolasCastillo, LilianaEncinales, AlexandraPorras, Alejandro RicoMendoza, RebeccaLynch, AmyNemirovsky, GraceMantus, Roberta L.DeBiasi, Jeffrey M.Bethony, Gary L.Simon, Aileen Y.Chang
2018Dec1 | Pediatric Infectious Disease Journal, Vol. 37, No. 12
+ Detection of Zika virus infection among asymptomatic pregnant women in the North of Peru
ClaudiaWeilg, LucindaTroyes, ZoilaVillegas, WilmerSilva-Caso, FernandoMazulis, AmmyFebres, MarioTroyes, Miguel AngelAguilar-Luis, Juanadel Valle-Mendoza
18 May 2018 | BMC Research Notes, Vol. 11, No. 1
+ Femur-sparing pattern of abnormal fetal growth in pregnant women from New York City after maternal Zika virus infection
Christie L.Walker, Audrey A.Merriam, Eric O.Ohuma, Manjiri K.Dighe, MichaelGale, LakshmiRajagopal, Aris T.Papageorghiou, CynthiaGyamfi-Bannerman, Kristina M.Adams Waldorf
2018Aug1 | American Journal of Obstetrics and Gynecology, Vol. 219, No. 2
+ Differential diagnosis of pathological intracranial calcifications in patients with microcephaly related to congenital Zika virus infection
Alexandre Ferreira daSilva
1 August 2018 | Radiologia Brasileira, Vol. 51, No. 4
+ Sleep in Children with Congenital Malformations of the Central Nervous System
Jacqueline F.Yates, Matthew M.Troester, David G.Ingram
23 May 2018 | Current Neurology and Neuroscience Reports, Vol. 18, No. 7
+ Neurological complications of Zika virus infection
Francisco JavierCarod-Artal
26 April 2018 | Expert Review of Anti-infective Therapy, Vol. 16, No. 5
+ Neuroimaging findings associated with congenital Zika virus syndrome: case series at the time of first epidemic outbreak in Pernambuco State, Brazil
PedroPires, PatriciaJungmann, Jully MouraGalvão, AdrianoHazin, LuizaMenezes, RicardoXimenes, GabrieleTonni, EdwardAraujo Júnior
5 December 2017 | Child's Nervous System, Vol. 34, No. 5
+ Fetal Neuropathology in Zika Virus-Infected Pregnant Female Rhesus Monkeys
Amanda J.Martinot, PeterAbbink, OnurAfacan, Anna K.Prohl, RoderickBronson, Jonathan L.Hecht, Erica N.Borducchi, Rafael A.Larocca, Rebecca L.Peterson, WilliamRinaldi, MelissaFerguson, Peter J.Didier, DeborahWeiss, Mark G.Lewis, Rafael A.De La Barrera, EdwardYang, Simon K.Warfield, Dan H.Barouch
2018May1 | Cell, Vol. 173, No. 5
+ Nervous System Injury and Neuroimaging of Zika Virus Infection
ShanshanWu, YuZeng, AlexanderLerner, BoGao, MengLaw
23 April 2018 | Frontiers in Neurology, Vol. 9
+ Imaging of congenital central nervous system infections
IlanaNeuberger, JacquelynGarcia, Mariana L.Meyers, TamaraFeygin, Dorothy I.Bulas, David M.Mirsky
17 March 2018 | Pediatric Radiology, Vol. 48, No. 4
+ Clinical assessment and brain findings in a cohort of mothers, fetuses and infants infected with ZIKA virus
MagdalenaSanz Cortes, Ana MariaRivera, MayelYepez, Carolina V.Guimaraes, IsraelDiaz Yunes, AlexanderZarutskie, IvanDavila, AnilShetty, ArunMahadev, Saray MariaSerrano, NicolasCastillo, WesleyLee, GregoryValentine, MichaelBelfort, GuidoParra, CarrieMohila, KjerstiAagaard, MiguelParra Saavedra
2018Apr1 | American Journal of Obstetrics and Gynecology, Vol. 218, No. 4
+ Adverse outcomes of pregnancy-associated Zika virus infection
William J.Britt
2018Apr1 | Seminars in Perinatology, Vol. 42, No. 3
+ Asymptomatic Prenatal Zika Virus Infection and Congenital Zika Syndrome
Enny SPaixao, Wei-YeeLeong, Laura CRodrigues, AnneliesWilder-Smith
7 April 2018 | Open Forum Infectious Diseases, Vol. 5, No. 4
+ Gottesfeld–Hohler Memorial Foundation Zika Virus Think Tank Summary
John C.Hobbins, Lawrence D.Platt, Joshua A.Copel, Anna G.Euser, YaldaAfshar, Roxanna A.Irani, DeborahLevine, MagdaSanz Cortes, AlfredAbuhamad, Stephanie L.Gaw, KarenHarris, MauricioHerrera, LaurenLynch, AdrianaMelo, LisaNoguchi, RenatoAguiar, Jeanne S.Sheffield, Katherine K.Minton
2018Apr1 | Obstetrics & Gynecology, Vol. 131, No. 4
+ Using immunocompromised mice to identify mechanisms of Zika virus transmission and pathogenesis
Clayton W.Winkler, Karin E.Peterson
19 January 2018 | Immunology, Vol. 153, No. 4
+ Congenital Zika syndrome and neuroimaging findings
Bruno Niemeyer de FreitasRibeiro
1 April 2018 | Radiologia Brasileira, Vol. 51, No. 2
+ Two Infants with Presumed Congenital Zika Syndrome, Brownsville, Texas, USA, 2016–2017
AshleyHoward, JohnVisintine, JaimeFergie, MiguelDeleon
2018Apr1 | Emerging Infectious Diseases, Vol. 24, No. 4
+ What we know and what we don’t know about perinatal Zika virus infection: a systematic review
AntoniSoriano-Arandes, IreneRivero-Calle, EleniNastouli, MariaEspiau, MAFrick, AnaAlarcon, FedericoMartinón-Torres
15 February 2018 | Expert Review of Anti-infective Therapy, Vol. 16, No. 3
+ Congenital involvement of the central nervous system by the Zika virus in a child without microcephaly – spectrum of congenital syndrome by the Zika virus
Bruno Niemeyerde Freitas Ribeiro, Bernardo CarvalhoMuniz, Emerson LeandroGasparetto, EdsonMarchiori
2018Mar1 | Journal of Neuroradiology, Vol. 45, No. 2
+ Neurogenic bladder findings in patients with Congenital Zika Syndrome: A novel condition
Lucia MariaCosta Monteiro, Glaura Nisya de OliveiraCruz, Juliana MarinFontes, Tania Regina DiasSaad Salles, Marcia Cristina BastosBoechat, Ana CarolinaMonteiro, Maria Elizabeth LopesMoreira, MargotDamaser
1 March 2018 | PLOS ONE, Vol. 13, No. 3
+ Zika Virus Alters DNA Methylation of Neural Genes in an Organoid Model of the Developing Human Brain
SylvieJanssens, MichaelSchotsaert, RahulKarnik, VinodBalasubramaniam, MarionDejosez, AlexanderMeissner, AdolfoGarcía-Sastre, Thomas P.Zwaka, Jack A.Gilbert
2018Feb27 | mSystems, Vol. 3, No. 1
+ The pathogenesis of microcephaly resulting from congenital infections: why is my baby’s head so small?
L. D.Frenkel, F.Gomez, F.Sabahi
5 October 2017 | European Journal of Clinical Microbiology & Infectious Diseases, Vol. 37, No. 2
+ Identification of novel small-molecule inhibitors of Zika virus infection
Ewa D.Micewicz, RonikKhachatoorian, Samuel W.French, PiotrRuchala
2018Feb1 | Bioorganic & Medicinal Chemistry Letters, Vol. 28, No. 3
+ Motor Abnormalities and Epilepsy in Infants and Children With Evidence of Congenital Zika Virus Infection
AndréPessoa, Vanessavan der Linden, MarshalynYeargin-Allsopp, Maria Durce Costa GomesCarvalho, Erlane MarquesRibeiro, KimVan Naarden Braun, Maureen S.Durkin, Daniel M.Pastula, Jazmyn T.Moore, Cynthia A.Moore
1 February 2018 | Pediatrics, Vol. 141, No. Supplement_2
+ Advances in Diagnosis, Surveillance, and Monitoring of Zika Virus: An Update
Raj K.Singh, KuldeepDhama, KumaragurubaranKarthik, RuchiTiwari, RekhaKhandia, AshokMunjal, Hafiz M. N.Iqbal, Yashpal S.Malik, RubénBueno-Marí
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29 January 2018 | Biology Open, Vol. 7, No. 1
+ Neurological manifestations of congenital Zika virus infection
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10 November 2017 | Child's Nervous System, Vol. 34, No. 1
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27 November 2017 | Child's Nervous System, Vol. 34, No. 1
+ Congenital Zika virus infection: a neuropathological review
L.Chimelli, E.Avvad-Portari
22 November 2017 | Child's Nervous System, Vol. 34, No. 1
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2018Jan1 | Canadian Journal of Infectious Diseases and Medical Microbiology, Vol. 2018
+ Zika virus infection as a cause of congenital brain abnormalities and Guillain-Barré syndrome: From systematic review to living systematic review
Michel JacquesCounotte, DianneEgli-Gany, MauraneRiesen, MillionAbraha, Teegwendé ValériePorgo, JingyingWang, NicolaLow
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1 January 2018 | Revista da Associação Médica Brasileira, Vol. 64, No. 1
+ The Recent Epidemic Spread of Zika Virus Disease
Chang-KwengLim
2018Jan1 | Uirusu, Vol. 68, No. 1
+ Zika Virus Infection in Pregnancy: Maternal, Fetal, and Neonatal Considerations
Carmen DZorrilla, InésGarcía García, LourdesGarcía Fragoso, AlbertoDe La Vega
16 December 2017 | The Journal of Infectious Diseases, Vol. 216, No. suppl_10
+ Placental Histopathology and Clinical Presentation of Severe Congenital Zika Syndrome in a Human Immunodeficiency Virus-Exposed Uninfected Infant
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7 December 2017 | Frontiers in Immunology, Vol. 8
+ How Does Imaging of Congenital Zika Compare with Imaging of Other TORCH Infections?
Deborah Levine
,- Jacques C. Jani
,- Ilse Castro-Aragon
,- Mieke Cannie
,
20 November 2017 | Radiology, Vol. 285, No. 3
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KarenSohan, Cathy A.Cyrus
22 September 2017 | International Journal of Gynecology & Obstetrics, Vol. 139, No. 3
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RajendraGharbaran, LatchmanSomenarain
2017Dec1 | Pathology, Vol. 49, No. 7
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MaliniDeSilva, Flor M.Munoz, ErickSell, HelenMarshall, AlisonTse Kawai, AlisaKachikis, PaulHeath, Nicola P.Klein, James M.Oleske, FyezahJehan, HansSpiegel, MirjanaNesin, Beckie N.Tagbo, AnjuShrestha, Clare L.Cutland, Linda O.Eckert, SonaliKochhar, AzucenaBardají
2017Dec1 | Vaccine, Vol. 35, No. 48
+ Impact of the Zika Virus for Anesthesiologists: A Review of Current Literature and Practices
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2017Dec1 | Journal of Cardiothoracic and Vascular Anesthesia, Vol. 31, No. 6
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15 September 2017 | Journal of Venomous Animals and Toxins including Tropical Diseases, Vol. 23, No. 1
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3 October 2017 | Maternal Health, Neonatology and Perinatology, Vol. 3, No. 1
+ The emerging radiological features of Zika virus infection
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2017Dec1 | Radiologia Brasileira, Vol. 50, No. 6
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Amy B.Rosenfeld, David J.Doobin, Audrey L.Warren, Vincent R.Racaniello, Richard B.Vallee
31 October 2017 | Proceedings of the National Academy of Sciences, Vol. 114, No. 46
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Nelci AdrianaCicuto Ferreira Rocha, Ana Carolinade Campos, FellipeCicuto Ferreira Rocha, FernandaPereira dos Santos Silva
2017Nov1 | Infant Behavior and Development, Vol. 49
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20 October 2017 | MMWR. Morbidity and Mortality Weekly Report, Vol. 66, No. 41
+ Brain MRI in Infants after Maternal Zika Virus Infection during Pregnancy
MehdiMejdoubi, AliceMonthieux, TiphaineCassan, CatherineLombard, OlivierFlechelles, ClaraAdenet
2017Oct5 | New England Journal of Medicine, Vol. 377, No. 14
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Bianca Guedes Ribeiro
,- Heron Werner
,- Flávia P. P. L. Lopes
,- L. Celso Hygino da Cruz, Jr
,- Tatiana M. Fazecas
,- Pedro A. N. Daltro
,- Renata A. Nogueira
,
11 October 2017 | RadioGraphics, Vol. 37, No. 6
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Deborah Levine
,
11 October 2017 | RadioGraphics, Vol. 37, No. 6
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1 October 2017 | Arquivos de Neuro-Psiquiatria, Vol. 75, No. 10
+ Congenital Zika syndrome and neuroimaging findings: what do we know so far?
Bruno Niemeyer de FreitasRibeiro, Bernardo CarvalhoMuniz, Emerson LeandroGasparetto, NinaVentura, EdsonMarchiori
2017Oct1 | Radiologia Brasileira, Vol. 50, No. 5
+ Rapid, Point‐of‐Care, Paper‐Based Plasmonic Biosensor for Zika Virus Diagnosis
QishengJiang, Yatin J.Chandar, SisiCao, Evan D.Kharasch, SrikanthSingamaneni, Jeremiah J.Morrissey
10 August 2017 | Advanced Biosystems, Vol. 1, No. 9
+ Ultrasound imaging for identification of cerebral damage in congenital Zika virus syndrome: a case series
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2017Sep1 | The Lancet Child & Adolescent Health, Vol. 1, No. 1
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Sjirk JanWestra
2017Sep1 | The Lancet Child & Adolescent Health, Vol. 1, No. 1
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Juan-CarlosSaiz, Miguel A.Martín-Acebes, RubénBueno-Marí, Oscar D.Salomón, Luis C.Villamil-Jiménez, JorgHeukelbach, Carlos H.Alencar, Paul K.Armstrong, Tania M.Ortiga-Carvalho, RosaliaMendez-Otero, Paulo H.Rosado-de-Castro, Pedro M.Pimentel-Coelho
22 August 2017 | Frontiers in Microbiology, Vol. 8
+ Serial Head and Brain Imaging of 17 Fetuses With Confirmed Zika Virus Infection in Colombia, South America
MiguelParra-Saavedra, JennitaReefhuis, Juan PabloPiraquive, Suzanne M.Gilboa, Martina L.Badell, Cynthia A.Moore, MarcelaMercado, DianaValencia, Denise J.Jamieson, MauricioBeltran, MagdaSanz-Cortes, Ana MariaRivera-Casas, MayelYepez, GuidoParra, MarthaOspina Martinez, Margaret A.Honein
2017Jul1 | Obstetrics & Gynecology, Vol. 130, No. 1
+ Nonmicrocephalic Infants with Congenital Zika Syndrome Suspected Only after Neuroimaging Evaluation Compared with Those with Microcephaly at Birth and Postnatally: How Large Is the Zika Virus “Iceberg”?
M.F.V.V.Aragao, A.C.Holanda, A.M.Brainer-Lima, N.C.L.Petribu, M.Castillo, V.van der Linden, S.C.Serpa, A.G.Tenório, P.T.C.Travassos, M.T.Cordeiro, C.Sarteschi, M.M.Valenca, A.Costello
18 May 2017 | American Journal of Neuroradiology, Vol. 38, No. 7
+ Association between Zika virus and fetopathy: a prospective cohort study in French Guiana
L.Pomar, G.Malinger, G.Benoist, G.Carles, Y.Ville, D.Rousset, N.Hcini, C.Pomar, A.Jolivet, V.Lambert
4 May 2017 | Ultrasound in Obstetrics & Gynecology, Vol. 49, No. 6
+ The spectrum of neuropathological changes associated with congenital Zika virus infection
LeilaChimelli, Adriana S. O.Melo, ElyzabethAvvad-Portari, Clayton A.Wiley, Aline H. S.Camacho, Vania S.Lopes, Heloisa N.Machado, Cecilia V.Andrade, Dione C. A.Dock, Maria ElisabethMoreira, FernandaTovar-Moll, Patricia S.Oliveira-Szejnfeld, Angela C. G.Carvalho, Odile N.Ugarte, Alba G. M.Batista, Melania M. R.Amorim, Fabiana O.Melo, Thales A.Ferreira, Jacqueline R. L.Marinho, Girlene S.Azevedo, Jeime I. B. F.Leal, Rodrigo F. Madeiroda Costa, StevensRehen, Monica B.Arruda, Rodrigo M.Brindeiro, RodrigoDelvechio, Renato S.Aguiar, AmilcarTanuri
22 March 2017 | Acta Neuropathologica, Vol. 133, No. 6
+ Viral infection, proliferation, and hyperplasia of Hofbauer cells and absence of inflammation characterize the placental pathology of fetuses with congenital Zika virus infection
David A.Schwartz
11 April 2017 | Archives of Gynecology and Obstetrics, Vol. 295, No. 6
Supplemental Material
Abbreviations
| | |
--- |
| Abbreviations: | |
| CNS | central nervous system |
| IPESQ | Instituto de Pesquisa in Campina Grande State Paraiba |
| RT-PCR | reverse transcription polymerase chain reaction |
| TORCH | toxoplasmosis, syphilis, varicella-zoster virus, parvovirus B19, rubella, cytomegalovirus, and herpes virus |
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10234 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2993982/ | Principles of sample size calculation - PMC
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Indian J Ophthalmol
. 2010 Nov-Dec;58(6):517–518. doi: 10.4103/0301-4738.71692
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Principles of sample size calculation
Nithya J Gogtay
Nithya J Gogtay
1 Department of Clinical Pharmacology, Seth GS Medical College and KEM Hospital, Parel, Mumbai, Maharashtra, India
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1 Department of Clinical Pharmacology, Seth GS Medical College and KEM Hospital, Parel, Mumbai, Maharashtra, India
✉
Correspondence to: Dr. Nithya Gogtay, Department of Clinical Pharmacology, Seth GS Medical College and KEM Hospital, Parel, Mumbai – 400 012, India. njgogtay@hotmail.com
Received 2010 Aug 31; Accepted 2010 Aug 31.
© Indian Journal of Ophthalmology
This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC2993982 PMID: 20952836
Abstract
In most areas in life, it is difficult to work with populations and hence researchers work with samples. The calculation of the sample size needed depends on the data type and distribution. Elements include consideration of the alpha error, beta error, clinically meaningful difference, and the variability or standard deviation. The final number arrived at should be increased to include a safety margin and the dropout rate. Over and above this, sample size calculations must take into account all available data, funding, support facilities, and ethics of subjecting patients to research.
Keywords: Alpha error, beta error, clinically meaningful difference, variability
In most areas in life, it is very difficult to work with populations. During elections for instance, news channels interview a few hundred people and predict results based on their choices. Similarly, in a factory manufacturing light bulbs, a few bulbs are chosen at random to assess their quality. Likewise, in research, while it is ideal to work with the entire population, it is almost impossible to do so. Hence researchers choose to work with samples. Sample size calculations enable researchers to draw strong robust conclusions from the limited amount of information and also permit generalization of results. It is however important to remember that since it is very difficult to predict the outcome of any clinical study or lab experiment, sample size calculations will always remain approximate.
The estimation of the minimum sample size required for any study is not a single unique method, but the concepts underlying most methods are similar. The determination of the sample size is critical in planning clinical research because this is usually the most important factor determining the time and funding to perform the experiment. In most studies, there is a primary research question that the researcher wants to investigate. Sample size calculations are based on this question. Sample size calculations must take into account all available data, funding, support facilities, and ethics of subjecting patients to research. The present paper outlines the principles of sample size calculation for randomized controlled trials (RCTs) with a few solved examples.
Elements in Sample Size Calculation
Sample size calculations begin with an understanding of the type of data and distribution we are dealing with. Very broadly, data are divided into quantitative (numerical) and categorical (qualitative) data. For the former, information on the mean responses in the two groups’ u 1 and u 2 are required as also the common standard deviation for the two groups. For categorical data, p 1 and p 2 or information on proportions of successes in the two groups is needed. This information is usually obtained either from the published literature, a pilot study, or at times guesstimated. The other two key components are the alpha and beta error. Because the estimated sample size represents the minimum number of subjects required for the study, a “safety factor” should be added. The size of the safety factor is again an educated guess. Additions for drop-outs/attrition during the course of the study should also be made. Apart from this, an understanding of whether the data are normally distributed (follows the Gaussian or bell-shaped curve) or otherwise is also needed.
Understanding of Key Terms
The calculation of sample size based on power considerations requires that an investigator specify the points given below. The first three items are under the control of the investigator:
The size of the effect that is clinically worthwhile to detect (d). This for numerical data is the difference between u1 and u2 for quantitative data and p1 and p2 for categorical data. This is also called the clinical meaningful difference, which will make the physician change his or her practice.
The probability of falsely rejecting a true null hypothesis (α-error). This is also called the false positive error and is the probability of finding a difference where none exists. This error is perceived to be the more dangerous of the two errors, since it can impact clinical practice. It is also called the regulator’s error. The alpha error is linked to the P-value or probability value and is conventionally set at 5%.
The probability of failing to reject a false null hypothesis (β-error). This is also called the false negative error and is the probability of NOT finding a difference when one actually exists. This is conventionally set either at 10% or 20% and is also called the investigator’s error.
The standard deviation of the population being studied (SD or σ). This is the variability or spread associated with quantitative data.
Standard values of the alpha and beta error are given in the solved examples and can be found in most statistics books. The examples below can be solved by hand using simple or scientific calculators. The website and the pdf file, provide an easy tool of how to use online sample size calculators.
Few Solved Examples
(A) Sample size for one mean, normal distribution
Problem: An emergency medicine physician wants to know if the mean heart rate after a particular type of trauma differs from the healthy population rate of 72 beats/min. He considers a mean difference of 6 beats/min to be clinically meaningful. He also chooses 9.1 beats/min as the variation based on a previously published study. How many patients will be needed to carry out the study at 5% significance and 80% power?
In this example, the following data are given to us:
the size of the effect that is clinically worthwhile to detect (_d_) = 6 beats/min
the probability of falsely rejecting a true null hypothesis (α) = 0.05, Z α = 1.96
the probability of failing to reject a false null hypothesis (β) = 0.80, Z β = 0.84
the standard deviation of the population being studied (SD or σ) = 9.1 beats/min.
n = 18
or
18 patients with a particular type of trauma need to be studied by the physician.
(B) Sample size for two means, quantitative data
where d = u 1 - u 2/2.
Problem: A new treatment for hypertension is being compared with placebo. How many patients will be required at 90% power and 5% significance to detect an average difference of 5 mmHg between the Rx group and placebo group assuming a standard deviation (a measure of interpatient variability) to be 10 mm?
In this example, the following data are given to us:
the size of the effect that is clinically worthwhile to detect (_d_) = 5 mm
the probability of falsely rejecting a true null hypothesis (α) = 0.05, Z α = 1.96
the probability of failing to reject a false null hypothesis (β) = 0.80, Z β = 1.282
the standard deviation of the population being studied (SD or σ) = 10 mm.
Solution:
where d = 5/10
thus, n = 84 patients per group.
(C) Sample size for two proportions, categorical data
The BASIL study-bypass versus angioplasty: The statistical calculations were based on the 3-year survival value of 50% in the angioplasty and 65% in the bypass group. At 5% significance and 90% power, how many patients would be needed to detect a difference between the two groups? (Lancet 2005;366:1925-34).
In this example, the following data are given to us:
the size of the effect that is clinically worthwhile to detect (_d_) = 15% or 0.15
the probability of falsely rejecting a true null hypothesis (_α_) = 0.05, Z α = 1.96
the probability of failing to reject a false null hypothesis (_β_) = 0.80, Z β = 1.282.
where, d is
p 1 – p 2
√p (1-p).
And p = p 1 + p 2/2
Solution:
p 1 = 0.65, p 2=.50, p = 0.575,
i.e., there will be 233 patients per group.
The sample size calculation should be done with the help of a statistician. However, the present article provides the basic understanding of the principles behind the sample size calculation. This would help in providing the required inputs to the clinicians while interacting with the statistician.
Footnotes
Source of Support: Nil
Conflict of Interest: None declared
References
1.Julios SA. Sample sizes for clinical trials with normal data. Stats Med. 2004;23:1921–86. doi: 10.1002/sim.1783. [DOI] [PubMed] [Google Scholar]
2.Devane D, Begley CM, Clarke M. How many do I need? Basic principles of sample size estimation. J Adv Nursing. 2004;47:297–302. doi: 10.1111/j.1365-2648.2004.03093.x. [DOI] [PubMed] [Google Scholar]
3.Karlsson J, Engebretsen L, Dainty K. ISAKOS scientific committee Considerations on sample size and power calculations in randomized clinical trials. Arthroscopy. 2003;19:997–9. doi: 10.1016/j.arthro.2003.09.022. [DOI] [PubMed] [Google Scholar]
Articles from Indian Journal of Ophthalmology are provided here courtesy of Wolters Kluwer -- Medknow Publications
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10235 | https://www.youtube.com/watch?v=zy-uFwhOUYY | Combinatorial Proofs: C(n,k)=C(n-1,k-1)+C(n-1,k)
Mathispower4u
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Posted: 4 Jul 2022
This video provides two combinatorial proofs for a binomial identity.
6 comments
Transcript:
Welcome to an introduction to combinatorial
proofs. Explanatory proofs are typically called combinatorial proofs. In general to give
a combinatorial proof for a binomial identity say a equals b, we do the following. Number one
we find a counting problem to answer in two ways. Number two we explain why one
answer to the counting problem is a, and then three, we explain why the other
answer to the counting problem is b. Since both a and b are the answers to the same
question we must have a equals b. The tricky part is coming up with the question. In this lesson
we'll take a look at two combinatorial proofs for the binomial identity and choose k is
equal to n minus 1 choose k minus 1 plus n minus 1 choose k. In the previous lesson we
proved this is true using an algebraic proof. One way to see the equation is true is to consider
bit strings. n choose k is the number of bit strings of length n containing k 1's. Some of the
bit strings start with a one and the rest start with a zero. First consider all the bit strings
which start with a one. After the one there must be n minus 1 more bits to get a total length of
n and exactly k minus 1 of them must be ones, as we already have 1 one and we need a total
of k ones. How many strings are like that. Well there are exactly n minus one choose k
minus one such bit strings. Of all the length n bit strings containing k 1's, n minus one choose k
minus one of them start with a 1. Similarly there are n minus 1 choose k, which start with a 0.
We need n minus 1 bits and now k of them must be ones. Since there are n minus 1 choose k bit
strings containing n minus one bits with k 1's, that is the number of n bit strings with k
1's which start with a zero. We answer the same question in two different ways so the two
answers must be the same. Therefore, n choose k is equal to n minus 1 choose k minus 1 plus n
minus 1 choose k. So this would be one possible combinatorial proof for the binomial identity.
Let's take a look at a second combinatorial proof. So here's another way to explain why the equation
is true. How many ways can you select k pizza toppings from a menu containing n choices. One
way to do this is to evaluate and choose k. Another way to answer the same question is to
first decide whether or not you want anchovies. If you do want anchovies, you still need to
pick k minus 1 toppings from now n-1 choices. This can be done in n minus 1 choose k minus 1
ways. If you do not want anchovies, then you still need to pick k toppings from n minus 1 choices.
The anchovies are out. You can do that in n minus 1 choose k ways. Since the choices with anchovies
are disjoint from the choices without anchovies, the total choices are n minus 1 choose k minus 1
plus n minus 1 choose k. So once again we answered the same question in two different ways. So
the two answers must be the same, n choose k equals n minus 1 choose k minus 1 plus n minus
1 choose k. I hope you found this helpful. |
10236 | https://artofproblemsolving.com/wiki/index.php/Decimal?srsltid=AfmBOop0Bo56wbEcBKPfuEiHp3RE1EPc_-75tadvGTp4fGH-K3AQ2Z2e | Art of Problem Solving
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Decimal
This article focuses on the non-integer real-numbers. Decimal can also refer to base ten number system.
A decimal is a number that is not an integer and is expressed in neither fraction or percent form. The whole number portion is separated from the fractional portion by a decimal point (.) which looks identical to a period.
Contents
[hide]
1 Converting Decimals
2 Comparing Decimals
3 Operations on Decimals
3.1 Addition and Subtraction
3.2 Multiplication
3.2.1 Powers of 10
3.2.2 General Rule
3.3 Division
3.3.1 Powers of 10
3.3.2 General Rule
4 Problems
4.1 Introductory
4.2 Intermediate
Converting Decimals
When converting fractions, note that the first digit after the decimal point represents tenths, the next digit after the decimal point represents hundredths, and so on. So we can convert each part accordingly. For example, 3.5 would equal 35/10 or 350%. 5.68 would equal , or .
For a quicker strategy, note that a tenth is 10 hundredths or 100 thousandths. Thus, we can go from straight to .
Comparing Decimals
Comparing decimals is similar to comparing whole numbers -- start with the largest digit and compare accordingly. For instance, hundreds to hundreds, tens to tens, ones to ones, tenths to tenths, hundredths to hundredths, thousandths to thousandths, etc.
Operations on Decimals
Addition and Subtraction
Addition and subtraction is similar to adding and subtracting whole numbers, but knowing place value is important. In particular, we have to make sure that tenths are added to tenths, hundredths are added to hundredths, and so on. This is done by lining up the decimal point.
Multiplication
Powers of 10
When multiplying by 10, note that the quantity of tenths become the quantity of ones, the quantity of hundredths become the quantity of tenths, and so on. Thus, multiplying by 10 means moving the decimal point one place to the right. Similarly, multiplying by 100 means moving the decimal point 2 places to the right, and multiplying by (where is an integer) means moving the decimal point places to the right.
When multiplying by 0.1, note that the quantity of ones become the quantity of tenths, the quantity of tenths become the quantity of hundredths, and so on. Thus, multiplying by 0.1 means moving the decimal point one place to the left. Similarly, multiplying by 0.01 means moving the decimal point 2 places to the left, and so on.
General Rule
Consider the expression .
First, note that (25 tenths) and (12 hundredths). Thus, because of the Commutative Property, .
Then we can treat part of the problem as a multiplication problem -- . Thus, .
Finally, we locate the decimal point accordingly, and . Therefore, .
In similar problems, we can extract whole numbers, multiply in a way similar to integers, and count decimal places to get the final answer.
Division
Powers of 10
When dividing by 10, note that the quantity of units become the quantity of tenths, the quantity of hundredths become the quantity of thousandths, and so on. Thus, dividing by 10 means moving the decimal point one place to the left. Similarly, dividing by 100 means moving the decimal point 2 places to the left, and dividing by (where is an integer) means moving the decimal point places to the left.
When dividing by 0.1, note that a tenth is one tenth of a unit, a hundredth is one tenth of a tenth, and so on. Thus, dividing by 0.1 is the same as multiplying by 10, so the decimal point is moved one place to the right. Similarly, dividing by 0.01 means moving the decimal point 2 places to the right, and so on.
General Rule
Dividing decimals by integers is similar to dividing integers by integers -- place value needs to be kept track. In particular, when doing the long division set up, the decimal point should be lined up.
As for dividing decimals by decimals, a common tactic is moving the decimal point some places to the right for both the dividend and the divisor. This can be done because both the numerator and denominator is multiplied by a power of 10. Afterward, we can proceed like a decimal divided by an integer.
Problems
Introductory
Practice Problems on Alcumus
Decimal Arithmetic (Prealgebra)
Rounding Decimals (Prealgebra)
Decimals and Fractions (Prealgebra)
Repeating Decimals (Prealgebra)
Intermediate
2002 AIME I Problems/Problem 7
2006 AIME I Problems/Problem 6
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Data Science
Intuition of the Arithmetic, Geometric and Harmonic Mean
What is the "average" and how do we find it? By forgetting the formula – and getting better at math.
7 min read
Thoughts on Probability and Statistics
How to Understand the Idea of "Average"
Formulas Are For Calculation, Not For Understanding
When people teach you about statistical concepts, you usually get a equation which is a formula for some quantity, like the arithmetic mean. Formulas are fine, but they are designed with calculation in mind. Usually, the equation will put the unknown on one side, and all known quantities on the other.
Like this.
Unfortunately, this view does not help students develop a good understanding of concepts like the "average". And as a result, it is not difficult to find people misapplying statistics, for example, using the arithmetic mean on financial returns data when the geometric mean makes more sense.
By rearranging the equation, you can get an alternative perspective which makes the concept of "average" clearer and more generalisable for different types of data.
For example, if we have quantities which we add to produce a total outcome (eg. getting $3 today, and $5 tomorrow gives 3+5 = $8), then we can use an arithmetic mean to tell us the quantity that, when added together repeatedly, gives us the same total outcome.
And if we have quantities to multiply to produce the final figure (eg. monthly returns on a stock portfolio), then it should become clear that the geometric mean is more appropriate.
For example, if you have $1, and one day it doubles to $2, and the next day, it grows by a multiple of 8 to $16, it would be the same if it grew at the geometric mean for 2 days. The geometric mean g satisfies gg=28=16, so g=4.
Context is important!
It is important to know that the appropriate "average" only make sense in context. If you were asked to "find the average" of 2 and 8, it should not make sense to you. The average of 2 and 8 might be 5 if its the average number of dollars received on two days, since 2+8 = 5+5. But it could also be 4 if it’s day-on-day growth rate, since 2 8 = 4 4. We don’t know which one is correct if we don’t know where these numbers are coming from.
The Harder Average: Harmonic Mean
What is the harmonic mean, and when is it used?
A Motivating Example: The Average Speed of a Road Trip
Two people are driving across the country. Since long-distance driving can be tiring, the two agree to take turns driving.
The two tables below indicate two possible trips, with different ways of splitting distances and times. The question: what is the average speed of travel?
(1) The two drivers drive an equal distance, but for different periods of time.
To figure out the average speed, we see that the speed of driver one and two are 100 and 66.666 km/h. We take the average of this figure and get 83.333 km/h. Right?
But maybe you’ve read something about how to average rates. You know that the arithmetic mean is incorrect, and that the harmonic mean needs to be used. (You can read the Wikipedia page on the harmonic mean to confirm this). You recall the formula for the harmonic mean…
… and you calculate that H = 2/(1/100 + 1/66.666) = 80 km/h.
(2) The two drivers drive different distances, for an equal amount of time.
After seeing the last example, you know not to use the arithmetic mean, because the "harmonic mean is correct for calculating average rates."
You find that the average speed is 2/(1/80 + 1/90) = 84.71 km/h.
But you will soon find out that this is actually incorrect. This is why you should not blindly follow formulas.
How to find the average speed
To figure out the average speed, I take the same approach as I took for the arithmetic and geometric mean. I construct a parallel world, where everyone is identical and acts identically, and I want the action in this parallel world to produce the same outcome that I have in my data. If we construct two imaginary drivers who both drive at the average speed, at what speed would they have to drive so that they would travel the same distance in the same period of time?
(1) They need to cover 1000 km in 12.5 hours, so this is 1000/12.5= 80 km/h. Note that this is the harmonic mean that we calculated earlier, and also note that the two drivers cover the same distance.
Average speed = harmonic mean
(2) They need to cover 1020 km in 12 hours, so 1020/12 = 85 km/h. Note that this is NOT the harmonic mean that was calculated earlier. This is the arithmetic mean of 80 and 90, and note that the two drivers drive for the same duration of time.
Average speed = arithmetic mean (?)
When is the harmonic mean appropriate?
It is commonly said that the harmonic mean is appropriate for finding the average of rates. Even Wikipedia states that "it is appropriate for situations when the average of rates is desired." But the example I give above has rates in both scenarios, and the harmonic mean is not appropriate for the second example.
I claim that the harmonic mean is equivalent to a weighted arithmetic mean, and the key issue to averaging rates is how the different rates are weighted.
Basing the following derivations on the road trip example, consider the following equations.
The first line is just the intuitive view of average speed.
In the second line, the numerator is the sum of distances for each driver. You can confirm this with dimensional analysis (L = length, T = time): speed has dimension L/T and time has dimension T, so the multiplication of speed and time is giving a quantity with dimension L. So far so good?
If we now separate out the terms, we can get the third line. The time fractions are dimensionless (T/T) and are giving the proportion of time spent at the speed of the iᵗʰ driver. It is a weighted arithmetic mean of speeds, weighted by time! And if the time spent at each speed is equal, then it is the arithmetic mean of speeds. Note that example (2) had equal times for both drivers, and this is why the arithmetic mean was correct.
We now do a second derivation, but we first take the reciprocal of all speeds, so that the units are inverted (hours per kilometre).
The first line is from the intuitive view of average speed.
Second line, we split the numerator, but this time we need to make sure that the unit is time. The reciprocal of speed has units of hours per km, and multiplying this by distance gives us a unit of hours.
We split up the fraction to get the third line. This time, it is a weighted average of reciprocal speeds, weighted by the proportion of the total distance traveled at that reciprocal speed.
If we now let all n distances be the same length…
…then the reciprocal average speed becomes the arithmetic mean of the reciprocal speeds. Taking the reciprocal again to restore the units to L/T, we get the harmonic mean.
Back to the original question: What is the harmonic mean, and when is it used?
The harmonic mean is the arithmetic mean of rate data with inverted units. When is it used? See below.
Practical Takeaways for Rate Data
Luckily, many data sets in the real world will have identical values in some dimension. For example, people might work at different rates, but they are made to perform the same tasks, travel the same distances, etc.
Identical in the numerator unit – take the harmonic mean. (eg. all travel at different speeds, but over the same distance).
Identical in the denominator unit – take the arithmetic mean (eg. all travel at different speeds, but for the same time).
Not identical in either unit – constructed a weighted arithmetic/harmonic mean using some of the ideas I have fleshed out.
I hope my thoughts have helped you think more critically about maths and data.
Written By
Andy
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Data, Data Science, Mathematics, Physics, Statistics
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10238 | https://www.kenhub.com/en/library/anatomy/the-myelin-sheath-and-myelination | What's new?
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ArticlesHistologyTypes of tissuesNervous tissueMyelin sheath and myelination
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ArticlesHistologyTypes of tissuesNervous tissueMyelin sheath and myelination
Myelin sheath and myelination
Author:
Alexandra Osika
•
Reviewer:
Francesca Salvador, MSc
Last reviewed: October 26, 2023
Reading time: 13 minutes
Myelin sheath
Stratum myelini
1/4
Synonyms: none
Axons are a key component of a neuron, they conduct electrical signals in the form of an action potential from the cell body of the neuron to its axon terminal where it synapses with another neuron. An axon is insulated by a myelin sheath throughout its length to increase the velocity of these electrical signals allowing signals to propagate quickly.
Axons which are covered by a myelin sheath, a multilayer of proteins and lipids, are said to be myelinated. If an axon is not surrounded by a myelin sheath, it is unmyelinated. Myelination is the formation of a myelin sheath.
Key facts of the myelin sheath and myelination
Table quiz
| Myelination | Produced by Schwann cells for peripheral axons Produced by oligodendrocytes for central axons |
| Myelin sheath function | Insulates axons allowing for rapid action potential conduction Separates axons from surrounding extracellular components |
| Brain myelination | Mature at 2 years of age |
| Clinical aspects | Demyelination, Schwannoma |
This article will discuss the structure and histology of myelin sheaths, their function, and the process of brain myelination.
Contents
Overview
Myelination
Myelin
Schwann cells
Oligodendrocytes
Myelin sheath function
Brain myelination
Clinical aspects
Demyelination
Schwannoma
Sources
Show all
Overview
To understand myelination, we must first understand the cellular structure of the nervous system. Recall that the nervous system is composed of two types of cells: neurons and neuroglia (also simply known as glia or glial cells). Neurons conduct signals throughout the nervous system, while neuroglia provide a supporting structural and metabolic role for neurons by protecting and nourishing neurons, as well as maintaining the surrounding interstitial fluid. This is why they are known as the “glue” of the nervous system (“glia” is Greek for “glue”).
Each neuron has four specialized regions to perform different functions:
Dendrites receive incoming information. They are part of the receptive segment of a neuron.
The cell body (also called soma or perikaryon) also receives incoming information and integrates information together. Depending on the type of neuron, various extensions or processes will extend from the cell body such as dendrites and an axon. As the cell body receives information, it is part of the receptive segment of a neuron.
The axon then conducts information from the cell body of a neuron to the axon terminal. An axon makes up the conductive segment of a neuron.
Axon terminals are the presynaptic component of a synapse, the site of intercellular communication, where a neuron transmits its signal to another neuron. Axon terminals are the transmissive segment of a neuron.
It is the axon of a neuron which is myelinated.
Before you go any further, why not test how well you know the different parts of a neuron and neuron types?
Myelination
Myelination is the formation of a myelin sheath. Myelin sheaths are made of myelin, and myelin is produced by different types of neuroglia: oligodendrocytes and Schwann cells, where oligodendrocytes myelinate axons in the central nervous system, and Schwann cells myelinate axons in the peripheral nervous system. So which cells form myelin in the spinal cord? Since the spinal cord is part of the central nervous system, oligodendrocytes form this myelin. Functionally, oligodendrocytes and Schwann cells perform the same role, but structurally they are different.
Remember these cells and their location with the mnemonic "COPS" (Central - Oligodendrocytes, Peripheral - Schwann).
Myelin
Myelin is made up of lipids and proteins, a fatty substance with a whitish appearance. It is made up of many concentric layers of plasma membrane to make up the myelin sheath around axons. Myelin sheath and myelin function are therefore the same, to increase the speed of nerve impulses.
The amount of myelin in the body increases throughout development, from fetal development up until maturity, with the myelination in the prefrontal cortex being the last to complete in the 2nd or 3rd decade. The more myelin and myelination an individual has, the quicker their response is to stimuli because myelin sheaths increase the speed of nerve impulses. Think of a baby that is still learning to walk– their response to stimuli is slow and uncoordinated compared to a child, teenager, or adult. This is partly because myelination of axons during infancy is still progressing.
Schwann cells
Schwann cell
Schwannocytus
1/4
Synonyms:
Neurolemmocyte, Neurolemmocytus
Schwann cells (also known as neurolemmocytes) are flat cells which make up myelin sheaths on axons of the peripheral nervous system. Each Schwann cell myelinates only one axon, where one peripheral axon will have multiple Schwann cells myelinating its length as one Schwann cell wraps a lipid-rich membrane layer around approximately 1 mm of an axon’s length. However, in a different arrangement, a Schwann cell can enclose many (up to 20) unmyelinated axons. In this way, the unmyelinated axons go through the Schwann cell, but the Schwann cell does not produce a myelin sheath for these axons.
Schwann cells will first start to myelinate axons during fetal development, wrapping its lipid-rich membrane around it many times until there are multiple layers surrounding the axon. As the wrapping continues, the nucleus and cytoplasm of the Schwann cell are gradually squeezed out. Once myelination is complete, the Schwann cell’s nucleus and cytoplasm finish in the outermost layer. The myelin sheath itself is the inner portion of these wrappings (approximately 100 layers of plasma membrane), and the outermost layer that contains the nucleus and cytoplasm is the neurilemma (also called the neurolemma, sheath of Schwann, and Schwann’s sheath).
Myelin sheath gap
Nodus interruptionis myelini
1/3
Synonyms:
Node of Ranvier, Nodus Ranvieri
Along an axon, there are gaps between Schwann cells and the myelin sheath called the nodes of Ranvier. Here, electrical impulses are formed more quickly and allow the signal to jump from node to node through the myelin sheath. In unmyelinated axons, the electrical signal travels through each part of the cell membrane which slows the speed of signal conduction.
Schwann cells also play a role in forming connective tissue sheaths in neuron development and axon regeneration, providing chemical and structural support to neurons. The neurilemma assists in regeneration of an axon when it is damaged by forming a regeneration tube to stimulate and guide its regeneration.
Oligodendrocytes
Oligodendrocyte
Oligodendrocytus
Synonyms:
none
Oligodendrocytes (or oligodendroglia) are star-shaped neuroglia that form the myelin sheaths on axons of the central nervous system. A single oligodendrocyte has about 15 flat, broad, arm-like processes coming out of the cell body. With these they can myelinate multiple axons by spiraling around them to form a myelin sheath. The cell body and nucleus of oligodendrocytes remain separate from the myelin sheath, and so there is no neurilemma (that is, a cell body and nucleus enveloping an axon) present in oligodendrocytes, unlike in Schwann cells. However, like in Schwann cells, nodes of Ranvier are also present on the axons myelinated by oligodendrocytes, but there are far fewer of them.
Once an axon in the central nervous system is injured, there is little regrowth unlike axons in the peripheral nervous system. It is uncertain why this is but it is thought to be because of a combination of an inhibitory influence on regrowth from oligodendrocytes and lack of neurolemma.
Myelin sheath function
Myelin sheath gap
Nodus interruptionis myelini
1/3
Synonyms:
Node of Ranvier, Nodus Ranvieri
Since the myelin sheath surrounds the axon, one of its functions is to separate the axon from surrounding extracellular components. Its main function, however, is to insulate the axon and increase the velocity of action potential propagation.
Myelin has properties of low capacitance and high electrical resistance which means it can act as an insulator. Therefore, myelin sheaths insulate axons to increase the speed of electrical signal conduction. This allows myelinated axons to conduct electrical signals at high speeds.
Nodes of Ranvier (gaps in myelination) contain clusters of voltage-sensitive sodium and potassium ion channels (approximately 1000 per µm2) whereas their distribution and numbers under myelin in the internodal axon membrane are spare. This creates an uneven distribution of ion channels, and the action potential in myelinated axons will “leap” from one node to the next in saltatory conduction. This type of conduction has important consequences:
Increased conduction velocity
Reduced metabolic cost of conduction as the amount of energy needed in myelinated fibers to conduct the impulse is less
The conduction velocity of an axon can be linked to the diameter. Myelinated axons are quite large in diameter, ranging from 1 - 13 µm. Unmyelinated axons on the other hand have a small diameter– generally less than 0.2 µm in the central nervous system and less than 1 µm in the peripheral nervous system. In unmyelinated axons, the conduction velocity is proportional to its (diameter)½ while the conduction velocity in myelinated axons increases linearly. This means that myelinated axons that are the same diameter as unmyelinated axons can conduct signals much faster.
Brain myelination
Myelination in the human brain is a continuous process from birth and is not mature until about 2 years of age. At this stage, motor and sensory systems are mature and myelination of the cerebral hemispheres is largely complete. There are, however, some processes which myelinate later in life: some thalamic radiations will be mature at about 5 - 7 years of age; and myelination of intracortical connections between association cortices continues into the 20s and 30s.
Brain myelination begins in utero, developing quite prominently from the 24th week of gestation. At birth, the myelination process continues to progress, and completes at about 2 years of age. It’s progression is predictable, and correlates with developmental milestones such as learning to walk.
Throughout the first year of life, myelin will spread throughout the whole brain in an orderly manner. Generally, myelination will start in the brainstem and progress to the cerebellum and basal ganglia, then will continue rostrally to the cerebrum, and rostrally from the occipital and parietal lobes to the frontal and temporal lobes. The progression typically follows the order from central to peripheral, caudal to rostral (inferior to superior), and dorsal to ventral (posterior to anterior).
In the cerebrum, myelination progresses from the lower order cortices to higher order cortices. Primary cortical areas such as the primary motor cortex myelinate first, followed by secondary cortices, such as the premotor and supplementary motor cortices, and finally tertiary cortical areas such as the prefrontal cortex.
Take a deeper look at myelinating cells and their functions with this study unit:
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Clinical aspects
Demyelination
Myelin sheaths around an axon can be destroyed or lost in demyelination. This may lead to a deterioration of the nerve because of the loss of myelin protecting the axon. It can be caused by various disorders like Tay-Sachs or multiple sclerosis and also by damaging medical treatments like chemotherapy and radiotherapy.
Schwannoma
A Schwannoma (neurilemmoma) is a benign peripheral nerve tumor that originates from Schwann cells. These tumors are most commonly found in the head and neck region. When a tumor arises from the vestibulocochlear nerve (CN VIII), it is known as an acoustic neuroma. It usually occurs in the middle years of life and interferes with the function of the cochlear root of the vestibulocochlear nerve– impairing hearing, causing unilateral hearing loss, and sometimes symptoms of tinnitus. If the tumor expands into the internal acoustic meatus, function of the facial nerve (CN VII) can also be affected, as the facial nerve runs adjacent to the vestibulocochlear nerve in the internal acoustic meatus. This can result in facial paralysis, swallowing difficulty, and a loss of sensation in the face. Schwannomas are typically treated by surgical removal with an excellent prognosis in most cases.
Sources
All content published on Kenhub is reviewed by medical and anatomy experts. The information we provide is grounded on academic literature and peer-reviewed research. Kenhub does not provide medical advice. You can learn more about our content creation and review standards by reading our content quality guidelines.
References:
Diepenbrock, N.H. (2012) Schwannoma. In Quick Reference To Critical Care. (2011). Wolters Kluwer Health, Lippincott Williams & Wilkins.
Martin, J. H. (2012). Neuroanatomy: Text and Atlas (4th ed.). New York: McGraw-Hill Medical.
Rapp, Peter R., & Bachevalier, Jocelyne. (2013). Chapter 43 - Cognitive Development and Aging. In Fundamental Neuroscience (pp. 919–945). Retrieved from
Ross, M. H., & Pawlina, W. (2011). Histology: A Text and Atlas (6th ed.). Philadelphia, PA: Lippincott Williams & Wilkins.
Tortora, G. J., & Nielsen, M. (2014). Principles of Human Anatomy (13th ed.). Hoboken, NJ: Wiley.
Waxman, S. G. (2010). Clinical neuroanatomy (26th ed.). New York: McGraw-Hill.
Article, review and layout:
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10239 | https://www.reddit.com/r/learnmath/comments/t04ooq/when_p_12_why_does_the_binomial_formula_become_2n/ | When p = 1/2, why does the binomial formula become 2^n : r/learnmath
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10240 | https://www.biophysics.org/Portals/0/BPSAssets/Articles/beard.pdf | A Molecular Modeler’s Guide to Statistical Mechanics Course notes for BIOE575 Daniel A. Beard Department of Bioengineering University of Washington Box 3552255 dbeard@bioeng.washington.edu (206) 685 9891 April 11, 2001 Contents 1 Basic Principles and the Microcanonical Ensemble 2 1.1 Classical Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 1.2 Ensembles and Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 1.2.1 An Ensembles of Particles . . . . . . . . . . . . . . . . . . . . . . . . . .
3 1.2.2 Microscopic Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . .
4 1.2.3 Formalism for Classical Systems . . . . . . . . . . . . . . . . . . . . . . .
7 1.3 Example Problem: Classical Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . .
8 1.4 Example Problem: Quantum Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . .
10 2 Canonical Ensemble and Equipartition 15 2.1 The Canonical Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 2.1.1 A Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 2.1.2 Another Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 2.1.3 One More Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 2.2 More Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 2.3 Formalism for Classical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 2.4 Equipartition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 2.5 Example Problem: Harmonic Oscillators and Blackbody Radiation . . . . . . . . .
21 2.5.1 Classical Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 2.5.2 Quantum Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 2.5.3 Blackbody Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 2.6 Example Application: Poisson-Boltzmann Theory . . . . . . . . . . . . . . . . . .
24 2.7 Brief Introduction to the Grand Canonical Ensemble . . . . . . . . . . . . . . . .
25 3 Brownian Motion, Fokker-Planck Equations, and the Fluctuation-Dissipation Theo-rem 27 3.1 One-Dimensional Langevin Equation and Fluctuation-Dissipation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 3.2 Fokker-Planck Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29 3.3 Brownian Motion of Several Particles . . . . . . . . . . . . . . . . . . . . . . . .
30 3.4 Fluctuation-Dissipation and Brownian Dynamics . . . . . . . . . . . . . . . . . .
32 1 Chapter 1 Basic Principles and the Microcanonical Ensemble The first part of this course will consist of an introduction to the basic principles of statistical mechanics (or statistical physics) which is the set of theoretical techniques used to understand microscopic systems and how microscopic behavior is reflected on the macroscopic scale. In the later parts of the course we will see how the tool set of statistical mechanics is key in its application to molecular modeling. Along the way in our development of basic theory we will uncover the principles of thermodynamics. This may come as a surprise to those familiar with the classical engineering paradigm in which the laws of thermodynamics appear as if from the brain of Jove (or from the brain of some wise old professor of engineering). This is not the case. In fact, thermodynamics arises naturally from basic principles. So with this foreshadowing in mind we begin by examining the classical laws of motion1.
1.1 Classical Laws of Motion Recall Newton’s famous second law of motion, often expressed as , where is the force acting to accelerate a particle of mass with the acceleration . For a collection of particles located at Cartesian positions the law of motion becomes (1.1.1) where are the forces acting on the particles2.
We shall see that in the absence of external fields or dissipation the Newtonian equation of motion preserves total energy: "!$#%'& ( ) + -, -, !$#/.
0 - 212 (1.1.2) 1This course will be concerned primarily with classical physics. Much of the material presented will be applicable to quantum mechanical systems, and occasionally such references will be made.
2A note on notation: Throughout these notes vectors are denoted by bold lower case letters (e.g.
354 , 674 ). The notation 8 3
4 denotes the time derivative of 354 , i.e., 8 354:9<;3
4>=5;? , and @ 354:9<;AB354C=D; ?EA .
2 Chapter 1 – Basic Principles 3 where # is some potential energy function and FHGJIF#LKMI and is the kinetic energy.
Another way to pose the classical law of motion is the Hamiltonian formulation, defined in terms of the particle positions NO P and momenta N5Q SR O P . It is convenient to adopt the notation (from quantum mechanics) QUT momenta and VWT positions, and to consider the scalar quantities Q and V , which denote the entries of the vectors X and Y . For a collection of particles X[Z]_^ and YZ]\_^ are the collective positions and momenta vectors listing all aS entries.
The so called Hamiltonian function is an expression of the total energy of a system: b ^ ) c Q ( !d#/.
V -Ve V ^ f1f (1.1.3) Hamilton’s equations of motion are written as: R V g I b I Q (1.1.4) R Q h G I b I V (1.1.5) Hamilton’s equations are equivalent to Newton’s: R V F Q EKM R Q HGJIi#fKSI V jlke (1.1.6) So why bother with Hamilton when we are already familiar with Newton? The reason is that the Hamiltonian formulation is often convenient. For example, starting from the Hamiltonian formulation, it is straightforward to prove energy conservation: m b m n ^ ) c o I b I V R V :!
I b I Q R Q Cp ^ ) + q o I b I V p o I b I Q pdG o I b I Q p o I b I V pLr (1.1.7) 1.2 Ensembles and Thermodynamics With our review of the equations of classical mechanics complete, we undertake our study of sta-tistical physics with an introduction to the concepts of statistical thermodynamics. In this section thermodynamics will be briefly introduced as a consequence of the interaction of ensembles of large numbers of particles. The material loosely follows Chapter 1 of Pathria’s Statistical Mechan-ics , and additional information can be found in that text.
1.2.1 An Ensembles of Particles Consider a collection of particles confined to a volume s , with total internal energy . A system of this sort is often referred to as an NVE system, as , s , and are the three thermodynamic variables that are held fixed. [In general three variables are necessary to define the thermodynamic state of a system. Other thermodynamic properties, such as temperature for example, cannot be assigned in an NVE ensemble without changing at least one of the variables , s , or .] We will refer to the thermodynamic state as the macrostate of the system.
Chapter 1 – Basic Principles 4 For a given macrostate, there is likely to be a large number of possible microstates, which cor-respond to different microscopic configurations of the particles in the system. According to the principles of quantum mechanics there is a finite fixed number of microscopic states that can be adopted by our NVE system. We denote this number of states as t .
< Dsu 1 3. For a classical system, the microstates are of course not discrete and the number of possible states for a fixed vs ensemble is in general not finite. To see this imagine a system of a single particle ( & ) travelling in an otherwise empty box of volume s . There are no external force fields acting on the particle so its total energy is w . The particle could be found in any location within the box, and its velocity could be directed in any direction without changing the thermodynamic macrostate defined by the fixed values of , s , and . Thus there are an infinite number of al-lowable states. Let us temporarily ignore this fact and move on with the discussion based on a finite (yet undeniably large) t .
x Dsy 1 . This should not bother those of us familiar with quantum mechanics. For classical applications we shall see that bookkeeping of the state space for classi-cal systems is done as an integration of the continuous state space rather than a discrete sum as employed in quantum statistical mechanics.
At this point don’t worry about how you might go about computing t .
< Dsu 1 , or how t might depend on , s , and for particular systems. We’ll address these issues later. For now just appreciate that the quantity t .
x 5su 1 exists for an NVE system.
1.2.2 Microscopic Thermodynamics Consider two such so called NVE systems, denoted system 1 and system 2, having macrostates defined by ( z , sF , ) and ( { , s| , ), respectively.
N1, V1, E1 N2, V2, E2 Figure 1.1: Two NVE systems in thermal contact.
Next, bring the two systems into thermal contact (see Fig. 1.1). By thermal contact we mean that the systems are allowed to exchange energy, but nothing else. That is and may change, but z , } , sF , and s| remain fixed. Of course the total energy remains fixed as well, that is, _~ ! J (1.2.8) if the two systems interact only with one another.
Now we introduce a fundamental postulate of statistical mechanics: At any time, system 1 is equally likely to be in any one of its tJ microstates and system 2 is equally likely to be in any one of its tf microstates (more on this assumption later). Given this assumption, the composite system is equally likely to be in an one of its t ~ . 5 1 possible microstates. The number t ~ . 51 can be expressed as the multiplication: t ~ . D1 tJ . -1tf . 1f (1.2.9) 3The number corresponds to the number of independent solutions to the Schr¨ odinger equation that the system can adopt for a given eigenvalue of the Hamiltonian.
Chapter 1 – Basic Principles 5 Next we look for the value of (or equivalently, ) for which the number of microstates t ~ . D 1 achieves its maximum value. We will call this achievement equilibrium, or more specifically thermal equilibrium. The assumption here being that physical systems naturally move from improbable macrostates to more probable macrostates4. Due to the large numbers with which we deal on the macro-level ( &e ^ ), the most probable macrostate is orders of magnitude more probable than even closely related macrostates. That means that for equilibrium we must maximize t ~ . D 1 under the constraint that the sum ~ ! remains constant.
At the maximum I t ~ KMI , or I/ tJ . 1tf . 1B I q I tJ tf !
tJ I I I tf I r||
: (1.2.10) where . { { 1 denote the maximum point.
Since I KSI G & from Eq. (1.2.8), Equa-tion (1.2.10) reduces to: & tJ I tJ I . 1 & tf} I tf I . 1f (1.2.11) which is equivalent to I I 7 tJ . 1 I I tf . 1f (1.2.12) To generalize, for any number of systems in equilibrium thermal contact, I I 7 t ] constant (1.2.13) for each system.
Let us pause and think for a moment: From our experience, what do we know about systems in equilibrium thermal contact? One thing that we know is that they should have the same temper-ature. Most people have an intuitive understanding of what temperature is. At least we can often gauge whether or not two objects are of equal or different temperatures. You might even know of a few ways to measure temperature. But do you have a precise physical definition of temperature?
It turns out that the constant is related to the temperature ¡ via ] & K ¢ ¡ (1.2.14) where ¢ is Boltzmann’s constant. Therefore temperature of the NVE ensemble is expressed as ¡ & ¢ I I 7 t .
< Dsy 1 (1.2.15) Until now some readers may have had a murky and vague mental picture of what the thermody-namic variable temperature represents. And now we all have a murky and vague mental picture of what temperature represents. Hopefully the picture will become more clear as we proceed.
Next consider that systems 1 and 2 are not only in thermal contact, but also their volumes are allowed to change in such a way that the total volume s ~ si !
s£ remains constant. For this 4Again, the term macrostate refers to the thermodynamic state of the composite system, defined by the variables 2¤ , S¤ , ¥¤ , and A , A , A . A more probable macrostate will be one that corresponds to more possible microstates than a less probable macrostate.
Chapter 1 – Basic Principles 6 example imagine a flexible wall separates the two chambers – the wall flexes to allow pressure to equilibrate between the chambers, but the particles are not allowed to pass. Thus z and { remain fixed. For such a system we find that maximizing t ~ .
sF Ds|D1 yields I¦ tJ .
sF 1tf .
s|D1B I sF q I tJ sF tf !
tJ I s£ I si I tf I s| r|§ § § § (1.2.16) or & tJ I tJ I sF .
s 1 & tf I tf I s| .
s 12 (1.2.17) or I I si 7 tJ .
s 1 I I s| tf .
s 1f (1.2.18) or I I s t %¨© constant (1.2.19) We shall see that the parameter ¨ is related to pressure, as you might expect. But first we have one more case to consider, that is mass equilibration. For this case, imagine that the partition between the chambers is perforated and particles are permitted to freely travel from one system to the next. The equilibrium statement for this system is I I 7S t ª« constant (1.2.20) To summarize, we have the following: 1. Thermal (Temperature) Equilibrium: ¬ ¬ 7 t .
2. Volume (Pressure) Equilibrium: ¬ ¬ § t ¨ .
3. Number (Concentration) Equilibrium: ¬ ¬ 7S t ª .
How do these relationships apply to the macroscopic world with which we are familiar? Recall the fundamental expression from thermodynamics: m ¡ m¯® G±° m s !³² m (1.2.21) which tells us how to relate changes in energy to changes in the variables entropy ® , volume s , and number of particles , occurring at temperature ¡ , pressure ° , and chemical potential ² . Equation (1.2.21) arose as an empericism which relates the three intrinsic thermodynamic properties ¡ , ° , and ² to the three extrinsic properties , s , and . In developing this relationship, it was necessary to introduce a novel idea, entropy, which we will try to make some sense of below.
For constant s and Equation (1.2.21) gives us o I ® I p §v & ¡ (1.2.22) Chapter 1 – Basic Principles 7 Going back to Equation (1.2.13) we see that ® ¢ 7S t% (1.2.23) which makes sense if we think of entropy as a measure of the total disorder in a system. The greater the number of possible states, the greater the entropy. For pressure and chemical potential we find the following relationships: For constant and we arrive at o I ® I s pJ ° ¡ or I I s t ¨ and ¨¦ ° ¢ ¡ (1.2.24) For constant and s we obtain o I ® I p § ´G ² ¡ or I I t ª and ª/´G ² ¢ ¡ (1.2.25) For completeness we repeat: o I ® I p §] & ¡ or I I 7S t % and v & ¢ ¡ (1.2.26) Through Eqs. (1.2.24)-(1.2.26) the internal thermodynamic parameters familiar to our everyday experience – temperature, pressure, and chemical potential – are related to the microscopic world of , s , and . The key to this translation is the formula ® µ¢ 7S t . As Pathria puts it, this formula “provides a bridge between the microscopic and the macroscopic” .
After introducing such powerful theory it is compulsory that we work out some example prob-lems in the following sections. But I recommend that readers tackling this subject matter for the first time should pause to appreciate what they have learned so far. By asserting that entropy (the most mysterious property to arise in thermodynamics) is simply proportional to the log of the num-ber of accessible microstates, we have derived direct relationships between the microscopic to the macroscopic worlds.
Before moving on to the application problems I should point out one more thing about the number t – that is its name. The quantity t is commonly referred to as the microcanonical partition function, a partition function being a statistically weighted sum over the possible states of a system. Since t is a non-biased enumeration of the microstates, we refer to it as microcanonical.
Similarly, another name for the NVE ensemble is the microcanonical ensemble. Later we will meet the canonical (NVT) and grand canonical ( ² VT) ensembles.
1.2.3 Formalism for Classical Systems The microcanonical partition function for a classical system is proportional to the volume of phase space accessible by the system. For a system of particles the phase space is a 6 -dimensional space encompassing the a variables V and the a variables Q , and the partition function is Chapter 1 – Basic Principles 8 proportional to the integral: t .
x 5su 1u·¶¸ .
b .
QF 5 Q|e Q ^ _¹
V V -V ^ f1 G 1 m QF m Q|j m Q ^ m V m Vj m V ^ (1.2.27) or using vector shorthand t .
< Dsy 1u ¶ ¸ .
b .
Xº 5Y»1 G 1 m ^ X m ^ Y (1.2.28) where the notation m ^ reminds us that the integration is over a -dimensional space.
In Eqs. (1.2.27)-(1.2.28) the delta function restricts the integration to the the constant energy hypersurface defined by b .
X 5Yi1 constant. [In general we won’t be integrating this difficult-looking delta function directly. Just think of it as a mathematical shorthand for restricting the phase space to a constant-energy subspace.] We notice that Equation (1.2.28) lacks a constant of proportionality that allows us to replace the proportionality symbol with the equality symbol and compute t .
x 5su 1 . This constant comes from relating a given volume of the classical phase space to a discrete number of quantum mi-crostates. It turns out that this constant of proportionality is & K ¼¾½ ^ , where ½ is Planck’s constant.
Thus t .
x Dsy 1 & ¿¼À½ ^ ¶£Á m ^ X m ^ Y (1.2.29) where the integration à Á is over the subspace defined by b .
Xº 5Y»1 .
From where does the constant ¿¼À½Ä^ come? We know from quantum mechanics that to specify the position of a particle, we have to allow its momentum to lose coherence. Similarly, when we specify the momentum with increasing certainty, the position loses coherence. If we consider ÅzV and ÅQ to be the fundamental uncertainties in position an momenta, then Planck’s constant tells us how these uncertainties depend upon one another: ½ÆlÅQ£Å¦V¦ (1.2.30) Thus the minimal discrete volume element of phase space is approximately ½ for a single particle in one dimension, or ½ ^ when there are a degrees of freedom. This explains (heuristically at least) the factor of ½ ^ . From where does the ¼ come? We shall see when we enumerate the quantum states of the ideal gas, the indistinguishability of the particles further reduces the partition function by a factor of ¿¼ , which fixes t as the number of distinguishable microstates.
1.3 Example Problem: Classical Ideal Gas A system of noninteracting monatomic particles is referred to as the ideal gas. For such a system the kinetic energy is the only contribution to the Hamiltonian b ) Q K ( , and t .
x 5su 1 & ¿¼À½ ^ ¶ § m ^ Y{¶ Á m ^ X (1.3.31) Chapter 1 – Basic Principles 9 where à § represents integration over the volume of the container. [The integral can be split into à § and à Á components because b .
X1 does not depend on particle positions in any way.] Therefore t .
x Dsy 1 & ¿¼À½ ^ s ¶ÄÁ m ^ X[ (1.3.32) It turns out that knowing tÇ·s is enough to derive the ideal gas law: ° ¢ ¡ I t I s s (1.3.33) or ° s ¢ <¡ ÈiÉ ¡· (1.3.34) where ÉÊ´¢ }Ë is the gas constant, È is the number of particles in moles, and }Ë is Avogadro’s number. For other properties (like energy and entropy) we need to do something with the à Á integral in Equation (1.3.32).
We approach this integral by first noticing that the constant energy surface ^ ) + Q ( defines a sphere of radius Éʵ.
( 1 ÌÍ in a -dimensional space. We can find the volume and surface area of such a sphere from a handbook of mathematical functions. In \ ^ the volume and surface area of a sphere are given by: s ^ . É 1 Î ^ ÌÍ .
a K ( 1D¼ É ^ and ® ^ .ÏÉ 1 ( Î ^ ÌÍ .
aS K ( G & 1 ¼ É ^ uÐ| (1.3.35) [One may wonder what to do in the case where a K ( is non-integral. Specifically, how would one define the .
a K ( 1 ¼ and .
a K ( G & 1D¼ factorials? We could use gamma functions Ñ .
a K ( !
& 1 and Ñ .
aS K ( 1 , where Ñ .ϲ 1 is a generalization of the factorial function: Ñ .E² 1 ¶ÇÒ ~ Ó ÐÕÔÖ|×Ð| m Ö . It turns out that Ñ .E² 1 Ø.E²G & 1D¼ for ²ÂÙ . So in the above equations for surface area and volume we are using the generalized factorial function ² ¼ ¶ Ò ~ Ó ÐÕÔÖ|× m Ö , which is the same as the regular factorial function for non-negative integer arguments.] Returning to the task at hand: we wish to evaluate the integral à Á m ^ X over the constant energy surface in \_^ defined by ^ ) + Q ( . One way to do this is to take ¶ÄÁ m ^ X ® ^ .-.
( 1 ÌÍ 1 (1.3.36) which gives us t .
x Dsy 1 a ¿¼ .
a K ( 1 ¼ q s ½ ^ .
( Î 1 ^ ÌÍ r .
( 1 ÌÍ (1.3.37) Taking the of this function, we will employ Stirling’s approximation, that ¿¼¯Æl G , for large . Thus 7S t .
x Dsy 1 7 q s ½ ^ .
( Î 1 ^ ÌÍ r !<Ú ( G vÛ o aS ( p ^ ÌÍ-Ü !
.
a v1 G & ( 7 .
( 1L (1.3.38) Chapter 1 – Basic Principles 10 In the limit of large , we know that the first three terms grow faster than the last two. So com-bining the 7 .
1 terms and keeping only terms of order and .
Ý1 results in 7 t .
x Dsy 1uÆl ]Û s ½ ^ oyÞ Î a p ^ ÌÍ Ü !
Ú ( (1.3.39) or the entropy ® .
x 5su 1yÆ ¢ 7 vÛ s ½ ^ o Þ Î a p ^ ÌÍ-Ü !ÊÚ ¢ ( (1.3.40) Using our thermodynamic definition of temperature, o I ® I p §Ý a ( ¢ & & ¡ or (1.3.41) a ( ¢ ¡ and ¡ ( a ¢ o p (1.3.42) As you can see, the internal energy is proportional to the temperature and, as expected, the number of particles. Inserting a ¢ ¡ K ( into Equation (1.3.40), we get: ® .
< Dsy -¡1 ¢ o s p !
a ( ¢ q Ú a !
7 o ( Î <¢ ¡ ½ pr (1.3.43) which is the Sackur-Tetrode equation for entropy.
We should note that if, instead of taking the integral à Á m ^ X to be the surface area of the constant-energy sphere, we had allowed the energy to vary within some small range, we would have arrived at the same results. In fact we shall see that, for the quantum mechanical ideal gas, that is precisely what we will have to do.
1.4 Example Problem: Quantum Ideal Gas As we saw for the classical ideal gas, analysis of the quantum mechanical ideal gas will hinge on the enumeration of the partition function, and not on the analysis of the underlying equations of motion. Nevertheless, it is necessary to introduce some quantum mechanical ideas to understand the ideal gas from the perspective of quantum mechanics. It will be worthwhile to go through this exercise to appreciate how statistical mechanics naturally applies to the discrete states observed in quantum systems.
First we must find the quantum mechanical state, or wave function ß .
Ö -à 5á 1 , of a single par-ticle living in an otherwise empty box. The equation describing the shape of the constant-energy wave function for a single particle in the presence of no potential field is G ½ â Î ³ã ß .
Ö à 5á01 ´G ½ â Î q I I Ö !
I I à !
I I á r ß .
Ö à á012 (1.4.44) We solve Equation (1.4.44) , a form of Schr¨ odinger’s equation, with the condition that ß .
Ö à 5á01yä on the walls of the container. The constant energy (the subscript “1” reminds us that this is Chapter 1 – Basic Principles 11 the energy of a single particle) is an eigenvalue of the Hamiltonian operator on the left hand side of Equation (1.4.44). Under these conditions the single-particle wave function has the form: ß .
Ö à 5á01 o ( å p ^ ÌÍæç7èé È Ô Î Ö å ê æ-çèé Èë Î à å ê æç7èé Èì Î á å ê (1.4.45) where the È Ô , Èë , and ÈFì can be any of the positive integers (1, 2, 3, . . . ). Here the box is assumed to be a cube with sides of length å . The energy is related to these numbers via ½ â å «í È Ô !±È ë !±È ìDî (1.4.46) If energy is fixed then the number of possible quantum states is equal to the number of sets N È Ô Èë ÈìP for which í È Ô !ïÈ ë !³È ìDî â s BÌ ^ ½ (1.4.47) where s BÌ ^ å . For a system of noninteracting particles, we have such sets of three integers, and the energy is the sum of the energies from each particle: ^ ) + È â s BÌ ^ ½ (1.4.48) where now represents the total energy of the system, and is a nondimensionalization of energy.
The similarities between the classical ideal gas and Equation (1.4.48) are striking. As in the classical system, the constant energy condition limits the quantum phase space to the surface of a sphere in a dimensional space. The important difference is that for the quantum mechanical system the phase space is discrete because the N È P are integers. This discrete nature of the phase space means that t .
x 5su 1 can be more difficult to pin down than it was for the classical case.
To see this imagine the regularly spaced lattice in a dimensional space which is defined by the set of positive integers N È P . The number t .
x Dsy 1 is equal to the number of lattice points which fall on the surface of the sphere defined by Equation (1.4.48)–this number is an irregular function of .
x 5su 1 . As an illustration, return to the single particle case. There is one possible quantum state for a ½ â s BÌ ^ and three possible states for a0½ Þ s BÌ ^ . Yet there are no possible states for energies falling between these two energies. Thus the distinct microstates can be difficult to enumerate. We shall see that as and become large, the discrete spectrum becomes more regular and smooth and easier to handle.
Consider the number ð .
x 5su 1 which we define to be the number of microstates with energy less than or equal to . In the limit of large and large , ð .
< Dsy 1 is equal to the volume of the “positive compartment” of a a dimensional sphere. Recalling Equation (1.3.35) gives ð . 1 o & ( p ^ q Î ^ ÌÍ .
a K ( 1 ¼ ^ ÌÍ r (1.4.49) [The factor .
& K ( 1B^ comes from limiting ð . « 1 to the volume spanned by the positive values of N È P .] Plugging in « â s BÌ ^ K ½ results in ð .
x Dsy 1 o s ½ ^ p .
( Î 1^ ÌÍ .
aS K ( 1 ¼ (1.4.50) Chapter 1 – Basic Principles 12 Next we calculate t .
x 5su 1 from ð .
x 5su 1 by assuming that the energy varies over some small range ´ñ Å , where Åóò . The enumeration of microstates within this energy range can be calculated as t .
x 5su ¹5Å©1uÆlÅ I ð .
x Dsy 1 I (1.4.51) which is valid for small Å (relative to ). From Equation (1.4.50), we have I ð I o a ( p ð ô (1.4.52) and thus t .
x Dsy ¹5Å©1 o a ( p ð Å (1.4.53) and t .
x 5su ¹5Å©1 7 vÛ s ½ ^ o Þ Î a p ^ ÌÍ-Ü !
a ( !
7 o aS ( p !
o Å p (1.4.54) As for the classical ideal gas, we take terms of order and order which grow much faster than 7 and the constant terms. Thus 7 t .
x Dsy 1 vÛ s ½ ^ o Þ Î a p ^ ÌÍ-Ü !
a ( (1.4.55) From Equation (1.4.55) we could derive the thermodynamics of the system, just as we did for the classical ideal gas. However we notice that the entropy, which is given by ® .
x 5su 1 l¢ ]Û s ½ ^ o Þ Î a p ^ ÌÍ-Ü !
a ¢ ( (1.4.56) is not equivalent to the Sackur-Tetrode expression, Equation (1.3.43). [The difference is a factor of & K ¼ in the partition function, which is precisely the factor that we added to the classical partition function, Equation (1.2.29), with no solid justification.] In fact, one might notice that the entropy, according to Equation (1.4.56) is not an extensive measure! If we increase the volume, energy, and number of particles by some fixed proportion, then the entropy will not increase by the same proportion. What have we done wrong? How can we recover the missing factor of & K ¿¼ ?
To justify this extra factor, we need to consider that the particles making up the ideal gas system are not only identical, they are also indistinguishable. We label the possible states that a given particle can be in as state 1, state 2, etc., and denote the number that exist in each state at a given instant as õ , õL , etc. Thus there are õ particles in state 1, and õL particles in state 2, and so on. Since the particles are indistinguishable, we can rearrange the particles of the system (by switching the states of the individual particles) in any way as long as the numbers Nõ ÏP remain unchanged, and the microstate of the system is unchanged. The number of ways the particles can be rearranged is given by ¿¼ õ D¼öõL¼÷÷ Chapter 1 – Basic Principles 13 Introducing another assumption, that if the temperature is high enough that the number of possible microstates of a single particle is so fantastically large that each possible single particle state is represented by, at most, one particle, then õ ¼ & (because each õ is either 1 or 0). Thus we need to correct the partition function by a factor & K `¼ , and as a result Equation (1.4.55) reduces to Equation (1.3.39).
Chapter 1 – Basic Principles 14 Problems 1. (Warm-up Problem) Invert Equation (1.3.40) to produce an equation for .
® Dsu v1 . Us-ing this equation and our basic thermodynamic definitions, derive the pressure-volume law (equation of state). How does this compare with Equation (1.3.34)?
2. (Particle in a box) Verify that Equation (1.4.45) is a solution to Equation (1.4.44). Evaluate the integral à § ß m Ö m à m á , for the one-particle system. What are the 6 lowest possible ener-gies of this system? For each of the 6 lowest energies count the number of corresponding quantum states. Are the energy levels equally spaced? Does the number of quantum states increase monotonically with E?
3. (Gas of finite size particles) Consider a gas of particles that do not interact in any way except that, each particle occupies a finite volume wø which cannot be overlapped by other particles.
What consequences does this imply for the ideal gas law? [Hint: return to the relationship t .
s/1{ùÃ m ^ Y . You might try assuming that each particle is a solid sphere.] Plot ° vs.
s for both the ideal gas law and the ° -s relationship for the finite-volume particles. (Use ¡ a K and ÈU & mole.) Discuss the following questions: Where do the curves differ?
Where are they the same? Why?
4. (Phase space of simple harmonic oscillator) Consider a system made up of a single particle of mass attached to a linear spring, with spring constant ú . One end of the spring is attached to the particle, the other is fixed in space, and the particle is free to move in one dimension, V . What is the Hamiltonian b .
Qi V 1 for this system? Plot the phase space for b .
Q» VS1 . Find an expression for the entropy ® . 1 of this system. You can assume that energy varies over some small range: ûñ Å , Åüò . Using & K ¡ I ® KSI , derive an expression for the “temperature” of this system. We saw that the ideal gas has an internal temperature of a ¢ ¡ K ( per particle. How does the energy as a function of temperature for the simple harmonic oscillator compare to that for the ideal gas? Does it make sense to calculate the “temperature” of a one-particle system? Why or why not?
Chapter 2 Canonical Ensemble and Equipartition In Chapter 1 we studied the statistical properties of a large number of particles interacting within the microcanonical ensemble – a closed system with fixed number of particles, volume, and inter-nal energy. While the microcanonical ensemble theory is sound and useful, the canonical ensemble (which fixes the number of particles, volume, and temperature while allowing the energy to vary) proves more convenient than the microcanonical for numerous applications. For example, consider a solution of macromolecules stored in a test tube. We may wish to understand the conformations adopted by the individual molecules. However each molecule exchanges energy with its environ-ment, as undoubtedly does the entire system of the solution and its container. If we focus our attention on a smaller subsystem (say one molecule) we adopt a canonical treatment in which variations in energy and other properties are governed by the statistics of an ensemble at a fixed thermodynamic temperature.
2.1 The Canonical Distribution 2.1.1 A Derivation Our study of the canonical ensemble begins by treating a large heat reservoir thermally coupled to a smaller system using the microcanonical approach. The energy of the heat reservoir denoted Lý and the energy of the smaller subsystem, . The system is assumed closed and the total energy is fixed: ! Lý ø constant.
This system is illustrated in Fig. (2.1). For a given energy of the subsystem, the reservoir can obtain t ý . ø2G 1 microstates, where t ý is the microcanonical partition function. According to our standard assumption that the probability of a state is proportional to the number of microstates available: ° lt ý . Lý 1 t . øºG 1f (2.1.1) We take the 7 of the microcanonical partition function and expand about 7 t ý . ø 1 : ° 7 t ý . ø 1 G I t ý I Lý þ þ þ þ ¯ÿ ¥. 1 !
O . 12 (2.1.2) 15 Chapter 2 – Canonical Ensemble 16 Er E Figure 2.1: A system with energy thermally coupled to a large heat reservoir with energy Lý .
For large reservoirs ( ò ø ) the higher order terms in Equation (2.1.2) vanish and we have 7 ° t ý constant G ¢ ¡ (2.1.3) where we have used the microcanonical definition of thermodynamic temperature ¡ . Thus ° Ó Ð (2.1.4) where & K ¢ ¡ has been defined previously. Equation 2.1.4) is the central result in canonical ensemble theory. It tells us how the probability of a given energy of a system depends on its energy.
2.1.2 Another Derivation A second approach to the canonical distribution found in Feynman’s lecture notes on statistical mechanics is also based on the central idea from microcanonical ensemble theory that the probability of a microstate is proportional to t . Thus °. 1 °. D1 t . øºG -1 t . øºG 1 (2.1.5) where again ø is the total energy of system and a heat reservoir to which the system is coupled.
The energies and are possible energies of the system and t is the microcanonical partition function for the reservoir. (The subscript O has been dropped.) Next Feynman makes us of the fact that energy is defined only up to an additive constant. In other words, there is no absolute energy value, and we can always add a constant, say , so long as we add the same to all relevant values of energy. Without changing its physical meaning Equation (2.1.5) can be modified: ° . 1 ° . D1 t . øuG !
1 t . øuG !
1 (2.1.6) Next we define the function .
Ö 1 t . ø¥G !
Ö 1 . Equating the right hand sides of Eqs. (2.1.5) and (2.1.6) results in t . øuG -1t . øuG !
51 t . øuG 1t . øuG !
1 (2.1.7) or . G -1 .
51 .
1 .
! G -12 (2.1.8) Chapter 2 – Canonical Ensemble 17 Equation (2.1.8) is uniquely solved by: .
51 .
1 Ó Ð (2.1.9) where is some constant. Therefore the probability of a given energy is proportional to Ó Ð Ì , which is the result from Section 2.1.1. To take the analysis one step further we can normalize the probability: °. 1 Ó Ð (2.1.10) where û ) Ó Ð (2.1.11) is the canonical partition function and Equation (2.1.10) defines the canonical distribution func-tion. [Feynman doesn’t go on to say why " & K ¢ ¡ ; we will see why later.] Summation in Equation (2.1.11) is over all possible microstates. Equation (2.1.11) is equation #1 on the first page of Feynman’s notes on statistical mechanics . Feynman calls Equation (2.1.10) the “sum-mit of statistical mechanics, and the entire subject is either the slide-down from this summit...or the climb-up.” The climb took us a little bit longer than it takes Feynman, but we got here just the same.
2.1.3 One More Derivation Since the canonical distribution function is the summit, it may be instructive to scale the peak once more from a different route. In particular we seek a derivation that stands on its own and does not rely on the microcanonical theory introduced earlier.
Consider a collection of identical systems which are thermally coupled and thus share en-ergy at a constant temperature. If we label the possible states of the system & ( and denote the energy these obtainable microstates as , then the total number is system is equal to the summation, ) ÈF (2.1.12) where the È are the number of systems which correspond to microstate . The total energy of the ensemble can be computed as ) È F # (2.1.13) where # is the average internal energy of the systems in the ensemble.
Eqs. (2.1.12) and (2.1.13) N ÈÍP represent constraints on the ways microstates can be distributed amongst the members of the ensemble. Analogous to our study of microcanonical statistics, here we assume that the probability of obtaining a given set N È P of numbers of systems in each mi-crostate is proportional to the number of ways this set can be obtained. Imagine the numbers È to represent bins count the number of systems at a given state. Since the systems are identical, they can be shuffled about the bins as long as the numbers È remain fixed. The number of possible ways to shuffle the states about the bins is given by: .
N È P 1 H¼ È D¼ È e¼ (2.1.14) Chapter 2 – Canonical Ensemble 18 One way to arrive at the canonical distribution is via maximizing the number under the constraints imposed by Eqs. (2.1.12) and (2.1.13). At the maximum value, ã ¸ (2.1.15) where the gradient operator is ã ¬ ¬ !
¬ ¬ !
, and ¸ is a vector which represents a direction allowed by the constraints.
[The occasional mathematician will point out the hazards of taking the derivative of a dis-continuous function with respect to a discontinuous variable. Easy-going types will be satisfied with the explanation that for astronomically large numbers of possible states, the function and the variables N È P are effectively continuous. Sticklers for mathematical rigor will have to find satisfaction elsewhere.] We can maximize the number by using the method of Lagrange multipliers. Again, it is convenient to work with the 7S of the number , which allows us to apply Stirling’s approxima-tion.
7S 7 .
H¼À1 G ) È 7S .ÏÈ ¼À1L (2.1.16) This equation is maximized by setting ã G ã ) ÈG ã ) È F (2.1.17) where and are the unknown Lagrange multipliers. The second two terms in this equations are the gradients of the constraint functions. Evaluating Equation (2.1.17) results in: G ÈG & GvG i (2.1.18) in which the entries of the gradients in Equation (2.1.17) are entirely uncoupled. Thus Equa-tion (2.1.18) gives us a straightforward expression for the optimal È : ÈF Ó Ð (2.1.19) where the unknown constants and can be obtained by returning to the constraints.
The probability of a given state follows from the first constraint (2.1.12) °! L Ó Ð " # Ó Ð (2.1.20) which is by now familiar as the canonical distribution function. As you might guess, the parameter will once again turn out to be & K ¢ ¡ when we examine the thermodynamics of the canonical ensemble.
[Note that the above derivation assumed that the numbers of states N È P assumes the most probable distribution, e.g., maximizes . For a more rigorous approach which directly evaluates the expected values of È see Section 3.2 of Pathria .] Chapter 2 – Canonical Ensemble 19 2.2 More Thermodynamics With the canonical distribution function defined according to Equation (2.1.20), we can calculate the expected value of a property of a canonical system $ k&%y # ke Ó Ð (2.2.21) where $ k&% is the expected value of some observable property k , and k is the value of k corre-sponding to the (') state. For example, the internal energy of a system in the canonical ensemble is defined as the expected, or average, value of : # # Ó Ð # Ó Ð HG I I| 7 ]Û ) Ó Ð Ü HG I I| (2.2.22) The Helmholtz free energy + is defined as + g#·G ¡ ® , and incremental changes in + can be related to changes in internal energy, temperature, and entropy by m + m #·G ¡ m-, G ®ym ¡ .
Substituting our basic thermodynamics accounting for the internal energy m # ¡ m¯® G$° m s !
² m , results in: m + ´G ®um ¡ G±° m s !±² m ô (2.2.23) Thus, the internal energy # + !
¡ ® can be expressed as: # + G ¡ o I + I ¡ p §]´G ¡ q I I ¡ o + ¡ p_r §v q I .
+ K ¡1 I¥.
& K ¡1 r § (2.2.24) We can equate Eqs. (2.2.22) and (2.2.24) by setting h & KS¢ ¡ . [So far we have still not shown that ¢ is the same constant (Boltzmann’s constant) that we introduced in Chapter 1; here ¢ is assumed to be some undetermined constant.] The Helmholtz free energy can be calculated directly from the canonical partition function: + ´G¢ ¡ (2.2.25) How do we equate the constant ¢ of this chapter to Boltzmann’s constant of the previous chap-ter? We know that the probability of a given state in the canonical ensemble is given by: ° F Ó Ð K. (2.2.26) Next we take the expected value of the log of this quantity: $ 7 °j/%uHG G[ $ 0%º´G G[ # .
+ GÇ# 1f (2.2.27) [You might think that in the study of statistical mechanics, we are terribly eager to take loga-rithms of every last quantity that we derive, perhaps with no a priori justification. Of course, the justification is sound in hindsight. So when in doubt in statistical mechanics, try taking a logarithm.
Maybe something useful will appear!] Chapter 2 – Canonical Ensemble 20 A useful relationship follows from Equation (2.2.27). Since + Gï# HG ¡ ® , ® HG¢ $ ° % .
The expected value of .
7 °j 1 is straightforward to evaluate: ® ûG¢ ) ° 7 ° (2.2.28) From this equation, we can make a connection to the microcanonical ensemble, and the ¢ from Chapter 1. In a microcanonical ensemble, each state is equally likely. Therefore °j t Ð| , and Equation (2.2.28) becomes ® ¢ ) t Ð| 7 t ¢ ) m m t 7S t ¢ 7S t% (2.2.29) which should look familiar. Thus the ¢ of Chapter 2 is identical to the ¢ of Chapter 1, Boltzmann’s constant.
2.3 Formalism for Classical Systems As in the construction of the classical microcanonical partition function, in defining the canonical partition function for classical systems we make use of the correction factor described in Chapter 1 which relates the volume of classical phase space to a distinct number of microstates. An elemen-tary volume of classical phase space m ^ Y m ^ X is assumed to correspond to m ^ Y m ^ X K ¿¼À½Ä^ distinguishable microstates. The partition function becomes: û & ¼¾½ ^ ¶ Ó Ð Á m ^ Y m ^ X[ (2.3.30) and mean values of a physical property k are expresses as: $ k&%u à k.
Y X1 Ó Ð Á 1 2 m ^ Y m ^ X Ã Ó Ð Á 1 2 m ^ Y m ^ X (2.3.31) 2.4 Equipartition The study of molecular systems often makes use of the equipartition theorem, which describes the correlation structure of the variables of a Hamiltonian system in the canonical ensemble. Recalling that the classical Hamiltonian of a system b is a function of 3 independent momentum and position coordinates. We denote these coordinates by Ö and seek to evaluate the ensemble average: $ Ö I b I Ö %u Ã é ¬ Á ¬ Ô " ê Ó Ð Á m4 65 Ã Ó Ð Á m 4 5 (2.4.32) where the integration is over all possible values of the 3 Ö coordinates. The Hamiltonian b depends on the internal coordinates although the dependence is not explicitly stated in Equa-tion (2.4.32).
Chapter 2 – Canonical Ensemble 21 Using integration by parts in the numerator to carry out the integration over the Ö coordinate produces: $ Ö I b I Ö %u à q é G Ô Ó Ð Á ê þ þ þ Ô87 " Ô9 " !
Ã é ¬ Ô ¬ Ô " ê Ó Ð Á m Ö r m:4 uÐ| 5 Ã Ó Ð Á m 4 5 (2.4.33) where the integration over m:4 uÐ| 5 indicates integration over all Ö coordinates excluding Ö . The notation ÖiÐ and Ö; indicates the extreme values accessible to the coordinate Ö . Thus for a momen-tum coordinate these extreme values would be ñ=< , while for a position coordinate the extreme values would come from the boundaries of the container. In either case, the first term of the nu-merator in Equation (2.4.33) vanishes because the Hamiltonian is expected to become infinite at the extreme values of the coordinates.
Equation (2.4.33) can be further simplified by noting that since the coordinates are independent, I Ö KSI Ö © ¸ > , where ¸ > is the usual Kronecker delta function. [ ¸ > z & for @? ; ¸ > z for BA C? .] After simplification we are left with $ Ö I b I Ö %ul¢ ¡J¸ > (2.4.34) which is the general form of the equipartition theorem for classical systems. It should be noted that this theorem is only valid when all coordinates of the system can be freely and independently excited, which may not always be the case for certain systems at low temperatures. So we should keep in mind that the equipartition theorem is rigorously true only in the limit of high temperature.
Equipartition tells us that for any coordinate $ Ö ¬ Á ¬ Ô %u ¢ ¡ . Applying this theorem to a momen-tum coordinate, Q , we find, $ Q I b I Q % $ Q ¯R V /%º ¢ ¡l (2.4.35) [Remember the basic formulation of Hamiltonian mechanics.] Similarly, $ V :R Q /%º´G¢ ¡· (2.4.36) From Equation (2.4.35), we see that the average kinetic energy associated with the (') coor-dinate is $ w K ( %/ù¢ ¡ K ( . For a three dimensional system, the average kinetic energy of each particle is specified by a ¢ ¡ K ( . If the potential energy of the Hamiltonian is a quadratic function of the coordinates, then each degree of freedom will contribute ¢ ¡ K ( energy, on average, to the internal energy of the system.
2.5 Example Problem: Harmonic Oscillators and Blackbody Radiation A classical problem is statistical mechanics is that of a blackbody radiation. What is the equilib-rium energy spectrum associated with a cavity of a given volume and temperature?
Chapter 2 – Canonical Ensemble 22 2.5.1 Classical Oscillator The vibrational modes of a simple material can be approximated by modeling the material as a collection of simple harmonic oscillators, with Hamiltonian (for the case of classical mechanics): b ) + q ED ø ( V !
& ( Q r (2.5.37) where each of the identical oscillators vibrates with one degree of freedom. The natural fre-quency of the oscillators is denoted by D¥ø . The partition function for such a system is expressed as: · & ¼¾½ ¶GF8HJI K GJ ) + q ED ø ( V !
& ( Q rML¼ q & ½ ¶NFOHJIQP G o ED ø ( V !
& ( Q pQR m V m Q r (2.5.39) Using the identity for Gaussian distributions Ã Ó ÐÕÔ m Ö TS Î , Equation (2.5.39) is reduced to û & ¼ Û & ½ o ( Î jED ø p ÌÍ o ( Î p ÌÍ Ü (2.5.40) or û & ¼ q ( Î ½ D¥ø r (2.5.41) Remember that the factor & K ¿¼ corrects for the fact that the particles in the system are indis-tinguishable. If the particles in the system are distinguishable, then the partition function is given by: û q ( Î ½ D¥ø r (2.5.42) which is the single-particle partition function raised to the power.
2.5.2 Quantum Oscillator The one-dimensional Schr¨ odinger wave equation for a particle in a harmonic potential is: GVU ( m ß m Ö !
& ( ED ø Ö£ ß ßJ (2.5.43) where the constant U is equal to ½ K ( Î , D¥ø is the angular frequency associated with the classical oscillator, and is the energy eigenvalue of the Schr¨ odinger operator. This equations has, for quantum numbers È% & ( , energy values of U D¥ø
K ( za U D¥ø-K ( Ú U Dø
K ( . The so-called Planck oscillator excludes the ÈÝ eigenvalue. [For a complete analysis and associated wave functions, see any introductory quantum physics text, such as French and Taylor .] Chapter 2 – Canonical Ensemble 23 Thus the single-particle partition function is given by (for the Schr¨ odinger oscillator): Ò ) ~ Ó Ð ÌÍ WX (2.5.44) which can be simplified Ó Ð WYX ÌÍ & G Ó Ð WX (2.5.45) For distinguishable oscillators the partition function becomes Z Ó Ð: WYX ÌÍ .
& G Ó Ð WYX 1 (2.5.46) From our thermodynamic analysis, we calculate internal energy of the -particle system as #´G I I| q U D¥ø ( !
U D¥ø Ó WYX G & r (2.5.47) The Planck analysis of this system (excluding the zero-point energy È¿ eigenvalue), results in a mean-energy per oscillator of: $ %º # U D¥ø Ó WYX G & (2.5.48) .
2.5.3 Blackbody Radiation Consider a large box or cavity with length dimensions å Ô , å ë , and å ì , in which radiation is re-flected off the six internal walls. [It is assumed that radiation is absorbed and emitted by the container, resulting in thermal equilibrium of the photons.] In this cavity, a given frequency D corresponds to wavenumber ¢z[D2K.\ , where \ is the speed of light and ¢ is wavenumber measured in units of inverse length. Wavenumbers obtainable in the rectangular cavity are specified by the Cartesian components ¢ Ô ( Î È Ô K å Ô , ¢Së{ ( Î ÈëeK å ë , and ¢ ì ( Î ÈFì K å ì , where È Ô , Èë , and ÈFì are integers and ¢. ¢ Ô !d¢ ë !d¢ ì 1 ÌÍ . Angular frequency, expressed in terms of the integers È Ô , Èë , and Èì , is: D¥ø2 ( Î \^]>.EÈ Ô K å Ô1 !.ÏÈëK å ë 1 ! .EÈFì K å ì 1 _ ÌÍ (2.5.49) The total number of modes corresponding to a given frequency range, as DøbacD , can be calculated from the integral (in the continuous limit): ¶ X ed X m È Ô m Èë m ÈFì (2.5.50) Changing integration variables to f|Ô È Ô K å Ô , f ë%ÈëeK å ë , f ìL%Èì K å ì , yields å Ô å ë å ì ¶ g h ig j g k mlÏ d X ÌÍ npo m fÄÔ m f ë m f ë (2.5.51) Chapter 2 – Canonical Ensemble 24 Evaluating this integral gives the number of modes with frequency of less than D : å Ô å ë å ì oÞ a p Î é D ( Î \ ê ^ (2.5.52) and the number with frequencies between D and Dv!
m D is å Ô å ë å ìeD ( Î \ ^ m D or s D ( Î \ ^ m D (2.5.53) where s is the volume of the cavity. Multiplying by a factor of 2, for the two possible opposite polarizations of a given mode, we obtain: s D Î \ ^ m D (2.5.54) for the number of obtainable states for a photon of frequency between D and D]!
m D . Multiplying by the Planck expression for $ % the mean energy per oscillator, we get m X s U D ^ m D Î \ ^ .
Ó WYX G & 1 (2.5.55) the radiation energy (sum total of energy of the photons) in the frequency range.
2.6 Example Application: Poisson-Boltzmann Theory As an example application of canonical ensemble theory to biomolecular systems, we next consider the distribution of ions around a solvated charged macromolecule. If the electric field q can be expressed as the gradient of a potential q HG ãEr , the Gauss’ Law can be expressed ã .
M1 ãsr .
M1 ´GBt.
M1f (2.6.56) where .
M1 is the position-dependent permittivity, and t.
S1 is the charge density. This electrostatic approximation is valid if the length scale of the system is much smaller than the wavelengths of the electromagnetic radiation.
We can split the charge density t.
S1 into two contributions: t.
M1 utwv.
M1 !xty.
M1f (2.6.57) where twv.
M1 is the charge density associated with the ionized residues on the macromolecule, and ty.
M1 is the charge density of the salt ions surrounding the molecule. For a mono-monovalent salt the mobile ions, distributed according to Boltzmann statistics (thermal equilibrium canonical distribution), have a mean-field concentration of ty.
M1 HG V{z-}Ë!| y.
Ó}~ Ì G Ó Ð }(~( Ì Y 12 (2.6.58) where | y is the bulk concentration of the salt, and {Ë is Avogadro’s number, and Vpz is the ele-mentary charge. This distribution assumes that ions interact with one another only through the electrostatic field, and thus is strictly valid only in the limit of dilute solutions.
Chapter 2 – Canonical Ensemble 25 The two terms on the right-hand side of Equation (2.6.58) correspond to concentrations of positive and negative valence ions. Substitution of Equation (2.6.58) into Gauss’ Law leads to the Poisson-Boltzmann equation: ã .
M1 ãsr .
M1 G ( Vpz{Ë| y æç7èi .
Vpze KS¢ ¡1 HG
twv.
M1f (2.6.59) a nonlinear partial differential equation for electrostatic potential surrounding a macromolecule.
Once the electrostatic potential is calculated, the ion concentration field is straightforward provided by Equation (2.6.58).
2.7 Brief Introduction to the Grand Canonical Ensemble Grand canonical ensemble theory is the statistical treatment of a system which exchanges not only energy, but also particles, with its environment in thermal equilibrium. Derivation of the basic probability distribution for the grand canonical distribution is similar to that of the canonical ensemble, except that both and are treated as statistically-varying quantities. The resulting probability distribution is of the form: ° L Ó × Ð " z.ϲ Dsu ¡1 (2.7.60) where each state is specified by number of particles , and energy .
The grand partition function is defined by summation over all and ? states: z.ϲ Dsy -¡1 ) ) Ó × Ð " (2.7.61) which is often written in a form like: z.
á Dsy -¡1 ) á ) Ó Ð " (2.7.62) where á Ó ×Ì is called the fugacity (the tendency to be unstable or fly away, from the Latin fugere meaning to flee according to the Oxford English Dictionary ).
Chapter 2 – Canonical Ensemble 26 Problems 1. (Derivation of canonical ensemble) Show that Equation (2.1.8) is uniquely solved by Equa-tion (2.1.9).
2. (Simple harmonic oscillators and blackbody radiation) Compare the classical oscillator with the Schr¨ odinger and Planck oscillators. (a) What is the energy per oscillator in the canonical ensemble for the classical case? Which oscillators (if any) obey equipartition? For those that do not, is there a limiting case in which equipartition is valid? [Hints: plot #fK as a function of temperature. Perhaps Taylor expansions of these expressions will be helpful.] (b) From Equation (2.5.55) obtain a nondimensional expression for energy per unit frequency spectrum, and plot the nondimensional energy distribution of blackbody radiation versus nondimensional frequency U D . At what frequency does the spectral energy distribution obtain a maximum?
3. (Electrical double layer) Consider a one-dimensional model of a metal electrode/solution electrolyte interface. The potential in the solution is governed by the Poisson-Boltzmann equation: m m Ö ( V z }Ë!| y ¢ ¡ æ-ç7è ] (a) Show that the above equation is the Poisson-Boltzmann equation in terms of the dimen-sionless potential Vpz r K ¢ ¡ . Show that this equation can be linearized as m m Ö Ý (b) Evaluate the Debye length ( & Kw ) for the case of 0.1 M and 0.0001 M solution of NaCl.
(c) Using the boundary condition q m m Ö r Ô ~ HG Vpz ¢ ¡ (where is the charge density on the surface of the electrode), find .
Ö 1 . Plot the concen-trations of Na and Cl as functions of Ö (assume a positively-charged electrode).
4. (Donnan equilibrium) Consider a gel which carries a certain concentration | B.
M1 of immobile charges and is immersed in an aqueous solution. The bulk solution carries mono-monovalent mobile ions of concentration |b .
M1 and |JÐ .
M1 . Away from the gel, the concentration of the salt ions achieves the bulk concentration, denoted | ø . What is the difference in electrical po-tential between the bulk solution and the interior of the gel? [Hint: assume that inside the gel, the overall concentration of negative salt ion balances immobile gel charge concentration.] Chapter 3 Brownian Motion, Fokker-Planck Equations, and the Fluctuation-Dissipation Theorem Armed with our understanding of the basic principles of microscopic thermodynamics, we are fi-nally ready to examine the motions of microscopic particles. In particular, we will study these motions from the perspective of stochastic equations, in which random processes are used to ap-proximate thermal interactions between the particles and their environment.
3.1 One-Dimensional Langevin Equation and Fluctuation-Dissipation Theorem Consider the following Langevin equation for the one-dimensional motion of a particle: ·R wF.
n 1 ! w.
n 1 k.y.
n 1 !Çk ý .
n 1 (3.1.1) where is the mass of the particle, is the coefficient of friction, k.y is the systematic (determinis-tic) force acting on the particle, and k ý is a random process used to induce thermal fluctuations in the energy of the particle. Equation (3.1.1) can be thought of as Newton’s second law with three forces acting on the particle: viscous damping, random thermal noise, and a systematic force.
Equation (3.1.1) can be factored Ó Ð ' Ì v m m0n ] Ó ' Ì v w _ k.yi!Çk ý (3.1.2) and has the general solution: w.
n 1 Ó Ð ' Ì v q w.
1 !
& ¶ ' ~¿Ó y Ì v . k.y.
, 1 !ïk ý .
, 1-1 m-, r (3.1.3) Using angled brackets $ % to denote averaging over trajectories, we can calculate the covariance 27 Chapter 3 – Thermal Motions 28 in particle velocity: $ w.
n 1 w.
n 1 %ô Ó Ð ' ' Ì v ] w .
1 !
w.
1 ¶ ' ~ Ó y Ì v $ k.y.
, 1 !Çk ý .
, -1 % m-, !
w.
1 ¶ ' ~ Ó y Ì v $ k.y.
, D1 !Çk ý .
, 1 % m-, !
& ¶ ' ~ ¶ ' ~ Ó e y y Ì v $ k ý .
, 1 k ý .
, D1 % mJ, m-, r (3.1.4) assuming that the deterministic forces have zero averages. Equation (3.1.4) can be further simpli-fied: $ w.
n 1 w.
n D1 %º%w .
1 Ó Ðe ' ' Ì v !
Ó Ðe ' ' Ì v ¶ ' ~ ¶ ' ~ Ó e y y Ì v $ k ý .
, 1 k ý .
, D1 % mJ, m-, » (3.1.5) And if we assume that the random force is a white noise process, then its correlation can be de-scribed by: $ k ý .
, 1 k ý .
, 1 %u +¸ .
, G , D1F (3.1.6) Integrating Equation (3.1.5) over , we obtain: $ w.
n 1 w.
n D1 %uw .
1 Ó Ðe ' ' Ì v !
Ó Ðe ' ' Ì v ¶ ' ~ + Ó Ä y Ì v .
n G , D1 m-, (3.1.7) where .
n 1 is the step function defined by .
n 1 > & n Ù & K ( n n .
(3.1.8) Finally, integration of Equation (3.1.7) yields: $ w.
n 1 w.
n D1 %w .
1 Ó Ðe ' ' Ì v !
+ ( ] Ó Ð ' Ð ' Ì v G Ó Ð ' ' Ì v _ (3.1.9) To obtain the mean kinetic energy, we take n n n : $ w .
n 1 %º%w .
1 Ó Ð¯ ' Ì v !
+ ( ] & G Ó Ð¯ ' Ì v _ (3.1.10) which approaches $ w .
n 1 %º + ( (3.1.11) at equilibrium. From equipartition we have $ w %º ¢ ¡ KM . So $ k ý .
, 1 k ý .
, 1 %º ( Õ¢ ¡J¸ .
, G , D1F (3.1.12) which is a statement of the fluctuation-dissipation theorem. To obtain thermal equilibrium, the strength of the random (thermal) noise must proportional to the frictional damping constant, as prescribed by Equation (3.1.12).
Chapter 3 – Thermal Motions 29 3.2 Fokker-Planck Equation Imagine integrating a stochastic differential equation such as Equation (3.1.1) a number of times so that the noisy trajectories converge into a probability density of states. Considering an -dimensional problem with the vector Ö .
n 1 representing the state space, we denote the probability distribution as .
Ö .
n 1 n 1 and introduce ° .
Ö .
n !x 1D n !x Ö .
n 1 n 1 as the probability of transition from state Ö .
n 1 at time n to Ö .
n ! 1 at time n ! .
From the transition probability it follows: .
Ö .
n ! 1D n ! 1 ¶ ° .
Ö .
n ! 1 n ! Ö .
n 1 n 1 .
Ö .
n 1 n 1 m Ö (3.2.13) Expanding the transition probability as a power series, we obtain .
Ö .
n ! 1D n ! 1 ¶ Û ° .
Ö .
n 1 n Ö .
n 1D n 1 !Ç!
Ò ) & È ¼ .
Ö .
n ! 1 G Ö .
n 1-1F .
Ö Ä.
n ! 1 G Ö .
n 11 I I Ö I Ö ° .
Ö .
n 1D n Ö .
n 1 n 1 r .
Ö .
n 1 n 1 m Ö (3.2.14) where the convention of summation over the indices e e is implied. Noting that °.
Ö .
n 1D n Ö .
n 1 n 1 ¸ .
Ö .
n 1 G Ö .
n 1-1 ¸ .
Ö .
n 1 G Ö .
n 11F¸ ..
Ö .
n 1 G Ö .
n 1-11F (3.2.15) we obtain .
Ö .
n ! 1 n ! 1 .
Ö .
n 1D n 1 !
¶ Û Ò ) & È ¼ .
Ö .
n ! 1 G Ö .
n 1-1F .
Ö .
n ! 1 G Ö .
n 1-1 I I Ö I Ö ¸ .
Ö .
n 1 G Ö .
n 1-1 r .
Ö .
n 1D n 1 m Ö (3.2.16) or .
Ö .
n ! 1D n ! 1 .
Ö .
n 1 n 1 !
¶ Û Ò ) & È ¼ .
Ö .
n ! 1 G Ö .
n 1-1F .
Ö .
n ! 1 G Ö .
n 11 I I Ö I Ö ¸ .
Ö .
n 1 G Ö .
n 1-1 r .
Ö .
n 1D n 1 m Ö (3.2.17) By successively integrating by parts, we can move the derivatives off of the delta functions: .
Ö .
n ! 1D n ! 1 .
Ö .
n 1 n 1 !
¶ ¸ .
Ö .
n 1 G Ö .
n 1-1 Û Ò ) .BG & 1 È ¼ I I Ö I Ö .
Ö .
n ! 1 G Ö .
n 11F .
Ö .
n ! 1 G Ö .
n 1-1 _ .
Ö .
n 1 n 1 m Ö (3.2.18) Chapter 3 – Thermal Motions 30 Equation (3.2.18) integrates to .
Ö .
n ! 1 n ! 1 .
Ö .
n 1D n 1 !
Ò ) .G & 1 È ¼ I I Ö I Ö ÷.
Ö .
n ! 1 G Ö .
n 11F .
Ö .
n ! 1 G Ö .
n 11 .
Ö .
n 1 n 1u (3.2.19) or in the limit ä , I I n .
Ö n 1 Ò ) .G & 1 È ¼ I I Ö I Ö q ç O ~ o & Ö .
n ! 1 G Ö .
n 10 Ö |.
n ! 1 G Ö .
n 1B p .
Ö .
n 1 n 1 r (3.2.20) Defining the Kramers-Moyal coefficients as ¡ ¢£¢£¢ & È ¼: ç ~ o & Ö .
n ! 1 G Ö .
n 10 Ö .
n ! 1 G Ö .
n 1B p (3.2.21) We obtain the Fokker-Planck equation for .
Ö n 1 : I I n .
Ö n 1 Ò ) .G & 1 I I Ö I Ö ¥¤ ¡ ¢£¢£¢ .
Ö .
n 1 n 1¦W (3.2.22) In the following section we evaluate the Kramers-Moyal expansion coefficients for a nonlinear -dimensional stochastic differential equation.
3.3 Brownian Motion of Several Particles Consider the general nonlinear Langevin equation for several variables § .
n 1 § .
n 1F § .
n 1 § .
n 1 : R § F ½ B.
§ .
n 1 n 1 !
£ S.
§ .
n 1 n 1ÄÑ .
n 1 (3.3.23) where Ñ M.
n 1 are uncorrelated white noise processes distributed according to: $ Ñ .
n 1 %u $ Ñ Í.
n -1Ñ .
n 1 %u ( ¸ > ¸ .
n G n D1f (3.3.24) In Equation (3.3.23) we use the Einstein convention of summation over repeated indices. Thus the matrix > describes the covariance structure of the random forces acting on § .
n 1 .
Equation (3.3.23)has the general solution: § .
n ! 1 G § .
n 1 ¶ ' ' ½ B.
§ .
n 1D n 1 !
> .
§ .
n 1 n 1ÄÑ .
n 1 m n (3.3.25) Chapter 3 – Thermal Motions 31 We can expand this integral by expanding the functions ½ and > ½ B.
§ .
n 1 n 1 ½ B.
§ .
n 1 n 1 ! § .
n 1 G § .
n 1 I I § ½ .
§ .
n 1D n 1 !
> .
§ .
n 1 n 1 £ M.
§ .
n 1 n 1 ! § .
n 1 G § .
n 1B I I § > .
§ .
n 1D n 1 !
(3.3.26) and inserting these expansions into Equation (3.3.25): § B.
n ! 1 G § B.
n 1 Ã ' ' ½ B.
§ .
n 1 n 1 m n !
à ' ' § .
n 1 G § .
n 1 ¬ ¬¨ © ½ .
§ .
n 1 n 1 m n !
!
à ' ' > .
§ .
n 1 n 1Ñ M.
n 1 m n !
à ' ' § .
n 1 G § .
n 1B ¬ ¬¨ © £ M.
§ .
n 1 n 1Ñ M.
n 1 m0n !
(3.3.27) We can expand the § .
n 1 G § .
n 1 terms in the above equation and recursively using Equa-tion (3.3.27): § .
n ! 1 G § .
n 1 Ã ' ' ½i .
§ .
n 1D n 1 m n !
à ' ' §«ª .
n 1 G §«ª .
n 1 ¬ ¬¨¬ ½i .
§ .
n 1 n 1 m n !
!
à ' ' ® ª .
§ .
n 1 n 1Ñ ª .
n 1 m0n !
à ' ' §«ª .
n 1 G §«ª .
n 1B ¬ ¬¨¬ ® ª .
§ .
n 1 n 1Ñ ª .
n 1 m0n !
(3.3.28) To compute $ § .
n !¯ 1 G § B.
n 1 % in the limit ä , we insert Equation (3.3.28) into Equa-tion (3.3.27), average over trajectories, and retain terms of first order: $ § B.
n !° 1 G § .
n 1 %2[ ½ .
§ .
n 1D n 1 !
¶ ' ' ¶ '± ' q I I § > .
§ .
n 1D n 1 r ® ª .
§ .
n 1 n 1 ( ¸ ª ¸ .
n G n 1 m n m n !
(3.3.29) where terms of order are not shown. To evaluate the integral in Equation (3.3.29), we use the identity: ¶ '± ' . ª .
§ .
n 1D n 1 ( ¸ ª ¸ .
n G n 1 m0n ® .
§ .
n 1D n 1F (3.3.30) and we get: ¡ ½ .
§ .
n 1D n 1 !
. .
§ .
n 1 n 1 I I § > .
§ .
n 1 n 1 (3.3.31) for the drift coefficients.
Evaluating $ § B.
n !² 1 G § .
n 1 § M.
n !² 1 G § M.
n 1 % , the only term that survives averaging and the limit ä is $ § .
n ! 1 G § B.
n 1B § .
n ! 1 G § .
n 1 %u ¶ ' ' ¶ ' ' > .
§ .
n 1 n 1J® ª .
§ .
n 1 n 1 ( ¸ ª ¸ .
n G n 1 m n m n ( > .
§ .
n 1 n 1J. .
§ .
n 1D n 1 (3.3.32) and ¡ À > .
§ .
n 1D n 1 .
§ .
n 1 n 1F (3.3.33) Chapter 3 – Thermal Motions 32 All higher-order coefficients are zero: ¡ ¢£¢£¢ & È ¼ ç O ~ $ § .
n ! 1 G § .
n 10 § Ä.
n ! 1 G § Ä.
n 1B %u Ȳ³ aL (3.3.34) The Fokker-Planck equation corresponding to Equation (3.3.23) is: I .
Ö .
n 1D n 1 I n I I Ö q G o ½ .
Ö .
n 1D n 1 !
. .
Ö .
n 1 n 1 I I Ö £ M.
Ö .
n 1 n 1 p .
Ö .
n 1 n 1 r !
I I Ö ÏI Ö .
Ö .
n 1D n 1 .
Ö .
n 1 n 1 .
Ö .
n 1D n 1Bu (3.3.35) 3.4 Fluctuation-Dissipation and Brownian Dynamics To examine how the fluctuation-dissipation theorem arises for the general nonlinear Langevin equation, we first examine the simple one-dimensional problem described by: R § ½ .
§ n 1 !
.
§ n 1Ñ .
n 1 $ Ñ .
n 1Ñ .
n 1 %u ( ¸ .
n G n D1F (3.4.36) If the systematic term is proportional to the gradient of a potential, ½ HG ¡ ã # , then the Fokker-Planck equation is expressed as: I .
Ö n 1 I n I I Ö ¡ .
ã # 1 !
I I Ö ] _ (3.4.37) If we further assume that Ó Ð ´ at equilibrium, then m m Ö q G ¡ .
m # m Ö 1 Ó Ð ´ r m m Ö ] Ó Ð ´ _ (3.4.38) It is straightforward to show that this condition is satisfied if ¡ ÌÍ , which is the fluctuation-dissipation theorem for this one-dimensional Brownian motion.
This relationship generalizes for the following -dimensional case: R § F ¡ > k{ y!
£ Ñ M.
n 1 (3.4.39) $ Ñ Í.
n -1Ñ .
n 1 %u ( ¸ > ¸ .
n G n D1 (3.4.40) k8 _ m m § #/.
§ .
n 1 n 1F (3.4.41) to { ¡ > (3.4.42) In the above equations, #/.
§ .
n 1 n 1 is the potential energy function, and ¡ > is the frictional interac-tion matrix which determines the hydrodynamic interactions between the particles in the system.
Thus the covariance structure of the random forcing is proportional to the hydrodynamic inter-action matrix. Given a diffusional matrix ¡ £ , generation of the random force vector requires the calculation of , the factorization of ¡ > . In fact, for Brownian dynamics simulations, this factorization represents the computational bottleneck that demands the majority of CPU resources.
Chapter 3 – Thermal Motions 33 Problems 1. (Brownian Motion) Consider the coupled Langevin equations: Ö !Cµ R Ö HG ã #ï!
b Ñ .
n 1 where is a diagonal matrix of particle masses, µ is the non-diagonal frictional interaction matrix, and the noise term Ñ is correlated according to: $ Ñ .
n -1Ñ .
n 1 %u (.¶ ¸ .
n G n 1 where ¶ is the identity matrix. The fluctuation-dissipation theorem tells us that: b b ¢ ¡ µ (a) Brownian Motion If we ignore inertia in the above system we get R Ö ´G & ¢ ¡ ¡ ã #ï!x· Ñ .
n 1 where ¡ h¢ ¡ µ Ð| is the diffusion matrix. Show that · ·h ¡ . Given a diffusion matrix ¡ , how would you calculate its factor · ?
(b) Brownian Dynamics Algorithm Under what circumstances do you think the inertial system will reduce to the over-damped (non-inertial) system? Assuming that the ran-dom and systematic forces remain constant over a time step of length Å n , find an ex-pression for w.
n !
Å n 1 in terms of w.
n 1 . [Hint: The equations can be decoupled by considering the eigen decomposition of + Ð| µ . Proceed by finding a matrix-vector equivalent to Equation (3.1.3).] Show that under a certain highly-damped limit, this expression reduces to the Brownian equation w.
n 1 ¡ ¢ ¡ systematic force + random force 2. (Numerical Methods) Devise a numerical propagation scheme for the Brownian equation.
Bibliography R. P. Feynman. Statistical Mechanics. A Set of Lectures. Addison-Wesley, Reading, MA, 1998.
A. P. French and E. F. Taylor. An Introduction to Quantum Physics. W. W. Norton and Co., New York, NY, 1978.
R. K. Pathria. Statistical Mechanics. Butterworth-Heinemann, Oxford, UK, second edition, 1996.
H. Risken. The Fokker-Planck Equation. Methods of Solution and Applications. Second edi-tion.
J. A. Simpson and E. S. C. Weiner, editors. Oxford English Dictionary Online
Oxford Univerisity Press, Oxford, UK, second edition, 2001.
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Tinea Cruris
Micah M. Pippin; Michael L. Madden; Moushumi Das.
Author Information and Affiliations
Authors
Micah M. Pippin1; Michael L. Madden2; Moushumi Das3.
Affiliations
1 LSUHSC Shreveport
2 LSUHSC / Rapides Regional MC
3 University of Auckland
Last Update: August 17, 2023.
Continuing Education Activity
Tinea cruris, also known as jock itch, is an infection involving the genital, pubic, perineal, and perianal skin caused by pathogenic fungi known as dermatophytes. The evaluation and treatment of tinea cruris are discussed in the activity. This activity reviews the role of the healthcare team in improving care for patients with this condition. The activity will also highlight evidence-based therapeutics and lifestyle modifications to avoid common mistakes and mistreatments associated with the management of dermatophyte infections.
Objectives:
Identify the etiology of tinea cruris.
Outline the evaluation of tinea cruris.
Review the treatment options available for tinea cruris.
Describe interprofessional team strategies to improve the management of patients affected by tinea cruris.
Access free multiple choice questions on this topic.
Introduction
Tinea cruris, also known as jock itch, is an infection involving the genital, pubic, perineal, and perianal skin caused by pathogenic fungi known as dermatophytes. These dermatophytes affect keratinized structures such as hair and the epidermis' stratum corneum resulting in a characteristic rash. Intertriginous areas are hospitable environments for fungus, with sweating, maceration, and alkaline pH being responsible for the groin's predilection for infection.
While tinea infections are often classified by the location of the body affected, they are also organized according to the responsible organism's primary source and mode of transmission. Geophilic, zoophilic, and anthropophilic fungi are found in and transmitted by soil, animals, and humans, respectively. Autoinfection of dermatophytes is also possible and especially crucial in tinea cruris as foot-to-groin spread can occur.
Etiology
Tinea cruris is caused by dermatophytes belonging to three genera, Trichophyton, Epidermophyton, and Microsporum. Trichophyton rubrum has been isolated most commonly and remains the most frequent cause of tinea cruris worldwide; however, most studies do recognize the increasing prevalence of Trichophyton mentagrophytes and other organisms in certain regions. Several risk factors have been identified that predispose an individual to tinea cruris, including excessive perspiration, occlusive clothing, improper hygiene, diabetes mellitus, immunocompromise, and lower socioeconomic status.
Athletes, especially those involved in contact sports, may be more likely to contract tinea infections. Genetics can also make a patient more susceptible to dermatophytes. Of all these factors, perspiration appears to be the most influential variable in the development of infection. In India, an area affected disproportionately often with dermatophytes, a study was conducted in response to the increasing frequency of and decreased treatment efficacy for local tinea infections. Diabetes mellitus, family members with tinea, and personal history of cooking food were found to be positively associated with chronic and relapsing disease.
Epidemiology
Cutaneous mycoses, including tinea cruris, affect 20 to 25 percent of the world's population. Developing and tropical countries have an increased prevalence of dermatophyte infections secondary to high temperatures and increased humidity. In the United States, there have been an estimated 29.4 million cases of superficial fungal infections and over 51 million reported physician visits. Adolescent and adult males comprise the majority of patients seen for tinea cruris and are affected by the disorder with increased frequency. Worldwide increases in the occurrence of dermatophytoses and the discovery of recalcitrant infections have caused global concern.
Pathophysiology
A simplified explanation of the complex and not well-understood pathophysiology of dermatophytes includes the organism's use of proteinases to digest keratin found in the skin's stratum corneum.
Histopathology
While not always necessary for diagnosis, a potassium hydroxide (KOH) preparation may reveal histology consistent with dermatophyte infections. Characteristic findings include branching or non-branching septate hyphae and possible arthroconidiospores.
History and Physical
Patients with tinea cruris present complaining of a pruritic rash involving the groin. The area may be irritated and painful if maceration is present, and secondary infections may result in inflammation and discomfort. Duration of symptoms, previous occurrences, similar rashes in other locations, and past treatments should be elucidated. Individuals should be questioned about any history of diabetes, immunocompromise, renal disease, or hepatic dysfunction. Clinicians should inquire about excessive sweating, wardrobe changes, and personal hygiene habits. A review of the patient's environmental and occupation exposures, including people, pets, animals, and contaminated soil, may be contributory.
On physical examination, an erythematous, scaly, annular plaque with a raised leading edge and central clearing can be visualized, extending anywhere from the groin, upper thigh, and perineum to the perianal region.
Evaluation
In most cases, tinea cruris can be diagnosed clinically; however, several tests exist to investigate a rash of the groin with unknown etiology. Potassium hydroxide (KOH) preparations, skin biopsy with periodic acid-Schiff (PAS) stain, and fungal cultures on Sabouraud’s agar media can be utilized when the diagnosis is in question or the case of recurrent or recalcitrant episodes.
In general, the sample should be obtained from the leading edge of the lesion to ensure an adequate collection of infected scales. Potassium hydroxide (KOH) mounts are generally procured with a scalpel scraping technique; however, new studies have indicated that utilizing a cellophane adhesive tape method may simplify the collection process, facilitate transportation, provide a higher quality sample, and improve slide preservation time.
Treatment / Management
Antifungals utilized in treating dermatophytoses, including tinea cruris, target the synthesis of ergosterol, a vital component of fungal plasma membranes. Management strategies are similar worldwide; however, some countries have specific guidelines based on their region's fungal profile. Topical therapies are effective and usually preferred. Allylamines (terbinafine, butenafine, naftifine) and azoles (clotrimazole, miconazole, sulconazole, oxiconazole, econazole, ketoconazole) are the mainstays of topical treatment regimens. They are generally prescribed once or twice daily for two to four weeks.
Deciding which agent to use should be based on patient compliance, cost, and medication accessibility, as there is insufficient data to directly compare the effectiveness of individual drugs and classes. Allylamines have a potentially shorter treatment timeline, have demonstrated lower relapse rates, and their metabolism is independent of the cytochrome p450 system. Azoles are not as costly as allylamines but often require a longer treatment duration. One newer topical azole, luliconazole, requires only once daily application for one week and may improve patient compliance through a more convenient dosing schedule. Ciclopirox olamine is an older topical preparation with a unique mechanism of action compared to the commonly used allylamines and azoles. Recent studies have demonstrated a number of benefits to ciclopirox therapy; however, it remains an underutilized antifungal medication.
Oral preparations exist to manage tinea cruris and are indicated for chronic, recurrent, and recalcitrant disease. Extensive or diffuse rashes and patients with immunocompromise may also require systemic treatments. Stratum corneum penetration and concentration maintenance, keratin adherence, patient tolerance, and a minimal drug interaction profile are hallmarks of the ideal systemic medication for dermatophytoses. Oral terbinafine and itraconazole have favorable characteristics for dermatophyte management and are the most often prescribed. Fluconazole has demonstrated efficacy in treating tinea cruris; however, it is not preferred due to its poor keratin adherence and prolonged treatment duration. Griseofulvin has similar pharmacokinetic limitations as fluconazole, and it is more appropriately utilized in the management of tinea capitis than tinea cruris. Due to its potential for hepatotoxicity, oral ketoconazole is no longer recommended for dermatophytoses. Topical antifungal therapies may be used as an adjunct in patients requiring systemic treatment. Topical and oral antibiotics can be administered when a secondary bacterial infection is present.
A commonly used alternative treatment known as Whitfield's ointment has insufficient evidence of benefit. Nystatin, a frequently utilized treatment for cutaneous candida infections, is ineffective for managing dermatophytoses such as tinea cruris. Combined topical corticosteroid and antifungal therapy remain controversial. Some studies have demonstrated improved cure rates with concomitant steroid and antifungal topical applications; however, these results were based on low-quality evidence. While steroids may improve acute inflammation and itching, they may also fortify dermatophyte's plasma membranes rendering antifungal medications less effective. Steroids also activate fungal metabolism and can potentially facilitate the worsening of the primary infection. Tinea incognito is another possible complication of steroid administration where the typical presentation of tinea is masked, and diagnosis is delayed. Currently, topical steroids are not recommended as part of an evidence-based tinea cruris treatment regimen.
Differential Diagnosis
The differential diagnosis for tinea cruris includes several other dermatologic conditions affecting the groin with similar presentations. Candidiasis, erythrasma, psoriasis, and seborrheic dermatitis exhibit comparable signs and symptoms and are most commonly confused with the fungal groin infection. Unlike tinea cruris, candidal intertrigo frequently affects women, and the rash may involve the scrotum and penis in males. Satellite lesions and erythema without central clearing are indicative of candida as opposed to tinea. The rash of erythrasma lacks an active border and demonstrates coral-red fluorescence on Wood lamp examination. Psoriasis will likely manifest in other areas in addition to the crural region. Seborrheic dermatitis presents with greasy scales on an erythematous base.
Prognosis
Patients with tinea cruris who undergo an appropriate treatment course experience cure rates ranging from 80 to 90 percent.
Complications
Failure of therapy and recurrence are the most likely complications of tinea cruris. They have been attributed to reinfection from close contacts, autoinfection from separate body locations, infection by uncommon species such as zoonoses, misdiagnosis, drug resistance, and non-adherence to the management plan. Steroid use may suppress the physical signs of tinea cruris, making the diagnosis more difficult. Also, chronic application can result in skin atrophy and telangiectasias. Secondary bacterial infection is another potential complication of tinea cruris. Majocchi’s granuloma is an uncommon complication of cutaneous fungal infections in which dermatophytes disseminate into the subcutaneous tissue secondary to skin breakdown, immunosuppression, or topical steroid use resulting in a deep, inflammatory disease. A dermatophytid reaction may result in an allergic response at a separate location from the original tinea site.
Consultations
Consultation with dermatology or infectious disease may be warranted in recurrent or recalcitrant cases.
Deterrence and Patient Education
Patient education should focus on non-pharmacologic measures to treat and prevent recurrences of tinea cruris. Loose-fitting, non-restrictive garments should be encouraged, and clothing should not be donned until the underlying skin is completely dried. Because autoinfection originating from tinea pedis may occur, patients should avoid walking barefoot, and protective footwear should be used in public facilities. Identification and treatment of potentially infected contacts, whether human or animal, should be undertaken. Self-treatment with over-the-counter antifungals and steroids should be discouraged as this may result in resistant or chronic infections and can hinder a clinician's ability to make an accurate and timely diagnosis.
Enhancing Healthcare Team Outcomes
Tinea cruris is a prevalent pathology with a worldwide distribution and an extensive history of affecting human populations. Despite our familiarity with the condition, there has been limited research specific to this subset of dermatophyte infections. Renewed interest has emerged with the development of recalcitrant infections and concern over fungal resistance. Self-treatment with easy-to-access over-the-counter topical antifungal and steroid preparations have been implicated as a probable cause of the observed decrease in treatment efficacy. In the context of fungal resistance, newly developed formulations such as luliconazole and underutilized, older agents like ciclopirox may be beneficial.
With regard to corticosteroids, there is conflicting data on the appropriateness of their use for tinea cruris. Current management principles and guidelines label them as mistreatment; however, continued investigations of their utility are underway. If a practitioner believes topical steroids may benefit a patient, close supervision should be maintained throughout the treatment course with continued consideration of known adverse event potential for the individual and public health outcomes.
With an understanding of the changing landscape of this common condition, and by implementing an interprofessional healthcare team approach including primary care clinicians (including PAs and NPs), dermatologists, infectious disease specialists, pharmacists, and nursing, patient care and public health may be improved through targeted, conscientious mycological treatment, patient education, and antifungal stewardship. [Level 2] This interprofessional paradigm should lead to the best outcomes with minimal to no adverse events. [Level 5]
Review Questions
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Tinea Cruris DermNet New Zealand
References
1.
: Gupta AK, Chaudhry M, Elewski B. Tinea corporis, tinea cruris, tinea nigra, and piedra. Dermatol Clin. 2003 Jul;21(3):395-400, v. [PubMed: 12956194]
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: Ely JW, Rosenfeld S, Seabury Stone M. Diagnosis and management of tinea infections. Am Fam Physician. 2014 Nov 15;90(10):702-10. [PubMed: 25403034]
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: Sahoo AK, Mahajan R. Management of tinea corporis, tinea cruris, and tinea pedis: A comprehensive review. Indian Dermatol Online J. 2016 Mar-Apr;7(2):77-86. [PMC free article: PMC4804599] [PubMed: 27057486]
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: Khurana A, Sardana K, Chowdhary A. Antifungal resistance in dermatophytes: Recent trends and therapeutic implications. Fungal Genet Biol. 2019 Nov;132:103255. [PubMed: 31330295]
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: Sardana K, Kaur R, Arora P, Goyal R, Ghunawat S. Is Antifungal Resistance a Cause for Treatment Failure in Dermatophytosis: A Study Focused on Tinea Corporis and Cruris from a Tertiary Centre? Indian Dermatol Online J. 2018 Mar-Apr;9(2):90-95. [PMC free article: PMC5885633] [PubMed: 29644192]
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: Gupta AK, Foley KA, Versteeg SG. New Antifungal Agents and New Formulations Against Dermatophytes. Mycopathologia. 2017 Feb;182(1-2):127-141. [PubMed: 27502503]
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: Odom R. Pathophysiology of dermatophyte infections. J Am Acad Dermatol. 1993 May;28(5 Pt 1):S2-S7. [PubMed: 8496407]
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: Nadalo D, Montoya C, Hunter-Smith D. What is the best way to treat tinea cruris? J Fam Pract. 2006 Mar;55(3):256-8. [PubMed: 16510062]
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: Singh S, Verma P, Chandra U, Tiwary NK. Risk factors for chronic and chronic-relapsing tinea corporis, tinea cruris and tinea faciei: Results of a case-control study. Indian J Dermatol Venereol Leprol. 2019 Mar-Apr;85(2):197-200. [PubMed: 30719987]
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: Ameen M. Epidemiology of superficial fungal infections. Clin Dermatol. 2010 Mar 04;28(2):197-201. [PubMed: 20347663]
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: Vena GA, Chieco P, Posa F, Garofalo A, Bosco A, Cassano N. Epidemiology of dermatophytoses: retrospective analysis from 2005 to 2010 and comparison with previous data from 1975. New Microbiol. 2012 Apr;35(2):207-13. [PubMed: 22707134]
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: Samdani AJ. Dermatophyte growth and degradation of human stratum corneum in vitro (pathogenesis of dermatophytosis). J Ayub Med Coll Abbottabad. 2005 Oct-Dec;17(4):19-21. [PubMed: 16599028]
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: Drake LA, Dinehart SM, Farmer ER, Goltz RW, Graham GF, Hardinsky MK, Lewis CW, Pariser DM, Skouge JW, Webster SB, Whitaker DC, Butler B, Lowery BJ, Elewski BE, Elgart ML, Jacobs PH, Lesher JL, Scher RK. Guidelines of care for superficial mycotic infections of the skin: tinea corporis, tinea cruris, tinea faciei, tinea manuum, and tinea pedis. Guidelines/Outcomes Committee. American Academy of Dermatology. J Am Acad Dermatol. 1996 Feb;34(2 Pt 1):282-6. [PubMed: 8642094]
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: Raghukumar S, Ravikumar BC. Potassium hydroxide mount with cellophane adhesive tape: a method for direct diagnosis of dermatophyte skin infections. Clin Exp Dermatol. 2018 Dec;43(8):895-898. [PubMed: 29845649]
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: Gupta AK, Cooper EA. Update in antifungal therapy of dermatophytosis. Mycopathologia. 2008 Nov-Dec;166(5-6):353-67. [PubMed: 18478357]
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: van Zuuren EJ, Fedorowicz Z, El-Gohary M. Evidence-based topical treatments for tinea cruris and tinea corporis: a summary of a Cochrane systematic review. Br J Dermatol. 2015 Mar;172(3):616-41. [PubMed: 25294700]
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: Sonthalia S, Agrawal M, Sehgal VN. Topical Ciclopirox Olamine 1%: Revisiting a Unique Antifungal. Indian Dermatol Online J. 2019 Jul-Aug;10(4):481-485. [PMC free article: PMC6615394] [PubMed: 31334080]
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: El-Gohary M, van Zuuren EJ, Fedorowicz Z, Burgess H, Doney L, Stuart B, Moore M, Little P. Topical antifungal treatments for tinea cruris and tinea corporis. Cochrane Database Syst Rev. 2014 Aug 04;2014(8):CD009992. [PMC free article: PMC11198340] [PubMed: 25090020]
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: Hay R. Therapy of Skin, Hair and Nail Fungal Infections. J Fungi (Basel). 2018 Aug 20;4(3) [PMC free article: PMC6162762] [PubMed: 30127244]
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: Trocoli Drakensjö I, Vassilaki I, Bradley M. Majocchis Granuloma Caused by Trichophyton mentagrophytes in 2 Immunocompetent Patients. Actas Dermosifiliogr. 2017 Jan-Feb;108(1):e6-e8. [PubMed: 26952202]
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: Disclosure: Micah Pippin declares no relevant financial relationships with ineligible companies.
: Disclosure: Michael Madden declares no relevant financial relationships with ineligible companies.
: Disclosure: Moushumi Das declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
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Pippin MM, Madden ML, Das M. Tinea Cruris. [Updated 2023 Aug 17]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
In this Page
Continuing Education Activity
Introduction
Etiology
Epidemiology
Pathophysiology
Histopathology
History and Physical
Evaluation
Treatment / Management
Differential Diagnosis
Prognosis
Complications
Consultations
Deterrence and Patient Education
Enhancing Healthcare Team Outcomes
Review Questions
References
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Spring Constant Calculator
Spring Constant Calculator
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
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Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
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Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
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Table of Contents:
What is the Spring Constant Calculator
The best spring calculator to calculate the spring constant in compression, extension and torsion springs. By entering a few basic spring dimensions such as: wire diameter, outer diameter, free length and total coils the spring constant calculator calculates the spring constant of your spring. Find out your spring constant, your springs maximum load, maximum travel and so much more. Our spring constant calculator can calculate the spring constant in english lbs/in or metric N/mm. Download 3D Blueprints, spec sheets for all your spring designs with the best spring constant calculator.
How to use the spring constant calculator, example:
Let's assume your Compression Spring has a:
| | |
--- |
| Wire diameter: | 0.043 inches |
| Outer Diameter: | 0.250 inches |
| Free Length: | 1.043 inches |
| Total Coils: | 13.7 |
| End Type: | Closed and Squared |
| Material Type: | Music Wire ASTM A228 |
Go to the Spring Constant Calculator and choose compression springs then enter the following specifications
Click Calculate
In the 1st Tab of “Preview Springs and Specs”
The Spring Constant calculator has calculated the
The spring constant calculator will also generate a 3D Custom spring CAD file along with a blueprint / Spec Sheet with all the information you will need to fabricate this spring.
Finally observe the Online Spring Force Tester Tab which shows the springs force and load in a real time animation which shows hooke's law in action.
The Spring Constant Calculator has over 70 Trillion Spring Configurations.
Example based on Acxess Springs Part Number PC043-250-13700-MW-1043-C-N-IN
Here is the complete Spring Constant Calculator Instructions page
What is Spring Constant?
The concept of the spring constant, commonly denoted as k, or spring rate is integral to understanding the mechanics of springs and their behavior under various loads. This parameter measures the stiffness of a spring and defines the amount of force necessary to move the spring's length, either by compressing, extending or rotating the spring by a unit distance. Essentially, it quantifies how resistant a spring is to being compressed, stretched or twisted, which is a critical aspect in fields ranging from mechanical engineering to physics. Compression springs compress, extension springs extend and torsion springs rotate or twist.
Hooke’s Law, which is foundational in the study of spring dynamics, states that the force exerted by a spring is proportional to the distance it is stretched, compressed or twisted. This relationship is mathematically represented by F=kx where F is the force applied to the spring, x is the displacement from the spring's original length, and k is the spring constant. This law highlights the linear behavior of springs within the elastic limit, where the deformation is reversible, and the spring returns to its original shape once the force is removed.
The units of the spring constant, such as Newtons per millimeter (N/mm) or pounds per inch (lbs/in), depend on the system of measurement used and indicate the force required per unit of length change in the case of compression and extension springs. Torsion springs constant unit of measure in imperial units is inch-lbs/degree and metric is N-mm/degree . A higher spring constant indicates a stiffer spring, requiring more force to produce the same displacement compared to a spring with a lower spring constant.
Understanding the spring constant is crucial not only for designing mechanical systems like vehicle suspensions and electronic devices but also for applications in biomechanics and architecture. It allows engineers and designers to predict how springs will behave under load, optimize performance, and ensure safety and comfort in their applications. Accurate calculation and application of the spring constant can lead to innovations in product design and improvements in a wide range of technological and industrial fields.
Can I change My Spring Constant?
Yes, The AI behind our spring constant calculator software is different, it designs stronger, weaker and almost identical spring designs at the time of inputs so you don't need to re-plug in more numbers. Just look at the results tab and you will get thousands of similar springs designs that are weaker, stronger and very similar so you can change your spring constant at any time by picking a different spring constant from the thousands of spring options to choose from.
How to Find Your Spring Constant Easily
Go to Spring Creator 5.0 and input just a few spring dimensions and get your spring constant in seconds.
G-Values for Common Spring Materials
Understanding the shear modulus (G) of various materials is crucial for accurately calculating the compression spring constant. The shear modulus, often denoted a G, is a measure of a material's ability to resist shear deformation and is a fundamental property in determining a spring's stiffness. Below is a breakdown of the G-values for common spring materials:
| | | |
---
| Music Wire ASTM A228 = 11.5 x 10^6 | Stainless Steel 302 ASTM A313 = 11.2 x 10^6 | Phosphor Bronze ASTM B 159 = 5.9 x 10^6 |
| Monel 400 AMS 7233 = 9.6 x 10^6 | Inconel X-750 AMS 5698,5699 = 11.5 x 10^6 | Copper = 6.5 x 10^6 |
| Beryllium Copper ASTM B 197 = 6.9 x 10^6 | | |
Music Wire ASTM A228 =
11.5 x 10^6
Stainless Steel 302 ASTM A313 =
11.2 x 10^6
Phosphor Bronze ASTM B 159 =
5.9 x 10^6
Monel 400 AMS 7233 =
9.6 x 10^6
Inconel X-750 AMS 5698,5699 =
11.5 x 10^6
Copper =
6.5 x 10^6
Beryllium Copper ASTM B 197 =
6.9 x 10^6
These values reflect the different properties of common spring materials for spring wire, influencing their performance in specific applications. For instance, materials like Music Wire and Inconel have higher G values, indicating a higher stiffness, which is ideal for springs that require significant resistance to deformation under load.
Compression Spring Constant
Compression spring constant or compression spring rate is defined as the rate of force per inch of compression or in metric its Newtons per millimeter of compression. The compression spring constant is a crucial factor in mechanical engineering, playing a pivotal role in applications where springs are compressed by an external load. It quantifies the stiffness of a compression spring, indicating how much force is required to compress the spring by a unit of length.
Formula for Compression Spring Constant
The fundamental formula to calculate the compression spring constant
k = Gd^4 ÷ (8D^3 n)
Where:
Example of Calculating the Compression Spring Constant
In this example, we calculate the compression spring constant k for a spring made from Music Wire, a common material known for its high strength and durability. The specifications for the spring are as follows: a wire diameter (d) of 0.035 inches, an outer diameter (OD) of 0.500 inches, a mean diameter (D) of 0.465 inches, and a total of 8 active coils (N). The spring's free length (FL) is 1.000 inch. Given these parameters, we use the spring constant formula k = Gd^4 ÷ (8D^3 n).
For Music Wire, the shear modulus (G) is 11.5×10^6 psi. Substituting all the known values into the formula, we perform the calculation as follows:
Finally, the spring constant k is computed to be 2.68 pounds per inch. This value signifies that a force of 2.68 pounds is required to compress this specific spring by one inch of distance. In other words the spring needs a force of 2.68 lbs to compress one inch of distance. This calculation not only helps in understanding the physical characteristics of the spring but also assists in predicting how the spring will perform under various load conditions, crucial for applications requiring precise mechanical movements and load-bearing capacities.
Example used is Part Number: AC035-500-10000-MW-1000-C-N-IN
Blueprint for Part Number AC035-500-10000-MW-1000-C-N-IN
Formula for Determining the Spring Constant from Load and Travel
Formula for Determining the Spring Constant from Load and Travel
The basic formula to calculate the spring constant for compression springs when you have predetermined the load and the distance traveled it needs to achieve is relatively straightforward:
k = Load / Distance Traveled
This equation derives from Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement caused by that force. Here, k represents the spring constant, indicating the force per unit of distance traveled.
Practical Example
Suppose you are designing a spring mechanism where a specific load needs to result in a precise amount of compression to function correctly. Consider a scenario where:
| |
| Load needed is 10 pounds Distance Traveled or compression needed is 4 inches |
Using the formula:
| |
| k = 10 lbs / 4 inches = 2.5 lbs / in Spring Rate per / inch of travel or compression |
Example spring used for this formula is part number:
AC047-536-26000-MW-6300-C-N-IN
Blueprint for part number: AC047-536-26000-MW-6300-C-N-IN
This spring constant calculator shows that the compression spring constant k is 2.5 pounds per inch. In practical terms, this means that for every inch that the spring is compressed, it requires 2.5 pounds of force to do so.
Extension Spring Constant
The extension spring constant is a measure of the stiffness of an extension spring, indicating how much force is required to extend the spring by a unit of distance. This characteristic is crucial for ensuring that extension springs function correctly within their intended applications, such as in machinery where parts must return to a specific position after being pulled.
Formula for Extension Spring Constant
The extension spring constant k is calculated using a formula similar to that of compression springs, reflecting Hooke’s Law, which states that the force exerted by a spring is proportional to the displacement of that spring. The formula for the extension spring constant is:
k = Gd^4 ÷ (8D^3 n)
Where:
This formula provides the basis for calculating how stiff the spring is per unit of distance traveled which is lbs/in or N/mm of extension, this translates to how much force is needed to extend it by a unit of travel.
Formula for Extension Spring Constant Based on Load and Travel
The formula to calculate the extension spring constant when the desired load and the resulting extension are specified, while also considering any initial tension in the spring, is given by:
k = Load−Initial Tension / Distance Traveled
This formula allows for a more accurate calculation of the extension spring constant by factoring in the initial tension, which is the force already present in the spring when it is at its relaxed length.
Example of Calculating the Extension Spring Constant
Consider an extension spring that needs to extend under a specific load and within a particular range of extension. For instance, suppose you have a spring with an initial tension of 2.5 pounds, and you want to achieve a load of 10 pounds when the spring is extended by 0.568 inches:
Using the formula provided:
k = 10 lbs − 2.5 lbs / 0.568 inches = 7.5 lbs / 0.568 inches = 13.204 lbs/in
This result, 13.204 pounds per inch extension spring constant, indicates the extension spring constant. It tells us that each inch of extension requires 13.204 pounds of force, then you factor in the initial tension of 2.5 lbs thus giving you a total extension spring load of 10 lbs at 0.568 inches of extended travel.
Example spring used for this formula is part number:
PE051-386-17000-SST-1344-MH-N-IN
Blueprint of part number PE051-386-17000-SST-1344-MH-N-IN
Torsion Spring Constant
The torsion spring constant or torsion spring rate defines the stiffness of a rotational torsion spring, indicating the torque required to rotate the spring by one degree of travel. The torsion spring constant in inches is: inch-lbs/degree and metric is N-mm/degree. Unlike compression or extension springs, which involve linear movement, torsion springs work on rotational motion. They are crucial in applications that require rotational force, such as hinges, automotive trunk lids, and various machinery components.
Formula for Torsion Spring Constant
The torsion spring constant is calculated using a unique formula due to the nature of rotational motion:
k = Ed^4 / 10.8DN
Where:
Example: Calculating Torsion Spring Constant
Let's take a practical example of calculating the torsion spring constant, using a spring made of Music Wire. Assume the spring parameters are as follows:
Substituting these values into the formula gives:
k = 30×10^6×(0.035)^4 / 10.8×0.465×3
Breaking down the calculation:
This calculation indicates that the torsion spring constant is approximately 2.988 inch pounds per 360 degrees. Let's break it down further and take 2.988 inch-lbs/360 degrees = 0.008 Torsion spring rate per degree of radial travel.
This means if we twist or rotate the torsion spring 90 degrees we will get 0.753 inch-lbs of torque. Formula is torsion spring rate of 0.008 inch-lbs x 90 degrees of radial travel = 0.753 inch-lbs of torque.
Formula to determine torsion spring constant k :
k = torque / Distance Traveled in degrees.
So in the above example take the torque of 0.753 inch-lbs / 90 degrees = 0.008 inch-lbs per degree of torsion spring rate. This is how to calculate torsion spring constant by torque or load and knowing the distance in degrees of travel.
Example spring used for this formula is part number:
AT035-500-3000-MW-RH-0750-N-IN
Modulus of Elasticity (E) for Common Spring Wire Materials
The modulus of elasticity, often represented as E, is a crucial property for spring wires because it measures the stiffness of a material when subjected to mechanical stress. This parameter, typically expressed in millions of pounds per square inch (psi × 10^6), provides a reference point for predicting how well a spring will return to its original shape after being flexed or compressed. Different materials have unique modulus values, influencing their behavior in various applications.
Renowned for its strength and high tensile properties, music wire is a popular choice for precision springs that require significant force and high resistance to deformation.
Stainless steel provides an excellent balance between strength and corrosion resistance, making it suitable for springs exposed to harsh environments or high humidity.
Chrome vanadium is prized for its resilience and durability, providing good resistance to fatigue, which makes it ideal for high-stress applications.
Chrome silicon, known for its high tensile strength and stability at elevated temperatures, is a great choice for automotive and aerospace applications.
Phosphor bronze offers excellent flexibility and fatigue resistance, while being more corrosion-resistant than many steel alloys. It is often used in electrical and marine applications.
These modulus values give insight into the mechanical properties of various spring wires, allowing engineers to select the most suitable material for specific performance requirements. Whether designing springs for precision equipment or industrial machinery, understanding the modulus of elasticity is essential to ensure the springs maintain their shape and function over time.
Our spring constant calculator is based on the formulas on this page.
Conclusion
In conclusion, understanding the spring constant calculator is crucial for designing and optimizing mechanical systems. Whether you're working with compression, extension, or torsion springs, accurately calculating the spring constant on our spring constant calculator ensures that your springs perform reliably and meet the desired specifications under varying loads. The modulus of elasticity (E) and shear modulus (G) are fundamental properties that influence spring stiffness, affecting how much force is needed to compress, extend, or rotate the spring. Given the diverse applications and importance of these calculations, utilizing a reliable and user-friendly tool is essential.
Spring Creator 5.0 simplifies these complex calculations, providing accurate spring constants for various spring types and materials. By inputting your spring parameters into the spring constant calculator, you can quickly and easily determine the spring constant you need for your specific design requirements. Embrace the efficiency and accuracy of Spring Creator 5.0 to streamline your spring design process, ensuring that your springs deliver consistent performance in every application.
Created by Alfonso Jaramillo Jr
President Acxess Spring
Over 40 Years of Experience in Spring Engineering and Manufacturing
Create the right spring with Spring Creator 5.0
Are you an engineer or an inventor looking for the right spring? Spring Creator 5.0 is the tool you need. Test and design compression, extension, and torsion springs, configuring every detail to your liking. Additionally, our tool provides you with a 3D blueprint containing all the necessary information for its manufacturing and allows you to visualize your spring in 3D CAD to complement your design. Discover our tool and start creating today!"
Created by Alfonso Jaramillo Jr
President Acxess Spring
Over 40 Years of Experience in Spring Engineering and Manufacturing
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10243 | https://www.doubtnut.com/qna/18920749 | What is the degree of unsaturation of cubane?
More from this Exercise
The correct Answer is:5
To find the degree of unsaturation of cubane (C8H8), we can follow these steps: Step 1: Understand the Formula for Degree of Unsaturation The degree of unsaturation (DU) can be calculated using the formula: DU=(2C+2+N−H−X)2 where: - C = number of carbon atoms - H = number of hydrogen atoms - N = number of nitrogen atoms (if any) - X = number of halogen atoms (F, Cl, Br, I) Step 2: Identify the Number of Atoms in Cubane For cubane, the molecular formula is C8H8. Therefore: - C=8 - H=8 - N=0 (no nitrogen) - X=0 (no halogens) Step 3: Substitute the Values into the Formula Now, substitute the values into the degree of unsaturation formula: DU=(2(8)+2+0−8−0)2 Step 4: Simplify the Equation Calculate the values inside the parentheses: DU=(16+2−8)2 DU=102 DU=5 Conclusion The degree of unsaturation of cubane is 5. ---
To find the degree of unsaturation of cubane (C8H8), we can follow these steps:
Step 1: Understand the Formula for Degree of Unsaturation
The degree of unsaturation (DU) can be calculated using the formula:
DU=(2C+2+N−H−X)2
where:
- C = number of carbon atoms
- H = number of hydrogen atoms
- N = number of nitrogen atoms (if any)
- X = number of halogen atoms (F, Cl, Br, I)
Step 2: Identify the Number of Atoms in Cubane
For cubane, the molecular formula is C8H8. Therefore:
- C=8
- H=8
- N=0 (no nitrogen)
- X=0 (no halogens)
Step 3: Substitute the Values into the Formula
Now, substitute the values into the degree of unsaturation formula:
DU=(2(8)+2+0−8−0)2
Step 4: Simplify the Equation
Calculate the values inside the parentheses:
DU=(16+2−8)2
DU=102
DU=5
Conclusion
The degree of unsaturation of cubane is 5.
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10244 | https://pubmed.ncbi.nlm.nih.gov/22910569/ | Skin biopsy: a pillar in the identification of cutaneous Mycobacterium tuberculosis infection - PubMed
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. 2012 Aug 21;6(8):626-31.
doi: 10.3855/jidc.2729.
Skin biopsy: a pillar in the identification of cutaneous Mycobacterium tuberculosis infection
Alejandro Hernández Solis1,Norma Estela Herrera González,Fernando Cazarez,Patricia Mercadillo Pérez,Hiram Olivera Olivera Diaz,Alejandro Escobar-Gutierrez,Ileana Cortés Ortíz,Heleodora González González,Arturo Reding-Bernal,Raúl Cícero Sabido
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1 Facultad de Medicina, UNAM, Hospital General de México, Mexico. drhernandezsolis@yahoo.com.mx
PMID: 22910569
DOI: 10.3855/jidc.2729
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Skin biopsy: a pillar in the identification of cutaneous Mycobacterium tuberculosis infection
Alejandro Hernández Solis et al. J Infect Dev Ctries.2012.
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. 2012 Aug 21;6(8):626-31.
doi: 10.3855/jidc.2729.
Authors
Alejandro Hernández Solis1,Norma Estela Herrera González,Fernando Cazarez,Patricia Mercadillo Pérez,Hiram Olivera Olivera Diaz,Alejandro Escobar-Gutierrez,Ileana Cortés Ortíz,Heleodora González González,Arturo Reding-Bernal,Raúl Cícero Sabido
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1 Facultad de Medicina, UNAM, Hospital General de México, Mexico. drhernandezsolis@yahoo.com.mx
PMID: 22910569
DOI: 10.3855/jidc.2729
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Introduction: The present study aimed to establish the frequency and clinical characteristics of cutaneous tuberculosis among Mexican adult patients.
Methodology: Ninety-five patients with clinically compatible lesions to cutaneous tuberculosis participated in the study. All patients were HIV negative and none of them had previous anti-TB treatment. A skin biopsy was taken from every patient suspected of having tuberculosis, and a histopathologic examination was performed as follows: Ziehl-Neelsen staining; culturing of mycobacteria by Löwenstein-Jensen (L-J) medium; Mycobacteria Growth Indicator Tube detection via BACTEC (MGIT-360); and polymerase chain reaction (PCR) with the sequence of insertion IS6110 for Mycobacterium tuberculosis complex.
Results: Tuberculosis was confirmed in 65 out of 95 cases (68.4%). Identified lesions were scrofuloderma (42 cases, 64.6%); lupus vulgaris (12 cases, 18.4%); warty tuberculosis (six cases, 9.2%); and papulonecrotic tuberculoid (five cases; 7.7%). The Ziehl-Neelsen staining was positive for acid fast bacilli in nine cases (13.8%) and 48 patients were positive for the PCR amplification (73.8%). All skin biopsies resulted positive for tuberculosis. A positive clinical response to the specific treatment was considered a confirmation for tuberculosis. The noninfectious etiology corresponded to 30 cases (31.6%).
Conclusions: Tuberculosis in developing countries is still an important cause of skin lesions which must be studied via histopathological examination and culture due to their low bacillary load. A PCR test is necessary to obtain faster confirmation of the disease and to establish an early, specific and effective treatment.
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Goldman_equation
Goldman equation
The Goldman-Hodgkin-Katz voltage equation, more commonly known as the Goldman equation is used in cell membrane physiology to determine the potential across a cell's membrane taking into account all of the ions that are permeant through that membrane.
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The discoverers of this are David E. Goldman of Columbia University, and the English Nobel laureates Alan Lloyd Hodgkin and Bernard Katz.
The equation
The GHK voltage equation for N monovalent positive ionic species and M negative:
This results in the following if we consider a membrane separating two K x N a 1 − x C l-solutions:
It is "Nernst-like" but has a term for each permeant ion. The Nernst equation can be considered a special case of the Goldman equation for only one ion:
E m = The membrane potential
P i o n = the permeability for that ion
[i o n]o u t = the extracellular concentration of that ion
[i o n]i n = the intracellular concentration of that ion
R = The ideal gas constant
T = The temperature in kelvins
F = Faraday's constant
The first term, before the parenthesis, can be reduced to 61.5 log for calculations at human body temperature (37 C)
Note that the ionic charge determines the sign of the membrane potential contribution.
The usefulness of the GHK equation to electrophysiologists is that it allows one to calculate the predicted membrane potential for any set of specified permeabilities. For example, if one wanted to calculate the resting potential of a cell, they would use the values of ion permeability that are present at rest (e.g. ). If one wanted to calculate the peak voltage of an action potential, one would simply substitute the permeabilities that are present at that time (e.g. ).
See also
GHK current equation
Category: Physical chemistry
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Goldman_equation". A list of authors is available in Wikipedia.
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10246 | https://en.wikipedia.org/wiki/Liouville_number | Jump to content
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Contents
(Top)
1 The existence of Liouville numbers (Liouville's constant)
1.1 Notes on the proof
2 Irrationality
3 Liouville numbers and transcendence
4 Uncountability
5 Liouville numbers and measure
6 Structure of the set of Liouville numbers
7 Irrationality measure
8 See also
9 References
10 External links
Liouville number
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From Wikipedia, the free encyclopedia
Class of irrational numbers
In number theory, a Liouville number is a real number with the property that, for every positive integer , there exists a pair of integers with such that
The inequality implies that Liouville numbers possess an excellent sequence of rational number approximations. In 1844, Joseph Liouville proved a bound showing that there is a limit to how well algebraic numbers can be approximated by rational numbers, and he defined Liouville numbers specifically so that they would have rational approximations better than the ones allowed by this bound. Liouville also exhibited examples of Liouville numbers thereby establishing the existence of transcendental numbers for the first time. One of these examples is Liouville's constant
in which the nth digit after the decimal point is 1 if is the factorial of a positive integer and 0 otherwise. It is known that π and e, although transcendental, are not Liouville numbers.
The existence of Liouville numbers (Liouville's constant)
[edit]
Liouville numbers can be shown to exist by an explicit construction.
For any integer and any sequence of integers such that for all and for infinitely many , define the number
In the special case when , and for all , the resulting number is called Liouville's constant:
It follows from the definition of that its base- representation is
where the th term is in the th place.
Since this base- representation is non-repeating it follows that is not a rational number. Therefore, for any rational number , .
Now, for any integer , and can be defined as follows:
Then,
Therefore, any such is a Liouville number.
Notes on the proof
[edit]
The inequality follows since ak ∈ {0, 1, 2, ..., b−1} for all k, so at most ak = b−1. The largest possible sum would occur if the sequence of integers (a1, a2, ...) were (b−1, b−1, ...), i.e. ak = b−1, for all k. will thus be less than or equal to this largest possible sum.
The strong inequality follows from the motivation to eliminate the series by way of reducing it to a series for which a formula is known. In the proof so far, the purpose for introducing the inequality in #1 comes from intuition that (the geometric series formula); therefore, if an inequality can be found from that introduces a series with (b−1) in the numerator, and if the denominator term can be further reduced from to , as well as shifting the series indices from 0 to , then both series and (b−1) terms will be eliminated, getting closer to a fraction of the form , which is the end-goal of the proof. This motivation is increased here by selecting now from the sum a partial sum. Observe that, for any term in , since b ≥ 2, then , for all k (except for when n=1). Therefore, (since, even if n=1, all subsequent terms are smaller). In order to manipulate the indices so that k starts at 0, partial sum will be selected from within (also less than the total value since it is a partial sum from a series whose terms are all positive). Choose the partial sum formed by starting at k = (n+1)! which follows from the motivation to write a new series with k=0, namely by noticing that .
For the final inequality , this particular inequality has been chosen (true because b ≥ 2, where equality follows if and only if n=1) because of the wish to manipulate into something of the form . This particular inequality allows the elimination of (n+1)! and the numerator, using the property that (n+1)! − n! = (n!)n, thus putting the denominator in ideal form for the substitution .
Irrationality
[edit]
Here the proof will show that the number where c and d are integers and cannot satisfy the inequalities that define a Liouville number. Since every rational number can be represented as such the proof will show that no Liouville number can be rational.
More specifically, this proof shows that for any positive integer n large enough that [equivalently, for any positive integer )], no pair of integers exists that simultaneously satisfies the pair of bracketing inequalities
If the claim is true, then the desired conclusion follows.
Let p and q be any integers with Then,
If then
meaning that such pair of integers would violate the first inequality in the definition of a Liouville number, irrespective of any choice of n .
If, on the other hand, since then, since is an integer, we can assert the sharper inequality From this it follows that
Now for any integer the last inequality above implies
Therefore, in the case such pair of integers would violate the second inequality in the definition of a Liouville number, for some positive integer n.
Therefore, to conclude, there is no pair of integers with that would qualify such an as a Liouville number.
Hence a Liouville number cannot be rational.
Liouville numbers and transcendence
[edit]
No Liouville number is algebraic. The proof of this assertion proceeds by first establishing a property of irrational algebraic numbers. This property essentially says that irrational algebraic numbers cannot be well approximated by rational numbers, where the condition for "well approximated" becomes more stringent for larger denominators. A Liouville number is irrational but does not have this property, so it cannot be algebraic and must be transcendental. The following lemma is usually known as Liouville's theorem (on diophantine approximation), there being several results known as Liouville's theorem.
Lemma: If is an irrational root of an irreducible polynomial of degree with integer coefficients, then there exists a real number such that for all integers with ,
Proof of Lemma: Let be a minimal polynomial with integer coefficients, such that .
By the fundamental theorem of algebra, has at most distinct roots. Therefore, there exists such that for all we get .
Since is a minimal polynomial of we get , and also is continuous. Therefore, by the extreme value theorem there exists and such that for all we get .
Both conditions are satisfied for .
Now let be a rational number. Without loss of generality we may assume that . By the mean value theorem, there exists such that
Since and , both sides of that equality are nonzero. In particular and we can rearrange:
Proof of assertion: As a consequence of this lemma, let x be a Liouville number; as noted in the article text, x is then irrational. If x is algebraic, then by the lemma, there exists some integer n and some positive real A such that for all p, q
Let r be a positive integer such that 1/(2r) ≤ A and define m = r + n. Since x is a Liouville number, there exist integers a, b with b > 1 such that
which contradicts the lemma. Hence a Liouville number cannot be algebraic, and therefore must be transcendental.
Establishing that a given number is a Liouville number proves that it is transcendental. However, not every transcendental number is a Liouville number. The terms in the continued fraction expansion of every Liouville number are unbounded; using a counting argument, one can then show that there must be uncountably many transcendental numbers which are not Liouville. Using the explicit continued fraction expansion of e, one can show that e is an example of a transcendental number that is not Liouville. Mahler proved in 1953 that π is another such example.
Uncountability
[edit]
Consider the number
: 3.1400010000000000000000050000....
3.14(3 zeros)1(17 zeros)5(95 zeros)9(599 zeros)2(4319 zeros)6...
where the digits are zero except in positions n! where the digit equals the nth digit following the decimal point in the decimal expansion of π.
As shown in the section on the existence of Liouville numbers, the same is true of the set of all Liouville numbers.
Moreover, the Liouville numbers form a dense subset of the set of real numbers.
Liouville numbers and measure
[edit]
From the point of view of measure theory, the set of all Liouville numbers is small. More precisely, its Lebesgue measure, , is zero. The proof given follows some ideas by John C. Oxtoby.: 8
For positive integers and set:
then
Observe that for each positive integer and , then
Since
and then
Now
and it follows that for each positive integer , has Lebesgue measure zero. Consequently, so has .
In contrast, the Lebesgue measure of the set of all real transcendental numbers is infinite (since the set of algebraic numbers is a null set).
One could show even more - the set of Liouville numbers has Hausdorff dimension 0 (a property strictly stronger than having Lebesgue measure 0).
Structure of the set of Liouville numbers
[edit]
For each positive integer n, set
The set of all Liouville numbers can thus be written as
Each is an open set; as its closure contains all rationals (the from each punctured interval), it is also a dense subset of real line. Since it is the intersection of countably many such open dense sets, L is comeagre, that is to say, it is a dense Gδ set.
Irrationality measure
[edit]
Main article: Irrationality measure
The Liouville–Roth irrationality measure (irrationality exponent, approximation exponent, or Liouville–Roth constant) of a real number is a measure of how "closely" it can be approximated by rationals. It is defined by adapting the definition of Liouville numbers: instead of requiring the existence of a sequence of pairs that make the inequality hold for each —a sequence which necessarily contains infinitely many distinct pairs—the irrationality exponent is defined to be the supremum of the set of for which such an infinite sequence exists, that is, the set of such that is satisfied by an infinite number of integer pairs with .: 246 For any value , the infinite set of all rationals satisfying the above inequality yields good approximations of . Conversely, if , then there are at most finitely many with that satisfy the inequality. If is a Liouville number then .
See also
[edit]
Brjuno number
Markov constant
Diophantine approximation
References
[edit]
^ Joseph Liouville (May 1844). "Mémoires et communications". Comptes rendus de l'Académie des Sciences (in French). 18 (20, 21): 883–885, 910–911.
^ Baker, Alan (1990). Transcendental Number Theory (paperback ed.). Cambridge University Press. p. 1. ISBN 978-0-521-39791-9.
^ Baker 1990, p. 86.
^ Kurt Mahler, "On the approximation of π", Nederl. Akad. Wetensch. Proc. Ser. A., t. 56 (1953), p. 342–366.
^ Oxtoby, John C. (1980). Measure and Category. Graduate Texts in Mathematics. Vol. 2 (Second ed.). New York-Berlin: Springer-Verlag. doi:10.1007/978-1-4684-9339-9. ISBN 0-387-90508-1. MR 0584443.
^ Bugeaud, Yann (2012). Distribution modulo one and Diophantine approximation. Cambridge Tracts in Mathematics. Vol. 193. Cambridge: Cambridge University Press. doi:10.1017/CBO9781139017732. ISBN 978-0-521-11169-0. MR 2953186. Zbl 1260.11001.
External links
[edit]
The Beginning of Transcendental Numbers
| v t e Irrational numbers |
| Chaitin's (Ω) Liouville Prime (ρ) Omega Cahen Logarithm of 2 Dottie Lemniscate (ϖ) Twelfth root of 2 Apéry's (ζ(3)) Cube root of 2 Plastic ratio (ρ) Square root of 2 Supergolden ratio (ψ) Erdős–Borwein (E) Golden ratio (φ) Square root of 3 Supersilver ratio (ς) Square root of 5 Silver ratio (σ) Square root of 6 Square root of 7 Euler's (e) Pi (π) Square root of 10 |
| Schizophrenic Transcendental Trigonometric |
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The National World War II Museum Uses Unique Ways to Teach History
The National World War II Museum in New Orleans, Louisiana, isn’t your traditional museum. Leaders pride themselves in using unique, innovative methods to share history with visitors.
“It’s a great place to inspire young people through the stories of a generation who sacrificed and saved the world from tyranny,” said Becky Mackie, Executive Vice President and Chief Operating Officer.
It’s not just pictures, panels to read, and artifact displays, but instead the museum uses immersive environments, interactive exhibits, and cinematic experiences to educate visitors on the war.
“When you walk into our Guadalcanal exhibit you sense that you are walking into a jungle, you hear the sounds and see the scenery of the jungle while learning WWII stories through first person accounts and curated exhibits,” Mackie explained. “When you sit in the Solomon Victory Theater watching the 4-D movie Beyond All Boundaries, the harsh weather our troops encountered is emphasized as snow falls from the ceiling during the Battle of the Bulge.”
Originally created as the National D-Day Museum, the museum was officially designated the National World War II Museum by an act of Congress in 2004.
Today, the seven-acre campus in a former warehouse district in New Orleans, continues to grow and attract thousands of visitors every year. Three separate EDA investments have supported the museum, helping to boost attendance, create new jobs for the community, and support the economic driver.
In 2007, an EDA grant helped to fund the museum’s first major expansion. A $1.2 million investment supported the technology components of the Solomon Victory Theatre.
“It highlights the history of World War II in 45 minutes with a dynamic, 4-D theatre experience. In the year after we premiered Beyond All Boundaries, our attendance grew 200%,” Mackie said.
In 2010, EDA again supported the museum through a $2 million grant for the Kushner Restoration Pavilion. The building features glass exterior walls that allow the public a behind-the-scenes view of the restoration and preservation of World War II artifacts, including PT-305, a patrol-torpedo boat. In addition to creating a place to expand restoration work, the pavilion created an additional 20 jobs, increasing the museum’s average wage by $3,000 per person.
The most recent, EDA investment was an $824,000 grant for infrastructure for the museum’s new Expressions of America experience. The exhibit will use mapped projection lighting on surfaces of the buildings to tell the story of the war through first-person experiences of individuals who served at home and abroad, stories told through their letters, and the story of Bob Hopes gift of humor to the troops during the war. It’s scheduled to open on Veteran’s Day later this year.
“It’s an opportunity for us to engage younger audiences as well as those looking for a unique experience, and it will open the campus for a visitor experience at night,” Mackie said.
As the museum completes its brick-and-mortar exhibit and educational facilities, it is also expanding its focus to make exhibits accessible to the audience that isn’t able to visit the museum in person. The museum is digitizing its archives and collections and making the collection accessible on the museum’s website.
It’s all in an effort to make sure our country’s past isn’t soon forgotten. “As a nation we always want to learn and know the history behind what made us,” Mackie said.
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From Wikipedia, the free encyclopedia
Graph made from a subset of another graph's nodes and their edges
In graph theory, an induced subgraph of a graph is another graph, formed from a subset of the vertices of the graph and all of the edges, from the original graph, connecting pairs of vertices in that subset.
Definition
[edit]
Formally, let be any graph, and let be any subset of vertices of G. Then the induced subgraph is the graph whose vertex set is and whose edge set consists of all of the edges in that have both endpoints in . That is, for any two vertices , and are adjacent in if and only if they are adjacent in . The same definition works for undirected graphs, directed graphs, and even multigraphs.
The induced subgraph may also be called the subgraph induced in by , or (if context makes the choice of unambiguous) the induced subgraph of .
Examples
[edit]
Important types of induced subgraphs include the following.
Induced paths are induced subgraphs that are paths. The shortest path between any two vertices in an unweighted graph is always an induced path, because any additional edges between pairs of vertices that could cause it to be not induced would also cause it to be not shortest. Conversely, in distance-hereditary graphs, every induced path is a shortest path.
Induced cycles are induced subgraphs that are cycles. The girth of a graph is defined by the length of its shortest cycle, which is always an induced cycle. According to the strong perfect graph theorem, induced cycles and their complements play a critical role in the characterization of perfect graphs.
Cliques and independent sets are induced subgraphs that are respectively complete graphs or edgeless graphs.
Induced matchings are induced subgraphs that are matchings.
The neighborhood of a vertex is the induced subgraph of all vertices adjacent to it.
Computation
[edit]
The induced subgraph isomorphism problem is a form of the subgraph isomorphism problem in which the goal is to test whether one graph can be found as an induced subgraph of another. Because it includes the clique problem as a special case, it is NP-complete.
References
[edit]
^ Diestel, Reinhard (2006), Graph Theory, Graduate texts in mathematics, vol. 173, Springer-Verlag, pp. 3–4, ISBN 9783540261834.
^ Howorka, Edward (1977), "A characterization of distance-hereditary graphs", The Quarterly Journal of Mathematics, Second Series, 28 (112): 417–420, doi:10.1093/qmath/28.4.417, MR 0485544.
^ Chudnovsky, Maria; Robertson, Neil; Seymour, Paul; Thomas, Robin (2006), "The strong perfect graph theorem", Annals of Mathematics, 164 (1): 51–229, arXiv:math/0212070, doi:10.4007/annals.2006.164.51, MR 2233847.
^ Johnson, David S. (1985), "The NP-completeness column: an ongoing guide", Journal of Algorithms, 6 (3): 434–451, doi:10.1016/0196-6774(85)90012-4, MR 0800733.
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10249 | http://www.biostathandbook.com/variabletypes.html | Kinds of variables - Handbook of Biological Statistics
Handbook of Biological Statistics
John H. McDonald
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Contents
Basics
Introduction
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Kinds of biological variables
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Power analysis
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One-sample t–test
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Homoscedasticity
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Kruskal–Wallis test
Nested anova
Two-way anova
Paired t–test
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Introduction to SAS
Choosing the right test
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Table of Contents
Types of biological variables
Summary
There are three main types of variables: measurement variables, which are expressed as numbers (such as 3.7 mm); nominal variables, which are expressed as names (such as "female"); and ranked variables, which are expressed as positions (such as "third"). You need to identify the types of variables in an experiment in order to choose the correct method of analysis.
Introduction
One of the first steps in deciding which statistical test to use is determining what kinds of variables you have. When you know what the relevant variables are, what kind of variables they are, and what your null and alternative hypotheses are, it's usually pretty easy to figure out which test you should use. I classify variables into three types: measurement variables, nominal variables, and ranked variables. You'll see other names for these variable types and other ways of classifying variables in other statistics references, so try not to get confused.
Isopod crustacean (pillbug or roly-poly), Armadillidium vulgare.
You'll analyze similar experiments, with similar null and alternative hypotheses, completely differently depending on which of these three variable types are involved. For example, let's say you've measured variable X in a sample of 56 male and 67 female isopods (Armadillidium vulgare, commonly known as pillbugs or roly-polies), and your null hypothesis is "Male and female A. vulgare have the same values of variable X." If variable X is width of the head in millimeters, it's a measurement variable, and you'd compare head width in males and females with a two-sample t–test or a one-way analysis of variance (anova). If variable X is a genotype (such as AA, Aa, or aa), it's a nominal variable, and you'd compare the genotype frequencies in males and females with a Fisher's exact test. If you shake the isopods until they roll up into little balls, then record which is the first isopod to unroll, the second to unroll, etc., it's a ranked variable and you'd compare unrolling time in males and females with a Kruskal–Wallis test.
Measurement variables
Measurement variables are, as the name implies, things you can measure. An individual observation of a measurement variable is always a number. Examples include length, weight, pH, and bone density. Other names for them include "numeric" or "quantitative" variables.
Some authors divide measurement variables into two types. One type is continuous variables, such as length of an isopod's antenna, which in theory have an infinite number of possible values. The other is discrete (or meristic) variables, which only have whole number values; these are things you count, such as the number of spines on an isopod's antenna. The mathematical theories underlying statistical tests involving measurement variables assume that the variables are continuous. Luckily, these statistical tests work well on discrete measurement variables, so you usually don't need to worry about the difference between continuous and discrete measurement variables. The only exception would be if you have a very small number of possible values of a discrete variable, in which case you might want to treat it as a nominal variable instead.
When you have a measurement variable with a small number of values, it may not be clear whether it should be considered a measurement or a nominal variable. For example, let's say your isopods have 20 to 55 spines on their left antenna, and you want to know whether the average number of spines on the left antenna is different between males and females. You should consider spine number to be a measurement variable and analyze the data using a two-sample t–test or a one-way anova. If there are only two different spine numbers—some isopods have 32 spines, and some have 33—you should treat spine number as a nominal variable, with the values "32" and "33," and compare the proportions of isopods with 32 or 33 spines in males and females using a Fisher's exact test of independence (or chi-square or G–test of independence, if your sample size is really big). The same is true for laboratory experiments; if you give your isopods food with 15 different mannose concentrations and then measure their growth rate, mannose concentration would be a measurement variable; if you give some isopods food with 5 mM mannose, and the rest of the isopods get 25 mM mannose, then mannose concentration would be a nominal variable.
But what if you design an experiment with three concentrations of mannose, or five, or seven? There is no rigid rule, and how you treat the variable will depend in part on your null and alternative hypotheses. If your alternative hypothesis is "different values of mannose have different rates of isopod growth," you could treat mannose concentration as a nominal variable. Even if there's some weird pattern of high growth on zero mannose, low growth on small amounts, high growth on intermediate amounts, and low growth on high amounts of mannose, a one-way anova could give a significant result. If your alternative hypothesis is "isopods grow faster with more mannose," it would be better to treat mannose concentration as a measurement variable, so you can do a regression. In my class, we use the following rule of thumb:
—a measurement variable with only two values should be treated as a nominal variable;
—a measurement variable with six or more values should be treated as a measurement variable;
—a measurement variable with three, four or five values does not exist.
Of course, in the real world there are experiments with three, four or five values of a measurement variable. Simulation studies show that analyzing such dependent variables with the methods used for measurement variables works well (Fagerland et al. 2011). I am not aware of any research on the effect of treating independent variables with small numbers of values as measurement or nominal. Your decision about how to treat your variable will depend in part on your biological question. You may be able to avoid the ambiguity when you design the experiment—if you want to know whether a dependent variable is related to an independent variable that could be measurement, it's a good idea to have at least six values of the independent variable.
Something that could be measured is a measurement variable, even when you set the values. For example, if you grow isopods with one batch of food containing 10 mM mannose, another batch of food with 20 mM mannose, another batch with 30 mM mannose, etc. up to 100 mM mannose, the different mannose concentrations are a measurement variable, even though you made the food and set the mannose concentration yourself.
Be careful when you count something, as it is sometimes a nominal variable and sometimes a measurement variable. For example, the number of bacteria colonies on a plate is a measurement variable; you count the number of colonies, and there are 87 colonies on one plate, 92 on another plate, etc. Each plate would have one data point, the number of colonies; that's a number, so it's a measurement variable. However, if the plate has red and white bacteria colonies and you count the number of each, it is a nominal variable. Now, each colony is a separate data point with one of two values of the variable, "red" or "white"; because that's a word, not a number, it's a nominal variable. In this case, you might summarize the nominal data with a number (the percentage of colonies that are red), but the underlying data are still nominal.
Ratios
Sometimes you can simplify your statistical analysis by taking the ratio of two measurement variables. For example, if you want to know whether male isopods have bigger heads, relative to body size, than female isopods, you could take the ratio of head width to body length for each isopod, and compare the mean ratios of males and females using a two-sample t–test. However, this assumes that the ratio is the same for different body sizes. We know that's not true for humans—the head size/body size ratio in babies is freakishly large, compared to adults—so you should look at the regression of head width on body length and make sure the regression line goes pretty close to the origin, as a straight regression line through the origin means the ratios stay the same for different values of the X variable. If the regression line doesn't go near the origin, it would be better to keep the two variables separate instead of calculating a ratio, and compare the regression line of head width on body length in males to that in females using an analysis of covariance.
Circular variables
One special kind of measurement variable is a circular variable. These have the property that the highest value and the lowest value are right next to each other; often, the zero point is completely arbitrary. The most common circular variables in biology are time of day, time of year, and compass direction. If you measure time of year in days, Day 1 could be January 1, or the spring equinox, or your birthday; whichever day you pick, Day 1 is adjacent to Day 2 on one side and Day 365 on the other.
If you are only considering part of the circle, a circular variable becomes a regular measurement variable. For example, if you're doing a polynomial regression of bear attacks vs. time of the year in Yellowstone National Park, you could treat "month" as a measurement variable, with March as 1 and November as 9; you wouldn't have to worry that February (month 12) is next to March, because bears are hibernating in December through February, and you would ignore those three months.
However, if your variable really is circular, there are special, very obscure statistical tests designed just for circular data; chapters 26 and 27 in Zar (1999) are a good place to start.
Nominal variables
Nominal variables classify observations into discrete categories. Examples of nominal variables include sex (the possible values are male or female), genotype (values are AA, Aa, or aa), or ankle condition (values are normal, sprained, torn ligament, or broken). A good rule of thumb is that an individual observation of a nominal variable can be expressed as a word, not a number. If you have just two values of what would normally be a measurement variable, it's nominal instead: think of it as "present" vs. "absent" or "low" vs. "high." Nominal variables are often used to divide individuals up into categories, so that other variables may be compared among the categories. In the comparison of head width in male vs. female isopods, the isopods are classified by sex, a nominal variable, and the measurement variable head width is compared between the sexes.
Nominal variables are also called categorical, discrete, qualitative, or attribute variables. "Categorical" is a more common name than "nominal," but some authors use "categorical" to include both what I'm calling "nominal" and what I'm calling "ranked," while other authors use "categorical" just for what I'm calling nominal variables. I'll stick with "nominal" to avoid this ambiguity.
Nominal variables are often summarized as proportions or percentages. For example, if you count the number of male and female A. vulgare in a sample from Newark and a sample from Baltimore, you might say that 52.3% of the isopods in Newark and 62.1% of the isopods in Baltimore are female. These percentages may look like a measurement variable, but they really represent a nominal variable, sex. You determined the value of the nominal variable (male or female) on 65 isopods from Newark, of which 34 were female and 31 were male. You might plot 52.3% on a graph as a simple way of summarizing the data, but you should use the 34 female and 31 male numbers in all statistical tests.
It may help to understand the difference between measurement and nominal variables if you imagine recording each observation in a lab notebook. If you are measuring head widths of isopods, an individual observation might be "3.41 mm." That is clearly a measurement variable. An individual observation of sex might be "female," which clearly is a nominal variable. Even if you don't record the sex of each isopod individually, but just counted the number of males and females and wrote those two numbers down, the underlying variable is a series of observations of "male" and "female."
Ranked variables
Ranked variables, also called ordinal variables, are those for which the individual observations can be put in order from smallest to largest, even though the exact values are unknown. If you shake a bunch of A. vulgare up, they roll into balls, then after a little while start to unroll and walk around. If you wanted to know whether males and females unrolled at the same time, but your stopwatch was broken, you could pick up the first isopod to unroll and put it in a vial marked "first," pick up the second to unroll and put it in a vial marked "second," and so on, then sex the isopods after they've all unrolled. You wouldn't have the exact time that each isopod stayed rolled up (that would be a measurement variable), but you would have the isopods in order from first to unroll to last to unroll, which is a ranked variable. While a nominal variable is recorded as a word (such as "male") and a measurement variable is recorded as a number (such as "4.53"), a ranked variable can be recorded as a rank (such as "seventh").
You could do a lifetime of biology and never use a true ranked variable. When I write an exam question involving ranked variables, it's usually some ridiculous scenario like "Imagine you're on a desert island with no ruler, and you want to do statistics on the size of coconuts. You line them up from smallest to largest...." For a homework assignment, I ask students to pick a paper from their favorite biological journal and identify all the variables, and anyone who finds a ranked variable gets a donut; I've had to buy four donuts in 13 years. The only common biological ranked variables I can think of are dominance hierarchies in behavioral biology (see the dog example on the Kruskal-Wallis page) and developmental stages, such as the different instars that molting insects pass through.
The main reason that ranked variables are important is that the statistical tests designed for ranked variables (called "non-parametric tests") make fewer assumptions about the data than the statistical tests designed for measurement variables. Thus the most common use of ranked variables involves converting a measurement variable to ranks, then analyzing it using a non-parametric test. For example, let's say you recorded the time that each isopod stayed rolled up, and that most of them unrolled after one or two minutes. Two isopods, who happened to be male, stayed rolled up for 30 minutes. If you analyzed the data using a test designed for a measurement variable, those two sleepy isopods would cause the average time for males to be much greater than for females, and the difference might look statistically significant. When converted to ranks and analyzed using a non-parametric test, the last and next-to-last isopods would have much less influence on the overall result, and you would be less likely to get a misleadingly "significant" result if there really isn't a difference between males and females.
Some variables are impossible to measure objectively with instruments, so people are asked to give a subjective rating. For example, pain is often measured by asking a person to put a mark on a 10-cm scale, where 0 cm is "no pain" and 10 cm is "worst possible pain." This is not a ranked variable; it is a measurement variable, even though the "measuring" is done by the person's brain. For the purpose of statistics, the important thing is that it is measured on an "interval scale"; ideally, the difference between pain rated 2 and 3 is the same as the difference between pain rated 7 and 8. Pain would be a ranked variable if the pains at different times were compared with each other; for example, if someone kept a pain diary and then at the end of the week said "Tuesday was the worst pain, Thursday was second worst, Wednesday was third, etc...." These rankings are not an interval scale; the difference between Tuesday and Thursday may be much bigger, or much smaller, than the difference between Thursday and Wednesday.
Just like with measurement variables, if there are a very small number of possible values for a ranked variable, it would be better to treat it as a nominal variable. For example, if you make a honeybee sting people on one arm and a yellowjacket sting people on the other arm, then ask them "Was the honeybee sting the most painful or the second most painful?", you are asking them for the rank of each sting. But you should treat the data as a nominal variable, one which has three values ("honeybee is worse" or "yellowjacket is worse" or "subject is so mad at your stupid, painful experiment that they refuse to answer").
Categorizing
It is possible to convert a measurement variable to a nominal variable, dividing individuals up into a two or more classes based on ranges of the variable. For example, if you are studying the relationship between levels of HDL (the "good cholesterol") and blood pressure, you could measure the HDL level, then divide people into two groups, "low HDL" (less than 40 mg/dl) and "normal HDL" (40 or more mg/dl) and compare the mean blood pressures of the two groups, using a nice simple two-sample t–test.
Converting measurement variables to nominal variables ("dichotomizing" if you split into two groups, "categorizing" in general) is common in epidemiology, psychology, and some other fields. However, there are several problems with categorizing measurement variables (MacCallum et al. 2002). One problem is that you'd be discarding a lot of information; in our blood pressure example, you'd be lumping together everyone with HDL from 0 to 39 mg/dl into one group. This reduces your statistical power, decreasing your chances of finding a relationship between the two variables if there really is one. Another problem is that it would be easy to consciously or subconsciously choose the dividing line ("cutpoint") between low and normal HDL that gave an "interesting" result. For example, if you did the experiment thinking that low HDL caused high blood pressure, and a couple of people with HDL between 40 and 45 happened to have high blood pressure, you might put the dividing line between low and normal at 45 mg/dl. This would be cheating, because it would increase the chance of getting a "significant" difference if there really isn't one.
To illustrate the problem with categorizing, let's say you wanted to know whether tall basketball players weigh more than short players. Here's data for the 2012-2013 men's basketball team at Morgan State University:
| Height (inches) | Weight (pounds) |
--- |
| 69 | 180 |
| 72 | 185 |
| 74 | 170 |
| 74 | 190 |
| 74 | 220 |
| 76 | 200 |
| 77 | 190 |
| 77 | 225 |
| 78 | 215 |
| 78 | 225 |
| 80 | 210 |
| 81 | 208 |
| 81 | 220 |
| 86 | 270 |
Height and weight of the Morgan State University men's basketball players.
If you keep both variables as measurement variables and analyze using linear regression, you get a P value of 0.0007; the relationship is highly significant. Tall basketball players really are heavier, as is obvious from the graph. However, if you divide the heights into two categories, "short" (77 inches or less) and "tall" (more than 77 inches) and compare the mean weights of the two groups using a two-sample t–test, the P value is 0.043, which is barely significant at the usual P<0.05 level. And if you also divide the weights into two categories, "light" (210 pounds and less) and "heavy" (greater than 210 pounds), you get 6 who are short and light, 2 who are short and heavy, 2 who are tall and light, and 4 who are tall and heavy. The proportion of short people who are heavy is not significantly different from the proportion of tall people who are heavy, when analyzed using Fisher's exact test (P=0.28). So by categorizing both measurement variables, you have made an obvious, highly significant relationship between height and weight become completely non-significant. This is not a good thing. I think it's better for most biological experiments if you don't categorize.
Likert items
Social scientists like to use Likert items: they'll present a statement like "It's important for all biologists to learn statistics" and ask people to choose 1=Strongly Disagree, 2=Disagree, 3=Neither Agree nor Disagree, 4=Agree, or 5=Strongly Agree. Sometimes they use seven values instead of five, by adding "Very Strongly Disagree" and "Very Strongly Agree"; and sometimes people are asked to rate their strength of agreement on a 9 or 11-point scale. Similar questions may have answers such as 1=Never, 2=Rarely, 3=Sometimes, 4=Often, 5=Always.
Strictly speaking, a Likert scale is the result of adding together the scores on several Likert items. Often, however, a single Likert item is called a Likert scale.
There is a lot of controversy about how to analyze a Likert item. One option is to treat it as a nominal variable with five (or seven, or however many) items. The data would then be summarized by the proportion of people giving each answer, and analyzed using chi-square or G–tests. However, this ignores the fact that the values go in order from least agreement to most, which is pretty important information. The other options are to treat it as a ranked variable or a measurement variable.
Treating a Likert item as a measurement variable lets you summarize the data using a mean and standard deviation, and analyze the data using the familiar parametric tests such as anova and regression. One argument against treating a Likert item as a measurement variable is that the data have a small number of values that are unlikely to be normally distributed, but the statistical tests used on measurement variables are not very sensitive to deviations from normality, and simulations have shown that tests for measurement variables work well even with small numbers of values (Fagerland et al. 2011).
A bigger issue is that the answers on a Likert item are just crude subdivisions of some underlying measure of feeling, and the difference between "Strongly Disagree" and "Disagree" may not be the same size as the difference between "Disagree" and "Neither Agree nor Disagree"; in other words, the responses are not a true "interval" variable. As an analogy, imagine you asked a bunch of college students how much TV they watch in a typical week, and you give them the choices of 0=None, 1=A Little, 2=A Moderate Amount, 3=A Lot, and 4=Too Much. If the people who said "A Little" watch one or two hours a week, the people who said "A Moderate Amount" watch three to nine hours a week, and the people who said "A Lot" watch 10 to 20 hours a week, then the difference between "None" and "A Little" is a lot smaller than the difference between "A Moderate Amount" and "A Lot." That would make your 0-4 point scale not be an interval variable. If your data actually were in hours, then the difference between 0 hours and 1 hour is the same size as the difference between 19 hours and 20 hours; "hours" would be an interval variable.
Personally, I don't see how treating values of a Likert item as a measurement variable will cause any statistical problems. It is, in essence, a data transformation: applying a mathematical function to one variable to come up with a new variable. In chemistry, pH is the base-10 log of the reciprocal of the hydrogen activity, so the difference in hydrogen activity between a ph 5 and ph 6 solution is much bigger than the difference between ph 8 and ph 9. But I don't think anyone would object to treating pH as a measurement variable. Converting 25-44 on some underlying "agreeicity index" to "2" and converting 45-54 to "3" doesn't seem much different from converting hydrogen activity to pH, or micropascals of sound to decibels, or squaring a person's height to calculate body mass index.
The impression I get, from briefly glancing at the literature, is that many of the people who use Likert items in their research treat them as measurement variables, while most statisticians think this is outrageously incorrect. I think treating them as measurement variables has several advantages, but you should carefully consider the practice in your particular field; it's always better if you're speaking the same statistical language as your peers. Because there is disagreement, you should include the number of people giving each response in your publications; this will provide all the information that other researchers need to analyze your data using the technique they prefer.
All of the above applies to statistics done on a single Likert item. The usual practice is to add together a bunch of Likert items into a Likert scale; a political scientist might add the scores on Likert questions about abortion, gun control, taxes, the environment, etc. and come up with a 100-point liberal vs. conservative scale. Once a number of Likert items are added together to make a Likert scale, there seems to be less objection to treating the sum as a measurement variable; even some statisticians are okay with that.
Independent and dependent variables
Another way to classify variables is as independent or dependent variables. An independent variable (also known as a predictor, explanatory, or exposure variable) is a variable that you think may cause a change in a dependent variable (also known as an outcome or response variable). For example, if you grow isopods with 10 different mannose concentrations in their food and measure their growth rate, the mannose concentration is an independent variable and the growth rate is a dependent variable, because you think that different mannose concentrations may cause different growth rates. Any of the three variable types (measurement, nominal or ranked) can be either independent or dependent. For example, if you want to know whether sex affects body temperature in mice, sex would be an independent variable and temperature would be a dependent variable. If you wanted to know whether the incubation temperature of eggs affects sex in turtles, temperature would be the independent variable and sex would be the dependent variable.
As you'll see in the descriptions of particular statistical tests, sometimes it is important to decide which is the independent and which is the dependent variable; it will determine whether you should analyze your data with a two-sample t–test or simple logistic regression, for example. Other times you don't need to decide whether a variable is independent or dependent. For example, if you measure the nitrogen content of soil and the density of dandelion plants, you might think that nitrogen content is an independent variable and dandelion density is a dependent variable; you'd be thinking that nitrogen content might affect where dandelion plants live. But maybe dandelions use a lot of nitrogen from the soil, so it's dandelion density that should be the independent variable. Or maybe some third variable that you didn't measure, such as moisture content, affects both nitrogen content and dandelion density. For your initial experiment, which you would analyze using correlation, you wouldn't need to classify nitrogen content or dandelion density as independent or dependent. If you found an association between the two variables, you would probably want to follow up with experiments in which you manipulated nitrogen content (making it an independent variable) and observed dandelion density (making it a dependent variable), and other experiments in which you manipulated dandelion density (making it an independent variable) and observed the change in nitrogen content (making it the dependent variable).
References
Fagerland, M. W., L. Sandvik, and P. Mowinckel. 2011. Parametric methods outperformed non-parametric methods in comparisons of discrete numerical variables. BMC Medical Research Methodology 11: 44.
MacCallum, R. C., S. B. Zhang, K. J. Preacher, and D. D. Rucker. 2002. On the practice of dichotomization of quantitative variables. Psychological Methods 7: 19-40.
Zar, J.H. 1999. Biostatistical analysis. 4th edition. Prentice Hall, Upper Saddle River, NJ.
Picture of isopod from Australian Insect Common Names.
⇐ Previous topic|Next topic ⇒
Table of Contents
This page was last revised December 4, 2014. Its address is It may be cited as:
McDonald, J.H. 2014. Handbook of Biological Statistics (3rd ed.). Sparky House Publishing, Baltimore, Maryland. This web page contains the content of pages 6-13 in the printed version.
©2014 by John H. McDonald. You can probably do what you want with this content; see the permissions page for details. |
10250 | https://www.andrews.edu/~rwright/Precalculus-RLW/Text/04-04.html | 4-04 Right Triangle Trigonometry and Identities
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1: Functions and Graphs
2: Polynomial Functions
3: Exponential and Logarithmic Functions
4: Trigonometry
5: Analytic Trigonometry
6: Additional Trigonometric Topics
7: Analytic Geometry and Conic Sections
8: Systems of Equations and Inequalities
9: Matrices
10: Sequences and Series
11: Analytic Geometry in 3-D
12: Introduction to Calculus
Precalculus by Richard Wright
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4-04 Right Triangle Trigonometry and Identities
Mr. Wright teaches the lesson.
Summary: In this section, you will:
Use basic trigonometric identities.
Use right triangles to solve real world problems.
SDA NAD Content Standards (2018): PC.5.1, PC.5.3
Radio towers. (pixabay/susie)
Using right triangles and measurements of angles of elevation and depression, heights of towers and buildings can be calculated. But first, an introduction to trigonometric identities.
Trigonometric Identities
Trigonometric identities are equations that are true for every value of the variable in the domain. Identities provide ways rewrite and simplify complex trigonometric expressions and equations.
Basic Trigonometric Identities
Reciprocal Identities
sin u=1 csc u sin u=1 csc u cos u=1 sec u cos u=1 sec u tan u=1 cot u tan u=1 cot u
csc u=1 sin u csc u=1 sin u sec u=1 cos u sec u=1 cos u cot u=1 tan u cot u=1 tan u
Quotient Identities
tan u=sin u cos u tan u=sin u cos u cot u=cos u sin u cot u=cos u sin u
Pythagorean Identities
sin 2 u + cos 2 u = 1 1 + tan 2 u = sec 2 u cot 2 u + 1 = csc 2 u
Note: sin 2 u = (sin u)2.
Cofunction Identities
sin (90° − u) = cos u cos (90° − u) = sin u
tan (90° − u) = cot u cot (90° − u) = tan u
sec (90° − u) = csc u csc (90° − u) = sec u
The reciprocal identities were already introduced in previous lessons. Refer back to lessons 4-02 and 4-03. The other identities are explained below.
Quotient Identities
On the unit circle, sin θ = y and cos θ = x. Divide these equations.
sin θ cos θ=y x=tan θ sinθ cosθ=y x=tanθ
Thus,
tan θ=sin θ cos θ tanθ=sinθ cosθ
And, the reciprocal is also true.
cot θ=cos θ sin θ cotθ=cosθ sinθ
Pythagorean Identities
Think of drawing a right triangle on the unit circle so that one leg is on the x-axis, the other leg is vertical, and the hypotenuse is a radius of the circle.
Right triangle on the unit circle.
The horizontal leg is the x-coordinate of the point, but on the unit circle cos θ = x. The vertical leg is the y-coordinate of the point, but on the unit circle sin θ = y. The hypotenuse is a radius of the unit circle, so its length is 1. Apply the Pythagorean Theorem.
x 2 + y 2 = 1 2
cos 2 θ + sin 2 θ = 1
Divide this by either sin 2 θ or cos 2 θ to get the other two Pythagorean identities.
Cofunction Identities
Imagine a right triangle where one acute angle measure is u. The two acute angles in a right triangle are complementary. If the other acute angle is v, see figure 3, then
u + v = 90°
Solve for v.
v = 90° − u
Right triangle
If the sides of the triangle are a, b, and c as in figure 3, then sin u=a c sinu=a c and cos v=a c cosv=a c. Thus sin u = cos v. Since v = 90° − u,
sin u = cos (90° − u)
The other cofunction identities have similar logic.
Use Trigonometric Identities
Let θ be an acute angle and cos θ = 0.3. Find the values of a) sin θ and b) tan θ using trigonometric identities.
Solution
We know cosine and want to find sine, so pick an identity that has both:
sin 2 θ + cos 2 θ = 1
sin 2 θ + (0.3)2 = 1
sin 2 θ = 0.91
sin θ ≈ 0.9539
We know cosine and want to find tangent. From part a we also know sine, so we can use a quotient identity.
tan θ=sin θ cos θ tan θ=0.9539 0.3 tan θ≈3.1783 tan θ=sin θ cos θ tan θ=0.9539 0.3 tan θ≈3.1783
Use Trigonometric Identities
Let α be an acute angle and csc α = 2. Find a) sin α and b) cot α using trigonometric identities.
Solution
We know cosecant and want to find sine, so use an identity with both of those.
sin α=1 csc α sin α=1 2 sin α=1 csc α sin α=1 2
We know cosecant and want to find cotangent, so use an identity with both of them.
cot 2 α+1=csc 2 α cot 2 α+1=2 2 cot α=3–√cot 2 α+1=csc 2 α cot 2 α+1=2 2 cot α=3
Let β be an acute angle and tan β = ½. Find a) cot β and b) sec β.
Answers
2; 5√2 5 2
Applications with Right Triangles
Some real-world problems can be solved by drawing right triangles and finding unknown lengths. Other problems use angles of elevation and depression.
Angles of Elevation and Depression
The angle of elevation is the angle between the horizontal up to an object.
The angle of depression is the angle between the horizontal down to an object.
Angles of Elevation and Depression
Solve a Problem with a Right Triangle
A 8-foot step ladder actually is not 8-feet high. The size of a step ladder is actually the length of the rails that the steps are attached to. When the ladder is in use the rails are slanted so the height is less. If the rails of an 8-foot step ladder make an angle of 50° with the ground, how high is the top of the ladder from the ground?
Solution
Draw a right triangle using the rails of the ladder as the hypotenuse. The hypotenuse is 8 feet long and the angle with the ground is 50°.
The hypotenuse and one angle are known and the opposite leg is the unknown. The formula of sine from lesson 4-03 has those three parts. Use sine to solve the problem.
sin L=opp hyp sin 50°=h 8 8 sin 50°=h h≈6.12 sin L=opp hyp sin 50°=h 8 8 sin 50°=h h≈6.12
The height of the ladder is actually about 6.12 feet.
Find the length of side a in figure 6.
Answer
8.57
Angle of Elevation
To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of elevation of 57° between a line of sight to the top of the tree and the ground. Find the height of the tree.
Solution
The adjacent leg and angle are known. The height is the unknown. Use a trigonometric formula that has those three quantities.
tan θ=opp adj tan 57°=h 30 30 tan 57°=h h≈46.20 tan θ=opp adj tan 57°=h 30 30 tan 57°=h h≈46.20
The height of the tree is about 46 feet.
Angle of Elevation
A hiker is standing on mesa 80 feet above the desert floor. The angle of depression to a creek is 50°. Farther away, the angle of depression to a horse is 30°. How far apart are the horse and the creek?
Solution
Draw a picture to visualize the situation.
α = 30° and β = 50°
This will be a two part problem. First, since AG¯¯¯¯¯AG‾ and DF¯¯¯¯¯DF‾ are parallel, then ∠α and ∠ADF are congruent. So are ∠β and ∠AEF.
Start by finding the length of DF using ΔADF.
tan u=opp adj tan 30°=80 DF DF=80 tan 30°DF≈138.56 tan u=opp adj tan 30°=80 DF DF=80 tan 30°DF≈138.56
Now find the length of EF using ΔAEF.
tan u=opp adj tan 50°=80 EF EF=80 tan 50°EF≈67.13 tan u=opp adj tan 50°=80 EF EF=80 tan 50°EF≈67.13
Now subtract to find x.
x = DF − EF
x = 138.56 − 67.13 = 71.43
The horse is approximately 71 feet from the creek.
A student is observing a radio tower. Two sets of guy wires are attached at the same spot on the ground 50 feet away from the tower. The other ends of the guy wires are attached to the tower at different points. If the angles of elevation of the guy wires are 50° and 70°, how far apart are they attached on the tower?
Answer
77.8 ft
Lesson Sumamry
Basic Trigonometric Identities
Reciprocal Identities
sin u=1 csc u sin u=1 csc u cos u=1 sec u cos u=1 sec u tan u=1 cot u tan u=1 cot u
csc u=1 sin u csc u=1 sin u sec u=1 cos u sec u=1 cos u cot u=1 tan u cot u=1 tan u
Quotient Identities
tan u=sin u cos u tan u=sin u cos u cot u=cos u sin u cot u=cos u sin u
Pythagorean Identities
sin 2 u + cos 2 u = 1 1 + tan 2 u = sec 2 u cot 2 u + 1 = csc 2 u
Note: sin 2 u = (sin u)2.
Cofunction Identities
sin (90° − u) = cos u cos (90° − u) = sin u
tan (90° − u) = cot u cot (90° − u) = tan u
sec (90° − u) = csc u csc (90° − u) = sec u
Angles of Elevation and Depression
The angle of elevation is the angle between the horizontal up to an object.
The angle of depression is the angle between the horizontal down to an object.
Angles of Elevation and Depression
Helpful videos about this lesson.
Mr. Wright Teaches the Lesson (
Pythagorean Identities (
Practice Exercises
Explain the cofunction identity.
Let θ be an acute angle. Use the given function value with trigonometric identities to evaluate the given function.
If sin θ = 0.9, find a) cos θ and b) csc θ.
If sin θ = 0.25, find a) sin(90° − θ) and b) tan θ.
If sec θ = 1.45, find a) cos θ and b) tan θ.
If cos θ = 0.6, find a) sin θ and b) cot θ.
If csc θ = 10, find a) sin θ and b) csc (90° − θ).
Problem Solving
A 20-ft ladder leans against a building so that the angle between the ladder and the ground is 75°. How high up the building does the ladder reach?
A 30-ft ladder leans against a building so that the angle between the ladder and the ground is 70°. How far from the building is the base of the ladder?
The angle of elevation to the top of the Willis Tower is 33.2° when you are a half-mile from the base of the tower. How high is the tower?
If the Empire State Building is 1250 ft high and the angle of the elevation to the top is 52°, how far from the building are you?
A group of civil engineers wants to build a bridge over a canyon, but they do not know how wide the canyon is. They raise different tall objects up beside the canyon until one of them casts a shadow to the other side of the canyon. The height of the object is 80 ft and they estimate the angle of elevation of the sun is 35°. Roughly, how wide is the canyon? (Ben P)
A tall pine tree grows vertically. If Sam is 50 feet from the tree and measures the angle of elevation as 80°, how tall is the tree?
A large advertising banner hangs on the side of a building. Duane works in a neighboring building 75 feet away and measures to angle of elevation to the top of the banner as 50° and the angle of depression to the bottom as 20°. How long is the banner?
Marie is standing on a platform waiting to ride a roller coaster. She measures the angle of depression to the bottom of the long hill as 13° and the angle of elevation to the top of the hill as 52°. If she is 110 feet away, how high is the hill?
A steeple is on top of a church. Marco stands 52 ft from the church and measures the angle of elevation to the base of the steeple as 44°. He measures the angle of elevation to the top of the steeple as 56°. How tall is the steeple?
Philip is standing on Inspiration Point in Arcadia Scenic Turnout 800 feet above Lake Michigan. He can see two ship, one behind the other. If the angle of depression to the closer ship is 18° and the farther ship is 15°, how far apart are the ships?
Mixed Review
(4-03) Use a special right triangle to evaluate a) tan 30° and b) sec π 4 sec π 4.
(4-03) Evaluate the six trigonometric functions for the given angle.
22. (4-02) Evaluate the six trigonometric functions for 4 π 3 4 π 3 using the unit circle.
23. (4-01) A car with a 30-inch diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?
24. (3-02) What is the intensity of a loud stereo blaring music at 95 dB?
Answers
For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement.
0.4359; 10⁄9
0.9682; 0.2582
0.6897; 1.05
0.8; 0.75
0.1; 1.0050
19.3 ft
10.3 ft
1728 ft
977 ft
114 ft
284 ft
116.7 ft
166.2 ft
26.9 ft
523 ft
3√3 3 3; 2–√2
sin θ=3 34√34,cos θ=5 34√34,tan θ=3 5,csc θ=34√3,sec θ=34√5,cot θ=5 3 sin θ=3 34 34,cos θ=5 34 34,tan θ=3 5,csc θ=34 3,sec θ=34 5,cot θ=5 3
sin 4 π 3=−3√2,cos 4 π 3=−1 2,tan 4 π 3=3–√,csc 4 π 3=−2 3√3,sec 4 π 3=−2,cot 4 π 3=3√3 sin 4 π 3=-3 2,cos 4 π 3=-1 2,tan 4 π 3=3,csc 4 π 3=-2 3 3,sec 4 π 3=-2,cot 4 π 3=3 3
3520 rad/min; 560.2 rev/min
0.00316 W/m 2
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10251 | https://math.stackexchange.com/questions/1685840/if-you-colored-every-point-of-a-circle-1-of-2-colors-is-there-always-2-same-col | combinatorics - If you colored every point of a circle 1 of 2 colors, is there always 2 same-colored points of distance $R$ apart? - Mathematics Stack Exchange
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If you colored every point of a circle 1 of 2 colors, is there always 2 same-colored points of distance R R apart?
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If every point on a circle of radius R R in R 2 R 2 were colored one of two colors, is there necessarily two points that are of the same color and of distance R R apart? what about >2>2 colors?
On the 2 2-color case, I'm thinking that if we have two points of different colors on the circle, say red and blue, then consider the chord of length R R from the blue point to another point on the circle. This new point must then be red. Continuing, we get a hexagon if I'm not mistaken, which doesn't help. But we can place another hexagon inscribed in the circle with the same colorings, ad infinitum. Intuitively, this leaves me to believe that there must be two points of the same color of distance R R apart on the circle due to some induction or something. I'm not sure! Does this hold for more than two colors?
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asked Mar 6, 2016 at 18:24
FreddieFreddie
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Given the point (cos ϕ,sin ϕ)(cosϕ,sinϕ), colour it red if ϕ∈[0,π 6[∪[π 3,π 2[∪[2 π 3,5 π 6[ϕ∈[0,π 6[∪[π 3,π 2[∪[2 π 3,5 π 6[ and blue otherwise. This is just your hexagon repeated over an interval.
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answered Mar 6, 2016 at 18:28
Hagen von EitzenHagen von Eitzen
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The answer to the question in the title is No. The circle is the disjoint union of six-element sets consisting of regular hexagon vertices. You can color the vertices of any such hexagon alternatively red and blue. Any two vertices of different hexagons cannot have distance R R; therefore the chosen colorings do not interfer.
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answered Mar 6, 2016 at 19:17
Christian BlatterChristian Blatter
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10252 | https://www.freemathhelp.com/forum/threads/simplifying-trig-expression-using-identities.70628/ | Simplifying trig expression using Identities | Free Math Help Forum
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Geometry and Trig
Simplifying trig expression using Identities
Thread starterLauren_avery
Start dateApr 24, 2011
L
Lauren_avery
New member
Joined Apr 24, 2011 Messages 1
Apr 24, 2011
#1
Hi there, I've been trying to work this problem for awhile now, and I'm getting close I think to answering it, I had a friend sort of help with the work, but had to leave partway through.
Simplify the given expression:
cos(x + y)cosY + Sin(x + y)SinY I've used the product identity cosXsinY = 1/2[cos(x+y) - cos(X-Y)]
This is what I have so far, using some substitution: U= x+y
(1/2)[cos(u-y) + cos(u+y)] + (1/2)[cos(u-y) + cos(u+y)]
My friend skippd the step here since he was in a rush told me to just resub x+y from u and go from there. Then he gave me:
(1/2)[cosX + cos(X+2Y)] + (1/2)[cosX - cos(X+2Y)]
I know the answer is supposed to come out to cosX. But either I'm doing the x+y from u substitution wrong, or I did the wrong Identity.
Thanks for the help!
S
soroban
Elite Member
Joined Jan 28, 2005 Messages 5,586
Apr 24, 2011
#2
Hello, Lauren_avery!
Simplify:cos(x+y)cosy+sin(x+y)siny\displaystyle \text{Simplify: }\:\cos(x + y)\cos y + \sin(x + y)\sin y Simplify:cos(x+y)cos y+sin(x+y)sin y
I’ve used the product identity:cosx siny=1 2[cos(x+y)−cos(x−y)]\displaystyle \text{I've used the product identity: }\:\cos x\sin y \:=\: \tfrac{1}{2}\left[\cos(x+y) - \cos(x-y)\right]I’ve used the product identity:cos x sin y=2 1[cos(x+y)−cos(x−y)] . . . . no
Click to expand...
You have:cosu cosy+sinu siny\displaystyle \text{You have: }\:\cos u\cos y + \sin u\sin y You have:cos u cos y+sin u sin y
Doesn’t that suggest:cos(A−B)=cosA cosB+sinA sinB?\displaystyle \text{Doesn't that suggest: }\:\cos(A - B) \:=\:\cos A\cos B + \sin A\sin B\,?Doesn’t that suggest:cos(A−B)=cos A cos B+sin A sin B?
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10253 | https://www.bytelearn.com/math-grade-6/lesson-plan/writing-ratios | Sign inSign up
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Grade 6Ratios And Rates
Intro To Ratios Lesson Plan
Lesson plan
Intro to Ratios Lesson Plan
Overview
In this lesson, we’ll introduce the concept of writing ratios from simple images, verbal descriptions, tables or tape diagrams to 6th graders. You can expect this lesson to take one 45-minute class period.
Grade 6
Ratios And Rates
6.RP.A.1
Step-by-step help
ByteLearn gives students targeted feedback and hints based on their specific mistakes
Preview step-by-step-help
Objective
Students will be able to understand the concept of ratios and use ratio language to describe the relationship between two quantities.
Materials
Teacher Slideshow
Online Practice
How to Teach Ratios
Notice and Wonder Warm-up
Students love notices and wonders. This activity will give students the opportunity to notice parts and wholes of ratios without even knowing it! Display slide 1 of the teacher slideshow. Ask students to jot down a few noticings and a few wonderings before sharing with their neighbor.
Copy these Google Slides for free
Here are some things you might hear:
Notices:
There are 4 suns.
There are 3 lightning bolts.
There are 5 moons.
There are more suns than lightning bolts.
There are more moons than suns and lightning bolts.
There are 12 total shapes.
Wonders:
Why are there more moons than there are suns and lightning bolts?
What is the comparison between each figure?
Warm-up discussion
The warm-up is a good intro into this lesson as it allows for students to start using ratio language by identifying how many of each shape there are and understanding that there is a total as well. Share with students that we have parts and a whole in this example. Explain to students that they are going to use this concept to describe ratios.
What is a ratio?
On slide 2, discuss with the class the definition of a ratio, as well as different representations of ratios, such as using the word “to”, using a colon, or writing as a fraction. Give students some time to write down the definition as well as the three forms of representation we’ll talk about today.
Writing a part to part ratio from images
This first example contains the same images from the warm-up. Since students are already a bit familiar, allow them some time to try writing the ratio on their own. If students need help getting started, ask, “How many suns are there? How many moons are there?”. They can consult with a partner to check their answers. You’ll want to make sure students write all three forms of the ratio. While discussing students’ answers, be sure to verbally explain the ratio, like the sentence at the bottom of the slide. “For every 4 suns, there are 5 moons.” This helps students to understand what the ratio representations really mean.
Congratulate students for writing their first set of ratios! Explain to students that this type of ratio is a “part to part” ratio because we’re comparing a part of the total to another part of the total.
Writing a part to whole ratio from images
For the next set of ratios, we’ll use the same images, but this time have students write the ratio of the number of moons to the total number of shapes. Again, since students are familiar with the numbers for this specific example, they should be able to write the ratio in the 3 different forms. Explain to students that this type of ratio is a “part to whole” ratio because we’re comparing a part of the total to the total.
After you’ve gone over the 3 forms of the ratio, ask if anyone can explain verbally what these ratios mean. Hopefully, students will explain that for every 5 moons, there are 12 total shapes. Explain to students that we will write ratios based on images (like these first examples), tables, and tape diagrams.
Write a ratio from a table
Some students will find writing ratios even easier from a table than from images, because they don’t have to count! Allow students to look at the table to find the number of students who play soccer and the number of students who play football. Once they find the correct values, they can write their ratio in the 3 forms.
Extend the example
Now ask students, “If I wanted to know the ratio of students who play football to students who play soccer, would anything change?” The idea behind this is to get students to recognize that the order of the ratio matters. Be sure to continuously reinforce this concept for students.
Additional questions to ask
What is the ratio of students who play baseball to the total number of students who play fall sports?
For students who get stuck with this one, remind them that they will need to add all of the numbers to get the total number of students who play a sport.
What is the ratio of students who play baseball to students who play football or soccer?
The key idea here is for students to recognize that they need to add to find the second amount, but they are not adding all the numbers, just the number of students who play football and the number of students who play soccer.
Write a ratio from a tape diagram
This last example shows a tape diagram. We’re asking students to write the ratio of cats to dogs. Remind students that the boxes in tape diagrams are all the same size, so they represent the same amount. We can count the number of boxes for cats, 7, and the number of boxes for dogs, 10. Once we’ve counted the necessary values, students should be really familiar with writing the ratios all 3 ways at this point.
Intro to Ratios Practice
After you’ve completed the examples with the whole class, it’s time for some independent practice! ByteLearn gives you access to tons of mild, medium, and spicy problems for introduction to ratios practice. Check out the online practice and assign to your students for classwork and/or homework!
Intro to Ratios Practice
Problem 1 of 7
The tape diagram below represents the number of hockey sticks and soccer balls that Coach Myers has ordered for the school's sports inventory
What is the ratio of hockey sticks to soccer balls?
What is the ratio of soccer balls to hockey sticks?
View this practice
Related Lessons
Converting Measurements Lesson Plan
Ratio Table Strategies Lesson Plan
Unit Rates Lesson Plan |
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Table of Contents
Table of Contents
What Is the Concentration Ratio?
How It Works
Formula and Interpretation
Example Calculation
Herfindahl-Herschman Index
FAQs
The Bottom Line
Concentration Ratio: Definition, Formula, and How to Calculate
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Definition
The concentration ratio is a metric that measures the market share of the largest firms in an industry.
What Is the Concentration Ratio?
The concentration ratio, in economics, is a ratio that indicates the size of firms in relation to their industry as a whole. Low concentration ratio in an industry would indicate greater competition among the firms in that industry, compared to one with a ratio nearing 100%, which would be evident in an industry characterized by a true monopoly.
Key Takeaways
The concentration ratio compares the size of firms in relation to their industry as a whole.
Low concentration ratio indicates greater competition in an industry, compared to one with a ratio nearing 100%, which would be a monopoly.
An oligopoly is apparent when the top five firms in the market account for more than 60% of total market sales, according to the concentration ratio.
Understanding the Concentration Ratio
The concentration ratio indicates whether an industry is comprised of a few large firms or many small firms. The four-firm concentration ratio, which consists of the market share of the four largest firms in an industry, expressed as a percentage, is a commonly used concentration ratio. Similar to the four-firm concentration ratio, the eight-firm concentration ratio is calculated for the market share of the eight largest firms in an industry. The three-firm and five-firm are two more concentration ratios that can be used.
Concentration Ratio Formula and Interpretation
The concentration ratio is calculated as the sum of the market share percentage held by the largest specified number of firms in an industry. The concentration ratio ranges from 0% to 100%, and an industry's concentration ratio indicates the degree of competition in the industry. A concentration ratio that ranges from 0% to 50% may indicate that the industry is perfectly competitive and is considered a low concentration.
A rule of thumb is that an oligopoly exists when the top five firms in the market account for more than 60% of total market sales. If the concentration ratio of one company is equal to 100%, this indicates that the industry is a monopoly.
Important
If the concentration ratio for the top five firms is over 60%, the market is considered an oligopoly. If the concentration ratio for one company is nearly 100%, it is considered a monopoly.
Example Calculation
Assume that ABC Inc., XYZ Corp., GHI Inc., and JKL Corp. are the four largest companies in the biotechnology industry, and an economist aims to calculate the degree of competition. For the most recent fiscal year, ABC Inc., XYZ Corp., GHI Inc., and JKL Corp. have market shares of 10%, 15%, 26%, and 33%, respectively.
Consequently, the biotech industry's four-firm concentration ratio is 84%. Therefore, the ratio indicates that the biotech industry is an oligopoly. The same could be calculated for more or less than four of the top companies in the industry. The concentration ratio only indicates the competitiveness of the industry and whether an industry follows an oligopolistic market structure.
Herfindahl-Herschman Index
The Herfindahl-Herschman Index (HHI) is an alternative indicator of firm size, calculated by squaring the percentage share (stated as a whole number) of each firm in an industry, then summing these squared market shares to derive an HHI. The HHI has a fair amount of correlation to the concentration ratio and can be a better measure of market concentration.
Which Industry Has the Highest Concentration Ratio?
The most concentrated industries are secondary market financing and other depository credit intermediation, according to Census data from 2017 (latest data available) analyzed by Statista. In those industries, the top four firms had 100% of the market share. Deep sea passenger transportation, home centers, guided missile and space propulsion, and pipeline transportation were also extremely concentrated, with the top four firms controlling over 95% of market share.
Does the U.S. Have Any Monopolies?
The U.S. has some legal monopolies, such as regional utilities and the electromagnetic bandwidth for radio and television broadcasts. Since it is impossible to have two competing companies offering the same service in the same region, these services operate as regulated monopolies, with their operations and prices closely monitored by government bodies.
Is a High Concentration Ratio Bad?
High levels of industry concentration can be bad if it results in worse services or prices for consumers. An increase in the concentration ratio generally means that there is less competition and consumers have fewer choices for that service. However, a high concentration ratio can sometimes be beneficial, if larger companies can take advantage of economies of scale to reduce costs.
The Bottom Line
A concentration ratio measures the dominance of large firms in an industry. A low concentration ratio indicates a highly competitive market, where no single player has a significant advantage. A high concentration ratio may indicate an oligopoly, where large firms can leverage their size to restrict competition.
Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
Reed N. Olsen, via Missouri State University. "ECO 165 Principles of Microeconomics: Market Structure: Oligopoly (Imperfect Competition)," Page 5.
Federal Reserve Bank of St. Louis. "What Makes a Market an Oligopoly?"
Statista. "Highest Four Firm Concentration Ratio (CR4) by Industry in the United States in 2017."
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10255 | http://olivernash.org/2018/07/08/poring-over-poncelet/assets/pdfs/DelCentina-2014a.pdf | Arch. Hist. Exact Sci.
DOI 10.1007/s00407-015-0163-y Poncelet’s porism: a long story of renewed discoveries, I Andrea Del Centina1 Received: 26 September 2014 © Springer-Verlag Berlin Heidelberg 2015 Abstract In 1813, J.-V. Poncelet discovered that if there exists a polygon of n-sides, which is inscribed in a given conic and circumscribed about another conic, then infi-nitely many such polygons exist. This theorem became known as Poncelet’s porism, and the related polygons were called Poncelet’s polygons. In this article, we trace the history of the research about the existence of such polygons, from the “prehistorical” work of W. Chapple, of the middle of the eighteenth century, to the modern approach of P. Griffiths in the late 1970s, and beyond. For reasons of space, the article has been divided into two parts, the second of which will appear in the next issue of this journal.
Contents General introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Introduction to part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 The prehistory: from Chapple to Steiner . . . . . . . . . . . . . . . . . . . . . .
2 The theorems and methods of Poncelet . . . . . . . . . . . . . . . . . . . . . .
3 Jacobi and the use of the elliptic functions . . . . . . . . . . . . . . . . . . . .
4 Trudi: the forgotten work . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Cayley’s explicit conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 An algebraic approach through invariants . . . . . . . . . . . . . . . . . . . . .
7 Other contributions from 1850 to 1875 . . . . . . . . . . . . . . . . . . . . . .
To all my teachers, friends and colleagues.
Communicated by: Jeremy Gray.
B Andrea Del Centina a.delcentina@unife.it 1 Dipartimento di Matematica, Università di Ferrara, Via Machiavelli 45, 44100 Ferrara, Italy 123 A. Del Centina 8 (2, 2)-Correspondences and closure problems . . . . . . . . . . . . . . . . . . .
9 The theorems of Darboux . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 Poncelet polygons in Halphen’s treatise . . . . . . . . . . . . . . . . . . . . . .
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
General introduction In 1822, Jean-Victor Poncelet published the Traité sur les propriétés projectives des figures. In this fundamental work, he gave a synthetic geometric proof of the following theorem, which became known as Poncelet’s closure theorem,1 or Poncelet’s porism2: let two smooth (real) conics be given in the plane, if there exists a polygon of n sides, which is inscribed in one conic and circumscribed about the other, then there are infinitely many such polygons, and every point of the first conic is vertex of one of them (Poncelet 1822, sections 565–567) (Fig. 1a, b).
This theorem, which undoubtedly is one of the most important and beautiful theo-rems of projective geometry, was discovered by Poncelet in 1813 during his captivity in Russia as a prisoner of war.
Any polygon inscribed in C and circumscribed about D is called Poncelet’s polygon (related to C and D), and sometimes inter-scribed polygon, or even in-and-circumscribed polygon (to C and D).
Poncelet proved the theorem as a corollary of a more general one, to which we refer as Poncelet’s general theorem and that can be stated as follows: let C, D1, D2, . . . , Dn be smooth conics from a pencil, if there exists a polygon of n sides whose vertices lie on C, and each side is tangent to one of the others D1, D2, . . . , Dn, then infinitely many such polygons exist.
The proofs that Poncelet gave of his theorems were heavily based on the unproved “principle of continuity,” and for this he was criticized by some contemporaries, and especially by Cauchy.
Six years after the publication of the Traité, Jacobi recognized the existence of a relation between these theorems and the elliptic function “amplitude.” In his elegant paper (Jacobi 1828), he gave a new proof of the theorem, in the “particular” case of two circles lying within each other, by applying some recursion formulae that arise in the iterated addition of a constant to an elliptic integral of the first kind.
Numerous mathematicians have been inspired to further studies in order to extend Jacobi’smethodtoconicsandtofindtheconditionsontwoconicsC and D allowingthe existence of a polygon, of a given number of sides, inter-scribed to them. At the same time, since the problem was originally an algebraic one (indeed, conics are algebraic curves, and the conditions of intersection and tangency are algebraic conditions too), some geometers considered Jacobi’s transcendental solution somehow unsatisfactory and looked for a purely algebraic–geometric approach to the problem.
1 From the German word Schliessungstheorem see Hurwitz (1879), also Dilgeldey (1903, p. 46).
2 From the Greek word πoρiσμoν: a proposition affirming the possibility of finding the conditions under which a certain problem becomes indeterminate or capable of infinite solutions. See for instance (Cayley 1853b).
123 Poncelet’s porism, I P P P1 P2 P3 P4 P5 P6 (b) (a) Fig. 1 Poncelet’s closure theorem, a for n = 5, and b for n = 7 These themes, especially until the last decade of the nineteenth century, produced a huge literature concerning variants, alternative proofs and generalizations of the Poncelet theorem. The major works of this period are due to Nicola Trudi, Arthur Cayley, George Salmon, Adolf Hurwitz, Gaston Darboux and George H. Halphen.
Around 1920, two new important contributions appeared. In 1919, Fran-cesco Ger-baldi applied what he called Halphen’s continued fractions, in order to get the bidegree of the covariant whose vanishing guarantees the existence of an inter-scribed n-gon (Gerbaldi 1919), and 2years later, Henry Lebesgue gave an elegant geometrical proof of the general Poncelet theorem and a re-interpretation of Cayley’s result (Lebesgue 1921).
After the 1920s, Poncelet’s theorem seemed less appealing, and it almost fell into oblivion. Only a few isolated papers on the subject were published in the following 50years, and, with the exception of Todd (1948), they were either not very relevant or considered Poncelet’s problem only for circles. It was only in the late 1970s that Phillip Griffiths and Joseph Harris, with their papers (1977, 1978a), renewed the interest of mathematicians (and physicists) in Poncelet polygons and related questions.3 Then in the nineties, W. Barth and J. Michel presented in a modern algebraic–geometric setting some result of Halphen and Gerbaldi (Barth and Michel 1993).
Over the years, several historical contributions on the subject have been published.
We mention here only those we believe to be of historical relevance—others will be quoted in the text. (Loria 1889a), a detailed report on the papers that were published on this subject after (Jacobi 1828) until the last decade of the nineteenth century; (Bos et al. 1987), which extensively discusses the works of Poncelet, Jacobi and Griffiths– Harris; (Flatto 2009), the first monograph devoted to Poncelet’s theorem; (Dragovi´ c 2011), which presents, in a modern setting, some of the major works on the subject, and explains the mechanical significance of the Poncelet closure theorem.4 3 The trajectory of a free particle [called billiard, see Birkhoff (1927)], which moves along a straight line inside an ellipse and reflects at the boundary according to the law for a light ray, is tangent to a confocal conic (Sinai 1976).
4 The book offers an excellent insight into the applications of Poncelet’s closure theorem and its gener-alizations, to the theory of integrable system, billiard dynamics, PDEs and statistical mechanics, with an extended bibliography. Moreover, chapter four is a very good summary of such geometrical topics as pencils of conics, polarity, invariants of pairs of conics, duality, etc.
123 A. Del Centina Leibniz, in the first page of Historia et origo calculi differentialis, wrote: Utilissimum est cognosci veras inventionum memorabilium origines, presertim earum non casu, sed meditandi innotuere. Id enim non eo tantum prodest, ut historia literaria suum cuique tribuat et alii ad pares laudes invitentur, sed etiam ut augeatur ars inveniendi, cognita methodo illustribus exemplis [It is most valuable to know the true origins of memorable inventions, particularly of those revealed not by chance, but through the force of reasoning. Its use is not just that History may give everyone his due and that others may look forward to similar praise, but also that the art of discovery be promoted and its method known through illustrious examples.5] (Leibniz 1846).
This work aims at giving a thorough historical account of these studies, spanning two centuries before the true nature and all facets of the problem were unveiled, on the occasion of the bicentennial jubilee of Poncelet’s famous theorem.6 For reasons of space, the article, which has sixteen sections, has been divided into two parts. Part I, sections 1–10, deals with the history of the research developed on the subject until the end of the nineteenth century; Part II, sections 1–6, takes into account those developed in the twentieth century. For the convenience of the reader, each part has been equipped with its own list of references.
Part II will be published in the next issue of this journal.
Introduction to part I The first section of Part I is devoted to the “prehistory,” i.e., to the studies relating to the existence of triangles, and other polygons of a small number of sides, inter-scribed to two circles, carried out from the middle of the seventeenth century until about 1822.
We count contributions by William Chapple, John Landen, Leonhard Euler, Nicolas Fuss, Simon A.J. Lhuilier and Jakob Steiner.
In the second section, we present the theorems of Poncelet and the proofs he gave.
His methods were those proper of the school of Monge, but that Poncelet perfected, by formalizing the method of central projection and introducing some new concepts and tools, such as the controversial “principle of continuity.” This, roughly speaking, can be stated as follows: A projective property that holds for a particular position of the figures involved holds true for any position of the figures, even if some of them disappear becoming imaginary. This principle allowed him to prove the theorems for circles and then pass to conics by a central projection.
Section three is devoted to illustrating the proof of the closure theorem for two circles, one lying inside the other, that Jacobi gave by means of the elliptic function “amplitude” (Jacobi 1828), and gives a glimpse of the “elliptic nature” of the problem.
Jacobi ended his paper by saying that it would be of great interest to make similar considerations directly for a system of two conics, so avoiding the use of the principle of continuity, but he never returned to this subject.
5 For the last sentence we have adopted the translation due to André Weil (Weil 1980, p. 226).
6 In the same occasion (Dragovi´ c and Radnovi´ c 2014) offers the current state of the art on billiard dynamics.
123 Poncelet’s porism, I His program was carried out by Trudi and, independently, by Cayley 25years later.
The work of Trudi is presented in the fourth section. In his main memoir (1853), he used the algebraicity of the complete integral of Euler’s differential equation, and the addition theorem for elliptic integrals of the first kind, to prove the closure theorem.
Through his method, the role that symmetric (2, 2)-correspondences will play in clo-sure problems can be seen to emerge. Unfortunately, Trudi’s paper remained almost unknown outside the Kingdom of Naples.
Section five is mainly devoted to the explicit conditions that allow the existence of an inter-scribed n-gon to two conics, first found by Cayley in (1853b), and further detailed in (1861), by using Abel’s addition theorem. His method revealed, in the form of the “Cayley cubic” y2 −□(x) = 0 (□(x) is the discriminant of the pencil generated by the two conics), the existence of an elliptic curve closely connected with the problem.
Section six is dedicated to Salmon’s algebraic approach, developed in his work (1857), that uses the projective invariants of a pair of conics.
In the third quarter of the nineteenth century, many papers were published concern-ing new proofs of the Poncelet theorem and its generalizations, to curves of higher degree or to quadrics surfaces in space. We illustrate this literature in section seven, remarking that some of these works, such as (Weyr 1870) and (Darboux 1870a,b, 1873a,b), strongly inspired mathematicians and physicists many decades later.
Around the 1870, the deep connection between Poncelet’s closure theorem and symmetric (2, 2)-correspondences emerged clearly. This aspect of the story, which reached its height in Hurwitz (1879), is discussed in section eight.
In section nine, we present, in some detail, the work of Darboux, who devoted a large part of his mathematical studies to questions connected with Poncelet polygons.
The new system of plane coordinates that Darboux introduced, now called “Darboux coordinates,” and the symmetric (2, 2)-correspondences were the main tools that he used for developing the theory.
In1888,thesecondvolumeofHalphen’streatiseonellipticfunctionswaspublished.
In it, he applied the theory developed in the first volume in terms of the Weierstrass ℘and σ functions, to several questions of geometry, mechanics and geodesy. Chapter ten of the book was expressly devoted to the Poncelet polygons, but other results on the same subject were inserted in chapters nine and fourteen. In section ten of the present article, we illustrate the content of these three chapters and discuss, at some length, what he called the “elliptic representations of point of the plane,” and his use of the development in continued fractions of √ X, X a polynomial of degree 3 or 4, to provide a new proof of Poncelet’s theorem.
1 The prehistory: from Chapple to Steiner Properties of triangles, which are inscribed in, or circumscribed about, a given circle, have been known since the Hellenistic period. For instance, it was known how to express the area of the inscribed triangle in terms of its sides and the radius and that among all the inscribed triangles the equilateral has maximal area. It was also known that given two concentric circles, of radii r, R with r < R, a triangle inscribed in the larger and circumscribed about the smaller exists, only if r = R/2, and that, in this 123 A. Del Centina case, all the (infinitely many) triangles that can be inter-scribed to the two circles are equilateral. Similar results were also known for other regular polygons.
In this section, we present some results on triangles, and other polygons, inscribed in one circle and circumscribed about another eccentric to the first, that were discovered from about the middle of the eighteenth century until few years after the publication of Poncelet’s treatise.
1.1 Chapple It seems Chapple was the first to study the problem of the existence of inter-scribed polygons, specifically triangles, to two non-concentric circles.7 His essay (1746) starts as follows: The following enquiry into the properties of triangles inscrib’d in, and circum-scrib’d about given circles, has let me to the discovery of some things relating to them, which I presume have not been hitherto taken notice of, having not met with them in any author; though an ingenious correspondent of mine, in the isle of Scilly, to whom I communicated some of the propositions herein after demonstrated, informs me that he had begun to consider it some years ago, but did not go thro’ with it; however I must acknowledge that a query of his to me, relating thereto, gave me the first hint, and induc’d to pursue the subject with more attention, than perhaps otherwise I should have done.
In his paper, which probably remained unknown to professional mathematicians, Chapple stated that if there exists a triangle which is inscribed in one circle C of radius R and is circumscribed about another circle c of radius r (lying inside the first), then the distance a between the centers of the circles must satisfy the equation a2 = R2 −2r R.
(1.1) This formula is sometimes called “Chapple’s formula.” J.S. Mackay called attention on Chapple’s essay only in 1887, but in his historical note (Mackay 1887) he gave just a partial transcription of it without commenting.8 To illustrate the work of Chapple, whose arguments are often confused and whose logic is very poor, even for the standard of his time, is not easy especially when trying to keep as faithful as possible to his thought.
7 From the obituary published in the Exeter Flying Post we learn that William Chapple (1718–1781), who served for 40years as Secretary of the Devon and Exeter Hospital, was an enthusiastic amateur of mathemat-ics who studied John Ward’s, The Young Mathematician’s Guide:/Being a Plain and Easy/Introduction/to the/Mathematicks/in five Parts, whose fourth edition appeared in London in 1724. Chapple was capable of using fluxions and contributed several articles to the English periodicals the Gentlemen’s Magazine, Miscellanea Curiosa Mathematica and Ladies’ Diary.
8 Chapple was quoted in Chapple (1901, pp. 552–553) and in Dingeldey (1903, p. 47) but not in Kötter (1901), whose sections 8–11 of chap. XVIII were devoted to the history of the Poncelet closure theorem. Let us remark that Chapple’s paper also escaped to Gino Loria (see Loria 1889a,b, 1896). To our knowledge, the work of Chapple has been discussed in some depth only in Bos et al. (1987).
123 Poncelet’s porism, I Chapple considered two (real) circles one lying inside the other. He first noticed (see his propositions I and II) that, if there exists an inter-scribed triangle having area A and whose sides have length x, y, z, one must have A = r(x + y + z) 2 = xyz 4R , (1.2) from which he deduced 2r R = xyz x + y + z · (1.3) Chapple recalled (see his proposition III) that if the two given circles are concentric, then an inter-scribed triangle exists only if 2r = R, and, in this case, it is always equilateral because—he observed—the circumcenter and incenter must coincide. He also remarked that the equilateral triangles have largest area among those inscribed in the exterior circle. Moreover, Chapple asserted (see his proposition IV, but the proof is very muddled) that, if the two circles are not concentric, then they may admit an inter-scribed triangle only if 2r ≤R.
Next Chapple stated two propositions: (V) (If there exists an inter-scribed triangle, then) “An infinite number of trian-gles may be drawn, which shall inscribe and circumscribe the same two circles, provided their diameters, with respect to each other, be limited, as in the two last propositions,” (VI) “The nearest distance of the peripheries of the two given circles, or, which amount the same, the distance of their centers, in order to render it possible to inscribe and circumscribe triangles, is fixed, and will be always the same.” To prove (V), Chapple argued as follows. From (1.3), it is plain that, if x, y and z are required to be found from the given r, R, the question is capable of innumerable solu-tions. In fact, he remarked one side, at least, of the inter-scribed triangle can be chosen at pleasure, provided that it does not exceed the longest segment that can be drawn within the larger circle and tangent to the smaller circle inside. Hence, he continued, if the mutual position of the two circles is fixed so that a triangle may be inter-scribed to them, innumerable triangles may be inscribed and circumscribed to the same two circles (1746, p. 120). Let us notice that Chapple missed one more relation which the data R,r, x, y, z must satisfy when an inter-scribed triangle exists, the one derived from Heron’s formula A2 = p(p −x)(p −y)(p −z), being p the half perimeter of the triangle.
To show (VI), Chapple proceeded in a very complicated way, but his reasoning was essentially as follows. He considered the two circles positioned so that an inter-scribed triangle exists and supposed that a is not fixed, i.e., that the inner circle can move freely while still allowing an inter-scribed triangle. He called AB the chord of the outside circle parallel to the direction of the motion and tangent to the inner circle, drew an inter-scribed triangle with this chord as one of its sides (so Chapple assumed the closure theorem, and accordingly his argument was circular), and denoted by E the third vertex of this triangle (Fig. 2a). He observed that moving the inner circle, from the initial position along the fixed direction AB, the vertex E also moves and leaves the outside circle, because the altitude of E decreases moving from the periphery of 123 A. Del Centina A B E E A B C o O G F R r (b) (a) Fig. 2 a Let AE B be an inter-scribed triangle to two circles one inside the other. Chapple observed that moving the inner circle along the chord AB the vertex E leaves the outside circle. b Chapple’s use of isosceles triangles to prove formula (1.1) the chord toward the middle. So, concluded Chapple, the inner circle cannot move while still allowing an inter-scribed triangle, unless a remains the same.
Chapple used propositions (V) and (VI) (i.e., his unproved closure theorem) to derive the formula (1.1). For, he first stated the following proposition (VII) “Of the innumerable triangles that may be inscrib’d and circumscrib’d in and about two given (eccentric) circles, to must of course isoscelar, the vertexes of which will be in the common diameter of those circles, which will cut their bases at right angles; now the content of that isoscelar triangles which hath the least base, and the greatest altitude, will be the greatest, and that of the other the least of all the triangles that can be inscribed and circumscrib’d in the given circles.” He proved the proposition through a number of geometrical lemmas that we will not consider here. At the end of his proof, he observed:“the proposition is every way demonstrated; and tho’ the method herein taken seems a little tedious and intricate, it is perhaps more concise and less troublesome than any which Fluxions would have afforded us.” Finally (see pp. 123–124), by using the existence of the isosceles triangles he proved formula (1.1). Chapple argued as follows. He put (see Fig. 2b) AF = x, FC = y, hence GF = 2R −x and y = √2R −x; then, since sin ̸ C AF = r x−r = y √ x2+y2 , with a simple computation he found x = R + r + √ R2 −2Rr, and this, being a = x −R −r, is equivalent to formula (1.1).
There is no doubt that, despite the many failures in his proofs and logic, Chapple grasped some fundamental aspects of the problem.
In order to draw attention to his paper, Chapple proposed to prove formula (1.1) as “prize question” in the Ladies’ Diary for the year 1746.
Robert Heath, editor of the journal, answered somewhat unsatisfactorily the year following (see The mathematical Questions 1817, pp. 393–395). In his answer, Heath 123 Poncelet’s porism, I d n m o F E D p C A B Δ δ Γ H G D H A B F E d p o n m G C Δ δ Γ (b) (a) Fig. 3 The first case of Landen’s question: the circle δ falls within ABC. a The circle δ is inside the circle , and b the circle δ in outside the circle also mentioned the solution he had received from J. Landen, a non-professional mathe-matician, that a few years later would be tackling the problem from a new point of view.
1.2 Landen Landen9 devoted Part I of his Mathematical Lucubrations to the following question (Landen 1755, p. 1)10: The two circles ABH and mno whose centers are D and d, respectively, being given in magnitude and position; let any given chord AB in the circle ABH touch the circle mno at o; and, from the extremities of that chord, let two other tangents be drawn to the circle mno, touching it at m and n, and intersecting each other at C: It is proposed to find the radius AE of the circle ABG circumscribing the triangle ABC (see Figs. 3a, b, 4a, b).
In the following, it is convenient to denote the circles mno, ABH and ABG, respec-tively, δ, and .
9 Although considered a mathematician of high rank, John Landen (1719–1790) was never a professional one. He began contributing to the Miscellanea Curiosa and to the mathematical problem section of the Ladies’Diaryfrom1744.In1754,hepublishedthefirstofhiseightpapersinthe PhilosophicalTransactions, and the following year he published the Mathematical Lucubrations. His name is often associated with an important transformation giving a relation between certain Eulerian integrals.
10 This work of Landen, which was summarized in Mackay (1887), is also quoted in Dingeldey (1903, p.
47) but not in Loria (1889a,b, 1896), Cantor (1901), Kötter (1901) and Bos et al. (1987).
123 A. Del Centina D d p n o m A B C G E F H δ Δ Γ A B o d δ Δ Γ F D G E p H C m n (b) (a) Fig. 4 a The second case of Landen’s question: the circle δ falls without ABC and o falls between A and B. b The third case: the circle δ falls without ABC but o does not fall between A and B Landen denoted by R and r the radii of and δ, respectively, and called d the distance between their centers. He considered the axis l of the given chord and called F the intersection of AB with l. Moreover, he drew the parallel to AB passing through d and denoted p its intersection with l. He also denoted E the center of the circle circumscribing the triangle ABC and put b = AF, x = AE. Always considering δ as standing upon AB (see the figures), he found the quotes of D above F, and above p, to be, respectively, ± √ R2 −b2, and ± √ R2 −b2 −r2. Then, he got dp = d2 −R2 + b2 −r2 ± 2r R2 −b2, Ao = b + d2 −R2 + b2 −r2 ± 2r R2 −b2, Bo = b − d2 −R2 + b2 −r2 ± 2r R2 −b2.
At this point, Landen wrote: “To proceed with perspicuity, it will convenient to consider distinctly the different cases of the circle mno falling within or without the triangle (ABC),” namely 1. δ falls within the triangle ABC (Fig. 3a, b), 2. δ falls without ABC but o falls between A and B (Fig. 4a), 3. δ falls without ABC and o does not fall between A and B (Fig. 4b).
123 Poncelet’s porism, I Case 1. Landen drew the segment AG, G being the point above AB where the line FH intersects , and noticed that the triangles AFG and dmC are similar, in fact they are both right and AGF = mCd. Therefore, ± √ x2 −b2 and x ± √ x2 −b2 being the quotes of E and G above F, respectively, the proportion b : x ± x2 −b2 = r : Cm, holds true, and so Cm = Cn = rx±r √ x2−b2 b . Then, AC = b + d2 −R2 + b2 −r2 ± 2r R2 −b2 + rx ± r √ x2 −b2 b , BC = b − d2 −R2 + b2 −r2 ± 2r R2 −b2 + rx ± r √ x2 −b2 b , b in this case being always greater than oF.11 The perimeter of the triangle ABC is 4b + 2rx±2r √ x2−b2 b ; hence, its area A is equal to 2br + r2x±r2√ x2−b2 b . From the known formula A = 1 2(AC × BC) sin AC B it follows that AC × BC = 4rx + 2r2x2 ± 2r2x √ x2 −b2 b , then, by equating this value with that obtaining multiplying the values of AC and BC above, and solving with respect to x, Landen got x = R2 −d2 ∓2r √ R2 −b2 4r + b2r R2 −d2 ∓2r √ R2 −b2 · From this expression is clear that, if d2 = R2 −2Rr, whatever b may be, one has x = R.
In the other two cases, that here for brevity we omit to discuss, Landen proceeded similarly and he found that x = ±2r √ R2 −b2 + d2 −R2 4r + b2r ±2r √ R2 −b2 + d2 −R2 · Again he observed that if d2 = R2 + 2r R, then, whatever b may be, is x = R.
11 In all cases, the proper sign of R2 −b2 is + or −, according if the center D is on the same, or on the contrary, side of AB with the center d; and the proper sign of x2 −b2 is + or −, according if the center E is on the same, or on the contrary, side of AB with the center d (Landen 1755, p. 5). See Figs. 3, and 4.
123 A. Del Centina Finally, as a corollary, he stated (Landen 1755, p. 5): It follows from what has been said that, d being equal to √ R2 −2r R or √ R2 + 2r R, whatever b may be, E will fall in D, and the circle circumscribing the triangle always coincides with the given circle ABH; a thing very remarable!
It seems to us that all this amounts to the following: given the two circles and δ, if the distance d between their centers is given by R2 −2r R (or by R2 + 2r R in the second and third case), then there exists a triangle ABC which is at the same time inscribed in and circumscribed about δ, and moreover, the triangle can be constructed starting from any point A on , the condition being independent of 2b, i.e., the length of the chord AB.
This means that, not only condition (1.1) is sufficient for the existence of an inter-scribed triangle to the two circles, but also that, in this case, the closure theorem is proved since the chord AB can be arbitrarily chosen.
Landen was aware of this, although he did not explicitly state it at this point. On the other hand, as we have noticed at the end of the previous subsection, Landen had been aware since 1747 that (1.1) represents a necessary condition, and he had certainly read Chapple’s (1746), so he knew that the existence of an inter-scribed triangle implies the existence of infinitely many others. This is well shown by the sequel of his memoir.
After having stated the corollary, Landen wrote: For this to happen and the circle mno fall within the triangle, it is obvious R must be no less than 2r, for, if it be, √ R2 −2r R, the quantity to which d ought to be equal, will be imaginary.
But, that the circle falling without the triangle, the same thing may happen though R be less than 2r, so that R be greater than r/4. The reason why R, in this case, must be greater than r/4 appears from this consideration. The distance of the center d from that point in the periphery ABH which is the farthest from the center is d + R = √ R2 −2r R + R, whose distance must be greater than r; otherwise, the circle ABH will fall entirely within the circle mno, and no chord in that can be tangent to this. Therefore, since √ R2 −2r R+R must be greater than r, √ R2 −2r R must be greater than r −R, R2+2r R greater than r2−2r R+ R2, and R greater than r/4.
Consequently, since r must be less than 4R, d must be less than √ R2 −2r R, or its equal 3R.
Here, Landen observed that, by a orthographic projection, i.e., a affine parallel projection, two circles—in a same plane—and their tangents can be mapped into two similar ellipses, i.e., having the same eccentricity, and their tangents (Landen 1755, p.
6). Then, he stated the following (see Fig. 5a, b): If within or without any ellipsis whose transverse axis is T,12 a second concentric similar ellipsis be described with its transverse axis t, in the same direction with T, and a third ellipsis be described, similar to the other two, with its center any where in the periphery of the second ellipsis, and having its transverse axis equal to [τ =] T 2−t2 2T ,13 12 For “transverse axis” he meant the “half of the major axis”.
13 This condition translates formula (1.1) in the present case, where t stands for a, T for R, and τ for r.
123 Poncelet’s porism, I δ δ δ γ Δ (a) (b) Fig. 5 a Landen observed that if there exists an inter-scribed triangle to the circles and δ, then there exist such triangles for all circles δ′, δ′′,... whose centers have distance from the center of equal to the distance of the center of δ. b By using orthographic projection, Landen extended the property illustrated in (a) to the case of concentric similar ellipsis and parallel to the transverse axes of the other ellipses; any tangent being drawn to this third ellipsis and continued both ways till it intersects the periphery of the first ellipsis in two points, and two other tangents being drawn to the same third ellipsis from those points of intersection, the locus where these last tangents continued to intersect each other will always be in the periphery of the first ellipsis.
The drawing of the tangents in that manner will be impossible unless t be less than 3T.
It is worthy underlining that Landen wrote: “any tangent being drawn to this third ellipsis and continued both ways till it intersects the periphery of the first ellipsis in two points, and two other tangents being drawn to the same third ellipsis from those points of intersection, the locus where these last tangents continued to intersect each other will always be in the periphery of the first ellipsis,” clearly this is the closure theorem for triangles and circles, or ellipses, in the previous configurations.
Landen concluded the first part of his essay by saying: “Other conclusions of a like nature may be drawn from what is done above and a consideration of other projections, but I have no inclination to pursue the speculation farther.” The projective methods of the school of Monge were still far in the future; nevertheless, these words show how Landen had perception of the projective nature of the question.
We can safely state that Landen’s work, although unfortunately it remained unknown outside England, contained in germ some ideas that Poncelet was to develop 60years later.
1.3 Euler Some nineteenth-century authors, first of all Steiner (see below), attributed formula (1.1) to Leonhard Euler. In his paper (1765), which can be considered a milestone in triangle geometry, he studied the positions and mutual distances of barycenter, 123 A. Del Centina orthocenter, incenter and circumcenter.14 In particular, he found that if a triangle, with side lengths x, y, z, is inter-scribed to two circles, one inside the other, then the distance a between their centers satisfies the following equation: a2 = (xyz)2 16A2 − xyz x + y + z · (1.4) Surprisingly enough in his article, Euler did not investigate the relation between a and the radii r, R of the two circles, and in fact this formula does not express a in terms of r, R. We notice that one can get formula (1.1) from (1.4), by taking into account (1.2) and (1.3), which were known to Euler. Nevertheless, we should add that N. Fuss, who was certainly well acquainted with Euler’s work, did not attribute formula (1.1) to him (see below). Moreover, as stressed in Bos et al. (1987, p. 295), formula (1.4) is not a formula by which a closure theorem could be detected, since a is a function of the sides of the triangle, which depend on it, and not of the radii, which are fixed.
1.4 Fuss Nicolaus Fuss, in his paper (Fuss 1797), studied first several problems concerning quadrangleswhichareinscribedin,orcircumscribedabout,agivencircle.Forinstance, he determined the radius R of the circumscribed circle, and the radius r of the inscribed circle, as functions of the length of the sides a, b, c, d of the quadrangle, for which he found, respectively: R = 1 4 (ab + cd)(ac + bd)(ad + bc) abcd , r = √ abcd a + c .
Only at the end of his paper did he consider the following problem (Fuss 1797, section 30): Datis radiis circolorum quadrilatero ABCD inscripti et circumscripti, invenire dis-tantia centrorum [Find the distance between the centers of the circumscribed circle and of the inscribed circle to the quadrangle ABCD].
To solve the problem Fuss proceeded as follows (see Fig. 6a). He denoted 2α, 2β, 2γ, 2δ the four angles at the vertexes A, B, C, D of the quadrangle. From the above formulae for R and r, he got R = r√1 + 2 sin 2α sin 2β sin 2α sin 2β , 14 In this paper, Euler discovered the nine point circle (i.e., the circle on which lie the three midpoints of the sides, the three feet of the altitudes, and the three midpoints of the line segments from each vertex to the orthocenter), and the line now referred to as Euler line, to which belong barycenter, orthocenter and circumcenter.
123 Poncelet’s porism, I O o A B C D T d A B C D o O T P F V E (a) (b) Fig. 6 a Fuss’ construction for the quadrangle. b Illustration of how Fuss proceeded in order to get a direct proof of formula (1.1) and, putting m = cos(α + β), and n = cos(α −β), he wrote R in the form R = r √ 1 + n2 −m2 n2 −m2 · He called d the distance between the centers O and o and denoted by T the inter-section between the line through o orthogonal to AB and the line through O orthogonal to the previous one. He found that T O = r √ 1 −n2 n −m , T o = r(n2 + m2 −1) n2 −m2 , d = r √ 1 + m2 −n2 n2 −m2 · Taking into account the last formula for R and computing R2 −d2, R2 + d2, he obtained (R2 −d2)2 = 2r2(R2 −d2) and so d2 = R2 + r2 ± r 4R2 + r2, which clearly generalizes formula (1.1) to the case of quadrangles.
In the subsequent, final, section, Fuss gave a direct proof of formula (1.1), and for this he argued as follows (see Fig. 6b).
He considered a triangle ABC inscribed in the circle of center O and circum-scribed about the circle δ of center o, lying inside the first. He observed that the line through A and o meets in a new point D and CB in the point F; moreover, 123 A. Del Centina Ao and Co bisect B AC and AC B, respectively; the line through D and O meets in a new point E; the diameter DE meets perpendicularly in V the side C B of the triangle. Then, he let T and P be the points of intersection with the diam-eter E D and the side CB of the respective perpendicular through o. He found that d2 = R2 −2R · DT + Do2. Moreover, CVD = DBV = π/2, and BED = DCV since they belong to the same chord. It follows that the triangles CVD and EBD are similar, and so DV /C D = BD/DE; therefore, C D × BD = DV × DE = 2R · DV .
Clearly, CoD = DAC + oCA and oCD = oCF + DCF.15 Since oCF = oCA and DCF = DCB = DEB = DAB = oAC, is CoD = oCD. It follows that Do = C D = BD and Do2 = C D × BD = 2R · DV , substituting this last value in the above formula for d2, he found d2 = R2 −2R(DT −DV ), and, since DT −DV = T V = oP = r, Fuss finally got d2 = R2 −2Rr.
Letusremarkthatthiswasthefirstoccurrenceofformula (1.1)inawidelycirculated journal known to an international mathematical public.
In a subsequent paper (Fuss 1802), Fuss considered the same question for polygons having more than four sides. However, the great difficulties he encountered in dealing with the general problem forced him to limit his analysis to polygona symmetrice irregularia [symmetrically irregular polygons], i.e., irregular inter-scribed polygons that are divided into two equal parts by the line of centers of the circles.
He let R,r, a be as above, and put p := R + a, q := R −a and s := pq/r. Then, using trigonometric formulae with great skill and a large amount of work, he got the following conditions for n = 5, 6, 7, 8, respectively: p2q2 −r2(p2 + q2) p2q2 + r2(p2 −q2) = ± q −r p + q , 3p4q4 −2p2q2r2(p2 + q2) = r4(p2 −q2), ±(s2 −s(p −q) −2pq) √p(p + q)(s −q) 2 ± (s2 −p2 −q2) √q(p + q)(s −p) 2 = ± s −p + q 2 (s2 + p2 −q2), p2r q2 −r2 + q2r p2 −r2 = p2r2 − (p2 −r2)(q2 −r2).
Since Fuss was unaware of any closure theorem, he did not realize that his results were true in general, i.e., without restriction to symmetrical irregular polygons.
1.5 Lhuilier Although the results of Fuss were published in a prestigious journal, many mathe-maticians failed to notice them. In fact, in the first volume of Gergonne’s Annales 15 In (Fuss 1797, p. 124), it is erroneously written oCD = oCE + DCE, and then (in the line below) oCE = oCA, DCE = oAC which is clearly false. Since our figure is the same as (Fuss 1797, Fig. 4) we may argue that the letter E and F were interchanged when printing the paper.
123 Poncelet’s porism, I de mathématiques pures et appliqués for the years 1810–1811, the following two questions were proposed: I) Un cercle étant donné et un point étant donné arbitrairement sur son plan et dans son intérieur, il y a toujours une longueur, et une seule longueur, laquelle étant prise pour rayon d’une nouveau cercle ayant pour centre le point donné, il arrivera qu’un même triangle pourra être à la fois inscrit au premier des deux cercle, et circonscrit au second; I I) Un cercle étant donné et un point étant donné arbitrairement sur son plan, il y a toujours une longueur, et une seule longueur, laquelle étant prise pour rayon d’une nouveau cercle ayant pour centre le point donné, il arrivera qu’un même triangle pourra être à la fois circonscrit au premier des deux cercle, et inscrit au second (Gergonne 1810, pp. 62–64).
The two questions can be reformulated as follows: I) Given a circle C and a point P inside it, there is exactly one length r such that if D is the circle of center P and radius r, there exists a triangle which is inscribed in C and circumscribed about D; I I) Given a circle D and a point P, there is exactly one length R such that if C is the circle of center P and radius R, there exists a triangle which is circumscribed about D and inscribed in C.
It seems that the proposer knew the formula (1.1) (in fact it was so), and had noticed that, fixed a, there is only one R for any given r, and viceversa.
These questions were soon solved by Lhuilier16 in his memoir (Lhuilier 1810).17 From a footnote in the first page of Lhuilier’s paper, the editors explain that they had known the theorem since 1807. It was communicated to them by Monsieur Mahieu, professor of mathematics in Alais, who learned it from Monsieur Maisonneuve, a mines engineer.
In the same footnote, the editors presented the proof given by Maisonneuve and that we think worthy enough to be reproduced here below.
If a denote the distance between the centers and x, y, z are the length of the sides of the inter-scribed triangle, one has:18 a2 = (xyz)2 (x + y + z)(y + z −x)(x + z −y)(x + y −z) − xyz x + y + z , on the other side, as well known, the following relations hold true 16A2 = (x + y + z)(y + z −x)(x + z −y)(x + y −z), 2A = r(x + y + z), 4RA = xyz.
16 Simon Antoine Jean Lhuilier, sometime written L’Huilier (1750–1840), Swiss mathematician, professor at the University of Geneva and member of several European Academies. He is mainly known for his studies in Analysis and for having generalized the Euler formula to planar graphs.
17 The questions posed by Gergonne, and the solution given by Lhuilier, completely escaped Loria but were cited in Lhuilier (1901, p. 151).
18 Maisonneuve probably deduced this formula from (1.4) and Heron’s formula.
123 A. Del Centina From the first two relations, it follows that xyz x + y + z = 2r R, and also, by equating the square of the third with the first, it follows that (xyz)2 (x + y + z)(y + z −x)(x + z −y)(x + y −z) = R2.
Then, taking into account the above expression of a2, from the last two formulae immediately follows a2 = R2 −2r R.
Lhuilier expressed formula (1.1) in the form Zz2 = R(R −2r), being Z and z the centers of the circles. The proof he submitted was quite complicated, but it can be seen as a trigonometric translation of Maisonnneve’s proof. We will not present it here, preferring to insist on Lhuilier’s clear perception of the closure theorem.
After having proved formula (1.1), he wrote: La relation entre la distance des centres de deux cercles et les rayons R et r de ces cercles, étant telle qu’ill vient d’être dit; si on circonscrit au cercle dont le rayon est r un triangle dont un des côtés soit une corde de l’autre cercle, ce triangle sera inscrit à ce dernier cercle; et reciproquement, si l’on inscrit au cercle dont le rayon est R un triangle dont un des côtés soit tangent à l’autre cercle, ce triangle sera circonscrit à ce dernier cercle. Il y a donc un nombre illimité de triangles qui peuvent être à la fois inscrits à un cercle et circonscrits à une autre cercle. Lorsque les rayons de ce cercles et la disatnce de leurs centres sont liés par l’équation Zz2 = R(R −2r) [If the relation of the distance between the centers of the circles and the radii R and r of these circles is as we have just said, then if we circumscribe about the circle of radius r a triangle having a side which is a chord of the other circle, this triangle will be inscribed in the last circle and vice versa. If we inscribe in the circle of radius R a triangle having a side which is tangent to the other circle, this triangle will be circumscribed about this last circle. Hence, there are infinitely many triangles which are inscribed in a given circle and circumscribed about another if the distance between their centers satisfies the equation Zz2 = R(R −2r)].
So, similarly to Chapple, it was the degree of freedom implicit in the equation (1.1) that led Lhuilier to the clear enunciation of the closure theorem for triangles inter-scribed to two circles, one inside the other. Nevertheless, his proof, based exclusively on the above argument, was not complete.
1.6 Steiner and the attribution to Euler of formula (1.1) In the second volume of Crelle’s journal, in the section Aufgaben und Lehrsät-ze, Jakob Steiner asked the reader to solve the following problem (Steiner 1827, p. 96): 123 Poncelet’s porism, I 3. Aufgabe. Wenn ein gegebenrs (irreguläres)19 Vieleck (n Eck) so beschaffen ist, dafs sowohl in als um dasselbe ein Kreis beschrieben werden kann, so soll man zwischen des Radien (r, R) der beiden Kreise und dem Abstande (a) ihrer Mittelpuncte von einander eine Gleichung finden. (Für das Dreieck ist diese zuerst von Euler gefundene Gleichung bekanntlich a2 = R2 −2r R). [If a (irreg-ular) polygon (n-gon) is given such that a circle can be drawn in and around it, it is required to find the equation relating the radii (r, R) of the two circles and the distance (a) between their centers. For triangles this equation, that was found for the first time by Euler, is a2 = R2 −2r R)].20 Following Steiner, other authors attributed formula (1.1) to Euler; see for instance (Jacobi 1828; Loria 1889a, 1896; White 1916).21 It seems that Steiner was not aware of Fuss (1797).
In a subsequent note, Steiner gave, without proofs, the equation that R,r, a must satisfy for the existence of an inter-scribed n-gon when n = 4, 5, 6, 8, respectively (Steiner 1827, p. 289): (R2 −a2)2 = 2r2(R2 + a2), r(R −a) = (R + a) (R −r + a)(R −r −a) + (R −r −a)2R , 3(R2 −a2)4 = 4r2(R2 + a2)(R2 −a2) + 16r4a2R2, (R2 + a2) (R2 −a2)4 + 4r4a2R2 −8r2a2R2(R2 −a2) ×8r2 (R2 −a2)2 −r2(R2 + a2) = (R2 −a2)4 −4r4a2R2 2 .
We do not know how Steiner obtained them, but, since he had Poncelet’s treatise handy, we may argue that he proved the formulae for symmetrically irregular polygons and extended their validity to irregular polygons via Poncelet’s theorem.
The lack of proofs was deplored by Jacobi (1828, p. 376). In this paper, that will be discussed in section 3, Jacobi felt compelled to challenge the paternity on behalf of Fuss, who had recently died, the results claimed by Steiner. He also affirmed that the case n = 7, that omitted by Steiner, was the most difficult to solve, and comparing the formulae of the two authors he established their equivalence in all cases except for n = 8.22 19 It is quite probable that Steiner was referring to irregular polygons as “completely irregular polygons,” and not to “symmetrically irregular polygons” in the sense of Fuss.
20 Steiner did not quote any paper by Euler.
21 No reference to Euler’s papers were provided in Jacobi (1828) and White (1916). Loria quoted Novi Comm. Ac. Sci. Imp. Petropolitanae, vol. 2 (1749), published in 1750, but, although this volume contains two memoirs by Euler, none of them presents results which may lead to formula (1.1).
22 In fact, Steiner’s formula for n = 8 is not correct [see for instance (Dingeldey 1903, p. 47)].
123 A. Del Centina 2 The theorems and methods of Poncelet In 1822, Jean-Victor Poncelet published his Traité sur les propriétés projectives des figures (Poncelet 1822).23 In this treatise, Poncelet adopted a highly synthetical approach and introduced two concepts that would be crucial in the setup of his entire book: the ideal chord and the principle of continuity (see below). Although Cauchy and Gergonne criticized the use of the principle of continuity,24 the treatise was well received by contemporaries, and the tools developed therein were adopted for decades in the nineteenth century [see (Chasles 1837; Kötter 1901) and also (Kline 1972, pp. 163–165), (Gray 2007, chap. 4)].
In his extremely rich book, largely conceived between March 1813 and June 1814 during his captivity as a prisoner of war in Saratov25 Poncelet formulated the follow-ing theorem that became known as the Poncelet closure theorem (PCT for short) or Poncelet’s porism (Fig. 7a): Theorem PCT Let C and D be two smooth conics in the projective plane, if there exists a polygon of n sides which is inscribed in C and circumscribed about D, then for every point P ∈C there is one such polygon having P as one of its vertices.
Poncelet obtained this theorem as a corollary of a more general result, which we will refer to as Poncelet general theorem (PGT for short), that can be formulated as follows (Fig. 7b): Theorem PGT Let C, D1, D2, . . . , Dn−1 be conics from a pencil F. Consider a n-gon P, P1, . . . , Pn−1 inscribed in C and having the side P P1 tangent to D1, the side P1P2 tangent to D2 and so on until Pn−2Pn−1 tangent to Dn−1. Then, if P move along C in such a way the sides P P1, P1P2 etc. remain tangent, respectively, to D1, D2 etc., the nth side Pn−1P envelops a conic belonging to F.
23 A second enlarged edition of this work appeared in two volumes more than 40years later (Poncelet 1865–1866).
24 See Cauchy’s report on the paper that Poncelet presented to the Academy in 1820 (Cauchy 1820), also reproduced in Poncelet (1822).
25 Poncelet took part, as Lieutenant of Engineers, in Napoleon’s Russian campaign in 1812. After the retreat of the French army, following the defeat at Borodino (September 1812), Poncelet was left for dead on the battlefield of Krasnoi (November 1812), here he was found by the enemy soldiers. As a prisoner of war, he was forced to march for almost five months until he was imprisoned in Saratov, on the banks of the river Volga (Didion 1870). During his captivity, Poncelet wrote seven notebooks, called by him “Cahiers de Saratoff,” where on the basis of what he had learnt from Carnot and Brianchon at the École polytechnique, he developed the projective theory of conic sections. Poncelet wrote the seventh notebook with the intent of presenting it to the Academy of Sciences of St. Petersbourg, with the hope, if accepted, of being called to Moscow until a peace agreement was reached between France and Russia. The events of 1814 interrupted this project. Poncelet based the development of his treatise (Poncelet 1822) on the Cahiers, of which the seventh might be regarded as a first attempt at a redaction. After his return to France, Poncelet published some of the results he had obtained in Saratov. In particular, in 1820, he presented to the Paris Academy the paper Mémoire relatif aux propriétés projective des sections coniques in which, for the first time, he presented the “principle of continuity” as a tool for solving difficult problems concerning conics. But it was only in 1862 that he published the seven notebooks in their entirety, as a part of the first volume of his new treatise Applications d’analyse et de géométrie (Poncelet 1862, vol. 1). For a detailed analysis of the content of the Cahiers (see Belhoste 1998).
123 Poncelet’s porism, I P C D P P P1 P2 P3 P4 D1 D2 D3 D4 C (a) (b) Fig. 7 a Poncelet’s closure theorem for n = 5. b Poncelet’s procedure in order to prove the general theorem for circles belonging to a same pencil The proof of PGT presented in the Traité was a slight variation of one that he had written during his imprisonment, but which was published much later (Poncelet 1862, sixth Cahier). In both variants, Poncelet first proved the theorem for circles, proceeding by induction on n, and then he extended its validity to conics using the method of projection (see below). The two proofs differ only in the first step of the induction, i.e., in the case of three circles. In fact, in Saratov, Poncelet had proven it by an enormous straightforward computation, while in the Traité he proceeded by the synthetical method, developing several preliminary geometrical lemmas, and applying an ad hoc reasoning, heavily based on the principle of continuity. We will discuss these proofs later on.
In an article published in 1817, Poncelet announced new methods in geometry. The problems that he said were solvable by these methods included the following (Poncelet 1817, p. 154): Deux section coniques ètant tracées su un même plane, construire un polygone de tant de côtés qu’on voudra qui soit, à la fois, inscrit à l’une d’elles et circonscrit à l’autre, en ne faisant usage que de la règle seulement [given two conics in traced in the same a plane, it is asked to construct a polygon, inscribed in one of them and circumscribed about the other, having whatever number of sides making use only of the ruler].
The formulation of the problem does not allude at all to the existence of a closure theorem, but rather suggests the existence of such a polygon for any pair of given conics. This fact led G. Loria to doubt that Poncelet knew the PCT much before the publication of the Traité (Loria 1889a, pp. 9–10).26 For this reason, many historians were unaware that Poncelet had achieved the proof of the theorem in 1813.
It is worthwhile stressing that, although in the Traité Poncelet did not explicitly express the link between PCT and the existence of conditional equations, such as 26 This is curious because, in footnote n.4 Loria quoted the first volume of Applications d’analyse et de géométrie, where the general theorem is stated and proved, and on its title page is written: “Sept cahiers manuscrits rédigés à Saratoff dans les prisons de Russie (1813–1814).” 123 A. Del Centina formula (1.1) or those given by Fuss and Steiner, he had been clear about this since his imprisonment in Saratov. In fact, in Poncelet (1862, pp. 357–358) he affirmed that if a polygon of n sides is inter-scribed to two circles of radii R, r, an equation f (R,r, a; n) = 0, being a the distance between the centers, must necessarily hold.
2.1 Poncelet’s methods Poncelet’s geometry always concerned the real plane and real space extended with the elements at infinity. In this setting, he aimed at deducing properties for systems of lines and conics by projectively generalizing properties proved for systems of lines and circles.
With this goal in mind, during his captivity Poncelet developed systematically the method of central projection, introduced by Brianchon (1810) and that he based on five fundamental principles.
The first three, which he gave without proof, being already largely accepted by geometers, affirm the projective equivalence of a circle and a conic, and the projective equivalence of a pencil of parallel lines and a pencil of intersecting lines.
The last two, which are more delicate, state, respectively: the projective equivalence of a system of a conic and a line at finite distance with a system of one circle and the line at infinity, and the projective equivalence of a system of two conics with the system of two circles [(see Poncelet 1862, vol. 1, p. 122; pp. 287–307; pp. 380–388) also (Poncelet 1822, Art.s 109–111; 121)]. Precisely: Theorem (4th principle) Let C be a conic and l be a line in the real plane at finite distance. Then, C and l are projective images of a circle and of the line at infinity.
Theorem (5th principle). Any pair of conics is the projective image of a pair of circles.
As already remarked, Poncelet considered only real objects in real planes and spaces. For this reason, he was able to prove the fourth and fifth principle only for certain positions of the figures involved. So, in order to extend their validity for all positions of the elements, even when some of them disappear becoming “imaginary,” he invoked the principle of continuity.
Before we discuss Poncelet’s proofs of the last two principles, it is useful to recall his concept of “ideal chord.” The concept of “ideal chord” In his early studies at the École Polytechnique, Poncelet had already met imaginary elements in connection with the intersection of a conic and a line, or the intersection of two conics. To take into account these elements, without actually extending the real plane with complex points,27 he introduced the concept of “ideal chord.” Let C be a (real) conic, and let l be a (real) line which intersects C in two points P1 and P2 (Fig. 8a). The segment P1P2 is the chord of C corresponding to l. Let d be 27 Complex projective geometry entered into the mathematical scene only around the middle of the nine-teenth century.
123 Poncelet’s porism, I C P1 P2 Q1 Q2 Q l d C Q1 Q2 d l M R1 R2 (a) (b) Fig. 8 a The real chord P1 P2 of the conic C. b The ideal chord R1 R2 of the conic C the diameter of C conjugated to l, let Q1, Q2 be the points of intersection of C with d, and say Q the intersection l ∩d. Then, the one has (QP1)2 = c(Q1Q)(Q2Q), where c is a real number that does not change for any lines parallel to l intersecting C.
When l′ is a line parallel to l, but “exterior” to C (Fig. 8b), Poncelet, unaware of the complex elements, associated with l′ and C not the pair of conjugate points that constitute the intersection l′ ∩C in the complex projective plane, but a segment in the real plane. He did this by simply extending the above relation to the present case, precisely: he let M = d ∩l′ and let R1 and R2 be the two points on l′ such that (M R1)2 = c(Q1M)(Q2M), and M R1 = M R2. Then, Poncelet defined the segment R1R2 to be the ideal chord intercepted by (the ideal secant) l′ on C.
In Poncelet’s mind, the ideal chord was a “real” justification for the “imaginary.” His thinking is well expressed in (Poncelet 1822, Art.s 50, 54): En supposant qu’on ne veuille pas créer des termes nouveaux pour désigner la droite mn [l] et ce qui lui appartient, et qu’on persiste à la regarder comme une secante de la courbe quand elle cesse de la rencontrer, nous dirons, a fin de conserver l’analogie entre les idées et le langage, que ses points d’intersections avec la courbe, et par conséquent la corde correspondente, sont imaginaires, qu’elle est elle-même sécante idéale de cette courbe... et on pourra regarder M′N ′ [R1R2] comme représentant, d’une manière fictive, la corde imaginaire qui correspond à la droit m′n′ [l′] considérée comme secante de la courbe [Let us supposethatwedonotwanttocreatenewtermsforthelinel andforwhatpertains to it and that we continuing to look at this line as a secant of the curve even when the intersection no longer exists; we will say, with the objective of preserving the analogy btween the ideas and the language, that the points of intersection with the curve, and consequently the corresponding chord, are imaginary, it is itself an ideal secant of that curve …and we can look at the segment R1R2 as 123 A. Del Centina (a) (b) Fig. 9 Poncelet defined the radical axis of a pencil of circles as the locus of points from which tangent segments drawn to all circles of the pencil have equal length. a The construction for the case of intersecting circles and b the construction in case of not intersecting circles a fictional representative of the imaginary chord corresponding to l′ considered as a secant of the curve].
Let us observe that if T1 = (a + ib, c + id), T2 = (a −ib, c −id) are the two complex conjugate intersection points of l′ and C in the complex plane, then we can write Tk = S ± i D where S = (a, c) and D = (b, d) represented real points in the plane. It is easy to see that R1 = S + D and R2 = S −D and S is the middle point of the segment R1R2. Vice versa it is also clear that from R1 and R2 one can find T1, T2.
So, by the concept of ideal chord, Poncelet was able to develop many features that nowadays are introduced by embedding the real plane into the complex one. In particular, he showed that two (real) conics always have two common chords (real or ideal), i.e., two conics always intersect in four points (real or imaginary) (Poncelet 1822, Art.s 58–59).
Let us to stress that Poncelet was aware that endpoints and chords may lie on the line at infinity. In particular, he noticed the existence of the two points at infinity through which all circles pass, i.e., of those points that later were called circular points (Poncelet 1822, Art. 94).
Poncelet introduced the concept of a “system of conics having two common chords” (real or ideal), which corresponds to the modern concept of a pencil of conics. In particular, a pencil of circles was defined by Poncelet as the system of circles having a real or ideal common chord. He also defined the radical axis of a pencil of circles as the locus of points from which tangent segments drawn to all circles of the pencil have equal length (Poncelet 1822, Art.s 71–77).28 This allowed Poncelet to extend the concept of radical axis to the case of non-intersecting circles (Fig. 9a, b); in fact, this 28 This definition, as Poncelet remarked, had already been introduced in Gaultier (1813). J. Steiner called it line of equal power (Steiner 1826). Defining power of a point P with respect to a circle the quantity h = d2 P −r2, where dP is the distance of P from the center of the circle of radius r, the radical axis of a pencil of circles is the locus of points P whose power is the same with respect to all circles of the pencil.
123 Poncelet’s porism, I line coincides with the real line on which lies the real or ideal common chord of all circles of the pencil.
Poncelet’s “principle of continuity” In its essence, the principle of continuity may be stated as follows: if for a figure in the plane certain properties have been deduced from the given data and theorems, these properties remain valid after a continuous deformation of the figure, even if during the deformation some aspects of the figure disappear becoming imaginary.
This principle, that “was admitted without saying by many geometers” as Poncelet recalled,29 enabled him to avoid the use of the “imaginary,” and to deduce results as if he was working into the complex projective space.
He formulated the principle of continuity in various ways and discussed it at length in several places of his treatise (see (Poncelet 1822, pp. xiii–xiv; Art.s 135–140)).
When in Saratov, Poncelet expressed this principle as follows (Poncelet 1862, p. 379): Quand on se proposera de découvrir quelque propriété générale de position d’une figure, on pourra imaginer que cette figure soit projetée sur un nouveau plane (d’après la manière indiquée ci-après), de telle sorte qu’ne ou plusieurs parties de cette figure soient réduites à des circonstances plus simples; on aura ainsi une nouvelle figure qui pourra remplacer la première, sinon pour toutes les dis-positions possibles au moins en général; on raisonnera sur cette figure comme tenant lieu de la première d’où l’on est parti, et les propriétés, les conséquences géérales qu’on en déduira seront également applicables à cette figure, quoiqu’il arrive des cas où la projection soit imaginaire [When it is proposed to discover certain general position properties of a figure, we can imagine that figure pro-jected on a new plane (in the way specified here below), so that one or several parts of it are reduced to more simple situations; thus, we will have a new figure that can replace the first, if not in all cases at least in general; we will argue on this figure as on the first from which it came, and the properties, the general consequences that can be deduced will be equally applicable to this figure, even if obtained by an imaginary projection].
We clarify Poncelet’s use of the principle of continuity by shortly discussing his proofs of the fourth and fifth fundamental principles mentioned above.
To prove the fourth principle, Poncelet proceeded as follows.
Let C and l be, respectively, a conic and a line in a plane π, which is thought of as immersed in real space. If l does not intersect C, one can find a point S / ∈π and a plane π′ such that, under the projection from S in π′, the conic C is mapped onto a circle and l onto the line at infinity. In Saratov, Poncelet gave an analytical proof of this fact, while in the Traité he provided a synthetic proof (Poncelet 1822, Art.s 109–111).
If l intersects C, such S and π′ cannot be found in the real space, then Poncelet claimed, on the basis of the principle of continuity, the validity of the fourth principle also in this case.
29 See Poncelet (1862, p. 124). The principle was used only occasionally until G. Monge revived it estab-lishing certain theorems of descriptive geometry (Kline 1972, pp. 163–165). A similar principle was also used by Carnot in his Géométrie de position (1802).
123 A. Del Centina Poncelet proceeded to prove the fifth principle similarly.
If C1 and C2 have at most two real points in common, then they have an ideal common chord along a certain line l not intersecting them. Hence, according to the fourth principle, C1 and l are the projective image of a circle C and of the line at infinity l∞. Then, the projective image of C2 passes through C ∩l∞(i.e., the circular points) and so is a circle.
If the two conics have more than two real points in common, this program is not realizable by a real projection, in fact two circles with three points in common coincide.
Then, Poncelet extended the validity of the fifth principle by invoking the principle of continuity.
Poncelet enounced and proved also the following (Poncelet 1822, n. 131) Theorem B Two conics C1, C2 which are tangent to each other in two different points, are the projective image of two concentric circles.
His reasoning was as follows. By the fifth principle, the two conics are the projective images of two circles and, at the same time, the two points of contact are the images of the circular points. Because the two circles are tangent to each other at the circular points, the line at infinity has the same pole with respect to them, and so they are concentric.
Let us remark that the fourth and fifth principle and theorem B are actually correct in the extended complex plane, so, ultimately the principle of continuity led Poncelet to correct results.
It can be observed (see Bos et al. 1987, p. 303) that these three results are succes-sively more counterintuitive, in the context of the real geometry. The first two can still be seen to be correct in certain cases, but the third does not apply at all in the real case: two real circles which are tangent one each other in two points coincide. This shows how daring Poncelet’s use of ideal chords and of the principle of continuity really was.
2.2 The analytical proof PGT Poncelet first considered the case of two circles (C) and (C′), of centers c and c′ and radii r and R, respectively (the first lying inside the second). He denoted by a the distance between the two centers and wrote the equations of the circles as: (C) : x2 + y2 = r2, (C′) : (x −a)2 + y2 = R2.
(2.1) From a point α := (α, β) on (C′), so that β2 + (α −a)2 = R2, (2.2) he drew two chords of (C′) tangent to (C), say αx1 and αx′′ (Fig. 10a), and showed that, if α moves along (C′), then the chord x1x′′ varies and envelopes a circle (C′′) from the pencil determined by (C) and (C′). Poncelet proved this result by a direct laborious computation. He first found the equations of αx1 and αx′′, then he obtained the equation for x1x′′: 123 Poncelet’s porism, I (C) (C ) (C ) α x1 x (C) (C ) (C ) α x (C ) x2 1x 1x2 (a) (b) Fig. 10 Illustration of how Poncelet proceeded in order to prove his general theorem in case of two circles (a) and of three circles (b) β(R2−a2)y+[a(R2−a2)+α(R2+a2)]x+(R2−a2)−2R2r2+a(R2−a2)α =0.
Taking into account the relation (2.2), he was able to eliminate α and β from this equation, and for (C′′) he obtained the following equation: x2(R2 −a2)2 + y2(R2 −a2)2 −2a (R2 −a2)2 −4R2r2 x = R2(R2 −a2 −2r2)2 −a2(R2 −a2)2 (2.3) (Poncelet 1862, pp. 314–319). Then, Poncelet proved that (C′′) belongs to the pencil determined by (C) and (C′) by showing that the ideal common chord of (C) and (C′′) coincides with the ideal common chord of (C′) and (C′′) (Poncelet 1862, pp.
319–320).
Let us observe that the equation (2.3) reduces to that of (C) when one puts a2 = R2 −2r R.
After having examined certain special cases in which (C′′) may degenerate, Pon-celet proceeded to the case of three circles (C), (C′) and (C′′) from a same pencil (Poncelet 1862, pp. 323–339).
He represented these circles by the following equations (C) : x2+y2 = R2, (C′) : (x−a)2+y2 = r2, (C′′) : (x −a′)2 + y2 = r′2.
From a point α := (α, β) on (C) he drew four chords of (C), say αx′′, αx2, α1x′′ and α1x2, the first and the third tangent to (C′), the second and fourth tangent to (C′′) (Fig. 10b).
Through extremely long and involved calculations, Poncelet was able to find the equation of the chord x′′x2,30 but it proved too complicated to admit calculating the envelope by eliminating the parameters α, β.
30 See for instance pp. 327, 330, 336 where the formulae to be written down require sheets folding out to the width of some quarto pages.
123 A. Del Centina α x x x x(4) (C) (C ) (C ) (C ) (C(4)) α x x x x(4) (C) (C ) (C ) (C ) (C(4)) (a) (b) Fig. 11 Illustrations of how Poncelet applied the main lemma in order to prove the general theorem for n > 3 Then, he proved the existence of a point on the line of centers, whose distance from the chord x′′x2 is the same whatever is the point α on (C), and therefore, that the enveloped curve of this chord is a circle (C′′′), for which he got the following equation ⎛ ⎜ ⎜ ⎝x − R2(a −a′) r′ −r a′ a (R2 −aa′) r′ + r a′ a ⎞ ⎟ ⎟ ⎠ 2 + y2 = R2 (R2 −a′2) + r′(R2 −a2) 2 (R2 −aa′)2 r′ + r a′ a 2 · Finally, arguing as in the previous case, he proved that (C′′′) is from the same pencil that (C), (C′) and (C′′) belong to. He also showed that the chord 1x′′ 1 x1 envelops the same circle (C′′′).
We will call this result main lemma.
At this point, Poncelet generalized these results to the case of n > 3 circles.
He considered n circles (C), (C′), (C′′), . . . , (C(n−1)) from a same pencil F, and a transversal αx′x′′x′′′ . . . inscribed in (C), and whose sides are tangent, in some order, to the inner circles (C′), (C′′), . . . , (C(n−1)). Clearly, without loss of gener-ality, one can suppose that αx′, x′x′′, . . . , x(n−2)x(n−1) are, respectively, tangent to (C′, (C′′), . . . , (C(n−1)) (see Fig. 11a). By the previous result, the chord αx′′ envelops a circle from F. Similarly, since the chord αx′′ and the chord x′′x′′′ are tangent to two circles from F, the chord αx′′′ envelops a circle from F, and so on (see Fig. 11b).
Clearly, x(n−1)α envelops a circle from the pencil F.
Poncelet used projection and the principle of continuity to extend these results to conics, but we will discuss this generalization later on, when we will comment the proof of PCT he gave in Poncelet (1822).
After having proved PGT, Poncelet enounced the closure theorem as follows (Pon-celet 1862, p. 355): 123 Poncelet’s porism, I Il est impossible, généralement parlant, d’inscrire à une courbe donnée du deuxiè-me degré un polygone qui soit en même temps circonscrit à une autre courbe de ce dedré, et quand la disposition particulière de ces courbes sera telle que l’inscription et la circonscription simultanées soient possibles pour un seul polygone essayé à volonté, il y aura, par la même, une infinité jouissant de cette propriété à l’égard des coniques données. Pour démontrer ce théorème directe-ment, soient deux ligne quelconque du second degré; d’après nos principes, ces ligne pourront, en général, être projectées suivant deux circonférences de cercle, bie que, dans des cas particuliers cela puisse devenir illusoire... [In general it is impossible to inscribe in a conic a polygon which is at the same time cir-cumscribed to another conic, but when, for the particular disposition of the two conics, it can be proved that this is possible for a particular polygon, then there will exist infinitely many polygons having the same property with respect to the given conics. For a direct proof of this theorem, let two curves of second degree be given, from our principles these curves can, in general, be projected onto two circles, although in some cases this may become illusory...].
Poncelet considered two circles, (C) and (C′) (the second lying entirely in the interior of the first), and a polygon α, x, x′, x′′, . . . , α′ inscribed in (C), whose sides are all tangent to (C′) except α′α, which will be tangent only for particular positions of the two given circles. Then, in order to prove PCT, he reasoned as follows.
Suppose that one can deform the polygon, maintaining the same number of sides, in such a way that it assumes all possible positions around (C′) while remaining inscribed in (C). If, by chance, there exists a polygon of the same number of sides which is inter-scribed to (C) and (C′), it is evident that there is a position of α, x, x′, x′′, . . . , α′ in which α′α is tangent to (C′). This is impossible, unless α′α is tangent to (C′) for any position of the polygon. In fact, α′α envelops a circle (C′′) (from the same pencil of (C) and (C′)) and if, for a certain position of the polygon, α′α is tangent to (C′), then (C′′) necessarily coincides with (C′) and all polygons α, x, x′, x′′, . . . , α′ will be inter-scribed to the two circles (C) and (C′).
Hence, on the basis of the fundamental principles, Poncelet concluded (Poncelet 1862, p. 357): on peut conclure des principes posés au commencement du IIIe Cahier, que la double proposition d’abord énoncée est vrai quelle que soit la situation relative des deux cercles donné (C) e (C′) [from the principles given at the beginning of the third Notebook, it follows that the proposition stated above holds true for every mutual position of the two given circles (C) e (C′)] andfinally, heextendedthevalidityof thetheoremtothecaseof conics byprojection (Poncelet 1862, p. 364): Le deux cercles (C) et (C′) peuvent être considérés comme la projection de deux courbes quelconques du second degré, au moins en général; car, pour des positions particulières de ces courbes, la projection peut devenir imaginaire, impossible geométriquement [the two circles (C) et (C′) can be considered as the projection of two arbitrary curves of second degree, at least in general, 123 A. Del Centina C A B C A B R R t t γ A B C P P t γ R (a) (b) Fig. 12 Poncelet first proved Proposition 1 for two concentric circles (a), then, by applying the principle of continuity, he extended the proposition to the case of two bitangent conics (b) because, for particular positions of these curves, the projection may become imaginary, geometrically impossible].
2.3 The synthetical proof of PGT Poncelet presented the proof of PCT in Art. 534 of his treatise, as a corollary of the PGT. Much of the proof of this theorem rests on the following proposition that he inserted in Art. 53131: Proposition (Main lemma) Let (c), (c′), (c′′) be three circles having a common chord mn, real or ideal (i.e., from the same pencil F. Let ABC be a triangle inscribed in (c′′) whose sides AB and AC are tangent, respectively, to (c′) and (c). If we move A on (c′′) in such a way the sides AB and AC remain tangent, respectively, to (c′) and (c), then the third side BC of the triangle will envelop a circle (c′′′) having the same common chord with the given circles (i.e., from the pencil F) As we have seen, Poncelet had already produced an analytical proof of the first part of this proposition in the notebooks of Saratov. In the Traité, in accordance with the spirit of the whole book, he wanted to give a synthetical proof.
Before we proceed to discuss this proof, it is useful to briefly review what Poncelet had shown in the previous Art.s 431–439 of his treatise.
In Art. 431, he stated the following: Proposition 1 Let ABC be an inscribed triangle to a given conic γ . If C moves along γ in such a way that the triangle remains inscribed in γ and the sides C A and C B rotate, respectively, around two fixed points P and P′, arbitrarily chosen on them, then the side AB will envelop a conic γ ′ which is bitangent to γ at the two points (real or ideal) where the line P, P′ meets γ . Moreover, if t is the point where AB touches γ ′, then P B and P′A met in a point R on Ct (see Fig. 12b).
31 Here and in the following we maintain Poncelet’s notation.
123 Poncelet’s porism, I To prove it, Poncelet first considered (Art. 433) the simpler case of concentric circles. Let (C) and (C′) be concentric circles with center in O, the second lying inside the first (Fig. 12a). Suppose ABC is a triangle inscribed in (C) with the side AB touching (C′) at t. If C moves along (C), in such a way that the sides CA and CB remain parallel to the original direction, then the side AB rotates around (C′). This is evident, claimed Poncelet, because all the chords AB have the same length and so have the same distance from O. Moreover, considering the parallelogram ABCR with AR//BC and BR//AC, it follows that the line Ct passes through R.
To complete the proof Poncelet argued as follows. By theorem B, the conic γ and the line through P, P′ are the projective image of a circle and of the line at infinity. In this way, the present configuration (Fig. 12b) is the projective image of the previous one (Fig. 12a), and then, the claim follows from what has been proved above for two concentric circles.
Unfortunately, the assumption that P, P′ are real points is not correct, in fact they are the images of the circular points, and, as known, under a projective map, either all the real points of a line have real images, or at most two of them have real images. So in general P and P′ will not be real (see Bos et al. 1987, p. 308).32 In Art. 434, Poncelet stated the reciprocal of the proposition above33: Proposition 2 Let ABC be a triangle inscribed in a conic γ . If C moves along γ so that the triangle remains inscribed in γ , the side AB moves remaining tangent to a same conic γ ′ having a double contact with γ along the direction T T ′, and the side AC rotates around a fixed point P placed on T T ′, then the third side will rotate around another point P′ placed on T T ′.
Then, he observed (Art. 437) that the conic γ , enveloped by the side AB, reduces to the point O, pole of the line P P′, when the points P and P′ are placed so that the polar line of one of them passes through the other.
By applying propositions 1 and 2, in Art. 439 Poncelet proved the following gen-eralization: Proposition 3 If a triangle inscribed in a conic moves, remaining inscribed in it, so that a first side passes through a fixed point, and a second side envelops a conic having a double contact with the first, then the third side will envelope another conic having double contact with the first.
We will return on these propositions later on, when we will discuss certain gener-alizations due to A. Cayley.
Now let see how Poncelet proved the main lemma.
First of all he found the point A′ at which C B touches the enveloped curve. He supposed the given circles have a real common chord M N on the line mn (see Fig. 13).
32 According to Bos et al. (1987, p. 308), it is possible to prove this part of the theorem along the lines implicit in Poncelet’s approach, by using cross-ratios to generalize the concept of parallelism and angles from the real to the complex case. This procedure, however, being rather laborious, these authors offered, in section 8.5 and specifically lemma 8.5 of their work, a modern alternative proof by means of closed conditions on Zariski-dense sets.
33 This proposition will be useful in section five.
123 A. Del Centina (c ) (c) (c ) A B C F A M N m n γ G H C B Fig. 13 Illustration of how Poncelet proceeded in order to prove the main lemma He observed that an infinitesimal displacement of the triangle ABC causes an infini-tesimal displacement of its sides, which, he affirmed, may be considered the same as the displacement that would occur if the points of contact B′ and C′ of the chords AB and AC with (c) and (c′) were fixed and the chords rotate around them.34 Then, by what he had observed above, it follows that the chord BC envelopes a conic γ having a double contact with (c′′). Hence, the point A′, where BC touches γ , is the same point where it touches the curve unknown (c′′′). Therefore, he concluded, if D is the point of intersection of BB′ and CC′, the line AD will intersect AB in the point A′.
At this point, Poncelet considered the intersection points F, G, H of the line mn, respectively, with the lines AC, AB and BC. From well-known properties of the circle, he deduced that: FB′2 = F M · F N = F A · FC, GC′2 = GM · GN = G A · GB.
From these relations, by applying what he had already shown in Art.s 162–163, Pon-celet deduced that H A′2 = H B · HC = H M · H N holds true. Finally, he remarked that this relation characterizes the circles passing through M and N. Thus, he had proved the proposition in case of circles having real intersections.35 34 Poncelet gave no further argument. His reasoning can be made rigorous by using the modern theory of deformation (see Bos et al. 1987, section 8).
35 For a deeper analysis of the arguments exposed in this subsection, we refer to Bos et al. (1987 sections.
4, 8).
123 Poncelet’s porism, I For extending the proof to circles in any position, Poncelet used the principle of continuity as follows: Ce raisonnement suppose, il est vair, que les cercles (c), (c′) et (c′′) aient deux points communs réels; mais, en vertu du principe de continuité, on peut l’étendre directement à celui où ces poins deviennent imaginaires, et où par conséquent la droit mn est une sécante idéale commune aux cercles proposés. Ainsi notre théorème est général et comprend tous les cas... [This reasoning supposes, of course, that the circles (c), (c′) and (c′′) have a real common chord; but, by the principle of continuity, it can be directly extended to the case in which the points of intersections are all imaginary, and the line mn is an ideal secant of the given circles. So our theorem is general and includes all cases...] Poncelet proved the PGT, in Art. 534, and the PCT, in Art.s 565–567, in the same way he had done in the Cahiers de Saratoff.
The work of Poncelet was appreciated by Dupin, Hachette, Malus and others former students at École Polytechnique, and partly also by Gergonne, Cauchy and Chasles.
But it was in Germany that Poncelet had major followers: Plücker, Steiner and Von Staudt.
In 1834, Poncelet was elected to the Académie des sciences. In the following years, he virtually abandoned geometry in favor of more experimental studies, in particular mechanics, a discipline in which he pointed out the central role of geometry.
The year after his death the Poncelet Prize was established. It was to be awarded by the French Academy of Sciences for the advancement of the science. Darboux and Halphen were among the first recipient of the prize.
3 Jacobi and the use of the elliptic functions In the extended Note historique in Poncelet (1862, pp. 480–498), Poncelet traced the history of the research on polygons inter-scribed to two conics in the 40years since 1822. He recalled that Carl Gustav Jacob Jacobi went to Paris in 1829, shortly after the publication of Jacobi (1828) and that he had several meetings with him, during which they exchanged ideas on that subject. At page 485, we read: M. Jacobi... m’apprit qu’au début de ses études sur ce suject, il avait aussi imag-iné de faire varier un tel polygone, d’une quantité infiniment petite, de manière que les arcs élémentaires décrits respectivement par les extrémités de l’un quel-conquedesescôtes,divisésparlalongueurdessegmentscorrespondantesformés sur sa direction, à partir du point de contact avec le cercle auquel il est tan-gent, représentaient autant de différentielles elliptiques de la première espèce, et fournissaient ainsi, par leur comparaison relative à chacun des côtés, autant d’équations distinctes, les mêmes qu’Euler Lagrange et Legendre avaient primi-tivement intégrées sous une forme rationelle, dans leurs mirables recherches sur la matière [Mr. Jacobi told me that at the beginning of his studies on this subject, he had also thought to move such a polygon by an infinitesimal displacement, in such a way that the elementary arches described by the extremities of each of 123 A. Del Centina P c C Q Q P T (C) (c) P c C Q Q P T 2ϕ (C) (c) O 2ϕ (a) (b) Fig. 14 Two steps of Jacobi’s thought on the use elliptic functions in the proof of Poncelet’s theorem in case of two circles, one inside the other. An infinitesimal displacement of the chord QP gives rise to an elliptic differential of the first kind. This idea was confirmed in the letter he wrote to Hermite on the 6th of August 1845 its sides, divided by the length of the corresponding segments along its direc-tion, represent as many elliptic differentials of first kind, which, by their relative comparison with each sides, give as many distinct equations, equal to those that Euler, Lagrange, and Legendre had previously integrated in rational form...].
All this seems to be confirmed by what Jacobi wrote to Hermite in a letter dated 6 of August 1845 (Jacobi 1846, pp. 178–179) (see Fig. 14a): Je suis aussi parvenu à étendre au théoréme d’Abel ma construction de l’addition des fonctions elliptiques. Dans cette dernière, la corde P Q d’un cercle touche constamment un autre cercle. Soit T le point d’intersection de deux positions consécutives de la droite; les deux angles Q′QT at P P′T étant égaux d’après une propriété du cercle, on aura P P′/PT = QQ′/Q′T ce qui est l’équation differentielle, dont par la construction de la droite inscrite à l’un et circonscrite à l’autre cercle on trouve l’intégtrale complète et algébrique, la même qui à été donnée par Euler [I have been also able to extend my construction for the addition of elliptic functions to the theorem of Abel. In this latter case, the chord P Q of a circle constantly touch another circle. Let T be the point of intersection of two consecutive positions of the line; the two angles Q′QT and P P′T being equal by a property of the circle, one has P P′/PT = QQ′/Q′T which is the differential equation, for which, by the construction of the line inscribed in one and circumscribed about the other circle, one finds the complete integral to be algebraic, the same that was given by Euler].
It is likely that Jacobi argued as follows (see Fig. 14b). The triangles PT P′ and QT Q′ are similar. So setting δ := PT , ds := P P′ and δ′ = P′T , ds′ = QQ′, it follows that: ds δ = ds′ δ′ = 2Rdϕ δ = 2Rdϕ′ δ′ , 123 Poncelet’s porism, I where 2ϕ = ̸ OC P and 2ϕ′ = ̸ OC P′. If a denotes the distance between the centers C and c, one has δ2 = Pc2 −r2. Therefore, since Pc2 = R2 + a2 + 2Ra cos 2ϕ, one gets δ2 = (R + a)2 −r2 −4Ra sin2 ϕ, and hence s 0 ds δ = ϕ 0 2Rdϕ (R + a)2 −r2 −4Ra sin2 ϕ · Putting k2 = 4Ra((R + a)2 −r2)−1 2 , this integral can be written 2R (R + a)2 −r2 ϕ 0 dϕ 1 −k2 sin2 ϕ which is readily seen to be an elliptic integral of the first kind, of modulus k (see Legendre 1825).
Summing up: Jacobi was lead to consider the elliptic function amplitude by looking at “the rolling” of a side of the polygon on the inner circle.
From the historical note of Poncelet, we also learn that Steiner suggested to Jacobi the use of elliptic functions, in fact at page 481 we read: D’après ce qu’a bien voulu me faire savoir plus tard M. Steiner,... ce serait par ses encouragements propres, ses avis éclairés que Jacobi, ayant pris connaissance du Traité des propriété projectives des figures, aurait été conduit à appliquer la théorie des fonctions elliptiques à démonstration des théorèmes (p. 322 et suiv. de cet ouvrage) sur les polygones simultanément inscrits et circonscrits à plusieurs cercles [Later Mr. Steiner kindly informed me that it was through his enlightened advice and encouragement that Jacobi, when he learned of the Traité des propriété projectives des figures, was led to apply the theory of elliptic functions in order to prove theorems (p. 322 and following of this work) on polygons simultaneously inscribed and circumscribed about several circles].36 3.1 Jacobi’s proof of PGT for circles Let us see, in some detail, how Jacobi proved the general theorem of Poncelet. He considered two circles (C), (c), the second within the first, respectively, of centers C, c and radii R, r, and a polygonal line (or transversal) AA′A′′A′′′ . . . inter-scribed to them (see Fig. 15). He put ̸ PC A = 2ϕ, ̸ PC A′ = 2ϕ′, ̸ PC A′′ = 2ϕ′′, etc. and denoted by a the distance between the two centers.
36 From November 1822 to August 1824, Steiner attended courses at the University of Berlin. Jacobi, 8years younger than Steiner, was at that time also a student in Berlin and soon they became friends.
123 A. Del Centina A A A A (C) (c) P c C t 2ϕ 2ϕ H K Fig. 15 Jacobi’s procedure in order to prove PGT for circles Then, one has the following equations ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ R cos(ϕ′ −ϕ) + a cos(ϕ′ −ϕ) = r R cos(ϕ′′ −ϕ′) + a cos(ϕ′′ −ϕ′) = r, R cos(ϕ′′′ −ϕ′′) + a cos(ϕ′′′ −ϕ′′) = r, .
.
.
which can be put in the form ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ (R + a) cos ϕ′ cos ϕ + (R −a) sin ϕ′ sin ϕ = r, (R + a) cos ϕ′′ cos ϕ′ + (R −a) sin ϕ′′ sin ϕ′ = r, (R + a) cos ϕ′′′ cos ϕ′′ + (R −a) sin ϕ′′′ sin ϕ′′ = r, .
.
.
(3.1) By subtracting each of these equations from the following one, since cos x−cos y sin y−sin x = tan x+y 2 , he got the following system of equations ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ tan ϕ′′+ϕ 2 = R−a R+a tan ϕ′, tan ϕ′′′+ϕ′ 2 = R−a R+a tan ϕ′′, .
.
.
about which, at page 35 of his paper, Jacobi wrote: In dieser Form der Gleichungen springt es sogleich in die Augen, dass sie mit denjenigen übersinkmmen, welche sur Vervielfachung der elliptischen Tran-scendenten aufgestellt werden [In this form, it is plain to see that these equations are the same as those for the multiplication of elliptic transcendentals].
123 Poncelet’s porism, I With this in mind, he considered the elliptic integral of first kind u = F(ϕ) = ϕ 0 dx 1 −k2 sin2 x , (3.2) with its inverse function (amplitude) ϕ = am(u).
Then, chosen any angle α := am(t), he put ϕ′ = am(u + t), ϕ′′ = am(u + 2t), and from the basic results on elliptic functions [see for instance (Legendre 1825, pp.
19–25)] he deduced tan ϕ′′ + ϕ 2 = am(t) tan ϕ′, (3.3) where am(t) := 1 −k2 sin2 α.
ThiscorrelationsuggestedtoJacobithepossibilityofdeterminingk in (3.2)insucha way that the successive values of ϕ, ϕ′, ϕ′′ . . . correspond to the vertices A, A′, A′′, . . .
of the polygonal line. He proceeded as follows.
From another basic relation in the theory of elliptic integrals (Legendre 1825, p. 19), for ϕ = am(u), ϕ′ = am(u + t) and α = am(t), one has cos ϕ cos ϕ′ + sin ϕ sin ϕ′am(t) = cos α, and since equation (3.1) can be put in the form cos ϕ cos ϕ′ + sin ϕ sin ϕ′ R −a R + a = r R + a , by comparing the last two equations he obtained cos α = r R + a , and 1 −k2 sin2 α = (R −a)2 (R + a)2 , which yield k2 = 4Ra (R + a)2 −r2 , and consequently R = r(1 + am(t)) 2 cos α , a = r(1 −am(t)) 1 + am(t) , r = 2R cos α 1 + am(t)· Jacobi observed that the quantities k and α do not depend on ϕ and u. This is very important, as Jacobi remarked, because in this way, starting from any point A on (C) and ̸ AC P/2 = ϕ = am(u), ̸ A′C P/2 = ϕ′ = am(u + t), the line AA′ is tangent to the circle determined by a and r as above. Moreover, the line AA′′ is tangent to the circle determined by 123 A. Del Centina a = R 1 −(2) 1 + (2) , r = 2R cos α(2) 1 + (2) , where α(2) = am(2t), (2) = 1 −k2 sin2 α(2), and in general the line AA(n), which closes the polygonal line, will be tangent to a circle determined by a = R 1 −(n) 1 + (n) , r = 2R cos α(n) 1 + (n) , where α(n) = am(nt), (n) = 1 −k2 sin2 α(n).
Clearly, all these circles have centers on the line C P.
Here, Jacobi made the crucial remark: all the circles belong to the same pencil determined by (C) and (c), and all lead to the same modulus k: Wir wollen jetzt beweisen, dass diese Kreise ein System bilden, welche dieselbe Linie zum Orte der gleichen Tangenten haben, welche zweckmässige Benennung Herr Steiner in einen geometrischen Arbeiten in diesem Journal eingführt hat [Let us prove that all these circles form a system of circles having the same line as locus of equal tangents, a suitable name introduced by Mr. Steiner in a geometrical paper in this journal].37 To prove this, he reasoned in the following way (see Fig. 16).
For a point Q on the line C P let d = QC, so that its distance from c is d −a.
Then, the tangential distances of Q from the two circles (C) and (c), are, respectively, √ d2 −r2 and (d −a)2 −r2). Comparing these, he deduced d = (R + a)2 −r2 2a −R, and, since (R + a)2 −r2 2a = 2R k2 , he finally found d = 2R k2 −R.
37 See Steiner’s paper Einige geometrische Betrachtungen (J. für die reihe und Ang. Math., 1 (1826), 161– 184), p. 165. This line is the radical axis of the pencil. As already seen, for a system of circles to have the same line of equal distances or to have a common (real or imaginary) chord is the same they must belong to a same pencil. See the note at p. 386 in Jacobi (1828).
123 Poncelet’s porism, I A A A A(n) (C) (c) Q P C Fig. 16 In his construction, Jacobi found a series of circles that he proved belonging to a same pencil. To prove this he showed, following Steiner, that all these circles have the same line as locus of equal tangents Then, Jacobi wrote: Wir sehen dass α in dem Ausdruck für d gar nicht vorkommt, sondern dass es bloss von k abhängt. Für alle jene Kreise aber ist dieses k dasselbe, und nur im α unterscheiden sie sich. Hätten wir daher für C [(C)] und irgend einen anderen Kreis den Ort ihrer gleichen Tangenten gesucht, se hätten wir denselben Aus-druck für D [d] gefunden, so dass also alle jene Kreise einen gemeinschaftlichen Ort der gleichen Tangenten haben [We see that α does not appear in the expres-sion for d, which depends only on k. For all these circles, k is the same, and it changes only for α. Thus, if we had searched the locus of equal tangents for (C) and another circle, we will have found the same expression for d; hence, all these circles have the same locus of equal tangents].38 The perpendicular to C P passing through the point Q, whose distance from C is as above, is the so-called line of “equal tangents.” Jacobi was now in the position to prove the PGT.
He considered a sequence (c),(c(1)), . . . , (c(n− 1)) of circles, with centers c,c(1), . . . , c(n−1) and radii r,r(1), . . . ,r(n−1), all lying within (C) and all belonging to the same pencil together with (C). He put a(i), i = 1, . . . , n −1, the distance of c(i) from C, and determined the angles α’s by cos α = r R + a , cos αi = r(i) R + a(i) , i = 1, . . . , n −1.
Jacobi remarked that k2 is the same for all inner circles, and so that the same function “am” occurs for each of them. Then, he found t, t(1), t(2), . . . by putting α = am(t), 38 One may argue as follows. It can be supposed (C), D1, D2 be given, respectively, by x2 + y2 = R2, (x −a1)2 + y2 = r2 1 and (x −a2)2 + y2 = r2 2. Then, the abscissas of the points of intersection are xi = (r2 i −R2 −a2 i )/2ai, and for circles belonging to the same pencil we must have x1 = x2 = x. From k2 = 2R R−r2−R2−a2 2a , it follows that k2 = 2R R−x which proves the claim.
123 A. Del Centina αi = am(t(i)) for i = 1, . . . , n −1. From a point A on (C), he drew the tangent AA(1) to the circle (c) (proceeding counterclockwise), then he drew the tangent A(1)A(2) to (c(1)), and proceeded similarly until he had drawn the tangent A(n−1)A(n) to (c(n−1)).
Putting ̸ AC P = 2ϕ, ̸ A(i)C P = 2ϕ(i), i = 1, . . . , n, he obtained ϕ = am(u), ϕ(i) = am(u + t + · · · + t(i−1)), i = 1, . . . , n.
Then, Jacobi observed that, with s := t +· · ·+t(n−1), the line A(n)A (which closes the transversal) envelops a circle, which is determined by r(n) = 2R cos am(s) 1 + am(s) , a(n) = R(1 −am(s) 1 + am(s) , being r(n) its radius and a(n) the distance of its center from C. This circle, being associated with the same modulus k, necessarily belongs to the same pencil that(C), (c), (c(1)), . . . , (c(n−1)) belong to.
3.2 Jacobi’s condition allowing inter-scribed n-gons to two circles By using his approach, Jacobi was able not only to prove PCT but also to find a necessaryandsufficientconditionthatallowstheexistenceofapolygon,ofanynumber n of sides, inter-scribed to two given circles(C) and (c), the second lying within the first. To do this, he proceeded as follows.
Suppose that starting from A0 on (C) the transversal A0, A(1) 0 , . . . , A(n) 0 , con-structed as above, closes, i.e., that A0 = A(n) 0 , after having turned i times around c. Let ̸ PC A0 = 2ϕ0 and ̸ PC A(n−1) 0 = 2ψ, then one has ψ = am(u + nc) = ϕ0 + iπ = am(u) + iπ.
Defined K := π/2 0 dx 1 −k2 sin2 x , one gets am(K) = π/2, am(u + 2K) = am(u) + π, and am(u + 2i K) = iπ + am(u) for any integer i. From above, it follows that am(u + nt) = am(u + 2i K), hence u + nt = u + 2i K, 123 Poncelet’s porism, I and then t = 2i K n · (3.4) This equation is clearly a necessary and sufficient condition on (C) and (c) for the existence of a n-gon inter-scribed to them. Since this condition does not depend on ϕ0, it follows that, if satisfied, the inter-scribed polygonal line AA(1)A(2) . . . always closes after n steps (and i turns), whatever is the point A from which one starts.
We stress that Jacobi proved PCT for a pair of circles in a particular position (one lying within the other), not for any pair of circles. Although he was aware of Poncelet’s projection methods and principle of continuity, he did not apply them to generalize his results to conics; possibly because he was not totally convinced of the validity of that principle.
Jacobi’s paper immediately attracted attention. Shortly after its publication, Legen-dre inserted Jacobi’s result as a section of the third volume of his Traité des fonctions elliptiques (Legendre 1828, pp. 174–180). Two years later Jacobi’s pupil F.J. Riche-lot,39 by applying Legendre’s duplication law for elliptic integrals, gave a recursive formula for determining the relation among the radii and the distance between the centers of two circles, allowing the existence of an inter-scribed 2n-gon, knowing the analogous relation for a n-gon. Defined p := R + a r , q := R −a r and denoted by R′,r′, a′, p′, q′ the analogous quantities for a 2n-gon, he found (Richelot 1830, p. 27): p = p′2 + p′2q′2 −q′2 p′2 −p′2q′2 + q′2 , q = q′2 + q′2 p′2 −p′2 q′2 −q′2 p′2 + p′2 · At the end of his paper, feeling his work was not fully completed, Jacobi wrote: Es dürfte nict ohne Interesse für die Theoria der elliptischen Functionen sein, ähnliche Betrachtungen unmittelbar für das System zweier Kegelschnitte anzustellen. Das Integral dürste dann in einer complicirteren Form erscheinen, die sichjedoch auf jene einfachere reduciren lassen muss. Vielleicht nehme ich später Gelagenheit, hieraus wieder zurückzukommen [It would be not without interestforthetheoryofellipticfunctions,tomakesimilarconsiderationsdirectly for a system of two conics. The integral may appear in a more complicated form, which must, however, be reduced to the simplest form found above. I will return on this subject on another occasion] but, to judge by what appears in his published works, he did not do so. Jacobi’s pro-gram was carried out by Nicola Trudi and, independently, by Arthur Cayley, 25years later.
39 Friedrich Julius Richelot (1808–1875), was a student of Jacobi at Könisberg. He graduated in 1831 with a thesis on the subdivision of the circle in 257 equal parts. In 1844, he succeeded to Jacobi at the University of Könisberg.
123 A. Del Centina 4 Trudi: the forgotten work In 1839, Nicola Trudi,40 stimulated by his teacher Vincenzo Flauti, became interested in questions related to the existence of polygons inscribed in, or circumscribed about, a conic and satisfying given conditions. Two years later he gathered the results of his studies in his extended memoir (Trudi 1841). The last part of the memoir was devoted to finding the algebraic relation, among the coefficients of two conics, which guaran-tees the existence of an in-and-circumscribed triangle to them. This work, published at Flauti’s expense, remained unknown outside the borders of the Kingdom of the Two Sicilies.41 In the spring of 1843, Trudi read a paper on this subject at the Royal Academy of Naples, but only a four-page summary, written in third person, was published in the Rendiconti of the Academy (Trudi 1843). This summary clearly indicates that: (1) he solved the question for the triangle; (2) he trusted his method in order to find the conditional relation for the existence of an inter-scribed polygon of any number of sides and that only the difficulties encountered with a problem of elimination led him to temporarily abandon the general question; (3) he had been induced to deal with that question as a continuation of the first of three Flauti’s research proposals; (4) he had been hitherto unaware of the interest raised by this topic, because of his late discovery of Jacobi’s paper.42 We extract from (Trudi 1843, p. 93) what follows: In ultimo ei ritorna al caso generale delle sezioni coniche per mostrare come questo metodo si applichi alla ricerca delle relazioni pei poligoni di qualsivoglia numero di lati iscrittibili tra esse: metodo che vincendo tutta la difficoltà che cir-conda la quistione, di cui trattasi, finisce per non recare altra pena, che quella di scrivere le formule corrispondenti. Intanto, mosso il Trudi dalle savie indicazioni dell’illustre Jacobi,43 promette di ritornare su questo argomento, per guardare la questione sott’altro punto di vista, e propriamente in rapporto all’utile, che può trarsene nella teorica delle funzioni ellittiche [Finally he returns to the general case of conic sections, to show how this method is applicable in order to find the relations for polygons of any number of sides to be inter-scribed to them: the method, which overcomes all the difficulty which surround the question that we are concerned with, gives without further effort the corresponding formu-lae. Trudi, following the wise guidance of Jacobi,44 promises to return to this 40 Nicola Trudi (1811–1884), was born in Campobasso and studied in the University of Naples. In 1851, he became professor of infinitesimal calculus in that University and then member of the local Royal Academy of Sciences. He contributed to the theory of elliptic functions, and to the theory of determinants with the publication of the treatise Teoria dei determinanti e loro applicazioni, Napoli 1862. For a biographical note and information on his scientific production (see Amodeo 1924, part two, pp. 190–213).
41 This memoir of Trudi, with three others by him, appeared at the end of Part II of (Flauti 1840–1844).
42 This is to say that Trudi read (Jacobi 1828) with great delate.
43 At page 90, Trudi explicitly refer to the phrase “ Es dürfte nicht onhe Interesse...” that we have transcribed from (Jacobi 1828) at the end of the previous section.
44 Idem.
123 Poncelet’s porism, I argument, looking at the question from another point of view, and properly in relation to the advantage which may result to the theory of elliptic functions.] Years later, having overcome all these difficulties, Trudi published the paper (Trudi 1853) answering to the general question, and completing Jacobi’s program. Unfortu-nately, this work also remained almost unknown outside the Neapolitan milieu.
4.1 Trudi’s first approach of 1841 In section 59 of his paper, Trudi raised the question of the construction of a triangle which is inscribed in a conic C and circumscribed about another conic C′. He observed that, if one takes as axis x a diameter of C, and for axis y the tangent to C at one of its intersection points with the chosen diameter, then C can be represented by the equation y2 = m2x2 + 2nx.
Moreover, he put Ay2 + 2Bxy + 2Cy + Dx2 + 2Ex + F = 0, the equation of C′. For (z, v) and (z′, v′) (general) points on C, he set r := v/z and r′ := v′/z′. With respect to these parameters, the chord of C joining the two points has equation y(r + r′) −x(rr′ + m) = 2n.
He also observed that the line y = ax + b is tangent to C′, if and only if, A′ + a2B′ + 2bC + 2abD′ + 2aE′ + b2F′ = 0 (4.1) where A′ = E2 −DF, B′ = C2 −AF, C′ = BE −C D D′ = AE −BC, E′ = C E −BF, F′ = B2 −AD.
If RR′R′′ is a triangle inscribed in C and circumscribed about C′, then the three sides RR′, R′R′′ and R′′R have, respectively, equation y(r + r′) −x(rr′ + m) = 2n, y(r′ + r′′) −x(r′r′′ + m) = 2n, y(r′′ + r) −x(r′′r + m) = 2n.
So, taking into account (4.1), Trudi was led to the following equations, which link the parameters r,r′,r′′ and the coefficients of C and C′: 123 A. Del Centina A′(r + r′)2 + B′(rr′ + m)2 + 4mC′(r + r′) + 4nD′(rr′ + m) + 2E′(r + r′)(rr′ + m) + 4n2F′ = 0, A′(r′ + r′′)2 + B′(r′r′′ + m)2 + 4mC′(r′ + r′) + 4nD′(r′r′′ + m) + 2E′(r′ + r′′)(r′r′′ + m) + 4n2F′ = 0, A′(r′′ + r)2 + B′(r′′r + m)2 + 4mC′(r′′ + r) + 4nD′(r′′r + m) + 2E′(r′′ + r)(rr′ + m) + 4n2F′ = 0.
By eliminating r,r′ and r′′ among them, he got the equation A′2 + 2mB′A′ + 4nA′D′ −8nC′E′ −4mE′2 + 4mnD′B′ +4n2B′F′ + m2B′2 = 0 (4.2) which expresses a necessary condition for the existence of an inter-scribed triangle to C and C′.
Trudi remarked that, as this equation does not depend on the parameter of the initial point, if such a triangle exists, then infinitely many others exist. In case C and C′ are circles, he also observed that his result leads to formula (1.1), that he attributed to Lhuilier.
At this point, Trudi tried to apply his method to the case of an inter-scribed quad-rangle, but the exceedingly long computation forced him to limit himself to the case of circles, that he represented by the equations y2 + x2 −2nx = 0, y2 + x2 −2ax + a2 −n′2 = 0, the first with center in (n, 0) and radius n, the second with center in (a, 0) and radius n′. In this case, he found that for the existence of an inter-scribed quadrangle to the two given circles the equation n′4 = (n′2 −a2)[n′2 −(2n −a)2] must hold. Trudi unaware of Fuss (1797) did not compare his formula with that of Fuss.
Nevertheless, an easy computation shows that the two formulae are in fact identical.
Fromafootnoteonpage97,welearnthatTruditriedtoapplyhismethodtopolygons with n ≥3 sides, but that he was unable, using standard procedures, to eliminate the n parameters among the n equations that can be deduced in this case. He expressed the desire to return on this subject as being of great importance.
Trudi kept his promise. In the spring of 1843 at the Royal Academy of Naples, he read the new memoir Delle relazioni fra i determinanti di due sezioni coniche l’una iscritta l’altra circoscritta ad un poligono irregolare [On the relations between the determinants (coefficients) of two conic sections, one inscribed and the other circumscribed to an irregular polygon]. As we have already said, a four-page summary of it was published in the reports of the Academy (Trudi 1843). Here, besides relation 4.2, the conditional equations for the existence of inter-scribed polygons to two circles, of 3, 6, 12, 24 sides and of 4, 8, 16, 32 sides are published. Moreover, since the volume of Crelle’s Journal for the year 1828, containing Jacobi’s famous paper, had recently arrived in Naples, Trudi was able to compare his formulae with those of Fuss (Trudi 123 Poncelet’s porism, I 1843, p. 90).45 Finally, he returned to the general case of two conics and showed how his method could be applied to any polygons.
As we will see below, it was through the relation 4.2 that Trudi perceived a link between the existence of inter-scribed polygons and the complete integral of Euler’s differential equation dx αx4 + βx3 + γ x2 + δx + ε = dy αy4 + βy3 + γ y2 + δy + ε .
Later in 1843, Jacobi went to Italy. He joined Steiner in Rome, and in April of 1844 they both visited Naples. Trudi had the opportunity to meet them and to talk with Jacobi about his studies on Poncelet’s theorem.46 According to Trudi, Jacobi manifested real interest in his results and encouraged him to pursue research in this field (Trudi 1863a, p. 4).47 In 1845, the seventh Congress of the Italian Scientists was held in Naples. On this occasion, Trudi read a paper titled Sull’eliminazione fra le equazioni algebriche ese-guita per mezzo della differenziazione e della integrazione [On the elimination among algebraic equations by means of differentiation and integration].48 In particular, Trudi announced to have easily deduced many theorems of Poncelet and got the relations among the coefficients of the two conics as requested by Jacobi. Trudi felt to be on the right path, and he wanted to further pursue his studies.
In 1853, at the Academy of Naples he presented his most important paper Su una rappresentazione geometrica immediata dell’equazione fondamentale nella teorica delle equazioni ellittiche On an immediate geometrical representation of the funda-mentalequationinthetheoryofellipticfunctions,printedonlyin1856.49 In a footnote on page 65, added when the paper went to press, Trudi wrote that he had just been informed by Joseph Sylvester that, in 1853, Arthur Cayley had published some notes on the same subject.50 Trudi specified that he had not yet had the opportu-nity to read them, because the Philosophical Magazine—the journal in which Cayley published his notes—was not among those available in the Royal Library of Naples.
4.2 The important paper of 1853 In (1853) Trudi solved, through geometrical constructions, the question of addition and multiplication of elliptic integrals of the form 45 Probably Trudi knew of the formulae given by Fuss only through the paper of Jacobi.
46 As recorded in the Rendiconto delle adunanze e dei lavori della reale Accademia delle Scienze di Napoli, 3 (1844) pp. 196–197, Jacobi and Steiner attended at the meeting of this Academy for the 23 and 30 of April 1844. At the first meeting, Trudi read a memoir on a problem of elimination among algebraic equations in several variables by means of differentiation and integration.
47 Here Trudi confused the year in which Jacobi visited Naples: he wrote “1845” instead of “1844.” 48 The proceedings of the Congress were published in 1846, we refer to as (Congresso Scienziati 1845).
49 The first volume of the Memorie della Reale Accademia delle Scienze di Napoli includes all papers presented to the Academy during the years 1852–1854.
50 Sylvester visited Naples in February of 1856 (see Hunger 2006b, p. 110).
123 A. Del Centina T O U S S U1 U2 x y Fig. 17 Description of how Trudi proceeded in order to prove his Theorem T1 du √ψ(u), where ψ(u) is a polynomial of degree four. Then, he applied the derived theorems and formulae to find the necessary and sufficient conditions under which two conics admit an inter-scribed n-gon.
First of all, Trudi proved the following (pp. 66–67).
Theorem T1 Let S, S′ be two conics and UU1 any chord of S tangent to S′. Let x be a diameter of S and O one its intersection points with S, and define u := tan U Ox, u1 := tan U1Ox. Then, u and u1 satisfy an equation of the form Au2u2 1 + 2B(u + u1)uu1 + C(u + u1)2 + 2D(u + u1) + 2Euu1 + F = 0 where A, B, C, D, E, F are constants depending on the minors of order 2 of the matrices of the two conics.51 Trudi chose as x-axis the line x and as y-axis the tangent to S at O (Fig. 17), so that he represented the two conics S, S′, respectively, by y2 = 2rx + mx2, ay2 + 2bxy + cx2 + 2dy + 2ex + f = 0.
Under this assumption, if U = (x, y) and U1 = (x1, y1), is u = x/y and u1 = x1/y1, and, since both U,U1 are on S, it follows x = 2r u2−m , x1 = 2r u2 1−m y = 2ru u2−m , y1 = 2ru1 u2 1−m (4.3) 51 Trudi’s statement says “costanti dipendenti dai determinanti delle due coniche” [constants depending on the determinants of the two conics].
123 Poncelet’s porism, I He chose X, Y for the coordinates on the line UU1 and wrote its equation in the form (Y −y)(x −x1) = (X −x)(y −y1), or, what is the same, (u −u1)Y = (uu1 + m)X + 2r.
Then, a line Y = pX + q will touch S′ if (and only if) (d2−af )p2+2(de−bf )p+(e2+cf )+2(ae−bd)pq+(b2−ac)q2+2(be−cd)q =0 holds true. Since p = uu1 + m u + u1 , q = 2r u + u1 , by substituting these expressions into the above relation, and setting A = d2 −af, D = m((dc −bf ) + 2r(bc −cd), B = de −bf, E = m(d2 −af ) + 2r(ae −bd), C = e2 −cf, F = m2(d2 −af ) + 4mr(ae −bd) + 4r2(b2 −ac), (4.4) he obtained the result.
Trudi also proved the converse of the above theorem (pp. 70–71), precisely: Theorem T2 Suppose that between two variables u and u1 there exists a relation of the form Au2u2 1 + 2B(u + u1)uu1 + C(u + u1)2 + 2D(u + u1) + 2Euu1 + F = 0, (4.5) and that a conic S of equation y2 = 2rx + mx2 is given. Then, a unique conic S′ can be found such that, for any chord UU1 of the first conic that touch the second, the trigonometric tangents tan U Ox and tan U1Ox always satisfy the relation (4.5).
In order to prove the theorem, Trudi proceeded as follows. He observed that, if such a conic S′ with equation ay2 + 2bxy + cx2 + 2dy + 2ex + f = 0 exists, then, setting u := tan U Ox andu1 := tan U1Ox asabove,fromtheoremT1itfollowsthatarelation similar to the (4.5), and whose coefficients are expressed by (4.3), holds true. There-fore, by comparing the coefficients of this relation with the given one, he expressed b, c, d, e, f in terms of A, B, C, D, E, F and a, so getting for S′ the equation (AF −E)2y2 + 2 [(BF −DE) + m(AD −BE)] xy + (C F −D2) + 2m(BD −C E) + m2(AC −B2) x2 + 4r(AD −BE)y + 4r (BD −C E) + m(AC −B2) x + 4r2(AC −B2) = 0.
(4.6) 123 A. Del Centina He also remarked that S′ is the envelope of all chords of S whose elements u, u1 satisfy the relation of theorem T1.
Then, Trudi proved the following (pp. 72–73): Theorem T3 The differentials du and du1 satisfy a relation of the form du αu4 + βu3 + γ u2 + δu + ε = ± du1 αu4 1 + βu3 1 + γ u2 1 + δu1 + ε , where α, β, γ, δ, ε are constants depending solely on the coefficients in the equations of S and S′.
To prove this, he differentiated relation (4.5) with respect to u and u1, and squared the two members so obtained, and setting C + E = G he obtained the equation du2 ψ(u) = du2 1 ψ(u1), where ψ(u) = [(Au2 1 + 2Bu1 + c)u + Bu2 1 + Gu + D]2, ψ(u) = [(Au2 + 2Bu + c)u1 + Bu2 + Gu + D]2.
Taking the square roots, he wrote the two differential equations du √ψ(u) = ± du1 √ψ(u1)· From here, with some computation, and setting ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ α = B2 −AC β = 2(BE −AD) γ = E2 −AF + 2(C E −BD) δ = 2(DE −BF) ε = D2 −C F (4.7) he finally derived the theorem.
Trudi knew [see his note at p. 74, where he referred to Euler (1794)] that the differential equation dx αx4 + βx3 + γ x2 + δx + ε = dy αy4 + βy3 + γ y2 + δy + ε (4.8) always admits a complete integral of the form Ax2y2 + 2B(x + y)xy + C(x + y)2 + 2D(x + y) + 2Exy + F = 0 (4.9) 123 Poncelet’s porism, I where ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ A = 4α(K + γ ) −β2 B = 2αδ + Kβ C = 4αε −K 2 D = 2βε + δK E = βδ + 2(K + γ )K F = 4ε(K + γ ) −δ2 and K is an arbitrary constant. This means, Trudi remarked, that in theorem T2 instead of the relation (4.4) one can consider the differential equation (4.8), and so the equation (4.5) of S′ can be put in the form (K +γ )y2+(mβ+δ)xy+(m2α−mK +ε)x2+2rβy+2r(2mα−K)x+4rα2 =0.
Next, from his theorem 4 (p. 77), Trudi showed that the conic S, and all the conics described by the previous equations, belong to the same pencil (theorem 5, p. 78).52 He also brought to light a remarkable property of the relation (4.5), precisely he proved (see theorem 6, pp. 79–81): Theorem T4 By eliminating the variables u1, u2, . . . , un−1 in the following system of equations in n + 1 variables ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ Au2u2 1 + 2B(u + u1)uu1 + C(u + u1)2 +2D(u + u1) + 2Euu1 + F = 0, Au2 1u2 2 + 2B(u1 + u2)u1u2 + C(u1 + u2)2 +2D(u1 + u2) + 2Eu1u2 + F = 0, .
.
.
Au2 n−1u2 n + 2B(un−1 + un)un−1u2 + C(un−1 + un)2 +2D(un−1 + un) + 2Eun−1un + F = 0 (4.10) the resulting equation in u, un is still of the form (4.5).
He used Euler’s differential equation to perform the elimination. By applying to each equation the procedure used in the proof of theorem T3, he obtained du2 ψ(u) = du2 1 ψ(u1) = du2 2 ψ(u2) = · · · = du2 n ψ(un) where ψ(ui) = αu4 i + βu3 i + γ u2 i + δui + ε with α, β, γ, δ, ε given by (4.7). Hence, one has du2 ψ(u) = du2 n ψ(un).
52 Trudi wrote “hanno le stesse secanti comuni, reali o ideali (secondo la denominazione dell’illustre Poncelet” [they have the same real or ideal chords (according the definition of the illustrious Poncelet).
123 A. Del Centina Then, by integrating, he obtained an equation in u and un (necessarily) of the (required) form A′u2u2 n + 2B′(u + un)uun + C′(u + un)2 + 2D′(u + un) + 2E′uun + F′ = 0, (4.11) with ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ A′ = 4α(K + γ ) −β2 B′ = 2αδ + Kβ C′ = 4αε −K 2 D′ = 2βε + δK E′ = βδ + 2(K + γ )K F′ = 4ε(K + γ ) −δ2, (4.12) and where K, no longer an arbitrary constant, depends on the number of equations and their coefficients. Trudi remarked that the value of K can be found by equating the two values of un obtained, from one side, by setting u = 0 (or u = ∞) in the equations (4.10), and from the other, by setting u = 0 (or u = ∞) in the (4.11).
The coefficients of (4.11), Trudi also noticed, are in general different from the coefficients of the equations (4.10), but he proved (pp. 81–82) that: (⋆) if K = −2(C E −BD), then the coefficients A′, B′, C′, D′, E′, F′ are, respectively, equal (up to a constant) to the coefficients A, B, C, D, E, F.
In the last part of his memoir, Trudi gave geometrical and analytical applications of his results; among the first he proved Poncelet’s theorem and for the second, he only considered the addition and multiplication of elliptic functions.
He started by solving the following problem (p. 83): Let S and S’ be two non-singular conics, and inscribed in S any polygon whose sides, but one, are tangent to S’, find the curve enveloped by the free side.
He considered a polygon U,U1,U2, . . . ,Un inscribed in a conic S and whose sides are all tangent to another conic S′, except UnU. He observed that there is no loss of generality in supposing the two conics have equations as in theorem T4. Then, if u, u1, u2, . . . , un, defined as above, correspond to the vertices of the polygon, the conditions of tangency of the first n sides give n equations like (4.10), whose coeffi-cients are expressed by means of (4.3). So, by eliminating the intermediate variables u1, u2, . . . , un−1, he got, via theorem 4, that the equation satisfied by the parameters of the extremities of the free side of the polygon will be of the form A′u2u2 n + 2B′(u + un)uun + C′(u + un)2 + 2D′(u + un) + 2E′uun + F′ = 0.
Hence, he applied the corollary to theorem T2, in order to get the conic envelop of UnU, and found (K +γ )y2+(mβ+δ)xy+(m2α+ε−mK)x2+2rβy+2r(2mα−K)x+4r2α =0, where K is determined as prescribed in theorem T4.
123 Poncelet’s porism, I Then, the previous results, and in particular the formulae (4.7) and (4.4), together with some computation, allowed Trudi to write the equation of the enveloped conic in terms of the coefficients of S and S′, which resulted in (a + μ)y2 + 2bxy + (c −mμ)x2 + 2dy + 2(e −rμ)x + f = 0 (4.13) where μ = K + 2(C E −BD) 4r2 , and = −det(S′).
Trudi remarked that from equation (4.13) it follows that the enveloped conic of the free side meets the conics S and S′ exactly in their common points (p. 85), that is, he added, “le due coniche date e la conica inviluppo hanno le stesse secanti comuni (reali, o ideali)”[the two given conics and the conic envelop have the same common chords (real, or ideal)].
He had proved: The envelop of the free side of any polygon inscribed in a conic S, whose side are all but one tangent to another conic S’, is a conic I, belonging to the pencil determined by S and S’, whose equation is given by (4.13).
Moreover, from the same equation, it follows that U,U1,U2, . . . ,Un,U is circum-scribed about S′, if μ = 0, i.e., if K = −2(C E −BD), and, since he showed that the reciprocal also holds true, he had proved that K = −2(C E −BD) is a necessary and sufficient condition for the existence of a polygon of n sides which is inscribed in S and circumscribed about S′.
This, Trudi stressed (p. 86), gives a proof of the closure theorem of Poncelet,53 and provides the condition on the coefficients of S and S′ that allows the existence of a polygon of n + 1 sides inter-scribed to the two conics.
When the condition above is satisfied, by virtue of the remark (⋆), (4.10) becomes ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ Cu2 1 + 2Du1 + F = 0 Au2 1u2 2 + 2B(u1 + u2)u1u2 + C(u1 + u2)2 + 2D(u1 + u2) + 2Eu1u2 + F = 0, Au2 2u2 3 + 2B(u2 + u3)u1u2 + C(u2 + u3)2 + 2D(u2 + u3) + 2Eu2u3 + F = 0, .
.
.
Cu2 n + 2Dun + F = 0 (4.14) so the condition on the two conics S and S′ that allowing the existence of a n-gon inter-scribed to them, is the equation obtained by eliminating the n variables from the above system of n + 1 equations.
53 In fact, the condition he found does not depend on any of the vertices of the inscribed polygon.
123 A. Del Centina Summing up, Trudi had succeeded in completing Jacobi’s program.
At this point, in order to illustrate his method, Trudi considered in detail the cases n = 3, 4 and 5, for which he obtained explicit formulae.
In case n = 3, the system above reduces to ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ Cu2 1 + 2Du1 + F = 0, Au2 1u2 2 + 2B(u1 + u2)u1u2 + C(u1 + u2)2 + 2D(u1 + u2) + 2Eu1u2 + F = 0, Cu2 n + 2Dun + F = 0.
From the first and the third equation, it follows that u1u2 = F/C and u1 + u2 = −2D/C, so substituting these values in the second one gets AF + C2 + 2EC −4BD = 0 (4.15) which, by the (4.3), gives (4.1): a necessary and sufficient condition for the existence of an inter-scribed triangle expressed in terms of the 2 × 2 minors of the matrices of S and S′.
When the two conics are circles, then B = D = 0, and (4.15) reduces to AF + C2 + 2EC = 0, that is readily seen to be equivalent to condition (1.1).
For the sake of space, we omit here the other two cases. Suffice to say that Trudi’s formulae for circles are equivalent to those found by Fuss.
Ten years later, shortly after Poncelet had published the first volume of Applications d’analyse et de géométrie, Trudi returned to the subject with two memoirs that we will comment in section seven.
5 Cayley’s explicit conditions Arthur Cayley54 had become interested in elliptic functions early in the 1840s. He approached that theory through Jacobi’s Fundamenta Nova Theoriae Functionum Ellipticarum (1829), that he had mastered immediately after his degree (Cayley 1895, pp. xi–xii).55 The celebrated Mémoire sur une propriété générale d’une classe très étendue de fonctions transcendantes authored by Abel was published posthumously in 1841.
Here, he proved the so-called addition theorem for Abelian integrals that provided the basis for the development of the modern algebraic geometry. This memoir attracted Cayley’s attention quite early in his scientific career.
54 For a comprehensive biography of Arthur Cayley, we refer to Crilly (2006).
55 The prominent position occupied in the Fundamenta by the theory of transformation naturally attracted his interest. As early as 1844, he wrote short notes on the subject. Cayley always maintained Jacobi’s point of view, and he particularly appreciated the algebraic approach. When he published the treatise on elliptic functions (Cayley 1876), where the theory of transformation is discussed at considerable length, it was already an old-fashioned work (Crilly 2006, p. 337).
123 Poncelet’s porism, I An Abelian integral is an integral of type R(x, y)dx, where R(x, y) is a rational function of x, y, being y(x) the algebraic function defined by χ(x, y) = 0 with χ(x, y) an irreducible polynomial.
Abel’s addition theorem affirms that, if θ(x, y; t) = 0 is a family of plane curves, depending rationally upon the parameter t, and (xi(t), yi(t)), i = 1, . . . , N, are the intersection points of the curves of the family with the curve χ(x, y) = 0, then N 1 xi(t) x0 R(x, y(x))dx = V (t) + log W(t), where V (t) and W(t) are rational functions of the parameter t. In particular, if the integral is of the first kind, i.e., if its value remains finite when integration is carried out along any path from the initial point x0 to the final point xi, the above sum reduces to a constant [see for instance (Markushevich 1992; Del Centina 2003; Bottazzini and Gray 2013), and for a very careful analysis (Kleiman 2004)]. Since any elliptic integral is an Abelian integral, this theorem clearly generalizes Euler’s addition theorem, which, as well known, holds for elliptic integrals dx √ X , where X is a polynomial of degree 3 and 4 without multiple roots, i.e., the integral is an elliptic integral of the first kind (Euler 1768).
The geometrical significance of the theorem of Abel extends to that of Euler: for instance, if X has degree 3, then one has x1 x0 dx √ X + x2 x0 dx √ X + x3 x0 dx √ X = cost., for any triple (xi, yi(xi)), i = 1, 2, 3, of collinear points on the cubic curve y2 = X.
LessgeneralversionofAbel’sadditiontheoremhadalreadyappearedinAbel(1828, 1829), where he considered the particular case of hyperelliptic functions (χ(x, y) = y2 −X, with X a polynomial of degree ≥5).
Some years later Jacobi explained the theorem for hyperelliptic functions in his paper (Jacobi 1832) and reformulated it as follows (p. 396): let X be a polynomial in x of degree 2m or 2m −1, and set (x) := x 0 A + A1x + · · · + Akxk √ X dx, then, given m values x0, x1, . . . , xm−1 of the variable, m −1 quantities a0, a1, . . . , am−2 can be algebraically determined from these (they are roots of an algebraic 123 A. Del Centina equation of degree m −1, whose coefficients are rationally expressed in terms of x0, x1, . . . , xm−1 and √X(x0), . . . , X(xm−1)), such that (x0) + (x1) + · · · + (xm−1) = (a0) + (a1) + · · · + (am−2).
This geometrical feature of the addition theorem represented a powerful inspiration for Cayley in developing the program that Jacobi had drawn in 1828, i.e., to find the conditions on two conics U and V for the existence of an inter-scribed n-gon. In a series of notes, published in 1853 and in 1854, by using Abel’s theorem and the development in power series of √det(ξU + V ) as main tools, he proved the closure theorem of Poncelet and found the required conditions on U and V . At the end of the first note (Cayley 1853a), referring to Jacobi (1828), he wrote: The preceding investigations were, it is hardly necessary to remark, suggested by a well-known memoir of the late illustrious Jacobi,56 and contain, I think, the extension which he remarks it would be interesting to make of the principles in such memoir to a system of two conics.
Cayley proved PCT for triangles in the first two notes. In the third one, he extended the results to polygons of n > 3 sides, and in the fourth he gave explicit conditions when n = 4, 5.
A few year later he returned to the subject with two papers concerning the porism of an in-and-circumscribed triangle (Cayley 1857, 1858). The aim of the first was to extend what we called “main lemma,” from the case of three circles, to the case of a conic and two curves of higher degree, while the aim of the second was to give new proofs of Poncelet’s propositions 1–3 (here stated in subsection 2.4). We will comment these papers, which are of pure geometrical character, at the end of this section.
In (1861), Cayley republished the results that he had already achieved on polygons inter-scribed to two conics in the years 1853–1854, in a more organized and complete form. This paper (Cayley 1861) became the standard reference for Cayley’s discoveries in this field.
Four years later, in a short note, he considered the problem of a triangle inscribed in and circumscribed about a (real) quartic curve (Cayley 1865).
At the meeting of the Royal Society in Liverpool in 1870, Cayley reconsidered the problem of the in-and-circumscribed polygons. It was one of his favorite problems, and he picked it up at the same point he had left it years before. He published two papers on this topic (Cayley 1871a,b). In the first, he introduced the (2, 2)-correspondences in the study of that problem; in the second, he raised, and solved, the question of the number of triangles which can be inscribed in and circumscribed about curves of degree higher than two.
We will discuss the two last-mentioned papers in section eight.
56 Jacobi died in 1851.
123 Poncelet’s porism, I 5.1 Cayley’s first notes (1853–1854) In the first note, Cayley considered two non-singular conics U and V , in the projective complex plane, meeting in four distinct points. It was known that, in this case, U and V admit a unique common self-polar triangle, i.e., a triangle ABC such that each vertex is the pole of the opposite side of the triangle, with respect to both U and V .57 Then, by choosing projective coordinates x, y, z so that A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1), Cayley was able to represent U and V , respectively, by the equations:58 x2 + y2 + z2 = 0, ax2 + by2 + cz2 = 0.
Let F to denote the pencil of conics mU +V = 0. He observed that for any tangent tk to the conic Uk := kU + V = (a + k)2x2 + (b + k)2y2 + (c + k)2z2 = 0 there is another conic from F which is tangent to tk, say pU + V = 0 (Fig. 18a).
Then, p can be taken as parameter for the tangent tk as well as for its point of contact Tk with Uk. A general point Tk on Uk has coordinates √b −c√a + p √a + k , √c −a√b + p √b + k , √a −b√c + p √c + k and tk is represented by the equation x √ b −c√a + p √ a + k+y √ c −a b + p √ b + k+z √ a −b√c + p √ c + k =0.
If tk meets U in the points P and P′ corresponding to the parameters θ, ∞and θ′, ∞, respectively, one has P = √ b −c √ a + θ, √ c −a √ b + θ, √ a −b √ c + θ .
By substituting these values in the equation of tk, Cayley got the equation (b −c)√a + k√a + p√a + θ + (c −a)√b + k√b + p√b + θ+ (a −b)√c + k√c + p√c + θ = 0, (5.1) connecting p and θ. He rationalized this equation by putting (a + k)(a + p)(a + θ) = λ + μa, (b + k)(b + p)(b + θ) = λ + μb, (c + k)(c + p)(c + θ) = λ + μc, 57 This result appears for instance in Poncelet (1822, p. 193).
58 We can only say that this was Cayley’s reasoning behind the choice of these equations; in fact, he did not at all justify it. See also next subsection, where we will comment on (Cayley 1861).
123 A. Del Centina tk pU + V = 0 Tk P P Uk P P Uk P V Uk U Uk (a) (b) Fig. 18 Cayley’s procedure for his proof of PCT in case n = 3. b if a triangle P P′P′′ exists, which is inscribed in V and whose sides are respectively tangent to the conics Uk,Uk′ and Uk′′ in the pencil, then (k) + (k′) + (k′′) = 0 values which, evidently, satisfy the equation in question. Squaring and eliminating λ and μ, he obtained [bc + ca + ab −(pθ + kp + kθ)]2+ −4(a + b + c + k + p + θ)(abc + kpθ) = 0 (5.2) which is the rational form of (5.1).
Cayley made the important observation that, due to the symmetry of (5.1) the same equation would have been obtained by eliminating L, M from the equations (ζ + a)(ζ + b)(ζ + c) = L + Mζ, for ζ = k, p, θ. Then, invoking Abel’s theorem,59 that if (x) := x ∞ dx √(x + a)(x + b)(x + c), then the algebraic relation (5.1) is equivalent to (θ) = (p) −(k).
There is of course a similar equation for θ′ with (k) taken with opposite sign: (θ′) = (p) + (k).
59 He also noticed that the result might be verified by means of Euler’s addition theorem for elliptic integrals.
123 Poncelet’s porism, I The elimination of (p) between the two equations gives (θ′) −(θ) = 2(k).
This means, remarked Cayley, that if the points P, P′ on V are such that their para-meters θ, θ′ satisfy this equation, then the line P P′ will always be tangent to the conic Uk.
If a triangle P P′P′′ exists, which is inscribed in V and whose sides P P′, P′P′′ and P′′P are, respectively, tangent to the conics Uk,Uk′ and Uk′′ in the pencil F (see Fig. 18b), one must have: (θ′) −(θ) = 2(k), (θ′′) −(θ′) = 2(k′), (θ) −(θ′′) = 2(k′′), hence, by adding, one gets (k) + (k′) + (k′′) = 0.
(5.3) Cayley observed that when (5.3) holds true, there are infinitely many triangles inscribed in U, the sides of which touch the three conics. So one has Proposition C Equation (5.3) is a necessary and sufficient condition, on the para-meters k, k′, k′′ of three conics Uk,Uk′,Uk′′ in F, for the existence of a triangle, and therefore of infinitely many, which is inscribed in V and whose sides are, respectively, tangent to Uk,Uk′,Uk′′.
Cayley, without explanation, also added that the same holds for a polygon of any number of sides.
In the subsequent paper (Cayley 1853b), he went further, giving an algebraic inter-pretation of (5.3).60 He let □ξ denote the determinant (ξ +a)(ξ +b)(ξ +c) of the conic ξU +V = 0, and noticed that according to Abel’s theorem k, k′, k′′ are the abscissae of the intersection points of the curve y2 = □x with a line y + β0x + β1 = 0. Substituting √ □x in the last equation he obtained √ □x + β0x + β1 = 0.
From here, it is clear that there exists a triangle inscribed U whose sides are, respec-tively, tangent to Uk,Uk′,Uk′′ if and only if 1 k √ □k 1 k′ √ □k′ 1 k′′ √ □k′′ = 0.
(5.4) 60 When Cayley returned to London from his summer’s travels in the Wales, he wrote to W.R. Hamilton thanking him for sending the compendious Lecture on Quaternions and enthusiastically describing his work (Cayley 1853b), that he had submitted to the Philosophical Transactions in July (Crilly 2006, p. 184–185).
123 A. Del Centina In order to get an explicit condition on the coefficients of the two conics, Cayley considered the development in power series of the square root of the determinant □ξ: √□ξ = A + Bξ + Cξ2 + Dξ3 + Eξ4 + · · · . He substituted the corresponding expressions for √ □k, √ □k′, √ □k′′, and wrote equation (5.4) in the form 1 k k2 1 k′ k′2 1 k′′ k′′2 (C + terms multiplied by k, k′, k′′).
Then, if k, k′, k′′ are all different, the above equation is equivalent to (C + terms multiplied by k, k′, k′′) = 0. This implies, when k = k′ = k′′ = 0, i.e., Uk = Uk′ = Uk′′ = U, that C = 0, is the necessary, and sufficient, condition for the existence of a triangle inscribed in V and circumscribed about U.
At this point of the paper (p. 101), Cayley claimed that the same reasoning applies to polygons of any number of sides, and he stated the erroneous (see below) theorem: (∗) the vanishing of the coefficient of ξn−1, in the development of √□ξ, is the condition for the existence of a n-gons inscribed in V and circumscribed about U.
Cayley applied his result in order to write explicit conditions for n = 3 and n = 4, and he found, respectively61: a2 + b2 + c2 −2bc −2ca −2ab = 0, (b + c −a)(c + a −b)(a + b −c) = 0.
He also claimed that similar relations hold for the pentagon, the hexagon, etc. Finally, he considered the case of two circles,62 that he wrote x2 + y2 −R2 = 0, (x −a)2 + y2 −r2 = 0, where a denotes the distance between their centers. He reformulated the above theorem in this case and verified that for a triangle his condition leads to equation (1.1).
Cayley corrected theorem (∗) in the two pages note (1853c). He began by writing: The two theorems in my “Note on the Porism of the in-and-circumscribed Poly-gon” are erroneous,63 the mistake arising from my having inadvertently assumed a wrong formula for the addition of elliptic integrals. The first of two theorems (which, in fact, includes the other as particular case) should be as follows.
61 We observe that Cayley in finding these conditions, inverted the role of U and V , and consequently their equations, see his footnote at p. 101.
62 The second inside the first.
63 Cayley was referring to theorem (∗) and to the equivalent form of it in case of circles.
123 Poncelet’s porism, I Theorem C The condition that there may be an infinity of n-gons, which are inscribed in the conic U = 0 and circumscribed about the conic V = 0, depends upon the development of □ξ = A + Bξ + Cξ2 + Dξ3 + Eξ4 + Fξ5 + Gξ6 + Hξ7 + · · · , precisely for n = 3, 5, 7, . . ., respectively, the conditions are: C = 0, C D D E = 0, C D E D E F E F G = 0, . . .
and for n = 4, 6, 8, . . ., respectively, the conditions are: D = 0, D E E F = 0, D E F E F G F G H = 0, . . .
After having stated the corrected version, Cayley remarked that the two examples (n = 3, 4) he had given in the previous paper were correct, being, respectively, equal to C = 0 and D = 0.
In the second page, he worked out the case of two concentric circles (a = 0), along the line of his new theorem. In particular, putting for brevity α = R2/r2, he verified that the condition C E −D2 = 0 for the pentagon is equivalent to α2 −12α +16 = 0, i.e., the well known r/R = cos π 5 = ( √ 5 + 1)/4.
In the note (Cayley 1854), he explicitly computed the conditional equations when the conics are circles up to n = 8 and compared them with those of Fuss and Steiner for n = 4, 5. Healsoconsideredthecaseof twoconcentriccircles, andsetting M = R2/r2 he established the following conditional equations: M −2 = 0,64 M −4 = 0, M2 −12M + 16 = 0, respectively, for n = 3, 4, 5. Cayley also remarked that the geometrical properties of the polygons inter-scribed to two conics having a double contact are obtained from the case of concentric circles.65 The papers on the in-and-circumscribed triangle (1857–1858) In the paper (Cayley 1857), he considered a triangle abc inscribed in a conic S, whose sides ac and bc are tangent to fixed curves A, B, and sought to find the curve C that is enveloped by the free side ab, when c moves on S. He recognized that ab, and its contact point γ with C, could be constructed in the same way that Poncelet had done in his proof of the “main lemma” (see Fig. 19, and also Fig. 13), but he preferred to follows another route “which it may be modified so as to be applicable to curves S of any order,” as he wrote at p. 344.
He first computed that the class of C, i.e., the degree of the dual C∗, is 2mn where m and n are, respectively, the class of A and B. Then, via an ingenious geometrical construction, he computed the number of bitangents to C, i.e., the number of ordinary 64 In a misprint, see p. 343, M −2 is written M + 2 = 0.
65 Since Poncelet it was known that two concentric circles are the projective image of two bitangent conics.
123 A. Del Centina c S A B C H γ a b α β Fig. 19 In his paper (1857) Cayley considered a more general situation: a triangle inscribed in a conic S having two sides tangent to other two fixed curves A and B, which are not necessarily conics. Cayley showed that, when the triangle abc moves remaining inscribed in S and whose sides ac and bc remain tangents to curves A and B, respectively, the third side envelops a third curve C double points of C∗, which turned ut to be mn(2mn −m −n + 1). He also proved that, in general, C does not have stationary tangents, i.e., that C∗does not have cusps.
Then, through the Plücker formula,66 he found that C has degree 2mn(2mn −1) −2mn(2mn −m −n + 1) = 2mn(m + n −1).
Since Cayley had already examined, through a number of lemmas, the cases in whichthecurvesandA,B S areinparticularpositions,atp.352hestatedthefollowing: Theorem C1 If a triangle abc is inscribed in a conic S, and the sides ac and bc are tangent to fixed curves A, B of class m and n, respectively, the side ab will envelope a curve C of the class 2mn, with in general mn(2mn −m −n) double tangents, but not stationary tangents (i.e., not tangent at inflexion points), and therefore of the order 2mn(m + n −1). If the curve A touch the conic S, each point of contact will give rise to n double tangents of the curve S, and so if the curve B touch the conic S, each point of contact will give rise to m double tangents of the curve C. Moreover, if A and B intersect on the conic S, each such intersection will give rise to a double tangent of the curve C. The curve C in general touches the conic S in the points in which it is intersected by any common tangent of the curves A and B; but if the points of contact be harmonically situated with respect to the conic S, then C does not pass through the points of intersection, but the tangents to S at the points of intersection are stationary tangents of C. There is of course in the above-mentioned special cases a corresponding reduction in the order of C.
Cayley applied the theorem above to the particular case in which the curves A and B are conics. In this case, the envelope C is of class 8 and, in general, of degree 24.
Moreover, he considered two special cases of great interest (p. 353–354): first, A and B both have a double contact with the conic S; second, A, B and S all pass through the same four points.
66 For a curve C of degree d and class d′, with δ nodes and κ cusps, one has d′ = d(d −1) −2δ −3κ.
123 Poncelet’s porism, I In the first case, Cayley showed that the curve C has degree 8 and splits into four conics, each having a double contact with the conic S. Attending only to one of these four conics, he obtained what he called “porism (homographic) of the in-and-circumscribed triangle”: If a triangle abc is inscribed in a conic, and two of the sides touch conics having double contact with the circumscribed conic to abc, then will the third side touch a conic having double contact with the circumscribed conic.
We observe that this is an extension of proposition 1 of subsection 2.4 above (Pon-celet 1822, Art. 433), in which two of the sides of the triangle pass through fixed points, and the remaining side envelops a conic having a double contact with the circumscribed conic.
In the second case, he showed that the curve C has order 4 and splits into two conics, each passing through the point of intersection of A, B and S. Attending only to one of these two conics, Cayley obtained what he called “porism (allographic) of the in-and-circumscribed triangle”: If a triangle abc is inscribed in a conic, and two of the sides touch conics meeting the circumscribed conic to abc in the same four points, the remaining side will touch a conic meeting the circumscribed conic in the four points.
In case of circles, the last claim is the “main lemma” of subsection 2.4.67 The following year Cayley published another paper on the subject, here he wrote (Cayley 1858, p. 31): In my former paper “On the Porism of the In-and-Circumscribed Triangle” [(Cayley 1857)], the two porisms (the homographic and the allographic) were established a priori, i.e., by means of an investigation of the order of the curve enveloped by the third side of a triangle. I propose in the present paper to give the a posteriori demonstration of these two porisms; first according to Poncelet, and then in a form not involving (as do his demonstration) the principle of pro-jections.68 My objection to the employment of the principle may be stated as follows: viz. that in a systematic development of the subject, the theorems relat-ing to a particular case and which are by the principle in question extended to the general case, are not in anywise more simple or easier to demonstrate than are the theorems for the general case; consequently, that the circuity of the method can and ought to be avoided.
Likely these words were not appreciated by Poncelet.
Cayley gave two proofs of both the porisms, one according to Poncelet and one independent from the principle of continuity, that for sake of space we will not com-ment.
The memoir of 1861 Cayley published a complete proof of Theorem C years later in Cayley (1861). We present this proof here below, developing some details.
67 In section eight of Bos et al. (1987), where the infinitesimal argument that Poncelet used in the proof of the main lemma is developed according to the modern theory of deformations, the authors showed that Poncelet’s argument applies not only to conics but to algebraic curves in general.
68 That is the principle of continuity.
123 A. Del Centina He considered the conics U = ax2 + by2 + cz2 = 0, V = x2 + y2 + z2 = 0, the pencil U + ξV = 0, and for n = 3 he proceeded as in the first notes (proposition C above). We remark that, at p. 229, he felt the need of justify the choice of these equations for U and V , by saying: The foregoing demonstration relates to the particular formsU = ax2+by2+cz2, V = x2 + y2 + z2; but observing that the function √(ξ + a)(ξ + b)(ξ + c), which enters under the integral sign in the transcendental function ξ is the square root of the discriminant of U + ξV , the theory of covariants shows at once that the conclusions apply to any form whatever of U, V .69 For an n-gon which is inscribed in V , and whose sides touch in the order the conics Uk1, . . . ,Ukn, Cayley directly wrote the condition (k1) + (k2) + · · · + (kn) = 0.
(5.5) “By Abel’s theorem,” Cayley noticed (p. 230), “this transcendental equation is equivalent to an algebraical one.” In fact, the k1, k2, . . . , kn are the abscissae of the intersection points of the curve y2 = □x with some algebraic curve θ(x, y) = 0.
Then, extracting Cayley’s paper, “if ϕ(x) and χ(x) are polynomial in x with arbitrary coefficients, and if ϕ(x)2 + χ(x)2□x = A(x −k1)(x −k2) · · · (x −kn), which implies that for n even the degrees of ϕ(x) and ϕ(x) are, respectively, n/2 and (n −4)/2, and for n odd are, respectively, (n −1)/2 and (n −3)/2, the algebraical equation is that obtaining b the elimination of the arbitrary coefficients from the system of equations ϕ(k1) + χ(k1)□k1 = 0 ϕ(k2) + χ(k2)□k2 = 0 .
.
.
ϕ(kn) + χ(kn)□kn = 0 or, what is the same, for n = 2p −1 it is {1, θ, . . . , θ p−1, √ □θ, . . . , θ p−2√ □θ} = 0, and for n = 2p it is {1, θ, . . . , θ p, √ □θ, . . . , θ p−2√ □θ} = 0, where the expressions in { } denote, respectively, the determinants formed by sub-stituting for θ the values k1, k2, . . . , kn, respectively. Thus, for n = 3 the equation is 69 It seems to us that this amount to say: given two (non-singular) conics in general position, by a suitable projective transformation of the plane, their equations always can be put that form. A remark that he missed to do in his early notes.
123 Poncelet’s porism, I 1 k1 √□k1 1 k2 √□k2 1 k3 √□k3 = 0 and for n = 4 it is 1 k1 k2 1 √□k1 1 k2 k2 2 √□k2 1 k3 k2 3 √□k3 1 k4 k2 4 √□k4 = 0 and so on.
Suppose □ξ = A + Bξ + Cξ2 + Dξ3 + Eξ4 + · · · , then substituting the corresponding expressions for √□k1, √□k2, etc., the determinant will divide by {1, θ, θ2, . . . , θn−1}, and it may be seen without difficulty that the resulting equation, on putting therein k1 = k2 = · · · = kn = 0, will, according as n = 3, 4, 5, 6 etc., be C = 0, D = 0, C D D E = 0, D E E F = 0, C D E D E F E F G = 0, etc., which is the theorem above referred to.” Let us remark that, there exists a n-gon inscribed in V and circumscribed about U, if, and only, it is possible to find coefficients of the polynomial ϕ(x) and χ(x) so that ϕ(x) + χ(x)□x has 0 as a root of multiplicity n.
In the remaining of the paper, Cayley applied his result to get the condition in polynomial form up to n = 9.
It is convenient here to change Cayley’s notation. If √ □x = A+ Bx +C2+C3x2+ · · · , then, for n = 2m the condition above is equivalent to C3 C4 · · · Cm+1 C4 C5 · · · Cm+2 .
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Cm+1 Cm+2 · · · C2m−1 = 0, (5.6) for n = 2m, and for n = 2m + 1 to the condition C2 C3 · · · Cm+1 C3 C4 · · · Cm+2 .
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Cm+1 Cm+2 · · · C2m = 0.
(5.7) 123 A. Del Centina It is worth calling attention to the fact that Cayley did not try to explain the geo-metrical meaning of the above equations, nowadays called Cayley’s conditions. This was done more than 100years later by Griffiths and Harris.
In the following of the paper, Cayley applied his result to find explicit formulae for the existence of an in-and-circumscribed n-gon to U and V up to n = 9. Moreover, he considered the particular case of two circles, determining the required conditional equations that he compared with those found by J. Mention the year before (see section seven).
In his paper of (1861), Cayley did not quote Trudi. Could it be that Sylvester had not informed him of Trudi’s results?
6 An algebraic approach through invariants If two conics U and V are such that a n-gon inter-scribed to them there exists, it is obvious that a certain relation must hold among the invariants (and covariants) of the two conics. This remark induced George Salmon to produce an “elementary” proof of Cayley’s result. What he had in mind was a proof built on the basics of the theory of invariants of a pair of conics, such as he had developed in his treatise (Salmon 1855), that avoids the use of the elliptic functions.
An invariant of an algebraic form f (x), in two or more variables, is a polynomial I (a) of the coefficients of f (x), that, under a linear transformation of the variables of determinant , remains unaltered up to a power of , i.e., I (a′) = k I (a). If k = 0, the invariant is said absolute. A covariant of f (x), is a polynomial I (a, x) of the coefficients and the variables of f (x), which, under a linear transformation as above is such that I (a′, x′) = k I (a, x). The theory of invariants and covariants of algebraic forms, which began to be developed in the early 1840s, with the pioneering work of George Boole, Cayley, James J. Sylvester, and Salmon, who formulated the basic concepts and developed the key techniques. Salmon also codified the theory in high-level textbooks. For the early history of the theory of invariants, we refer to Crilly (1986) and Hunger (1989, 2006a,b).
Salmon realized his program with the paper (Salmon 1857), that was published, divided into three distinct parts, in a single issue of the Philosophical Magazine. Here below we illustrate the content of this paper, extracting directly from it in the hope of keeping its original flavor, but also inserting some detail from Salmon’s treatise on conic sections (Salmon 1855).
Let λU + V = 0 be the general conic of the pencil generated by the pair of conics U, V . Its determinant, as a polynomial in λ, can be written det(λU + V ) = λ3 + λ2 + ′λ + ′, where and ′ are, respectively, the determinant of U and V , and ′ are, respec-tively, tr(U · adj(V )) and tr(V · adj(U)). We explicitly remark that , , ′, are of degree 3 in the coefficients of U and V . For a general theorem of the theory of invari-123 Poncelet’s porism, I ants, all the projective invariants of the pair of conics U, V , are rational functions of , , ′ and ′.70 In the first part of the paper (pp. 190–191), Salmon posed the following problem (a simplified version of Poncelet’s main lemma): find the envelop of the third side of the triangle inscribed in the conic U, and two of whose sides touch the conic V.
To this end he argued as follows. The condition that λU + V = 0 represents a pair of line is expressed by the condition λ3 + λ2 + λ + ′ = 0. Salmon wrote: Since the value of λ plainly cannot depend on the particular axes to which the equations are referred, it follows that no matter how the equations are trans-formed, the ratios of the coefficients of the powers of λ in the equation just written remain unaltered. Let now the sides of the triangle in any position be x, y, z, then the equations of the conics admit of being transformed into: U = 2xy + 2yz + 2xz = 0, V = l2x2 + m2y2 + n2z2 −2lmxy −2lnxz −2mnyz −2Axy = 0; and it is plain that the equation AU + V = 0 represents a conic that the third side z touches.
But in this case we find, if p : l + m + n and r := lmn, = 2, = −p2 −2A, ′ = 2p(2r + An), ′ = −(2r + Ar)2, whence 4′ −′2 = 8A(2r + An)2, and the equation AU + V = 0 can be written (4′ −′2)U −4′V = 0.
The coefficients in this equation being invariants, it follows that the conic which we have proved is touched by the third side is a fixed conic. which is an equation depending on the coefficients of the two given conics.
Let us remark that, if the condition 4′ −′2 = 0 is satisfied, is clear that the envelope of the third side of the triangle coincides with the conic V .
Next Salmon asked the question: find the locus of the third vertex of a triangle circumscribed about V when the other two vertices move along U.
He continued by observing: 70 For a modern approach to the theory of invariants of a pair of conics (see Sommerville 1933; Todd 1947 or Dolgachev 2012).
123 A. Del Centina In this case, the equations of the conics U and V can be transformed into U = 2xy + 2yz + 2xz + Az2 = 0, V = l2x2 + m2y2 + n2z2 −2lmxy −2lnxz −2mnyz = 0, and we have = 2 −A, = −p2 + 2lm A, ′ = 4pr, ′ = −4r2.
Again, let F = 0 be the equation of the covariant conic which passes through the points of contact of the common tangents to U and V (see my “Conics,” pp.
268, 288),71 the coefficient of z2 in its equation will be −4rn(1 −A).
It is useful here to recall some facts pertaining to the theory of reciprocal (dual) conics. Let S and S′ be two general conics of equation S = ax2 + a′y2 + a′′z2 + 2bzy + 2b′yx + 2b′′xz = 0, S′ = Ax2 + A′y2 + A′′z2 + 2Bzy + 2B′yx + 2B′′xz = 0, then, the reciprocal conic of the conic S + λS′ has equation + λ + λ2′ = 0, (6.1) where and ′ are, respectively, the reciprocal conics of S and S′, and is the polynomial (a′A′′ + a′′A′ −2bB)x2 + (a′′A + aA′′ −2b′B′)y2 + (aA′ + a′A −2b′′B′′)z2 +2(b′B′′ + b′′B′ −aB −bA)yz + 2(b′′B + bB′′ −a′B′ −b′A′)xz +2(bB′ + b′B −a′′B′′ −b′′A′′)xy.
Since the conics of the original system pass through four fixed points, the conics of the reciprocal system always touch four fixed lines.
The form of equation (6.1) shows that the reciprocal always touches 4′ −2 = 0. This means that this last equation is the equation of the four common tangents to and ′ and to the other conics of the reciprocal system. The form of 4′ = 2 also shows that is touched by those four lines and that passes through the points of contact. Since the same holds for ′, it follows that the eight point of contact of the four common tangents to the two conics and ′, all lie on the conic = 0.
The reciprocal of the system + λ′ = 0 is the system S + λF + λ2′S′ = 0, where F is what becomes when the coefficients of are written in terms of the elements of the polynomials and ′, a, a′, . . . , b′′, and A, A′, . . . , B′′: 71 Salmon was referring to Salmon (1855).
123 Poncelet’s porism, I Fig. 20 The conic of equation F = 0 passing through the eight points of contact of the four common tangents to two conics S and S′ (a′A′′ + a′′A′ −2bB)x2 + (a′′A + aA′′ −2b′B′)y2 + (aA′ + a′A −2b′′B′′)z2 +2(b′B′′ + b′′B′ −aB −bA)yz + 2(b′′B + bB′′ −a′B′ −b′A′)xz +2(bB′ + b′B −a′′B′′ −b′′A′′)xy.
From the above, it follows that the conic F = 0 passes through the eight points of contact of the four common tangents to S and S′ (see Fig. 20).
Moreover, it is easy to check that aA′ + a′A −2b′′B′′ is equal to −4rn(1 −A) when S = U and S′ = V .
At this point, Salmon wrote: it can be seen that the coefficient of z2 vanishes identically in the equation 16′2U −4(4′ −′2)′F + (4′ −′2)2V = 0, which is therefore the equation of the locus required. Its form shows that this conic is tangent to the four common tangents to U and V .
If4′−′2 = 0,theenvelopereducesto V ,andthelocustoU,inconformity to Mr. Cayley’s theorem. It does not seem impracticable to obtain the equation of the locus by the same method when the two sides touch different conics.
In the second part of his paper (pp. 267–269), Salmon considered the problem of finding the locus of the fourth vertex of a quadrilateral, whose other three vertices move on V , and whose sides touch U. He put for brevity α := 4′, β := 2 −4′, γ := 2α + β, 123 A. Del Centina V U aU + bV Fig. 21 Salmon’s question for n = 4 and reduced the problem to finding the locus of the third vertex of a triangle two of whose vertices move on V , two of whose sides touch U, and the third touches αU + βV . Proceeding as in the first part of his paper, he obtained for the required locus the following equation α2γ 2U + αγβ2F + ′β4V = 0, which clearly reduces to V if γ = 0.
Finally, in the third part (pp. 337–338), Salmon considered the question: find the locus of the free vertex of a polygon, whose sides all touch U, and whose vertices all but one move on V.72 On p. 337, he wrote: This [the question above] is immediately reduced to the last question, since the line joining the two vertices of the polygon adjacent to that whose locus is sought, touches a conic whose equation is of the form aU + bV = 0 [see Fig. 21]. The locus will therefore always be of the form ′λ2V + λμF + μ2U = 0, So Salmon proceeded by induction, similarly to Poncelet in his proof of the general theorem.
The procedure allowed him to find a recursive procedure for determining the condi-tionthatshouldmakeitpossibletodescribeapolygoninscribedin V andcircumscribed about U. Precisely: if λ′, μ′ are the values for a polygon of n −1 sides, and λ′′, μ′′ those for a polygon of n sides, then the values for a polygon of n + 1 sides are λ′′′ = μ′μ′2, μ′′′ = ′λ′λ′′(αμ′′ −′βλ′′).
Since for a triangle one has λ′ = α, μ′ = ′β, for quadrilateral one has λ′′ = β2, μ′′ = αγ , the required conditions are: 72 Let us recall that Landen had determined this locus when U and V are real circles.
123 Poncelet’s porism, I triangle, β = 0, quadrilateral, γ = 0, pentagon, δ = 0, δ := α2γ −′β3, hexagon, ϵ = 0, ϵ := δ −′γ 2, heptagon, φ = 0, φ := α2γ ϵ −δ2, octagon, ψ = 0, ψ := δφ −′β3ϵ2, .
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Salmon concluded by stating: “I suppose these values will be found to coincide with those found by a different way by Mr. Cayley in a former Number of this Journal, but I have not verified this.” Two years after Salmon’s paper was printed, Francesco Brioschi proved that the formulae of Cayley and Salmon were equivalent, by showing that both descend from a common principle (Brioschi 1857).
To show this, he supposed the conic U is circumscribed to the triangle abc whose sides are given by x = 0, y = 0, z = 0. This allowed him to put the equation of U in the form U = αyz + βzx + γ xy = 0.
Then, he considered another conic V = l2x2 + m2y2 + n2z2 −ayz −bzx −cxy = 0, and observed that the lines x = 0, y = 0, z = 0 will be, respectively, tangents to the conics k1U −V = 0, k2U −V = 0 and k3U −V = 0 if and only if a = 2mn −αk1, b = 2ln −βk2, c = 2lm −γ k3.
Brioschi denoted 2(k) = a0k3 + a1k2 + a2k + a3 the discriminant of the “function” kU −V . Setting p = lα + mβ + nγ, q = 4r −lαk1 −mβk2 −nγ k2, r = lmn, he got a0 = αβγ, a1 = p2 −αβγ (k1 + k2 + k3), a2 = 2pq + αβγ (k1k2 + k3k1 + k2k3), a3 = q2 −αβγ k1k2k3.
Hence, if the equation k3 + Ak2 + Bk + C = 0 has solutions k1, k2, k3, it will be a1 −a0 A = p2, a2 −a0B = 2pq, a3 −a0C = q2.
(6.2) 123 A. Del Centina Multiplying these equations, respectively, for k2 1, k1, 1, adding up, and taking into account the previous equation, he got a0k3 1 + a1x2 1 + a2x1 + a3 = (pk1 + q)2.
Proceeding similarly for k2 and k3, he finally found that a0k3 + a1k2 + a2k + a3 −(pk + q)2 = a0(k −k1)(k −k2)(k −k3), (6.3) and therefore, if ψ(x) := dk/(k), by Abel’s theorem, as in Cayley (1853a), he got that k1, k2, k3 must satisfy the transcendental equation ϵ1ψ(k1) + ϵ2ψ(k2) + ϵ3ψ(k3) = C, which is equivalent to the following (irrational) algebraic equation 1 k1 (k1) 1 k2 (k2) 1 k3 (k3) = 0.
(6.4) Brioschi noticed that (6.2) also leads to 4(a1 −a0 A)(a3 −a0C) −(a2 −a0B)2 = 0.
(6.5) Now, relation (6.4) constitutes the result of Cayley (n = 3), while relation (6.5) constitutes the result of Salmon (n = 3). From (6.3), it follows that p = (k1) −(k2) k1 −k2 , q = k1(k2) −k2(k1) k1 −k2 , then, setting (k) = A0 + A1k + A2k2 + · · · the development in power series of the discriminant, he obtained q = A0 −k1k2P, where P = A2 + A3(k1 + k2) + A4(k2 1 + k1k2 + k2 2) + · · · From the relations above, it follows that k3 = a−1 0 (k1k2P2 −2A0P), which, if k1 = k2 = 0, i.e., the first two sides of the triangle are tangent to V , gives k3 = −2A0 A2 a0 = a2 2 −4a1a3 4a0a3 , 123 Poncelet’s porism, I This means, remarked Brioschi, that the triangle abc is circumscribed to V if and only if a2 2 −4a1a3 = 0, conditions equivalent, respectively, to (Cayley’s) C = 0 and to (Salmon’s) 4′ − ′2 = 0.
By considering diagonals and intermediate triangles, as Poncelet and Jacobi had done in the case of a pencil of circles (recall Fig. 11b), Brioschi showed that the same holds true first for quadrilateral, and then for a polygon of any number of sides.
Almost 100years after the publication of Salmon’s paper, John A. Todd revisited the same subject in Todd (1948). We will return on this argument in section thirteen.
7 Other contributions from 1850 to 1875 In the third quarter of the nineteenth century, many papers related to Poncelet’s theorem and its generalizations appeared. In this section, we present and discuss only those which, in our opinion, are the most interesting both per se and from a historical point of view. We have divided them into two major branches along which the theory developed: (1) find new proofs of PCT and simplify those already known; (2) extend the theorems in higher dimension and prove other “closure theorems.” In this section, we also aim to introduce the reader to the topics discussed in the subsequent sections.73 7.1 New proofs of PCT In 1849, a new paper by Jacobi’s pupil Richelot was printed. In his (1849) he obtained, in the case of two nested circles, an algebraic condition for the existence of an in-and-circumscribed p-gon, p being a prime. He also gave a method for solving the problem in case of a polygon of n sides knowing the solution for polygons of n −1 and n + 1 sides.
In 1860, J. Mention published Essai sur le problème de Fuss (Mention 1860).74 He called “problem of Fuss,” the problem of determining the relation between the data R,r, δ, n for the existence of a n-gon inter-scribed to two circles, respectively, of radii R, r, being δ the distance between their centers. To solve the problem, Mention argued, very ingeniously, as follows.
Let AB be a chord of the circle C of radius R and center O, which is tangent to the circle c of center I and radius r. Denote by !
A, and !
B, the angle between the chord and 73 For a more extended review of the literature of this period, we refer to Loria (1889a,b, 1896), but alert the reader that many references therein are incorrectly dated or have page numbering wrong, or even present a misleading indiction of the journal that should contain the quoted paper.
74 The memoir was read at the Academy of Saint Petersburg the 13th of Mai 1859. Very little is known about J. Mention (1821–?), probably a Russian mathematician. In the years 1845–1865, he published several short notes in Nouvelles Annales de Mathematique.
123 A. Del Centina O I A B r R δ ˆ A ˆ B Fig. 22 Illustration of how Mention proceeded in order to determine the relation among R,r, δ, n for the existence of a n-gon inter-scribed to two circles of radii R and r, being δ the distance between their centers the other tangent to c drawn, respectively, from A, and B (Fig. 22). We have that O A = r sin !
A 2 , O B = r sin !
B 2 · Then, considering the triangles I AO, I BO and by using some trigonometry one has δ2 = R2 + r2 sin2 !
A 2 −2Rr sin !
A 2 cos B AO − !
A 2 , δ2 = R2 + r2 sin2 !
B 2 −2Rr sin !
B 2 cos ABO − !
B 2 .
By subtracting the two equations above, and using some trigonometry, Mention got the following equation r 2R cot !
A 2 + cot !
B 2 = cos BAO.
Now, by adding the same two equations above, and taking into account the previous one, he got R2 + r2 −δ2 2Rr −r 2R cot !
A 2 cot !
B 2 = sin BAO.
123 Poncelet’s porism, I By squaring and adding, he finally obtained 4R2r2 −(R2 + r2 −δ2)2 r4 = cot2 !
A 2 + cot2 !
B 2 + cot2 !
A 2 cot2 !
B 2 + −2 cot !
A 2 cot !
B 2 R2 −δ2 r2 · Setting ν = 2R2r2 + 2R2δ2 + 2r2δ2 −R4 −r4 −δ4 r4 , i = R2 −δ2 r2 , and x1 = cot !
A/2, x2 = cot !
B/2, Mention wrote the equation above in the form ν = x2 1x2 2 + x2 1 + x2 −2ix1x2.
If a polygon of n sides is inscribed in C and circumscribed about c, denoting x1, x2, x3, . . . , xn the cotangents of the half of the angles at the respective vertices, one has the following system: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ν = x2 1x2 2 + x2 1 + x2 2 −2ix1x2 ν = x2 2x2 3 + x2 2 + x2 3 −2ix2x3 .
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ν = x2 n−1x2 n + x2 n−1 + x2 n −2ixn−1xn ν = x2 nx2 1 + x2 n + x2 1 −2ixnx1.
(7.1) These equations, Mention affirmed, will be compatible only if a certain relation among ν, i and n holds true, and, vice versa, if such a relation is satisfied it will be possible to give to any initial angle an arbitrary value. Therefore, if a polygon of n-sides can be inscribed in C and circumscribed about c, then there are infinitely many such polygons.
For a triangle, one has ν = x2 1x2 2 + x2 1 + x2 2 −2ix1x2 ν = x2 2x2 3 + x2 2 + x2 3 −2ix2x3 ν = x2 3x2 1 + x2 3 + x2 1 −2ix3x1, from the first and the last equations one gets x2 + x3 = 2ix1 1 + x2 1 , x2x3 = x2 1 −ν 1 + x2 1 .
By substituting these values into the second, after some computation, one is lead to ν + 2i + 1 = 0, which is readily seen to be equivalent to δ2 = R2 ± 2Rr, which includes (1.1).
123 A. Del Centina Proceeding similarly, Mention found the conditional equation that allows the exis-tence of an inter-scribed n-gon for n up to 11, but for higher value of n, the difficulties in the elimination become insurmountable, and he was able to give only some recursive formula.
Let us remark that the system (7.1) can be deduced from the more general (4.10) found by Trudi.
In 1862, Poncelet published the first volume of Applications d’analyse et de géométrie. In the Note historique, inserted at the end of the book, he described the development of the theory until then.75 Recalling (Jacobi 1828), he expressed his disagreement with the term “géométrie élémentaire” that Jacobi used in the title in connection with PCT. Then, Poncelet quoted the results of Fuss, Steiner, Richelot, Cay-ley, Brioschi and Mention. This latter, according to him, had the great merit of having tackled the question directly and geometrically, and, although he had only solved the problem for n ≤11, he had highlighted the scale of relation between the polygons of n, n −1 and n −2 sides (see Poncelet 1862, p. 483). He continued by saying: C’est d’ailleurs une question de savoir si le problème, si mal résolut par Fuss en 1792, l’a été mieux depuis par d’autre, notamment en Angleterre par M.
Cayley, qui, ignorant sans doute mes publications de 1817 et 1822 citées plus haut, a attribué gratuitement à cet ancien et estimable géomètre, sous le nome de porisme, le théorème de la p. 364 sur les cercles. Parmi les nombreux Mémoire de M, Cayley, écrits dans une langue mathématique pour moi doublement étrangère, j’entrevois bien, en effet, de belles méthodes algébriques pour passer d’un terme à un autre de la série des polygones, mais non pour franchir, sans calculs intermé-diaires, l’intervalle qui sépare entre eux deux termes de rang quelconque. Ainsi, par exemple, dans son dernier Mémoire résumé, de mars 1861, il n’arrive à la for-mule de l’ennéagone, obtenue par M. Mention et relative au cas simple de deux cercles, qu’après avoir laborieusement calculé toutes celles qui appartiennent aux polygones d’ordre inférieur [This is also a question of whether the problem, so badly solved by Fuss in 1792, has been better solved by others later, as in England by M. Cayley, who, undoubtedly ignorant of my publications of 1817 and 1822 quoted above, assigned gratuitously to this estimated geometer, under the name of porism, the theorem at p. 364 on the circles. Among the memoirs of M. Cayley, which are written in a mathematical language doubly extraneous to me, I see, indeed, beautiful algebraic methods for passing from one term to another of the series of polygons, but which do not cross, without intermedi-ate calculations, the interval between two of any rank. So, for example, in his latest and concluding memoir of March 1861, he arrives at the formula for the enneagon, obtained by M. Mention and relative to the simple case of two circles, only after having painstakingly calculated all those belonging to polygons of a lesser number of sides].
Cayley replied to Poncelet’s remarks with a letter in which, after having denied wish to attribute the theorem in question to Fuss, he dealt with the criticism levelled 75 This note makes for very interesting reading in many ways.
123 Poncelet’s porism, I at his paper of 1861. He briefly reviewed his method, displayed the formulae for n up to 8 and stressed that the condition was actually and explicitly found for a polygon of any number of sides, underlying, “sans passer par celles qui appartiennent aux polygones d’ordre inférieur” [without passing through those belonging to polygons of a less number of sides]. Cayley’s letter to Poncelet ended with this meaningful statement: Comme j’attache, je l’avoue, un peu d’importance à cette solution (laquelle selon l’explication que je viens de donner ne parait pas mériter la critique que vous en faites) je serais bien aise si vous voulez bien communiquer cette lettre à l’Académie [As I attach, I confess, some importance to this solution (which, according to the explanation I have just given, does not seem to deserve your criticism) I will be glad if you could communicate this letter to the Academy].
The letter was quickly published in the Comptes rendus (Cayley 1862).
In his historical note, Poncelet failed to quote Trudi, whose memoir of 1853 he had probably not read. This lack of recognition somewhat annoyed Trudi, who regarded his results as being more general and deeper than those of Mention.
The following year Trudi published the long memoir (Trudi 1863a), on which, returning to the question related to the existence of inter-scribed polygons to two conics, he claimed priority for the complete analytical proof of Poncelet’s closure theorem. In the Notizie storiche, that he inserted at the beginning of his work, Trudi suggested that he had not been mentioned because the title of his memoir of 1853, Rappresentazione geometrica immediata dell’equazione fondamentale nella teorica delle funzioni ellittiche [Immediate geometrical representation of the fundamental equation in the theory of elliptic functions], had not alluded at all to Poncelet’s theorem and related questions. To endorse his priority, he also mentioned the first studies of 1841, the memoir of 1843 he had read at the Congresso degli Scienziati held in Naples in 1845, and the encouragements he had received from Jacobi that led him to write the memoir of 1853.76 Then, he added: Risulta da questi fatti che noi possiamo pretendere alla piccola gloria di aver dato i primi una dimostrazione analitica compiuta e diretta dei teoremi di Poncelet, e di aver dato anche i primi un metodo per la ricerca della relazione, affiché un poligono di qualsivoglia numero di lati possa esser iscritto e circoscritto a due coniche [From these facts, it appears that we can expect little glory for being the first to give a direct and complete analytical proof of the theorems of Poncelet, and for being also the first to give a method to obtain the relation under which a polygon of any number of sides can be inscribed in and circumscribed about to two conics].
The same year, Trudi published another paper on the same subject, that he entitled, more explicitly, Su’ teoremi di Poncelet relativi a’ poligoni iscritti e circoscritti alle 76 In passing we note that Trudi’s memoir of 1853 was cited by Angelo Genocchi in his paper on a construction of the theorem of Abel, in relation to the addition of elliptic functions, published in the first volume of the just founded Annali di Matematica Pura e Applicata (Genocchi 1858, p. 36).
123 A. Del Centina coniche [On Poncelet’s theorems related to inscribed and circumscribed polygons to conics] (Trudi 1863b).
These two memoirs do not add much to the previous one of 1853, so for sake of space we avoid comments. Suffice to say that Trudi (1853) and Trudi (1863a) were mentioned by Loria (1889b, 1896), while Dingeldey only quoted the second (Dingeldey 1903, p. 47). These memoirs and (Trudi 1863b) were cited in Gerbaldi (1919, p. 97), where Trudi’s method was shortly presented (see our section eleven).
Since then it seems that Trudi’s work on the theorem of Poncelet has been forgotten until very recently (see Dragovi´ c 2011, p. 105).
Poncelet’s book quoted above contained, as an appendix, a memoir by Théodore Moutard,77 titled Recherhes analytiques sur les polygones simultanément inscrits et circonscrits a deaux coniques (Moutard 1862). By means of algebraic methods, Moutard wrote an equation of the curve enveloped by the last side of a n-gon inscribed in a conic A0, and whose first n −1 sides are tangent to another conic A, when its vertices move along A0. By means of elegant geometrical considerations, he found a simple recursive law for the formation of the conditional equations relative to the cases 3, 4, . . . , n. The study of this law comes down to the study of certain functional equa-tions, whose solution leads directly to the transcendental functions , H of Jacobi, of which the elliptic functions sn(u), cn(u) and dn(u) are simple rational expressions.
Moutard observed how many of the properties of these functions were related to the theorem of Poncelet.
Jakob Rosanes and Moritz Pasch, with their joint work (Rosanes and Pash 1865), also completed Jacobi’s project. Generalizing the method used by Jacobi, they were able to write the relation that the coefficients of two conics, A and B, must satisfy for the existence of a polygon of n sides inter-scribed to them. In the introduction to their paper, after having recalled Euler, Fuss, Steiner, Jacobi, they quoted (Cayley 1853a,b), (Moutard 1862), briefly summarized the results therein, and added: Die gegenwärtige Abhandlung, deren Verfasser von den letztgenannten beiden Arbeiten bis vor kurzer Zeit keine Kenntniss hatten, scheint von diesen sowohl in Bezug auf den eingeschlagenen Weg, als die Form der Resultate, welche grosse Ächnlichkeit mit den von Jacobi gefundenen Formeln aufweist, sosehr verschieden, dass die Veröffentlichung derselben wohl gerechtfertigt erscheinen dürfe [the present memoir, whose authors until recently had no knowledge of the last works mentioned, which seem to have great resemblance, both in terms of path as in the shape of the results, with the formulae found by Jacobi, are indeed rather different, so that the publication of it probably could be justified.] By performing a projective change of coordinates, Rosanes and Pasch put the equa-tions of A and B in the simple form x2 + y2 + z2 = 0, αx2 + βy2 + γ z2 = 0, 77 Théodore Florentin Moutard (1827–1901), engineer. His mathematical work was primarily in the theory of algebraic surfaces, differential geometry and differential equations. He taught mechanics at the École des mines.
123 Poncelet’s porism, I and considered separately four cases, according to the behavior of the intersections and the common tangents of the two conics are real or imaginary, and for each of these cases they obtained the relative condition allowing the existence of an inter-scribed polygon. Finally, they computed these relations in terms of the coefficients of the cubic polynomial δ0λ3 + δ1λ2 + δ2λ + δ3 = det(B −λA) for n = 3, 4, getting, respectively δ2 0 = 4δ1δ2, δ3 2 + 8δ0δ2 3 = 4δ1δ2δ3.
As we know, these relations were already obtained by Salmon years before.
Four years later, in their paper (Rosanes and Pash 1869), Rosanes and Pasch recog-nized that the above question (and in fact an entire class of geometrical problems) could be put in the following form.
Let there be given a symmetric doubly quadratic equation f (t0, t1) = at2 0t2 1 + 2bt0t1(t2 0 + t2 1) + c(t0 + t1)2 + 2dt0t1 + 2e(t0 + t1) + f.
For a fixed value of t1, there are two values of the first variable satisfying the equation, say t0 and another value t2. Fixed t2, there is another value other than t1, say t3, satis-fying the same equation. Proceeding in this way one gets a sequence t0, t1, t2, . . . , tn, such that f (t0, t1) = f (t1, t2) = · · · = f (tn−1, tn) = 0. The question is: for a given n > 2 is t0 = tn and at the same time tn+1 = t1, and in general tn+h = th?
At this point (p. 169), the two authors affirmed that t0 and tn satisfy an equation of the same form as above, that is, by eliminating the intermediate variables t1, . . . , tn−1 one obtains an−1t2 0t2 n + 2bn−1t0tn(t2 0 + t2 n)+cn−1(t0 + tn)2+2dn−1t0tn + 2e(t0 + tn)+ fn−1.
Let us remark that this does not appear completely justified, and in fact, Trudi felt the need to prove it by means Euler’s differential equation.
Rosanes and Pasch asked for the condition under which t0 = tn for a certain n > 2.
By means of an elaborate algebraic computation, they proved that a necessary and sufficient condition is given by the vanishing of the function qn, defined recursively as follows: q0 = 0, q1 = q2 = 1, q3 = σ qn−2qn+2 + qn−1qn+1 = σq2 n, qn−2q2 n+1 + qn+2q2 n−1 = qn(λq2 n + 2δqn−1qn+1), if n is even, or qn−2qn+2 + λqn−1qn+1 = σq2 n, qn−2q2 n+1 + qn+2q2 n−1 = qn(q2 n + 2δqn−1qn+1), if n is odd.
123 A. Del Centina Clearly, qn is an entire function of the coefficients of f (t0, t1), of degree n2/4 −1 or (n2 −1)/4 according if n is even or odd.
Here (p. 173) Rosanes and Pasch, without further explanation, claimed that: if P0, P1, . . . , Pn is a polygonal line inscribed in A and circumscribed about B, the corresponding parameters t0, t1, . . . , tn of these points satisfy, in pairs (t0, t1), (t1, t2), etc., to a symmetric biquadratic equation of the previous type, whose coefficients depend on those of A and B. Hence, the condition for the closure of the polygonal line, i.e, for the existence of a n-gon inter-scribed to A and B, is given by qn = 0. Then, expressing λ, σ and δ in terms of the coefficients of the discriminant det(B −αA) = δ0α3 + δ1α2 + δ2α + δ3 [as in Rosanes and Pash (1865), section 9], they wrote down, respectively, for a triangle a quadrangle and a pentagon, the following conditions: q3 = δ2 2 −4δ1δ3 = 0, q4 = 2[8δ0δ2 3 + (δ2 2 −4δ2δ3)] = 0, q5 = q2 3 −16δ0δ2 3q4 = 0, which are readily seen to be equivalent to that given by Salmon.
Hence, on the base of induction, Rosanes and Pasch claimed: qn is an entire function of the coefficients of the two conics, which is of degree 3 n2 4 −1 or 3n2 −1 4 , (7.2) according if n is even or odd.
The question of determining the degree of the invariant whose vanishing guaranties the existence of an inter-scribed n-gon to the two given conics, was to be studied in depth by Gerbaldi 50years later.
In a footnote, Rosanes and Pasch gave notice of dissertation (Simon 1867) by Max Simon.78 In his thesis, Simon presented a new proof of PCT by means of the emerging theory of the Weierstrass ℘-function, instead of the classical elliptic functions of Jacobi, and expressed the conditional equation in terms of the invariant of the pencil of conics. He also noticed the relation between PCT and biquadratic binary equations (Simon 1867, pp. 8–12). It is worth to say that in 1864–1865, while still student in Berlin, Simon participated in a seminar dealing with these topics. An enlarged version of the thesis was published years later (Simon 1876). This approach to PCT was later codified by Halphen, in the second volume of his treatise on elliptic functions (Halphen 1888). We will return on this in section ten.
7.2 New closure theorems Several closure theorems were proposed by Steiner after 1832 [see the appendix of Steiner (1832)]. In one of these, he considered two (real) circles C1 and C2, the second lying inside the interior of the first, and a sequence of circles c1, c2, . . . , cn 78 After graduating, Max Simon (1844–1918) moved in Strasburg where he taught from 1871 until 1912.
His research dealt mainly with the history of mathematics.
123 Poncelet’s porism, I c1 c2 C c cn−1 cn C c c1 c2 c3 c12 (a) (b) Fig. 23 Steiner’s closure theorem for circles (1832) such that each of them is tangent to both C1 and C2, and ci is tangent to ci−1 for every i = 2, . . . , n (see Fig. 23a).
Steiner claimed that either the chain never closes whatever n is, i.e., cn is never tangent to c1 for any n, or the chain closes, i.e., cn is tangent to c1 for some n. In this case, the same happens for any similar chain of n circles whatever is the first circle c1 one considers (see Fig. 23b).
Denoting R1, R2 the radii of the two given circles, A the distance between their centers, and m the number of times the chain wraps around C2, Steiner gave the following conditional equation, Bedingungsgleichung (Steiner 1832, pp. 318–320), allowing the existence of the relative closed chain: (R1 −R2)2 −4R1R2 tan2 m n π = A2.
To prove the claim when C1 and C2 are concentric poses no difficulty. Since in this case the figure is completely symmetric, it is enough to apply some elementary geometries (Fig. 24a), and it follows that all circles in the chain have the same diameter.
It is also clear that does not matter from which position one starts: if one chain closes, then all chains close.
A suitable circular inversion allows us to pass from the case of concentric circles to the general case proposed by Steiner. In fact, such a transformation map circles into circles, lines into circles, and preserve tangency and angles (Fig. 24b).
Steiner proposed a new closure problem in Steiner (1846). Let E be a non-singular plane cubic and P, Q be two fixed point of it. Chosen a point A1 ̸= P, Q on E, the line P A1 meets E in a third point A2. The line Q A2 meets E, other then in A2 and Q, in another point A3. Similarly, the line P A3 meets E, other than in P and A3, in another point A4. Continuing on this way one gets a transversal A1A2 A3A4 . . . A2n A2n+1 inscribed in E. Then, he stated: there are two possibilities, either the polygonal line never closes, or it closes, i.e., A2n+1 = A1, forming a polygon of 2n sides inscribed in the cubic; in this case, the same holds true whatever is the initial point A1.
123 A. Del Centina θ θ (a) (b) Fig. 24 a The case of concentric circles is easy to solve. b By circular inversion one pass to the general case The theorem was proved by Clebsch as follows (see Clebsch, 1864, p. 106). If the polygonal line closes, argued Clebsch, then, according to Abel’s addition theorem, one has the following equivalences (in the group law on the cubic E): ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ A1 + A2 + P + u0 ≡0 A2 + A3 + Q + u0 ≡0 .
.
.
A2n−1 + A2n + P + u0 ≡0 A2n + A1 + Q + u0 ≡0 where u0 is a constant. Adding the first, third, fifth, …and penultimate equation, and subtracting from this sum the sum of the second, fourth, sixth,…and last equation, one gets n(P −Q) ≡0, which expresses the condition on the points P, Q which allows the existence of an inscribed 2n-gon.
Since this condition does not depend on the choice of the initial point A1, it follows that if P, Q satisfy it, then the transversal closes in 2n steps whatever is the initial point. Let us observe that if n = 3, then the above condition means that P, Q are both flexes of E.
A second proof of Steiner’s theorem was given by the Czech Eduard Weyr in his paper (Weyr 1870). In the same paper, Weyr also proved a similar theorem for a plane curve C of degree four with two ordinary double points P, Q (see Fig. 25). Precisely: given A1 ∈C {P, Q} the line P A1 meets C in a point B1, the line QB1 meets C in a point A2, and keep going on this way one has a sequence of points A1, A2, . . . A2n+1.
Then, Weyr proved that, if A2n+1 = A1 for some n, then the sequence always closes, after 2n steps, for any choice of A1 on C. This problem, as we will see later on, was revisited in Griffiths (1976, pp. 346–347).
123 Poncelet’s porism, I P Q A1 B1 A2 B2 A3 B3 A4 Fig. 25 Closure theorem that Steiner proposed in (1846) for a plane quartic curve with two nodes The first attempt to extend Poncelet’s theorem from conics to quadrics was made by Cayley. In his paper (Cayley 1853d), he tried to extend the reasoning of Cayley (1853a) to the case of two quadrics Q) x2 + y2 + z2 + w2 = 0 and Q′) ax2 + by2 + bz2 + dw2 = 0, by taking chords of Q touching Q′ in order to construct an inter-scribed polygonal line to the two quadrics. He was lead to consider hyperelliptic integrals of type " x = dx √(x + a)(x + b)(x + c)(x + d)(x −k)(x −k′), where k, k′ are two values of the parameter λ in λQ + Q′ = 0. After having developed some transcendental equations, Cayley soon realized that the theory of Poncelet poly-gons for conics could not be extended to quadrics in the simple way that one might be led to suppose.
Michel Chasles investigated the properties of n-gons which are inscribed in a given (real) ellipse (Chasles 1865). He found that among all such n-gons, there are infinitely many having maximum perimeter, and these all have their sides tangent to a second ellipse confocal to the first. Similarly, he proved that among all the n-gons circum-scribed about a given ellipse, there are infinitely many having minimum perimeter, and all these are inscribed in a second ellipse confocal to the first. These results were generalized to (real) ellipsoids by Darboux.
In the short note of 1870, Darboux stated three theorems regarding polygons inscribed in an ellipsoid and circumscribed about another, which extend to ellipsoids 123 A. Del Centina the results of Chasles. A polygonal line P1, P2, . . . , Pn+1 is said inscribed in an ellip-soid A, if all its vertices are on A; is said circumscribed about an ellipsoid B, if all its sides are tangent to B. If P1 = Pn+1, the polygonal line closes in a polygon of n-sides, which may be inscribed in A or/and circumscribed about B. The first and the third are: There are infinitely many polygons of n sides inscribed in an ellipsoid A having maximum perimeter, and all them are circumscribed to two ellipsoids B, B1 confocal to A; There are infinitely many polygons of n sides circumscribed about an ellipsoid B having minimum perimeter, and all them are inscribed in an ellipsoid A and (simulta-neously) circumscribed to another ellipsoids B1 confocal to B and A.
The second, and more important, theorem, can be stated as follows: Let B, B1 and A1, A2, . . . , An be confocal ellipsoids. In general, there are no n-gons having their vertices on A1, A2, . . . , An, whose sides are tangent to B and B1. If one such polygon exists, then infinitely many others exist enjoining the same property.
Darboux remarked that to prove the theorems was necessary the use of hyper-elliptic functionswithfourperiods.TheproofswereactuallygivenbymeansofAbel’saddition theorem for hyperelliptic integrals in the second volume of his treatise Leçons sur la théorie générale des surfaces (Darboux 1889, pp. 303–307).79 More than 100years later, these theorems, especially the second, became of great interest for physicists [see for instance (Dragovi´ c 2011) and the references therein]. In his note, perhaps foreshadowing the future, Darboux had written:“Un rayon lumineux qui se réfléchirait à l’intérieur de l’ellipsoide décrira ces polygones s’il est d’abord dirigé suivant le premier côté” [a ray of light that is reflected within an ellipsoid, will describe one of these polygons if is first directed along the first side].
In the years following Darboux continued to work on Poncelet’s theorem, producing the very interesting results that we will present in section nine.
A new type of Poncelet theorem for quadrics was proved by Weyr. He considered the smooth curve E intersection of two quadric Q and Q′ of rank ≥3 in P3. He fixed a ruling S on Q and a ruling on Q′. If A1 is any point of E, the line in S from A1 intersects E in another point A2, and the line in from A2 intersects E in another point A3. By proceeding in this manner, one gets a skew polygonal line connecting the points A1, A2, . . . A2n+1 on E. Weyr stated that if for some n the polygonal line closes, i.e., A2n+1 = A1, then also the polygonal line constructed starting from any other point of E closes after 2n steps (Weyr 1870, p. 28). He obtained the result as a corollary of the analogous theorem for plane quartics with two nodes that we have recalled above.
This last theorem can be seen as the historical origin of the Poncelet theorem in space of Griffiths and Harris that will be discussed in section fourteen of this paper.
79 At p. 307, Darboux quoted O. Staude, who in 1883 had proved the theorems by the use of the theta function with four periods (Staude 1883).
123 Poncelet’s porism, I 8 (2, 2)-Correspondences and closure problems From what we have seen above it is clear that Trudi and Cayley must be credited for having pursued Jacobi’s plan ahead of others, and for investigating the difficult determination of the relation between the coefficients of two conics when they admit an in-and-circumscribed n-gon. If Cayley had the great merit of having given this relation in explicit form, Trudi had that, as we will be clear shortly, of having brought to light the role that biquadratic binary equations play in the study of the Poncelet polygons and related questions.
The theory of algebraic correspondences, that arose in the 1850s in the school of Chasles, developed gradually in the second half of the nineteenth century in Chasles (1864), De Morgan (1865), Cayley (1866), Cremona (1867), Zeuthen (1871), Brill (1873) and others.80 Here, we shortly recall some basic facts of this theory that will be useful later.
Let f (x, y) be a polynomial of degree m in x and of degree n in y. The equation f (x, y) = 0, determines an (m, n)-correspondence between the variables x and y, in the sense that, to any value of x correspond n values y1, . . . , yn of y, while to any values of y correspond m values x1, . . . , xm of x. We may think of x, y as parameters fixing two points P, Q, respectively, on a line l and on a line l′, or, more generally, on two unicursal (i.e., rational) curves C and C′.81 A coincidence of a point P with one of its correspondent points Q1, . . . , Qn, occurs when one of the yi is equal to the x from which it arises. Therefore, such coincidences are given by the equation f (x, x) = 0, which is of degree m + n. The principle of correspondence (Chasles 1864) affirms that there are, in general, m + n coincidences.82 A branch point is a point such that two (or more) of its n corresponding points coincide. If we write f (x, y) = X0yn + X1yn−1+· · ·+ Xn, where the Xi have degree m in x, this equation in y has two coincident roots if ∂f ∂y = nX0yn−1+· · ·+ Xn−1 = 0.
Therefore, in general a (m, n)-correspondence has 2m(n −1) branch points.
A correspondence is said symmetric, if for any pair of corresponding points P, Q also Q, P is a pair of corresponding points. This means that the two polynomials f (x, y) and f (y, x) are identical.
80 For a historical study of this concept (see Segre 1892; Coolidge 1940).
81 The term unicursal was coined by Cayley, who also derived the fundamental properties of these curves (see Cayley 1866).
82 This principle is also referred as Chasles’ principle of correspondence, or even as the Chasles–Cayley– Brill principle of correspondence.
123 A. Del Centina Symmetric (2, 2)-correspondences are associated with biquadratic (sometime also called doubly quadratic) equations of the following type f (x, y) = ax2y2 + bxy(x + y) + c(x2 + y2) + dxy + e(x + y) + f = 0.
From above it follows that, a general symmetric (2, 2)-correspondence has four coincidence (or fixed points), corresponding to the roots of ax4 + 2bx3 + (2c + d)x2 + 2ex + f = 0, (8.1) and four branch points corresponding to the roots of D(x) = Q(x)2 −4P(x)R(x) = 0, where D(x) is the discriminant of the polynomial f (x, y) = P(x)y2 + Q(x)y + R(x) (as polynomial in y).
8.1 Cayley’s papers of 1871 As we have seen, the relation between Poncelet’s closure theorem and symmetric (2, 2)-correspondences emerged in part through the works of Trudi (1853, 1863a,b), and of Rosanes and Pash (1869).
Cayley at the beginning of his paper (1871a) wrote: The porism of the in-and-circumscribed polygon has its foundation in the theory of the symmetrical (2, 2) correspondence of points on a conic; viz. a (2, 2) correspondence is such that to any given position of either point there correspond two positions of the other point; in a symmetrical (2, 2) correspondence either point indifferently may be considered as a first point and the other of them will be the second point of the correspondence. Or, what is the same thing, if x, y are the parameters which serve to determine the two points, then x, y are connected by an equation of the form83 (∗) (x, 1)2(y, 1)2 = 0, which is symmetrical with respect to the parameters x, y.
It seems it is here that, for the first time, symmetrical (2, 2)-correspondences are explicitly associated with the construction of Poncelet’s polygons.
Before continuing with the exposition of Cayley’s paper, we explain what he meant by the above.
Let C and D be two non-singular conics in a plane π. One may suppose C rationally parameterized by a parameter s, so that to any point on C corresponds a value of s and vice versa. Let P be a point on C corresponding to the value x of the parameter. In 83 This is the symbolic form that Cayley used to write a doubly quadratic equation.
123 Poncelet’s porism, I the construction of Poncelet, to the point P correspond two points Q and Q′ on C, of parameter value, respectively, y and y′, such that the lines P Q and P Q′ are tangent to D. This construction gives a (2, 2)-correspondence on C, which is clearly symmetric, since any one of Q and Q′ can be chosen as first correspondent of P. Hence, the parameter values of corresponding points are connected by an equation of the form ax2y2 + bxy(x + y) + c(x2 + y2) + dxy + e(x + y) + f = 0.
Cayley showed that any symmetric (2, 2)-correspondence on a conic is defined in this way. One may suppose C to be given, in parametric equations, by (1, s, s2), so if X, Y, Z are the coordinates in the plane, the line lxy joining two points (1, x, x2) and (1, y, y2) on C is expressed by the equation X Y Z 1 x x2 1 y y2 = 0, that is xyX −(x + y)Y + Z = 0. In the dual plane π∗, this line is represented by the point (xy, −(x + y), 1). Let D be any non-singular conic, represented by its tangential equation (i.e., by the dual conic) of equation Aα2 + Bβ2 + Cγ 2 + Dβγ + Eαγ + Fαβ = 0.
The line lxy will be tangent to D if and only if Ax2y2 + B(x + y)2 + C −D(x + y) + Exy −Fxy(x + y) = 0, and this represents a general symmetric (2, 2)-correspondence.
Cayley showed that, whether symmetric or not, a (2, 2)-correspondence always leads to a differential equation of the form dx √ X ± dy √ Y = 0, where X, Y are polynomials of degree 4, respectively, in x and y, having the same coefficientsifthecorrespondenceissymmetric.Inthiscase,healsonoticedthat(p.85): if A and B are corresponding points, the corresponding points of B are A and a new point C; those of C are B and a new pointD, and so on; so that the points form a series A, B, C, D, E, F, . . .; and the porismatic property is that, if for a given position of A this series closes at a certain term, for instance, if F = A; then it will always thus close, whatever be the position of A.
This follows at once, observed Cayley, from the consideration of the differential equation dx √ X = dy √ Y and its complete integral of the form (∗).84 In fact, since this differential equation is integrable in the form 84 It is useful to recall (4.8) and (4.9).
123 A. Del Centina (y) −(x) = (k), by forming the equations for the corresponding points B, C; C, D;. . . and, assuming that the series closes after n steps, one has (z) −(y) = (k), .
.
.
(x) −(u) = (k), where (x) in the last equation must differ by a period of the integral from that in the first. Hence, by adding, Cayley got = n(k), or (k) = 1 n , which gives the condition on the coefficients of the equation (∗) for the series to close after n steps. This condition is independent of x, i.e., from the position of A.
We have to say that, although Cayley had showed how (2, 2)-correspondences come into play, in Cayley (1871a) he did not use the principle of correspondence in order to prove PCT. Nevertheless, this principle was used in another paper, which appeared in the same year 1871, to solve the following problem: compute the number of the in-and-circumscribed triangles to given curves. In that paper, he wrote (Cayley 1871b, p. 369): The problem of the in-and-circumscribed triangle is a particular case of that of the in-and-circumscribed polygon: the last-mentioned problem may be thus stated—to find a polygon such that the angles [vertices] are situate in and the sides touch a given curve or curves. And we may in the first instance inquire as to the number of such polygons. In the case where the curves containing the angle [vertices] and touched by the sides, respectively, are all of them distinct curves, the number of polygons is obtained very easily and has a simple expression …But when several of the curves become one and the same curve, and in particular when the angles [vertices] are all of them situate in and the sides all touch one and the same curve, it is a much more difficult problem to find the number of polygons.
Cayley considered a triangle of vertices a, c, e, respectively, on the three curves a, c, e, whose sides B, D, F are tangent, respectively, to the three curves B, D, F (see Fig. 26a). He computed, for 52 possible cases of coincidence among the six curves involved, the number of in-and-circumscribed triangles to them (the results were listed in a table, eight pages long, inserted in the paper), by using the theory of correspondences.85 He applied the principle of correspondence as follows (art. 1): consider the unclosed trilateral aBcDeFg (see Fig. 26b), where the points a and g 85 We have not verified these results except for conics, and only in the following cases: a = c = e and B = D; a = c = e; no conditions. In these cases, the number of triangles showed in the table seems to be correct. It would be of interest to reinterpret Cayley’s results in light of the modern algebraic geometry.
123 Poncelet’s porism, I B a c e D F B a c e g D F g (a) (b) Fig. 26 In his paper (1871b) Cayley used the principle of correspondence in order to prove Poncelet’s general theorem. The figures illustrate his procedure in order to get the proof are on the same curve a. Starting from an arbitrary point a on the curve a, let ac be any one of the tangents to the curve B, touching this curve, say at the point B, and intersecting the curve c in the point c. The same for the tangents ce and eg. Suppose that for a position of the point a there correspond χ positions g, g′, etc.. Similarly, suppose that starting from a position of g there correspond χ′ positions of a. Then, the points a, g, over the curve a, are in a (χ, χ′)-correspondence. When one of the points g coincides with one of the points a, the point a = g is a coincidence point of the correspondence, and the trilateral in question becomes an in-and-circumscribed triangle. Thus, the numbers of triangles are equal to that of coincidence points. By the general theory of correspondences this numbers is, in several of the cases but not in all, equal to χ + χ′. Cayley was able to express χ and χ′ in terms of the order, and of the class of the curves a, c, e and B, D, F, that he denoted, respectively, a, c, e, b, d, f , and A, C, E, B, D, F. Then, via the principle of correspondence and an accurate analysis of the possible situations of tangency (flexes, double tangents, etc.), he computed the required number.
8.2 Hurwitz’s general view Adolf Hurwitz, referring to the closure theorems, began his note (Hurwitz 1879) with the following words: Es giebt in der Geometrie eine grosse Anzahl von Sätzen, die aussagen, dass ein gewisses Ereiguiss unendlich oft Statt hat, sobald es nur ein Mal oder endlich oft eintritt [There is a large number of theorems in Geometry, affirming that if a certain event happens once then it happens an infinite number of times].
Hurwitz recognized that all these theorems (as those recalled in the previous sec-tion) have the common feature of being linked to (2, 2)-correspondences, and saw in the principle of correspondence the explanation for the existence of infinitely many solutions of the problem, when at least one solution exists.
123 A. Del Centina P1 = P7 P3 = P5 P2 = P6 A = P4 P1 = P6 P2 = P5 P3 = P4 (a) (b) Fig. 27 Hurwitz’s second example in (1879): a the case n = 6, b the case n = 5 To illustrate his thinking, Hurwitz examined various examples, and the first he considered it was that of Steiner’s chain of circles (Fig. 23a, b).
Let c1, c2, . . . , cn+1 be such a chain and denote by t1, t2, . . . , tn+1 the respec-tive points of tangency with the circle C. The law that associates t1 to tn+1 is a (2, 2)-correspondence on C, which in general has 4 coincidences. If the chain closes, remarked Hurwitz, i.e., for some n is tn+1 = t1, each point ti is a coincidence point.
Hence, if n > 2, the correspondence has more than 4 coincidence points, and then, equation (8.1) is an identity, i.e., every point on C is a coincidence point. This means that any chain of tangent circles closes in the same way, whatever is the starting point t1 on C.
The subsequent example was Poncelet’s closure theorem. As it is known, the con-struction of an inter-scribed transversal P1, P2, . . . , Pn+1 to two conics K1 and K2 leads to a symmetric (2, 2)-correspondence on K1. Hurwitz observed that, for any even number n = 2m and any point A ∈K1 ∩K2, one can construct a polygon of n sides, which is inscribed in K1 and circumscribed about K2. He proceeded as follows (Fig. 27a, illustrates the case n = 6). Let Pm+1 = A, the tangent from Pm+1 to K2 meets K1 in Pm+1 and another point Pm, then the tangent to K2 from this last point meets K1 in Pm and another point Pm−1, and continuing in this way one gets a point P1 on K1. It is clear that taking P1 as starting point of Poncelet’s construction, one gets a closed polygon of n sides inter-scribed to K1 and K2. Similarly, for any odd number n = 2m + 1 and for any of the four contact points on K1 of the four common tangents to the two conics, one can construct a polygon of n sides inter-scribed to them (Fig. 27b illustrates the procedure for n = 5). Hence, for any n there are always 4 coincidences of the correspondence. If there exists a proper polygon of n sides which is inter-scribed to K1 and K2, then, since each vertex is a coincidence point of the correspondence (which must be counted twice), there are in total 4+2n coincidences, and then, as above, it follows that every point of K1 is a coincidence point. This brief but conclusive reasoning gave Poncelet’s problem its true setting.
Hurwitz also examined the problem of Steiner polygons (Fig. 28a, b).
123 Poncelet’s porism, I P1 P5 = P6 P3 = P8 P2 = P9 P Q P4 = P7 P1 P4 = P5 P3 = P6 P2 = P7 P Q (a) (b) Fig. 28 Hurwitz’s third example in (1879): a the case n = 4, b the case n = 3 Let E be a non-singular plane cubic, and P, Q be two points on it. Starting from any point P1 on E, one can draw a polygonal line of vertices P1, P2, P3, . . . , P2n+1, which is inscribed in E and whose sides alternatively pass through P and Q. The map P P1 →P P2n+1 defines a (2, 2)-correspondence among the lines of the pencil through P. Hurwitz noticed that if n is even, and the line P Pn+1 is tangent to E at Pn+1, then P1 and P2n+1 coincide, i.e., one has a coincidence point (Fig. 27a illustrates the case n = 4)). The same happens if n is odd and the line QPn+1 is tangent to E at Pn+1 (Fig. 27b illustrates the case n = 3). Then, Hurwitz remarked, whatever the parity of n may be, the correspondence always has 4 coincidences, being 4 the tangents than can be drawn from each of the points P, Q. If another (proper) polygonal line closes forming an inscribed 2n-gon to E, then every edge of it is a coincidence point, and therefore, there are 4 + 2n > 4 coincidence points. This means that every point of E is vertex of a similar inscribed 2n-gon.
Finally, Hurwitz considered the problem proposed in Weyr (1870, p. 28). Let C4 be a quartic of first species, i.e., the smooth intersection of two quadrics surfaces Q1, Q2 in P3. Let s and σ be two secants of C4. The plane through s containing a point 1 of C4 (not on s) intersects C4 in another point 2 (we are using Hurwitz notation); the plane through σ containing 2 intersects C4 in another point 3; the plane through s containing 3 intersects C4 in another point 4, and so on. In this way, it is determined a polygonal line 1, 2, 3, 4, . . . , 2n + 1 of 2n-sides which is inscribed in C4 and whose sides alternatively meet s and σ (the “odd numbered” meet s, and the “even numbered” meet σ). By associating with the plane s, 1 the plane s, 2n + 1, one gets a symmetric (2, 2)-correspondence among the planes of the pencil through s. This correspondence always have 4 coincidences. They come, similarly to the case of the Steiner polygons, from the contact points of the 4 planes through s, or σ, that are tangent to C4. If the 123 A. Del Centina polygonal line closes, i.e., the point 2n + 1 coincides with 1, the point 1, and then, every vertex of the constructed polygon is a new coincidence point. Therefore, by the principle of correspondence, every point of C4 enjoys this property, and so every point of C4 is vertex of a 2n-gon ,inscribed in C4, whose sides alternatively meet s and σ.
It is evident that, by projecting C4 from one of its points into a plane, one has the Steiner theorem for a smooth cubic. Moreover, C4 projects doubly on a conic from the vertex S of one of the four cones in the pencil Q1+λQ2 = 0, and since the two rulings of a non-singular quadric are projected from S onto the tangents to another conic, it is clear that any Poncelet 2n-gon inter-scribed to these two conics is the projection from S of a 2n-gon inscribed in C4 of the type above.
Hurwitz concluded by saying (Hurwitz 1879, p. 15): Scliesslich sei noch darauf hingrwiesen, dass unser Kriterium immer nur das Resultat ergiebt, dass gewisse Aufgaben unendlich Lösungen haben, wenn sie Eiene oder eine endliche Anzahl von Lösungen haben besitzen; nicht aber auch die Möglich-keit, dass dieser Umstand wirklich eintreiten kann, was in vielen Fällen nicht selbstverständlich ist [Finally, we call attention on the fact that our criterion always only gives the certitude that if a problem admits a solution, or a finite number of solutions, then the problem admits infinitely many solutions, but it does not give the possibility to verify that at least one solution actually exists; this in many cases is not obvious at all].
If Cayley was the first to explicitly recognize the link between Poncelet’s polygons and symmetric (2, 2)-correspondences, Hurwitz was the first to complete the algebraic explanation of the “porismatic character” of certain questions. Loria in his remark-able work (Loria 1896), reporting on Hurwitz’s paper, at the end of the paragraph concerning the closure theorems, wrote: Non sappiamo se più ammirare la vastità di vedute o la perfezione della forma, e colla quale poniamo termine a questa digressione, alla quale invano cercheremo chiusa più degna [we do not know whether to wonder more at the breadth of views or at the perfection of the form, and so with this we bring to an end this digression, for which we should seek in vain a close more worthy.] 8.3 Geometric interpretation of (2, 2)-correspondences We end this section by introducing a geometrical interpretation of the (2, 2)-correspondences, suggested by the examples above, and that will be useful to have at hand in the sequel.
Any non-singular quadric is projectively equivalent to the Segre quadric S, embed-ding of P1 × P1 into P3 via the Segre map: (1, u) × (1, v) →(1, u, v, uv). If (z0, z1, z2, z3) are homogeneous coordinates in P3, then S is defined by z0z3 −z1z2 = 0.
A (2, 2)-correspondence au2v2 + bu2v + b′uv2 + cuv + du2 + d′v2 + eu + e′v + f = 0, 123 Poncelet’s porism, I is obtained intersecting S with the quadric Q defined by the equation az2 3 + bz3z1 + b′z3z2 + cz3z0 + dz2 1 + d′z2 2 + ez1z0 + e′z2z0 + f z2 0 = 0.
Conversely, the intersection of any quadric Q with the Segre quadric defines a (2, 2)-correspondence.
Any pair of corresponding points (P, Q) under a (2, 2)-correspondence is then associated with the intersection point of the x-line corresponding to P and the y-line corresponding to Q. So, to any (2, 2)-correspondence is associated a curve E = S ∩Q of bi-degree (2, 2) and vice versa. Generally, the curve E is non-singular, hence has genus 1, i.e., is an elliptic curve. In fact, the projection of E ⊂S ≃P1 × P1, on the first factor is a 2 to 1 morphism, ramified at the four points of the intersection of E with the plane z1 −z2 = 0 (corresponding to the condition u = v), and by the Riemann–Hurwitz formula it follows that E has genus 1.86 We remark that the branch points of the (2, 2)-correspondence are associated with the lines of the two rulings of S which are tangent to E.
9 The theorems of Darboux Gaston Darboux likely started to work on Poncelet’s theorems and related questions in 1868. At the end of that year, he presented a memoir to the Académie des Sciences on an important class of curves (and surfaces) of degree four. These curves were those resulting from the intersection of a sphere with a quadric, which he proposed to call cycliques (Darboux 1869, p. 1311). He had extended to these curves many of the more important properties of the circle, and, as a consequence, he had found a new proof of the Poncelet closure theorem: On obtient, comme conséquence de ces propriétés, une demonstration, nouvelle et indépendente de la théorie de fonctions elliptiques, du théorème del Poncelet sur les polygones inscrits et circonscrits. On démontre de même un théorème qui est un peu plus général que le théorème de Poncelet [We obtain, as a consequence of these properties, a new proof, independent of the theory of elliptic functions, of the theorem of Poncelet on inscribed and circumscribed polygons. We also prove a theorem which is somewhat more general than the theorem of Poncelet].
For some reason,87 the printing of this memoir was delayed, and Darboux decided to publish it elsewhere. The first part appeared in 1870, in the Memoirs of the Academy of Bordeaux (Darboux 1870b). In 1872, Darboux published the paper Sur un nouveau 86 See for instance (Griffiths and Harris 1978b). This formula was stated by Riemann and proved by Hurwitz in (1891). For the present case: if f : X →Y is a map of degree two from the curve X, of genus # g, onto the curve Y, of genus g, then 2# g −2 = 2(2g −2) + N, where N is the number of branch points, i.e., of those points p ∈Y such that f −1(p) contains only one point of X. Here, since g = 0 and N = 4, one has # g = 1.
87 See the foreword of Darboux (1873a).
123 A. Del Centina système de coordonnées et sur les polygones circonscrits aux conique. Here, he wrote (Darboux 1872, p. 100)88: Dans un mémoire présenté en 1868 à l’Académie des sciences, j’ai été conduit à une démonstration indirecte des théorèmes de Poncelet… Cette démonstration m’avait paru mériter d’être développée parce qu’elle donnait, sans l’emploi des coordonnées elliptiques, et au moyen d’une transformation analytique des plus simples, la proposition fondamentale de Poncelet… Depuis en examinant la méthode employée, j’ai reconnu qu’elle était susceptible d’extension, et que, par sa nature même, elle conduisait à des théorèmes ayant la plus grande analogie avec ceux de Poncelet, et qu’on peut considérer comme des généralisations des propositions de l’illustre géomètre. Si, après tant de belles démonstrations, soit analytiques, soit géométriques de ces propositions, je me permets d’en proposer une nouvelle, c’est que cell-ci me parait réellement se distinguer par quelques principes qui n’ont pas encore été employés dans l’étude de cette question [In a memoir presented in 1868 to the Academy of Sciences, I have been lead to an indirect proof of the Poncelet theorems …This proof seemed to me worthy of being developed, because it gives, without the use of elliptic functions, and by means of very simple analytical transformations, Poncelet’s fundamental proposition of …After having examined the method used, I have recognized that it was capable of extension, and that, by its own nature, it led to theorems having a greatanalogywiththoseofPonceletandthatcanbeconsideredasgeneralizations of the propositions of the illustrious geometer. If, after the many beautiful proofs, both analytic and geometric, of these propositions I allow myself to present here a new one, it is because it actually seems to me to be distinguished for some principles that have not yet being used in the study of that question].
The second part of the memoir that he had presented to the Paris Academy of Sciences followed in 1873, published in the same journal where the first part had appeared. The same year, a new redaction of the whole memoir was printed, in the form of a book, by Gauthier–Villars (Darboux 1873a). Here, in the Notes et Additions, severalin-depthstudieswereincluded.NoteII,titledSurunedémonstrationanalytique des théorèmes de Poncelet, et sur un nouveaux système de coordonnées dans le plane, which we refer to as (Darboux 1873b), contained in its first five sections the paper Sur un nouveau système de coordonnées etc., published the year before.
The new proofs of Poncelet’s theorems were based on the properties of certain curves, today known as “Poncelet curves”or “Poncelet–Darboux curves”: curves of degreen passingthroughtheintersectionpointsofn+1tangentstoagivenconic.89 The key point in Darboux’s approach was the introduction of a new system of coordinates, by which every point of the plane is seen as the point of intersection of two tangents to a fixed conic. This new system of coordinates, today called “Darboux coordinates” [see 88 Darboux was referring to the proof in art. 38 of his memoir (see Darboux 1873a, p. 99).
89 These curves were called “Poncelet curves” in Böhmer (1985), but recently the name of “Poncelet– Darboux curves” seems to be preferred, see for instance (Dragovi´ c 2011). We will adopt the second name, in honor of Darboux who introduced them.
123 Poncelet’s porism, I for instance (Dragovi´ c 2011)], was suggested to Darboux by the Chasles representation of a quadric as a double plane (see Darboux 1917, pp. 236–237).90 9.1 Darboux coordinates and Poncelet–Darboux curves All tangents to a non-singular conic (K) can be obtained by varying the parameter m in the equation αm2 + βm + γ = 0, (9.1) where α, β, γ are linear functions of the coordinates in the plane.91 Then, since from any point on (K) only one tangent can be drawn to it, the conic (K) can be represented by the equation β2 −4αγ = 0.
(9.2) If a tangent to (K) passes through a point (α′, β′, γ ′), then m must satisfy the equation α′m2 + β′m + γ ′ = 0. So, putting ρ, ρ′ its roots, one has α′ = β′ ρ + ρ′ = γ ′ ρρ′ · (9.3) Vice versa if ρ, ρ′ are given, the equations (9.3) determine the point (α′, β′, γ ′).
Darboux considered (ρ, ρ′) as new coordinates in the plane (Darboux 1873, p.
184). These are called Darboux coordinates. Clearly, with respect to them, the conic (K) has equation (ρ −ρ′)2 = 0.
(9.4) Moreover, any conic having two double contacts with (K), i.e., given by K −L2 = 0, where L is a linear form, can be written d(ρ −ρ′) = aρρ′ + b(ρ + ρ′) + c, where a, b, c, d are constants and that is of the form Aρρ′ + Bρ + Cρ′ + D = 0.
An algebraic equation, of degree m in ρ and of degree m′ in ρ′, f (ρ, ρ′) = 0 defines an algebraic curve in the plane. It is easy to see that this curve is of degree m or m +m′, according as f is symmetric or not with respect to ρ and ρ′. Moreover, a count 90 The projection of a non-singular quadric Q on a plane π from a point A / ∈Q, not passing through A, gives a birational map of degree two, from Q onto π, which is branched along the conic (K) := Q ∩π.
Under this map, all the lines of the two rulings of Q are mapped into the tangents to the conic (K).
91 The tangent to the conic : f (x, y, z) = 0 at P ∈ is given by (∂x f )P x + (∂y f )P y + (∂z f )P z = 0.
If the conic is given by y2 −xz, and P is (m2, m, 1), the tangent at P has equation m2x + my + z = 0.
Since (K) can be mapped onto by a suitable projective transformation of the plane, it is clear that by applying the inverse map the claim follows.
123 A. Del Centina of constants shows that any curve of degree m is represented by an equation of this type, which is symmetric with respect to ρ and ρ′, and depends on (m + 1)(m + 2)/2 arbitrary constants (p. 185–186).
Here, Darboux wrote:“Nous pouvons, à l’aide de ces seules remarques, démontrer plusieurs théorèmes généraux sur les polygones inscrits et circonscrits” [By means of only these remarks, we can prove several general theorems on inscribed and circum-scribed polygons].
We will focus only on some of these theorems, the first of which is the following: Theorem D1 If a curve of degree n passes through the n2 points of intersection of two systems of n tangents to the conic (K), then it contains infinitely many sets of n2 points, each of them constituting the intersection locus of two systems of n tangents to (K).
Let A1, A2, . . . , An and B1, B2, . . . , Bn be two system of n lines in the plane. The intersection points Ai ∩B j, i, j = 1, . . . , n, constitute a set of n2 points. Any curve of degree n passing through these n2 points is given by A1A2 · · · An −kB1B2 · · · Bn = 0.
(9.5) If all the lines are tangent to the conic (K), and then ⎧ ⎨ ⎩ Ai = αa2 i + βai + γ = α(ai −ρ)(ai −ρ′) Bi = αb2 i + βbi + γ = α(bi −ρ)(bi −ρ′), (9.6) for I = 1, . . . , n, by setting ⎧ ⎨ ⎩ ϕ(ρ) = (ρ −a1)(ρ −a2) · · · (ρ −an) ψ(ρ) = √ k(ρ −b1)(ρ −b2) · · · (ρ −bn), (9.7) then the equation of the curve becomes ϕ(ρ)ϕ(ρ′) = ψ(ρ)ψ(ρ′), or ϕ(ρ) ψ(ρ) = ψ(ρ′) ϕ(ρ′) .
(9.8) Darboux observed (as he had already done in section n. 28 of his book), that this equation can be written (ρ) (ρ) = (ρ′) (ρ′), (9.9) where (u) = mϕ(u) + nψ(u), (u) = nϕ(u) + mψ(u), 123 Poncelet’s porism, I which is of the same form as (9.8), but the roots of the polynomials are different.
Since equation (9.9) contains a new parameter that can assume any arbitrary value, Darboux declared the theorem proved (p. 187).
As example he observed that, if a conic contains the four vertices of a quadrangle circumscribed to another conic, then it contains the vertices of infinitely many other quadrangles circumscribed to the same conic.
The second theorem is the following Theorem D2 If a curve of degree n contains the vertices of a (n + 1)-gon, whose sides are tangent to a given conic (K), then it contains the vertices of infinitely many (n + 1)-gons whose sides are tangent to (K).
Let A0, A1, A2, . . . , An be n + 1 tangents to (K). Darboux observed that every curve C of degree n, which passes through the n(n + 1)/2 points of intersection of the tangents, is represented by an equation of the form a0 A0 + a1 A1 + · · · + an An = 0, (9.10) where a0, a1, . . . , an denote arbitrary constants.92 Each line Ai, being tangent to (K), has equation Ai : α(bi −ρ)(bi −ρ′) = 0, and then (9.10) can be written ai (bi −ρ)(bi −ρ′) = 0, (9.11) or, multiplying by ρ −ρ′, ai (bi −ρ) = ai (bi −ρ′), (9.12) and this equation is readily seen to be of the form f (ρ) ϕ(ρ) = f (ρ′) ϕ(ρ′) .
(9.13) Vice versa, all equations of this type can be reduced to the form (9.12), which represents a curve of degree n, containing all vertices of the polygon circumscribed to (K), and whose sides are defined by ϕ(ρ) = 0.
Darboux concluded that, since equation (9.13) can be put in the form f (ρ) ϕ(ρ) + k f (ρ) = f (ρ′) ϕ(ρ′) + k f (ρ′), (9.14) where k is an arbitrary constant, the theorem is proved.
92 This can be easily proved by induction on n (see Darboux 1873a, pp. 191–192).
123 A. Del Centina The curves that satisfy the conditions of theorem D2 are called Poncelet–Darboux curves of degree n related to the conic (K).
Darboux deduced the PCT as a corollary of theorem D2, by arguing as follows.
Suppose that a conic (C) contains the n + 1 vertices of a polygon of n + 1 sides A0, A1, . . . , An, which is circumscribed to the conic (K). One may fix the coefficients a0, a1, . . . , an in the equation (9.10) so that the curve C intersects (C) in other n points, besides the n + 1 vertices of the inscribed polygon. Then, by Bezout’s theorem, the curve C having at least 2n + 1 points in common with C decomposes in the conic (C) and another curve (C′) of degree n −2. Therefore, from theorem D2, the curve C = (C)∪(C′) contains the vertices of ∞1 polygons circumscribed to (K). It follows that the conic (C) is circumscribed to every such polygons, and the Poncelet closure theorem is proved.
Actually Darboux went further, and he showed that the curve (C′) completely decomposes into conics, or in conics and a line, according to the parity of n. The conic (C) has an equation of the form A(ρ2 + ρ′2) + Bρρ′ + Cρρ′(ρ + ρ′) + D(ρ + ρ′) + Eρ2ρ′2 + F = 0, (9.15) a tangent ρ = a intersects (C) in two points, determined by the values ρ1, ρ2 of ρ.
These two values are the coordinates of another vertex of the polygon, and since these two must satisfy an equation of the form above, the new vertex will move along a conic. Continuing in this way, it became clear that (C′) decomposes in k conics, or in k −1 conics and a line, according as n = 2(k + 1) or n = 2k + 1.
Up to this point, we have discussed the first five sections of Darboux (1873b), i.e Darboux (1872). In the remaining sections of his paper, Darboux proved others interesting results, such as those on hyperelliptic integrals or quartic curves with two nodes. He also gave a new proof of Chasles’s theorem on confocal conics and showed a connection between Euler’s differential equation and equations of type (9.15).
In his paper (Darboux 1880), he exhibited the connection between the existence of a Poncelet n-gons, and the rational transformations of elliptic integrals.
In 1917, Darboux published his last treatise Principes de géométrie analytique.93 In the third part of the book, titled Les théorèmes de Poncelet, he gathered the many results on Poncelet’s theorems and related questions that he had obtained over the years. Probably, apart from some improvements or new proofs that he could have obtained later, Darboux achieved many of these results before 1880. So we think that it is not too long a chronological leap to present them here, before commenting on the contributions to the theory that were made in the twentieth century.
9.2 Biquadratic equations and a new proof of PGT Darboux divided part III into three chapters and devoted the first section of the first chapter to the definition of his new system of coordinates. In doing so, he introduced a few changes with respect to Darboux (1872) that it will be convenient to state here.
93 In it Darboux presented, in coordinated form, the lectures he had delivered at intervals since 1872, either at the Sorbonne and at the École Normale.
123 Poncelet’s porism, I He considered the conic (K) given by y2 −xz = 0, so that any tangent to (K) is represented by the equation m2x −2my + z = 0, where m is a parameter. Then, if (x, y, z) are the homogeneous coordinates of a point P determined by the roots ρ, ρ1 of the equation above, it follows that 2y = x(ρ + ρ1), z = xρρ1, and the equation of (K) becomes (ρ −ρ′)2 = 0. Since the above formulae are symmetric with respect to ρ and ρ1, a curve F(x, y, z) = 0 of degree m is represented in the new system of coordinates by the equation F 1, ρ + ρ1 2 , ρ1 = 0, which is symmetric and of degree m in ρ, ρ1. Vice versa, any such equation in ρ, ρ1 represents a curve of degree m. For instance Aρρ1 + B(ρ + ρ1) + C = 0 represents the line Az + 2By + Cx = 0, and Aρ2ρ2 1 + Bρρ1(ρ + ρ1)) + C(ρ + ρ1)2 + 2Dρρ1 + E(ρ + ρ1) + F = 0 represents the conic Az2 + 2Byz + 4Cy2 + 2Dzx + 2Exy + Fx2 = 0.
In the subsequent sections of the first chapter, Darboux gave theorems D1, D2 and other theorems already published in Darboux (1873a), also giving them alternative proofs.
In the second chapter, he proved the following: Theorem D3 Let (C) and (K) be two conics. Suppose that from a point M on the conic (C) are drawn the two tangents to the conic (K) and that from the two new points M1, M−1 of intersection of these tangents with the first conic, new tangents to (K) are drawn, and so on. It results from this construction a polygonal line inter-scribed to the two conics . . . M−h . . . M−1M M1 . . . Mh . . ., which can be prolonged in two directions, such that the parameters ρi of Mi and ρi+k of Mi+k, (i > 0), satisfy a symmetric biquadratic equation fk−1(ρi, ρi+k) = 0.
123 A. Del Centina To prove this theorem, he proceeded by induction on k (Darboux 1917, n. 160). He observed that any pair of consecutive parameters ρi, ρi+1 satisfies an equation of the form: $ f (ρi, ρi+1) = Aρi 2ρ2 i+1 + Bρiρi+1()ρi + ρi+1) + C(ρi 2 + ρ2 i+1) +Dρi2ρ2 i+1 + E(ρi 2 + ρ2 i+1) + F = 0.
(9.16) By eliminating ρi from f (ρi, ρi+1) = 0 and f (ρi, ρi−1) = 0, he obtained an equation of bidegree 4 containing the factor (ρi−1−ρi+1)2. Dividing the resultant by this factor, he obtained a biquadratic equation f1(ρi−1, ρi+1) = 0, symmetric with respect the two variables. Since by replacing i by i +1, it follows that ρ, ρi+2 satisfy the equation f1(ρ, ρi+2) = 0, he had proved the theorem for k = 2. Then, he supposed that for a certain value of k the following holds f (ρi, ρi+1) = 0, f1(ρ, ρi+2) = 0, . . . , fk−1(ρi, ρi+k) = 0.
The resultant of the elimination of ρi+k between fk−1(ρi, ρi+k) = 0, f (ρi+k, ρi+k+1) = 0, (9.17) is an equation (ρi, ρi+k+1) = 0 of degree 4 with respect to both the variables.
Since the second of the equations (9.17) is verified when ρi+k+1 is replaced by ρi+k−1, Darboux affirmed that the same holds true for the previous equation. From this, he deduced that the first member must contain a factor fk−2(ρi, ρi+k−1), which equated to zero gives the relation between ρ1 and ρi+k−1. Hence, = fk−2(ρi, ρi+k+1) fk(ρi, ρi+k+1), where fk(ρi, ρi+k+1) is a biquadratic polyno-mial with respect to the two variables ρi, ρi+k+1. Then, the required relation is fk(ρi, ρi+k+1) = 0. Since fk is of the same form of fk−1, fk−2, . . . the second step of the induction holds true, and the theorem is completely proved.
Darboux observed that together with fk−1(ρi, ρi+k) = 0, also holds fk−1(ρ−i, ρ−i−k) = 0, and, since i is any integer, by changing i with −i−k it follows that fk−1(ρi+k, ρi) = 0, i.e., the primitive relation is symmetric.
From this, he deduced the following (n. 161): Corollary All equations fk−1(ρi, ρi+k) = 0 have equal roots for the same values of ρi.
Then, he gave a geometrical interpretation of the above results. Since the equa-tions fk−1(ρi, ρi+k) = 0 are symmetric with respect to the variables, the point of coordinates ρi, ρi+k will describe a conic (Ck−1).
On the other hand, the points of this conic for which the two values of ρi+k coincide are those where it is touched by one of the common tangents with (K). Since these points always correspond to equal values of ρi, he could claim (Fig. 29a): 123 Poncelet’s porism, I A1 A2 A3 A4 A5 A6 A7 (K) (C) l2 l5 l3 l6 A1 A1 A2 A3 A4 A5 A6 A7 A2 A3 A4 A5 A6 A7 (a) (b) Fig. 29 a Theorem D4. b Darboux applied Theorem D4 in order to prove the Poncelet closure theorem for n ≥5. The figure illustrates the case n = 7 Theorem D4 If a polygonal line moves while remaining inscribed in (C) and cir-cumscribed about (K), the intersection point of any two sides whose indexes differ by k, for instance the ith and (i + k)th, always describes a conic inscribed into the quadrilateral of their common tangents.
In light of this theorem, Darboux gave the following new proof of PCT.
Since he had already proved the theorem for n = 3, 4 (n. 86, 150), he supposed n ≥5. He let A1A2 . . . An be an inter-scribed polygon to (C) and (K), and let A′ 1A′ 2 . . . A′ n+1 be a transversal constructed as above, starting from any point A′ 1 on (C) (see Fig. 29b). In view of the above theorem, the intersection point of the two sides A′ 1A′ 2, A′ n A′ n+1 describes a conic (C′) which must pass through all the vertices of the inter-scribed polygon, and, there being at least five of, these the conic (C′) must coincide with (C).
Darboux stressed that this proof, although less simple than the one that could be achieved directly from the previous theorem, had the advantage of highlighting the following corollary (Fig. 30): Corollary If a polygon moves remaining inter-scribed to the conics (C) and (K), the conic (Ck−1), described by the intersection point of two sides indexed by i and i + k, is inscribed in the quadrilateral of the common tangents to (C) and (K). Moreover, the diagonals of the polygon envelop conics belonging to the pencil defined by (C) and (K).
Darboux devoted chapter three to discuss the general Poncelet theorem. He pre-sented essentially two proofs of it, the first based on the theory of conic envelopes (n.167–169),thesecondonthepropertiesofEuler’sdifferentialequation(n.170–171).
Here, for brevity, we will comment only the second, as it is of greater interest from a historical point of view.
He considered two conic ( f ) and (ϕ), represented by their tangential equations: f = a0u2 + a2v2 + a4w2 + 2a3vw + 2a2uw + 2a1uv = 0, (9.18) ϕ = v2 −4uw = 0.
(9.19) 123 A. Del Centina (C) (K) (C1) Fig. 30 An illustration of the corollary to Theorem D4 for n = 5 Then, the equation of the tangential pencil f + mϕ = 0 is F = − 0 x y z x a0 a1 a2 −2m y a1 a2 + m a3 z a2 −2m a3 a1 = 0.
(9.20) He put this equation in the form F = H + Km + Lm2 = 0, (9.21) where H, K, L are polynomials of degree 2 in the variables x, y, z. By adopting the coordinates ρ, ρ1, he expressed H, K, L in terms of ρ, ρ1 and observed that equation (9.21) becomes of degree 2 with respect to both the new variables. Putting f (ρ) = a0ρ4 + 4a1ρ3 + 6a2ρ2 + 4a3ρ + a4, the discriminant of (9.21) is of the form K 2 −4H L = f (ρ) f (ρ1).
(9.22) Considering (9.20) as an equation in ρ1, that is F = Pρ2 1 +Qρ1+ R = 0, he computed its discriminant, getting Q2 −4P R = f (ρ)[4m2 −im −j], (9.23) where i = aoa1 −4a1a3 + 3a2 2 and j = a0a2a4 + 2a1a2a3 −a0a2 3 −a4a2 1 −a2 2.
For F = P1ρ2 + Q1ρ + R1 = 0, it follows Q2 1 −4P1R1 = f (ρ1)[4m2 −im −j].
(9.24) 123 Poncelet’s porism, I After setting (m) := 4m2 −im −j, he differentiated equation (9.21) with respect to m, ρ, ρ1, obtaining (2Lm + K)dm + (2Pρ1 + Q)dρ1 + (2P1ρ + Q1)dρ = 0.
(9.25) Since ⎧ ⎨ ⎩ 2Lm + K = ±√f (ρ) f (ρ1) 2Pρ1 + Q = ±√f (ρ)(m) 2P1ρ + Q1 = ±√f (ρ1)(m) (9.26) equation (9.25) gives dm √(m) = ± dρ √f (ρ) ± dρ1 √f (ρ1), (9.27) which reduces to dρ √f (ρ) ± dρ1 √f (ρ1) = 0, (9.28) if one moves on one of the two conics of the pencil which pass through the point (ρ, ρ1).94 Darboux showed that the converse also holds true; hence, he had proved the following theorem: Theorem D5 If ρ and ρ1 vary so that they satisfy one or the other (according the sign) of the above differential equations, the point (ρ, ρ1) describes one of the conics of the pencil, and vice versa.
This established, Darboux proceed to give a new proof of the PGT.
His reasoning was as follows. Let A1A2 . . . An be a n-gon circumscribed about the base conic of the tangential pencil, and suppose that it moves so that all its vertices but one describe other conics of the pencil. He put ρ, ρ1, . . . , ρn be the parameters of the different sides, and supposed, without loss of generality, that the n −1 vertices (ρ1, ρ2), (ρ2, ρ3),…,(ρn−1, ρn) describe conics of the pencil. Hence, by the theorem above, the following relations must hold: dρ1 √f (ρ1) = ± dρ2 √f (ρ2), . . .
, dρn−1 f (ρn−1) = ± dρn √f (ρn)· Then, by eliminating the intermediate variables ρ2, . . . , ρn−1, he obtained dρ1 √f (ρ1) = ± dρn √f (ρn) which shows that the free vertex (ρ1, ρn) of the polygon also moves along a conic of the pencil. Thus, he had proved (see Fig. 31): Theorem D6 If a polygon moves while remaining circumscribed about a conic (K) in such a way that all its vertices except one, describe the conics (K1), . . . , (Kn−1) 94 We recall that the pencil considered is a tangential pencil.
123 A. Del Centina (K) (K1) (K2) (K3) Fig. 31 An illustration of Theorem D6 for n = 6 all inscribed in a quadrangle circumscribed about (K), i.e., belonging to the same tangential pencil as (K), then also the last vertex of the polygon describes a conic belonging to this pencil.
Finally, Darboux observed that by a transformation by reciprocal polars (i.e., by duality), this amounts to Poncelet’s general theorem.
We stress the similarity with the proof given by Trudi. Darboux never quoted Trudi, whose papers he had probably not read.
10 Poncelet polygons in Halphen’s treatise Georges Henri Halphen became interested in Poncelet polygons in the late 1870s (Halphen 1878, 1879a,b). In Halphen (1878), as an application of the results devel-oped therein, he computed the number of conics, from a given system whose first characteristic is μ,95 containing the vertices of a triangle, or of a quadrangle, which in turn is circumscribed about a fixed conic from the same system. Halphen used Salmon’s conditions β = 0, γ = 0 (see section 6 above),96 to show that in the first case α = 2, β = 0, and in the second α = 3, β = 0. So he found, respectively, 2μ and 3μ, and then for a pencil these numbers are 2 and 3.
One year later, in the short note (Halphen 1879a), he wrote: On sait,... pour que deux coniques A, B soient ainsi, la première inscrite, la secondecirconscriteàunpolygonedem côtés,ilfautetilsuffitqueleurséléments satisfassent à une seule relations. Cette relation a été explicitement formée par divers géomètres pour les nombres m les plus simples, sans qu’on ait jusqu’à 95 A system of conics S is given by an equation of second degree % ai j(λ)xi x j = 0, whose coefficients depend on a parameter. Chasles defined first characteristic of the system the number μ of conics in S which pass through a point, and second characteristic of the system the number ν of conics in S which are tangent to a line. Perfecting Chasles’ theory (see Halphen 1878, pp. 27–31), Halphen proved that the number of conics in a system of characteristics (μ, ν) which satisfy a projective condition, is, under certain hypothesis, αμ + βν, where α and β are positive integers depending on the condition. In particular, if the system is a pencil μ = 1 and ν = 2.
96 Halphen quoted Salmon, Higher Algebra, in the French translation by Bazin, p. 203.
123 Poncelet’s porism, I présent découvert quelle en est la loi. Cette loi est certaiment fot compliquée et, comme on le sait d’après Jacobi, n’est autre que la loi des polyno[m]es naissant de la multiplication des fonctions elliptiques…Si l’on suppose donnée la conique B et que l’on astreigne la conique A à faire partie d’un système S, il y a parmi les coniques de ce système plusieurs solutions A. On demande le nombre. [It is known,…for two conics A and B such that the first is inscribed in, and the second circumscribed about, a polygon of m sides, it is necessary and sufficient that their coefficients satisfy only one condition. This relation has been explicitly found by several geometers for the more simple [the first] numbers m, without having yet discovered what the [general] law is. This law is certainly very complicated, and after Jacobi as is well known, it is nothing but the law of polynomials arising from the multiplication of the elliptic functions…If one supposes that the conic B be given and the conic A is forced to belong to a system S, then among the conics of this system there are several solutions A. One asks for the number.] His claim“…without having yet discovered what the law is” sounds rather strange.
Since Halphen was well aware of Cayley’s result, it could be that he was only referring to the conditional equations expressed in terms of invariants, as in Salmon’s Conics.
Anyway, Halphen was here looking for the number of conics A, in a same system as a given conic B, such that there is n-gon which is inscribed in B and circumscribed about A.
According to modern literature (see Barth and Michel 1993), the conics A will be said n-inscribed in the conic B, and, reversing the situation, the conic B is said n-circumscribed about the conic A. So Halphen wanted to find the number of conics in a pencil which are n-inscribed in a given one from the same pencil. The question he was considering was an essentially new, and difficult, problem in the landscape of Poncelet’s polygons.
After having recalled the result for triangles and quadrilaterals as above, Halphen continued by saying: Des considérations tirées de la théorie des caractéristiques conduisent aisément à conclure que, pour le cas général, le nombre cherché est toujours de la forme Mμ, M étant un nombre qui ne dépend que de m. Mais la détermination de ce nombre M n’est pas sans difficulté. Il ma fallu de faire une étude assez approfondie de la rélation générale, dont la loi n’est pas explicitement connue, pour lever cette difficulté. J’y suis parvenu, et je peux actuellement donner le théorème suivant [Considerations deduced from the theory of characteristics easily lead to the conclusion that, for the general case, the required number is always of the form Mμ, M being a number depending only on m.97 But the determination of this number M is not at all without difficulty. It took me an in-depth study of the general relation, whose law is not yet known explicitly, to overcome this difficulty. I succeeded and now I can formulate the following theorem] 97 Here M has the meaning of the previous α. In particular, Halphen asserted that, in these cases, one has always β = 0.
123 A. Del Centina then Halphen stated that the number M is given by the following formula M = 1 4m2 1 −1 p2 1 −1 q2 1 −1 r2 · · · , (10.1) where p, q,r, . . . are the primes in the prime factorization of n = pαqβrγ · · · .
He neither proved nor explained this formula, which fits well in the cases m = 3, 4 seen above. However, since 4M equals the number of the primitive m-th part of the periods of an elliptic function (with additive group of periods !), i.e., of those w such that mw ≡0(mod !) but kw ̸= 0(mod !) for any k which divides m, precisely: T (m) = (p2 −1)p2(α−1)(q2 −1)q2(β−1)(r2 −1)r2(γ −1) · · · , (10.2) we may think that Halphen used the multiplication of the argument to write down (10.2) and (10.1). In fact, this was so, as we will see in a while.
The first of the three volumes constituting Halphen’s Traité des fonctions elliptiques et de leurs applicationswas printed in 1886. With this fundamental work, Halphen pre-sented the theory of elliptic functions, in terms of the new functions introduced by Weierstrass, mainly the ℘(u) and the σ(u), that he found more suitable, especially in dealing with applications, than Jacobi’s sn(u) and cn(u). The second volume, devoted to the applications to mechanics, physics, geodesy, geometry and integral calculus, followed in Halphen (1888). The third was printed posthumously in 1891. This last volume, instead of the theories of the modular equation and of the complex multi-plication together with a historical survey, as Halphen had projected, contained only some unpublished manuscripts on the division of periods, an article already published on the complex multiplication, and some fragments (Halphen 1891).
In the treatise, Poncelet polygons make their first entry at the end of chapter I of the first volume, where, in discussing the geometrical interpretation of the addition formulae for the Jacobian elliptic functions, Halphen forwarded again Jacobi’s proofs of the theorems of Poncelet for circles.
Halphen devoted the whole chapter X of the second volume to the same subject.
In fact, he entitled it “Les polygones de Poncelet.” Here, he presented the numerous results on this topic that he had probably been collecting since 1878. Among other things, he proved Poncelet’s theorems by means of symmetric (2, 2)-correspondences, determined the closure conditions and re-obtained Cayley’s formulae. Then, he intro-duced the “elliptic representation” of points of the plane to study the problem of finding the number of conics from a pencil that are n-circumscribed about a given one belonging to the same pencil: the question he had considered 10years before.
Halphen returned on Poncelet’s theorems in chapter XIV, where he gave new proofs based on the development in continued fractions of √ X, with X is a polynomial in one variable of degree 3 or 4.
Gino Loria, in his historical account, did not comment on the results on Poncelet polygons contained in Halphen’s treatise, published the year before, partly because, as he wrote, “facendo parte di un’opera voluminosa, non si può fotografare in poche frasi con sufficiente chiarezza” [being part of a voluminous work, it cannot be photographed in a few sentences with sufficient clarity] (Loria 1889a, p. 20).
123 Poncelet’s porism, I As we have said above, the third volume of Halphen’s treatise contained some unpublished fragments. Among these, we can find the proof of formula (10.2) that we summarize here below (Halphen 1891, pp. 194–201).
He denoted by wn a (nonzero) nth part of a period of the Weierstrass’s function ℘, with n a prime number. In this case, modulo periods, all wn, are defined by the formula wn = 2pω + 2p′ω′ n = (p, p′), where ω and ω′ are half-periods, and 0 ≤p, p′ ≤n −1 are integers such that p2 + p′2 ̸= 0. Their number is n2 −1, and they form a group, partitioned in n + 1 cyclic group of n −1 elements. Halphen observed that, when n is any positive integer, the formula above gives a nth part of period if and only if p, p′ and n are relatively prime (i.e., (p, p′, n) = 1). Moreover, a cyclic group if formed by multiplying an element wn by the integers less than n and relatively prime to n.
Then, Halphen supposed n = aα, with a a prime number. In this case, wn is a n-part of a period if and only if one—at least—of p, p′ is prime with a. So the number of wn is n2 −1 − &n a 2 −1 ' = n2 1 −1 a2 .
Now, mwn is a nth part of a period if m is not divisible by a. Then, taking m in the sequence 1, 2, . . . , n −1, the numbers a, 2a, 3a, . . . , (aα−1 −1)a must be excluded, and so only aα −1 −(aα−1 −1) numbers m remain. He set ϕ(n) = aα −aα−1 = n 1 −1 a .
Reasoning as above, he concluded that the number of groups is T (n) = n 1 + 1 a .
Finally, Halphen supposed that n = aαbβcγ · · · , and he put p n = p1 ah + p2 bk + p3 cl + · · · ; p′ n = p′ 1 ah′ + p′ 2 bk′ + p′ 3 cl′ + · · · In this way, observed Halphen, the wn are given as sum of elements of the form p1ω ah + p′ 1ω′ ah′ ; p2ω bk + p′ 2ω′ bk′ ; . . .
123 A. Del Centina Then, wn will be a n-part of a period if and only if one of the exponents h, h′ is equal to α, one of k, k is equal to β, and so on. Hence, wn is the sum of elements wn1, wn2, . . ., with n1 = aα, n2 = bβ, n3 = cγ , etc. Moreover, there will also be elements wn belonging to different groups, because the condition pr′ −rp′ ≡ 0(mod n) decomposes in p1r′ 1 −r1 p′ 1 ≡0(mod n1), p2r′ 2 −r2 p′ 2 ≡0(mod n2), one of which, at least, does not hold by the hypothesis. Therefore, it follows that the number of groups is T (n) = T (n1)T (n2) · · · = n 1 + 1 a 1 + 1 b · · · , the number of elements in each group is ϕ(n) = ϕ(n1)ϕ(n2) · · · = n 1 −1 a 1 −1 b · · · , and then the number of wn is ϕ(n)T (n) = n2 1 −1 a2 1 −1 b2 · · · , where ϕ(n), he concluded, “est bien connue en Arithmétique, comme dénom-brant les nombres premiers à n et inférieurs à n” [is well known in Arithmetic, as the number countingthosenumberswhicharelessthann andprimeton]:theEulertotientfunction.
10.1 Doubly quadratic equations and closure conditions Before entering these questions, we briefly recall the content of chapter IX, titled “Equation d’Euler.” Here, Halphen introduced doubly quadratic equations (i.e., (2, 2)-correspondences), that he represented in the form F = (m, n)xm yn = 0, where m, n ∈{0, 1, 2} and (m, n) denote the coefficient of the monomial xm yn. He also wrote F = Ay2 + 2By + C = A′x2 + 2B′x + C′, with A, B, C and A′, B′, C′ polynomials of degree 2, respectively, in x and y, and put X = B2 −AC, Y = B′2 −A′C′. The equation F = 0 implies the differential equation dx √ X ± dy √ Y = 0.
If F is symmetric, i.e., (0, 1) = (1, 0), (1, 2) = (2, 1) and (0, 2) = (2, 0), the two polynomials X,Y, of degree four, are the same except for the variable, i.e., xk and yk 123 Poncelet’s porism, I have the same coefficient for any k = 4, . . . , 0. In this case, the above differential equation is Euler’s differential equation (Euler 1768, 1794), previously considered by Trudi.
To any polynomial X of degree 4, there is an associated elliptic function f (u) having (only) two simple poles (i.e., of degree two), such that putting x = f (u) one has dx √ X = du.
The same holds for y = f (u1), so that the above differential equation becomes du ± du1 = 0 and gives u = ±u1 + c where c is a constant. Hence, observed Halphen, every symmetric doubly quadratic equation expresses the relation between f (u) and f (u + U), where f is an elliptic function of degree two of the variable u and U is a constant.
In the preamble of chapter X, he wrote: Dès le début du Tome I, on a vu (p. 13) l’addition des arguments réprésentée par une construction géométrique au moyen de deux cercles. A chaque point de l’un des cercles, on fait correspondre un argument elliptique: la corde qui joit deux points, dont la différence des arguments est constante, enveloppe le seconde cercle.
Cette construction de l’addition peut être modifiée de façon que, au lieu de deux cercles, on ait à considérer deux coniques quelconques. On n’en surait douter, d’après les enseignements de la Géométrie projective. Mai il convient de présenter directement cette construction sous sa forme générale. C’est à quoi se prête merveileusement la considération des équations doublement quadra-tiques, object principal du Chapitre précédent [At the beginning of volume I, we have seen (p. 13) the addition of the arguments represented geometrically by means of two circles. To each point of one of these circles, there corresponds an elliptic argument: the chord joining two points, whose arguments differ by a constant, envelops the second circle. This construction of the addition can be modified in such a way that, instead of the two circles, one has two conics whatever. This is indubitable by the principles of the projective geometry. But it is convenient to present this construction in a more general form. For doing this, the doubly quadratic equations, the main object of the previous chapter, fit wonderfully].
Let F(x, y) = 0 be a symmetric doubly quadratic equation. Fixing any value for x, the equation gives two values for y, say x1 one of these. For x = x1, the equation gives other two values, one of which is x and the other x2. To the latter correspond x1 and a new one x3, and so on. Denoting x−1 the second correspondent of x, x−2 the second correspondent of x−1, keep doing this way it is established a sequence . . . , x−2, x−1, x, x1, x2, x3 . . .
Halphen fixed a non-singular conic C and denoted x a parameter that rationally determines the points on C.98 He observed that, if in the above sequence two con-98 Halphen referred to C as “unicursal curve,” see also back in section five.
123 A. Del Centina x x1 x2 x3 x−1 x−2 x−3 Fig. 32 Every symmetric doubly-quadratic equation F(x, y) = 0, once fixed a value for x, gives a sequence . . . x−2, x−1, x, x1, x2, . . . which determines an inter-scribed polygon to two conics. The figure illustrates this situation secutive values represent the end points of a chord of C, then the envelope of these chords is another conic D: in fact, for any point of C there are exactly two tangents to the enveloped curve, which is necessarily of the second class, i.e., a conic. Hence, Halphen claimed: every symmetric doubly quadratic equation translates the relation among the ending points of a variable chord inscribed in a conic C and enveloping another conic D.
In particular, any sequence as above determined a polygonal line, represented . . . , x−2, x−1, x, x1, x2, x3 . . ., which is inscribed in the conic C and, at the same, time circumscribed about D (see Fig. 32).
There are four particular values α0, α1, α2, α3 of the parameter x, to each of them corresponds a double root of y; say β0, β1, β2, β3 these double roots. The tangents to D from the point (of parameter) α0 coincide, so α0 ∈D. Hence, α0, α1, α2, α3 are the parameters on C of the four points C ∩D, and β0, β1, β2, β3 are the parameters of the same points on D. Then, he proved (Halphen 1888, pp. 340, 374) that symmetric doubly quadratic equations are characterized by the two invariants: α = (α0 −α1)(α3 −α2) (α0 −α2)(α3 −α1), γ = (β0 −β1)(β3 −β2) (β0 −β2)(β3 −β1), corresponding to the cross-ratios of the four points C ∩D, taken in the same order, on C and on D.
Among the polygonal lines . . . , x−2, x−1, x, x1, x2, x3 . . ., there are those obtained starting from a point (of parameter) α belonging to C ∩D, or from a point of C where a common tangent to C and D touch C. These polygonal lines, Halphen observed, can be prolonged only in one direction. If α1, α2, . . . are the other vertices of the transversal, then, by taking x as initial point of the polygonal line, the points αn and αn−1, αn−2, etc. are such that the polygonal line folds up on itself, and xn = α, xn+1 = xn−1, xn+2 = xn−2, . . .
x2n = αn.
In general, this does not occur drawing the polygonal line in the other direction. The same holds when α′ is a point of contact of a common tangent to the two conics. To 123 Poncelet’s porism, I x x1 x2 x3 C D C D x x1 x2 x3 (a) (b) Fig. 33 Folded polygons inter-scribed to the conics C and D: a of the first kind, b of the second kind these, Halphen gave the names of folded polygonal lines, respectively, of first and second kind (see Fig. 33a, b).
The closure condition for a polygonal line, in order to give a polygon of m sides, is x = xm.
In chapter IX, Halphen had shown that the two roots x, x1 of F = 0, corresponding to a certain value of y, can be considered, respectively, equal to ℘(u) and ℘(u + U).
In this way, the vertices . . . , x−2, x−1, x, x1, x2, x3 . . . of a polygonal line have para-meters given by the values of ℘for . . . , u −2U, u −U, u, u +U, u +2U, u +3U, . . ..
Then, according to what he had already proved, the parameters y corresponding to x = ℘(u) are ℘(u ± U/2). With this “elliptic representation of the polygonal line,” the closure condition is translated into the condition that mU must be a period.
By means of the theory of the function ℘(u) developed in the first volume of his treatise, Halphen was able to express the condition above through the invariants α and γ . Precisely, he found that the condition for mU to be a period can be expressed by the vanishing of a polynomial involving the invariants x = −[α2 −2γ (2γ 2 −3γ + 2)α + γ 4]2 28α2(α −1)2γ 4(γ −1)4 , y = −(γ 2 −α)(γ 2 −2γ + α)(γ 2 −2αγ + α) 23α(α −1)γ 2(γ −1)2 , (to be not confused with the variables x, y of the doubly quadratic equation F = 0).
Then, he wrote these equations explicitly for m = 3, . . . , 11 (p. 377): m = 3 . . .
x = 0, m = 4 . . .
y = 0, m = 5 . . .
y −x = 0, m = 6 . . .
y −x −y2 = 0, m = 7 . . .
(y −x)x −y3 = 0, m = 8 . . .
(y −x)(2x −y) −xy2 = 0, m = 9 . . .
y3(y −x −y2) −(y −x)3) = 0, m = 10 . . .
y2(xy −x2 −y3) −x(y −x −y2)2 = 0, m = 11 . . .
(xy −x2 −y3)(y −x)3 −xy(y −x −y2)2 = 0.
123 A. Del Centina 10.2 On the Cayley conditions At page 387 of the second volume of his treatise, Halphen commented: Le calcul de la condition pour l’existence des polygones à m côtés se fait, au moyen des invariants x, y, comme il a été indiqué précédemment. C’est ce qu’on a de plus simple sur ce suject. On ne surait némmoins omettre un autre moyen de faire le calcul, infiniment moins commode, mais extêmement élégant. Il à été trouvé par M. Cayley [The computation of the condition for the existence of polygons of m sides is carried out by means of the invariants x, y, as previously shown. This is the simplest one we have on the subject. But we cannot omit another method of doing this computation, which is infinitely less handy to deal with, but very much more elegant. It was found by Mr. Cayley].
Halphen, recognizing the elegance of Cayley’s method, also expressed the convic-tion that it was more difficult to handle than that he had just expounded. The reason for his opinion is not clear, but probably it was because Cayley called upon the theory of Abelian integrals, while he preferred to treat the question algebraically.
The condition for which mU is a period, observed Halphen, is equivalent to the existence of a polynomial function M(℘(u)) + N(℘(u))℘′(u) having a zero of order m for u = U. He had already noticed (p. 344) that the discriminant F(s) of the pencil sX + Y has roots proportional to ℘(U + e1), ℘(U + e2), ℘(U + e3), where e1, e2, e3 are such that ℘′(u)2 = 4(℘(u)−e1)(℘(u)−e2)(℘(u)−e3), and this allowed him to put the above polynomial function in the form M(s) + N(s)√F(s). Since the value u = U corresponds to s = 0, the existence of the root U of order m requires that the development in power series of M(s) + N(s)√F(s) around s = 0 begins with a term of degree m. Let F(s) = p0 + p1s + p2s2 + p3 3 + · · · , (10.3) he observed that, since s has degree 2, and M(s) + N(s)√F(s) must be of degree m, it was convenient to consider two cases according as m is even or odd.
For m = 2n, one has M = a0 + a1s + · · · + ansn and N = b0 + b1s + · · · + bn−2sn−2; then, by equating to zero the terms of degree 0, 1, . . . , n in (10.3) equations are obtained containing the coefficients of M, while for the following terms only the coefficients of N occur: bn−2 p3 + bn−2 p4 + · · · + b0 pn+1 = 0, bn−2 p4 + bn−3 p5 + · · · + b0 pn+2 = 0, · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · bn−2 pn+1 + bn−3 pn+2 + · · · + b0 p2n−1 = 0.
For m = 2n + 1, M is the same but N has one more term bn−1sn−1, and then the equations above become: 123 Poncelet’s porism, I bn−1 p2 + bn−2 p3 + · · · + b0 pn+1 = 0, bn−1 p3 + bn−2 p4 + · · · + b0 pn+2 = 0, · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · bn−1 pn+1 + bn−2 pn+2 + · · · + b0 p2n = 0.
It follows that the condition for the existence of a Poncelet’s polygon of 2n sides is p3 p4 · · · pn+1 p4 p5 · · · pn+2 .
.
.
.
.
.
· · · .
.
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pn+1 pn+2 · · · p2n−1 = 0, while for a polygon of 2n + 1 sides is p2 p3 · · · pn+1 p3 p4 · · · pn+2 .
.
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.
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· · · .
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pn+1 pn+2 · · · p2n = 0.
Hence, he observed that for a triangle the condition is just p2 = 0, for the quad-rangle is p3 = 0, and then, referring to the invariant x and y above, he added: “on vérifiera aisément la concordance de ces conditions avec celles qui on été trouvées précédemment” [one can easily check that these conditions are the same than those previously determined].
Halphen also affirmed that, at that point, the link with the continued fractions was evident, but that he reserved this study for a subsequent chapter.
10.3 The “elliptic representation of the plane” and the curve m From the elliptic representation of the polygonal line inter-scribed to two conics C and D, Halphen knew that the closure condition was equivalent to “mU is a period,” and here he wanted to go more in depth into the question.
In the first volume of his treatise (Halphen 1886, pp. 96–103), Halphen defined the polynomial functions of ℘(u) and ℘′(u): ψn(u) = ℘′(u) & −1 2n(℘(u)) n2−4 2 + · · · ' , for n even, ψn(u) = & n(℘(u)) n2−1 2 + · · · ' , for n odd.
These functions vanish exactly at the nth parts of the periods (n ≥2). Then, he defined the irrational functions γn(u) = ψn(u)ψ2(u)−n2−1 3 , 123 A. Del Centina γ1 = γ2 = 1, and showed that they satisfy the recursive formula γm+nγm−n = γm+1γm−1γ 2 n −γn+1γn−1γ 2 m, and consequently that γ2n+1 = γn+2γ 3 n −γn−1γ 3 n+1, γ2n = γn(γn+2γ 2 n−1 −γn−2γ 2 n+1).
Putting x(u) = γ 3 3 (u), y(u) = γ4(u), it follows that γ5 = y −x, γ6 = y −x −y2, γ7 = (y −x)x −y3, γ8 = (y[(y −x)(2x −y) −xy2], γ9 = x 1 3 [y3(y −x −y2) −(y −x)3)], .
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In chapter X of the second volume of his treatise (pages 392–404), by means of the duplication formula, Halphen expressed the functions x(2u) and y(2u) as combinants of the pencil generated by two conics f and ψ, i.e., as rational functions over P2 invariant under the action of the symmetric group 4.99 In particular, he proved that the expression of γn(2u) does not depend on the conics f and ψ but only on the four points f ∩ψ.
At p. 404, Halphen wrote: Ici se present tout naturellement la considération du lieu géométrique défini par l’équation ψn(2u) = 0. Nous devons donner de ce lieu géométrique une définition indépendante des fonctions elliptique et mettre en lunière un fait bien remarquable: ce lieu se décompose en plusieurs lignes distinctes…[It is natural here to consider the geometrical locus defined by the equation ψn(2u) = 0.
We must give to this geometrical locus a definition independent of the elliptic functions and highlight a very remarkable fact: This locus decomposes in several distinct components…] To this end, Halphen proceeded as follows.
He considered α0, α1, α2, α3, which are four points in the plane in general position, i.e., no three of them are on a line, as base points of a pencil of conics, and he associated with them the half-period ωi, i = 0, 1, 2, 3, with ω0 = 0. For any given point z in the plane, he considered in the pencil with base points α0, α1, α2, α3 the conic f passing through z. The cross-ratio α of the four points α0, α1, α2, α3 on f (which 99 A combinant of the pair f, ψ, according to Sylvester (1853), is a covariant of the forms that, besides having the ordinary character of invariance when linear substitutions are applied to the variables, possesses the same character of invariance when linear substitutions are applied to their linear combinations.
123 Poncelet’s porism, I coincides with the cross-ratio of the four lines zα0, zα1, zα2 and zα3), wrote Halphen, defines the absolute invariant of the elliptic functions. Clearly, he was referring to the invariant J = g3 2 g3 2−27g2 3 , which can be expressed as rational function of anyone of the six cross-ratios of the four points ∞, e1, e2, e3, where e1, e2, e3 are the roots of the equation ℘′2 = 4℘3 −g2℘−g3. In this way, observed Halphen, the function ℘(u) is a parameter for the points on f , which assumes the values ∞, e1, e2, e3 for the values α0, α1, α2, α3 of the argument. Let γ be the cross-ratio of the four lines through α0 containing, in this order, z, α1, α2, α3. From the assumption above, it follows that γ = ℘(u) −e1 ℘(u) −e2 e3 −e2 e3 −e1 , and at the same time α = e3 −e2 e3 −e1 · At this point, Halphen affirmed (p. 405): Voilà donc les fonctions elliptiques et l’argument définis, pour chaque point du plan, par les rapports anharmoniques de deux faisceaux de droits. Il s’agit maintenant de reconnaître la propriété géometrique des points qui répondent à des parties aliquotes de périodes [here then are the elliptic functions and the argument defined, for every point of the plane, by means of the cross-ratios of two pencils of lines. It is now time to recognize the geometrical property of those points which correspond to the nth parts of the periods].
Then, he considered the line α0z through α0 and z, and the conic ψ of the pencil which is tangent (necessarily at α0) to α0z. For the pair f , ψ, he found that the argument u of the point z coincides with U. So he claimed: Voici donc la propriété géometrique de tout point z dont l’argument u est une m-ième partie de période: si, parz, on mène une conique f du faisceau et que, tangentiellement à α0z, on prenne une autre conique ψ du faicseau, il existe des polygones de m côtés, inscrits dans f et circonscrits à ψ. Pour chaque entier m, il y a un lieu du point z; c’est ce lieu qui doit être examiné. [Here is the geometrical property of each point z whose argument u is the mth part of a period: if through z is drawn a conic f of the pencil and, tangentially to α0z, another conic of pencil is taken, then there exists a polygon of m sides inscribed in F and circumscribed about ψ. For each integer m, there is a locus of the point z; it is this locus that has to be examined.] It will be useful, for the future, to denote this locus m and let ′ m be the sub-locus of primitive mth parts of periods (as defined at the beginning of this section). The intersections of m with the tangent to ψ at α0 define the conics in the pencil which admit an inscribed polygon of m sides which, at the same time, is circumscribed about ψ. So the degree of m (′ m) is the number of those conics in the pencil which are m-circumscribed (properly m-circumscribed) about the conic ψ.
123 A. Del Centina ψ α0 z y f α1 α2 α3 Fig. 34 Halphen’s construction for doubling the argument of z on f : the corresponding point on f is the point y which is the second intersection of the second tangent to ψ from z Halphen first considered the multiplication of the argument by 2. He knew that, f and ψ being as above, the point of the plane having the same modulus and argument twice the argument of z, is precisely the second intersection with f of the second tangent to ψ from z (Fig. 34).
Aiming to express this construction in formulae, he let α0 = (1, 1, 1), α1 = (−1, 1, 1), α2 = (1, −1, 1), α3 = (1, 1, −1).
Then, denoting by (x1, x2, x3) the variable coordinates in the plane and putting z = (z1, z2, z3), he was led to the following equations for the conics fz and ψz (p. 406): fz = (z2 2 −z2 3)x2 1 + (z2 3 −z2 1)x2 2 + (z2 1 −z2 2)x2 3 = 0, ψz = (z2 −z3)x2 1 + (z3 −z1)x2 2 + (z1 −z2)x2 3 = 0.
He also found that the point of contact x of the second tangent from z to the conic φz has coordinates x1 = z2z3 −z1z2 −z1z3, x2 = z3z1 −z2z3 −z2z1, x3 = z1z2 −z3z1 −z3z2, and y has coordinates y1 = z2 2z2 3 −z2 1z2 2 −z2 1z2 3, y2 = z2 3z2 1 −z2 2z2 3 −z2 1z2 2, y3 = z2 1z2 2 −z2 3z2 1 −z2 2z2 3.
From these formulae, it follows y1z2 −z1y2 (z1 −z2)x3 = y2z3 −z2y3 (z2 −z3)x1 = y3z1 −z3y1 (z3 −z1)x2 = z1z2 + z2z3 + z3z1.
123 Poncelet’s porism, I α0 α1 α2 α3 Π3 z3 f ψ Π3 Fig. 35 An illustration of the locus 3 of points z whose argument is a third period: a conic He observed that the points z such that y = z are those points whose argument is a third of a period, so for the locus 3 he found the equation z1z2 + z2z3 + z3z1 = 0, i.e., a conic (see Fig. 35).
Similarly, Halphen showed that the condition z1z2z3 = 0 characterizes the points of the plane whose argument is a quarter period; hence, 4 decomposes into the three coordinate lines (see Fig. 36).
For 6 he argued as follows. Let (2nω + 2n′ω′)/6 the argument of z, then the integers n and n′ cannot be both even; otherwise, it would be a point 3. Consider the point whose argument is nω + n′ω′ is one of the αi ̸= α0, then the difference between the arguments of αi and z is (2nω + 2n′ω′)/3. Then, if one considers the conic ψ tangent to the line αiz, the corresponding argument U is a third of a period. Hence, the point z has, with respect to αi, the same property that each point in 3 has with respect to α0. By changing the sign of the coordinates, the points α0, α1, α2, α3 are permuted among them, and this change the conic A0 : z1z2 + z2z3 + z3z1 = 0 into the conics z2z3 −z1z2 −z1z3 = 0, z3z1 −z2z3 −z2z1 = 0, z1z2 −z3z1 −z3z2 = 0.
Denoting these conics, respectively, A1, A2, A3, it follows that ′ 6 is the locus A1 ∪ A2 ∪A3, so it has degree 6.
123 A. Del Centina α0 α1 α2 α3 Π4 z f ψ Π4 P1 P2 P3 P4 Fig. 36 The locus 4 of points z whose argument is a quarter period: the locus is decomposed into the three coordinate lines Ifinthe A1, A2, A3 the z’sarereplacedbythe y’sasabove,oneobtainstheequations of the three curves constituting the locus ′ 12, each of them of degree 8, so ′ 12 has degree 24. Continuing in this way, one see that deg′ 24 = 96, etc.
At this point (p. 409) Halphen recalled that the roots of the function ψm(2u) are all the nth parts of the periods, being n any divisor of 2m.
Then, he claimed: if m is odd, the locus m contains n and 2n, in total four curves, for each divisor n of m; if m = 2am′, where m′ is odd, ′ m decomposes in 3a + (3a + 4) % n′ curves, where by % n′ he denoted the number of the divisors n′ ̸= 1 of m′, in particular if m = 2a the number of distinct curves is only 3a.
We may interpret Halphen’s reasoning as follows. If m is not a prime, then m decomposes in ′ m and a number of curve ′ n, not necessarily irreducible, one for each n|m, n ≥3. Since 3 and 4 have, respectively, degree 2 and 3, the degrees of m and ′ m can be recursively computed for many values of m. We have already seen that ′ 6 decomposes into 3 conics, so is degree is 6, and 6 = ′ 6 ∪3, so that it has degree 8. One can easily see that ′ 8 has 3 components of degree 4, and 8 = ′ 8 ∪4, so 8 has in total 4 components, and total degree 15. Similarly, ′ 12 has 3 components of degree 8, and 12 = ′ 12 ∪′ 6 ∪′ 4 ∪′ 3, so it has in total 10 components and degree 35. This process can be continued.
In Gruson (1992, p. 193), it is suggested that Halphen here proved the formula deg′ n = 1 4n2 " p|n,p prime 1 −p−2 , but in the second volume of Halphen’s treatise there is no trace of it. Nevertheless, taking (10.2), the above reasoning leads to that formula. Moreover, one has degn = (n2 −1)/4, if n is odd, and degn = n2/4 −1, if n is even.
The construction above has clear formulation in Barth and Michel (1993), where the two authors, unaware of Halphen’s result, proved the above formula in a modern algebraic–geometric setting. We will return to this in the penultimate section of our paper.
123 Poncelet’s porism, I 10.4 Continued fractions and Poncelet polygons At p. 388 of the second volume of his treatise, Halphen noticed that Cayley’s method for determining the closure condition “se rattache, de la manière la plus directe, à la théorie des fractions continues, ainsi qu’on le verra dans un Chapitre ultérieur” [this method, as it will see in a subsequent chapter, is connected, in the more direct way, to the theory of continued fractions].
He dealt with this question in chapter XIV, where he pursued the study, initiated by Abel and Jacobi, of the development in continued fractions of √X(x), being X(x) a polynomial in the variable x.
Abel, in his celebrated memoir of 1826, proved that the integral ρ(x)dx √X(x), where ρ(x) is a polynomial, can be expressed by means of rational functions and logarithms of algebraic functions if and only if √X(x) admits a periodic continued fraction development.
In his note (Jacobi 1831), Jacobi studied an algorithm for developing √X(x) in continued fractions. The difficulties that he encountered in the computation forced him to abandon the algebraic route, and to consider the use of elliptic functions in order to express the partial quotients of the continued fraction. He wrote down interesting formulae in the case X has at most degree four.100 Halphen considered the continued fractions development of the more general element √ X + √ Y x + y , where Y := X(y)(degX = 3, 4). His point of departure was the function Vm := Cm [σ(a −u)]2m−1σ(u + 2ma + mv −w) [σ(u)σ(u + v)]m , where σ is the Weierstrass σ-function, and m an integer. Vm is a doubly periodic function with respect to any of the four arguments a, u, v, w. Through the development of Vm/Vm−1 in continued fractions, Halphen established some general properties for the development of √ X+ √ Y x+y , e.g., symmetry and periodicity. Then, he concentrated on the development of √X(x) and deduced recursive formulae, simpler than those found by Jacobi, for the computation of the partial terms of the continued fraction.
Afterward, on page 600, Halphen returned shortly on Poncelet’s polygons. He recognized that if the polynomial X(x) is of degree 3, two relations that he had found studying the development of √X(x) were the same that “d’après M. Cayley” he had 100 Jacobi published these formulae without proof. They were proved by Borchardt (1854), who also extended them to the case of the continued fraction development of √X(x), with degX > 4.
123 A. Del Centina already discussed in chapter X at page 389. These two relations were those expressing the conditions for the existence of a Poncelet polygon of 2n, or 2n + 1, sides.
With this observation, Halphen brought to light a connection between Poncelet’s closure theorem and the development in continued factions of √X(x). Thirty years later this connection was investigated in depth by Gerbaldi, who dealt with the question from an algebraic point of view.
This seems a good place to temporarily stop our story about Poncelet’s porism, because with Gerbaldi and his studies we enter the twentieth century.
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123 |
10256 | https://pubmed.ncbi.nlm.nih.gov/30526054/ | Effects of Temperature Abuse on the Growth and Staphylococcal Enterotoxin A Gene (sea) Expression of Staphylococcus aureus in Milk - PubMed
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. 2019 Apr;16(4):282-289.
doi: 10.1089/fpd.2018.2544. Epub 2018 Dec 11.
Effects of Temperature Abuse on the Growth and Staphylococcal Enterotoxin A Gene (sea) Expression of Staphylococcus aureus in Milk
Milijana Babić1,Marija Pajić2,Miodrag Radinović2,Stanko Boboš2,Snežana Bulajić1,Aleksandra Nikolić3,Branko Velebit3
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Affiliations
1 1 Department of Food Hygiene and Technology, Faculty of Veterinary Medicine, University of Belgrade , Belgrade, Serbia .
2 2 Faculty of Agriculture, University of Novi Sad , Novi Sad, Serbia .
3 3 Institute of Meat Hygiene and Technology , Belgrade, Serbia .
PMID: 30526054
DOI: 10.1089/fpd.2018.2544
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Effects of Temperature Abuse on the Growth and Staphylococcal Enterotoxin A Gene (sea) Expression of Staphylococcus aureus in Milk
Milijana Babić et al. Foodborne Pathog Dis.2019 Apr.
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Foodborne Pathog Dis
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. 2019 Apr;16(4):282-289.
doi: 10.1089/fpd.2018.2544. Epub 2018 Dec 11.
Authors
Milijana Babić1,Marija Pajić2,Miodrag Radinović2,Stanko Boboš2,Snežana Bulajić1,Aleksandra Nikolić3,Branko Velebit3
Affiliations
1 1 Department of Food Hygiene and Technology, Faculty of Veterinary Medicine, University of Belgrade , Belgrade, Serbia .
2 2 Faculty of Agriculture, University of Novi Sad , Novi Sad, Serbia .
3 3 Institute of Meat Hygiene and Technology , Belgrade, Serbia .
PMID: 30526054
DOI: 10.1089/fpd.2018.2544
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The aim of this study was to determine the effects of different temperatures and storage time on Staphylococcus aureus growth, sea gene expression, and synthesis of staphylococcal enterotoxin A (SEA) in the pasteurized and UHT-pasteurized milk. Pasteurized and UHT-pasteurized milk were inoculated with 3.98 log 10 CFU/mL of S. aureus (ATCC 13565). Inoculated milk samples were stored at 8°C, 15°C, and 22°C for 24, 48, and 72 h, respectively. SEA synthesis was detected with a fully automated miniVIDAS instrument using the Enzyme-Linked Fluorescent Assay (ELFA) technology. The patterns of gene regulation were detected by quantitative reverse transcriptase PCR. The 2-ΔΔCT method has been used as a relative quantification strategy for gene expression responses data analysis. The results indicated that growth rate, sea gene expression, and SEA synthesis were influenced by type of milk, storage time, and temperature. Incubation of milk at different temperatures (15°C and 22°C) and times was used to simulate inadequate transport and storage conditions. Storage of pasteurized milk at 22°C for 24 h significantly upregulated the expression of sea gene compared with milk stored at 8°C, which coincides with the achieved S. aureus number of 10 5 CFU/mL and detected amount of SEA. In addition, storage of UHT-pasteurized milk at 22°C for 24 h and at 15°C for 48 h significantly upregulated the sea gene expression compared with milk stored at 8°C, which coincides with the detected amount of SEA and the dynamics of S. aureus number change. It can, therefore, be concluded that implementing good hygiene practices to avoid pre- and post-heat treatment milk contamination and maintaining the cold chain at temperature <8°C throughout the entire dairy production chain are of paramount importance to decrease the risk of staphylococcal food poisoning.
Keywords: gene expression; Staphylococcus aureus; enterotoxin A synthesis; milk.
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Effect of Temperature on the Expression of Classical Enterotoxin Genes among Staphylococci Associated with Bovine Mastitis.Homsombat T, Boonyayatra S, Awaiwanont N, Pichpol D.Homsombat T, et al.Pathogens. 2021 Aug 2;10(8):975. doi: 10.3390/pathogens10080975.Pathogens. 2021.PMID: 34451439 Free PMC article.
Co-Occurrence of Staphylococcus aureus and Ochratoxin A in Pasteurized Milk.Zhang Z, Song Y, Ma L, Huang K, Liang Z.Zhang Z, et al.Toxins (Basel). 2022 Oct 21;14(10):718. doi: 10.3390/toxins14100718.Toxins (Basel). 2022.PMID: 36287986 Free PMC article.
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Given circle $C(O,R)$ with center $O$ and radius $R,$ we'll be looking into the problem of constructing $\Delta ABC$ inscribed into $C(O,R)$ and having inradius $r.$ Assuming $R\gt 2r,$ the triangle exists and is unique, provided either one of the angles or one of the sides is given. The two problems are equivalent since equal chords in a circle subtend equal angles and vice versa. There are two constructions; one is designated as $A, R, r,$ the other $a, R, r.$
$A, R, r$
Assume the problem has been solved. Draw the inscribed and circumscribed circles. Since angle A is given, we may also assume the side BC be known. The bisectors of angles B and C intersect at the center I of the inscribed circle. $\angle B + \angle C + \angle A = 180^{\circ}.$ $\angle IBC + \angle ICB + \angle BIC = 180^{\circ}.$ Therefore, $\angle BIC = 90^{\circ} + \angle A/2.$ This means that point I belongs to a locus of points from which segment BC is seen under angle $90^{\circ} + \angle A/2.$ This is a circular arc that is easily constructed.
On the other hand, $I$ lies on the line parallel to and at the distance $r$ from $BC.$ Therefore, the point $I$ is easily determined as the intersection of a circle and a straight line.
Once $I$ has been constructed, double the angles $IBC$ and $ICB.$ Alternatively, draw the circle $C(I,r)$ with center $I$ and radius $r$, and the tangents to $C(I,r)$ from $B$ and $C.$ Point $A$ lies at the intersection of thus obtained two rays.
The construction starts with drawing the circumscribed circle and any inscribed angle equal to $A.$ This determines the segment $BC.$ Then $I$ is constructed as above and then $A.$
For a proof of correctness, just note that, by the construction, $\angle BIC=90^{\circ}+A/2,$ and drawing the tangents from $B$ and $C$ makes $BI$ and $CI$ angle bisectors in $\Delta ABC,$ implying that $\angle BAC=180^{\circ}-2(90^{\circ}-A/2)=A,$ meaning that the rays $BA$ and $CA$ meet on the circumcircle $C(O,R).$
$a, R, r$
Professor René Sperb suggested a construction that makes use of Euler's formula for the distance $d$ between the incenter and circumcenter of a triangle. The three quantities are related by $2Rr = R^{2} - d^{2}.$ It follows that $d$ could be constructed as a leg of a right triangle with the other leg $r$ and the hypotenuse $R-r.$
Note that Euler's formula explains the necessity whereas the ultimate success of the construction the sufficiency, of the condition $R\gt 2r.$
Thus we draw circles $C(O,R)$ and $C(O,d)$ of radii $R$ and $d$ around the same point $O,$ place chord $BC=a$ in the former, and path a line, say, $l\parallel BC$ at distance $r.$ The intersection of $l$ and $C(O,d)$ gives the incenter $I$ and, with it, the incircle $C(I,r).$ The tangents to the incircle from $B$ and $C$ meet at point $A$ on the circumcircle. This was obvious for the preceding construction where $\angle BIC$ was such that insured the correct value of angle at $A;$ it may be less obvious for the present one.
A proof of correctness of this construction consists in showing that the two constructions produce exactly the same circle $C(I,r).$ But this is indeed so, since, as we have seen, the first construction is correct so that, the Euler formula for the resulting $\Delta ABC$ assures that the distance $d$ for that triangle is exactly the distance used in the second construction.
We can be more explicit and derive the Euler formula along the way. First note that the arc through $B$ and $C$ used in the first construction is centered at the midpoint (I'll call it $D)$ of the arc $BC$ of circle $C(O,R).$ (This is to be expected as that arc is part of $(D)$ - a circle through the incenter.) This is indeed so because (in the diagram below) $\angle BCD = \angle CBD = A/2$ so that $\angle BDC = 180^{\circ}-A,$ making $BIC=90^{\circ}+A/2,$ for any point $I$ on the arc.
We shall now show that the distance between the $I$ found in the first construction satisfies the Euler formula, as it should. Using the notations in the diagram,
$\begin{align} y &= R - r - h,\ x &= \sqrt{\rho^{2}-(h+r)^{2}},\ \rho^{2} &= h^{2}+(a/2)^{2},\ d^{2} &= x^{2}+y^{2}\ &= \rho^{2}-(h+r)^{2}+(R-r-h)^{2}\ &= h^{2}+(a/2)^{2}-(h+r)^{2}+(R-r-h)^{2}\ &= (a/2)^{2}-2Rr+(R-h)^{2},\ R^{2}&=(R-h)^{2}+(a/2)^{2} \end{align}$
Combining the last two we get $d^{2}=R^{2}-2Rr$ - the Euler formula!
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10258 | https://online.stat.psu.edu/stat505/book/export/html/635 | Lesson 2: Linear Combinations of Random Variables
Lesson 2: Linear Combinations of Random Variables
Overview
This lesson is concerned with linear combinations or if you would like linear transformations of the variables. Mathematically linear combinations can be expressed as shown in the expression below:
.
Here what we have is a set of coefficients through that is multiplied bycorresponding variables through . So, in the first term, we have times which is added to times and so on up to the variable . Mathematically this is expressed as the sum of j = 1, ... , p of the terms times . The random variables through are collected into a column vector X and the coefficient to are collected into a column vector c. Hence, the linear combination can be expressed as .
The selection of the coefficients through is very much dependent on the application of interest and what kinds of scientific questions we would like to address.
Later on in this course, when we learn about multivariate data reduction techniques, the interpretation of linear combinations will be of great importance.
Objectives
Upon completion of this lesson, you should be able to:
Interpret the meaning of a specified linear combination;
Compute the sample mean and variance of a linear combination from the sample means, variances, and covariances of the individual variables.
2.1 - Examples of Linear Combinations
2.1 - Examples of Linear Combinations
Example 2-1: Women’s Health Survey (Linear Combinations)
The Women's Health Survey data contains observations for the following variables:
calcium (mg)
iron (mg)
protein(g)
vitamin A(μg)
vitamin C(mg)
In addition to addressing questions about the individual nutritional component, we may wish to address questions about certain combinations of these components. For instance, we might want to ask what is the total intake of vitamins A and C (in mg). We note that in this case, Vitamin A is measured in micrograms while Vitamin C is measured in milligrams. There are a thousand micrograms per milligram so the total intake of the two vitamins, Y, can be expressed as the following:
In this case, our coefficients , and are all equal to 0 since the variables , and do not appear in this expression. In addition, is equal to 0.001 since each microgram of vitamin A is equal to 0.001 milligrams of vitamin A. In summary, we have
Example 2-2: Monthly Employment Data
Another example where we might be interested in linear combinations is in the Monthly Employment Data. Here we have observations on 6 variables:
Number people laid off or fired
Number of people resigning
Number of people retiring
Number of jobs created
Number of people hired
Number of people entering the workforce
Net employment decrease:
In looking at the net job increase, which is equal to the number of jobs created, minus the number of jobs lost.
In this case, we have the number of jobs created, (), minus the number of people laid off or fired, (), minus the number of people resigning, (), minus the number of people retired, (). These are all of the people that have left their jobs for whatever reason.
In this case
Because variables 5 and 6 are not included in this expression,
Net employment increase:
In a similar fashion, net employment increase is equal to the number of people hired, (), minus the number of people laid off or fired, (), minus the number of people resigning, (), minus the number of people retired, ().
In this case
Net unemployment increase:
Net unemployment increase is going to be equal to the number of people laid off or fired, (), plus the number of people resigning, (), plus the number of people entering the workforce, (), minus the number of people hired, ().
Unfilled jobs:
Finally, if we wanted to ask about the number of jobs that went unfilled, this is simply equal to the number of jobs created, (), minus the number of people hired, ().
In other applications, of course, other linear combinations would be of interest.
2.2 - Measures of Central Tendency
2.2 - Measures of Central Tendency
Overview
Because linear combinations are functions of random quantities, they also are random vectors, and hence have population means and variances. Moreover, if you are looking at several linear combinations, they will have covariances and correlations as well.
Therefore we are interested in knowing:
What is the population mean of Y?
What is the population variance of Y?
What is the population covariance between two linear combinations and ?
Population Mean
The population mean of a linear combination is equal to the same linear combination of the population means of the component variables. If
then
Mathematically you express this as the sum of j = 1 to p of times the corresponding mean of the variable. If the coefficient c's are collected into a vector c and the mean are collected into a mean vector you can express this as c transpose times .
We can estimate the population mean by replacing the population means with the corresponding sample means; that is replace all of the 's with 's so that equals times plus times and so on...
Population mean of a linear combination
Example 2-3: Women’s Health Survey (Population Mean)
The following table shows the sample means for each of the five nutritional components that we computed in the previous lesson.
| Variable | Mean |
--- |
| Calcium | 624.0 mg |
| Iron | 11.1 mg |
| Protein | 65.8 g |
| Vitamin A | 839.6 μg |
| Vitamin C | 78.9 mg |
If, as previously, we define Y to be the total intake of vitamins A and C (in mg) or:
Then we can work out the estimated mean intake of the two vitamins as follows:
mg.
2.3 - Population Variance
2.3 - Population Variance
Linear combinations not only have a population mean but they also have a population variance. The population variance of a linear combination is expressed as the following double sum of j = 1 to p and k = 1 to p over all pairs of variables.
In each term within the double sum, the product of the paired coefficients times is multiplied by the covariance between the and variables. If is the variance-covariance matrix of , then .
Expressions of vectors and matrices of this form are called quadratic forms.
When using this expression, the covariance between the variables and itself, or is simply equal to the variance of the variable, or .
The variance of the random variable y can be estimated by the sample variances or s squared Y. This is obtained by substituting the sample variances and covariances for the population variances and covariances as shown in the expression below.
A simplified calculation can be found below. This involves two terms.
Population variance of linear combinations
The first term involves summing over all the variables. Here we take the squared coefficients and multiply them by their respective variances. In the second term, we sum over all unique pairs of variables j less than k. Again take the product of times times the covariances between variables j and k. Since each unique pair appears twice in the original expression, we must multiply the sum by 2.
Example 2-4: Women’s Health Survey (Population Variance)
Looking at the Women's Nutrition survey data we obtained the following variance/covariance matrix as shown below from the previous lesson.
If we wanted to take a look at the total intake of vitamins A and C (in mg) remember we defined this earlier as:
Therefore the sample variance of Y is equal to times the variance for , plus the variance for , plus 2 times 0.001 times the covariance between and . The next few lines carry out the mathematical calculations using these values.
2.4 - Population Covariance
2.4 - Population Covariance
Sometimes we are interested in more than one linear combination or variable. In this case, we may be interested in the association between those two linear combinations. More specifically, we can consider the covariance between two linear combinations of the data.
Consider the pair of linear combinations:
Here and are two distinct linear combinations. Both variables and are going to be random and so they will be potentially correlated. We can assess the association between these variables using the covariance as the two vectors c and d are distinct.
The population covariance between and is obtained by summing over all pairs of variables. We then multiply respective coefficients from the two linear combinations as times times the covariances between j and k.
Population Covariance between two linear combinations
We can then estimate the population covariance by using the sample covariance. This is obtained by simply substituting the sample covariances between the pairs of variables for the population covariances between the pairs of variables.
Sample Covariance between two linear combinations
Correlation
The population correlation between variables and can be obtained by using the usual formula of the covariance between and divided by the standard deviation for the two variables as shown below.
Population Correlation between two linear combinations
This population correlation is estimated by the sample correlation where we simply substitute the sample quantities for the population quantities below
Sample Correlation between two linear combinations
Example 2-5: Women’s Health Survey (Pop. Covariance and Correlation)
Here is the matrix of the data as was shown previously.
We may wish to define the total intake of vitamins A and C in mg as before.
and we may also want to take a look at the total intake of calcium and iron:
Then the sample covariance between and can then be obtained by looking at the covariances between each pair of the component variables time the respective coefficients. So in this case we are looking at pairing and , and , and , and and . You will notice that in the expression below , , and all appear. The variables are taken from the matrix above and substituted into the expression and the math is carried out below.
You should be able at this point to be able to confirm that the sample variance of is 159,745.4 as shown below:
And, if we care to obtain the sample correlation between and , we take the sample covariance that we just obtained and divide it by the square root of the product of the two component variances, 5463.1, for , which we obtained earlier, and 159745.4, which we just obtained above. Following this math through, we end up with a correlation of about 0.235 as shown below.
2.5 - Summary
2.5 - Summary
In this lesson we learned about:
The definition of a linear combination of random variables;
Expressions of the population mean and variance of a linear combination and the covariance between two linear combinations;
How to compute the sample mean of a linear combination from the sample means of the component variables;
How to compute the sample variance of a linear combination from the sample variances and covariances of the component variables;
How to compute the sample covariance and correlation between two linear combinations from the sample covariances of the component variables.
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10261 | https://www.wyzant.com/resources/answers/894625/find-two-numbers-whose-difference-is-30-and-whose-product-is-a-minimum | WYZANT TUTORING
Ella P.
Find two numbers whose difference is 30 and whose product is a minimum.
3 Answers By Expert Tutors
TIM B. answered • 05/04/22
Mathematical Acumen for Hire
Write y = x + 30, or y − x = 30.
Next, if y = x + 30, then xy or x(x + 30) or x2 +30x
gives the product of the two numbers x & y.
Now write f(x) = x2 +30x. By The First Derivative Test, f'(x) = d(x2 +30x)/dx or 2x + 30 gives a critical point at x = -15 (to obtain 2x + 30 equal to 0).
Then The Second Derivative Test gives f''(x) as d2y/dx2 equal to d(2x + 30)/dx or 2 which is greater than zero. This positive value of 2 for f''(-15) indicates that f(x) = x2 +30x has a lowest point or minimum for
(x,[x2 +30x]) at (-15,[(-15)2 +30(-15)]) or (-15,-225).
Since -225 is obtained by (-15 times 15) and (15 minus -15) gives a difference of 30, the two numbers sought would be (x,y) equal to (-15,15).
Mark M.
answered • 05/04/22
Mathematics Teacher - NCLB Highly Qualified
Calculus is not always the most direct.
z = x2 - 30
From Algebra minimum is at x = -(-30) / 2(1)
Colin S. answered • 05/04/22
Master in all Calculus Topics
We can think of a function z(x,y) = xy, where x and y can be the two numbers. We can add the constraint that x - y = 30 (here we're saying x is the larger number arbitrarily).
Like any function, when we want to find its minimum we take its derivative, set it equal to zero and solve for the variable values at which this minimum occurs. This is annoying if z is a two variable function, but we can easily make it single variable by recognizing that y = x - 30. Therefore:
z = x(x-30)
z = x2 - 30x
dz/dx = 2x - 30
2x - 30 = 0
x = 15
This means y = -15, so the two numbers are -15 and 15.
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12/30/2011, 07:00 AM #1
Whitebeam
Registered Member
Join Date: Mar 2011
Location: Dorset, England.
Posts: 153ppm vs mg/l
I thought I understood this and that it was accepted that there is a direct equivalence between these quantities.
However, a major test kit supplier has just posted on another forum to point out that this should not be so. As ppm is a weight-weight ratio and mg/l is a weight-volume ratio, we should be adjusting by the specific gravity of the solution; i.e., if our tank water has an S.G. of (say) 1.025, 420mg/l should be expressed as 420/1.025 = 410ppm.
Thoughts/comments please?
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12/30/2011, 07:29 AM#2
HighlandReefer
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Posts: 14,591 Randy discusses this in the article below:
From it:
"mg/L (milligram per liter)
Milligram per L is a unit of concentration often used for chemicals in aquaria.�For chemical concentrations in freshwater, 1 mg/L is approximately 1 ppm (= 1 mg/kg).�For chemical concentrations in seawater, where 1 L weighs approximately 1.023 kg, 1 mg/L is approximately 0.978 ppm (= 0.978 mg/kg)."
and here:
"ppm (part per million)
ppm is a unit of proportion equal to 10-6. It is equal to 10-3 g/kg and 1 mg/kg, and is close to 10-3 g/L or 1 mg/L (in fresh water; in seawater, it equals about 1.023 x 10-3 g/L).�It is often used to measure the concentrations of different species in reef aquaria.�A calculator for various weight-related units can be found here."
Cliff Babcock
Intestests: Digital Microscopy; Marine Pest Control; Marine Plants & Macroalgae
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12/30/2011, 07:32 AM#3
HighlandReefer
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Posts: 14,591 It really depends on how many significant figures you want to take your calculations down to. One can easily round it off to a direct correlation which is usually close enough for our purposes.
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Intestests: Digital Microscopy; Marine Pest Control; Marine Plants & Macroalgae
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12/30/2011, 07:33 AM#4
Whitebeam
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Posts: 153 Quote: Originally Posted by HighlandReefer Randy discusses this in the article below: From it: "mg/L (milligram per liter) Milligram per L is a unit of concentration often used for chemicals in aquaria.�For chemical concentrations in freshwater, 1 mg/L is approximately 1 ppm (= 1 mg/kg).�For chemical concentrations in seawater, where 1 L weighs approximately 1.023 kg, 1 mg/L is approximately 0.978 ppm (= 0.978 mg/kg)." Cheers Cliff - thanks for confirming. It did all make sense the moment I thought about it, but it did seem to challenge the received wisdom of equivalence. Now the next doubt enters my mind ... my test kits all say they read in ppm, however as the method involves taking a given volume of tank water and testing it, I assume they are directly measuring in mg/l - I wonder if their titration equivalence tables have actually been adjusted for the density of the water being tested? Peter
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12/30/2011, 07:33 AM#5
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Posts: 19,447 Randy and a few others might comment on your question.
That supplier just posted the following on the other board:
PPM for solutes in liquids means mg/kg.
Scientific data on seawater solutes such as calcium are expressed in ppm.
The measure ppm is independent of temperature.
Whereas mg/L depends on temperature, the difference will be slight if the temperature difference is small, say a few degrees.
If one wants to convert a concentration from ppm (which is the basis) into mg/L then one has to apply the following equation:
mg/L = ppm x absolute density (absolute density in kg/L).
Neglecting differences as small as approx. 0.1%, one can use 1.025 as the density in the above conversion equation. That is for typical aquarium / seawater. For freshwater, for general work, the density is usually taken as 1.00 in the above conversion.
For example in seawater:
400 ppm calcium = 410 mg/L calcium
1290 ppm magnesium = 1322 mg/L magnesium
For more accurate comparisons and calibration, one has to take such things into account.
For general work, and to avoid making it overcomplicated for most hobbyists, it would suffice to say ppm is mg/L.
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12/30/2011, 07:40 AM#6
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Posts: 19,447 Cliff & Peter, it is not polite to post while someone else is typing his post.
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12/30/2011, 07:41 AM#7
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Posts: 153 Quote: Originally Posted by Habib Randy and a few others might comment on your question. That supplier just posted the following on the other board: Thanks Habib! I might have guessed you'd end up here too ;-) I usually bring my nerdy (and usually pretty much irrelevant ;-) technical questions here, as there tends to be slightly more scientific rigour applied to answers here than on the other forum. Peter
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12/30/2011, 07:46 AM#8
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Posts: 14,591 Quote: Originally Posted by Habib Cliff & Peter, it is not polite to post while someone else is typing his post. ______ Cliff Babcock Intestests: Digital Microscopy; Marine Pest Control; Marine Plants & Macroalgae Current Tank Info: 180 g. mixed reef system
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12/30/2011, 08:12 AM#9
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Posts: 153 Quote: Originally Posted by Habib Cliff & Peter, it is not polite to post while someone else is typing his post. Just type faster, my friend ;-) Thanks for your help, as always. Peter
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10263 | https://proofwiki.org/wiki/Euler%27s_Criterion | Euler's Criterion
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Contents
1 Theorem
1.1 Euler's Criterion for Quadratic Residue
2 Proof
3 Source of Name
4 Sources
Theorem
Let a be a residue order n of m, where a and m are coprime.
Then:
: aϕ(m)/d≡1(modm)
where:
: ϕ(m) denotes the Euler ϕ function of m
: d denotes the gretest common divisor of ϕ(m) and n
: ≡ denotes modulo congruence.
Euler's Criterion for Quadratic Residue
Let p be an odd prime.
Let a≢0(modp).
Then:
| | | | | | | | | | | | | | | |
--- --- --- --- --- --- ---
| | | | | | ap−12 | ≡ | | | | 1 | (modp) | | if and only if a is a quadratic residue of p | |
| | | | | | ap−12 | ≡ | | | | −1 | (modp) | | if and only if a is a quadratic non-residue of p. | |
Proof
| | |
--- |
| | This theorem requires a proof. You can help Pr∞fWiki by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Source of Name
This entry was named for Leonhard Paul Euler.
Sources
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Euler's criterion
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): residue: 2.
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Euler's criterion
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): residue: 2.
Retrieved from "
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10264 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_2?srsltid=AfmBOoq4syVYmVnh4Y7XqdVmdRmDCFymtp0HvP5YnNv8_pT75Ls8r8-N | Art of Problem Solving
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ResourcesAops Wiki 2021 AMC 12A Problems/Problem 2
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2021 AMC 12A Problems/Problem 2
Contents
[hide]
1 Problem
2 Solution 1 (Algebra)
3 Solution 2 (Algebra)
4 Solution 3 (Process of Elimination)
5 Solution 4 (Graphing)
6 Video Solution (Quick and Easy)
7 Video Solution by Aaron He
8 Video Solution by Hawk Math
9 Video Solution by OmegaLearn (Using Logic and Analyzing Answer Choices)
10 Video Solution by TheBeautyofMath
11 See also
Problem
Under what conditions is true, where and are real numbers?
It is never true.
It is true if and only if .
It is true if and only if .
It is true if and only if and .
It is always true.
Solution 1 (Algebra)
One can square both sides to get . Then, . Also, it is clear that both sides of the equation must be nonnegative. The answer is .
~Jhawk0224
Solution 2 (Algebra)
Complete the square of the left side by rewriting the radical to be From there it is evident for the square root of the left to be equal to the right, must be equal to zero. Also, we know that the equivalency of square root values only holds true for nonnegative values of , making the correct answer
~AnkitAmc
Solution 3 (Process of Elimination)
The left side of the original equation is the arithmetic square root, which is always nonnegative. So, we need which refutes and Next, picking refutes and picking refutes By POE (Process of Elimination), the answer is
~MRENTHUSIASM
Solution 4 (Graphing)
If we graph then we get the union of:
positive -axis
positive -axis
origin
Therefore, the answer is
The graph of is shown below. ~MRENTHUSIASM (credit given to TheAMCHub)
Video Solution (Quick and Easy)
~Education, the Study of Everything
Video Solution by Aaron He
Video Solution by Hawk Math
Video Solution by OmegaLearn (Using Logic and Analyzing Answer Choices)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
See also
2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 1Followed by
Problem 3
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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10265 | https://proofwiki.org/wiki/Product_of_4_Consecutive_Integers_is_One_Less_than_Square | Product of 4 Consecutive Integers is One Less than Square - ProofWiki
Product of 4 Consecutive Integers is One Less than Square
From ProofWiki
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Contents
1 Theorem
2 Proof 1
2.1 Lemma
3 Proof 2
3.1 Lemma
4 Sources
Theorem
Let a a, b b, c c and d d be consecutive integers.
Then:
∃n∈Z:a b c d=n 2−1∃n∈Z:a b c d=n 2−1
That is, the product of a a, b b, c c and d d is one less than a square.
Proof 1
Lemma
Let a a, b b, c c and d d be consecutive integers.
Let us wish to prove that the product of a a, b b, c c and d d is one less than a square.
Then it is sufficient to consider a a, b b, c c and d d all strictly positive.
□◻
As a a, b b, c c and d d are all consecutive, we can express them as:
a a, a+1 a+1, a+2 a+2 and a+3 a+3
where a≥1 a≥1.
Hence:
a(a+1)(a+2)(a+3)+1 a(a+1)(a+2)(a+3)+1==a 4+6 a 3+11 a 2+6 a+1 a 4+6 a 3+11 a 2+6 a+1
==(a 2+3 a+1)2(a 2+3 a+1)2 by inspection
Hence the result.
Proof 2
Lemma
Let a a, b b, c c and d d be consecutive integers.
Let us wish to prove that the product of a a, b b, c c and d d is one less than a square.
Then it is sufficient to consider a a, b b, c c and d d all strictly positive.
□◻
As a a, b b, c c and d d are all consecutive, we can express them as:
a a, a+1 a+1, a+2 a+2 and a+3 a+3
where a≥1 a≥1.
Then:
a(a+3)a(a+3)==a 2+3 a a 2+3 a
(a+1)(a+2)(a+1)(a+2)==a 2+3 a+2 a 2+3 a+2
⇝⇝a(a+1)(a+2)(a+3)a(a+1)(a+2)(a+3)==(a 2+3 a)(a 2+3 a+2)(a 2+3 a)(a 2+3 a+2)
==(n−1)(n+1)(n−1)(n+1)where n=a 2+3 a+1 n=a 2+3 a+1
==n 2−1 n 2−1Difference of Two Squares
■◼
Sources
1980:David M. Burton: Elementary Number Theory(revised ed.)... (previous)... (next): Chapter 2 2: Divisibility Theory in the Integers: 2.2 2.2 The Greatest Common Divisor: Problems 2.2 2.2: 8(b)8(b)
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10266 | https://www.youtube.com/watch?v=3tYeRgemZVw | Evaluate Expressions with Parentheses
Mathispower4u
327000 subscribers
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Description
4402 views
Posted: 7 Feb 2017
This video explains how to evaluate expressions with parentheses.
1 comments
Transcript:
in this video we will evaluate expressions with parentheses sometimes it can be a little challenging to interpret the operation of an expression because there are so many ways to write equivalent expressions just remember if there's a single number inside the parentheses this just represents the value of the number itself and when an operation is being performed the operator will be given except for the case when we have implied multiplication we have implied multiplication here because we have a number inside parentheses next to a second number inside parenthesis and therefore this is implied multiplication this expression is equivalent [Music] to8 -2 and because a negative a negative is positive 8 -2 is equal to pos6 and there are also many other ways to write this product we can leave the 8 outside the parentheses and then put the -2 inside parenthesis this also means 8 -2 we can also include a multiplication sign as we did here with or without parentheses so we could write 8 -2 or8 -2 with no parentheses we can also use different symbols for multiplication but all these expressions are equivalent looking at the next expression notice how we have a division sign here before the -2 in parentheses this just means 8 / -2 a negative ID negative is equal to a positive this quotient is pos4 again the reason we have the parenthesis here is just to clarify that we're dividing by a -2 without the parentheses the quotient is not as clear we would have8 / by -2 and having the division sign right next to the negative sign does tend to be a little more confusing but it does mean the same thing and is equivalent and because a fraction bar means division we can write an equivalent quotient as8 / -2 again with the8 and or the -2 in parentheses all these expressions are equivalent and mean8 / -2 next we have a minus sign in front of the -2 in parentheses which means we have the expression 8us -2 and again the parentheses are used here to clarify what we're subtracting if we did not have parentheses the expression would be8 - -2 which is not incorrect but using the parenthesis it's more clear that we are subtracting -2 and subtracting -2 is equivalent to adding positive2 so this is equivalent to8 + 2 which equal -68 - -2 is equal to -6 next notice how we have a plus sign in front of the parentheses which means we were adding -2 our expression is8 + -2 which is equal to -10 again the parentheses are used to clarify that we are adding -2 if we do not have the parentheses we would have8 + -2 which looks like this which is equivalent but just not as clear when parentheses are included for the last two examples notice how we have a difference inside the parentheses when following the order of operations we know we need to simplify inside the parentheses first so for both of these last two expressions We Begin by determining -5 - 2 well5 - 2 is equivalent to -5 +2 which is equal to -7 so this first expression simplifies to -8 then in parentheses we have -7 and again this is implied multiplication this means 8 7 which is equal to POS 56 looking at our last expression we already know that -5 - 2 is = to -7 the expression simplifies to-8 -7 because of the minus sign here we know we subtract -7 from8 well8 - -7 is equivalent to8 + POS 7 which is equal to -1 I hope you found this helpful |
10267 | https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-6-for-ccss/section/8.2/primary/lesson/comparing-absolute-values-4424306-msm6-ccss/ | Skip to content
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8.2
Comparing Absolute Values
Written by:Katie Sinclear
Fact-checked by:The CK-12 Editorial Team
Last Modified: Aug 01, 2025
Lesson
Absolute Value
The absolute value of an integer is an important concept in math, often used to measure distance and change. Remember that the absolute value of a number is the distance of that number from zero.
Use the interactive below to see how absolute values are used to measure distance on a number line.
Progress
0 / 4
1.
|−9|<|9|
TRUE
FALSE
Sea Level
The lesson Comparing Integers looked at the elevations of various locations in the United States.
In the interactive, see how using the absolute values of integers can help determine distances from zero elevation.
The Stock Market
A common application of absolute value is in the stock market. Stock traders are more interested in the amount a stock changes by than in the actual value. Many stock boards show a decrease in value using the - symbol and the color red. An increase in value is shown using the + symbol and the color green.
The interactive below displays the changes in the value of several stocks over the course of a day. Through the following interactive, you will use absolute value to order the stocks from greatest to least change in price.
Absolute Value Puzzle
In the interactive below, practice comparing integers and their absolute values by completing a puzzle game.
Summary
The absolute value of a number is its distance from zero.
When ordering absolute value of numbers, ignore the signs because the absolute value of any number is always positive.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Overview
The absolute value of a number is its distance from zero.
When ordering absolute value of numbers, ignore the signs because the absolute value of any number is always positive.
Vocabulary
Zero
number line
Amount
order
Test Your Knowledge
Question 1
Compare the following integers using inequality symbol.
@$\begin{align}|-9| \ \underline{\;\;\;\;\;\;} \ |8|\end{align}@$
a
<
b
@$\begin{align}=\end{align}@$
c
By simplifying the absolute values, we know that @$\begin{align}|-9| = 9\end{align}@$ and @$\begin{align}|8| = 8\end{align}@$.
Since @$\begin{align}9 \gt 8\end{align}@$ , then we can conclude @$\begin{align}|-9| \ \gt \ |8|\end{align}@$
Question 2
Compare the following integers using inequality symbol.
@$\begin{align}|-12| \ \underline{\;\;\;\;\;\;} \ 12\end{align}@$
a
b
@$\begin{align}=\end{align}@$
c
<
@$\begin{align}|-12| = 12\end{align}@$ , so @$\begin{align}12 = 12.\end{align}@$
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Related Content
Absolute Value of Integers
Position Displacement
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10268 | https://math.stackexchange.com/questions/2563956/find-all-natural-numbers-n-such-that-2n-divides-3n-1 | Find all natural numbers $n$ such that $2^n$ divides $3^n -1$ - Mathematics Stack Exchange
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Find all natural numbers n n such that 2 n 2 n divides 3 n−1 3 n−1 [duplicate]
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When does 2 n 2 n divide 3 n−1 3 n−1 (1 answer)
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Find all natural numbers n n such that 2 n 2 n divides 3 n−1 3 n−1
I think that the only solutions are n=0,1,2,4 n=0,1,2,4, but I have no idea on how to prove it.
I tried to write 3 n−1 3 n−1 as 1+3+3 2+...+3 n−1 1+3+3 2+...+3 n−1 and manipulate the sum but found my self at the equally hard problem of finding the power of two dividing 3 k+1 3 k+1
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edited Dec 12, 2017 at 22:13
Math Lover
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asked Dec 12, 2017 at 22:09
user228054 user228054
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3 math.stackexchange.com/questions/1252716/… and links to LTE lemma there : artofproblemsolving.com/community/c6h324597zwim –zwim 2017-12-12 22:24:08 +00:00 Commented Dec 12, 2017 at 22:24
Induction: base case n=0 n=0 we have 1 1 divides 0 0, second case (optional base case) n=1 n=1 we have 2 2 divides 2 2. Now consider n+1 n+1. We have 2 n+1=2 n 2.2 n+1=2 n 2. We also have 3 n 3−1.3 n 3−1. Now we know that an odd number squared will always be odd, and multiplied by an odd number will also stay odd. But an odd number subtracted by 1 1 is always even and therefore divisible by 2 2.JohnColtraneisJC –JohnColtraneisJC 2017-12-12 22:27:03 +00:00 Commented Dec 12, 2017 at 22:27
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Well 3 n−1=(3−1)(1+3+3 2+...+3 n−1)=2(1+3+3 2+...+3 n−1)3 n−1=(3−1)(1+3+3 2+...+3 n−1)=2(1+3+3 2+...+3 n−1)
So 2 n|3 n−1 2 n|3 n−1 if and only if 2 n−1|(1+3+3 2+...+3 n−1)2 n−1|(1+3+3 2+...+3 n−1).
If n n is odd and greater than one (1+3+3 2+....+3 n−1)(1+3+3 2+....+3 n−1) is odd so we can assume n n is even.
Let n=2 m n=2 m then 2 2 m|3 2 m−1=(3 m−1)(3 m+1)2 2 m|3 2 m−1=(3 m−1)(3 m+1). So 3 m±1 3 m±1 are both even and only one of them is is divisible by 4 4.
So 2 2 m−1|3 m±1 2 2 m−1|3 m±1 so 2 2 m−1≤3 m±1 2 2 m−1≤3 m±1.
But 2 2 m−1=1 2∗4 m≤3 m±1 2 2 m−1=1 2∗4 m≤3 m±1
So (4 3)m≤2±2 3 m<2 2 3(4 3)m≤2±2 3 m<2 2 3
If m≥3 m≥3 then (4 3)m≥2 10 27>2 2 3 3≥2+2 3 m(4 3)m≥2 10 27>2 2 3 3≥2+2 3 m
So m<3 m<3
So if n>1 n>1 then n=2 m;m≤2 n=2 m;m≤2.
So solutions must be a subset of {0,1,2,4}{0,1,2,4}.
And you have already determined that {0,1,2,4}{0,1,2,4} are all solutions.
There's probably a more elegant way.
My first thought was FTL that as gcd(3,2 n)=1 gcd(3,2 n)=1 and ϕ(2 n)=2 n−1 ϕ(2 n)=2 n−1 then 3 2 n−1≡1 mod 2 n 3 2 n−1≡1 mod 2 n. So
If 3 m≡1 mod 2 n 3 m≡1 mod 2 n then m m is a multiple of a non-trivial factor of 2 n−1 2 n−1 .i.e. even
but that didn't really get me closer.
Likewise 3 n=(2+1)n=2 n+∑(n k)2 k 3 n=(2+1)n=2 n+∑(n k)2 k and for 2 n|∑(n k)2 k 2 n|∑(n k)2 k seemed like it should yeild something relevent but I wasn't able to put my finger on it exactly.
Similarly 3 n=(4−1)n 3 n=(4−1)n.
Its a enough to convince me the answers are related to powers of 2 2 but not enough to actually prove it.
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edited Dec 12, 2017 at 23:35
answered Dec 12, 2017 at 22:43
fleabloodfleablood
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The result is clear for n=0 n=0. For n=1,2,3…n=1,2,3… let the highest power of 2 2 that divides 3 n−1 3 n−1 be 2 p(n)2 p(n). If n n is odd, say n=2 m+1 n=2 m+1, then 3 n−1=(3−1)(3 2 m+3 2 m−1+⋯+1)3 n−1=(3−1)(3 2 m+3 2 m−1+⋯+1). The sum has an odd number of terms, so p(n)=1 p(n)=1. If n n is even, say n=2 m n=2 m, then 3 n−1=(3 m−1)(3 m+1)3 n−1=(3 m−1)(3 m+1). By induction, 3 m≡1 mod 8 3 m≡1 mod 8 if m m is even and 3 m≡3 mod 8 3 m≡3 mod 8 if m m is odd. Hence p(2 m)=p(m)+1 p(2 m)=p(m)+1 if m m is even, p(m)+2 p(m)+2 if m is odd. By applying this repeatedly we get that if n=2 a b n=2 a b, where a>0 a>0 and b b is odd, then p(n)=a+2 p(n)=a+2. It follows easily that for n>0 n>0 we have p(n)≥n p(n)≥n iff n=1,2,4 n=1,2,4.
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answered Dec 12, 2017 at 23:36
Michael BehrendMichael Behrend
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Let ν 2(n)=max{m∈N:2 m∣n}ν 2(n)=max{m∈N:2 m∣n}. We may tackle the given problem by finding an explicit form for (or, at least, an explicit recursive algorithm for the determination of) ν 2(3 n−1)ν 2(3 n−1). By letting a n=3 n−1 a n=3 n−1 we have
a 2 k+1=3⋅9 k−1≡2(mod 8)a 2 k+1=3⋅9 k−1≡2(mod 8)
from which ν 2(a 2 k+1)=1 ν 2(a 2 k+1)=1, and
a 2 k=a k(3 k+1)a 2 k=a k(3 k+1)
from which ν 2(a 2 k)=ν 2(a k)+ν 2(3 k+1)ν 2(a 2 k)=ν 2(a k)+ν 2(3 k+1). On the other hand
3 2 m+1+1=3⋅9 m+1≡4(mod 8)3 2 m+1+1=3⋅9 m+1≡4(mod 8)
3 2 m+1=9 m+1≡2(mod 8)3 2 m+1=9 m+1≡2(mod 8)
so ν 2(3 2 m+1+1)=2 ν 2(3 2 m+1+1)=2 and ν 2(3 2 m+1)=1 ν 2(3 2 m+1)=1. In particular
ν 2(a n)≤1+2 log 2(n)ν 2(a n)≤1+2 log 2(n)
and the only solutions of 2 n∣(3 n−1)2 n∣(3 n−1) are associated to n≤7 n≤7, hence they can be found by direct inspection. In explicit terms,
ν 2(a n)=2+ν 2(n)−1 odd(n).ν 2(a n)=2+ν 2(n)−1 odd(n).
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edited Dec 12, 2017 at 23:57
answered Dec 12, 2017 at 23:40
Jack D'AurizioJack D'Aurizio
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I write this with two ad-hoc invented notations:
define [n:m]=1[n:m]=1 if m|n m|n and [n:m]=0[n:m]=0 if m/|n m⧸|n ("Iverson-brackets")
define {n,m}=A{n,m}=A if n n contains m m to the power of A A; or "n=x⋅m A n=x⋅m A" (or "valuation" v m(n)v m(n))
Evaluating Fermat's little theorem and Euler's totient theorem and the concept of "order of cyclic subgroups modulo prime" we can find:
A={3 n−1,2}=1+[n:2]+{n:2}A={3 n−1,2}=1+[n:2]+{n:2}
which means also, that A A is logarithmic in n n and the number of solutions must be finite and occur in small A A and n n. Actual enumeration of small cases gives
n 0 1 2 3 4 5 6⋮3 n−1 0 2 8 26 80 242 728 A†1 3 1 4 1 3(=1+[n:2]+{n,2})†1+0+0 1+1+1 1+0+0 1+1+2 1+0+0 1+1+1 n 3 n−1 A(=1+[n:2]+{n,2})0 0††1 2 1 1+0+0 2 8 3 1+1+1 3 26 1 1+0+0 4 80 4 1+1+2 5 242 1 1+0+0 6 728 3 1+1+1⋮
( ††: this is not defined here and might be assumed zero or infinite)
From n=5 n=5 the smaller increase in A A than that of n n comes into account and thus no more equality A=n A=n can occur.
(For a more explicite discussion see a short text of mine)
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edited Dec 13, 2017 at 7:58
answered Dec 13, 2017 at 7:50
Gottfried HelmsGottfried Helms
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10269 | http://ndl.ethernet.edu.et/bitstream/123456789/88007/10/Ulaby-FundamentalsOfAppliedElectromagnetics-2014.pdf | FUNDAMENTALS OF APPLIED ELECTROMAGNETICS Seventh Edition Fawwaz T. Ulaby University of Michigan, Ann Arbor Umberto Ravaioli University of Illinois, Urbana–Champaign Pearson Boston · Columbus · Indianapolis · New York · San Francisco · Upper Saddle River · Amsterdam Cape Town · Dubai · London · Madrid · Milan · Munich · Paris · Montreal · Toronto Delhi · Mexico City · Sau Paula · Sydney · Hong Kong · Seoul · Singapore · Taipei · Tokyo Library of Congress Cataloging-in-Publication Data on File Vice President and Editorial Director, ECS: Marcia J. Horton Acquisitions Editor: Julie Bai Editorial Assistant: Sandra Rodriguez Managing Editor: Scott Disanno Production Editor: Rose Kernan Art Director: Marta Samsel Art Editor: Gregory Dulles Manufacturing Manager: Mary Fischer Manufacturing Buyer: Maura Zaldivar-Garcia Product Marketing Manager: Bram Van Kempen Field Marketing Manager: Demetrius Hall Marketing Assistant: Jon Bryant Cover Designer: Black Horse Designs c ⃝2015, 2010 Pearson Education, Inc.
Upper Saddle River, New Jersey 07458 All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages with, or arising out of, the furnishing, performance, or use of these programs.
Previous editions copyright c ⃝2007 by Pearson Education, Inc.
Pearson Education Ltd., London Pearson Education Australia Pty. Ltd., Sydney Pearson Education Singapore, Pte. Ltd.
Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada, Inc., Toronto Pearson Education de Mexico, S.A. de C.V.
Pearson Education–Japan, Tokyo Pearson Education Malaysia, Pte. Ltd.
Pearson Education, Inc., Upper Saddle River, New Jersey www.pearsonhighered.com 10 9 8 7 6 5 4 3 2 1 ISBN-13: 978-0-13-335681-6 ISBN-10: 0-13-335681-7 We dedicate this book to Jean and Ann Lucia.
This page intentionally left blank Preface to Seventh Edition Building on the core content and style of its predecessor, this seventh edition (7/e) of Applied Electromagnetics introduces new features designed to help students develop a deeper understanding of electromagnetic concepts and applications.
Prominent among them is a set of 52 web-based simulation modules that allow the user to interactively analyze and design transmission line circuits; generate spatial patterns of the electric and magnetic fields induced by charges and currents; visualize in 2-D and 3-D space how the gradient, divergence, and curl operate on spatial functions; observe the temporal and spatial waveforms of plane waves propagating in lossless and lossy media; calculate and display field distributions inside a rectangular waveguide; and generate radiation patterns for linear antennas and parabolic dishes.
These are valuable learning tools; we encourage students to use them and urge instructors to incorporate them into their lecture materials and homework assignments.
Additionally, by enhancing the book’s graphs and il-lustrations, and by expanding the scope of topics of the Technology Briefs, additional bridges between electromagnetic fundamentals and their countless engineering and scientific applications are established. In summary: NEW TO THIS EDITION • A set of 10 additional interactive simulation modules, bringing the total to 52 • Updated Technology Briefs • Enhanced figures and images • New/updated end-of-chapter problems • The interactive modules and Technology Briefs can be found at the Student Website on
ACKNOWLEDGMENTS As authors, we were blessed to have worked on this book with the best team of professionals: Richard Carnes, Leland Pierce, JaniceRichards, RoseKernan, andPaulMailhot. Weare exceedingly grateful for their superb support and unwavering dedication to the project.
We enjoyed working on this book.
We hope you enjoy learning from it.
Fawwaz T. Ulaby Umberto Ravaioli vi PREFACE CONTENT The book begins by building a bridge between what should be familiar to a third-year electrical engineering student and the electromagnetics (EM) material covered in the book. Prior to enrolling in an EM course, a typical student will have taken one or more courses in circuits. He or she should be familiar with circuit analysis, Ohm’s law, Kirchhoff’s current and voltage laws, and related topics. Transmission lines constitute a natural bridge between electric circuits and electromagnetics. Without having to deal with vectors or fields, the student uses already familiar concepts to learn about wave motion, the reflection and transmissionofpower, phasors, impedancematching, andmany of the properties of wave propagation in a guided structure. All of these newly learned concepts will prove invaluable later (in Chapters 7 through 9) and will facilitate the learning of how plane waves propagate in free space and in material media.
Transmission lines are covered in Chapter 2, which is preceded in Chapter 1 with reviews of complex numbers and phasor analysis.
The next part of the book, contained in Chapters 3 through 5, covers vector analysis, electrostatics, and magnetostatics. The electrostatics chapter begins with Maxwell’s equations for the time-varying case, which are then specialized to electrostatics and magnetostatics, thereby providing the student with an overall framework for what is to come and showing him or her why electrostatics and magnetostatics are special cases of the more general time-varying case.
Chapter 6 deals with time-varying fields and sets the stage for the material in Chapters 7 through 9.
Chapter 7 covers plane-wave propagation in dielectric and conducting media, and Chapter 8 covers reflection and transmission at discontinuous boundaries and introduces the student to fiber optics, waveguides and resonators.
In Chapter 9, the student is introduced to the principles of radiation by currents flowing in wires, such as dipoles, as well as Suggested Syllabi Two-semester Syllabus One-semester Syllabus 6 credits (42 contact hours per semester) 4 credits (56 contact hours) Chapter Sections Hours Sections Hours 1 Introduction: All 4 All 4 Waves and Phasors 2 Transmission Lines All 12 2-1 to 2-8 and 2-11 8 3 Vector Analysis All 8 All 8 4 Electrostatics All 8 4-1 to 4-10 6 5 Magnetostatics All 7 5-1 to 5-5 and 5-7 to 5-8 5 Exams 3 2 Total for first semester 42 6 Maxwell’s Equations All 6 6-1 to 6-3, and 6-6 3 for Time-Varying Fields 7 Plane-wave Propagation All 7 7-1 to 7-4, and 7-6 6 8 Wave Reflection All 9 8-1 to 8-3, and 8-6 7 and Transmission 9 Radiation and Antennas All 10 9-1 to 9-6 6 10 Satellite Communication All 5 None — Systems and Radar Sensors Exams 3 1 Total for second semester 40 Total 56 Extra Hours 2 0 PREFACE vii to radiation by apertures, such as a horn antenna or an opening in an opaque screen illuminated by a light source.
To give the student a taste of the wide-ranging applications of electromagnetics in today’s technological society, Chapter 10 concludes the book with overview presentations of two system examples: satellite communication systems and radar sensors.
The material in this book was written for a two-semester sequence of six credits, but it is possible to trim it down to generate a syllabus for a one-semester four-credit course. The accompanying table provides syllabi for each of these two options.
MESSAGE TO THE STUDENT The web-based interactive modules of this book were developed with you, the student, in mind.
Take the time to use them in conjunction with the material in the textbook.
Video animations can show you how fields and waves propagate in time and space, how the beam of an antenna array can be made to scan electronically, and examples of how current is induced in a circuit under the influence of a changing magnetic field. The modules are a useful resource for self-study.
You can find them at the Student Website link on Use them!
ACKNOWLEDGMENTS Special thanks are due to reviewers for their valuable comments and suggestions. They include Constantine Balanis of Arizona State University, Harold Mott of the University of Alabama, David Pozar of the University of Massachusetts, S. N. Prasad of Bradley University, Robert Bond of New Mexico Institute of Technology, Mark Robinson of the University of Colorado at Colorado Springs, and Raj Mittra of the University of Illinois. I appreciate the dedicated efforts of the staff at Prentice Hall and I am grateful for their help in shepherding this project through the publication process in a very timely manner.
Fawwaz T. Ulaby This page intentionally left blank List of Technology Briefs TB1 LED Lighting 20 TB2 Solar Cells 38 TB3 Microwave Ovens 82 TB4 EM Cancer Zappers 112 TB5 Global Positioning System 150 TB6 X-Ray Computed Tomography 164 TB7 Resistive Sensors 196 TB8 Supercapacitors as Batteries 214 TB9 Capacitive Sensors 218 TB10 Electromagnets 256 TB11 Inductive Sensors 268 TB12 EMF Sensors 292 TB13 RFID Systems 322 TB14 Liquid Crystal Display (LCD) 336 TB15 Lasers 368 TB16 Bar-Code Readers 382 TB17 Health Risks of EM Fields 424 This page intentionally left blank Contents Preface v List ofTechnology Briefs ix List of Modules xvii Photo Credits xix Chapter 1 Introduction: Waves and Phasors 1 1-1 Historical Timeline 3 1-1.1 EM in the Classical Era 3 1-1.2 EM in the Modern Era 3 1-2 Dimensions, Units, and Notation 11 1-3 The Nature of Electromagnetism 12 1-3.1 The Gravitational Force: A Useful Analogue 12 1-3.2 Electric Fields 13 1-3.3 Magnetic Fields 15 1-3.4 Static and Dynamic Fields 16 1-4 Traveling Waves 18 1-4.1 Sinusoidal Waves in a Lossless Medium 19 TB1 LED Lighting 20 1-4.2 Sinusoidal Waves in a Lossy Medium 28 1-5 The Electromagnetic Spectrum 30 1-6 Review of Complex Numbers 32 1-7 Review of Phasors 36 1-7.1 Solution Procedure 37 TB2 Solar Cells 38 1-7.2 Traveling Waves in the Phasor Domain 43 Chapter 1 Summary 43 Problems 44 Chapter 2 Transmission Lines 48 2-1 General Considerations 49 2-1.1 The Role of Wavelength 49 2-1.2 Propagation Modes 51 2-2 Lumped-Element Model 52 2-3 Transmission-Line Equations 56 2-4 Wave Propagation on a Transmission Line 57 2-5 The Lossless Microstrip Line 60 xii CONTENTS 2-6 The Lossless Transmission Line: General Considerations 65 2-6.1 Voltage Reflection Coefficient 66 2-6.2 Standing Waves 70 2-7 Wave Impedance of the Lossless Line 75 2-8 Special Cases of the Lossless Line 78 2-8.1 Short-Circuited Line 78 2-8.2 Open-Circuited Line 81 2-8.3 Application of Short-Circuit/ Open-Circuit Technique 81 TB3 Microwave Ovens 82 2-8.4 Lines of Length l = nλ/2 84 2-8.5 Quarter-Wavelength Transformer 84 2-8.6 Matched Transmission Line: ZL = Z0 85 2-9 Power Flow on a Lossless Transmission Line 86 2-9.1 Instantaneous Power 86 2-9.2 Time-Average Power 87 2-10 The Smith Chart 88 2-10.1 Parametric Equations 89 2-10.2 Wave Impedance 92 2-10.3 SWR, Voltage Maxima and Minima 93 2-10.4 Impedance to Admittance Transformations 96 2-11 Impedance Matching 101 2-11.1 Lumped-Element Matching 102 2-11.2 Single-Stub Matching 108 2-12 Transients on Transmission Lines 111 TB4 EM Cancer Zappers 112 2-12.1 Transient Response 115 2-12.2 Bounce Diagrams 118 Chapter 2 Summary 122 Problems 124 Chapter 3 Vector Analysis 133 3-1 Basic Laws of Vector Algebra 134 3-1.1 Equality of Two Vectors 135 3-1.2 Vector Addition and Subtraction 135 3-1.3 Position and Distance Vectors 136 3-1.4 Vector Multiplication 136 3-1.5 Scalar and Vector Triple Products 139 3-2 Orthogonal Coordinate Systems 140 3-2.1 Cartesian Coordinates 141 3-2.2 Cylindrical Coordinates 142 3-2.3 Spherical Coordinates 145 3-3 Transformations between Coordinate Systems 147 3-3.1 Cartesian to Cylindrical Transformations 147 TB5 Global Positioning System 150 3-3.2 Cartesian to Spherical Transformations 152 3-3.3 Cylindrical to Spherical Transformations 153 3-3.4 Distance between Two Points 153 3-4 Gradient of a Scalar Field 154 3-4.1 Gradient Operator in Cylindrical and Spherical Coordinates 155 3-4.2 Properties of the Gradient Operator 156 3-5 Divergence of a Vector Field 158 3-6 Curl of a Vector Field 162 TB6 X-Ray Computed Tomography 164 3-6.1 Vector Identities Involving the Curl 166 3-6.2 Stokes’s Theorem 166 3-7 Laplacian Operator 167 Chapter 3 Summary 169 Problems 171 Chapter 4 Electrostatics 178 4-1 Maxwell’s Equations 179 4-2 Charge and Current Distributions 180 4-2.1 Charge Densities 180 4-2.2 Current Density 181 4-3 Coulomb’s Law 182 4-3.1 Electric Field due to Multiple Point Charges 183 4-3.2 Electric Field due to a Charge Distribution 184 4-4 Gauss’s Law 187 4-5 Electric Scalar Potential 189 4-5.1 Electric Potential as a Function of Electric Field 189 4-5.2 Electric Potential Due to Point Charges 191 CONTENTS xiii 4-5.3 Electric Potential Due to Continuous Distributions 191 4-5.4 Electric Field as a Function of Electric Potential 192 4-5.5 Poisson’s Equation 193 4-6 Conductors 195 TB7 Resistive Sensors 196 4-6.1 Drift Velocity 198 4-6.2 Resistance 199 4-6.3 Joule’s Law 200 4-7 Dielectrics 201 4-7.1 Polarization Field 202 4-7.2 Dielectric Breakdown 203 4-8 Electric Boundary Conditions 203 4-8.1 Dielectric-Conductor Boundary 207 4-8.2 Conductor-Conductor Boundary 208 4-9 Capacitance 210 4-10 Electrostatic Potential Energy 213 TB8 Supercapacitors as Batteries 214 TB9 Capacitive Sensors 218 4-11 Image Method 223 Chapter 4 Summary 225 Problems 226 Chapter 5 Magnetostatics 235 5-1 Magnetic Forces and Torques 237 5-1.1 Magnetic Force on a Current-Carrying Conductor 238 5-1.2 Magnetic Torque on a Current-Carrying Loop 241 5-2 The Biot–Savart Law 244 5-2.1 Magnetic Field due to Surface and Volume Current Distributions 244 5-2.2 Magnetic Field of a Magnetic Dipole 248 5-2.3 Magnetic Force Between Two Parallel Conductors 250 5-3 Maxwell’s Magnetostatic Equations 251 5-3.1 Gauss’s Law for Magnetism 251 5-3.2 Ampere’s Law 252 TB10 Electromagnets 256 5-4 Vector Magnetic Potential 259 5-5 Magnetic Properties of Materials 260 5-5.1 Electron Orbital and Spin Magnetic Moments 261 5-5.2 Magnetic Permeability 261 5-5.3 Magnetic Hysteresis of Ferromagnetic Materials 262 5-6 Magnetic Boundary Conditions 264 5-7 Inductance 265 5-7.1 Magnetic Field in a Solenoid 265 5-7.2 Self-Inductance 267 TB11 Inductive Sensors 268 5-7.3 Mutual Inductance 270 5-8 Magnetic Energy 271 Chapter 5 Summary 272 Problems 274 Chapter 6 Maxwell’s Equations for Time-Varying Fields 281 6-1 Faraday’s Law 282 6-2 Stationary Loop in a Time-Varying Magnetic Field 284 6-3 The Ideal Transformer 288 6-4 Moving Conductor in a Static Magnetic Field 289 TB12 EMF Sensors 292 6-5 The Electromagnetic Generator 294 6-6 Moving Conductor in a Time-Varying Magnetic Field 296 6-7 Displacement Current 297 6-8 Boundary Conditions for Electromagnetics 299 6-9 Charge-Current Continuity Relation 299 6-10 Free-Charge Dissipation in a Conductor 302 6-11 Electromagnetic Potentials 302 6-11.1 Retarded Potentials 303 6-11.2 Time-Harmonic Potentials 304 Chapter 6 Summary 307 Problems 308 xiv CONTENTS Chapter 7 Plane-Wave Propagation 313 7-1 Time-Harmonic Fields 315 7-1.1 Complex Permittivity 315 7-1.2 Wave Equations 316 7-2 Plane-Wave Propagation in Lossless Media 316 7-2.1 Uniform Plane Waves 317 7-2.2 General Relation Between E and H 319 TB13 RFID Systems 322 7-3 Wave Polarization 324 7-3.1 Linear Polarization 325 7-3.2 Circular Polarization 326 7-3.3 Elliptical Polarization 328 7-4 Plane-Wave Propagation in Lossy Media 331 7-4.1 Low-Loss Dielectric 333 7-4.2 Good Conductor 334 TB14 Liquid Crystal Display (LCD) 336 7-5 Current Flow in a Good Conductor 339 7-6 Electromagnetic Power Density 343 7-6.1 Plane Wave in a Lossless Medium 343 7-6.2 Plane Wave in a Lossy Medium 344 7-6.3 Decibel Scale for Power Ratios 345 Chapter 7 Summary 346 Problems 348 Chapter 8 Wave Reflection and Transmission 352 8-1 Wave Reflection and Transmission at Normal Incidence 353 8-1.1 Boundary between Lossless Media 354 8-1.2 Transmission-Line Analogue 356 8-1.3 Power Flow in Lossless Media 357 8-1.4 Boundary between Lossy Media 359 8-2 Snell’s Laws 362 8-3 Fiber Optics 365 8-4 Wave Reflection and Transmission at Oblique Incidence 367 TB15 Lasers 368 8-4.1 Perpendicular Polarization 370 8-4.2 Parallel Polarization 374 8-4.3 Brewster Angle 375 8-5 Reflectivity and Transmissivity 376 8-6 Waveguides 380 TB16 Bar-Code Readers 382 8-7 General Relations for E and H 383 8-8 TM Modes in Rectangular Waveguide 384 8-9 TE Modes in Rectangular Waveguide 388 8-10 Propagation Velocities 388 8-11 Cavity Resonators 392 8-11.1 Resonant Frequency 393 8-11.2 Quality Factor 393 Chapter 8 Summary 395 Problems 397 Chapter 9 Radiation and Antennas 403 9-1 The Hertzian Dipole 406 9-1.1 Far-Field Approximation 408 9-1.2 Power Density 409 9-2 Antenna Radiation Characteristics 411 9-2.1 Antenna Pattern 411 9-2.2 Beam Dimensions 412 9-2.3 Antenna Directivity 414 9-2.4 Antenna Gain 416 9-2.5 Radiation Resistance 416 9-3 Half-Wave Dipole Antenna 417 9-3.1 Directivity of λ/2 Dipole 419 9-3.2 Radiation Resistance of λ/2 Dipole 419 9-3.3 Quarter-Wave Monopole Antenna 420 9-4 Dipole of Arbitrary Length 420 9-5 Effective Area of a Receiving Antenna 422 TB17 Health Risks of EM Fields 424 9-6 Friis Transmission Formula 427 9-7 Radiation by Large-Aperture Antennas 429 9-8 Rectangular Aperture with Uniform Aperture Distribution 432 9-8.1 Beamwidth 433 9-8.2 Directivity and Effective Area 434 9-9 Antenna Arrays 435 9-10 N-Element Array with Uniform Phase Distribution 442 CONTENTS xv 9-11 Electronic Scanning of Arrays 444 9-11.1 Uniform-Amplitude Excitation 445 9-11.2 Array Feeding 445 Chapter 9 Summary 450 Problems 452 Chapter 10 Satellite Communication Systems and Radar Sensors 457 10-1 Satellite Communication Systems 458 10-2 Satellite Transponders 460 10-3 Communication-Link Power Budget 462 10-4 Antenna Beams 463 10-5 Radar Sensors 464 10-5.1 Basic Operation of a Radar System 464 10-5.2 Unambiguous Range 465 10-5.3 Range and Angular Resolutions 466 10-6 Target Detection 467 10-7 Doppler Radar 469 10-8 Monopulse Radar 470 Chapter 10 Summary 473 Problems 474 Appendix A Symbols, Quantities, Units, and Abbreviations 475 Appendix B Material Constants of Some Common Materials 479 Appendix C Mathematical Formulas 483 Appendix D Answers to Selected Problems 485 Bibliography 491 Index 493 This page intentionally left blank List of Modules 1.1 Sinusoidal Waveforms 27 1.2 Traveling Waves 29 1.3 Phase Lead/Lag 31 2.1 Two-Wire Line 60 2.2 Coaxial Cable 61 2.3 Lossless Microstrip Line 64 2.4 Transmission-Line Simulator 73 2.5 Wave and Input Impedance 78 2.6 Interactive Smith Chart 101 2.7 Quarter-Wavelength Transformer 109 2.8 Discrete Element Matching 110 2.9 Single-Stub Tuning 111 2.10 Transient Response 121 3.1 Vector Addition and Subtraction 145 3.2 Gradient 158 3.3 Divergence 162 3.4 Curl 168 4.1 Fields due to Charges 194 4.2 Charges in Adjacent Dielectrics 207 4.3 Charges above Conducting Plane 209 4.4 Charges near Conducting Sphere 210 5.1 Electron Motion in Static Fields 238 5.2 Magnetic Fields due to Line Sources 246 5.3 Magnetic Field of a Current Loop 249 5.4 Magnetic Force Between Two Parallel Conductors 251 6.1 Circular Loop in Time-varying Magnetic Field 287 6.2 Rotating Wire Loop in Constant Magnetic Field 296 6.3 Displacement Current 300 7.1 Linking E to H 321 7.2 Plane Wave 324 7.3 Polarization I 331 7.4 Polarization II 332 7.5 Wave Attenuation 339 7.6 Current in a Conductor 342 8.1 Normal Incidence on Perfect Conductor 362 8.2 Multimode Step-Index Optical Fiber 367 8.3 Oblique Incidence 379 8.4 Oblique Incidence in Lossy Medium 380 8.5 Rectangular Waveguide 393 9.1 Hertzian Dipole (l ≪λ) 410 9.2 Linear Dipole Antenna 422 9.3 Detailed Analysis of Linear Antenna 423 9.4 Large Parabolic Reflector 435 9.5 Two-dipole Array 440 9.6 Detailed Analysis of Two-Dipole Array 441 9.7 N-Element Array 447 9.8 Uniform Dipole Array 449 This page intentionally left blank Photo Credits Page 2 (Fig 01-01): Line Art: 2-D LCD array, Source: Fawwaz Ulaby Page 4 (Ch 01-01A): Thales of Miletus (624–546 BC), Photo Researchers, Inc./Science Source Page 4 (Ch 01-01B): Isaac Newton, Mary Evans/Science Source Page 4 (Ch 01-01C): Benjamin West, Benjamin Franklin Drawing Electricity from the Sky, Painting/Alamy Page 4 (Ch 01-01D): Replica of the Voltaic pile invented by Alessandro Volta 1800, Clive Streeter/DK Images Page 4 (Ch 01-01E): Hans Christian Ørsted, Danish Physicist, Science Source Page 4 (Ch 01-01F): Andre-Marie Ampere, Nickolae/Fotolia Page 5 (Ch 01-01G): Michael Faraday, Nicku/Shutterstock Page 5 (Ch 01-01H): James Clerk Maxwell (1831–1879), SPL/Science Source Page 5 (Ch 01-01I): Heinrich Rudolf Hertz, Science Source Page 5 (Ch 01-01J): Nicola Tesla, Bain News Service/NASA Page 5 (Ch 01-01K): Early X-Ray of Hand, Bettmann/Corbis Page 5 (Ch 01-01M): Albert Einstein, Science Source Page 6 (Ch 01-02A): Telegraph, Morse apparatus, vintage engraved illustration, Morphart Creation/Shutterstock Page 6 (Ch 01-02B):ThomasAlva EdisonWith His ’Edison Effect’ Lamps, Education Images/Getty Images, Inc.
Page 6 (Ch 01-02C): Replica of an early type of telephone made by Scottish-born telephony pioneerAlexander Graham Bell (1847–1922), Science & Society Picture Library/Getty Images Page 6 (Ch 01-02D): Guglielmo Marconi, Pach Brothers/Library of Congress Prints and Photographs Division [LC-USZ62-39702] Page 6 (Ch 01-02E): De Forest seated at his invention, the radio-telephone, called theAudion, Jessica Wilson/Science Source Page 6 (Ch 01-02F): The staff of KDKA broadcast reports of the 1920 presidential election, Bettmann/Corbis Page 7 (Ch 01-02G): This bottle-like object is a Cathode Ray tube which forms the receiver of the new style television inventedbyDr. VladimirZworykin,Westinghouseresearch engineer, who is holding it, Bettmann/Corbis Page 7 (Ch 01-02H): Radar in operation in the Second World War, Library of Congress Department of Prints and Photographs [LC-USZ62-101012] Page 7 (Ch 01-02I): Shockly, Brattain, and Bardeen with an apparatus used in the early investigations which led to the invention of the transistor, Photo Researchers, Inc./Science Source xx PHOTO CREDITS Page 7 (Ch 01-02J): A Photograph of Jack Kilby’s Model of the First Working Integrated Circuit Ever Built circa 1958, Fotosearch/Archive Photos/Getty Images Page 7 (Ch 01-02K): Shown here is the 135-foot rigidized inflatable balloon satellite undergoing tensile stress test in a dirigible hanger at Weekesville, North Carolina, NASA Page 7 (Ch 01-02L): Pathfinder on Mars, JPL/NASA Page 8 (Ch 01-03A): Abacus isolated on white, Sikarin Sup-phatada/Shutterstock Page 8 (Ch 01-03B): Pascaline; a mechanical calculator invented by Blaise Pascal in 1642, Science Source Page 8 (Ch 01-03C): Original Caption: Portrait of American electrical engineer Vannevar Bush, Bettmann/Corbis Page 8 (Ch 01-03D): J. Presper Eckert and John W. Mauchly, are pictured with the Electronic Numerical Integrator and Computer (ENIAC) in this undated photo from the University of Pennsylvania Archives, University of Pennsylvania/AP images Page 8 (Ch 01-03E): Description: DEC PDP-1 computer, on display at the Computer History Museum, USA, Volker Steger/Science Source Page 9 (Ch 01-03F): Classic Antique Red LED Diode Calculator, James Brey/E+/Getty Images Page 9 (Ch 01-03G): Apple I computer.
This was released in April 1976 at the Homebrew Computer Club, USA, Volker Steger/Science Page 9 (Ch 01-03H): UNITED STATES—DECEMBER 07: The IBM Personal Computer System was introduced to the market in early 1981, SSPL/Getty Images, Inc.
Page 9 (Ch 01-03I): NEW YORK, UNITED STATES: Chess enthusiasts watch World Chess champion Garry Kasparov on a television monitor as he holds his head in his hands, Stan Honda/Getty Images, Inc.
Page 10 (Fig 01-02A): The Very Large Array of Radio Telescopes, VLA, NRAO/NASA Page 10 (Fig 01-02B): SCaN’s Benefits to Society—Global Posi-tioning System, Jet Propulsion Laboratory/NASA Page 10 (Fig 01-02C): Motor, ABB Page 10 (Fig 01-02D and Page 338 (Fig TF14-04)): TV on white background, Fad82/Fotolia Page 10 (Fig 01-02E): Nuclear Propulsion Through Direct Conver-sion of Fusion Energy, John Slough/NASA Page 10 (Fig01-02F):Trackingstationhasbird’seyeviewonVAFB, Ashley Tyler/US Air Force Page 10 (Fig 01-02G): Glass Fiber Cables, Kulka/Zefa/Corbis Page 10 (Fig 01-02H): Electromagnetic sensors, HW Group Page 10 (Fig 01-02I): Touchscreen smartphone, Oleksiy Mark/Shutterstock Page 10 (Fig 01-02J): Line Art: Electromagnetics is at the heart of numerous systems and applications:, Source: Based on IEEE Spectrum Page 20 (TF 01-01a): Lightbulb, Chones/Fotolia Page 20 (TF 01-01b): Fluorescent bulb, Wolf1984/Fotolia Page 20 (TF 01-01c): 3d render of an unbranded screw-in LED lamp, isolated on a white background, Marcello Bortolino/Getty Images, Inc.
Page 21 (TF 01-03): Line Art: Lighting efficiency, Source: Based on Courtesy of National Research Council, 2009 Page 27 (Mod 01-01): Screenshot: Sinusoidal Waveforms, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page 29 (Mod 01-02): Screenshot: TravelingWaves, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page 31 (Mod02-04): Screenshot: Phase Lead/Lag, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page 33 (Fig 01-17): Line Art: Individual bands of the radio spectrum and their primary allocations in the US. [See expandable version on CD.], Source: U.S. Department of Commerce Page 60 (Mod 02-01): Screenshot: Two-Wire Line, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page 61 (Mod 02-02): Screenshot: Coaxial Cable, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page 62 (Fig 02-10a): Line Art: Microstrip line: longitudinal view, Source: Prof. Gabriel Rebeiz, U. California at San Diego Page 62 (Fig 02-10b): Line Art: Microstrip line: Cross-sectional view, Source: Prof. Gabriel Rebeiz, U. California at San Diego Page 62 (Fig 02-10c): Circuit board, Gabriel Reibeiz Page 64 (Mod02-03): Screenshot: Lossless Microstrip Line, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page 73 (Mod02-04): Screenshot: Transmission-Line Simulator, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page 78 (Mod 02-05): Screenshot: Wave and Input Impedance, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page 83 (TF 03-02): Microwave oven cavity, Pearson Education, Inc.
Page101 (Mod 02-06): Screenshot: Interactive Smith Chart, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey PHOTO CREDITS xxi Page109 (Mod 02-07): Screenshot: Quarter-Wavelength Trans-former, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page110 (Mod 02-08): Screenshot: Discrete Element Matching, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page111 (Mod 02-09): Screenshot: Single-Stub Tuning, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page112 (TF 04-01): Microwave ablation for cancer liver treatment, Radiological Society of North America (RSNA) Page113 (TF 04-02): Setup for a percutaneous microwave ablation procedure shows three single microwave applicators connected to three microwave generators, Radiological Society of North America (RSNA) Page114 (TF 04-03): Line Art: Bryan Christie Design LLC Page121 (Mod 02-10): Screenshot: Transient Response, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page145 (Mod 03-01): Screenshot: Vector Addition and Subtrac-tion, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page150 (TF 05-01): Touchscreen smartphone with GPS navi-gation isolated on white reflective background, Oleksiy Mark/Shutterstock Page150 (TF 05-02): SCaN’s Benefits to Society—Global Position-ing System, Jet Propulsion Laboratory/NASA Page151 (TF 05-03): SUV, Konstantin/Fotolia Page158 (Mod 03-02): Screenshot: Gradient, Source: Graphics created with Wolfram Matematica® Page162 (Mod 03-03): Screenshot: Divergence, Source: Graphics created with Wolfram Matematica® Page164 (TF 06-01): X-ray of pelvis and spinal column, Cozyta/Getty Images, Inc.
Page164 (TF 06-02): CT scan advance technology for medical diagnosis, Tawesit/Fotolia Page165 (TF 06-03c): Digitally enhanced CT scan of a normal brain in transaxial (horizontal) section, Scott Camazine/Science Source Page168 (Mod 03-04): Screenshot: Curl, Source: Graphics created with Wolfram Matematica Page194 (Mod 04-01): Screenshot: Fields due to Charges, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page207 (Mod 04-02): Screenshot: Charges inAdjacent Dielectrics, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page209 (Mod 04-03): Screenshot: Charges above Conducting Plane, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page210 (Mod 04-04): Screenshot: Charges near Conducting Sphere, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page214 (TF 08-01): Various electrolytic capacitors, David J.
Green/Alamy Page214 (TF08-02A): High-speed train in motion, Metlion/Fotolia Page214 (TF08-02B): Cordless Drill, Derek Hatfield/Shutterstock Page214 (TF08-02C): The 2006 BMW X3 Concept Gasoline Electric Hybrid uses high-performance capacitors (or “Super Caps”) to store and supply electric energy to the vehicle’sActive Transmission, Passage/Car Culture/Corbis Page214 (TF 08-02D): LED Electric torch—laser Pointer isolated on white background, Artur Synenko/Shutterstock Page222 (TF 09-06): Line Art: Bryan Christie Design, LLC Page222 (TF 09-07): Line Art: Fingerprint representation, Source: Courtesy of Dr. M. Tartagni, University of Bologna, Italy Page238 (Mod 05-01): Screenshot: Electron Motion in Static Fields, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page246 (Mod 05-02): Screenshot: Magnetic Fields due to Line Sources, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page249 (Mod 05-03): Screenshot: Magnetic Field of a Current Loop, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page251 (Mod 05-04): Screenshot: Magnetic Force Between Two Parallel Plates, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page258 (TF 10-05A): CHINA—JUNE 20: A maglev train awaits departure in Shanghai, China, on Saturday, June 20, 2009, Qilai Shen/Bloomberg/Getty Images Page258 (TF 10-5b and c): Line Art: Magnetic trains—(b) internal workings of the Maglev train, Source: Amy Mast, Maglev trains are making history right now. Flux, volume 3 issue 1, National High Magnetic Field Laboratory Page287 (Mod 06-01): Screenshot: Circular Loop in Time-varying Magnetic Field, Source: Copyright c ⃝by Pearson Education, Upper Saddle River, New Jersey Page296 (Mod 06-02): Screenshot: Rotating Wire Loop in Constant Magnetic Field, Source: Copyright c ⃝by Pearson Education, Upper Saddle River, New Jersey Page300 (Mod 06-02): Screenshot: Displacement Current, Source: Copyright c ⃝by Pearson Education, Upper Saddle River, New Jersey Page321 (Mod 07-01): Screenshot: Linking E to H, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey xxii PHOTO CREDITS Page322 (TF 13-01): Jersey cow on pasture, Lakeview Im-ages/Shutterstock Page323 (TF 13-2): Line Art: How an RFID system works is illustrated through this EZ-Pass example: Tag, Source: Prof. C. F. Huang Page324 (Mod 07-02): Screenshot: PlaneWave, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page331 (Mod 07-03): Screenshot: Polarization I, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page332 (Mod 07-04): Screenshot: Polarization II, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page339 (Mod 07-05): Screenshot: Wave Attenuation, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page342 (Mod 07-06): Screenshot: Current in Conductor, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page362 (Mod 08-01): Screenshot: Normal Incidence on Perfect Conductor, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page367 (Mod 08-02): Screenshot: Multimode Step-Index Optical Fiber, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page368 (TF 15-01A): Optical Computer Mouse, William White-hurst/Cusp/Corbis Page368 (TF 15-01B): Laser eye surgery, Will & Deni McIn-tyre/Science Source Page368 (TF 15-01C): Laser Star Guide, NASA Page368 (TF 15-01D): Laser: TRUMPF GmbH + Co. KG Page379 (Mod 08-03): Screenshot: Oblique Incidence, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page380 (Mod 08-04): Screenshot: Oblique Incidence in Lossy Medium, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page393 (Mod 08-05): Screenshot: Rectangular Waveguide, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page410 (Mod 09-01): Screenshot: Hertzian Dipole (l ≪λ), Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page423 (Mod 09-03): Screenshot: Detailed Analysis of Linear Antenna, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page424 (TF 17-01A): Smiling woman using computer, Edbock-stock/Fotolia Page424 (TF 17-01B): Vector silhouette of Power lines and electric pylons, Ints Vikmanis/Alamy Page424 (TF 17-01C): Telecommunications tower, Poliki/Fotolia Page435 (Mod 09-04): Screenshot: Large Parabolic Reflector, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page436 (Fig 09-25): The AN/FPS-85 Phased Array Radar Facility in the Florida panhandle, near the city of Freeport, NASA Page440 (Mod 09-05): Screenshot: Two-dipole Array, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page447 (Mod 09-07): Screenshot: N-Element Array, Source: c ⃝ Pearson Education, Upper Saddle River, New Jersey Page449 (Mod 09-08): Screenshot: Uniform Dipole Array, Source: c ⃝Pearson Education, Upper Saddle River, New Jersey Page464 (Text 10-01): 1. Dipoles and helices at VHF...steering and scanning. (79 words/212 pages), Source: R. G. Meadows and A. J. Parsons, Satellite Communications, Hutchinson Publishers, London, 1989 C H A P T E R 1 Introduction: Waves and Phasors Chapter Contents Overview, 2 1-1 Historical Timeline, 3 1-2 Dimensions, Units, and Notation, 11 1-3 The Nature of Electromagnetism, 12 1-4 Traveling Waves, 18 TB1 LED Lighting, 20 1-5 The Electromagnetic Spectrum, 30 1-6 Review of Complex Numbers, 32 1-7 Review of Phasors, 36 TB2 Solar Cells, 38 Chapter 1 Summary, 43 Problems, 44 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Describe the basic properties of electric and magnetic forces.
2. Ascribe mathematical formulations to sinusoidal waves traveling in both lossless and lossy media.
3. Apply complex algebra in rectangular and polar forms.
4. Apply the phasor-domain technique to analyze circuits driven by sinusoidal sources.
2 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS LCD display Liquid crystal Unpolarized light Exit polarizer Entrance polarizer 2-D pixel array Molecular spiral 678 Figure 1-1 2-D LCD array.
Overview Liquid crystal displays have become integral parts of many electronic consumer products, ranging from alarm clocks and cell phones to laptop computers and television systems. LCD technology relies on special electrical and optical properties of a class of materials known as liquid crystals, which are neither pure solids nor pure liquids but rather a hybrid of both. The molecular structure of these materials is such that when light travels through them, the polarization of the emerging light depends on whether or not a voltage exists across the material.
Consequently, when no voltage is applied, the exit surface appears bright, and conversely, when a voltage of a certain level is applied across the LCD material, no light passes through it, resulting in a dark pixel. In-between voltages translate into a range of grey levels.
By controlling the voltages across individual pixels in a two-dimensional array, a complete image can be displayed (Fig. 1-1). Color displays are composed of three subpixels with red, green, and blue filters.
▶The polarization behavior of light in an LCD is a prime example of how electromagnetics is at the heart of electrical and computer engineering. ◀ The subject of this book is applied electromagnetics (EM), whichencompassesthestudyofbothstaticanddynamicelectric and magnetic phenomena and their engineering applications.
Primary emphasis is placed on the fundamental properties of dynamic (time-varying) electromagnetic fields because of their greater relevance to practical problems in many applications, including wireless and optical communications, radar, bioelec-tromagnetics, and high-speed microelectronics. We study wave propagationinguidedmedia, suchascoaxialtransmissionlines, optical fibers and waveguides; wave reflection and transmission at interfaces between dissimilar media; radiation by antennas; and several other related topics.
The concluding chapter is intended to illustrate a few aspects of applied EM through an ex-amination of design considerations associated with the use and operation of radar sensors and satellite communication systems.
We begin this chapter with a chronology of the history of electricity and magnetism. Next, we introduce the fundamental electric and magnetic field quantities of electromagnetics, as well as their relationships to each other and to the electric charges and currents that generate them. These relationships constitute the underpinnings of the study of electromagnetic phenomena. Then, in preparation for the material presented in Chapter 2, we provide short reviews of three topics: traveling waves, complex numbers, and phasors, all useful in solving time-harmonic problems.
1-1 HISTORICAL TIMELINE 3 1-1 Historical Timeline The history of EM may be divided into two overlapping eras.
In the classical era, the fundamental laws of electricity and magnetism were discovered and formulated. Building on these formulations, the modern era of the past 100 years ushered in the birth of the field of applied EM, the topic of this book.
1-1.1 EM in the Classical Era Chronology 1-1 provides a timeline for the development of electromagnetic theory in the classical era. It highlights those discoveries and inventions that have impacted the historical development of EM in a very significant way, even though the selected discoveries represent only a small fraction of those responsible for our current understanding of electromagnetics.
As we proceed through the book, some of the names highlighted in Chronology 1-1, such as those of Coulomb and Faraday, will appear again later as we discuss the laws and formulations named after them.
The attractive force of magnetite was reported by the Greeks some 2800 years ago. It was also a Greek, Thales of Miletus, who first wrote about what we now call static electricity: he described how rubbing amber caused it to develop a force that could pick up light objects such as feathers. The term “electric” first appeared in print around 1600 in a treatise on the (electric) force generated by friction, authored by the physician to Queen Elizabeth I, William Gilbert.
About a century later, in 1733, Charles-Fran¸ cois du Fay introduced the notion that electricity involves two types of “flu-ids,” one “positive” and the other “negative,” and that like-fluids repel and opposite-fluids attract. His notion of a fluid is what we today call electric charge. The invention of the capacitor in 1745, originally called the Leyden jar, made it possible to store significant amounts of electric charge in a single device.
A few years later, in 1752, Benjamin Franklin demonstrated that lightning is a form of electricity. He transferred electric charge from a cloud to a Leyden jar via a silk kite flown in a thunderstorm. The collective eighteenth-century knowledge about electricity was integrated in 1785 by Charles-Augustin de Coulomb, in the form of a mathematical formulation characterizing the electrical force between two charges in terms of their strengths and polarities and the distance between them.
The year 1800 is noted for the development of the first electric battery by Alessandro Volta, and 1820 was a banner year for discoveries about how electric currents induce magnetism. This knowledge was put to good use by Joseph Henry, who devel-oped one of the earliest electromagnets and dc (direct current) electric motors. Shortly thereafter, Michael Faraday built the firstelectricgenerator(theconverseoftheelectricmotor). Fara-day, in essence, demonstrated that a changing magnetic field induces an electric field (and hence a voltage). The converse re-lation, namely that a changing electric field induces a magnetic field, was first proposed by James Clerk Maxwell in 1864 and then incorporated into his four (now) famous equations in 1873.
▶Maxwell’s equations represent the foundation of classical electromagnetic theory. ◀ Maxwell’s theory, which predicted the existence of electromagnetic waves, was not fully accepted by the scientific community at that time, not until verified experimentally by means of radio waves by Heinrich Hertz in the 1880s. X-rays, another member of the EM family, were discovered in 1895 by Wilhelm R¨ ontgen. In the same decade, Nikola Tesla was the first to develop the ac (alternating current) motor, considered a major advance over its predecessor, the dc motor.
Despite the advances made in the 19th century in our understanding of electricity and magnetism and how to put them to practical use, it was not until 1897 that the fundamental carrier of electric charge, the electron, was identified and its properties quantified by Joseph Thomson. The ability to eject electrons from a material by shining electromagnetic energy, such as light, on it is known as the photoelectric effect.
▶To explain the photoelectric effect, Albert Einstein adopted the quantum concept of energy that had been advanced a few years earlier (1900) by Max Planck.
Symbolically, this step represents the bridge between the classical and modern eras of electromagnetics. ◀ 1-1.2 EM in the Modern Era Electromagnetics plays a role in the design and operation of every conceivable electronic device, including the diode, transistor, integrated circuit, laser, display screen, bar-code reader, cell phone, and microwave oven, to name but a few.
Given the breadth and diversity of these applications (Fig. 1-2), it is far more difficult to construct a meaningful timeline for the modern era than for the classical era. That said, one can develop timelines for specific technologies and link their milestone innovations to EM. Chronologies 1-2 and 1-3 present timelines for the development of telecommunications and computers, 4 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS ca. 900 Legend has it that while walking across a field in northern Greece, a shepherd named Magnus experiences a pull on the iron nails in his sandals by the black rock he is standing on. The region was later named Magnesia and the rock became known as magnetite [a form of iron with permanent magnetism].
ca. 600 Greek philosopher Thales describes how amber, after being rubbed with cat fur, can pick up feathers [static electricity].
ca. 1000 Magnetic compass used as a navigational device. 1600 William Gilbert (English) coins the term electric after the Greek word for amber (elektron), and observes that a compass needle points north-south because the Earth acts as a bar magnet.
1671 Isaac Newton (English) demonstrates that white light is a mixture of all the colors.
1733 Charles-François du Fay (French) discovers that electric charges are of two forms, and that like charges repel and unlike charges attract.
1745 Pieter van Musschenbroek (Dutch) invents the Leyden jar, the first electrical capacitor.
1752 Benjamin Franklin (American) invents the lightning rod and demonstrates that lightning is electricity.
1785 Charles-Augustin de Coulomb (French) demonstrates that the electrical force between charges is proportional to the inverse of the square of the distance between them.
1800 Alessandro Volta (Italian) develops the first electric battery.
1820 Hans Christian Oersted (Danish) demonstrates the interconnection between electricity and magnetism through his discovery that an electric current in a wire causes a compass needle to orient itself perpendicular to the wire.
1820 Andre-Marie Ampère (French) notes that parallel currents in wires attract each other and opposite currents repel.
1820 Jean-Baptiste Biot (French) and Félix Savart (French) develop the Biot-Savart law relating the magnetic field induced by a wire segment to the current flowing through it.
Chronology 1-1: TIMELINE FOR ELECTROMAGNETICS IN THE CLASSICAL ERA Electromagnetics in the Classical Era BC BC 1-1 HISTORICAL TIMELINE 5 1888 Nikola Tesla (Croatian-American) invents the ac (alternating current) electric motor.
1895 Wilhelm Röntgen (German) discovers X-rays. One of his first X-ray images was of the bones in his wife's hands. [1901 Nobel prize in physics.] 1897 Joseph John Thomson (English) discovers the electron and measures its charge-to-mass ratio. [1906 Nobel prize in physics.] 1905 Albert Einstein (German-American) explains the photoelectric effect discovered earlier by Hertz in 1887.
[1921 Nobel prize in physics.] 1827 Georg Simon Ohm (German) formulates Ohm's law relating electric potential to current and resistance.
1827 Joseph Henry (American) introduces the concept of inductance, and builds one of the earliest electric motors. He also assisted Samual Morse in the development of the telegraph.
1831 Michael Faraday (English) discovers that a changing magnetic flux can induce an electromotive force.
1873 James Clerk Maxwell (Scottish) publishes his Treatise on Electricity and Magnetism in which he unites the discoveries of Coulomb, Oersted, Ampère, Faraday, and others into four elegantly constructed mathematical equations, now known as Maxwell’s Equations.
1887 Chronology 1-1: TIMELINE FOR ELECTROMAGNETICS IN THE CLASSICAL ERA (continued) Electromagnetics in the Classical Era Heinrich Hertz (German) builds a system that can generate electromagnetic waves (at radio frequencies) and detect them.
1835 Carl Friedrich Gauss (German) formulates Gauss's law relating the electric flux flowing through an enclosed surface to the enclosed electric charge.
6 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS Chronology 1-2: TIMELINE FOR TELECOMMUNICATIONS Telecommunications 1825 1837 Samuel Morse (American) patents the electromagnetic telegraph, using a code of dots and dashes to represent letters and numbers. 1872 Thomas Edison (American) patents the electric typewriter.
1876 Alexander Graham Bell (Scottish-American) invents the telephone, the rotary dial becomes available in 1890, and by 1900, telephone systems are installed in many communities.
1887 Heinrich Hertz (German) generates radio waves and demonstrates that they share the same properties as light.
1887 Emil Berliner (American) invents the flat gramophone disc, or record.
Guglielmo Marconi (Italian) files his first of many patents on wireless transmission by radio. In 1901, he demonstrates radio telegraphy across the Atlantic Ocean. [1909 Nobel prize in physics, shared with Karl Braun (German).] 1897 Karl Braun (German) invents the cathode ray tube (CRT). [1909 Nobel prize with Marconi.] 1902 Reginald Fessenden (American) invents amplitude modulation for telephone transmission. In 1906, he introduces AM radio broadcasting of speech and music on Christmas Eve.
1912 Lee De Forest (American) develops the triode tube amplifier for wireless telegraphy. Also in 1912, the wireless distress call issued by the Titanic was heard 58 miles away by the ocean liner Carpathia, which managed to rescue 705 Titanic passengers 3.5 hours later.
1919 Edwin Armstong (American) invents the superheterodyne radio receiver.
1920 Birth of commercial radio broadcasting; Westinghouse Corporation establishes radio station KDKA in Pittsburgh, Pennsylvania.
1896 William Sturgeon (English) develops the multiturn electromagnet.
1-1 HISTORICAL TIMELINE 7 1958 Jack Kilby (American) builds first integrated circuit (IC) on germanium and, independently, Robert Noyce (American) builds first IC on silicon.
Echo, the first passive communication satellite is launched, and successfully reflects radio signals back to Earth. In 1963, the first communication satellite is placed in geosynchronous orbit.
1969 ARPANET is established by the U.S. Department of Defense, to evolve later into the Internet. 1979 Japan builds the first cellular telephone network: • 1983 cellular phone networks start in the United States.
• 1990 electronic beepers become common.
• 1995 cell phones become widely available.
• 2002 cell phone supports video and Internet.
1984 Worldwide Internet becomes operational.
1988 First transatlantic optical fiber cable between the U.S. and Europe.
1997 Mars Pathfinder sends images to Earth. 2004 Wireless communication supported by many airports, university campuses, and other facilities. 2012 Smartphones worldwide exceed 1 billion.
Vladimir Zworykin (Russian-American) invents television. In 1926, John Baird (Scottish) transmits TV images over telephone wires from London to Glasgow. Regular TV broadcasting began in Germany (1935), England (1936), and the United States (1939).
1926 Transatlantic telephone service between London and New York.
1932 First microwave telephone link, installed (by Marconi) between Vatican City and the Pope’s summer residence.
1933 Edwin Armstrong (American) invents frequency modulation (FM) for radio transmission.
1935 Robert Watson-Watt (Scottish) invents radar.
1938 H. A. Reeves (American) invents pulse code modulation (PCM).
1947 William Shockley, Walter Brattain, and John Bardeen (all Americans) invent the junction transistor at Bell Labs. [1956 Nobel prize in physics.] 1955 Pager is introduced as a radio communication product in hospitals and factories.
1955 Narinder Kapany (Indian-American) demonstrates the optical fiber as a low-loss, light-transmission medium.
1923 1960 Chronology 1-2: TIMELINE FOR TELECOMMUNICATIONS (continued) Telecommunications 8 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS 1941 Konrad Zuze (German) develops the first programmable digital computer, using binary arithmetic and electric relays.
1945 John Mauchly and J. Presper Eckert develop the ENIAC, the first all-electronic computer.
1950 Yoshiro Nakama (Japanese) patents the floppy disk as a magnetic medium for storing data.
1956 John Backus (American) develops FORTRAN, the first major programming language.
1958 Bell Labs develops the modem.
1960 Digital Equipment Corporation introduces the first minicomputer, the PDP-1, to be followed with the PDP-8 in 1965.
1964 IBM’s 360 mainframe becomes the standard computer for major businesses.
1965 John Kemeny and Thomas Kurtz (both American) develop the BASIC computer language.
Chronology 1-3: TIMELINE FOR COMPUTER TECHNOLOGY Computer Technology ca 1100 Abacus is the earliest known calculating device.
1614 John Napier (Scottish) develops the logarithm system.
Blaise Pascal (French) builds the first adding machine using multiple dials.
Gottfried von Leibniz (German) builds calculator that can do both addition and multiplication.
Charles Xavier Thomas de Colmar (French) builds the Arithmometer, the first mass-produced calculator.
1642 1671 1820 1885 Dorr Felt (American) invents and markets a key-operated adding machine (and adds a printer in 1889).
1930 Vannevar Bush (American) develops the differential analyzer, an analog computer for solving differential equations.
BC PRINT FOR Counter = 1 TO Items PRINT USING “##.”; Counter; LOCA TE , ItemColumn PRINT Item$(Counter); LOCA TE , PriceColumn PRINT Price$(Counter) NEXT Counter 1-1 HISTORICAL TIMELINE 9 Chronology 1-3: TIMELINE FOR COMPUTER TECHNOLOGY (continued) Computer Technology 1989 Tim Berners-Lee (British) invents the World Wide Web by introducing a networked hypertext system.
1991 Internet connects to 600,000 hosts in more than 100 countries.
1995 Sun Microsystems introduces the Java programming language.
1996 Sabeer Bhatia (Indian-American) and Jack Smith (American) launch Hotmail, the first webmail service.
1997 IBM’s Deep Blue computer defeats World Chess Champion Garry Kasparov.
2002 The billionth personal computer was sold, second billion reached in 2007.
2010 iPad introduced in 2010.
1968 1971 Texas Instruments introduces the pocket calculator.
1971 Ted Hoff (American) invents the Intel 4004, the first computer microprocessor.
1976 IBM introduces the laser printer.
1976 Apple Computer sells Apple I in kit form, followed by the fully assembled Apple II in 1977 and the Macintosh in 1984.
1980 Microsoft introduces the MS-DOS computer disk operating system. Microsoft Windows is marketed in 1985.
1981 IBM introduces the PC.
Douglas Engelbart (American) demonstrates a word-processor system, the mouse pointing device and the use of “windows.” 10 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS Microwave ablation for liver cancer treatment Electromagnetic sensors LCD Screen Optical fiber Plasma propulsion Global Positioning System (GPS) Motor Ultrasound transducer Ablation catheter Liver Ultrasound image Radar Astronomy: The Very Large Array of Radio Telescopes Cell phone Telecommunication Figure 1-2 Electromagnetics is at the heart of numerous systems and applications.
1-2 DIMENSIONS, UNITS, AND NOTATION 11 Table 1-1 Fundamental SI units.
Dimension Unit Symbol Length meter m Mass kilogram kg Time second s Electric charge coulomb C Temperature kelvin K Amount of substance mole mol Luminous intensity candela cd technologies that have become integral parts of today’s societal infrastructure. Some of the entries in these chronologies refer to specific inventions, such as the telegraph, the transistor, and the laser. The operational principles and capabilities of some of these technologies are highlighted in special sections called Technology Briefs, scattered throughout the book.
1-2 Dimensions, Units, and Notation The International System of Units, abbreviated SI after its French name Syst eme Internationale, is the standard system used in today’s scientific literature for expressing the units of physical quantities.
Length is a dimension and meter is the unit by which it is expressed relative to a reference standard. The SI system is based on the units for the seven fundamental dimensions listed in Table 1-1. The units for all other dimensions are regarded as secondary because they are based on, and can be expressed in terms of, the seven fundamental units. Appendix A contains a list of quantities used in this book, together with their symbols and units.
For quantities ranging in value between 10−18 and 1018, a set of prefixes, arranged in steps of 103, are commonly used to denote multiples and submultiples of units. These prefixes, all of which were derived from Greek, Latin, Spanish, and Danish terms, are listed in Table 1-2. A length of 5 × 10−9 m, for example, may be written as 5 nm.
In EM we work with scalar and vector quantities. In this book we use a medium-weight italic font for symbols denoting scalar quantities, such as R for resistance, and a boldface roman font for symbols denoting vectors, such as E for the electric field vector. A vector consists of a magnitude (scalar) and a direction, with the direction usually denoted by a unit vector.
For example, E = ˆ xE, (1.1) Table 1-2 Multiple and submultiple prefixes.
Prefix Symbol Magnitude exa E 1018 peta P 1015 tera T 1012 giga G 109 mega M 106 kilo k 103 milli m 10−3 micro μ 10−6 nano n 10−9 pico p 10−12 femto f 10−15 atto a 10−18 where E is the magnitude of E and ˆ x is its direction. A symbol denoting a unit vector is printed in boldface with a circumflex (ˆ) above it.
Throughout this book, we make extensive use of phasor representation in solving problems involving electromagnetic quantities that vary sinusoidally in time.
Letters denoting phasor quantities are printed with a tilde (∼) over the letter.
Thus, E is the phasor electric field vector corresponding to the instantaneous electric field vector E(t). This notation is discussed in more detail in Section 1-7.
Notation Summary • Scalar quantity: medium-weight italic, such as C for capacitance.
• Units: medium-weight roman, as in V/m for volts per meter.
• Vector quantities: boldface roman, such as E for electric field vector • Unit vectors: boldface roman with circumflex (ˆ) over the letter, as in ˆ x.
• Phasors: a tilde (∼) over the letter; E is the phasor counterpart of the sinusoidally time-varying scalar field E(t), and E is the phasor counterpart of the sinusoidally time-varying vector field E(t).
12 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS 1-3 The Nature of Electromagnetism Our physical universe is governed by four fundamental forces of nature: • The nuclear force, which is the strongest of the four, but its range is limited to subatomic scales, such as nuclei.
• The electromagnetic force exists between all charged particles. It is the dominant force in microscopic systems, such as atoms and molecules, and its strength is on the order of 10−2 that of the nuclear force.
• The weak-interaction force, whose strength is only 10−14 that of the nuclear force. Its primary role is in interactions involving certain radioactive elementary particles.
• The gravitational force is the weakest of all four forces, having a strength on the order of 10−41 that of the nuclear force. However, it often is the dominant force in macroscopic systems, such as the solar system.
This book focuses on the electromagnetic force and its consequences. Even though the electromagnetic force operates at the atomic scale, its effects can be transmitted in the form of electromagnetic waves that can propagate through both free space and material media.
The purpose of this section is to provide an overview of the basic framework of electromagnetism, which consists of certain fundamental laws governing the electric and magnetic fields induced by static and moving electric charges, the relations between the electric and magnetic fields, and how these fields interact with matter. As a precursor, however, we will take advantage of our familiarity with the gravitational force by describing some of its properties because they provide a useful analogue to those of the electromagnetic force.
1-3.1 The Gravitational Force: A Useful Analogue According to Newton’s law of gravity, the gravitational force Fg21 acting on mass m2 due to a mass m1 at a distance R12 from m2 (Fig. 1-3) is given by Fg21 = −ˆ R12 Gm1m2 R2 12 (N), (1.2) where G is the universal gravitational constant, ˆ R12 is a unit vector that points from m1 to m2, and the unit for force m1 m2 Fg12 Fg21 R12 R12 ˆ Figure 1-3 Gravitational forces between two masses.
is newton (N).
The negative sign in Eq. (1.2) accounts for the fact that the gravitational force is attractive. Conversely, Fg12 = −Fg21, where Fg12 is the force acting on mass m1 due to the gravitational pull of mass m2. Note that the first subscript of Fg denotes the mass experiencing the force and the second subscript denotes the source of the force.
▶The force of gravitation acts at a distance. ◀ The two objects do not have to be in direct contact for each to experience the pull by the other. This phenomenon of action at a distance has led to the concept of fields. An object of mass m1 induces a gravitational field ψ ψ ψ1 (Fig. 1-4) that does not physically emanate from the object, yet its influence exists at every point in space such that if another object of mass m2 were to exist at a distance R12 from the object of mass m1, then −R Gravitational field ψ1 m1 ˆ Figure 1-4 Gravitational field ψ ψ ψ1 induced by a mass m1.
1-3 THE NATURE OF ELECTROMAGNETISM 13 the object of mass m2 would experience a force acting on it equal to Fg21 = ψ ψ ψ1m2, (1.3) where ψ ψ ψ1 = −ˆ R Gm1 R2 (N/kg).
(1.4) In Eq. (1.4) ˆ R is a unit vector that points in the radial direction away from object m1, and therefore −ˆ R points toward m1. The force due to ψ ψ ψ1 acting on a mass m2, for example, is obtained from the combination of Eqs. (1.3) and (1.4) with R = R12 and ˆ R = ˆ R12. The field concept may be generalized by defining the gravitational field ψ ψ ψ at any point in space such that when a test mass m is placed at that point, the force Fg acting on it is related to ψ ψ ψ by ψ ψ ψ = Fg m .
(1.5) The force Fg may be due to a single mass or a collection of many masses.
1-3.2 Electric Fields The electromagnetic force consists of an electrical component Fe and a magnetic component Fm.
▶The electrical force Fe is similar to the gravitational force, but with two major differences: (1) the source of the electrical field is electric charge, not mass, and (2) even though both types of fields vary inversely as the square of the distance from their respective sources, electric charges may have positive or negative polarity, resulting in a force that may be attractive or repulsive. ◀ We know from atomic physics that all matter contains a mixture of neutrons, positively charged protons, and negatively charged electrons, with the fundamental quantity of charge being that of a single electron, usually denoted by the letter e.
The unit by which electric charge is measured is the coulomb (C), named in honor of the eighteenth-century French scientist Charles Augustin de Coulomb (1736–1806).
The magnitude of e is e = 1.6 × 10−19 (C).
(1.6) The charge of a single electron is qe = −e, and that of a proton is equal in magnitude but opposite in polarity: qp = e.
▶Coulomb’s experiments demonstrated that: (1) two like charges repel one another, whereas two charges of opposite polarity attract, (2) the force acts along the line joining the charges, and (3) its strength is proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance between them. ◀ These properties constitute what today is called Coulomb’s law, which can be expressed mathematically as Fe21 = ˆ R12 q1q2 4πϵ0R2 12 (N) (in free space), (1.7) where Fe21 is the electrical force acting on charge q2 due to charge q1 when both are in free space (vacuum), R12 is the distance between the two charges, ˆ R12 is a unit vector pointing from charge q1 to charge q2 (Fig. 1-5), and ϵ0 is a universalconstantcalledtheelectricalpermittivityoffreespace [ϵ0 = 8.854 × 10−12 farad per meter (F/m)]. The two charges are assumed to be isolated from all other charges. The force Fe12 acting on charge q1 due to charge q2 is equal to force Fe21 in magnitude, but opposite in direction: Fe12 = −Fe21.
+q1 +q2 Fe12 Fe21 R12 R12 ˆ Figure 1-5 Electric forces on two positive point charges in free space.
14 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS R Electric field lines +q ˆ Figure 1-6 Electric field E due to charge q.
The expression given by Eq. (1.7) for the electrical force is analogous to that given by Eq. (1.2) for the gravitational force, and we can extend the analogy further by defining the existence of an electric field intensity E due to any charge q as E = ˆ R q 4πϵ0R2 (V/m) (in free space), (1.8) where R is the distance between the charge and the observation point, and ˆ R is the radial unit vector pointing away from the charge.
Figure 1-6 depicts the electric-field lines due to a positive charge. For reasons that will become apparent in later chapters, the unit for E is volt per meter (V/m).
▶If any point charge q′ is present in an electric field E (due to other charges), the point charge will experience a force acting on it equal to Fe = q′E. ◀ Electric charge exhibits two important properties. The first is encapsulated by the law of conservation of electric charge, which states that the (net) electric charge can neither be created nor destroyed. If a volume contains np protons and ne electrons, then its total charge is q = npe −nee = (np −ne)e (C).
(1.9) Even if some of the protons were to combine with an equal number of electrons to produce neutrons or other elementary particles, the net charge q remains unchanged. In matter, the quantum mechanical laws governing the behavior of the protons inside the atom’s nucleus and the electrons outside it do not allow them to combine.
▶The second important property of electric charge is embodied by the principle of linear superposition, which states that the total vector electric field at a point in space due to a system of point charges is equal to the vector sum of the electric fields at that point due to the individual charges. ◀ This seemingly simple concept allows us in future chapters to compute the electric field due to complex distributions of charge without having to be concerned with the forces acting on each individual charge due to the fields by all of the other charges.
The expression given by Eq. (1.8) describes the field induced by an electric charge residing in free space. Let us now consider what happens when we place a positive point charge in a material composed of atoms. In the absence of the point charge, the material is electrically neutral, with each atom having a positivelychargednucleussurrounded byacloudofelectronsof equal but opposite polarity. Hence, at any point in the material not occupied by an atom the electric field E is zero. Upon placing a point charge in the material, as shown in Fig. 1-7, the atoms experience forces that cause them to become distorted.
The center of symmetry of the electron cloud is altered with respect to the nucleus, with one pole of the atom becoming positively charged relative to the other pole. Such a polarized atom is called an electric dipole, and the distortion process is called polarization. The degree of polarization depends on the distance between the atom and the isolated point charge, and the orientation of the dipole is such that the axis connecting + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − q + + − + − + − + − Figure 1-7 Polarization of the atoms of a dielectric material by a positive charge q.
1-3 THE NATURE OF ELECTROMAGNETISM 15 its two poles is directed toward the point charge, as illustrated schematically in Fig. 1-7. The net result of this polarization process is that the electric fields of the dipoles of the atoms (or molecules) tend to counteract the field due to the point charge.
Consequently, the electric field at any point in the material is different from the field that would have been induced by the point charge in the absence of the material. To extend Eq.(1.8)fromthefree-spacecasetoanymedium, wereplacethe permittivity of free space ϵ0 with ϵ, where ϵ is the permittivity of the material in which the electric field is measured and is therefore characteristic of that particular material. Thus, E = ˆ R q 4πϵR2 (V/m).
(1.10) (material with permittivity ϵ) Often, ϵ is expressed in the form ϵ = ϵrϵ0 (F/m), (1.11) where ϵr is a dimensionless quantity called the relative permittivity or dielectric constant of the material. For vacuum, ϵr = 1; for air near Earth’s surface, ϵr = 1.0006; and the values of ϵr for materials that we have occasion to use in this book are tabulated in Appendix B.
In addition to the electric field intensity E, we often find it convenient to also use a related quantity called the electric flux density D, given by D = ϵE (C/m2), (1.12) with unit of coulomb per square meter (C/m2).
▶These two electric quantities, E and D, constitute one of two fundamental pairs of electromagnetic fields.
The second pair consists of the magnetic fields discussed next. ◀ 1-3.3 Magnetic Fields As early as 800 B.C., the Greeks discovered that certain kinds of stones exhibit a force that attracts pieces of iron. These stones are now called magnetite (Fe3O4) and the phenomenon S N B Magnetic field lines Figure 1-8 Pattern of magnetic field lines around a bar magnet.
they exhibit is known as magnetism. In the thirteenth century, French scientists discovered that when a needle was placed on the surface of a spherical natural magnet, the needle oriented itself along different directions for different locations on the magnet. By mapping the directions indicated by the needle, it was determined that the magnetic force formed magnetic-field lines that encircled the sphere and appeared to pass through two points diametrically opposite to each other. These points, called the north and south poles of the magnet, were found to exist for every magnet, regardless of its shape. The magnetic-field pattern of a bar magnet is displayed in Fig. 1-8. It was also observed that like poles of different magnets repel each other and unlike poles attract each other.
▶The attraction-repulsion property for magnets is similar to the electric force between electric charges, except for one important difference: electric charges can be isolated, but magnetic poles always exist in pairs. ◀ If a permanent magnet is cut into small pieces, no matter how small each piece is, it will always have a north and a south pole.
The magnetic lines surrounding a magnet represent the magnetic flux density B.
A magnetic field not only exists around permanent magnets but can also be created by electric current. This connection between electricity and magnetism was discovered in 1819 by the Danish scientist Hans Oersted 16 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS B B B B z y x r B B B B I φ ˆ Figure 1-9 The magnetic field induced by a steady current flowing in the z direction.
(1777–1851), who observed that an electric current in a wire caused a compass needle placed in its vicinity to deflect and that the needle turned so that its direction was always perpendicular to the wire and to the radial line connecting the wire to the needle. From these observations, he deduced that the current-carrying wire induced a magnetic field that formed closed circularloopsaroundthewire(Fig.1-9). ShortlyafterOersted’s discovery, French scientists Jean Baptiste Biot and Felix Savart developed an expression that relates the magnetic flux density B at a point in space to the current I in the conductor. Application of their formulation, known today as the Biot–Savart law, to the situation depicted in Fig. 1-9 for a very long wire residing in free space leads to the result that the magnetic flux density B induced by a constant current I flowing in the z direction is given by B = ˆ φ φ φ μ0I 2πr (T), (1.13) where r is the radial distance from the current and ˆ φ φ φ is an azimuthal unit vector expressing the fact that the magnetic field direction is tangential to the circle surrounding the current (Fig. 1-9). The magnetic field is measured in tesla (T), named in honor of Nikola Tesla (1856–1943), a Croatian-American electricalengineerwhoseworkontransformersmadeitpossible to transport electricity over long wires without too much loss.
The quantity μ0 is called the magnetic permeability of free space [μ0 = 4π × 10−7 henry per meter (H/m)], and it is analogous to the electric permittivity ϵ0. In fact, as we will see in Chapter 2, the product of ϵ0 and μ0 specifies c, the velocity of light in free space: c = 1 √μ0ϵ0 = 3 × 108 (m/s).
(1.14) We noted in Section 1-3.2 that when an electric charge q′ is subjected to an electric field E, it experiences an electric force Fe = q′E. Similarly, if a charge q′ resides in the presence of a magnetic flux density B, it experiences a magnetic force Fm, but only if the charge is in motion and its velocity u is in a direction not parallel (or anti-parallel) to B. In fact, as we learn in more detail in Chapter 5, Fm points in a direction perpendicular to both B and u.
To extend Eq. (1.13) to a medium other than free space, μ0 should be replaced with μ, the magnetic permeability of the material in which B is being observed. The majority of natural materials are nonmagnetic, meaning that they exhibit a magnetic permeability μ = μ0. For ferromagnetic materials, such as iron and nickel, μ can be much larger than μ0. The magnetic permeability μ accounts for magnetization properties of a material. In analogy with Eq. (1.11), μ of a particular material can be defined as μ = μrμ0 (H/m), (1.15) where μr is a dimensionless quantity called the relative magnetic permeability of the material.
The values of μr for commonly used ferromagnetic materials are given in Appendix B.
▶We stated earlier that E and D constitute one of two pairs of electromagnetic field quantities. The second pair is B and the magnetic field intensity H, which are related to each other through μ: B = μH.
(1.16) 1-3.4 Static and Dynamic Fields In EM, the time variable t, or more precisely if and how electric and magnetic quantities vary with time, is of crucial importance. Before we elaborate further on the significance 1-3 THE NATURE OF ELECTROMAGNETISM 17 Table 1-3 The three branches of electromagnetics.
Branch Condition Field Quantities (Units) Electrostatics Stationary charges Electric field intensity E (V/m) (∂q/∂t = 0) Electric flux density D (C/m2) D = ϵE Magnetostatics Steady currents Magnetic flux density B (T) (∂I/∂t = 0) Magnetic field intensity H (A/m) B = μH Dynamics Time-varying currents E, D, B, and H (time-varying fields) (∂I/∂t ̸= 0) (E, D) coupled to (B, H) of this statement, it will prove useful to define the following time-related adjectives unambiguously: • static—describes a quantity that does not change with time. The term dc (i.e., direct current) is often used as a synonym for static to describe not only currents but other electromagnetic quantities as well.
• dynamic—refers to a quantity that does vary with time, but conveys no specific information about the character of the variation.
• waveform—refers to a plot of the magnitude profile of a quantity as a function of time.
• periodic—a quantity is periodic if its waveform repeats itself at a regular interval, namely its period T . Examples include the sinusoid and the square wave. By application of the Fourier series analysis technique, any periodic waveform can be expressed as the sum of an infinite series of sinusoids.
• sinusoidal—also called ac (i.e., alternating current), describes a quantity that varies sinusoidally (or cosinu-soidally) with time.
In view of these terms, let us now examine the relationship between the electric field E and the magnetic flux density B.
Because E is governed by the charge q and B is governed by I = dq/dt, one might expect that E and B must be somehow related to each other. They may or may not be interrelated, depending on whether I is static or dynamic.
Let us start by examining the dc case in which I remains constant with time.
Consider a small section of a beam of charged particles, all moving at a constant velocity. The moving charges constitute a dc current. The electric field due to that sectionofthebeamisdeterminedbythetotalchargeq contained in it. The magnetic field does not depend on q, but rather on the rate of charge (current) flowing through that section. Few charges moving very fast can constitute the same current as many charges moving slowly. In these two cases the induced magnetic field is the same because the current I is the same, but the induced electric field is quite different because the numbers of charges are not the same.
Electrostatics and magnetostatics refer to the study of EM under the specific, respective conditions of stationary charges and dc currents. They represent two independent branches, so characterized because the induced electric and magnetic fields do not couple to each other.
Dynamics, the third and more general branch of electromagnetics, involves time-varying fields induced by time-varying sources, that is, currents and associated charge densities. If the current associated with the beam of moving charged particles varies with time, then the amount of charge present in a given section of the beam also varies with time, and vice versa. As we see in Chapter 6, the electric and magnetic fields become coupled to each other in that case.
▶A time-varying electric field generates a time-varying magnetic field, and vice versa. ◀ Table 1-3 provides a summary of the three branches of electromagnetics.
The electric and magnetic properties of materials are characterized by the parameters ϵ and μ, respectively. A third 18 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS Table 1-4 Constitutive parameters of materials.
Parameter Units Free-Space Value Electrical permittivity ϵ F/m ϵ0 = 8.854 × 10−12 ≈ 1 36π × 10−9 Magnetic permeability μ H/m μ0 = 4π × 10−7 Conductivity σ S/m 0 fundamental parameter is also needed, the conductivity of a material σ, which is measured in siemens per meter (S/m).
The conductivity characterizes the ease with which charges (electrons) can move freely in a material. If σ = 0, the charges do not move more than atomic distances and the material is said to be a perfect dielectric. Conversely, if σ = ∞, the charges can move very freely throughout the material, which is then called a perfect conductor.
▶The parameters ϵ, μ, and σ are often referred to as the constitutive parameters of a material (Table 1-4).
A medium is said to be homogeneous if its constitutive parameters are constant throughout the medium. ◀ Concept Question 1-1: What are the four fundamental forces of nature and what are their relative strengths?
Concept Question 1-2: What is Coulomb’s law? State its properties.
Concept Question 1-3: What are the two important properties of electric charge?
Concept Question 1-4: What do the electrical permit-tivity and magnetic permeability of a material account for?
Concept Question 1-5: What are the three branches and associated conditions of electromagnetics?
1-4 Traveling Waves Waves are a natural consequence of many physical processes: waves manifest themselves as ripples on the surfaces of oceans and lakes; sound waves constitute pressure disturbances that travel through air; mechanical waves modulate stretched strings; and electromagnetic waves carry electric and magnetic fields through free space and material media as microwaves, light, and X-rays. All these various types of waves exhibit a number of common properties, including: • Moving waves carry energy.
• Waves have velocity; it takes time for a wave to travel from one point to another.
Electromagnetic waves in vacuum travel at a speed of 3 × 108 m/s, and sound waves in air travel at a speed approximately a million times slower, specifically 330 m/s. Sound waves cannot travel in vacuum.
• Many waves exhibit a property called linearity. Waves that do not affect the passage of other waves are called linear because they can pass right through each other. The total of two linear waves is simply the sum of the two waves as they would exist separately. Electromagnetic waves are linear, as are sound waves. When two people speak to one another, the sound waves they generate do not interact with one another, but simply pass through each other. Water waves are approximately linear; the expanding circles of ripples caused by two pebbles thrown intotwolocationsona lakesurfacedonotaffecteachother.
Although the interaction of the two circles may exhibit a complicated pattern, it is simply the linear superposition of two independent expanding circles.
Waves are of two types: transient waves caused by sudden disturbances and continuous periodic waves generated by a repetitive source. We encounter both types of waves in this book, but most of our discussion deals with the propagation of continuous waves that vary sinusoidally with time.
An essential feature of a propagating wave is that it is a self-sustaining disturbance of the medium through which it travels. If this disturbance varies as a function of one space variable, such as the vertical displacement of the string shown in Fig. 1-10, we call the wave one-dimensional. The vertical displacement varies with time and with the location along the length of the string.
Even though the string rises up into a second dimension, the wave is only one-dimensional because the disturbance varies with only one space variable.
1-4 TRAVELING WAVES 19 u Figure 1-10 A one-dimensional wave traveling on a string.
A two-dimensional wave propagates out across a surface, like the ripples on a pond [Fig. 1-11(a)], and its disturbance can be described by two space variables.
And by extension, a three-dimensional wave propagates through a volume and its (a) Circular waves (c) Spherical wave (b) Plane and cylindrical waves Plane wavefront Two-dimensional wave Cylindrical wavefront Spherical wavefront Figure 1-11 Examples of two-dimensional and three-dimensional waves: (a) circular waves on a pond, (b) a plane light wave exciting a cylindrical light wave through the use of a long narrow slit in an opaque screen, and (c) a sliced section of a spherical wave.
disturbance may be a function of all three space variables.
Three-dimensional waves may take on many different shapes; they include plane waves, cylindrical waves, and spherical waves. A plane wave is characterized by a disturbance that at a given point in time has uniform properties across an infinite plane perpendicular to its direction of propagation [Fig. 1-11(b)]. Similarly, for cylindrical and spherical waves, the disturbances are uniform across cylindrical and spherical surfaces [Figs. 1-11(b) and (c)].
In the material that follows, we examine some of the basic properties of waves by developing mathematical formulations that describe their functional dependence on time and space variables.
To keep the presentation simple, we limit our discussion to sinusoidally varying waves whose disturbances are functions of only one space variable, and we defer the discussion of more complicated waves to later chapters.
1-4.1 Sinusoidal Waves in a Lossless Medium Regardless of the mechanism responsible for generating them, all linear waves can be described mathematically in common terms.
▶A medium is said to be lossless if it does not attenuate the amplitude of the wave traveling within it or on its surface. ◀ 20 TECHNOLOGY BRIEF 1: LED LIGHTING Technology Brief 1: LED Lighting After lighting our homes, buildings, and streets for over 100 years, the incandescent light bulb created by Thomas Edison (1879) will soon become a relic of the past. Many countries have taken steps to phase it out and replace it with a much more energy-efficient alternative: the light-emitting diode (LED).
Light Sources The three dominant sources of electric light are the incandescent, fluorescent, and LED light bulbs (Fig. TF1-1). We examine each briefly.
Incandescent Light Bulb ▶Incandescence is the emission of light from a hot object due to its temperature. ◀ By passing electric current through a thin tungsten filament, which basically is a resistor, the filament’s temperature rises to a very high level, causing the filament to glow and emit visible light. The intensity and shape of the emitted spectrum depends on the filament’s temperature. A typical example is shown by the green curve in Fig. TF1-2. The tungsten spectrum is similar in shape to that of sunlight (yellow curve in Fig. TF1-2), particularly in the blue and green parts of the spectrum (400–550 nm). Despite the relatively strong (compared with sunlight) yellow light emitted by incandescent sources, the quasi-white light they produce has a quality that the human eye finds rather comfortable.
(a) (b) (c) Figure TF1-1 (a) Incandescent light bulb; (b) fluorescent mercury vapor lamp; (c) white LED.
TECHNOLOGY BRIEF 1: LED LIGHTING 21 25 50 100 75 0 0.4 Wavelength (micrometers) Energy (arbitrary units) 0.5 0.6 0.7 Noon sunlight White LED (with phosphor) Incandescent tungsten Fluorescent mercury Figure TF1-2 Spectra of common sources of visible light.
▶The incandescent light bulb is significantly less expensive to manufacture than the fluorescent and LED light bulbs, but it is far inferior with regard to energy efficacy and operational lifetime (Fig. TF1-7). ◀ Of the energy supplied to an incandescent light bulb, only about 2% is converted into light, with the remainder wasted as heat! In fact, the incandescent light bulb is the weakest link in the overall conversion sequence from coal to light (Fig. TF1-3).
Fluorescent Light Bulb To fluoresce means to emit radiation in consequence to incident radiation of a shorter wavelength. By passing a stream of electrons between two electrodes at the ends of a tube [Fig. TF1-1(b)] containing mercury gas (or the noble gases Coal Power plant E1 = 0.35 Transmission lines E2 = 0.92 Light E3 = 0.024 Overall efficiency for conversion of chemical energy to light energy is E1 × E2 × E3 = 0.35 × 0.92 × 0.024 ═ 0.8% Figure TF1-3 Lighting efficiency. (Source: National Research Council, 2009.) 22 TECHNOLOGY BRIEF 1: LED LIGHTING neon, argon, and xenon) at very low pressure, the electrons collide with the mercury atoms, causing them to excite their own electrons to higher energy levels. When the excited electrons return to the ground state, they emit photons at specific wavelengths, mostly in the ultraviolet part of the spectrum. Consequently, the spectrum of a mercury lamp is concentrated into narrow lines, as shown by the blue curve in Fig. TF1-2.
▶To broaden the mercury spectrum into one that resembles that of white light, the inside surface of the fluorescent light tube is coated with phosphor particles [such as yttrium aluminum garnet (YAG) doped with cerium]. The particles absorb the UV energy and then reradiate it as a broad spectrum extending from blue to red; hence the name fluorescent. ◀ Light-Emitting Diode The LED contained inside the polymer jacket in Fig.TF1-1(c) is a p-n junction diode fabricated on a semiconductor chip.
When a voltage is applied in a forward-biased direction across the diode (Fig.TF1-4), current flows through the junction and some of the streaming electrons are captured by positive charges (holes). Associated with each electron-hole recombining act is the release of energy in the form of a photon.
▶The wavelength of the emitted photon depends on the diode’s semiconductor material. The materials most commonly used are aluminum gallium arsenide (AIGaAs) to generate red light, indium gallium nitride (InGaN) to generate blue light, and aluminum gallium phosphide (AIGaP) to generate green light. In each case, the emitted energy is confined to a narrow spectral band. ◀ Electrons V Holes Photon Photon e _ I _ + p-type n-type Figure TF1-4 Photons are emitted when electrons combine with holes.
Wavelength (nm) 0 400 500 600 700 1 2 3 Spectral power (arbitrary units) Figure TF1-5 The addition of spectra from three monochromatic LEDs.
TECHNOLOGY BRIEF 1: LED LIGHTING 23 Two basic techniques are available for generating white light with LEDs: (a) RGB and (b) blue/conversion. The RGB approach involves the use of three monochromatic LEDs whose primary colors (red, green, and blue) are mixed to generate an approximation of a white-light spectrum. An example is shown in Fig. TF1-5. The advantage of this approach is that the relative intensities of the three LEDs, can be controlled independently, thereby making it possible to “tune” the shape of the overall spectrum so as to generate an esthetically pleasing color of “white.” The major shortcoming of the RGB technique is cost; manufacturing three LEDs instead of just one.
With the blue LED/phosphor conversion technique, a blue LED is used with phosphor powder particles suspended in the epoxy resin that encapsulates it. The blue light emitted by the LED is absorbed by the phosphor particles and then reemitted as a broad spectrum (Fig. TF1-6). To generate high-intensity light, several LEDs are clustered into a single enclosure.
Comparison ▶Luminous efficacy (LE) is a measure of how much light in lumens is produced by a light source for each watt of electricity consumed by it. ◀ Of the three types of light bulbs we discussed, the incandescent light bulb is by far the most inefficient and its useful lifespan is the shortest (Fig. TF1-7). For a typical household scenario, the 10-year cost—including electricity and replacement cost—is several times smaller for the LED than for the alternatives.
25 50 100 75 0 0.4 0.5 0.6 0.7 Wavelength (micrometers) Energy (arbitrary units) White LED (phosphor-based blue LED) Blue LED Figure TF1-6 Phosphor-based white LED emission spectrum.
Parameter Type of Light Bulb Luminous Efficacy (lumens/W) Incandescent ~12 ~40 ~70 ~150 Useful Lifetime (hours) ~1000 ~20,000 ~60,000 ~100,000 Purchase Price ~$1.50 ~$5 ~$10 ~$5 Estimated Cost over 10 Years ~$410 ~$110 ~$100 ~$40 Fluorescent White LED Circa 2010 Circa 2025 Figure TF1-7 Even though the initial purchase price of a white LED is several times greater than that of the incandescent light bulb, the total 10-year cost of using the LED is only one-fourth of the incandescent’s (in 2010) and is expected to decrease to one-tenth by 2025.
24 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS By way of an example, let us consider a wave traveling on a lake surface, and let us assume for the time being that frictional forces can be ignored, thereby allowing a wave generated on the water surface to travel indefinitely with no loss in energy.
If y denotes the height of the water surface relative to the mean height (undisturbed condition) and x denotes the distance of wave travel, the functional dependence of y on time t and the spatial coordinate x has the general form y(x, t) = A cos 2πt T −2πx λ + φ0 (m), (1.17) where A is the amplitude of the wave, T is its time period, λ is its spatial wavelength, and φ0 is a reference phase. The quantity y(x, t) can also be expressed in the form y(x, t) = A cos φ(x, t) (m), (1.18) where φ(x, t) = 2πt T −2πx λ + φ0 (rad).
(1.19) (a) y(x, t) versus x at t = 0 (b) y(x, t) versus t at x = 0 −A 0 T 2 T 3T 2 A T y(0, t) t At x = 0 −A 0 λ 2 3λ 2 A y(x, 0) x At t = 0 λ λ Figure 1-12 Plots of y(x, t) = A cos 2πt T −2πx λ as a function of (a) x at t = 0 and (b) t at x = 0.
The angle φ(x, t) is called the phase of the wave, and it should not be confused with the reference phase φ0, which is constant with respect to both time and space. Phase is measured by the same units as angles, that is, radians (rad) or degrees, with 2π radians = 360◦.
Let us first analyze the simple case when φ0 = 0: y(x, t) = A cos 2πt T −2πx λ (m).
(1.20) The plots in Fig. 1-12 show the variation of y(x, t) with x at t = 0 and with t at x = 0. The wave pattern repeats itself at a spatial period λ along x and at a temporal period T along t.
If we take time snapshots of the water surface, the height profile y(x, t) would exhibit the sinusoidal patterns shown in Fig. 1-13.
In all three profiles, which correspond to three 2 3λ 2 y(x, 0) y(x, T/4) y(x, T/2) A −A A −A A −A (a) t = 0 (b) t = T/4 (c) t = T/2 x x x P P P up λ λ 2 3λ 2 λ λ 2 3λ 2 λ λ Figure 1-13 Plots of y(x, t) = A cos 2πt T −2πx λ as a function of x at (a) t = 0, (b) t = T /4, and (c) t = T /2.
Note that the wave moves in the +x direction with a velocity up = λ/T .
1-4 TRAVELING WAVES 25 different values of t, the spacing between peaks is equal to the wavelength λ, even though the patterns are shifted relative to one another because they correspond to different observation times. Because the pattern advances along the +x direction at progressively increasing values of t, y(x, t) is called a wave traveling in the +x direction. If we track a given point on the wave, such as the peak P, and follow it in time, we can measure the phase velocity of the wave. At the peaks of the wave pattern, the phase φ(x, t) is equal to zero or multiples of 2π radians.
Thus, φ(x, t)= 2πt T −2πx λ =2nπ, n = 0, 1, 2, . . .
(1.21) Had we chosen any other fixed height of the wave, say y0, and monitored its movement as a function of t and x, this again would have been equivalent to setting the phase φ(x, t) constant such that y(x, t) = y0 = A cos 2πt T −2πx λ , (1.22) or 2πt T −2πx λ = cos−1 y0 A = constant.
(1.23) The apparent velocity of that fixed height is obtained by taking the time derivative of Eq. (1.23), 2π T −2π λ dx dt = 0, (1.24) which gives the phase velocity up as up = dx dt = λ T (m/s).
(1.25) ▶The phase velocity, also called the propagation velocity, is the velocity of the wave pattern as it moves across the water surface. ◀ The water itself mostly moves up and down; when the wave moves from one point to another, the water does not move physically along with it.
The frequency of a sinusoidal wave, f , is the reciprocal of its time period T : f = 1 T (Hz).
(1.26) Combining the preceding two equations yields up = f λ (m/s).
(1.27) The wave frequency f , which is measured in cycles per second, has been assigned the unit (Hz), named in honor of the German physicist Heinrich Hertz (1857–1894), who pioneered the development of radio waves.
Using Eq. (1.26), Eq. (1.20) can be rewritten in a more compact form as y(x, t) = A cos 2πf t −2π λ x = A cos(ωt −βx), (1.28) (wave moving along +x direction) where ω is the angular velocity of the wave and β is its phase constant (or wavenumber), defined as ω = 2πf (rad/s), (1.29a) β = 2π λ (rad/m).
(1.29b) In terms of these two quantities, up = f λ = ω β .
(1.30) 26 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS φ0 = π/4 φ0 = −π/4 T t T 2 3T 2 y −A A Leads ahead of reference wave Lags behind reference wave Reference wave (φ0 = 0) Figure 1-14 Plots of y(0, t) = A cos [(2πt/T ) + φ0] for three different values of the reference phase φ0.
So far, we have examined the behavior of a wave traveling in the +x direction.
To describe a wave traveling in the −x direction, we reverse the sign of x in Eq. (1.28): y(x, t) = A cos(ωt + βx).
(1.31) (wave moving along −x direction) ▶The direction of wave propagation is easily determined by inspecting the signs of the t and x terms in the expression for the phase φ(x, t) given by Eq. (1.19): if one of the signs is positive and the other is negative, then the wave is traveling in the positive x direction, and if both signs are positive or both are negative, then the wave is traveling in the negative x direction.
The constant phase reference φ0 has no influence on either the speed or the direction of wave propagation. ◀ We now examine the role of the phase reference φ0 given previously in Eq. (1.17).
If φ0 is not zero, then Eq. (1.28) should be written as y(x, t) = A cos(ωt −βx + φ0).
(1.32) A plot of y(x, t) as a function of x at a specified t or as a function of t at a specified x is shifted in space or time, respec-tively, relative to a plot with φ0 = 0 by an amount proportional to φ0. This is illustrated by the plots shown in Fig. 1-14. We observe that when φ0 is positive, y(t) reaches its peak value, or any other specified value, sooner than when φ0 = 0. Thus, the wave with φ0 = π/4 is said to lead the wave with φ0 = 0 by a phase lead of π/4; and similarly, the wave with φ0 = −π/4 is said to lag the wave with φ0 = 0 by a phase lag of π/4. A wave function with a negative φ0 takes longer to reach a given value of y(t), such as its peak, than the zero-phase reference function.
▶When its value is positive, φ0 signifies a phase lead in time, and when it is negative, it signifies a phase lag. ◀ Exercise 1-1: Consider the red wave shown in Fig. E1.1.
What is the wave’s (a) amplitude, (b) wavelength, and (c) frequency, given that its phase velocity is 6 m/s?
Figure E1.1 −2 −4 −6 6 4 2 0 2 1 3 4 5 6 7 8 9 10 x (cm) υ (volts) Answer: (a) A = 6 V, (b) λ = 4 cm, (c) f = 150 Hz.
1-4 TRAVELING WAVES 27 Module 1.1 Sinusoidal Waveforms Learn how the shape of the waveform is related to the amplitude, frequency, and reference phase angle of a sinusoidal wave.
Exercise 1-2: The wave shown in red in Fig. E1.2 is given by υ = 5 cos 2πt/8. Of the following four equations: (1) υ = 5 cos(2πt/8 −π/4), (2) υ = 5 cos(2πt/8 + π/4), (3) υ = −5 cos(2πt/8 −π/4), (4) υ = 5 sin 2πt/8, (a) which equation applies to the green wave? (b) which equation applies to the blue wave?
Figure E1.2 −5 5 0 t (s) υ (volts) 2 1 3 4 5 6 7 8 9 10 11 12 13 14 Answer: (a) #2, (b) #4.
28 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS −10 m −5 m 0 5 m 10 m y(x) y(x) 10e−0.2x Wave envelope x (m) 1 2 3 4 5 6 7 8 Figure 1-15 Plot of y(x) = (10e−0.2x cos πx) meters. Note that the envelope is bounded between the curve given by 10e−0.2x and its mirror image.
Exercise 1-3: The electric field of a traveling electromagnetic wave is given by E(z, t) = 10 cos(π × 107t + πz/15 + π/6) (V/m).
Determine (a) the direction of wave propagation, (b) the wave frequency f , (c) its wavelength λ, and (d) its phase velocity up.
Answer: (a) −z direction, (b) f = 5 MHz, (c) λ = 30 m, (d) up = 1.5 × 108 m/s.
1-4.2 Sinusoidal Waves in a Lossy Medium If a wave is traveling in the x direction in a lossy medium, its amplitude decreases as e−αx.
This factor is called the attenuation factor, and α is called the attenuation constant of the medium and its unit is neper per meter (Np/m). Thus, in general, y(x, t) = Ae−αx cos(ωt −βx + φ0).
(1.33) The wave amplitude is now Ae−αx, not just A. Figure 1-15 shows a plot of y(x, t) as a function of x at t = 0 for A = 10 m, λ = 2 m, α = 0.2 Np/m, and φ0 = 0. Note that the envelope of the wave pattern decreases as e−αx.
The real unit of α is (1/m); the neper (Np) part is a dimensionless, artificial adjective traditionally used as a reminder that the unit (Np/m) refers to the attenuation constant of the medium, α. A similar practice is applied to the phase constant β by assigning it the unit (rad/m) instead of just (l/m).
Concept Question 1-6: How can you tell if a wave is traveling in the positive x direction or the negative x direction?
Concept Question 1-7: How does the envelope of the wave pattern vary with distance in (a) a lossless medium and (b) a lossy medium?
Concept Question 1-8: Why does a negative value of φ0 signify a phase lag?
Example 1-1: Sound Wave in Water An acoustic wave traveling in the x direction in a fluid (liquid or gas) is characterized by a differential pressure p(x, t). The unit for pressure is newton per square meter (N/m2). Find an expression for p(x, t) for a sinusoidal sound wave traveling in the positive x direction in water, given that the wave frequency is 1 kHz, the velocity of sound in water is 1.5 km/s, the wave amplitude is 10 N/m2, and p(x, t) was observed to be at its maximum value at t = 0 and x = 0.25 m. Treat water as a lossless medium.
1-4 TRAVELING WAVES 29 Module 1.2 Traveling Waves Learn how the shape of a traveling wave is related to its frequency and wavelength, and to the attenuation constant of the medium.
Solution: According to the general form given by Eq. (1.17) for a wave traveling in the positive x direction, p(x, t) = A cos 2π T t −2π λ x + φ0 (N/m2).
The amplitude A = 10 N/m2, T = 1/f = 10−3 s, and from up = f λ, λ = up f = 1.5 × 103 103 = 1.5 m.
Hence, p(x, t) = 10 cos 2π × 103t −4π 3 x + φ0 (N/m2).
Since at t = 0 and x = 0.25 m, p(0.25, 0) = 10 N/m2, we have 10 = 10 cos −4π 3 0.25 + φ0 = 10 cos −π 3 + φ0 , which yields the result (φ0 −π/3) = cos−1(1), or φ0 = π/3.
Hence, p(x, t) = 10 cos 2π × 103t −4π 3 x + π 3 (N/m2).
30 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS Example 1-2: Power Loss A laser beam of light propagating through the atmosphere is characterized by an electric field given by E(x, t) = 150e−0.03x cos(3 × 1015t −107x) (V/m), where x is the distance from the source in meters.
The attenuation is due to absorption by atmospheric gases.
Determine (a) the direction of wave travel, (b) the wave velocity, and (c) the wave amplitude at a distance of 200 m.
Solution: (a) Since the coefficients of t and x in the argument of the cosine function have opposite signs, the wave must be traveling in the +x direction.
(b) up = ω β = 3 × 1015 107 = 3 × 108 m/s, which is equal to c, the velocity of light in free space.
(c) At x = 200 m, the amplitude of E(x, t) is 150e−0.03×200 = 0.37 (V/m).
Exercise 1-4: Consider the red wave shown in Fig. E1.4.
What is the wave’s (a) amplitude (at x = 0), (b) wavelength, and (c) attenuation constant?
Figure E1.4 −5 5 0 x (cm) υ (volts) (2.8, 4.23) (8.4, 3.02) 2 1 3 4 5 6 7 8 9 10 11 12 13 14 Answer: (a) 5 V, (b) 5.6 cm, (c) α = 0.06 Np/cm.
Exercise 1-5: The red wave shown in Fig. E1.5 is given by υ = 5 cos 4πx (V). What expression is applicable to (a) the blue wave and (b) the green wave?
Figure E1.5 −5 5 0 x (m) υ (volts) 0.25 0.5 0.75 1.0 1.25 5 V 3.52 V 1.01 V Answer: (a) υ = 5e−0.7x cos 4πx (V), (b) υ = 5e−3.2x cos 4πx (V).
Exercise 1-6: An electromagnetic wave is propagating in the z direction in a lossy medium with attenuation constant α = 0.5 Np/m.
If the wave’s electric-field amplitude is 100 V/m at z = 0, how far can the wave travel before its amplitude is reduced to (a) 10 V/m, (b) 1 V/m, (c) 1 μV/m?
Answer: (a) 4.6 m, (b) 9.2 m, (c) 37 m.
1-5 The Electromagnetic Spectrum Visible light belongs to a family of waves arranged according to frequency and wavelength along a continuum called the electromagnetic spectrum (Fig. 1-16). Other members of this family include gamma rays, X rays, infrared waves, and radio waves. Generically, they all are called EM waves because they share the following fundamental properties: • A monochromatic (single frequency) EM wave consists of electric and magnetic fields that oscillate at the same frequency f .
• The phase velocity of an EM wave propagating in vacuum is a universal constant given by the velocity of light c, defined earlier by Eq. (1.14).
1-5 THE ELECTROMAGNETIC SPECTRUM 31 Module 1.3 Phase Lead/Lag Examine sinusoidal waveforms with different values of the reference phase constant φ0.
• In vacuum, the wavelength λ of an EM wave is related to its oscillation frequency f by λ = c f .
(1.34) Whereas all monochromatic EM waves share these properties, each is distinguished by its own wavelength λ, or equivalently by its own oscillation frequency f .
The visible part of the EM spectrum shown in Fig. 1-16 covers a very narrow wavelength range extending between λ = 0.4 μm (violet) and λ = 0.7 μm (red).
As we move progressively toward shorter wavelengths, we encounter the ultraviolet, X-ray, and gamma-ray bands, each so named because of historical reasons associated with the discovery of waves with those wavelengths. On the other side of the visible spectrum lie the infrared band and then the microwave part of the radio region. Because of the link between λ and f given by Eq. (1.34), each of these spectral ranges may be specified in terms of its wavelength range or its frequency range. In practice, however, a wave is specified in terms of its wavelength λ if λ < 1 mm, which encompasses all parts of the EM spectrum except for the radio region, and the wave is specified in terms of its frequency f if λ > 1 mm (i.e., in the radio region). A wavelength of 1 mm corresponds to a frequency of 3 × 1011 Hz = 300 GHz in free space.
32 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS 1 fm 1 pm 1 nm 1 Å 1 EHz 1 PHz 1 THz 1 GHz 1 MHz 1 kHz 1 Hz 1 mm 1 mm 1 km 1 Mm 1 m 10-15 1023 1021 1018 1015 1012 109 106 103 1 10-12 10-10 10-9 10-6 10-3 103 106 108 1 Frequency (Hz) Wavelength (m) Visible Gamma rays Cancer therapy X-rays Medical diagnosis Ultraviolet Sterilization Infrared Heating, night vision Radio spectrum Communication, radar, radio and TV broadcasting, radio astronomy Atmospheric opacity 100% 0 Atmosphere opaque Optical window Infrared windows Radio window Ionosphere opaque Figure 1-16 The electromagnetic spectrum.
The radio spectrum consists of several individual bands, as shown in the chart of Fig. 1-17. Each band covers one decade of the radio spectrum and has a letter designation based on a nomenclature defined by the International Telecommunication Union.
Waves of different frequencies have different applications because they are excited by different mechanisms, and the properties of an EM wave propagating in a nonvacuum material may vary considerably from one band to another.
Although no precise definition exists for the extent of the microwave band, it is conventionally regarded to cover the full ranges of the UHF, SHF, and EHF bands. The EHF band is sometimes referred to as the millimeter-wave band because the wavelength range covered by this band extends from 1 mm (300 GHz) to 1 cm (30 GHz).
Concept Question 1-9: What are the three fundamen-tal properties of EM waves?
Concept Question 1-10: What is the range of frequen-cies covered by the microwave band?
Concept Question 1-11: What is the wavelength range of the visible spectrum?
What are some of the applications of the infrared band?
1-6 Review of Complex Numbers Any complex number z can be expressed in rectangular form as z = x + jy, (1.35) where x and y are the real (Re) and imaginary (Im) parts of z, respectively, and j = √−1. That is, x = Re(z), y = Im(z).
(1.36) Alternatively, z may be cast in polar form as z = |z|ejθ = |z|∠ θ (1.37) where |z| is the magnitude of z, θ is its phase angle, and ∠ θ is a useful shorthand representation for ejθ. Applying Euler’s identity, ejθ = cos θ + j sin θ, (1.38) 1-6 REVIEW OF COMPLEX NUMBERS 33 U .
S .
D E P A R T M E N T O F C O M M E R C E N A TI O N A L T E L E C O M M U N I C A T I O N S & I N F O R M A TI O N A D M I N IS T R A T I O N UNITED STATES THE RADIO SPECTRUM RADIO SERVICES COLOR LEGEND AERONAUTICAL MOBILE AERONAUTICAL MOBILE SATELLITE AERONAUTICAL RADIO-NAVIGATION AMATEUR INTER-SATELLITE LAND MOBILE LAND MOBILE SATELLITE MARITIME MOBILE RADIO ASTRONOMY RADIO-DETERMINATION SATELLITE RADIOLOCATION RADIOLOCATION SATELLITE AMATEUR SATELLITE BROAD-CASTING BROAD-CASTING SATELLITE EARTH EXPLORATION SATELLITE MARITIME MOBILE SATELLITE MARITIME RADIO-NAVIGATION METEORO-LOGICAL AIDS METEORO-LOGICAL SATELLITE RADIONAVIGATION RADIONAVIGATION SATELLITE SPACE OPERATION SPACE RESEARCH FIXED FIXED SATELLITE MOBILE MOBILE SATELLITE STANDARD FREQUENCY AND TIME SIGNAL STANDARD FREQUENCY AND TIME SIGNAL SATELLITE U.S. DEPARTMENT OF COMMERCE National Telecommunications and Information Administration Office of Spectrum Management October 2003 ALLOCATIONS VLF EHF SHF UHF VHF MF HF FREQUENCY MOBILE (AERONAUTICAL TELEMETERING) S) 5.68 5.73 5.90 5.95 6.2 6.525 6.685 6.765 7.0 7.1 7.3 7.35 8.1 8.195 8.815 8.965 9.040 9.4 9.5 9.9 9.995 10.003 10.005 10.1 10.15 11.175 11.275 11.4 11.6 11.65 12.05 12.10 12.23 13.2 13.26 13.36 13.41 13.57 13.6 13.8 13.87 14.0 14.25 14.35 14.990 15.005 15.010 15.10 15.6 15.8 16.36 17.41 17.48 17.55 17.9 17.97 18.03 18.068 18.168 18.78 18.9 19.02 19.68 19.80 19.990 19.995 20.005 20.010 21.0 21.45 21.85 21.924 22.0 22.855 23.0 23.2 23.35 24.89 24.99 25.005 25.01 25.07 25.21 25.33 25.55 25.67 26.1 26.175 26.48 26.95 26.96 27.23 27.41 27.54 28.0 29.7 29.8 29.89 29.91 30.0 NOT ALLOCATED RADIONAVIGATION FIXED MARITIME MOBILE FIXED MARITIME MOBILE FIXED MARITIME MOBILE Radiolocation RADIONAVIGATION FIXED MARITIME MOBILE Radiolocation FIXED MARITIME MOBILE FIXED MARITIME MOBILE AERONAUTICAL RADIONAVIGATION AERONAUTICAL RADIONAVIGATION Aeronautical Mobile Maritime Radionavigation (Radio Beacons) MARITIME RADIONAVIGATION (RADIO BEACONS) Aeronautical Radionavigation (Radio Beacons) 3 9 14 19.95 20.05 30 30 59 61 70 90 110 130 160 190 200 275 285 300 3 kHz 300 kHz 300 kHz 3 MHz 3 MHz 30 MHz 30 MHz 300 MHz 3 GHz 300 GHz 300 MHz 3 GHz 30 GHz Aeronautical Radionavigation (Radio Beacons) MARITIME RADIONAVIGATION (RADIO BEACONS) Aeronautical Mobile Maritime Radionavigation (Radio Beacons) AERONAUTICAL RADIONAVIGATION (RADIO BEACONS) AERONAUTICAL RADIONAVIGATION (RADIO BEACONS) Aeronautical Mobile Aeronautical Mobile RADIONAVIGATION AERONAUTICAL RADIONAVIGATION MARITIME MOBILE Aeronautical Radionavigation MOBILE (DISTRESS AND CALLING) MARITIME MOBILE MARITIME MOBILE (SHIPS ONLY) MOBILE AERONAUTICAL RADIONAVIGATION (RADIO BEACONS) AERONAUTICAL RADIONAVIGATION (RADIO BEACONS) BROADCASTING (AM RADIO) MARITIME MOBILE (TELEPHONY) MARITIME MOBILE (TELEPHONY) MOBILE (DISTRESS AND CALLING) MARITIME MOBILE LAND MOBILE MOBILE FIXED STANDARD FREQ. AND TIME SIGNAL (2500kHz) STANDARD FREQ. AND TIME SIGNAL Space Research MARITIME MOBILE LAND MOBILE MOBILE FIXED AERONAUTICAL MOBILE (R) STANDARD FREQ.
AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (OR) AERONAUTICAL MOBILE (R) FIXED MOBILE Radio-location FIXED MOBILE AMATEUR FIXED FIXED FIXED FIXED FIXED MARITIME MOBILE MOBILE MOBILE MOBILE STANDARD FREQ. AND TIME SIGNAL (5000 KHZ) AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (OR) STANDARD FREQ.
Space Research MOBILE AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (OR) FIXED MOBILE BROADCASTING MARITIME MOBILE AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (OR) FIXED Mobile AMATEUR SATELLITE AMATEUR AMATEUR FIXED Mobile MARITIME MOBILE MARITIME MOBILE AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (OR) FIXED BROADCASTING FIXED STANDARD FREQ. AND TIME SIGNAL (10,000 kHz) STANDARD FREQ.
Space Research AERONAUTICAL MOBILE (R) AMATEUR FIXED Mobile AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (OR) FIXED FIXED BROADCASTING MARITIME MOBILE AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (OR) RADIO ASTRONOMY Mobile AMATEUR BROADCASTING AMATEUR AMATEUR SATELLITE Mobile FIXED BROADCASTING STANDARD FREQ. AND TIME SIGNAL (15,000 kHz) STANDARD FREQ.
Space Research FIXED AERONAUTICAL MOBILE (OR) MARITIME MOBILE AERONAUTICAL MOBILE (OR) AERONAUTICAL MOBILE (R) FIXED FIXED BROADCASTING STANDARD FREQ.
Space Research FIXED MARITIME MOBILE Mobile FIXED AMATEUR AMATEUR SATELLITE BROADCASTING FIXED AERONAUTICAL MOBILE (R) MARITIME MOBILE FIXED FIXED FIXED Mobile MOBILE FIXED STANDARD FREQ. AND TIME SIGNAL (25,000 kHz) STANDARD FREQ.
Space Research LAND MOBILE MARITIME MOBILE LAND MOBILE MOBILE RADIO ASTRONOMY BROADCASTING MARITIME MOBILE LAND MOBILE FIXED MOBILE FIXED MOBILE MOBILE FIXED FIXED FIXED FIXED FIXED LAND MOBILE MOBILE AMATEUR AMATEUR SATELLITE MOBILE LAND MOBILE MOBILE MOBILE FIXED FIXED MOBILE MOBILE FIXED FIXED LAND MOBILE LAND MOBILE LAND MOBILE LAND MOBILE Radio Astronomy RADIO ASTRONOMY LAND MOBILE FIXED FIXED MOBILE MOBILE MOBILE LAND MOBILE FIXED LAND MOBILE FIXED FIXED MOBILE MOBILE LAND MOBILE AMATEUR BROADCASTING (TV CHANNELS 2-4) FIXED MOBILE FIXED MOBILE FIXED MOBILE FIXED MOBILE AERONAUTICAL RADIONAVIGATION BROADCASTING (TV CHANNELS 5-6) BROADCASTING (FM RADIO) AERONAUTICAL RADIONAVIGATION AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE AERONAUTICAL MOBILE AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (R) AERONAUTICAL MOBILE (R) MOBILE FIXED AMATEUR BROADCASTING (TV CHANNELS 7-13) MOBILE FIXED MOBILE FIXED MOBILE SATELLITE FIXED MOBILE SATELLITE MOBILE FIXED MOBILE SATELLITE MOBILE FIXED MOBILE AERONAUTICAL RADIONAVIGATION STD. FREQ. & TIME SIGNAL SAT. (400.1 MHz) MET. SAT.
(S-E) SPACE RES.
(S-E) Earth Expl.
Satellite (E-S) MOBILE SATELLITE (E-S) FIXED MOBILE RADIO ASTRONOMY RADIOLOCATION Amateur LAND MOBILE Meteorological Satellite (S-E) LAND MOBILE BROADCASTING (TV CHANNELS 14 - 20) BROADCASTING (TV CHANNELS 21-36) TV BROADCASTING RADIO ASTRONOMY RADIOLOCATION FIXED Amateur AERONAUTICAL RADIONAVIGATION MOBILE FIXED AERONAUTICAL RADIONAVIGATION Radiolocation Radiolocation MARITIME RADIONAVIGATION MARITIME RADIONAVIGATION Radiolocation Radiolocation Radiolocation RADIO-LOCATION RADIO-LOCATION Amateur AERONAUTICAL RADIONAVIGATION (Ground) RADIO-LOCATION Radio-location AERO. RADIO-NAV.(Ground) FIXED SAT. (S-E) RADIO-LOCATION Radio-location FIXED FIXED SATELLITE (S-E) FIXED AERONAUTICAL RADIONAVIGATION MOBILE FIXED MOBILE RADIO ASTRONOMY Space Research (Passive) AERONAUTICAL RADIONAVIGATION RADIO-LOCATION Radio-location RADIONAVIGATION Radiolocation RADIOLOCATION Radiolocation Radiolocation Radiolocation RADIOLOCATION RADIO-LOCATION MARITIME RADIONAVIGATION MARITIME RADIONAVIGATION METEOROLOGICAL AIDS Amateur Amateur FIXED FIXED SATELLITE (E-S) MOBILE FIXED SATELLITE (E-S) FIXED SATELLITE (E-S) MOBILE FIXED FIXED FIXED FIXED MOBILE FIXED SPACE RESEARCH (E-S) FIXED Fixed MOBILE SATELLITE (S-E) FIXED SATELLITE (S-E) FIXED SATELLITE (S-E) FIXED SATELLITE (S-E) FIXED SATELLITE (S-E) FIXED SATELLITE (E-S) FIXED SATELLITE (E-S) FIXED SATELLITE (E-S) FIXED SATELLITE (E-S) FIXED FIXED FIXED FIXED FIXED FIXED FIXED MET.
SATELLITE (S-E) Mobile Satellite (S-E) Mobile Satellite (S-E) Mobile Satellite (E-S) (no airborne) Mobile Satellite (E-S)(no airborne) Mobile Satellite (S-E) Mobile Satellite (E-S) MOBILE SATELLITE (E-S) EARTH EXPL.
SATELLITE(S-E) EARTH EXPL.
SAT. (S-E) EARTH EXPL.
SATELLITE (S-E) MET.
SATELLITE (E-S) FIXED FIXED SPACE RESEARCH (S-E) (deep space only) SPACE RESEARCH (S-E) AERONAUTICAL RADIONAVIGATION RADIOLOCATION Radiolocation Radiolocation Radiolocation Radiolocation MARITIME RADIONAVIGATION Meteorological Aids RADIONAVIGATION RADIOLOCATION Radiolocation RADIO-LOCATION Radiolocation Radiolocation Amateur Amateur Amateur Satellite RADIOLOCATION FIXED FIXED FIXED FIXED FIXED SATELLITE (S-E) FIXED SATELLITE (S-E) Mobile SPACE RESEARCH (Passive) EARTH EXPL.
SAT. (Passive) RADIO ASTRONOMY SPACE RESEARCH (Passive) EARTH EXPL.
SATELLITE (Passive) RADIO ASTRONOMY BROADCASTING SATELLITE AERONAUTICAL RADIONAV.
Space Research (E-S) Space Research Land Mobile Satellite (E-S) Radio-location RADIO-LOCATION RADIO NAVIGATION FIXED SATELLITE (E-S) Land Mobile Satellite (E-S) Land Mobile Satellite (E-S) Fixed Mobile FIXED SAT. (E-S) Fixed Mobile FIXED Mobile FIXED MOBILE Space Research Space Research Space Research SPACE RESEARCH (Passive) RADIO ASTRONOMY EARTH EXPL. SAT.
(Passive) Radiolocation RADIOLOCATION Radiolocation FX SAT (E-S) FIXED SATELLITE (E-S) FIXED FIXED FIXED MOBILE EARTH EXPL.
SAT. (Passive) MOBILE Earth Expl.
Satellite (Active) Standard Frequency and Time Signal Satellite (E-S) Earth Exploration Satellite (S-S) MOBILE FIXED MOBILE FIXED Earth Exploration Satellite (S-S) FIXED MOBILE FIXED SAT (E-S) FIXED SATELLITE (E-S) MOBILE SATELLITE (E-S) FIXED SATELLITE (E-S) MOBILE SATELLITE (E-S) Standard Frequency and Time Signal Satellite (S-E) Stand. Frequency and Time Signal Satellite (S-E) FIXED MOBILE RADIO ASTRONOMY SPACE RESEARCH (Passive) EARTH EXPLORATION SAT. (Passive) RADIONAVIGATION RADIONAVIGATION INTER-SATELLITE RADIONAVIGATION RADIOLOCATION Radiolocation SPACE RE.
.(Passive) EARTH EXPL.
SAT. (Passive) FIXED MOBILE FIXED MOBILE FIXED MOBILE Mobile Fixed FIXED SATELLITE (S-E) BROAD-CASTING BCST SAT.
FIXED MOBILE F X SAT(E-S) MOBILE FIXED EARTH EXPLORATION SATELLITE FI XED SATELLITE (E-S) MOBILE SATELLITE (E-S) MOBILE FIXED SPACE RESEARCH (Passive) EARTH EXPLORATION SATELLITE (Passive) EARTH EXPLORATION SAT. (Passive) SPACE RESEARCH (Passive) INTER-SATELLITE RADIO-LOCATION SPACE RESEARCH FIXED MOBILE FIXED MOBILE SATELLITE (E-S) MOBILE SATELLITE RADIO NAVIGATION RADIO-NAVIGATION SATELLITE EARTH EXPLORATION SATELLITE FIXED SATELLITE (E-S) MOBILE FIXED FIXED SATELLITE (E-S) AMATEUR AMATEUR SATELLITE AMATEUR AMATEUR SATELLITE Amateur Satellite Amateur RADIO-LOCATION MOBILE FIXED MOBILE SATELLITE (S-E) FIXED SATELLITE (S-E) MOBILE FIXED BROAD-CASTING SATELLITE BROAD-CASTING SPACE RESEARCH (Passive) RADIO ASTRONOMY EARTH EXPLORATION SATELLITE (Passive) MOBILE FIXED MOBILE FIXED RADIO-LOCATION FIXED SATELLITE (E-S) MOBILE SATELLITE RADIO-NAVIGATION SATELLITE RADIO-NAVIGATION Radio-location EARTH EXPL.
SATELLITE (Passive) SPACE RESEARCH (Passive) FIXED FIXED SATELLITE (S-E) SPACE RESEARCH (Passive) RADIO ASTRONOMY EARTH EXPLORATION SATELLITE (Passive) FIXED MOBILE MOBILE INTER-SATELLITE RADIO-LOCATION INTER-SATELLITE Radio-location MOBILE MOBILE SATELLITE RADIO-NAVIGATION RADIO-NAVIGATION SATELLITE AMATEUR AMATEUR SATELLITE Amateur Amateur Satellite RADIO-LOCATION MOBILE FIXED FIXED SATELLITE (S-E) MOBILE FIXED FIXED SATELLITE (S-E) EARTH EXPLORATION SATELLITE (Passive) SPACE RES.
(Passive) SPACE RES.
(Passive) RADIO ASTRONOMY FIXED SATELLITE (S-E) FIXED MOBILE FIXED MOBILE FIXED MOBILE FIXED MOBILE FIXED MOBILE FIXED SPACE RESEARCH (Passive) RADIO ASTRONOMY EARTH EXPLORATION SATELLITE (Passive) EARTH EXPLORATION SAT. (Passive) SPACE RESEARCH (Passive) INTER-SATELLITE INTER-SATELLITE INTER-SATELLITE INTER-SATELLITE MOBILE MOBILE MOBILE MOBILE SATELLITE RADIO-NAVIGATION RADIO-NAVIGATION SATELLITE FIXED SATELLITE (E-S) FIXED FIXED EARTH EXPLORATION SAT.
(Passive) SPACE RES.
(Passive) SPACE RESEARCH (Passive) RADIO ASTRONOMY EARTH EXPLORATION SATELLITE (Passive) MOBILE FIXED MOBILE FIXED MOBILE FIXED FIXED SATELLITE (S-E) FIXED SATELLITE(S-E) FIXED SATELLITE (S-E) EARTH EXPL.
SAT. (Passive) SPACE RES.
(Passive) Radio-location Radio-location RADIO-LOCATION AMATEUR AMATEUR SATELLITE Amateur Amateur Satellite EARTH EXPLORATION SATELLITE (Passive) SPACE RES. (Passive) MOBILE MOBILE SATELLITE RADIO-NAVIGATION RADIO-NAVIGATION SATELLITE MOBILE MOBILE FIXED RADIO-ASTRONOMY FIXED SATELLITE (E-S) FIXED 3.0 3.025 3.155 3.230 3.4 3.5 4.0 4.063 4.438 4.65 4.7 4.75 4.85 4.995 5.003 5.005 5.060 5.45 MARITIME MOBILE AMATEUR AMATEUR SATELLITE FIXED Mobile MARITIME MOBILE STANDARD FREQUENCY & TIME SIGNAL (20,000 KHZ) Space Research AERONAUTICAL MOBILE (OR) AMATEUR SATELLITE AMATEUR MET. SAT. (S-E) MOB. SAT. (S-E) SPACE RES. (S-E) SPACE OPN. (S-E) MET. SAT. (S-E) Mob. Sat. (S-E) SPACE RES. (S-E) SPACE OPN. (S-E) MET. SAT. (S-E) MOB. SAT. (S-E) SPACE RES. (S-E) SPACE OPN. (S-E) MET. SAT. (S-E) Mob. Sat. (S-E) SPACE RES. (S-E) SPACE OPN. (S-E) MOBILE FIXED FIXED Land Mobile FIXED MOBILE LAND MOBILE LAND MOBILE MARITIME MOBILE MARITIME MOBILE MARITIME MOBILE MARITIME MOBILE LAND MOBILE FIXED MOBILE MOBILE SATELLITE (E-S) Radiolocation Radiolocation LAND MOBILE AMATEUR MOBILE SATELLITE (E-S) RADIONAVIGATION SATELLITE MET. AIDS (Radiosonde) METEOROLOGICAL AIDS (RADIOSONDE) SPACE RESEARCH (S-S) FIXED MOBILE LAND MOBILE FIXED LAND MOBILE FIXED FIXED RADIO ASTRONOMY RADIO ASTRONOMY METEOROLOGICAL AIDS (RADIOSONDE) METEOROLOGICAL AIDS (Radiosonde) METEOROLOGICAL SATELLITE (s-E) Fixed FIXED MET. SAT.
(s-E) FIXED FIXED AERONAUTICAL MOBILE SATELLITE (R) (space to Earth) AERONAUTICAL RADIONAVIGATION RADIONAV. SATELLITE (Space to Earth) AERONAUTICAL MOBILE SATELLITE (R) (space to Earth) Mobile Satellite (S- E) RADIO DET. SAT. (E-S) MOBILESAT(E-S) AERO. RADIONAVIGATION AERO. RADIONAV.
AERO. RADIONAV.
RADIO DET. SAT. (E-S) RADIO DET. SAT. (E-S) MOBILE SAT. (E-S) MOBILE SAT. (E-S) Mobile Sat. (S-E) RADIO ASTRONOMY RADIO ASTRONOMY MOBILE SAT. (E-S) FIXED MOBILE FIXED FIXED (LOS) MOBILE (LOS) SPACE RESEARCH (s-E)(s-s) SPACE OPERATION (s-E)(s-s) EARTH EXPLORATION SAT. (s-E)(s-s) Amateur d e x i F E L I B O M RADIOLOCATION AMATEUR RADIO ASTRON.
SPACE RESEARCH EARTH EXPL SAT FIXED SAT. (S-E) FIXED MOBILE FIXED SATELLITE (S-E) FIXED MOBILE FIXED SATELLITE (E-S) FIXED SATELLITE (E-S) MOBILE FIXED SPACE RESEARCH (S-E) (Deep Space) AERONAUTICAL RADIONAVIGATION EARTH EXPL. SAT.
(Passive) 300 325 335 405 415 435 495 505 510 525 535 1605 1615 1705 1800 1900 2000 2065 2107 2170 2173.5 2190.5 2194 2495 2501 2502 2505 2850 3000 RADIO-LOCATION BROADCASTING FIXED MOBILE AMATEUR RADIOLOCATION MOBILE FIXED MARITIME MOBILE MARITIME MOBILE (TELEPHONY) MARITIME MOBILE LAND MOBILE MOBILE FIXED 30.0 30.56 32.0 33.0 34.0 35.0 36.0 37.0 37.5 38.0 38.25 39.0 40.0 42.0 43.69 46.6 47.0 49.6 50.0 54.0 72.0 73.0 74.6 74.8 75.2 75.4 76.0 88.0 108.0 117.975 121.9375 123.0875 123.5875 128.8125 132.0125 136.0 137.0 137.025 137.175 137.825 138.0 144.0 146.0 148.0 149.9 150.05 150.8 152.855 154.0 156.2475 157.0375 157.1875 157.45 161.575 161.625 161.775 162.0125 173.2 173.4 174.0 216.0 220.0 222.0 225.0 235.0 300 ISM – 6.78 ± .015 MHz ISM – 13.560 ± .007 MHz ISM – 27.12 ± .163 MHz ISM – 40.68 ± .02 MHz ISM – 24.125 ± 0.125 GHz 30 GHz ISM – 245.0 ± 1GHz ISM – 122.5 ± .500 GHz ISM – 61.25 ± .250 GHz 300.0 322.0 328.6 335.4 399.9 400.05 400.15 401.0 402.0 403.0 406.0 406.1 410.0 420.0 450.0 454.0 455.0 456.0 460.0 462.5375 462.7375 467.5375 467.7375 470.0 512.0 608.0 614.0 698 746 764 776 794 806 821 824 849 851 866 869 894 896 901901 902 928 929 930 931 932 935 940 941 944 960 1215 1240 1300 1350 1390 1392 1395 2000 2020 2025 2110 2155 2160 2180 2200 2290 2300 2305 2310 2320 2345 2360 2385 2390 2400 2417 2450 2483.5 2500 2655 2690 2700 2900 3000 1400 1427 1429.5 1430 1432 1435 1525 1530 1535 1544 1545 1549.5 1558.5 1559 1610 1610.6 1613.8 1626.5 1660 1660.5 1668.4 1670 1675 1700 1710 1755 1850 MARITIME MOBILE SATELLITE (space to Earth) MOBILE SATELLITE (S-E) RADIOLOCATION RADIONAVIGATION SATELLITE (S-E) RADIOLOCATION Amateur Radiolocation AERONAUTICAL RADIONAVIGATION SPA CE RESEARCH ( Passive) EARTH EXPL SAT (Passive) RADIO ASTRONOMY MOBILE MOBILE FIXED-SAT (E-S) FIXED FIXED FIXED LAND MOBILE (TLM) MOBILE SAT.
(Space to Earth) MARITIME MOBILE SAT.
(Space to Earth) Mobile (Aero. TLM) MOBILE SATELLITE (S-E) MOBILE SATELLITE (Space to Earth) AERONAUTICAL MOBILE SATELLITE (R) (space to Earth) 3.0 3.1 3.3 3.5 3.6 3.65 3.7 4.2 4.4 4.5 4.8 4.94 4.99 5.0 5.15 5.25 5.35 5.46 5.47 5.6 5.65 5.83 5.85 5.925 6.425 6.525 6.70 6.875 7.025 7.075 7.125 7.19 7.235 7.25 7.30 7.45 7.55 7.75 7.90 8.025 8.175 8.215 8.4 8.45 8.5 9.0 9.2 9.3 9.5 10.0 10.45 10.5 10.55 10.6 10.68 10.7 11.7 12.2 12.7 12.75 13.25 13.4 13.75 14.0 14.2 14.4 14.47 14.5 14.7145 15.1365 15.35 15.4 15.43 15.63 15.7 16.6 17.1 17.2 17.3 17.7 17.8 18.3 18.6 18.8 19.3 19.7 20.1 20.2 21.2 21.4 22.0 22.21 22.5 22.55 23.55 23.6 24.0 24.05 24.25 24.45 24.65 24.75 25.05 25.25 25.5 27.0 27.5 29.5 29.9 30.0 ISM – 2450.0 ± 50 MHz 30.0 31.0 31.3 31.8 32.0 32.3 33.0 33.4 36.0 37.0 37.6 38.0 38.6 39.5 40.0 40.5 41.0 42.5 43.5 45.5 46.9 47.0 47.2 48.2 50.2 50.4 51.4 52.6 54.25 55.78 56.9 57.0 58.2 59.0 59.3 64.0 65.0 66.0 71.0 74.0 75.5 76.0 77.0 77.5 78.0 81.0 84.0 86.0 92.0 95.0 100.0 102.0 105.0 116.0 119.98 120.02 126.0 134.0 142.0 144.0 149.0 150.0 151.0 164.0 168.0 170.0 174.5 176.5 182.0 185.0 190.0 200.0 202.0 217.0 231.0 235.0 238.0 241.0 248.0 250.0 252.0 265.0 275.0 300.0 ISM – 5.8 ± .075 GHz ISM – 915.0 ± 13 MHz INTER-SATELLITE RADIOLOCATION SATELLITE (E-S) AERONAUTICAL RADIONAV.
RADIO ASTRONOMY FIXED MARITIME MOBILE FIXED MARITIME MOBILE Aeronautical Mobile STANDARD FREQ. AND TIME SIGNAL (60 kHz) FIXED Mobile STAND. FREQ. & TIME SIG.
MET. AIDS (Radiosonde) Space Opn. (S-E) MOBILE.
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Figure 1-17 Individual bands of the radio spectrum and their primary allocations in the US. [See expandable version on book website: em.eecs.umich.edu.] 34 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS θ (z) y z x |z| x = |z| cos θ y = |z| sin θ θ = tan−1 (y/x) |z| = x2 + y2 + (z) Figure 1-18 Relation between rectangular and polar representations of a complex number z = x + jy = |z|ejθ.
we can convert z from polar form, as in Eq. (1.37), into rectangular form, z = |z|ejθ = |z| cos θ + j|z| sin θ.
(1.39) This leads to the relations x = |z| cos θ, y = |z| sin θ, (1.40) |z| = + x2 + y2 , θ = tan−1(y/x).
(1.41) The two forms are illustrated graphically in Fig. 1-18. When using Eq. (1.41), care should be taken to ensure that θ is in the proper quadrant. Also note that, since |z| is a positive quantity, only the positive root in Eq. (1.41) is applicable. This is denoted by the + sign above the square-root sign.
The complex conjugate of z, denoted with a star superscript (or asterisk), is obtained by replacing j (wherever it appears) with −j, so that z∗= (x + jy)∗= x −jy = |z|e−jθ = |z|∠ −θ . (1.42) The magnitude |z| is equal to the positive square root of the product of z and its complex conjugate: |z| = + √ z z∗.
(1.43) We now highlight some of the properties of complex algebra that will be encountered in future chapters.
Equality: If two complex numbers z1 and z2 are given by z1 = x1 + jy1 = |z1|ejθ1, (1.44) z2 = x2 + jy2 = |z2|ejθ2, (1.45) then z1 = z2 if and only if x1 = x2 and y1 = y2 or, equivalently, |z1| = |z2| and θ1 = θ2.
Addition: z1 + z2 = (x1 + x2) + j(y1 + y2).
(1.46) Multiplication: z1z2 = (x1 + jy1)(x2 + jy2) = (x1x2 −y1y2) + j(x1y2 + x2y1), (1.47a) or z1z2 = |z1|ejθ1 · |z2|ejθ2 = |z1||z2|ej(θ1+θ2) = |z1||z2|[cos(θ1 + θ2) + j sin(θ1 + θ2)].
(1.47b) Division: For z2 ̸= 0, z1 z2 = x1 + jy1 x2 + jy2 = (x1 + jy1) (x2 + jy2) · (x2 −jy2) (x2 −jy2) = (x1x2 + y1y2) + j(x2y1 −x1y2) x2 2 + y2 2 , (1.48a) 1-6 REVIEW OF COMPLEX NUMBERS 35 or z1 z2 = |z1|ejθ1 |z2|ejθ2 = |z1| |z2|ej(θ1−θ2) = |z1| |z2|[cos(θ1 −θ2) + j sin(θ1 −θ2)].
(1.48b) Powers: For any positive integer n, zn = (|z|ejθ)n = |z|nejnθ = |z|n(cos nθ + j sin nθ), (1.49) z1/2 = ±|z|1/2ejθ/2 = ±|z|1/2[cos(θ/2) + j sin(θ/2)].
(1.50) Useful Relations: −1 = ejπ = e−jπ = 1∠ 180◦, j = ejπ/2 = 1∠ 90◦, (1.51) −j = −ejπ/2 = e−jπ/2 = 1∠ −90◦, (1.52) j = (ejπ/2)1/2 = ±ejπ/4 = ±(1 + j) √ 2 , (1.53) −j = ±e−jπ/4 = ±(1 −j) √ 2 .
(1.54) Example 1-3: Working with Complex Numbers Given two complex numbers V = 3 −j4, I = −(2 + j3), (a) express V and I in polar form, and find (b) V I, (c) V I ∗, (d) V/I, and (e) √ I .
θI θV |V | V |I | I −2 −3 −4 3 Figure 1-19 Complex numbers V and I in the complex plane (Example 1-3).
Solution: (a) |V | = + √ V V ∗ = + √(3 −j4)(3 + j4) = + √9 + 16 = 5, θV = tan−1(−4/3) = −53.1◦, V = |V |ejθV = 5e−j53.1◦= 5∠ −53.1◦, |I| = + √ 22 + 32 = + √ 13 = 3.61.
Since I = (−2 −j3) is in the third quadrant in the complex plane [Fig. 1-19], θI = 180◦+ tan−1 3 2 = 236.3◦, I = 3.61∠ 236.3◦.
(b) V I = 5e−j53.1◦× 3.61ej236.3◦ = 18.03ej(236.3◦−53.1◦) = 18.03ej183.2◦.
36 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS (c) V I ∗= 5e−j53.1◦× 3.61e−j236.3◦ = 18.03e−j289.4◦ = 18.03ej70.6◦.
(d) V I = 5e−j53.1◦ 3.61ej236.3◦ = 1.39e−j289.4◦ = 1.39ej70.6◦.
(e) √ I = √ 3.61ej236.3◦ = ± √ 3.61 ej236.3◦/2 = ±1.90ej118.15◦.
Exercise 1-7: Express the following complex functions in polar form: z1 = (4 −j3)2, z2 = (4 −j3)1/2.
Answer: z1 = 25∠ −73.7◦, z2 = ± √ 5 ∠ −18.4◦.
[See EM (the “ EM” symbol refers to the book website: em.eecs.umich.edu).] Exercise 1-8: Show that √2j = ±(1 + j). (See EM.) 1-7 Review of Phasors Phasor analysis is a useful mathematical tool for solving problems involving linear systems in which the excitation is a periodic time function. Many engineering problems are cast in the form of linear integro-differential equations.
If the excitation, more commonly known as the forcing function, varies sinusoidally with time, the use of phasor notation to represent time-dependent variables allows us to convert a linear integro-differential equation into a linear equation with no sinusoidal functions, thereby simplifying the method of solution. After solving for the desired variable, such as the voltage or current in a circuit, conversion from the phasor domain back to the time domain provides the desired result.
The phasor technique can also be used to analyze linear systems when the forcing function is a (nonsinusoidal) periodic time function, such as a square wave or a sequence of pulses.
C R i υs(t) + − Figure 1-20 RC circuit connected to a voltage source υs(t).
By expanding the forcing function into a Fourier series of sinusoidal components, we can solve for the desired variable using phasor analysis for each Fourier component of the forcing function separately.
According to the principle of superposition, the sum of the solutions due to all of the Fourier components gives the same result as one would obtain had the problem been solved entirely in the time domain without the aid of the Fourier representation. The obvious advantage of the phasor–Fourier approach is simplicity. Moreover, in the case of nonperiodic source functions, such as a single pulse, the functions can be expressed as Fourier integrals, and a similar application of the principle of superposition can be used as well.
The simple RC circuit shown in Fig. 1-20 contains a sinusoidally time-varying voltage source given by υs(t) = V0 sin(ωt + φ0), (1.55) where V0 is the amplitude, ω is the angular frequency, and φ0 is a reference phase. Application of Kirchhoff’s voltage law gives the following loop equation: R i(t) + 1 C i(t) dt = υs(t).
(1.56) (time domain) Our objective is to obtain an expression for the current i(t). We can do this by solving Eq. (1.56) in the time domain, which is somewhat cumbersome because the forcing function υs(t) is a sinusoid. Alternatively, we can take advantage of the phasor-domain solution technique as follows.
1-7 REVIEW OF PHASORS 37 1-7.1 Solution Procedure Step 1: Adopt a cosine reference To establish a phase reference for all time-varying currents and voltages in the circuit, the forcing function is expressed as a cosine (if not already in that form). In the present example, υs(t) = V0 sin(ωt + φ0) = V0 cos π 2 −ωt −φ0 = V0 cos ωt + φ0 −π 2 , (1.57) where we used the properties sin x = cos(π/2 −x) and cos(−x) = cos x.
Step 2: Express time-dependent variables as phasors Any cosinusoidally time-varying function z(t) can be expressed as z(t) = Re Z ejωt , (1.58) where Z is a time-independent function called the phasor of the instantaneous function z(t). To distinguish instantaneous quantities from their phasor counterparts, a tilde (∼) is added over the letter denoting a phasor. The voltage υs(t) given by Eq. (1.57) can be cast in the form υs(t) = Re V0ej(ωt+φ0−π/2) = Re V0ej(φ0−π/2)ejωt = Re Vsejωt , (1.59) where Vs consists of the expression inside the square bracket that multiplies ejωt, Vs = V0ej(φ0−π/2).
(1.60) The phasor Vs, corresponding to the time function υs(t), contains amplitude and phase information but is independent of the time variable t. Next we define the unknown variable i(t) in terms of a phasor ˜ I, i(t) = Re( ˜ Iejωt), (1.61) and if the equation we are trying to solve contains derivatives or integrals, we use the following two properties: di dt = d dt Re( ˜ Iejωt) = Re d dt ( ˜ Iejωt) = Re[jω ˜ Iejωt], (1.62) and i dt = Re( ˜ Iejωt) dt = Re ˜ Iejωt dt = Re ˜ I jω ejωt .
(1.63) Thus, differentiation of the time function i(t) is equivalent to multiplication of its phasor ˜ I by jω, and integration is equivalent to division by jω.
Step 3: Recast the differential / integral equation in phasor form Upon using Eqs. (1.59), (1.61), and (1.63) in Eq. (1.56), we have R Re( ˜ Iejωt) + 1 C Re ˜ I jω ejωt = Re( Vsejωt).
(1.64) Combining all three terms under the same real-part (Re) operator leads to Re R + 1 jωC ˜ I − Vs ejωt = 0.
(1.65a) 38 TECHNOLOGY BRIEF 2: SOLAR CELLS Technology Brief 2: Solar Cells A solar cell is a photovoltaic device that converts solar energy into electricity. The conversion process relies on the photovoltaic effect, which was first reported by 19-year-old Edmund Bequerel in 1839 when he observed that a platinum electrode produced a small current if exposed to light. The photovoltaic effect is often confused with the photoelectric effect; they are interrelated, but not identical (Fig. TF2-1).
The photoelectric effect explains the mechanism responsible for why an electron is ejected by a material in consequence to a photon incident upon its surface [Fig. TF2-1(a)]. For this to happen, the photon energy E (which is governed by its wavelength through E = hc/λ, with h being Planck’s constant and c the velocity of light) has to exceed the binding energy with which the electron is held by the material. For his 1905 quantum-mechanical model of the photoelectric effect, Albert Einstein was awarded the 1921 Nobel Prize in physics.
Whereas a single material is sufficient for the photoelectric effect to occur, at least two adjoining materials with different electronic properties (to form a junction that can support a voltage across it) are needed to establish a photovoltaic current through an external load [Fig.TF2-1(b)]. Thus, the two effects are governed by the same quantum-mechanical rules associated with how photon energy can be used to liberate electrons away from their hosts, but the followup step of what happens to the liberated electrons is different in the two cases.
(a) Photoelectric effect (b) Photovoltaic effect e _ n-type p-type I Load Photon e _ Photon Metal Figure TF2-1 Comparison of photoelectric effect with the photovoltaic effect.
The PV Cell Today’s photovoltaic (PV) cells are made of semiconductor materials. The basic structure of a PV cell consists of a p-n junction connected to a load (Fig. TF2-2).
Typically, the n-type layer is made of silicon doped with a material that creates an abundance of negatively charged atoms, and the p-type layer also is made of silicon but doped with a different material that creates an abundance of holes (atoms with missing electrons). The combination of the two layers induces an electric field across the junction, so when an incident photon liberates an electron, the electron is swept under the influence of the electric field through the n-layer and out to the external circuit connected to the load.
The conversion efficiency of a PV cell depends on several factors, including the fraction of the incident light that gets absorbed by the semiconductor material, as opposed to getting reflected by the n-type front surface or transmitted through to the back conducting electrode. To minimize the reflected component, an antireflective coating usually is inserted between the upper glass cover and the n-type layer (Fig. TF2-2).
TECHNOLOGY BRIEF 2: SOLAR CELLS 39 Light photons Front-conducting electrode Antireflective coating Glass cover Back-conducting electrode p-n junction − + p-type layer (silicon) n-type layer (silicon) Figure TF2-2 Basic structure of a photovoltaic cell.
The PV cell shown in Fig. TF2-2 is called a single-junction cell because it contains only one p-n junction. The semiconductor material is characterized by a quantity called its band gap energy, which is the amount of energy needed to free an electron away from its host atom. Hence, for that to occur, the wavelength of the incident photon (which, in turn, defines its energy) has to be such that the photon’s energy exceeds the band gap of the material. Solar energy extends over a broad spectrum, so only a fraction of the solar spectrum (photons with energies greater than the band gap) is absorbed by a single-junction material. To overcome this limitation, multiple p-n layers can be cascaded together to form a multijunction PV device (Fig. TF2-3). The cells usually are arranged such that the top cell has the highest band gap energy, thereby capturing the high-energy (short-wavelength) photons, followed by the cell with the next lower band gap, and so on.
▶The multijunction technique offers an improvement in conversion efficiency of 2–4 times over that of the single-junction cell. However, the fabrication cost is significantly greater as well. ◀ Modules, Arrays, and Systems A photovoltaic module consists of multiple PV cells connected together so as to supply electrical power at a specified voltage level, such as 12 or 24 V. The combination of multiple modules generates a PV array (Fig.TF2-4). The amount of generated power depends on the intensity of the intercepted sunlight, the total area of the module or array, and the 40 TECHNOLOGY BRIEF 2: SOLAR CELLS conversion efficiencies of the individual cells. If the PV energy source is to serve multiple functions, it is integrated into an energy-management system that includes a dc-to-ac current converter and batteries to store energy for later use (Fig. TF2-5).
Wavelength (nm) IR 400 500 600 700 800 InGaP InGaAs Ge Figure TF2-3 In a multijunction PV device, different layers absorb different parts of the light spectrum.
PV cell PV module PV array Figure TF2-4 PV cells, modules, and arrays.
PV array dc dc ac dc/ac dc-to-ac inverter Battery storage system FigureTF2-5 Components of a large-scale photovoltaic system.
1-7 REVIEW OF PHASORS 41 Had we adopted a sine reference—instead of a cosine reference—to define sinusoidal functions, the preceding treatment would have led to the result Im R + 1 jωC ˜ I − Vs ejωt = 0.
(1.65b) Since both the real and imaginary parts of the expression inside the curly brackets are zero, the expression itself must be zero.
Moreover, since ejωt ̸= 0, it follows that ˜ I R + 1 jωC = Vs (phasor domain).
(1.66) The time factor ejωt hasdisappearedbecauseitwascontainedin all three terms. Equation (1.66) is the phasor-domain equivalent of Eq. (1.56).
Step 4: Solve the phasor-domain equation From Eq. (1.66) the phasor current ˜ I is given by ˜ I = Vs R + 1/(jωC) .
(1.67) Before we apply the next step, we need to convert the right-hand side of Eq. (1.67) into the form I0ejθ with I0 being a real quantity. Thus, ˜ I = V0ej(φ0−π/2) jωC 1 + jωRC = V0ej(φ0−π/2) ωCejπ/2 √ 1 + ω2R2C2 ejφ1 = V0ωC √ 1 + ω2R2C2 ej(φ0−φ1), (1.68) where we have used the identity j = ejπ/2. The phase angle associated with (1 + jωRC) is φ1 = tan−1(ωRC) and lies in the first quadrant of the complex plane.
Table 1-5 Time-domain sinusoidal functions z(t) and their cosine-reference phasor-domain counterparts Z, where z(t) = Re [ Zejωt].
z(t) Z A cos ωt A A cos(ωt + φ0) Aejφ0 A cos(ωt + βx + φ0) Aej(βx+φ0) Ae−αx cos(ωt + βx + φ0) Ae−αxej(βx+φ0) A sin ωt Ae−jπ/2 A sin(ωt + φ0) Aej(φ0−π/2) d dt (z(t)) jω Z d dt [A cos(ωt + φ0)] jωAejφ0 z(t) dt 1 jω Z A sin(ωt + φ0) dt 1 jω Aej(φ0−π/2) Step 5: Find the instantaneous value To find i(t), we simply apply Eq. (1.61). That is, we multiply the phasor ˜ I given by Eq. (1.68) by ejωt and then take the real part: i(t) = Re ˜ Iejωt = Re V0ωC √ 1 + ω2R2C2 ej(φ0−φ1)ejωt = V0ωC √ 1 + ω2R2C2 cos(ωt + φ0 −φ1).
(1.69) In summary, we converted all time-varying quantities into the phasor domain, solved for the phasor ˜ I of the desired instantaneous current i(t), and then converted back to the time domain to obtain an expression for i(t). Table 1-5 provides a summary of some time-domain functions and their phasor-domain equivalents.
42 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS Example 1-4: RL Circuit The voltage source of the circuit shown in Fig. 1-21 is given by υs(t) = 5 sin(4 × 104t −30◦) (V).
(1.70) Obtain an expression for the voltage across the inductor.
Solution: The voltage loop equation of the RL circuit is Ri + L di dt = υs(t).
(1.71) BeforeconvertingEq.(1.71)intothephasordomain, weexpress Eq. (1.70) in terms of a cosine reference: υs(t) = 5 sin(4 × 104t −30◦) = 5 cos(4 × 104t −120◦) (V).
(1.72) The coefficient of t specifies the angular frequency as ω = 4 × 104 (rad/s). Per the second entry in Table 1-5, The voltage phasor corresponding to υs(t) is Vs = 5e−j120◦ (V), and the phasor equation corresponding to Eq. (1.71) is R ˜ I + jωL ˜ I = Vs.
(1.73) R = 6 Ω L = 0.2 mH i υs(t) + − + − υL Figure 1-21 RL circuit (Example 1-4).
Solving for the current phasor ˜ I, we have ˜ I = Vs R + jωL = 5e−j120◦ 6 + j4 × 104 × 2 × 10−4 = 5e−j120◦ 6 + j8 = 5e−j120◦ 10ej53.1◦= 0.5e−j173.1◦ (A).
The voltage phasor across the inductor is related to ˜ I by VL = jωL ˜ I = j4 × 104 × 2 × 10−4 × 0.5e−j173.1◦ = 4ej(90◦−173.1◦) = 4e−j83.1◦ (V), and the corresponding instantaneous voltage υL(t) is therefore υL(t) = Re VLejωt = Re 4e−j83.1◦ej4×104t = 4 cos(4 × 104t −83.1◦) (V).
Concept Question 1-12: Why is the phasor technique useful? When is it used? Describe the process.
Concept Question 1-13: How is the phasor technique used when the forcing function is a nonsinusoidal periodic waveform, such as a train of pulses?
Exercise 1-9: A series RL circuit is connected to a voltage source given by υs(t) = 150 cos ωt (V).
Find (a) the phasor current ˜ I and (b) the instantaneous current i(t) for R = 400 , L = 3 mH, and ω = 105 rad/s.
Answer: (a) ˜ I = 150/(R + jωL) = 0.3∠ −36.9◦(A), (b) i(t) = 0.3 cos(ωt −36.9◦) (A). (See EM.) Exercise 1-10: A phasor voltage is given by V = j5 V.
Find υ(t).
Answer: υ(t) = 5 cos(ωt + π/2) = −5 sin ωt (V).
(See EM.) CHAPTER 1 SUMMARY 43 1-7.2 Traveling Waves in the Phasor Domain AccordingtoTable1-5, ifwesetφ0 = 0, itsthirdentrybecomes A cos(ωt + βx) Aejβx.
(1.74) From the discussion associated with Eq. (1.31), we concluded that A cos(ωt + βx) describes a wave traveling in the negative x direction.
▶In the phasor domain, a wave of amplitude A traveling in the positive x direction in a lossless medium with phase constant β is given by the negative exponential Ae−jβx, and conversely, a wave traveling in the negative x direction is given by Aejβx. Thus, the sign of x in the exponential is opposite to the direction of travel. ◀ Chapter 1 Summary Concepts • Electromagnetics is the study of electric and magnetic phenomena and their engineering applications.
• The International System of Units consists of the six fundamental dimensions listed in Table 1-1. The units of all other physical quantities can be expressed in terms of the six fundamental units.
• The four fundamental forces of nature are the nuclear, weak-interaction, electromagnetic, and gravitational forces.
• The source of the electric field quantities E and D is the electric charge q.
In a material, E and D are related by D = ϵE, where ϵ is the electrical permittivity of the material. In free space, ϵ = ϵ0 ≈(1/36π) × 10−9 (F/m).
• The source of the magnetic field quantities B and H is the electric current I. In a material, B and H are related by B = μH, where μ is the magnetic permeability of the medium. In free space, μ = μ0 = 4π × 10−7 (H/m).
• Electromagnetics consists of three branches: (1) electrostatics, which pertains to stationary charges, (2) magnetostatics, which pertains to dc currents, and (3) electrodynamics, which pertains to time-varying currents.
• A traveling wave is characterized by a spatial wavelength λ, a time period T , and a phase velocity up = λ/T .
• An electromagnetic (EM) wave consists of oscillating electric and magnetic field intensities and travels in free space at the velocity of light c = 1/ √ϵ0μ0 . The EM spectrum encompasses gamma rays, X-rays, visible light, infrared waves, and radio waves.
• Phasor analysis is a useful mathematical tool for solving problems involving time-periodic sources.
Mathematical and Physical Models Electric field due to charge q in free space E = ˆ R q 4πϵ0R2 Magnetic field due to current I in free space B = ˆ φ φ φ μ0I 2πr Plane wave y(x, t) = Ae−αx cos(ωt −βx + φ0) • α = 0 in lossless medium • phase velocity up = f λ = ω β • ω = 2πf ; β = 2π/λ • φ0 = phase reference Complex numbers • Euler’s identity ejθ = cos θ + j sin θ • Rectangular-polar relations x = |z| cos θ, y = |z| sin θ, |z| = + x2 + y2 , θ = tan−1(y/x) Phasor-domain equivalents Table 1-5 44 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS Important Terms Provide definitions or explain the meaning of the following terms: angular velocity ω attenuation constant α attenuation factor Biot–Savart law complex conjugate complex number conductivity σ constitutive parameters continuous periodic wave Coulomb’s law dielectric constant dynamic electric dipole electric field intensity E electric flux density D electric polarization electrical force electrical permittivity ϵ electrodynamics electrostatics EM spectrum Euler’s identity forcing function fundamental dimensions instantaneous function law of conservation of electric charge LCD liquid crystal lossless or lossy medium magnetic field intensity H magnetic flux density B magnetic force magnetic permeability μ magnetostatics microwave band monochromatic nonmagnetic materials perfect conductor perfect dielectric periodic phase phase constant (wave number) β phase lag and lead phase velocity (propagation velocity) up phasor plane wave principle of linear superposition reference phase φ0 relative permittivity or dielectric constant ϵr SI system of units static transient wave velocity of light c wave amplitude wave frequency f wave period T waveform wavelength λ PROBLEMS Section 1-4: Traveling Waves ∗1.1 A 2 kHz sound wave traveling in the x direction in air was observed to have a differential pressure p(x, t) = 10 N/m2 at x = 0 and t = 50 μs. If the reference phase of p(x, t) is 36◦, find a complete expression for p(x, t). The velocity of sound in air is 330 m/s.
1.2 For the pressure wave described in Example 1-1, plot the following: (a) p(x, t) versus x at t = 0 (b) p(x, t) versus t at x = 0 Be sure to use appropriate scales for x and t so that each of your plots covers at least two cycles.
∗1.3 A harmonic wave traveling along a string is generated by an oscillator that completes 180 vibrations per minute. If it is observed that a given crest, or maximum, travels 300 cm in 10 s, what is the wavelength?
∗Answer(s) available in Appendix D.
1.4 A wave traveling along a string is given by y(x, t) = 2 sin(4πt + 10πx) (cm), where x is the distance along the string in meters and y is the vertical displacement. Determine: (a) the direction of wave travel, (b) the reference phase φ0, (c) the frequency, (d) the wavelength, and (e) the phase velocity.
1.5 Two waves, y1(t) and y2(t), have identical amplitudes and oscillate at the same frequency, but y2(t) leads y1(t) by a phase angle of 60◦. If y1(t) = 4 cos(2π × 103t), write the expression appropriate for y2(t) and plot both functions over the time span from 0 to 2 ms.
∗1.6 The height of an ocean wave is described by the function y(x, t) = 1.5 sin(0.5t −0.6x) (m).
Determine the phase velocity and wavelength, and then sketch y(x, t) at t = 2s over the range from x = 0 to x = 2λ.
PROBLEMS 45 1.7 A wave traveling along a string in the +x direction is given by y1(x, t) = A cos(ωt −βx), where x = 0 is the end of the string, which is tied rigidly to a wall, as shown in Fig. P1.7.
x x = 0 Incident wave y Figure P1.7 Wave on a string tied to a wall at x = 0 (Problem 1.7).
When wave y1(x, t) arrives at the wall, a reflected wave y2(x, t) is generated. Hence, at any location on the string, the vertical displacement ys is the sum of the incident and reflected waves: ys(x, t) = y1(x, t) + y2(x, t).
(a) Write an expression for y2(x, t), keeping in mind its direction of travel and the fact that the end of the string cannot move.
(b) Generate plots of y1(x, t), y2(x, t) and ys(x, t) versus x over the range −2λ ≤x ≤0 at ωt = π/4 and at ωt = π/2.
1.8 Two waves on a string are given by the following functions: y1(x, t) = 4 cos(20t −30x) (cm) y2(x, t) = −4 cos(20t + 30x) (cm) where x is in centimeters.
The waves are said to interfere constructively when their superposition |ys| = |y1 + y2| is a maximum, and they interfere destructively when |ys| is a minimum.
∗(a) What are the directions of propagation of waves y1(x, t) and y2(x, t)?
(b) At t = (π/50) s, at what location x do the two waves interfere constructively, and what is the corresponding value of |ys|?
(c) At t = (π/50) s, at what location x do the two waves interferedestructively, andwhatisthecorrespondingvalue of |ys|?
1.9 Give expressions for y(x, t) for a sinusoidal wave traveling along a string in the negative x direction, given that ymax = 40 cm, λ = 30 cm, f = 10 Hz, and (a) y(x, 0) = 0 at x = 0 (b) y(x, 0) = 0 at x = 3.75 cm ∗1.10 An oscillator that generates a sinusoidal wave on a string completes 20 vibrations in 50 s. The wave peak is observed to travel a distance of 2.8 m along the string in 5 s. What is the wavelength?
1.11 The vertical displacement of a string is given by the harmonic function: y(x, t) = 2 cos(16πt −20πx) (m), where x is the horizontal distance along the string in meters.
Suppose a tiny particle were attached to the string at x = 5 cm.
Obtain an expression for the vertical velocity of the particle as a function of time.
∗1.12 Given two waves characterized by y1(t) = 3 cos ωt y2(t) = 3 sin(ωt + 60◦) does y2(t) lead or lag y1(t) and by what phase angle?
1.13 The voltage of an electromagnetic wave traveling on a transmission line is given by υ(z, t) = 5e−αz sin(4π × 109t −20πz) (V), where z is the distance in meters from the generator.
(a) Find the frequency, wavelength, and phase velocity of the wave.
(b) At z = 2 m, the amplitude of the wave was measured to be 2 V. Find α.
∗1.14 Acertainelectromagneticwavetravelinginseawaterwas observedtohaveanamplitudeof 98.02 (V/m)atadepthof10m, and an amplitude of 81.87 (V/m) at a depth of 100 m. What is the attenuation constant of seawater?
1.15 A laser beam traveling through fog was observed to have an intensity of 1 (μW/m2) at a distance of 2 m from the laser gun and an intensity of 0.2 (μW/m2) at a distance of 3 m. Given that the intensity of an electromagnetic wave is proportional to the square of its electric-field amplitude, find the attenuation constant α of fog.
46 CHAPTER 1 INTRODUCTION: WAVES AND PHASORS Section 1-5: Complex Numbers 1.16 Evaluate each of the following complex numbers and express the result in rectangular form: (a) z1 = 8ejπ/3 ∗(b) z2 = √ 3 ej3π/4 (c) z3 = 2e−jπ/2 (d) z4 = j3 (e) z5 = j−4 (f) z6 = (1 −j)3 (g) z7 = (1 −j)1/2 1.17 Complex numbers z1 and z2 are given z1 = 3 −j2 z2 = −4 + j3 (a) Express z1 and z2 in polar form.
(b) Find |z1| by first applying Eq. (1.41) and then by applying Eq. (1.43).
∗(c) Determine the product z1z2 in polar form.
(d) Determine the ratio z1/z2 in polar form.
(e) Determine z3 1 in polar form.
1.18 Complex numbers z1 and z2 are given by z1 = −3 + j2 z2 = 1 −j2 Determine (a) z1z2, (b) z1/z∗ 2, (c) z2 1, and (d) z1z∗ 1, all in polar form.
1.19 If z = −2 + j4, determine the following quantities in polar form: (a) 1/z (b) z3 ∗(c) |z|2 (d) Im{z} (e) Im{z∗} 1.20 Findcomplexnumbers t = z1+z2 ands = z1 −z2, both in polar form, for each of the following pairs: (a) z1 = 2 + j3 and z2 = 1 −j2 (b) z1 = 3 and z2 = −j3 (c) z1 = 3∠30◦and z2 = 3∠ −30◦ ∗(d) z1 = 3∠30◦and z2 = 3∠ −150◦ 1.21 Complex numbers z1 and z2 are given by z1 = 5∠ −60◦ z2 = 4∠ 45◦.
(a) Determine the product z1z2 in polar form.
(b) Determine the product z1z∗ 2 in polar form.
(c) Determine the ratio z1/z2 in polar form.
(d) Determine the ratio z∗ 1/z∗ 2 in polar form.
(e) Determine √z1 in polar form.
∗1.22 If z = 3 −j5, find the value of ln(z).
1.23 If z = 3 −j4, find the value of ez.
1.24 If z = 3ejπ/6, find the value of ez.
Section 1-6: Phasors ∗1.25 A voltage source given by υs(t) = 25 cos(2π × 103t −30◦) (V) is connected to a series RC load as shown in Fig. 1-20. If R = 1 M and C = 200 pF, obtain an expression for υc(t), the voltage across the capacitor.
1.26 Find the phasors of the following time functions: (a) υ(t) = 9 cos(ωt −π/3) (V) (b) υ(t) = 12 sin(ωt + π/4) (V) (c) i(x, t) = 5e−3x sin(ωt + π/6) (A) ∗(d) i(t) = −2 cos(ωt + 3π/4) (A) (e) i(t) = 4 sin(ωt + π/3) + 3 cos(ωt −π/6) (A) 1.27 Find the instantaneous time sinusoidal functions corresponding to the following phasors: (a) V = −5ejπ/3 (V) (b) V = j6e−jπ/4 (V) (c) ˜ I = (6 + j8) (A) ∗(d) ˜ I = −3 + j2 (A) (e) ˜ I = j (A) (f) ˜ I = 2ejπ/6 (A) PROBLEMS 47 1.28 A series RLC circuit is connected to a generator with a voltage υs(t) = V0 cos(ωt + π/3) (V).
(a) Write the voltage loop equation in terms of the current i(t), R, L, C, and υs(t).
(b) Obtain the corresponding phasor-domain equation.
(c) Solve the equation to obtain an expression for the phasor current ˜ I.
1.29 The voltage source of the circuit shown in Fig. P1.29 is given by υs(t) = 25 cos(4 × 104t −45◦) (V).
Obtain an expression for iL(t), the current flowing through the inductor.
R1 = 20 Ω, R2 = 30 Ω, L = 0.4 mH υs(t) L i R1 R2 iL iR2 A + − Figure P1.29 Circuit for Problem 1.29.
C H A P T E R 2 Transmission Lines Chapter Contents 2-1 General Considerations, 49 2-2 Lumped-Element Model, 52 2-3 Transmission-Line Equations, 56 2-4 Wave Propagation on a Transmission Line, 57 2-5 The Lossless Microstrip Line, 60 2-6 The Lossless Transmission Line: General Considera-tions, 65 2-7 Wave Impedance of the Lossless Line, 75 2-8 Special Cases of the Lossless Line, 78 TB3 Microwave Ovens, 82 2-9 Power Flow on a Lossless Transmission Line, 86 2-10 The Smith Chart, 88 2-11 Impedance Matching, 101 2-12 Transients on Transmission Lines, 111 TB4 EM Cancer Zappers, 112 Chapter 2 Summary, 122 Problems, 124 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Calculate the line parameters, characteristic impedance, and propagation constant of coaxial, two-wire, parallel-plate, and microstrip transmission lines.
2. Determine the reflection coefficient at the load-end of the transmission line, the standing-wave pattern, and the locations of voltage and current maxima and minima.
3. Calculate the amount of power transferred from the generator to the load through the transmission line.
4. Use the Smith chart to perform transmission-line calculations.
5. Analyze the response of a transmission line to a voltage pulse.
2-1 GENERAL CONSIDERATIONS 49 Sending-end port A ~ A' B B' Transmission line Load circuit Generator circuit Receiving-end port + − Vg Zg ZL Figure 2-1 A transmission line is a two-port network connecting a generator circuit at the sending end to a load at the receiving end.
2-1 General Considerations In most electrical engineering curricula, the study of electromagnetics is preceded by one or more courses on electrical circuits.
In this book, we use this background to build a bridge between circuit theory and electromagnetic theory.
The bridge is provided by transmission lines, the topic of this chapter. By modeling transmission lines in the form of equivalent circuits, we can use Kirchhoff’s voltage and current laws to develop wave equations whose solutions provide an understanding of wave propagation, standing waves, and power transfer. Familiarity with these concepts facilitates the presentation of material in later chapters.
Although the notion of transmission lines may encompass all structures and media that serve to transfer energy or information between two points, including nerve fibers in the human body and fluids and solids that support the propagation of mechanical pressure waves, this chapter focuses on transmission lines that guide electromagnetic signals.
Such transmission lines include telephone wires, coaxial cables carrying audio and video information to TV sets or digital data to computer monitors, microstrips printed on microwave circuit boards, and optical fibers carrying light waves for the transmission of data at very high rates.
Fundamentally, a transmission line is a two-port network, with each port consisting of two terminals, as illustrated in Fig. 2-1. One of the ports, the line’s sending end, is connected to a source (also called the generator). The other port, the line’s receiving end, is connected to a load. The source connected to the transmission line’s sending end may be any circuit generating an output voltage, such as a radar transmitter, an amplifier, or a computer terminal operating in transmission mode.
From circuit theory, a dc source can be represented by a Th´ evenin-equivalent generator circuit consisting of a generator voltage Vg in series with a generator resistance Rg, as shown in Fig. 2-1. In the case of alternating-current (ac) signals, the generator circuit is represented by a voltage phasor Vg and an impedance Zg.
The load circuit, or simply the load, may be an antenna in the case of radar, a computer terminal operating in the receiving mode, the input terminals of an amplifier, or any output circuit whose input terminals can be represented by an equivalent load impedance ZL.
2-1.1 The Role of Wavelength In low-frequency circuits, circuit elements usually are interconnected using simple wires.
In the circuit shown in Fig. 2-2, for example, the generator is connected to a simple RC load via a pair of wires. In view of our definition in the preceding paragraphs of what constitutes a transmission line, we pose the following question: Is the pair of wires between terminals AA′ and terminals BB′ a transmission line? If so, under what set of circumstances should we explicitly treat the C R i Vg VAA' l A A' B B' + + − − VBB' + − Transmission line Figure 2-2 Generator connected to an RC circuit through a transmission line of length l.
50 CHAPTER 2 TRANSMISSION LINES pair of wires as a transmission line, as opposed to ignoring their presence altogether and treating the circuit as only an RC-load connected to a generator Vg? The answer to the first question is: yes, the pair of wires does constitute a transmission line. And the answer to the second question is: the factors that determine whether or not we should treat the wires as a transmission line are governed by the length of the line l and the frequency f of the signal provided by the generator. (As we will see later, the determining factor is the ratio of the length l to the wavelength λ of the wave propagating on the transmission line between the source and load terminals AA′ and BB′, respectively.) If the generatorvoltageiscosinusoidalintime, thenthevoltageacross the input terminals AA′ is VAA′ = Vg(t) = V0 cos ωt (V), (2.1) where ω = 2πf is the angular frequency, and if we assume that thecurrentflowingthroughthewirestravelsatthespeedoflight, c = 3 × 108 m/s, then the voltage across the output terminals BB′ will have to be delayed in time relative to that across AA′ by the travel delay-time l/c. Thus, assuming no ohmic losses in the transmission line and ignoring other transmission line effects discussed later in this chapter, VBB′(t) = VAA′(t −l/c) = V0 cos [ω(t −l/c)] = V0 cos(ωt −φ0), (2.2) with φ0 = ωl c (rad).
(2.3) Thus, the time delay associated with the length of the line l manifests itself as a constant phase shift φ0 in the argument of the cosine.
Let us compare VBB′ to VAA′ at t = 0 for an ultralow-frequency electronic circuit operating at a frequency f = 1 kHz.
For a typical wire length l = 5 cm, Eqs. (2.1) and (2.2) give VAA′ = V0 and VBB′ = V0 cos(2πf l/c) = 0.999999999998 V0. Hence, for all practical purposes, the presence of the transmission line may be ignored and terminal AA′ may be treated as identical with BB′ so far as its voltage is concerned. On the other hand, had the line beena20kmlongtelephonecablecarryinga1kHzvoicesignal, then the same calculation would have led to VBB′ = 0.91V0, a deviation of 9%. The determining factor is the magnitude of φ0 = ωl/c. From Eq. (1.27), the velocity of propagation up of a traveling wave is related to the oscillation frequency f and the wavelength λ by up = f λ (m/s).
In the present case, up = c. Hence, the phase delay φ0 = ωl c = 2πf l c = 2π l λ radians.
(2.4) ▶When l/λ is very small, transmission-line effects may be ignored, but when l/λ ≳0.01, it may be necessary to account not only for the phase shift due to the time delay, but also for the presence of reflected signals that may have been bounced back by the load toward the generator. ◀ Power loss on the line and dispersive effects may need to be considered as well.
▶A dispersive transmission line is one on which the wave velocity is not constant as a function of the frequency f . ◀ This means that the shape of a rectangular pulse, which through Fourieranalysiscanbedecomposedintomanysinusoidalwaves of different frequencies, gets distorted as it travels down the line because its different frequency components do not propagate at the same velocity (Fig. 2-3). Preservation of pulse shape is very important in high-speed data transmission, not only between Dispersionless line Short dispersive line Long dispersive line Figure 2-3 A dispersionless line does not distort signals passing through it regardless of its length, whereas a dispersive line distorts the shape of the input pulses because the different frequency components propagate at different velocities.
The degree of distortion is proportional to the length of the dispersive line.
2-1 GENERAL CONSIDERATIONS 51 TEM Transmission Lines Higher-Order Transmission Lines Metal (g) Rectangular waveguide (h) Optical fiber Concentric dielectric layers Metal Dielectric spacing w h Metal 2a 2b Dielectric spacing (a) Coaxial line Metal strip conductor Dielectric spacing w h Metal ground plane (e) Microstrip line (c) Parallel-plate line (d) Strip line Metal Dielectric spacing Dielectric spacing Metal ground plane Metal (f) Coplanar waveguide d D Dielectric spacing (b) Two-wire line Figure 2-4 A few examples of transverse electromagnetic (TEM) and higher-order transmission lines.
terminals, but also across transmission line segments fabricated within high-speed integrated circuits. At 10 GHz, for example, the wavelength is λ = 3 cm in air but only on the order of 1 cm in a semiconductor material. Hence, even lengths between devices on the order of millimeters become significant, and their presence has to be accounted for in the design of the circuit.
2-1.2 Propagation Modes A few examples of common types of transmission lines are shown in Fig. 2-4. Transmission lines may be classified into two basic types: • Transverse electromagnetic (TEM) transmission lines: Waves propagating along these lines are characterized by electric and magnetic fields that are entirely transverse to the direction of propagation.
Such an orthogonal configuration is called a TEM mode. A good example is the coaxial line shown in Fig. 2-5: the electric field is in the radial direction between the inner and outer conductors, while the magnetic field circles the inner conductor, and neitherhasacomponentalongthelineaxis(thedirectionof wave propagation). Other TEM transmission lines include the two-wire line and the parallel-plate line, both shown in Fig. 2-4. Although the fields present on a microstrip line 52 CHAPTER 2 TRANSMISSION LINES Vg Rg RL Load Cross section Magnetic field lines Electric field lines Generator Coaxial line + − Figure 2-5 In a coaxial line, the electric field is in the radial direction between the inner and outer conductors, and the magnetic field forms circles around the inner conductor. The coaxial line is a transverse electromagnetic (TEM) transmission line because both the electric and magnetic fields are orthogonal to the direction of propagation between the generator and the load.
do not adhere to the exact definition of a TEM mode, the nontransverse field components are sufficiently small (in comparison to the transverse components) to be ignored, thereby allowing the inclusion of microstrip lines in the TEM class. A common feature among TEM lines is that they consist of two parallel conducting surfaces.
• Higher-order transmission lines: Waves propagating along these lines have at least one significant field component in the direction of propagation.
Hollow conducting waveguides, dielectric rods, and optical fibers belong to this class of lines (Chapter 8).
Only TEM-mode transmission lines are treated in this chapter.
This is because they are more commonly used in practice and, fortunately, less mathematical rigor is required for treating them than is required for lines that support higher-order modes. We start our treatment by representing the transmission line in terms of a lumped-element circuit model, and then we apply Kirchhoff’s voltage and current laws to derive a pair of equations governing their behavior, known as the telegrapher’s equations.
By combining these equations, we obtain wave equations for the voltage and current at any location along the line.
Solution of the wave equations for the sinusoidal steady-state case leads to a set of formulas that can be used for solving a wide range of practical problems. In the latter part of this chapter we introduce a graphical tool known as the Smith chart, which facilitates the solution of transmission-line problems without having to perform laborious calculations involving complex numbers.
2-2 Lumped-Element Model When we draw a schematic of an electronic circuit, we use specific symbols to represent resistors, capacitors, inductors, diodes, and the like. In each case, the symbol represents the functionality of the device, rather than its shape, size, or other attributes. We shall do the same for transmission lines.
▶A transmission line will be represented by a parallel-wire configuration [Fig. 2-6(a)], regardless of its specific shape or constitutive parameters. ◀ Thus, Fig. 2-6(a) may represent a coaxial line, a two-wire line, or any other TEM line.
Drawing again on our familiarity with electronic circuits, when we analyze a circuit containing a transistor, we mimic the functionality of the transistor by an equivalent circuit composed of sources, resistors, and capacitors.
We apply the same approach to the transmission line by orienting the line along the z direction, subdividing it into differential sections each of length z [Fig. 2-6(b)] and then representing each section by an equivalent circuit, as illustrated in Fig. 2-6(c).
This representation, often called the lumped-element circuit model, consists of four basic elements, with values that henceforth will be called the transmission line parameters. These are: • R′: The combined resistance of both conductors per unit length, in /m, 2-2 LUMPED-ELEMENT MODEL 53 R' Δz L' Δz R' Δz L' Δz R' Δz L' Δz R' Δz L' Δz Δz Δz Δz Δz (a) Parallel-wire representation (b) Differential sections each Δz long (c) Each section is represented by an equivalent circuit.
G' Δz C' Δz G' Δz C' Δz G' Δz C' Δz G' Δz C' Δz Δz Δz Δz Δz Figure 2-6 Regardless of its cross-sectional shape, a TEM transmission line is represented by the parallel-wire configuration shown in (a).
To obtain equations relating voltages and currents, the line is subdivided into small differential sections (b), each of which is then represented by an equivalent circuit (c).
• L′: The combined inductance of both conductors per unit length, in H/m, • G′: The conductance of the insulation medium between the two conductors per unit length, in S/m, and • C ′: The capacitance of the two conductors per unit length, in F/m.
Whereas the four line parameters are characterized by different formulas for different types of transmission lines, the equivalent model represented by Fig. 2-6(c) is equally applicable to all TEM transmission lines. The prime superscript is used as a reminder that the line parameters are differential quantities whose units are per unit length.
Expressions for the line parameters R′, L′, G′, and C ′ are given in Table 2-1 for the three types of TEM transmission lines diagrammed in parts (a) through (c) of Fig. 2-4.
For each of these lines, the expressions are functions of two sets of parameters: (1) geometric parameters defining the cross-sectional dimensions of the given line and (2) the electromagnetic constitutive parameters of the conducting and insulating materials. The pertinent geometric parameters are: • Coaxial line [Fig. 2-4(a)]: a = outer radius of inner conductor, m b = inner radius of outer conductor, m • Two-wire line [Fig. 2-4(b)]: d = diameter of each wire, m D = spacing between wires’ centers, m • Parallel-plate line [Fig. 2-4(c)]: w = width of each plate, m h = thickness of insulation between plates, m 54 CHAPTER 2 TRANSMISSION LINES Table 2-1 Transmission-line parameters R′, L′, G′, and C ′ for three types of lines.
Parameter Coaxial Two-Wire Parallel-Plate Unit R′ Rs 2π 1 a + 1 b 2Rs πd 2Rs w /m L′ μ 2π ln(b/a) μ π ln (D/d) + (D/d)2 −1 μh w H/m G′ 2πσ ln(b/a) πσ ln (D/d) + (D/d)2 −1 σw h S/m C ′ 2πϵ ln(b/a) πϵ ln (D/d) + (D/d)2 −1 ϵw h F/m Notes: (1) Refer to Fig. 2-4 for definitions of dimensions. (2) μ, ϵ, and σ pertain to the insulating material between the conductors. (3) Rs = √πf μc/σc. (4) μc and σc pertain to the conductors.
(5) If (D/d)2 ≫1, then ln (D/d) + (D/d)2 −1 ≈ln(2D/d).
▶The pertinent constitutive parameters apply to all three lines and consist of two groups: (1) μc and σc are the magnetic permeability and electrical conductivity of the conductors, and (2) ϵ, μ, and σ are the electrical permittivity, magnetic permeability, and electrical conductivity of the insulation material separating them. ◀ Appendix B contains tabulated values for these constitutive parameters for various materials.
For the purposes of the present chapter, we need not concern ourselves with the derivations leading to the expressions in Table 2-1.
The techniques necessary for computing R′, L′, G′, and C ′ for the general case of an arbitrary two-conductor configuration are presented in later chapters.
The lumped-element model shown in Fig. 2-6(c) reflects the physical phenomena associated with the currents and voltages on any TEM transmission line.
It consists of two in-series elements, R′ and L′, and two shunt elements, G′ and C ′. To explain the lumped-element model, consider a small section of a coaxial line, as shown in Fig. 2-7. The line consists of inner and outer conductors of radii a and b separated by a material with (μc, σc) b a (ε, μ, σ) Conductors Insulating material Figure 2-7 Cross section of a coaxial line with inner conductor of radius a and outer conductor of radius b. The conductors have magnetic permeability μc and conductivity σc, and the spacing material between the conductors has permittivity ϵ, permeability μ, and conductivity σ.
permittivity ϵ, permeability μ, and conductivity σ. The two metal conductors are made of a material with conductivity σc and permeability μc. When a voltage source is connected across the terminals connected to the two conductors at the sending 2-2 LUMPED-ELEMENT MODEL 55 end of the line, currents flow through the conductors, primarily along the outer surface of the inner conductor and the inner surface of the outer conductor. The line resistance R′ accounts for the combined resistance per unit length of the inner and outer conductors. The expression for R′ is derived in Chapter 7 and is given by Eq. (7.96) as R′ = Rs 2π 1 a + 1 b (coax line) (/m), (2.5) where Rs, which represents the surface resistance of the conductors, is given by Eq. (7.92a) as Rs = πf μc σc ().
(2.6) The surface resistance depends not only on the material properties of the conductors (σc and μc), but also on the frequency f of the wave traveling on the line.
▶For a perfect conductor with σc = ∞or a high-conductivity material such that (f μc/σc) ≪1, Rs approaches zero, and so does R′. ◀ Next, let us examine the line inductance L′, which accounts for the joint inductance of both conductors. Application of Ampere’s law in Chapter 5 to the definition of inductance leads to the following expression [Eq. (5.99)] for the inductance per unit length of a coaxial line: L′ = μ 2π ln b a (coax line) (H/m).
(2.7) The line conductance G′ accounts for current flow between the outer and inner conductors, made possible by the conductivity σ of the insulator.
It is precisely because the currentflowisfromoneconductortotheotherthat G′ appearsas a shunt element in the lumped-element model. For the coaxial line, the conductance per unit length is given by Eq. (4.76) as G′ = 2πσ ln(b/a) (coax line) (S/m).
(2.8) ▶If the material separating the inner and outer conductors is a perfect dielectric with σ = 0, then G′ = 0. ◀ The last line parameter on our list is the line capacitance C ′.
When equal and opposite charges are placed on any two noncontacting conductors, a voltage difference develops between them. Capacitance is defined as the ratio of the charge to the voltage difference. For the coaxial line, the capacitance per unit length is given by Eq. (4.117) as C ′ = 2πϵ ln(b/a) (coax line) (F/m).
(2.9) All TEM transmission lines share the following useful relations: L′C ′ = μϵ (all TEM lines), (2.10) and G′ C ′ = σ ϵ (all TEM lines).
(2.11) If the insulating medium between the conductors is air, the transmission line is called an air line (e.g., coaxial air line or two-wire air line). For an air line, ϵ = ϵ0 = 8.854 × 10−12 F/m, μ = μ0 = 4π × 10−7 H/m, σ = 0, and G′ = 0.
Concept Question 2-1: What is a transmission line?
When should transmission-line effects be considered, and when may they be ignored?
Concept Question 2-2: What is the difference between dispersive and nondispersive transmission lines? What is the practical significance of dispersion?
Concept Question 2-3: What constitutes a TEM trans-mission line?
Concept Question 2-4: What purpose does the lumped-element circuit model serve? How are the line parameters R′, L′, G′, and C ′ related to the physical and electromagnetic constitutive properties of the transmission line?
56 CHAPTER 2 TRANSMISSION LINES Exercise 2-1: Use Table 2-1 to evaluate the line parameters of a two-wire air line with wires of radius 1 mm, separated by a distance of 2 cm. The wires may be treated as perfect conductors with σc = ∞.
Answer: R′ = 0, L′ = 1.20 (μH/m), G′ = 0, C ′ = 9.29 (pF/m). (See EM.) Exercise 2-2: Calculate the transmission line parameters at 1 MHz for a coaxial air line with inner and outer conductor diameters of 0.6 cm and 1.2 cm, respectively.
The conductors are made of copper (see Appendix B for μc and σc of copper).
Answer: R′ = 2.07 × 10−2 (/m), L′ = 0.14 (μH/m), G′ = 0, C ′ = 80.3 (pF/m). (See EM.) 2-3 Transmission-Line Equations A transmission line usually connects a source on one end to a load on the other. Before considering the complete circuit, however, we will develop general equations that describe the voltage across and current carried by the transmission line as a function of time t and spatial position z.
Using the lumped-element model of Fig. 2-6(c), we begin by considering a differential length z as shown in Fig. 2-8. The quantities υ(z, t) and i(z, t) denote the instantaneous voltage and current at the left end of the differential section (node N), and similarly υ(z+z, t) and i(z+z, t) denote the same quantities at node (N + 1), located at the right end of the section. Application of R' Δz L' Δz Δz i(z + Δz, t) i(z, t) Node N + − + − G' Δz C' Δz Node N + 1 υ(z, t) υ(z + Δz, t) Figure 2-8 Equivalent circuit of a two-conductor transmission line of differential length z.
Kirchhoff’s voltage law accounts for the voltage drop across the series resistance R′z and inductance L′z: υ(z, t) −R′z i(z, t) −L′ z ∂i(z, t) ∂t −υ(z + z, t) = 0.
(2.12) Upon dividing all terms by z and rearranging them, we obtain − υ(z + z, t) −υ(z, t) z = R′ i(z, t) + L′ ∂i(z, t) ∂t .
(2.13) In the limit as z →0, Eq. (2.13) becomes a differential equation: −∂υ(z, t) ∂z = R′ i(z, t) + L′ ∂i(z, t) ∂t .
(2.14) Similarly, Kirchhoff’s current law accounts for current drawn from the upper line at node (N +1) by the parallel conductance G′ z and capacitance C ′ z: i(z, t) −G′ z υ(z + z, t) −C ′ z ∂υ(z + z, t) ∂t −i(z + z, t) = 0.
(2.15) Upon dividing all terms by z and taking the limit z →0, Eq. (2.15) becomes a second-order differential equation: −∂i(z, t) ∂z = G′ υ(z, t) + C ′ ∂υ(z, t) ∂t .
(2.16) The first-order differential equations (2.14) and (2.16) are the time-domain forms of the transmission-line equations, known as the telegrapher’s equations.
Except for the last section of this chapter, our primary interest is in sinusoidal steady-state conditions. To that end, we make use of the phasor representation with a cosine reference, as outlined in Section 1-7. Thus, we define υ(z, t) = Re[ V (z) ejωt], (2.17a) i(z, t) = Re[ ˜ I(z) ejωt], (2.17b) where V (z) and ˜ I(z) are the phasor counterparts of υ(z, t) and i(z, t), respectively, each of which may be real or complex.
Upon substituting Eqs. (2.17a) and (2.17b) into Eqs. (2.14) and 2-3 WAVE PROPAGATION ON A TRANSMISSION LINE 57 (2.16), and utilizing the property given by Eq. (1.62) that ∂/∂t in the time domain is equivalent to multiplication by jω in the phasor domain, we obtain the following pair of equations: −d V (z) dz = (R′ + jωL′) ˜ I(z), (2.18a) −d ˜ I(z) dz = (G′ + jωC ′) V (z).
(2.18b) (telegrapher’s equations in phasor form) 2-4 Wave Propagation on a Transmission Line Thetwofirst-ordercoupledequations(2.18a)and(2.18b)canbe combined to give two second-order uncoupled wave equations, one for V (z) and another for ˜ I(z). The wave equation for V (z) is derived by first differentiating both sides of Eq. (2.18a) with respect to z, resulting in −d2 V (z) dz2 = (R′ + jωL′)d ˜ I(z) dz .
(2.19) Then, upon substituting Eq. (2.18b) for d ˜ I(z)/dz, Eq. (2.19) becomes d2 V (z) dz2 −(R′ + jωL′)(G′ + jωC ′) V (z) = 0, (2.20) or d2 V (z) dz2 −γ 2 V (z) = 0, (2.21) (wave equation for V (z)) where γ = (R′ + jωL′)(G′ + jωC ′) .
(2.22) (propagation constant) Application of the same steps to Eqs. (2.18a) and (2.18b) in reverse order leads to d2 ˜ I(z) dz2 −γ 2 ˜ I(z) = 0.
(2.23) (wave equation for ˜ I(z)) The second-order differential equations (2.21) and (2.23) are called wave equations for V (z) and ˜ I(z), respectively, and γ is called the complex propagation constant of the transmission line. As such, γ consists of a real part α, called the attenuation constant of the line with units of Np/m, and an imaginary part β, called the phase constant of the line with units of rad/m.
Thus, γ = α + jβ (2.24) with α = Re(γ ) = Re (R′ + jωL′)(G′ + jωC ′) (Np/m), (attenuation constant) (2.25a) β = Im(γ ) = Im (R′ + jωL′)(G′ + jωC ′) (rad/m).
(phase constant) (2.25b) In Eqs. (2.25a) and (2.25b), we choose the square-root solutions that give positive values for α and β. For passive transmission lines, α is either zero or positive. Most transmission lines, and all those considered in this chapter, are of the passive type. The gain region of a laser is an example of an active transmission line with a negative α.
The wave equations (2.21) and (2.23) have traveling wave solutions of the following form: V (z) = V + 0 e−γ z + V − 0 eγ z (V), (2.26a) ˜ I(z) = I + 0 e−γ z + I − 0 eγ z (A).
(2.26b) 58 CHAPTER 2 TRANSMISSION LINES Vg + − Zg (V0 +, I0 +)e−γz ZL z Incident wave (V0−, I0−)eγz Reflected wave Figure 2-9 In general, a transmission line can support two traveling waves, an incident wave (with voltage and current amplitudes (V + 0 , I+ 0 )) traveling along the +z direction (towards the load) and a reflected wave (with (V − 0 , I− 0 )) traveling along the −z direction (towards the source).
As shown later, the e−γ z term represents a wave propagating in the +z direction while the eγ z term represents a wave propagating in the −z direction (Fig. 2-9). Verification that these are indeed valid solutions is easily accomplished by substituting the expressions given by Eqs. (2.26a) and (2.26b), as well as their second derivatives, into Eqs. (2.21) and (2.23).
In their present form, the solutions given by Eqs. (2.26a) and (2.26b) contain four unknowns, the wave amplitudes (V + 0 , I + 0 ) ofthe+zpropagatingwaveand(V − 0 , I − 0 )ofthe−zpropagating wave. We can easily relate the current wave amplitudes, I + 0 and I − 0 , to the voltage wave amplitudes, V + 0 and V − 0 , by using Eq. (2.26a) in Eq. (2.18a) and then solving for the current ˜ I(z).
The process leads to ˜ I(z) = γ R′ + jωL′ [V + 0 e−γ z −V − 0 eγ z].
(2.27) Comparison of each term with the corresponding term in Eq. (2.26b) leads us to conclude that V + 0 I + 0 = Z0 = −V − 0 I − 0 , (2.28) where Z0 = R′ + jωL′ γ = R′ + jωL′ G′ + jωC ′ (), (2.29) is called the characteristic impedance of the line.
▶It should be noted that Z0 is equal to the ratio of the voltage amplitude to the current amplitude for each of the traveling waves individually (with an additional minus sign in the case of the −z propagating wave), but it is not equal to the ratio of the total voltage V (z) to the total current ˜ I(z), unless one of the two waves is absent. ◀ It seems reasonable that the voltage-to-current ratios of the two waves V + 0 /I + 0 and V − 0 /I − 0 , are both related to the same quantity, namely Z0, but it is not immediately obvious as to why one of the ratios is the negative of the other. The explanation, which is available in more detail in Chapter 7, is based on a directional rule that specifies the relationships between the directions of the electric and magnetic fields of aTEM wave and its direction of propagation. On a transmission line, the voltage is related to the electric field E and the current is related to the magnetic field H. To satisfy the directional rule, reversing the direction of propagation requires reversal of the direction (or polarity) of I relative to V . Hence, V − 0 /I − 0 = −V + 0 /I + 0 .
In terms of Z0, Eq. (2.27) can be cast in the form ˜ I(z) = V + 0 Z0 e−γ z −V − 0 Z0 eγ z.
(2.30) According to Eq. (2.29), the characteristic impedance Z0 is determined by the angular frequency ω of the wave traveling along the line and the four line parameters (R′, L′, G′, and C ′).
These, in turn, are determined by the line geometry and its constitutive parameters.
Consequently, the combination of Eqs. (2.26a) and (2.30) now contains only two unknowns, namely V + 0 and V − 0 , as opposed to four.
In later sections, we apply boundary conditions at the source and load ends of the transmission line to obtain expressions for the remaining wave amplitudes V + 0 and V − 0 . In general, each is a complex quantity characterized by a magnitude and a phase angle: V + 0 = |V + 0 |ejφ+, (2.31a) V − 0 = |V − 0 |ejφ−.
(2.31b) After substituting these definitions in Eq. (2.26a) and using Eq. (2.24) to decompose γ into its real and imaginary parts, we can convert back to the time domain to obtain an expression 2-4 WAVE PROPAGATION ON A TRANSMISSION LINE 59 for υ(z, t), the instantaneous voltage on the line: υ(z, t) = Re( V (z)ejωt) = Re V + 0 e−γ z + V − 0 eγ z ejωt = Re[|V + 0 |ejφ+ejωte−(α+jβ)z + |V − 0 |ejφ−ejωte(α+jβ)z] = |V + 0 |e−αz cos(ωt −βz + φ+) + |V − 0 |eαz cos(ωt + βz + φ−).
(2.32) From our review of waves in Section 1-4, we recognize the first term in Eq. (2.32) as a wave traveling in the +z direction (the coefficients of t and z have opposite signs) and the second term as a wave traveling in the −z direction (the coefficients of t and z are both positive). Both waves propagate with a phase velocity up given by Eq. (1.30): up = f λ = ω β .
(2.33) Because the wave is guided by the transmission line, λ often is called the guide wavelength. The factor e−αz accounts for the attenuation of the +z propagating wave, and the factor eαz accounts for the attenuation of the −z propagating wave.
▶The presence of two waves on the line propagating in opposite directions produces a standing wave. ◀ To gain a physical understanding of what that means, we shall first examine the relatively simple but important case of a lossless line (α = 0) and then extend the results to the more general case of a lossy transmission line (α ̸= 0). In fact, we shall devote the next several sections to the study of lossless transmission lines because in practice many lines can be designed to exhibit very low-loss characteristics.
Example 2-1: Air Line An air line is a transmission line in which air separates the two conductors, which renders G′ = 0 because σ = 0.
In addition, assume that the conductors are made of a material with high conductivity so that R′ ≈0. For an air line with a characteristic impedance of 50 and a phase constant of 20 rad/m at 700 MHz, find the line inductance L′ and the line capacitance C ′.
Solution: The following quantities are given: Z0 = 50 , β = 20 rad/m, f = 700 MHz = 7 × 108 Hz.
With R′ = G′ = 0, Eqs. (2.25b) and (2.29) reduce to β = Im (jωL′)(jωC ′) = Im jω √ L′C ′ = ω √ L′C ′ , Z0 = jωL′ jωC ′ = L′ C ′ .
The ratio of β to Z0 is β Z0 = ωC ′, or C ′ = β ωZ0 = 20 2π × 7 × 108 × 50 = 9.09 × 10−11 (F/m) = 90.9 (pF/m).
From Z0 = L′/C ′, it follows that L′ = Z2 0C ′ = (50)2 × 90.9 × 10−12 = 2.27 × 10−7 (H/m) = 227 (nH/m).
Exercise 2-3: Verify that Eq. (2.26a) indeed provides a solution to the wave equation (2.21). (See EM.) Exercise 2-4: A two-wire air line has the following line parameters: R′ = 0.404 (m/m), L′ = 2.0 (μH/m), G′ = 0, and C ′ = 5.56 (pF/m). For operation at 5 kHz, determine (a) the attenuation constant α, (b) the phase constant β, (c) the phase velocity up, and (d) the characteristic impedance Z0. (See EM.) Answer: (a) α = 3.37 × 10−7 (Np/m), (b) β = 1.05 × 10−4 (rad/m), (c) up = 3.0 × 108 (m/s), (d) Z0 = (600 −j1.9) = 600∠ −0.18◦.
60 CHAPTER 2 TRANSMISSION LINES Module 2.1 Two-Wire Line The input data specifies the geometric and electric parameters of a two-wire transmission line.
The output includes the calculated values for the line parameters, characteristic impedance Z0, and attenuation and phase constants, as well as plots of Z0 as a function of d and D.
2-5 The Lossless Microstrip Line Because its geometry is well suited for fabrication on printed circuit boards, the microstrip line is the most common interconnect configuration used in RF and microwave circuits.
It consists of a narrow, very thin strip of copper (or another good conductor) printed on a dielectric substrate overlaying a ground plane (Fig. 2-10(a)).
The presence of charges of opposite polarity on its two conducting surfaces gives rise to electric field lines between them (Fig. 2-10(b)). Also, the flow of current 2-5 THE LOSSLESS MICROSTRIP LINE 61 Module 2.2 Coaxial Cable Except for changing the geometric parameters to those of a coaxial transmission line, this module offers the same output information as Module 2.1.
through the conductors (when part of a closed circuit) generates magnetic field loops around them, as illustrated in Fig. 2-10(b) for the narrow strip. Even though the patterns of E and B are not everywhere perfectly orthogonal, they are approximately so in the region between the conductors, which is where the E and B fields are concentrated the most. Accordingly, the microstrip line is considered a quasi-TEM transmission line, which allows us to describe its voltages and currents in terms of the one-dimensional TEM model of Section 2-4, namely Eqs. (2.26) through (2.33).
The microstrip line has two geometric parameters: the width of the elevated strip, w, and the thickness (height) of 62 CHAPTER 2 TRANSMISSION LINES (a) Longitudinal view (b) Cross-sectional view with E and B field lines (c) Microwave circuit Dielectric insulator (ε, μ, σ) Conducting strip (μc , σc) Conducting ground plane (μc , σc) w h E B Figure 2-10 Microstrip line: (a) longitudinal view, (b) cross-sectional view, and (c) circuit example.
(Courtesy of Prof. Gabriel Rebeiz, U. California at San Diego.) the dielectric layer, h.
We will ignore the thickness of the conducting strip because it has a negligible influence on the propagation properties of the microstrip line, so long as the strip thickness is much smaller than the width w, which is almost always the case in practice. Also, we assume the substrate material to be a perfect dielectric with σ = 0 and the metal strip and ground plane to be perfect conductors with σc ≈∞. These two assumptions simplify the analysis considerably without incurring significant error. Finally, we set μ = μ0, which is always true for the dielectric materials used in the fabrication of microstrip lines. These simplifications reduce the number of geometric and material parameters to three, namely w, h, and ϵ.
Electric field lines always start on the conductor carrying positive charges and end on the conductor carrying negative charges.
For the coaxial, two-wire, and parallel-plate lines shown in the upper part of Fig. 2-4, the field lines are confined to the region between the conductors. A characteristic attribute of such transmission lines is that the phase velocity of a wave traveling along any one of them is given by up = c √ϵr , (2.34) where c is the velocity of light in free space and ϵr is the relative permittivity of the dielectric medium between the conductors.
In the microstrip line, even though most of the electric field lines connecting the strip to the ground plane do pass directly through the dielectric substrate, a few go through both the air region above the strip and the dielectric layer [Fig. 2-10(b)].
This nonuniform mixture can be accounted for by defining an effective relative permittivity ϵeff such that the phase velocity is given by an expression that resembles Eq. (2.34), namely up = c √ϵeff .
(2.35) Methods for calculating the propagation properties of the microstrip line are quite complicated and beyond the scope of this text.
However, it is possible to use curve-fit approximations to rigorous solutions to arrive at the following set of expressions:† ϵeff = ϵr + 1 2 + ϵr −1 2 1 + 10 s −xy , (2.36) where s is the width-to-thickness ratio, s = w h , (2.37) †D. H. Schrader, Microstrip Circuit Analysis, Prentice Hall, 1995, pp. 31– 32.
2-5 THE LOSSLESS MICROSTRIP LINE 63 and x and y are intermediate variables given by x = 0.56 ϵr −0.9 ϵr + 3 0.05 , (2.38a) y = 1 + 0.02 ln s4 + 3.7 × 10−4s2 s4 + 0.43 + 0.05 ln(1 + 1.7 × 10−4s3).
(2.38b) The characteristic impedance of the microstrip line is given by Z0 = 60 √ϵeff ln 6 + (2π −6)e−t s + 1 + 4 s2 , (2.39) with t = 30.67 s 0.75 .
(2.40) Figure 2-11 displays plots of Z0 as a function of s for various types of dielectric materials.
Z0 (Ω) εr = 2.5 εr = 6 εr = 10 s s = w/h w = strip width h = substrate thickness Microstrip 2 4 6 8 10 0 50 100 150 Figure 2-11 Plots of Z0 as a function of s for various types of dielectric materials.
The corresponding line and propagation parameters are given by R′ = 0 (because σc = ∞), (2.41a) G′ = 0 (because σ = 0), (2.41b) C ′ = √ϵeff Z0c , (2.41c) L′ = Z2 0C ′, (2.41d) α = 0 (because R′ = G′ = 0), (2.41e) β = ω c √ϵeff .
(2.41f) The preceding expressions allow us to compute the values of Z0 and the other propagation parameters when given values for ϵr, h, and w. This is exactly what is needed in order to analyze a circuit containing a microstrip transmission line. To perform the reverse process, namely to design a microstrip line by selecting values for its w and h such that their ratio yields the required value of Z0 (to satisfy design specifications), we need to express s in terms of Z0.
The expression for Z0 given by Eq. (2.39) is rather complicated, so inverting it to obtain an expression for s in terms of Z0 is rather difficult. An alternative option is to generate a family of curves similar to those displayed in Fig. 2-11 and to use them to estimate s for a specified value of Z0. A logical extension of the graphical approach is to generate curve-fit expressions that provide high-accuracy estimates of s. The error associated with the following formulas is less than 2%: (a) For Z0 ≤(44 −2ϵr) , s = w h = 2 π (q −1) −ln(2q −1) + ϵr −1 2ϵr ln(q −1) + 0.29 −0.52 ϵr (2.42) with q = 60π2 Z0 √ϵr , and (b) for Z0 ≥(44 −2ϵr) , s = w h = 8ep e2p −2 , (2.43a) 64 CHAPTER 2 TRANSMISSION LINES Module 2.3 Lossless Microstrip Line The output panel lists the values of the transmission-line parameters and displays the variation of Z0 and ϵeff with h and w.
with p = ϵr + 1 2 Z0 60 + ϵr −1 ϵr + 1 0.23 + 0.12 ϵr .
(2.43b) The foregoing expressions presume that ϵr, the relative permittivity of the dielectric substrate, has already been specified.
For typical substrate materials including Duroid, Teflon, silicon, and sapphire, ϵr ranges between 2 and 15.
2-6 THE LOSSLESS TRANSMISSION LINE: GENERAL CONSIDERATIONS 65 Example 2-2: Microstrip Line A 50 microstrip line uses a 0.5 mm thick sapphire substrate with ϵr = 9. What is the width of its copper strip?
Solution: Since Z0 = 50 > 44 −18 = 32, we should use Eq. (2.43): p = ϵr + 1 2 × Z0 60 + ϵr −1 ϵr + 1 0.23 + 0.12 ϵr = 9 + 1 2 × 50 60 + 9 −1 9 + 1 0.23 + 0.12 9 = 2.06, s = w h = 8ep e2p −2 = 8e2.06 e4.12 −2 = 1.056.
Hence, w = sh = 1.056 × 0.5 mm = 0.53 mm.
To check our calculations, we use s = 1.056 to calculate Z0 to verify that the value we obtained is indeed equal or close to 50 . With ϵr = 9, Eqs. (2.36) to (2.40) yield x = 0.55, y = 0.99, t = 12.51, ϵeff = 6.11, Z0 = 49.93 .
The calculated value of Z0 is, for all practical purposes, equal to the value specified in the problem statement.
2-6 The Lossless Transmission Line: General Considerations According to the preceding section, a transmission line is fully characterized by two fundamental parameters, its propagation constant γ and its characteristic impedance Z0, both of which are specified by the angular frequency ω and the line parameters R′, L′, G′, and C ′. In many practical situations, the transmission line can be designed to exhibit low ohmic losses by selecting conductors with very high conductivities and dielectric materials (separating the conductors) with negligible conductivities. As a result, R′ and G′ assume very small values such that R′ ≪ωL′ and G′ ≪ωC ′. These conditions allow us to set R′ = G′ ≈0 in Eq. (2.22), which yields γ = α + jβ = jω √ L′C ′ , (2.44) which in turn implies that α = 0 (lossless line), β = ω √ L′C ′ (lossless line).
(2.45) For the characteristic impedance, application of the lossless line conditions to Eq. (2.29) leads to Z0 = L′ C ′ (lossless line), (2.46) which now is a real number. Using the lossless line expression for β [Eq. (2.45)], we obtain the following expressions for the guide wavelength λ and the phase velocity up: λ = 2π β = 2π ω √ L′C ′ , (2.47) up = ω β = 1 √ L′C ′ .
(2.48) Upon using Eq. (2.10), Eqs. (2.45) and (2.48) may be rewritten as β = ω√μϵ (rad/m), (2.49) up = 1 √μϵ (m/s), (2.50) where μ and ϵ are, respectively, the magnetic permeability and electrical permittivity of the insulating material separating the conductors. Materials used for this purpose are usually characterized by a permeability μ0 = 4π × 10−7 H/m (the permeability of free space). Also, the permittivity ϵ is often specified in terms of the relative permittivity ϵr defined as ϵr = ϵ/ϵ0, (2.51) 66 CHAPTER 2 TRANSMISSION LINES where ϵ0 = 8.854 × 10−12 F/m ≈(1/36π) × 10−9 F/m is the permittivity of free space (vacuum). Hence, Eq. (2.50) becomes up = 1 √μ0ϵrϵ0 = 1 √μ0ϵ0 · 1 √ϵr = c √ϵr , (2.52) where c = 1/√μ0ϵ0 = 3 × 108 m/s is the velocity of light in free space. If the insulating material between the conductors is air, then ϵr = 1 and up = c.
In view of Eq. (2.51) and the relationship between λ and up given by Eq. (2.33), the wavelength is given by λ = up f = c f 1 √ϵr = λ0 √ϵr , (2.53) where λ0 = c/f is the wavelength in air corresponding to a frequency f . Note that, because both up and λ depend on ϵr, the choice of the type of insulating material used in a transmission line is dictated not only by its mechanical properties, but by its electrical properties as well.
According to Eq. (2.52), if ϵr of the insulating material is independent of f (which usually is the case for commonly used TEM lines), the same independence applies to up.
▶If sinusoidal waves of different frequencies travel on a transmission line with the same phase velocity, the line is called nondispersive. ◀ This is an important feature to consider when digital data are transmitted in the form of pulses. A rectangular pulse or a series of pulses is composed of many Fourier components with different frequencies. If the phase velocity is the same for all frequency components (or at least for the dominant ones), then the pulse’s shape does not change as it travels down the line.
In contrast, the shape of a pulse propagating in a dispersive medium becomes progressively distorted, and the pulse length increases (stretches out) as a function of the distance traveled in the medium (Fig. 2-3), thereby imposing a limitation on the maximum data rate (which is related to the length of the individual pulses and the spacing between adjacent pulses) that can be transmitted through the medium without loss of information.
Table 2-2 provides a list of the expressions for γ , Z0, and up for the general case of a lossy line and for several types of lossless lines. The expressions for the lossless lines are based on the equations for L′ and C ′ given in Table 2-1.
Exercise 2-5: For a lossless transmission line, λ = 20.7 cm at 1 GHz. Find ϵr of the insulating material.
Answer: ϵr = 2.1. (See EM.) Exercise 2-6: A lossless transmission line uses a dielectric insulating material with ϵr = 4.
If its line capacitance is C ′ = 10 (pF/m), find (a) the phase velocity up, (b) the line inductance L′, and (c) the characteristic impedance Z0.
Answer: (a)up = 1.5×108 (m/s), (b)L′ = 4.45(μH/m), (c) Z0 = 667.1 . (See EM.) 2-6.1 Voltage Reflection Coefficient With γ = jβ for the lossless line, Eqs. (2.26a) and (2.30) for the total voltage and current become V (z) = V + 0 e−jβz + V − 0 ejβz, (2.54a) ˜ I(z) = V + 0 Z0 e−jβz −V − 0 Z0 ejβz.
(2.54b) These expressions contain two unknowns, V + 0 and V − 0 .
According to Section 1-7.2, an exponential factor of the form e−jβz is associated with a wave traveling in the positive z direction, from the source (sending end) to the load (receiving end). Accordingly, we refer to it as the incident wave, with V + 0 as its voltage amplitude. Similarly, the term containing V − 0 ejβz represents a reflected wave with voltage amplitude V − 0 , traveling along the negative z direction, from the load to the source.
To determine V + 0 and V − 0 , we need to consider the lossless transmission line in the context of the complete circuit, including a generator circuit at its input terminals and a load at its output terminals, as shown in Fig. 2-12. The line, of length l, is terminated in an arbitrary load impedance ZL.
▶For convenience, the reference of the spatial coordinate z is chosen such that z = 0 corresponds to the location of the load. ◀ 2-6 THE LOSSLESS TRANSMISSION LINE: GENERAL CONSIDERATIONS 67 Table 2-2 Characteristic parameters of transmission lines.
Propagation Phase Characteristic Constant Velocity Impedance γ = α + jβ up Z0 General case γ = (R′ + jωL′)(G′ + jωC ′) up = ω/β Z0 = (R′ + jωL′) (G′ + jωC ′) Lossless α = 0, β = ω√ϵr/c up = c/√ϵr Z0 = L′/C ′ (R′ = G′ = 0) Lossless coaxial α = 0, β = ω√ϵr/c up = c/√ϵr Z0 = 60/√ϵr ln(b/a) Lossless α = 0, β = ω√ϵr/c up = c/√ϵr Z0 = 120/√ϵr two-wire · ln[(D/d) + (D/d)2 −1] Z0 ≈ 120/√ϵr ln(2D/d), if D ≫d Lossless α = 0, β = ω√ϵr/c up = c/√ϵr Z0 = 120π/√ϵr (h/w) parallel-plate Notes: (1) μ = μ0, ϵ = ϵrϵ0, c = 1/√μ0ϵ0, and √μ0/ϵ0 ≈(120π) , where ϵr is the relative permittivity of insulating material. (2) For coaxial line, a and b are radii of inner and outer conductors. (3) For two-wire line, d = wire diameter and D = separation between wire centers. (4) For parallel-plate line, w = width of plate and h = separation between the plates.
At the sending end, at z = −l, the line is connected to a sinusoidal voltage source with phasor voltage Vg and internal impedance Zg. Since z points from the generator to the load, positive values of z correspond to locations beyond the load, and therefore are irrelevant to our circuit. In future sections, we will find it more convenient to work with a spatial dimension that also starts at the load, but whose direction is opposite of z.
We shall call it the distance from the load d and define it as d = −z, as shown in Fig. 2-12.
The phasor voltage across the load, VL, and the phasor current through it, ˜ IL, are related by the load impedance ZL as ZL = VL ˜ IL .
(2.55) The voltage VL is the total voltage on the line V (z) given by Eq. (2.54a), and ˜ IL is the total current ˜ I(z) given by Eq. (2.54b), both evaluated at z = 0: VL = V (z=0) = V + 0 + V − 0 , (2.56a) ˜ IL = ˜ I(z=0) = V + 0 Z0 −V − 0 Z0 .
(2.56b) Using these expressions in Eq. (2.55), we obtain ZL = V + 0 + V − 0 V + 0 −V − 0 Z0.
(2.57) Solving for V − 0 gives V − 0 = ZL −Z0 ZL + Z0 V + 0 .
(2.58) 68 CHAPTER 2 TRANSMISSION LINES Vg Ii Zg Z0 ZL ~ Vi ~ ~ + + VL ~ IL ~ + Transmission line Generator Load z = −l z = 0 z d = l d d = 0 − − − Figure 2-12 Transmission line of length l connected on one end to a generator circuit and on the other end to a load ZL.
The load is located at z = 0 and the generator terminals are at z = −l. Coordinate d is defined as d = −z.
▶The ratio of the amplitudes of the reflected and incident voltage waves at the load is known as the voltage reflection coefficient . ◀ From Eq. (2.58), it follows that = V − 0 V + 0 = ZL −Z0 ZL + Z0 = ZL/Z0 −1 ZL/Z0 + 1 = zL −1 zL + 1 (dimensionless), (2.59) where zL = ZL Z0 (2.60) is the normalized load impedance.
In many transmission-line problems, we can streamline the necessary computation by normalizing all impedances in the circuit to the characteristic impedance Z0.
Normalized impedances are denoted by lowercase letters.
In view of Eq. (2.28), the ratio of the current amplitudes is I − 0 I + 0 = −V − 0 V + 0 = −.
(2.61) ▶We note that whereas the ratio of the voltage amplitudes is equal to , the ratio of the current amplitudes is equal to −. ◀ The reflection coefficient is governed by a single parameter, the normalized load impedance zL. As indicated by Eq. (2.46), Z0 of a lossless line is a real number. However, ZL is in general a complex quantity, as in the case of a series RL circuit, for example, for which ZL = R + jωL. Hence, in general also is complex and given by = ||ejθr, (2.62) where || is the magnitude of and θr is its phase angle. Note that || ≤1.
▶A load is said to be matched to a transmission line if ZL = Z0 because then there will be no reflection by the load ( = 0 and V − 0 = 0). ◀ On the other hand, when the load is an open circuit (ZL = ∞), = 1 and V − 0 = V + 0 , and when it is a short circuit (ZL = 0), = −1 and V − 0 = −V + 0 (Table 2-3).
Example 2-3: Reflection Coefficient of a Series RC Load A 100 transmission line is connected to a load consisting of a 50 resistor in series with a 10 pF capacitor. Find the reflection coefficient at the load for a 100 MHz signal.
Solution: The following quantities are given (Fig. 2-13): RL = 50 , CL = 10 pF = 10−11 F, Z0 = 100 , f = 100 MHz = 108 Hz.
2-6 THE LOSSLESS TRANSMISSION LINE: GENERAL CONSIDERATIONS 69 Table 2-3 Magnitude and phase of the reflection coefficient for various types of load.
The normalized load impedance zL = ZL/Z0 = (R + jX)/Z0 = r + jx, where r = R/Z0 and x = X/Z0 are the real and imaginary parts of zL, respectively.
Reflection Coefficient = ||ejθr Load || θr Z0 ZL = (r + jx)Z0 (r −1)2 + x2 (r + 1)2 + x2 1/2 tan−1 x r −1 −tan−1 x r + 1 Z0 Z0 0 (no reflection) irrelevant Z0 (short) 1 ±180◦(phase opposition) Z0 (open) 1 0 (in-phase) Z0 jX = jωL 1 ±180◦−2 tan−1 x Z0 jX = −j ωC 1 ±180◦+ 2 tan−1 x The normalized load impedance is zL = ZL Z0 = RL −j/(ωCL) Z0 = 1 100 50 −j 1 2π × 108 × 10−11 = (0.5 −j1.59) .
From Eq. (2.59), the voltage reflection coefficient is = zL −1 zL + 1 = 0.5 −j1.59 −1 0.5 −j1.59 + 1 = −0.5 −j1.59 1.5 −j1.59 = −1.67ej72.6◦ 2.19e−j46.7◦= −0.76ej119.3◦.
This result may be converted into the form of Eq. (2.62) by replacing the minus sign with e−j180◦. Thus, = 0.76ej119.3◦e−j180◦= 0.76e−j60.7◦= 0.76∠ −60.7◦, or || = 0.76, θr = −60.7◦.
CL RL 50 Ω Z0 = 100 Ω 10 pF A A' Transmission line Figure 2-13 RC load (Example 2-3).
Example 2-4: || for Purely Reactive Load Show that || = 1 for a lossless line connected to a purely reactive load.
Solution: The load impedance of a purely reactive load is ZL = jXL.
70 CHAPTER 2 TRANSMISSION LINES From Eq. (2.59), the reflection coefficient is = ZL −Z0 ZL + Z0 = jXL −Z0 jXL + Z0 = −(Z0 −jXL) (Z0 + jXL) = − Z2 0 + X2 L e−jθ Z2 0 + X2 L ejθ = −e−j2θ, where θ = tan−1 XL/Z0. Hence || = | −e−j2θ| = [(e−j2θ)(e−j2θ)∗]1/2 = 1.
Exercise 2-7: A 50 lossless transmission line is terminated in a load with impedance ZL = (30−j200) .
Calculate the voltage reflection coefficient at the load.
Answer: = 0.93∠ −27.5◦. (See EM.) Exercise 2-8: A 150 lossless line is terminated in a capacitor with impedance ZL = −j30 . Calculate .
Answer: = 1∠ −157.4◦. (See EM.) 2-6.2 Standing Waves Using the relation V − 0 = V + 0 in Eqs. (2.54a) and (2.54b) yields V (z) = V + 0 (e−jβz + ejβz), (2.63a) ˜ I(z) = V + 0 Z0 (e−jβz −ejβz).
(2.63b) These expressions now contain only one, yet to be determined, unknown, V + 0 . Before we proceed to solve for V + 0 , however, let us examine the physical meaning underlying these expressions.
We begin by deriving an expression for | V (z)|, the magnitude of V (z). Upon using Eq. (2.62) in Eq. (2.63a) and applying the relation | V (z)| = [ V (z) V ∗(z)]1/2, where V ∗(z) is the complex conjugate of V (z), we have | V (z)| = V + 0 (e−jβz + ||ejθrejβz) · (V + 0 )∗(ejβz + ||e−jθre−jβz) 1/2 = |V + 0 | 1 + ||2 + ||(ej(2βz+θr) + e−j(2βz+θr)) 1/2 = |V + 0 | 1 + ||2 + 2|| cos(2βz + θr) 1/2 , (2.64) where we have used the identity ejx + e−jx = 2 cos x (2.65) for any real quantity x. To express the magnitude of V as a function of d instead of z, we replace z with −d on the right-hand side of Eq. (2.64): | V (d)| = |V + 0 | 1 + ||2 + 2|| cos(2βd −θr) 1/2 . (2.66) By applying the same steps to Eq. (2.63b), a similar expression can be derived for | ˜ I(d)|, the magnitude of the current ˜ I(d): | ˜ I(d)| = |V + 0 | Z0 [1 + ||2 −2|| cos(2βd −θr)]1/2.
(2.67) 2-6 THE LOSSLESS TRANSMISSION LINE: GENERAL CONSIDERATIONS 71 (b) |I(d)| versus d ~ 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 V |V(d)| ~ d d |I(z)| ~ 0 5 10 15 20 25 30 mA dmax |V|max ~ |V|min ~ |I|max ~ |I|min ~ Voltage min max max min Current dmin λ 3λ 4 λ 4 λ 2 λ 3λ 4 λ 4 λ 2 (a) |V(d)| versus d ~ Figure 2-14 Standing-wave pattern for (a) | V (d)| and (b) | ˜ I(d)| for a lossless transmission line of characteristic impedance Z0 = 50 , terminated in a load with a reflection coefficient = 0.3ej30◦.
The magnitude of the incident wave |V + 0 | = 1 V.
The standing-wave ratio is S = | V |max/| V |min = 1.3/0.7 = 1.86.
The variations of | V (d)| and | ˜ I(d)| as a function of d and the position on the line relative to the load (at d = 0), are illustrated in Fig. 2-14 for a line with |V + 0 | = 1 V, || = 0.3, θr = 30◦, and Z0 = 50 . The sinusoidal patterns are called standing waves and are caused by the interference of the two traveling waves. The maximum value of the standing-wave pattern of | V (d)| corresponds to the position on the line at which the incident and reflected waves are in-phase [2βd −θr = 2nπ in Eq. (2.66)] and therefore add constructively to give a value equal to (1 + ||)|V + 0 | = 1.3 V. The minimum value of | V (d)| occurswhenthetwowavesinterferedestructively, whichoccurs when the incident and reflected waves are in phase-opposition [2βd−θr = (2n+1)π]. In this case, | V (d)| = (1−||)|V + 0 | = 0.7 V.
▶Whereas the repetition period is λ for the incident andreflectedwavesconsideredindividually, therepetition period of the standing-wave pattern is λ/2. ◀ The standing-wave pattern describes the spatial variation of the magnitude of V (d) as a function of d.
If one were to observe the variation of the instantaneous voltage as a function of time at location d = dmax in Fig. 2-14, that variation would be as cos ωt and would have an amplitude equal to 1.3 V [i.e., υ(t) would oscillate between −1.3 V and +1.3 V]. Similarly, the instantaneous voltage υ(d, t) at any location d will be sinusoidal with amplitude equal to | V (d)| at that d. Interactive Module 2.4† provides a highly recommended simulation tool for gaining better understanding of the standing-wave patterns for V (d) and ˜ I(d) and the dynamic behavior of υ(d, t) and i(d, t).
Close inspection of the voltage and current standing-wave patterns shown in Fig. 2-14 reveals that the two patterns are in phase opposition (when one is at a maximum, the other is at a minimum, and vice versa). This is a consequence of the fact that the third term in Eq. (2.66) is preceded by a plus sign, whereas the third term in Eq. (2.67) is preceded by a minus sign.
The standing-wave patterns shown in Fig. 2-14 are for = 0.3 ej30◦. Thepeak-to-peakvariationofthepattern(| V |min to | V |max) depends on ||, which in general can vary between 0 and 1. For the special case of a matched line with ZL = Z0, we have || = 0 and | V (d)| = |V + 0 | for all values of d, as shown in Fig. 2-15(a).
▶With no reflected wave present, there are no interference and no standing waves. ◀ The other end of the || scale, at || = 1, corresponds to when the load is a short circuit ( = −1) or an open circuit ( = 1). The standing-wave patterns for those two cases are shown in Figs. 2-15(b) and (c); both exhibit maxima of 2|V + 0 | and minima equal to zero, but the two patterns are spatially shifted relative to each other by a distance of λ/4. A purely reactive load (capacitor or inductor) also satisfies the condition || = 1, but θr is generally neither zero nor 180◦(Table 2-3).
Exercise 2.9 examines the standing-wave pattern for a lossless line terminated in an inductor.
†At em.eecs.umich.edu 72 CHAPTER 2 TRANSMISSION LINES 0 |V(d)| ~ |V(d)| ~ |V(d)| d d d ~ 0 0 |V0 +| (a) ZL = Z0 (b) ZL = 0 (short circuit) (c) ZL = (open circuit) 2|V0 +| 2|V0 +| Matched line Short-circuited line Open-circuited line λ/2 λ/2 8 λ 3λ 4 λ 4 λ 2 λ 3λ 4 λ 4 λ 2 λ 3λ 4 λ 4 λ 2 Figure 2-15 Voltage standing-wave patterns for (a) a matched load, (b) a short-circuited line, and (c) an open-circuited line.
Now let us examine the maximum and minimum values of the voltage magnitude. From Eq. (2.66), | V (d)| is a maximum when the argument of the cosine function is equal to zero or a multiple of 2π. Let us denote dmax as the distance from the load at which | V (d)| is a maximum. It then follows that | V (d)| = | V |max = |V + 0 |[1 + ||], (2.68) when 2βdmax −θr = 2nπ, (2.69) with n = 0 or a positive integer. Solving Eq. (2.69) for dmax, we have dmax = θr + 2nπ 2β = θrλ 4π + nλ 2 , n = 1, 2, . . .
if θr < 0, n = 0, 1, 2, . . .
if θr ≥0, (2.70) where we have used β = 2π/λ.
The phase angle of the voltage reflection coefficient, θr, is bounded between −π and π radians.
If θr ≥0, the first voltage maximum occurs at dmax = θrλ/4π, corresponding to n = 0, but if θr < 0, the first physically meaningful maximum occurs at dmax = (θrλ/4π) + λ/2, corresponding to n = 1.
Negative values of dmax correspond to locations past the end of the line and therefore have no physical significance.
Similarly, the minima of | V (d)| occur at distances dmin for which the argument of the cosine function in Eq. (2.66) is equal to (2n + 1)π, which gives the result | V |min = |V + 0 |[1 −||], when (2βdmin −θr) = (2n + 1)π, (2.71) with −π ≤θr ≤π.
The first minimum corresponds to n = 0. The spacing between a maximum dmax and the adjacent minimum dmin is λ/4. Hence, the first minimum occurs at dmin = dmax + λ/4, if dmax < λ/4, dmax −λ/4, if dmax ≥λ/4.
(2.72) ▶The locations on the line corresponding to voltage maxima correspond to current minima, and vice versa. ◀ The ratio of | V |max to | V |min is called the voltage standing-wave ratio S, which from Eqs. (2.68) and (2.71) is given by S = | V |max | V |min = 1 + || 1 −|| (dimensionless).
(2.73) This quantity, which often is referred to by its acronym, VSWR, or the shorter acronym SWR, provides a measure of the mismatch between the load and the transmission line; for a matched load with = 0, we get S = 1, and for a line with || = 1, S = ∞.
2-6 THE LOSSLESS TRANSMISSION LINE: GENERAL CONSIDERATIONS 73 Module 2.4 Transmission-Line Simulator Upon specifying the requisite input data—including the load impedance at d = 0 and the generator voltage and impedance at d = l—this module provides a wealth of output information about the voltage and current waveforms along the trasmission line. You can view plots of the standing wave patterns for voltage and current, the time and spatial variations of the instantaneous voltage υ(d, t) and current i(d, t), and other related quantities.
Concept Question 2-5: The attenuation constant α represents ohmic losses. In view of the model given in Fig. 2-6(c), what should R′ and G′ be in order to have no losses? Verify your expectation through the expression for α given by Eq. (2.25a).
Concept Question 2-6: How is the wavelength λ of the wave traveling on the transmission line related to the free-space wavelength λ0?
Concept Question 2-7: When is a load matched to a transmission line? Why is it important?
Concept Question 2-8: What is a standing-wave pat-tern? Why is its period λ/2 and not λ?
Concept Question 2-9: Whatistheseparationbetween the location of a voltage maximum and the adjacent current maximum on the line?
Exercise 2-9: Use Module 2.4 to generate the voltage and current standing-wave patterns for a 50 line of length 1.5λ, terminated in an inductance with ZL = j140 .
74 CHAPTER 2 TRANSMISSION LINES Answer: See Module 2.4 display.
Example 2-5: Standing-Wave Ratio A 50 transmission line is terminated in a load with ZL = (100 + j50) . Find the voltage reflection coefficient and the voltage standing-wave ratio.
Solution: From Eq. (2.59), is given by = zL −1 zL + 1 = (2 + j1) −1 (2 + j1) + 1 = 1 + j1 3 + j1 .
Converting the numerator and denominator to polar form yields = 1.414ej45◦ 3.162ej18.4◦= 0.45ej26.6◦.
Using the definition for S given by Eq. (2.73), we have S = 1 + || 1 −|| = 1 + 0.45 1 −0.45 = 2.6.
Example 2-6: Measuring ZL A slotted-line probe is an instrument used to measure the unknown impedance of a load, ZL.
A coaxial slotted line contains a narrow longitudinal slit in the outer conductor. A small probe inserted in the slit can be used to sample the magnitude of the electric field and, hence, the magnitude | V (d)| ofthevoltageontheline(Fig.2-16). Bymovingtheprobealong Vg ~ 40 cm 30 cm 20 cm 10 cm Probe tip Slit Sliding probe To detector + Zg ZL − Figure 2-16 Slotted coaxial line (Example 2-6).
the length of the slotted line, it is possible to measure | V |max and | V |min and the distances from the load at which they occur. Use of Eq. (2.73), namely S = | V |max/| V |min, provides the voltage standing-wave ratio S. Measurements with a Z = 50 slotted line terminated in an unknown load impedance determined that S = 3. The distance between successive voltage minima was found to be 30 cm, and the first voltage minimum was located at 12 cm from the load. Determine the load impedance ZL.
Solution: The following quantities are given: Z0 = 50 , S = 3, dmin = 12 cm.
Since the distance between successive voltage minima is λ/2, λ = 2 × 0.3 = 0.6 m, and β = 2π λ = 2π 0.6 = 10π 3 (rad/m).
From Eq. (2.73), solving for || in terms of S gives || = S −1 S + 1 = 3 −1 3 + 1 = 0.5.
Next, we use the condition given by Eq. (2.71) to find θr: 2βdmin −θr = π, for n = 0 (first minimum), which gives θr = 2βdmin −π = 2 × 10π 3 × 0.12 −π = −0.2π (rad) = −36◦.
2-7 WAVE IMPEDANCE OF THE LOSSLESS LINE 75 Hence, = ||ejθr = 0.5e−j36◦= 0.405 −j0.294.
Solving Eq. (2.59) for ZL, we have ZL = Z0 1 + 1 − = 50 1 + 0.405 −j0.294 1 −0.405 + j0.294 = (85 −j67) .
Exercise 2-10: If = 0.5∠ −60◦and λ = 24 cm, find the locations of the voltage maximum and minimum nearest to the load.
Answer: dmax = 10 cm, dmin = 4 cm. (See EM.) Exercise 2-11: A 140 lossless line is terminated in a load impedance ZL = (280+j182) . If λ = 72 cm, find (a) the reflection coefficient , (b) the voltage standing-wave ratio S, (c) the locations of voltage maxima, and (d) the locations of voltage minima.
Answer: (a) = 0.5∠ 29◦, (b) S = 3.0, (c) dmax = 2.9 cm + nλ/2, (d) dmin = 20.9 cm + nλ/2, where n = 0, 1, 2, . . . . (See EM.) 2-7 Wave Impedance of the Lossless Line The standing-wave patterns indicate that on a mismatched line the voltage and current magnitudes are oscillatory with position along the line and in phase opposition with each other. Hence, the voltage to current ratio, called the wave impedance Z(d), must vary with position also. Using Eqs. (2.63a) and (2.63b) with z = −d, Z(d) = V (d) ˜ I(d) = V + 0 [ejβd + e−jβd] V + 0 [ejβd −e−jβd] Z0 = Z0 1 + e−j2βd 1 −e−j2βd = Z0 1 + d 1 −d (), (2.74) where we define d = e−j2βd = ||ejθre−j2βd = ||ej(θr−2βd) (2.75) asthephase-shiftedvoltagereflectioncoefficient, meaningthat d has the same magnitude as , but its phase is shifted by 2βd relative to that of .
▶Z(d) is the ratio of the total voltage (incident-and reflected-wave voltages) to the total current at any point d on the line, in contrast with the characteristic impedance of the line Z0, which relates the voltage and current of each of the two waves individually (Z0 = V + 0 /I + 0 = −V − 0 /I − 0 ). ◀ In the circuit of Fig. 2-17(a), at terminals BB′ at an arbitrary location d on the line, Z(d) is the wave impedance of the line when “looking” to the right (i.e., towards the load). Application of the equivalence principle allows us to replace the segment to the right of terminals BB′ with a lumped impedance of value Z(d), as depicted in Fig. 2-17(b).
From the standpoint of the input circuit to the left of terminals BB′, the two circuit configurations are electrically identical.
(a) Actual circuit + Vg ~ Zg A B A′ B′ (b) Equivalent circuit + Vg ~ Zg Z(d ) Z(d ) A B C A′ B′ C′ ZL Z0 d = l 0 d − − Figure 2-17 The segment to the right of terminals BB′ can be replaced with a discrete impedance equal to the wave impedance Z(d).
76 CHAPTER 2 TRANSMISSION LINES Of particular interest in many transmission-line problems is the input impedance at the source end of the line, at d = l, which is given by Zin = Z(l) = Z0 1 + l 1 −l .
(2.76) with l = e−j2βl = ||ej(θr−2βl).
(2.77) By replacing with Eq. (2.59) and using the relations ejβl = cos βl + j sin βl, (2.78a) e−jβl = cos βl −j sin βl, (2.78b) Eq. (2.76) can be written in terms of zL as Zin = Z0 zL cos βl + j sin βl cos βl + jzL sin βl = Z0 zL + j tan βl 1 + jzL tan βl .
(2.79) From the standpoint of the generator circuit, the transmission line can be replaced with an impedance Zin, as shown in Fig. 2-18. The phasor voltage across Zin is given by Vi = ˜ IiZin = VgZin Zg + Zin , (2.80) Simultaneously, from the standpoint of the transmission line, thevoltageacrossitattheinputofthelineisgivenby Eq.(2.63a) with z = −l: Vi = V (−l) = V + 0 [ejβl + e−jβl].
(2.81) Equating Eq. (2.80) to Eq. (2.81) and then solving for V + 0 leads to V + 0 = VgZin Zg + Zin 1 ejβl + e−jβl .
(2.82) This completes the solution of the transmission-line wave equations, given by Eqs. (2.21) and (2.23), for the special case of a lossless transmission line. We started out with the general solutions given by Eq. (2.26), which included four unknown amplitudes, V + 0 , V − 0 , I + 0 , and I − 0 . We then determined that Vg Ii Zg Zin A A′ A A′ Z0 ZL ~ Ii ~ IL ~ Vi ~ ~ + + + VL ~ + Transmission line Generator Load z = −l d = l z = 0 d = 0 Vg Zg Zin Vi ~ ~ + − − − − − Figure 2-18 At the generator end, the terminated transmission line can be replaced with the input impedance of the line Zin.
Z0 = V + 0 /I + 0 = −V − 0 /I − 0 , thereby reducing the unknowns to the two voltage amplitudes only. Upon applying the boundary condition at the load, we were able to relate V − 0 to V + 0 through , and, finally, by applying the boundary condition at the source, we obtained an expression for V + 0 .
Example 2-7: Complete Solution for υ(z, t) and i(z, t) A 1.05 GHz generator circuit with series impedance Zg = 10 and voltage source given by υg(t) = 10 sin(ωt + 30◦) (V) is connected to a load ZL = (100 + j50) through a 50 , 67 cm long lossless transmission line. The phase velocity of the line is 0.7c, where c is the velocity of light in a vacuum. Find υ(z, t) and i(z, t) on the line.
2-7 WAVE IMPEDANCE OF THE LOSSLESS LINE 77 Solution: From the relationship up = λf , we find the wavelength λ = up f = 0.7 × 3 × 108 1.05 × 109 = 0.2 m, and βl = 2π λ l = 2π 0.2 × 0.67 = 6.7π = 0.7π = 126◦, where we have subtracted multiples of 2π.
The voltage reflection coefficient at the load is = ZL −Z0 ZL + Z0 = (100 + j50) −50 (100 + j50) + 50 = 0.45ej26.6◦.
With reference to Fig. 2-18, the input impedance of the line, given by Eq. (2.76), is Zin = Z0 1 + l 1 −l = Z0 1 + e−j2βl 1 −e−j2βl = 50 1 + 0.45ej26.6◦e−j252◦ 1 −0.45ej26.6◦e−j252◦ = (21.9 + j17.4) .
Rewriting the expression for the generator voltage with the cosine reference, we have υg(t) = 10 sin(ωt + 30◦) = 10 cos(90◦−ωt −30◦) = 10 cos(ωt −60◦) = Re[10e−j60◦ejωt] = Re[ Vgejωt] (V).
Hence, the phasor voltage Vg is given by Vg = 10 e−j60◦= 10∠ −60◦ (V).
Application of Eq. (2.82) gives V + 0 = VgZin Zg + Zin 1 ejβl + e−jβl = 10e−j60◦(21.9 + j17.4) 10 + 21.9 + j17.4 · (ej126◦+ 0.45ej26.6◦e−j126◦)−1 = 10.2ej159◦ (V).
Using Eq. (2.63a) with z = −d, the phasor voltage on the line is V (d) = V + 0 (ejβd + e−jβd) = 10.2ej159◦(ejβd + 0.45ej26.6◦e−jβd), and the corresponding instantaneous voltage υ(d, t) is υ(d, t) = Re[ V (d) ejωt] = 10.2 cos(ωt + βd + 159◦) + 4.55 cos(ωt −βd + 185.6◦) (V).
Similarly, Eq. (2.63b) leads to ˜ I(d) = 0.20ej159◦(ejβd −0.45ej26.6◦e−jβd), i(d, t) = 0.20 cos(ωt + βd + 159◦) + 0.091 cos(ωt −βd + 185.6◦) (A).
78 CHAPTER 2 TRANSMISSION LINES Module 2.5 Wave and Input Impedance The wave impedance, Z(d) = V (d)/ ˜ I(d), exhibits a cyclical pattern as a function of position along the line. This module displays plots of the real and imaginary parts of Z(d), specifies the locations of the voltage maximum and minimum nearest to the load, and provides other related information.
2-8 Special Cases of the Lossless Line We often encounter situations involving lossless transmission lines with particular terminations or lines whose lengths lead to particularly useful line properties. We now consider some of these special cases.
2-8.1 Short-Circuited Line The transmission line shown in Fig. 2-19(a) is terminated in a short circuit, ZL = 0. Consequently, the voltage reflection coefficient defined by Eq. (2.59) is = −1, and the voltage standing-wave ratio given by Eq. (2.73) is S = ∞.
With 2-8 SPECIAL CASES OF THE LOSSLESS LINE 79 −1 1 0 0 0 (b) (a) −1 1 (c) (d) Voltage Current Impedance Z0 l d d d l 0 short circuit λ 3λ 4 λ 4 λ 2 λ 3λ 4 λ 4 λ 2 λ 3λ 4 λ 4 λ 2 2jV0 + Vsc(d) ~ ~ Isc(d) Z0 2V0+ jZ0 Zin sc Zin sc Figure 2-19 Transmission line terminated in a short circuit: (a) schematic representation, (b) normalized voltage on the line, (c) normalized current, and (d) normalized input impedance.
z = −d and = −1 in Eqs. (2.63a) and (2.63b), and = −1 in Eq. (2.74), the voltage, current, and wave impedance on a short-circuited lossless transmission line are given by Vsc(d) = V + 0 [ejβd −e−jβd] = 2jV + 0 sin βd, (2.83a) ˜ Isc(d) = V + 0 Z0 [ejβd + e−jβd] = 2V + 0 Z0 cos βd, (2.83b) Zsc(d) = Vsc(d) ˜ Isc(d) = jZ0 tan βd.
(2.83c) The voltage Vsc(d) is zero at the load (d = 0), as it should be for a short circuit, and its amplitude varies as sin βd. In contrast, the current ˜ Isc(d) is a maximum at the load and it varies as cos βd. Both quantities are displayed in Fig. 2-19 as a function of d.
Denoting Zsc in as the input impedance of a short-circuited line of length l, Zsc in = Vsc(l) ˜ Isc(l) = jZ0 tan βl.
(2.84) A plot of Zsc in/jZ0 versus l is shown in Fig. 2-19(d). For the short-circuited line, if its length is less than λ/4, its impedance is equivalent to that of an inductor, and if it is between λ/4 and λ/2, it is equivalent to that of a capacitor.
In general, the input impedance Zin of a line terminated in an arbitrary load has a real part, called the input resistance Rin, and an imaginary part, called the input reactance Xin: Zin = Rin + jXin.
(2.85) In the case of the short-circuited lossless line, the input impedance is purely reactive (Rin = 0).
If tan βl ≥0, the line appears inductive to the source, acting like an equivalent inductor Leq whose impedance equals Zsc in. Thus, jωLeq = jZ0 tan βl, if tan βl ≥0, (2.86) or Leq = Z0 tan βl ω (H).
(2.87) The minimum line length l that would result in an input impedance Zsc in equivalent to that of an inductor with inductance Leq is l = 1 β tan−1 ωLeq Z0 (m).
(2.88) 80 CHAPTER 2 TRANSMISSION LINES Similarly, if tan βl ≤0, the input impedance is capacitive, in which case the line acts like an equivalent capacitor with capacitance Ceq such that 1 jωCeq = jZ0 tan βl, if tan βl ≤0, (2.89) or Ceq = − 1 Z0ω tan βl (F).
(2.90) Since l is a positive number, the shortest length l for which tan βl ≤0 corresponds to the range π/2 ≤βl ≤π.
Hence, the minimum line length l that would result in an input impedance Zsc in equivalent to that of a capacitor of capacitance Ceq is l = 1 β π −tan−1 1 ωCeqZ0 (m).
(2.91) ▶These results imply that, through proper choice of the length of a short-circuited line, we can make them into equivalent capacitors and inductors of any desired reactance. ◀ Such a practice is indeed common in the design of microwave circuits and high-speed integrated circuits, because making an actual capacitor or inductor often is much more difficult than fabricating a shorted microstrip transmission line on a circuit board.
Example 2-8: Equivalent Reactive Elements Choose the length of a shorted 50 lossless transmission line (Fig.2-20)suchthatitsinputimpedanceat2.25GHzisidentical to that of a capacitor with capacitance Ceq = 4 pF. The wave velocity on the line is 0.75c.
Solution: We are given up = 0.75c = 0.75 × 3 × 108 = 2.25 × 108 m/s, Z0 = 50 , f = 2.25 GHz = 2.25 × 109 Hz, Ceq = 4 pF = 4 × 10−12 F.
Z0 Zin sc Zin sc short circuit l Zc = 1 jωCeq Figure 2-20 Shorted line as equivalent capacitor (Example 2-8).
The phase constant is β = 2π λ = 2πf up = 2π × 2.25 × 109 2.25 × 108 = 62.8 (rad/m).
From Eq. (2.89), it follows that tan βl = − 1 Z0ωCeq = − 1 50 × 2π × 2.25 × 109 × 4 × 10−12 = −0.354.
The tangent function is negative when its argument is in the second or fourth quadrants.
The solution for the second quadrant is βl1 = 2.8 rad or l1 = 2.8 β = 2.8 62.8 = 4.46 cm, and the solution for the fourth quadrant is βl2 = 5.94 rad or l2 = 5.94 62.8 = 9.46 cm.
We also could have obtained the value of l1 by applying Eq. (2.91). The length l2 is greater than l1 by exactly λ/2.
In fact, any length l = 4.46 cm + nλ/2, where n is a positive integer, also is a solution.
2-8 SPECIAL CASES OF THE LOSSLESS LINE 81 2-8.2 Open-Circuited Line With ZL = ∞, as illustrated in Fig. 2-21(a), we have = 1, S = ∞, and the voltage, current, and input impedance are given by Voc(d) = V + 0 [ejβd + e−jβd] = 2V + 0 cos βd, (2.92a) ˜ Ioc(d) = V + 0 Z0 [ejβd −e−jβd] = 2jV + 0 Z0 sin βd, (2.92b) Zoc in = Voc(l) ˜ Ioc(l) = −jZ0 cot βl.
(2.93) Plots of these quantities are displayed in Fig. 2-21 as a function of d.
2-8.3 Application of Short-Circuit/ Open-Circuit Technique A network analyzer is a radio-frequency (RF) instrument capable of measuring the impedance of any load connected to its input terminal. When used to measure (1) Zsc in, the input impedance of a lossless line when terminated in a short circuit, and (2) Zoc in , the input impedance of the line when terminated in an open circuit, the combination of the two measurements can be used to determine the characteristic impedance of the line Z0 and its phase constant β. Indeed, the product of the expressions given by Eqs. (2.84) and (2.93) gives Z0 = + Zsc in Zoc in , (2.94) and the ratio of the same expressions leads to tan βl = −Zsc in Zoc in .
(2.95) Because of the π phase ambiguity associated with the tangent function, the length l should be less than or equal to λ/2 to provide an unambiguous result.
−1 1 0 0 0 2V0 + Voc(d) ~ (b) (a) 0 −1 1 2jV0+ Ioc(d) Z0 ~ (c) jZ0 Zin oc (d) Z0 d d d l l Zin oc Voltage Current Impedance λ 3λ 4 λ 4 λ 2 λ 3λ 4 λ 4 λ 2 3λ 4 λ 4 λ 2 λ Figure 2-21 Transmission line terminated in an open circuit: (a) schematic representation, (b) normalized voltage on the line, (c) normalized current, and (d) normalized input impedance.
82 TECHNOLOGY BRIEF 3: MICROWAVE OVENS Technology Brief 3: Microwave Ovens Percy Spencer, while working for Raytheon in the 1940s on the design and construction of magnetrons for radar, observed that a chocolate bar that had unintentionally been exposed to microwaves had melted in his pocket. The process of cooking by microwave was patented in 1946 and by the 1970s, microwave ovens had become standard household items.
Microwave Absorption A microwave is an electromagnetic wave whose frequency lies in the 300 MHz–300 GHz range (see Fig. 1-16.) When a material containing water is exposed to microwaves, the water molecule reacts by rotating itself so as to align its own electric dipole along the direction of the oscillating electric field of the microwave. The rapid vibration motion creates heat in the material, resulting in the conversion of microwave energy into thermal energy. The absorption coefficient of water, α(f ), exhibits a microwave spectrum that depends on the temperature of the water and the concentration of dissolved salts and sugars present in it. If the frequency f is chosen such that α(f ) is high, the water-containing material absorbs much of the microwave energy passing through it and converts it to heat. However, it also means that most of the energy is absorbed by a thin surface layer of the material, with not much energy remaining to heat deeper layers. The penetration depth δp of a material, defined as δp = 1/2α, is a measure of how deep the power carried by an EM wave can penetrate into the material. Approximately 95% of the microwave energy incident upon a material is Food with 50% water Food with 20% water Chocolate bar 50 40 30 20 10 0 1 2 3 4 5 Microwave oven frequency (2.54 GHz) Frequency (GHz) Penetration Depth δp (cm) T = 20◦C Pure water 3δp 95% of energy absorbed in this layer 678 FigureTF3-1 Penetration depth as a function of frequency (1–5 GHz) for pure water and two foods with different water contents.
TECHNOLOGY BRIEF 3: MICROWAVE OVENS 83 absorbed by the surface layer of thickness 3δp. Figure TF3-1 displays calculated spectra of δp for pure water and two materials with different water contents.
▶The frequency most commonly used in microwave ovens is 2.54 GHz. The magnitude of δs at 2.54 GHz varies beween ∼2 cm for pure water and 8 cm for a material with a water content of only 20%. ◀ This is a practical range for cooking food in a microwave oven; at much lower frequencies, the food is not a good absorber of energy (in addition to the fact that the design of the magnetron and the oven cavity become problematic), and at much higher frequencies, the microwave energy cooks the food very unevenly (mostly the surface layer). Whereas microwaves are readily absorbed by water, fats, and sugars, they can penetrate through most ceramics, glass, or plastics without loss of energy, thereby imparting little or no heat to those materials.
Oven Operation To generate high-power microwaves (∼700 watts) the microwave oven uses a magnetron tube (Fig. TF3-2), which requires the application of a voltage on the order of 4000 volts. The typical household voltage of 115 volts is increased to the required voltage level through a high-voltage transformer. The microwave energy generated by the magnetron is transferred into a cooking chamber designed to contain the microwaves within it through the use of metal surfaces and safety Interlock switches.
▶Microwaves are reflected by metal surfaces, so they can bounce around the interior of the chamber or be absorbed by the food, but not escape to the outside. ◀ If the oven door is made of a glass panel, a metal screen or a layer of conductive mesh is attached to it to ensure the necessary shielding; microwaves cannot pass through the metal screen if the mesh width is much smaller than the wavelength of the microwave (λ ≈12 cm at 2.5 GHz). In the chamber, the microwave energy establishes a standing-wave pattern, which leads to an uneven distribution. This is mitigated by using a rotating metal stirrer that disperses the microwave energy to different parts of the chamber.
115 V Metal screen Magnetron Interlock switch Stirrer 4,000 V High-voltage transformer FigureTF3-2 Microwave oven cavity.
84 CHAPTER 2 TRANSMISSION LINES Example 2-9: Measuring Z0 and β Find Z0 and β of a 57 cm long lossless transmission line whoseinputimpedancewasmeasuredasZsc in = j40.42 when terminated in a short circuit and as Zoc in = −j121.24 when terminated in an open circuit. From other measurements, we know that the line is between 3 and 3.25 wavelengths long.
Solution: From Eqs. (2.94) and (2.95), Z0 = + Zsc in Zoc in = (j40.42)(−j121.24) = 70 , tan βl = −Zsc in Zoc in = 1 3 .
Since l is between 3λ and 3.25λ, βl = (2πl/λ) is between 6π radians and (13π/2) radians. This places βl in the first quadrant (0 to π/2) radians. Hence, the only acceptable solution for tan βℓ= √1/3 is βl = π/6 radians.
This value, however, does not include the 2π multiples associated with the integer λ multiples of l. Hence, the true value of βl is βl = 6π + π 6 = 19.4 (rad), in which case β = 19.4 0.57 = 34 (rad/m).
2-8.4 Lines of Length l = nλ/2 If l = nλ/2, where n is an integer, tan βl = tan [(2π/λ) (nλ/2)] = tan nπ = 0.
Consequently, Eq. (2.79) reduces to Zin = ZL, for l = nλ/2, (2.96) whichmeansthatahalf-wavelengthline(oranyintegermultiple of λ/2) does not modify the load impedance.
2-8.5 Quarter-Wavelength Transformer Another case of interest is when the length of the line is a quarter-wavelength (or λ/4 + nλ/2, where n = 0 or a positive integer), corresponding to βl = (2π/λ)(λ/4) = π/2.
From Eq. (2.79), the input impedance becomes Zin = Z2 0 ZL , for l = λ/4 + nλ/2.
(2.97) The utility of such a quarter-wave transformer is illustrated by Example 2-10.
Example 2-10: λ/4 Transformer A 50 lossless transmission line is to be matched to a resistive load impedance with ZL = 100 via a quarter-wave section as shown in Fig. 2-22, thereby eliminating reflections along the feedline. Find the required characteristic impedance of the quarter-wave transformer.
Solution: To eliminate reflections at terminal AA′, the input impedance Zin looking into the quarter-wave line should be equal to Z01, the characteristic impedance of the feedline. Thus, Zin = 50 . From Eq. (2.97), Zin = Z2 02 ZL , or Z02 = Zin ZL = √ 50 × 100 = 70.7 .
Whereas this eliminates reflections on the feedline, it does not eliminate them on the λ/4 line. However, since the lines are lossless, all the power incident on AA′ will end up getting transferred into the load ZL.
Z01 = 50 Ω ZL = 100 Ω Zin Z02 A A' λ/4 λ/4 transformer Feedline Figure 2-22 Configuration for Example 2-10.
2-8 SPECIAL CASES OF THE LOSSLESS LINE 85 In this example, ZL is purely resistive. To apply the λ/4 transformer technique to match a transmission line to a load with a complex impedance, a slightly more elaborate procedure is required (Section 2-11).
2-8.6 Matched Transmission Line: ZL = Z0 For a matched lossless transmission line with ZL = Z0, (1) the input impedance Zin = Z0 for all locations d on the line, (2) = 0, and (3) all the incident power is delivered to the load, regardless of the line length l. A summary of the properties of standing waves is given in Table 2-4.
Table 2-4 Properties of standing waves on a lossless transmission line.
Voltage maximum | V |max = |V + 0 |[1 + ||] Voltage minimum | V |min = |V + 0 |[1 −||] Positions of voltage maxima (also positions of current minima) dmax = θrλ 4π + nλ 2 , n = 0, 1, 2, . . .
Position of first maximum (also position of first current minimum) dmax = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ θrλ 4π , if 0 ≤θr ≤π θrλ 4π + λ 2 , if −π ≤θr ≤0 Positions of voltage minima (also positions of current maxima) dmin = θrλ 4π + (2n + 1)λ 4 , n = 0, 1, 2, . . .
Position of first minimum (also position of first current maximum) dmin = λ 4 1 + θr π Input impedance Zin = Z0 zL + j tan βl 1 + jzL tan βl = Z0 1 + l 1 −l Positions at which Zin is real at voltage maxima and minima Zin at voltage maxima Zin = Z0 1 + || 1 −|| Zin at voltage minima Zin = Z0 1 −|| 1 + || Zin of short-circuited line Zsc in = jZ0 tan βl Zin of open-circuited line Zoc in = −jZ0 cot βl Zin of line of length l = nλ/2 Zin = ZL, n = 0, 1, 2, . . .
Zin of line of length l = λ/4 + nλ/2 Zin = Z2 0/ZL, n = 0, 1, 2, . . .
Zin of matched line Zin = Z0 |V + 0 | = amplitude of incident wave; = ||ejθr with −π < θr < π; θr in radians; l = e−j2βl.
86 CHAPTER 2 TRANSMISSION LINES Concept Question 2-10: What is the difference be-tween the characteristic impedance Z0 and the input impedance Zin? When are they the same?
Concept Question 2-11: What is a quarter-wave trans-former? How can it be used?
Concept Question 2-12: A lossless transmission line of length l is terminated in a short circuit. If l < λ/4, is the input impedance inductive or capacitive?
Concept Question 2-13: What is the input impedance of an infinitely long line?
Concept Question 2-14: If the input impedance of a lossless line is inductive when terminated in a short circuit, will it be inductive or capacitive when the line is terminated in an open circuit?
Exercise 2-12: A 50 lossless transmission line uses an insulating material with ϵr = 2.25. When terminated in an open circuit, how long should the line be for its input impedance to be equivalent to a 10-pF capacitor at 50 MHz?
Answer: l = 9.92 cm. (See EM.) Exercise 2-13: A 300 feedline is to be connected to a 3 m long, 150 line terminated in a 150 resistor. Both lines are lossless and use air as the insulating material, and the operating frequency is 50 MHz.
Determine (a) the input impedance of the 3 m long line, (b) the voltage standing-wave ratio on the feedline, and (c) the characteristic impedance of a quarter-wave transformer were it to be used between the two lines in order to achieve S = 1 on the feedline. (See EM.) Answer: (a) Zin = 150 , (b) S = 2, (c) Z0 = 212.1 .
2-9 Power Flow on a Lossless Transmission Line Our discussion thus far has focused on the voltage and current attributes of waves propagating on a transmission line. Now we examine the flow of power carried by the incident and reflected waves. We begin by reintroducing Eqs. (2.63a) and (2.63b) with z = −d: V (d) = V + 0 (ejβd + e−jβd), (2.98a) ˜ I(d) = V + 0 Z0 (ejβd −e−jβd).
(2.98b) In these expressions, the first terms represent the incident-wave voltage and current, and the terms involving represent the reflected-wave voltage and current.
The time-domain expressions for the voltage and current at location d from the load are obtained by transforming Eq. (2.98) to the time domain: υ(d, t) = Re[ V ejωt] = Re[|V + 0 |ejφ+(ejβd + ||ejθre−jβd)ejωt] = |V + 0 |[cos(ωt + βd + φ+) + || cos(ωt −βd + φ+ + θr)], (2.99a) i(d, t) = |V + 0 | Z0 [cos(ωt + βd + φ+) −|| cos(ωt −βd + φ+ + θr)], (2.99b) where we used the relations V + 0 = |V + 0 |ejφ+ and = ||ejθr, both introduced earlier as Eqs. (2.31a) and (2.62), respectively.
2-9.1 Instantaneous Power The instantaneous power carried by the transmission line is equal to the product of υ(d, t) and i(d, t): P(d, t) = υ(d, t) i(d, t) = |V + 0 |[cos(ωt + βd + φ+) + || cos(ωt −βd + φ+ + θr)] × |V + 0 | Z0 [cos(ωt + βd + φ+) −|| cos(ωt −βd + φ+ + θr)] = |V + 0 |2 Z0 [cos2(ωt + βd + φ+) −||2 cos2(ωt −βd + φ+ + θr)] (W).
(2.100) 2-9 POWER FLOW ON A LOSSLESS TRANSMISSION LINE 87 Per our earlier discussion in connection with Eq. (1.31), if the signsprecedingωt andβd intheargumentofthecosinetermare both positive or both negative, then the cosine term represents a wave traveling in the negative d direction. Since d points from the load to the generator, the first term in Eq. (2.100) represents the instantaneous incident power traveling towards the load. This is the power that would be delivered to the load in the absence of wave reflection (when = 0). Because βd is preceded by a minus sign in the argument of the cosine of the second term in Eq. (2.100), that term represents the instantaneous reflected power traveling in the +d direction, away from the load. Accordingly, we label these two power components P i(d, t) = |V + 0 |2 Z0 cos2(ωt + βd + φ+) (W), (2.101a) P r(d, t) = −||2 |V + 0 |2 Z0 cos2(ωt −βd + φ+ + θr) (W).
(2.101b) Using the trigonometric identity cos2 x = 1 2(1 + cos 2x), the expressions in Eq. (2.101) can be rewritten as P i(d, t) = |V + 0 |2 2Z0 [1 + cos(2ωt + 2βd + 2φ+)], (2.102a) P r(d, t) = −||2 |V + 0 |2 2Z0 [1 + cos(2ωt −2βd + 2φ+ + 2θr)].
(2.102b) We note that in each case, the instantaneous power consists of a dc (non–time-varying) term and an ac term that oscillates at an angular frequency of 2ω.
▶The power oscillates at twice the rate of the voltage or current. ◀ 2-9.2 Time-Average Power From a practical standpoint, we usually are more interested in the time-average power flowing along the transmission line, Pav(d), than in the instantaneous power P(d, t).
To compute Pav(d), we can use a time-domain approach or a computationally simpler phasor-domain approach.
For completeness, we consider both.
Time-domain approach The time-average power is equal to the instantaneous power averaged over one time period T = 1/f = 2π/ω.
For the incident wave, its time-average power is P i av(d) = 1 T T !
0 P i(d, t) dt = ω 2π 2π/ω !
0 P i(d, t) dt.
(2.103) Upon inserting Eq. (2.102a) into Eq. (2.103) and performing the integration, we obtain P i av = |V + 0 |2 2Z0 (W), (2.104) which is identical with the dc term of P i(d, t) given by Eq. (2.102a). A similar treatment for the reflected wave gives P r av = −||2 |V + 0 |2 2Z0 = −||2P i av.
(2.105) ▶The average reflected power is equal to the average incident power, diminished by a multiplicative factor of ||2. ◀ Note that the expressions for P i av and P r av are independent of d, which means that the time-average powers carried by the incident and reflected waves do not change as they travel along the transmission line. This is as expected, because the transmission line is lossless.
The net average power flowing towards (and then absorbed by) the load shown in Fig. 2-23 is Pav = P i av + P r av = |V + 0 |2 2Z0 [1 −||2] (W). (2.106) 88 CHAPTER 2 TRANSMISSION LINES Vg Zg ZL ~ Transmission line + Pav i d = l d = 0 Pav = |Γ|2 Pav r i − Figure 2-23 The time-average power reflected by a load connected to a lossless transmission line is equal to the incident power multiplied by ||2.
Phasor-domain approach For any propagating wave with voltage and current phasors V and ˜ I, a useful formula for computing the time-average power is Pav = 1 2Re V · ˜ I ∗ , (2.107) where ˜ I ∗is the complex conjugate of ˜ I. Application of this formula to Eqs. (2.98a) and (2.98b) gives Pav = 1 2 Re V + 0 (ejβd + e−jβd) · V + 0 ∗ Z0 (e−jβd −∗ejβd) = 1 2 Re |V + 0 |2 Z0 (1 −||2 + e−j2βd −∗ej2βd) = |V + 0 |2 2Z0 {[1 −||2] + Re [||e−j(2βd−θr) −||ej(2βd−θr)]} = |V + 0 |2 2Z0 {[1 −||2] + ||[cos(2βd −θr) −cos(2βd −θr)]} = |V + 0 |2 2Z0 [1 −||2], (2.108) which is identical to Eq. (2.106).
Exercise 2-14: For a 50 lossless transmission line terminated in a load impedance ZL = (100 + j50) , determine the fraction of the average incident power reflected by the load.
Answer: 20%. (See EM.) Exercise 2-15: For the line of Exercise 2-14, what is the magnitude of the average reflected power if |V + 0 | = 1 V?
Answer: P r av = 2 (mW). (See EM.) Concept Question 2-15: According to Eq. (2.102b), the instantaneous value of the reflected power depends on the phase of the reflection coefficient θr, but the average reflected power given by Eq. (2.105) does not. Explain.
Concept Question 2-16: What is the average power deliveredbyalosslesstransmission linetoareactiveload?
Concept Question 2-17: What fraction of the incident power is delivered to a matched load?
Concept Question 2-18: Verify that 1 T T !
0 cos2 2πt T + βd + φ dt = 1 2 , regardless of the values of d and φ, so long as neither is a function of t.
2-10 The Smith Chart The Smith chart, developed by P. H. Smith in 1939, is a widely used graphical tool for analyzing and designing transmission-line circuits. Even though it was originally intended to facilitate calculationsinvolvingcompleximpedances, theSmithcharthas become an important avenue for comparing and characterizing the performance of microwave circuits. As the material in this and the next section demonstrates, use of the Smith chart not only avoids tedious manipulations of complex numbers, but it also allows an engineer to design impedance-matching circuits with relative ease.
2-10 THE SMITH CHART 89 Γr Γi θr = 90o θrA = 53o θr = 0o θr = 180o θrB = 202o θr = 270o or −90o −1 −1 −0.9 −0.7 −0.5 −0.3 0.1 0.1 −0.1 −0.2 −0.3 −0.4 −0.5 −0.6 −0.7 −0.8 −0.9 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.2 0.3 0.5 0.7 0.9 1 |ΓB| = 0.54 B |ΓA| = 0.5 |Γ| = 1 A D C Unit circle Open-circuit load Short-circuit load Figure 2-24 The complex plane. Point A is at A = 0.3 + j0.4 = 0.5ej53◦, and point B is at B = −0.5 −j0.2 = 0.54ej202◦. The unit circle corresponds to || = 1. At point C, = 1, corresponding to an open-circuit load, and at point D, = −1, corresponding to a short circuit.
2-10.1 Parametric Equations The reflection coefficient is, in general, a complex quantity composed of a magnitude || and a phase angle θr or, equivalently, a real part r and an imaginary part i, = ||ejθr = r + ji , (2.109) where r = || cos θr, (2.110a) i = || sin θr.
(2.110b) The Smith chart lies in the complex plane. In Fig. 2-24, point A represents a reflection coefficient A = 0.3 + j0.4 or, equivalently, |A| = [(0.3)2 + (0.4)2]1/2 = 0.5 and θrA = tan−1(0.4/0.3) = 53◦.
Similarly, point B represents B = −0.5 −j0.2, or |B| = 0.54 and θrB = 202◦ [or, equivalently, θrB = (360◦−202◦) = −158◦].
90 CHAPTER 2 TRANSMISSION LINES ▶When both r and i are negative, θr is in the third quadrant in the r–i plane.
Thus, when using θr = tan−1(i/ r) to compute θr, it may be necessary to add or subtract 180◦to obtain the correct value of θr. ◀ The unit circle shown in Fig. 2-24 corresponds to || = 1. Because || ≤1 for a transmission line terminated with a passive load, only that part of the r–i plane that lies within the unit circle is useful to us; hence, future drawings will be limited to the domain contained within the unit circle.
Impedances on a Smith chart are represented by their values normalized to Z0, the line’s characteristic impedance. From = ZL/Z0 −1 ZL/Z0 + 1 = zL −1 zL + 1 , (2.111) the inverse relation is zL = 1 + 1 − .
(2.112) The normalized load impedance zL is, in general, a complex quantity composed of a normalized load resistance rL and a normalized load reactance xL: zL = rL + jxL.
(2.113) Using Eqs. (2.109) and (2.113) in Eq. (2.112), we have rL + jxL = (1 + r) + ji (1 −r) −ji , (2.114) which can be manipulated to obtain explicit expressions for rL and xL in terms of r and i.
This is accomplished by multiplying the numerator and denominator of the right-hand side of Eq. (2.114) by the complex conjugate of the denominator and then separating the result into real and imaginary parts.
These steps lead to rL = 1 −2 r −2 i (1 −r)2 + 2 i , (2.115a) xL = 2i (1 −r)2 + 2 i .
(2.115b) Equation(2.115a)impliesthatthereexistmanycombinationsof values for r and i that yield the same value for the normalized load resistance rL.
For example, (r, i) = (0.33, 0) gives rL = 2, as does (r, i) = (0.5, 0.29), as well as an infinite number of other combinations.
In fact, if we were to plot in the r–i plane all possible combinations of r and i corresponding to rL = 2, we would obtain the circle labeled rL = 2 in Fig. 2-25. Similar circles can be obtained for other values of rL. After some algebraic manipulations, Eq. (2.115a) can be rearranged into the following parametric equation for the circle in the r–i plane corresponding to a given value of rL: r − rL 1 + rL 2 + 2 i = 1 1 + rL 2 .
(2.116) The standard equation for a circle in the x–y plane with center at (x0, y0) and radius a is (x −x0)2 + (y −y0)2 = a2.
(2.117) Comparison of Eq. (2.116) with Eq. (2.117) shows that the rL circle is centered at r = rL/(1+rL) and i = 0, and its radius is 1/(1+rL). It therefore follows that all rL-circles pass through thepoint(r, i) = (1, 0). ThelargestcircleshowninFig.2-25 is for rL = 0, which also is the unit circle corresponding to || = 1. This is to be expected, because when rL = 0, || = 1 regardless of the magnitude of xL.
A similar manipulation of the expression for xL given by Eq. (2.115b) leads to (r −1)2 + i −1 xL 2 = 1 xL 2 , (2.118) which is the equation of a circle of radius (1/xL) centered at (r, i) = (1, 1/xL).
The xL circles in the r–i plane are quite different from those for constant rL. To start with, the normalized reactance xL may assume both positive and negative values, whereas the normalized resistance cannot be negative (negative resistances cannot be realized in passive circuits). Hence, Eq. (2.118) represents two families of circles, one for positive values of xL and another for negative ones.
Furthermore, as shown in Fig. 2-25, only part of a given circle falls within the bounds of the || = 1 unit circle.
The families of circles of the two parametric equations given by Eqs. (2.116) and (2.118) plotted for selected values of rL and xL constitute the Smith chart shown in Fig. 2-26. The Smith chart provides a graphical evaluation of Eqs. (2.115a and b) and their inverses. For example, point P in Fig. 2-26 represents a normalized load impedance zL = 2 −j1, which corresponds to a voltage reflection coefficient = 0.45 exp(−j26.6◦). The 2-10 THE SMITH CHART 91 Γi Γr xL = 2 xL = 1 xL = 0.5 xL = 0 rL = 0 rL = 0.5 rL = 1 rL = 2 xL = −0.5 xL = −2 xL = −1 Figure 2-25 Families of rL and xL circles within the domain || ≤1.
magnitude || = 0.45 is obtained by dividing the length of the line between the center of the Smith chart and the point P by the length of the line between the center of the Smith chart and the edge of the unit circle (the radius of the unit circle corresponds to || = 1). The perimeter of the Smith chart contains three concentric scales.
The innermost scale is labeled angle of reflection coefficient in degrees. This is the scale for θr. As indicated in Fig. 2-26, θr = −26.6◦(−0.46 rad) for point P.
The meanings and uses of the other two scales are discussed next.
Exercise 2-16: Use the Smith chart to find the values of corresponding to the following normalized load impedances: (a) zL = 2 + j0, (b) zL = 1 −j1, (c) zL = 0.5 −j2, (d) zL = −j3, (e) zL = 0 (short circuit), (f) zL = ∞(open circuit), (g) zL = 1 (matched load).
Answer: (a) = 0.33, (b) = 0.45∠ −63.4◦, (c) = 0.83∠ −50.9◦, (d) = 1∠ −36.9◦, (e) = −1, (f) = 1, (g) = 0. (See EM.) 92 CHAPTER 2 TRANSMISSION LINES R P O 0 0 Inner scale: θr in degrees Middle scale: wavelengths toward load Outermost scale: wavelengths toward generator 0.25λ 0.25λ −26.6o zL = 2 − j1 Figure 2-26 Point P represents a normalized load impedance zL = 2 −j1. The reflection coefficient has a magnitude || = OP /OR = 0.45 and an angle θr = −26.6◦. Point R is an arbitrary point on the rL = 0 circle (which also is the || = 1 circle).
2-10.2 Wave Impedance From Eq. (2.74), the normalized wave impedance looking toward the load at a distance d from the load is z(d) = Z(d) Z0 = 1 + d 1 −d , (2.119) where d = e−j2βd = ||ej(θr−2βd) (2.120) is the phase-shifted voltage reflection coefficient. The form of Eq. (2.119) is identical with that for zL given by Eq. (2.112): zL = 1 + 1 − .
(2.121) This similarity in form suggests that if is transformed into d, zL gets transformed into z(d). On the Smith chart, the transformation from to d is achieved by maintaining || constant and decreasing its phase θr by 2βd, which corresponds to a clockwise rotation (on the Smith chart) over an angle 2-10 THE SMITH CHART 93 of 2βd radians. A complete rotation around the Smith chart corresponds to a phase change of 2π in .
The length d corresponding to this phase change satisfies 2βd = 2 2π λ d = 2π, (2.122) from which it follows that d = λ/2.
▶The outermost scale around the perimeter of the Smith chart (Fig. 2-26), called the wavelengths toward generator (WTG) scale, has been constructed to denote movement on the transmission line toward the generator, in units of the wavelength λ. That is, d is measured in wavelengths, and one complete rotation corresponds to d = λ/2. ◀ In some transmission-line problems, it may be necessary to move from some point on the transmission line toward a point closer to the load, in which case the phase of must be increased, which corresponds to rotation in the counterclockwise direction. For convenience, the Smith chart contains a third scale around its perimeter (in between the θr scaleandtheWTGscale)foraccommodatingsuchanoperation.
It is called the wavelengths toward load (WTL) scale.
To illustrate how the Smith chart is used to find Z(d), consider a 50 lossless transmission line terminated in a load impedance ZL = (100 −j50) . Our objective is to find Z(d) at a distance d = 0.1λ from the load. The normalized load impedance is zL = ZL/Z0 = 2 −j1, and is marked by point A on the Smith chart in Fig. 2-27. On the WTG scale, point A is located at 0.287λ.
Next, we construct a circle centered at (r, i) = (0, 0) and passing through point A. Since the center of the Smith chart is the intersection point of the r and i axes, all points on this circle have the same value of ||.
This constant-|| circle is also a constant-SWR circle. This follows from the relation between the voltage standing-wave ratio (SWR) and ||, namely S = 1 + || 1 −|| .
(2.123) ▶A constant value of || corresponds to a constant value of S, and vice versa. ◀ As was stated earlier, to transform zL to z(d), we need to maintain || constant, which means staying on the SWR circle, while decreasing the phase of by 2βd radians.
This is equivalent to moving a distance d = 0.1λ toward the generator on the WTG scale.
Since point A is located at 0.287λ on the WTG scale, z(d) is found by moving to location 0.287λ + 0.1λ = 0.387λ on the WTG scale.
A radial line through this new position on the WTG scale intersects the SWR circle at point B. This point represents z(d), and its value is z(d) = 0.6 −j0.66. Finally, we unnormalize z(d) by multiplying it by Z0 = 50 to get Z(d) = (30−j33) . This result can be verified analytically using Eq. (2.119). The points between points A and B on the SWR circle represent different locations along the transmission line.
If a line is of length l, its input impedance is Zin = Z0 z(l), with z(l) determined by rotating a distance l from the load along the WTG scale.
Exercise 2-17: Use the Smith chart to find the normalized input impedance of a lossless line of length l terminated in a normalized load impedance zL for each of the following combinations: (a) l = 0.25λ, zL = 1 + j0, (b) l = 0.5λ, zL = 1 + j1, (c) l = 0.3λ, zL = 1 −j1, (d) l = 1.2λ, zL = 0.5 −j0.5, (e) l = 0.1λ, zL = 0 (short circuit), (f) l = 0.4λ, zL = j3, (g) l = 0.2λ, zL = ∞(open circuit).
Answer: (a) zin = 1 + j0, (b) zin = 1 + j1, (c) zin = 0.76 + j0.84, (d) zin = 0.59 + j0.66, (e) zin = 0+j0.73, (f) zin = 0+j0.72, (g) zin = 0 −j0.32.
(See EM.) 2-10.3 SWR, Voltage Maxima and Minima Consider a load with zL = 2 + j1.
Figure 2-28 shows a Smith chart with an SWR circle drawn through point A, representing zL. The SWR circle intersects the real (r) axis at two points, labeled Pmax and Pmin. At both points i = 0 and = r. Also, on the real axis, the imaginary part of the load impedance xL = 0. From the definition of , = zL −1 zL + 1, (2.124) it follows that points Pmax and Pmin correspond to = r = r0 −1 r0 + 1 (for i = 0), (2.125) 94 CHAPTER 2 TRANSMISSION LINES Load Input 0.287λ 0.1λ 0.387λ S W R c i r c l e A B 0.1 00 λ zL = 2 − j1 z(d) Figure 2-27 Point A represents a normalized load zL = 2 −j1 at 0.287λ on the WTG scale. Point B represents the line input at d = 0.1λ from the load. At B, z(d) = 0.6 −j0.66.
where r0 is the value of rL where the SWR circle intersects the r axis. Point Pmin corresponds to r0 < 1 and Pmax corresponds to r0 > 1. Rewriting Eq. (2.123) for || in terms of S, we have || = S −1 S + 1 .
(2.126) For point Pmax, || = r; hence r = S −1 S + 1 .
(2.127) The similarity in form of Eqs. (2.125) and (2.127) suggests that S equals the value of the normalized resistance r0. By definition S ≥1, and at point Pmax, r0 > 1, which further satisfies the similarity condition.
In Fig. 2-28, r0 = 2.6 at Pmax; hence S = 2.6.
2-10 THE SMITH CHART 95 0.213λ 0.25λ Distance to voltage maximum from load Distance to voltage minimum from load dmax = 0.037λ r0 = 2.6 dmin = 0.287λ 0 S W R A Pmin Pmax zL = 2 + j1 dmax dmin Figure 2-28 Point A represents a normalized load with zL = 2+j1. The standing wave ratio is S = 2.6 (at Pmax), the distance between the load and the first voltage maximum is dmax = (0.25 −0.213)λ = 0.037λ, and the distance between the load and the first voltage minimum is dmin = (0.037 + 0.25)λ = 0.287λ.
▶S is numerically equal to the value of r0 at Pmax, the point at which the SWR circle intersects the real axis to the right of the chart’s center. ◀ Points Pmin and Pmax also represent locations on the line at which the magnitude of the voltage | V | is a minimum and a maximum, respectively.
This is easily demonstrated by considering Eq. (2.120) for d. At point Pmax, the total phase of d, that is, (θr −2βd), equals zero or −2nπ (with n being a positive integer), which is the condition corresponding to | V |max, as indicated by Eq. (2.69). Similarly, at Pmin the total phase of d equals −(2n + 1)π, which is the condition for | V |min. Thus, for the transmission line represented by the SWR circle shown in Fig. 2-28, the distance between the load and the nearest voltage maximum is dmax = 0.037λ, obtained by 96 CHAPTER 2 TRANSMISSION LINES moving clockwise from the load at point A to point Pmax, and the distance to the nearest voltage minimum is dmin = 0.287λ, corresponding to the clockwise rotation from A to Pmin. Since the location of | V |max corresponds to that of | ˜ I|min and the location of | V |min corresponds to that of | ˜ I|max, the Smith chart provides a convenient way to determine the distances from the load to all maxima and minima on the line (recall that the standing-wave pattern has a repetition period of λ/2).
2-10.4 Impedance to Admittance Transformations In solving certain types of transmission-line problems, it is often more convenient to work with admittances than with impedances. Any impedance Z is in general a complex quantity consisting of a resistance R and a reactance X: Z = R + jX ().
(2.128) The admittance Y is the reciprocal of Z: Y = 1 Z = 1 R + jX = R −jX R2 + X2 (S).
(2.129) The real part of Y is called the conductance G, and the imaginary part of Y is called the susceptance B. That is, Y = G + jB (S).
(2.130) Comparison of Eq. (2.130) with Eq. (2.129) reveals that G = R R2 + X2 (S), (2.131a) B = −X R2 + X2 (S).
(2.131b) A normalized impedance z is defined as the ratio of Z to Z0, the characteristic impedance of the line. The same concept applies to the definition of the normalized admittance y; that is, y = Y Y0 = G Y0 +j B Y0 = g+jb (dimensionless), (2.132) where Y0 = 1/Z0 is the characteristic admittance of the line and g = G Y0 = GZ0 (dimensionless), (2.133a) b = B Y0 = BZ0 (dimensionless).
(2.133b) The lowercase quantities g and b represent the normalized conductance and normalized susceptance of y, respectively.
Of course, the normalized admittance y is the reciprocal of the normalized impedance z, y = Y Y0 = Z0 Z = 1 z .
(2.134) Accordingly, using Eq. (2.121), the normalized load admit-tance yL is given by yL = 1 zL = 1 − 1 + (dimensionless).
(2.135) Now let us consider the normalized wave impedance z(d) at a distance d = λ/4 from the load.
Using Eq. (2.119) with 2βd = 4πd/λ = 4πλ/4λ = π gives z(d = λ/4) = 1 + e−jπ 1 −e−jπ = 1 − 1 + = yL.
(2.136) ▶Rotation by λ/4 on the SWR circle transforms z into y, and vice versa. ◀ In Fig. 2-29, the points representing zL and yL are diametrically opposite to each other on the SWR circle.
In fact, such a transformation on the Smith chart can be used to determine any normalized admittance from its corresponding normalized impedance, and vice versa.
The Smith chart can be used with normalized impedances or with normalized admittances. As an impedance chart, the Smith chart consists of rL and xL circles, the resistance and reactance of a normalized load impedance zL, respectively.
When used as an admittance chart, the rL circles become gL circles and the xL circles become bL circles, where gL and bL are the conductance and susceptance of the normalized load admittance yL, respectively.
2-10 THE SMITH CHART 97 Load impedance zL Load admittance yL B A Figure 2-29 Point A represents a normalized load zL = 0.6 + j1.4. Its corresponding normalized admittance is yL = 0.25 −j0.6, and it is at point B.
Example 2-11: Smith-Chart Calculations A 50 lossless transmission line of length 3.3λ is terminated by a load impedance ZL = (25 + j50) .
Use the Smith chart to find (a) the voltage reflection coefficient, (b) the voltage standing-wave ratio, (c) the distances of the first voltage maximumandfirstvoltageminimumfromtheload, (d)theinput impedance of the line, and (e) the input admittance of the line.
Solution: (a) The normalized load impedance is zL = ZL Z0 = 25 + j50 50 = 0.5 + j1, which is marked as point A on the Smith chart in Fig. 2-30. A radial line is drawn from the center of the chart at point O through point A to the outer perimeter of the chart.
The line crosses the scale labeled “angle of reflection coefficient in degrees” at θr = 83◦.
Next, measurements are made to determine lengths OA and OO′, of the lines between O and A and between points O and O′, respectively, where O′ is an 98 CHAPTER 2 TRANSMISSION LINES S = 4.26 A B C D E O O′ l = 0.3λ 0.435λ 0.135λ Location of |V|max ~ Location of |V|min zin yin zL ~ zL = 0.5 + j1 zin dmax dmin 3.3λ Figure 2-30 Solution for Example 2-11. Point A represents a normalized load zL = 0.5+j1 at 0.135λ on the WTG scale. At A, θr = 83◦ and || = OA/OO′ = 0.62. At B, the standing-wave ratio is S = 4.26. The distance from A to B gives dmax = 0.115λ and from A to C gives dmin = 0.365λ. Point D represents the normalized input impedance zin = 0.28 −j0.40, and point E represents the normalized input admittance yin = 1.15 + j1.7.
2-10 THE SMITH CHART 99 arbitrary point on the rL = 0 circle. The length OO′ is equal to the radius of the || = 1 circle. The magnitude of is then obtained from || = OA/OO′ = 0.62. Hence, = 0.62∠ 83◦.
(2.137) (b) The SWR circle passing through point A crosses the r axis at points B and C. The value of rL at point B is 4.26, from which it follows that S = 4.26.
(c) The first voltage maximum is at point B on the SWR circle, which is at location 0.25λ on the WTG scale. The load, represented by point A, is at 0.135λ on the WTG scale. Hence, the distance between the load and the first voltage maximum is dmax = (0.25 −0.135)λ = 0.115λ.
The first voltage minimum is at point C. Moving on the WTG scale between points A and C gives dmin = (0.5 −0.135)λ = 0.365λ, which is 0.25λ past dmax.
(d) The line is 3.3λ long; subtracting multiples of 0.5λ leaves 0.3λ. From the load at 0.135λ on the WTG scale, the input of the line is at (0.135 + 0.3)λ = 0.435λ. This is labeled as point D on the SWR circle, and the normalized impedance is zin = 0.28 −j0.40, which yields Zin = zinZ0 = (0.28 −j0.40)50 = (14 −j20) .
(e) The normalized input admittance yin is found by moving 0.25λ on the Smith chart to the image point of zin across the circle, labeled point E on the SWR circle. The coordinates of point E give yin = 1.15 + j1.7, and the corresponding input admittance is Yin = yinY0 = yin Z0 = 1.15 + j1.7 50 = (0.023 + j0.034) S.
Example 2-12: Determining ZL Using the Smith Chart This problem is similar to Example 2-6, except that now we demonstrate its solution using the Smith chart.
Given that the voltage standing-wave ratio S = 3 on a 50 line, that the first voltage minimum occurs at 5 cm from the load, and that the distance between successive minima is 20 cm, find the load impedance.
Solution: The distance between successive minima equals λ/2. Hence, λ = 40 cm. In wavelength units, the first voltage minimum is at dmin = 5 40 = 0.125λ.
Point A on the Smith chart in Fig. 2-31 corresponds to S = 3.
Using a compass, the constant S circle is drawn through point A.
Point B corresponds to locations of voltage minima. Upon moving 0.125λ from point B toward the load on the WTL scale (counterclockwise), we arrive at point C, which represents the location of the load. The normalized load impedance at point C is zL = 0.6 −j0.8.
Multiplying by Z0 = 50 , we obtain ZL = 50(0.6 −j0.8) = (30 −j40) .
Concept Question 2-19: The outer perimeter of the Smith chart represents what value of ||? Which point on the Smith chart represents a matched load?
100 CHAPTER 2 TRANSMISSION LINES S = 3.0 A Voltage min Load B 0.125λ C zL = ?
dmin 0.125λ Figure 2-31 Solution for Example 2-12. Point A denotes that S = 3, point B represents the location of the voltage minimum, and point C represents the load at 0.125λ on the WTL scale from point B. At C, zL = 0.6 −j0.8.
Concept Question 2-20: What is an SWR circle?
What quantities are constant for all points on an SWR circle?
Concept Question 2-21: What line length corresponds to one complete rotation around the Smith chart? Why?
Concept Question 2-22: Which points on the SWR circle correspond to locations of voltage maxima and minima on the line and why?
Concept Question 2-23: Given a normalized impe-dance zL, how do you use the Smith chart to find the corresponding normalized admittance yL = 1/zL?
2-11 IMPEDANCE MATCHING 101 Module 2.6 Interactive Smith Chart Locate the load on the Smith chart; display the corresponding reflection coefficient and SWR circle; “move” to a new location at a distance d from the load, and read the wave impedance Z(d) and phase-shifted reflection coefficient d; perform impedance to admittance transformations and vice versa; and use all of these tools to solve transmission-line problems via the Smith chart.
2-11 Impedance Matching A transmission line usually connects a generator circuit at one end to a load at the other.
The load may be an antenna, a computer terminal, or any circuit with an equivalent input impedance ZL.
▶The transmission line is said to be matched to the load when its characteristic impedance Z0 = ZL, in which case waves traveling on the line towards the load are not reflected back to the source. ◀ Since the primary use of a transmission line is to transfer power or transmit coded signals (such as digital data), a matched load ensuresthatallofthepowerdeliveredtothetransmissionlineby the source is transferred to the load (and no echoes are relayed back to the source).
The simplest solution to matching a load to a transmission line is to design the load circuit such that its impedance ZL = Z0. Unfortunately, this may not be possible in practice because the load circuit may have to satisfy other requirements.
An alternative solution is to place an impedance-matching 102 CHAPTER 2 TRANSMISSION LINES Zg ZL Vg ~ Feedline Generator Load Matching network Zin Z0 M M' A A' − + Figure 2-32 Thefunctionofa matching network is to transform theloadimpedanceZL suchthattheinputimpedanceZin looking into the network is equal to Z0 of the feedline.
network between the load and the transmission line as illustrated in Fig. 2-32.
▶The purpose of the matching network is to eliminate reflections at terminals MM′ for waves incident from the source. Even though multiple reflections may occur between AA′ and MM′, only a forward traveling wave exists on the feedline. ◀ Within the matching network, reflections can occur at both terminals (AA′ and MM′), creating a standing-wave pattern, but the net result (of all of the multiple reflections within the matching network) is that the wave incident from the source experiences no reflection when it reaches terminals MM′. This is achieved by designing the matching network to exhibit an impedance equal to Z0 at MM′ when looking into the network from the transmission line side. If the network is lossless, then all the power going into it will end up in the load.
▶Matching networks may consist of lumped elements, such as capacitors and inductors (but not resistors, because resistors incur ohmic losses), or of sections of transmission lines with appropriate lengths and terminations. ◀ The matching network, which is intended to match a load impedance ZL = RL +jXL to a lossless transmission line with characteristic impedance Z0, may be inserted either in series (between the load and the feedline) as in Fig. 2-33(a) and (b) or in parallel [Fig. 2-33(c) to (e)]. In either case, the network has to transform the real part of the load impedance from RL (at the load) to Z0 at MM′ in Fig. 2-32 and transform the reactive part from XL (at the load) to zero at MM′.
To achieve these two transformations, the matching network must have at least two degrees of freedom (that is, two adjustable parameters).
If XL = 0, the problem reduces to a single transformation, in which case matching can be realized by inserting a quarter-wavelength transformer (Section 2-8.5) next to the load [Fig. 2-33(a)].
▶For the general case where XL ̸= 0, a λ/4 transformer can still be designed to provide the desired matching, but it has to be inserted at a distance dmax or dmin from the load [Fig. 2-33(b)], where dmax and dmin are the distances to voltage maxima and minima, respectively. ◀ The design procedure is outlined in Module 2.7. The in-parallel insertion networks shown in Fig. 2-33(c)–(e) are the subject of Examples 2-13 and 2-14.
2-11.1 Lumped-Element Matching In the arrangement shown in Fig. 2-34, the matching network consists of a single lumped element, either a capacitor or an inductor, connected in parallel with the line at a distance d from the load. Parallel connections call for working in the admittance domain. Hence, the load is denoted by an admittance YL and the line has characteristic admittance Y0. The shunt element has admittance Ys. At MM′, Yd is the admittance due to the transmission-line segment to the right of MM′.
The input admittance Yin (referenced at a point just to the left of MM′) is equal to the sum of Yd and Ys: Yin = Yd + Ys.
(2.138) In general Yd is complex and Ys is purely imaginary because it represents a reactive element (capacitor or inductor). Hence, Eq. (2.138) can be written as Yin = (Gd + jBd) + jBs = Gd + j(Bd + Bs).
(2.139) When all quantities are normalized to Y0, Eq. (2.139) becomes yin = gd + j(bd + bs).
(2.140) 2-11 IMPEDANCE MATCHING 103 (b) If ZL = complex: in-series λ/4 transformer inserted at d = dmax or d = dmin (c) In-parallel insertion of capacitor at distance d1 (a) If ZL is real: in-series λ/4 transformer inserted at AA' Zin Z0 Z02 ZL M A M' A' Feedline λ/4 transformer (d) In-parallel insertion of inductor at distance d2 (e) In-parallel insertion of a short-circuited stub y(d2) Z0 Z0 ZL Zin L M A M' A' d2 Feedline ys(l1) ZL Z0 Z0 Z0 l1 M A M' A' Feedline d1 ZL Zin Z01 Z01 Z(d) Z02 M B A M' B' A' Feedline λ/4 d y(d1) ZL Zin Z0 Z0 C M A M' A' d1 Feedline Figure 2-33 Five examples of in-series and in-parallel matching networks.
(b) Equivalent circuit Ys Yd M M' Feedline Yin (a) Transmission-line circuit YL Yin Y0 Y0 Yd M M' Shunt element Load Feedline Ys d Figure 2-34 Inserting a reactive element with admittance Ys at MM′ modifies Yd to Yin.
104 CHAPTER 2 TRANSMISSION LINES To achieve a matched condition at MM′, it is necessary that yin = 1+j0, which translates into two specific conditions, namely gd = 1 (real-part condition), (2.141a) bs = −bd (imaginary-part condition).
(2.141b) The real-part condition is realized through the choice of d, the distance from the load to the shunt element, and the imaginary-part condition is realized through the choice of lumped element (capacitor or inductor) and its value. These two choices are the two degrees of freedom needed in order to match the load to the feedline.
Example 2-13: Lumped Element A load impedance ZL = 25 −j50 is connected to a 50 transmissionline. Insertashuntelementtoeliminatereflections towards the sending end of the line. Specify the insert location d (in wavelengths), the type of element, and its value, given that f = 100 MHz.
(a) Analytical Solution: The normalized load admittance is yL = Z0 ZL = 50 1 25 −j50 = 0.4 + j0.8.
Upon replacing zL with 1/yL in Eq. (2.124), the reflection coefficient at the load becomes = 1 −yL 1 + yL = 1 −(0.4 + j0.8) 1 + (0.4 + j0.8) = 0.62e−j82.9◦.
Equation (2.119) provides an expression for the input impedance at any location d from the end of the line. If we invert the expression to convert it to admittance, we obtain the following expression for yd, the admittance of the line to the right of MM′ in Fig. 2-34(a): yd = 1 −||ej(θr−2βd) 1 + ||ej(θr−2βd) = 1 −||ejθ′ 1 + ||ejθ′ , (2.142) where θ′ = θr −2βd.
(2.143) Multiplying the numerator and denominator of this expression by the complex conjugate of the denominator leads to yd = 1 −||ejθ′ 1 + ||ejθ′ 1 + ||e−jθ′ 1 + ||e−jθ′ = 1 −||2 1 + ||2 + 2|| cos θ′ −j 2|| sin θ′ 1 + ||2 + 2|| cos θ′ .
(2.144) Hence, the real and imaginary components of yd are gd = 1 −||2 1 + ||2 + 2|| cos θ′ , (2.145a) bd = −2|| sin θ′ 1 + ||2 + 2|| cos θ′ .
(2.145b) TosatisfythefirstconditionofEq.(2.141a), weneedtochoosed (which is embedded in the definition for θ′ given by Eq. (2.143)) such that 1 −||2 1 + ||2 + 2|| cos θ′ = 1, which leads to the solution cos θ′ = −||.
(2.146) Since cos θ′ is negative, θ′ can be in either the second or third quadrant. Given that || = 0.62, we obtain θ′ 1 = −128.3◦, or θ′ 2 = +128.3◦.
Each value of θ′ offers a possible solution for d. We shall label them d1 and d2.
2-11 IMPEDANCE MATCHING 105 (b) Second solution (a) First solution Feedline 50 pF d2 = 0.207λ 50 nH d1 = 0.063λ Y0 Yd1 Feedline Y0 Yd2 YL = (0.4 + j0.8)Y0 YL = (0.4 + j0.8)Y0 Figure 2-35 Solutions for Example 2-13.
Solution for d1 [Fig. 2-35(a)] With θ′ 1 = −128.3◦= −2.240 rad, θr = −82.9◦= −1.446 rad, and β = 2π/λ, solving Eq. (2.143) for d gives d1 = λ 4π (θr −θ′ 1) = λ 4π (−1.446 + 2.240) = 0.063λ.
Next, to satisfy the second condition in Eq. (2.141), we need to determine bs1 such that bs1 = −bd. Using Eq. (2.145b), we obtain bs1 = 2|| sin θ′ 1 + ||2 + 2|| cos θ′ = 2 × 0.62 sin(−128.3◦) 1 + 0.622 + 2 × 0.62 cos(−128.3◦) = −1.58.
The corresponding impedance of the lumped element is Zs1 = 1 Ys1 = 1 ys1Y0 = Z0 jbs1 = Z0 −j1.58 = jZ0 1.58 = j31.62 .
Since the value of Zs1 is positive, the element to be inserted should be an inductor and its value should be L = 31.62 ω = 31.62 2π × 108 = 50 nH.
The results of this solution have been incorporated into the circuit of Fig. 2-35(a).
Solution for d2 [Fig. 2-35(b)] Repeating the procedure for θ′ 2 = 128.3◦leads to d2 = λ 4π (−1.447 −2.239) = −0.293λ.
A negative value for d2 is physically meaningless because that would place it to the right of the load, but since we know that impedances repeat themselves every λ/2, we simply need to add λ/2 to the solution: d2 (physically realizable) = −0.293λ + 0.5λ = 0.207λ.
The associated value for bs is +1.58. Hence Zs2 = −j31.62 , which is the impedance of a capacitor with C = 1 31.62ω = 50 pF.
Figure 2-35(b) displays the circuit solution for d2 and C.
106 CHAPTER 2 TRANSMISSION LINES E A F C ys1 = −j1.58 gL = 1 circle First intersection of gL = 1 circle with SWR circle.
At C, yd1 = 1 + j1.58. Admittance of short-circuit stub (Example 2-14) d1 l1 = 0.09 Load zL Load yL 0.063λ 0.115λ B Figure 2-36 Solution for point C of Examples 2-13 and 2-14. Point A is the normalized load with zL = 0.5−j1; point B is yL = 0.4 + j0.8.
Point C is the intersection of the SWR circle with the gL = 1 circle. The distance from B to C is d1 = 0.063λ. The length of the shorted stub (E to F) is l1 = 0.09λ (Example 2-14).
(b) Smith-chart solution: The normalized load impedance is zL = ZL Z0 = 25 −j50 50 = 0.5 −j1, which is represented by point A on the Smith chart of Fig. 2-36.
Next, we draw the constant S circle through point A. As alluded to earlier, to perform the matching task, it is easier to work with admittances than with impedances. The normalized load admittance yL is represented by point B, obtained by rotating point A over 0.25λ, or equivalently by drawing a line from point A through the chart center to the image of point A on the S circle. The value of yL at B is yL = 0.4 + j0.8, 2-11 IMPEDANCE MATCHING 107 d2 = 0.207λ l2 = 0.410λ G E ys2 = j1.58 Second intersection of gL = 1 circle with SWR circle.
At D, yd2 = 1 − j1.58. Admittance of short-circuit stub (Example 2-14) Load zL Load yL D B A Figure 2-37 Solution for point D of Examples 2-13 and 2-14. Point D is the second point of intersection of the SWR circle and the gL = 1 circle. The distance B to D gives d2 = 0.207λ, and the distance E to G gives l2 = 0.410λ (Example 2-14).
and it is located at position 0.115λ on the WTG scale. In the admittance domain, the rL circles become gL circles, and the xL circles become bL circles. To achieve matching, we need to move from the load toward the generator a distance d such that the normalized input admittance yd of the line terminated in the load (Fig. 2-34) has a real part of 1. This condition is satisfied by either of the two matching points C and D on the Smith charts of Figs. 2-36 and 2-37, respectively, corresponding to intersections of the S circle with the gL = 1 circle. Points C and D represent two possible solutions for the distance d in Fig. 2-34(a).
Solution for Point C (Fig. 2-36): At C, yd = 1 + j1.58, which is located at 0.178λ on the WTG scale. The distance between points B and C is d1 = (0.178 −0.115)λ = 0.063λ.
108 CHAPTER 2 TRANSMISSION LINES Looking from the generator toward the parallel combination of the line connected to the load and the shunt element, the normalized input admittance at terminals MM′ is yin = ys + yd, where ys is the normalized input admittance of the shunt element. To match the feed line to the parallel combination, we need yin = 1 + j0. Thus, 1 + j0 = ys + 1 + j1.58, or ys = −j1.58.
This is the same result obtained earlier in the analytical solution, which led to choosing an inductor L = 50 nH.
Solution for Point D (Fig. 2-37): At point D, yd = 1 −j1.58, and the distance between points B and D is d2 = (0.322 −0.115)λ = 0.207λ.
The needed normalized admittance of the reactive element is ys = +j1.58, which, as shown earlier, corresponds to a capacitor C = 50 pF.
2-11.2 Single-Stub Matching The single-stub matching network shown in Fig. 2-38(a) consists of two transmission line sections, one of length d connecting the load to the feedline at MM′ and another of length l connected in parallel with the other two lines at MM′.
This second line is called a stub, and it is usually terminated in either a short or open circuit, and hence its input impedance and YL Yin Y0 Y0 Y0 Ys Yd M M' d Shorted stub (a) Transmission-line circuit (b) Equivalent circuit l Feedline Load Ys Yd Yin M M' Feedline Figure 2-38 Shorted-stub matching network.
admittance are purely reactive. The stub shown in Fig. 2-38(a) has a short-circuit termination.
▶The required two degrees of freedom are provided by the length l of the stub and the distance d from the load to the stub position. ◀ Because at MM′ the stub is added in parallel to the line (which is why it is called a shunt stub), it is easier to work with admittances than with impedances. The matching procedure consists of two steps. In the first step, the distance d is selected so as to transform the load admittance YL = 1/ZL into an admittance of the form Yd = Y0 + jB, when looking toward the load at MM′. Then, in the second step, the length l of the stub line is selected so that its input admittance Ys at MM′ is equal to −jB. The parallel sum of the two admittances at MM′ yields Y0, the characteristic admittance of the line. The procedure is illustrated by Example 2-14.
2-11 IMPEDANCE MATCHING 109 Module 2.7 Quarter-Wavelength Transformer This module allows you to go through a multi-step procedure to design a quarter-wavelength transmission line that, when inserted at the appropriate location on the original line, presents a matched load to the feedline.
Example 2-14: Single-Stub Matching Repeat Example 2-13, but use a shorted stub (instead of a lumped element) to match the load impedance ZL = (25 −j50) to the 50 transmission line.
Solution: In Example 2-13, we demonstrated that the load can be matched to the line via either of two solutions: (1) d1 = 0.063λ, and ys1 = jbs1 = −j1.58, (2) d2 = 0.207λ, and ys2 = jbs2 = j1.58.
The locations of the insertion points, at distances d1 and d2, remain the same, but now our task is to select corresponding lengths l1 and l2 of shorted stubs that present the required admittances at their inputs.
To determine l1, we use the Smith chart in Fig. 2-36. The normalized admittance of a short circuit is −j∞, which is represented by point E on the Smith chart, with position 0.25λ on the WTG scale. A normalized input admittance of −j1.58 is located at point F, with position 0.34λ on the WTG scale.
110 CHAPTER 2 TRANSMISSION LINES Module 2.8 Discrete Element Matching For each of two possible solutions, the module guides the user through a procedure to match the feedline to the load by inserting a capacitor or an inductor at an appropriate location along the line.
Hence, l1 = (0.34 −0.25)λ = 0.09λ.
Similarly, ys2 = j1.58 is represented by point G with position 0.16λ on theWTG scale of the Smith chart in Fig. 2-37.
Rotating from point E to point G involves a rotation of 0.25λ plus an additional rotation of 0.16λ or l2 = (0.25 + 0.16)λ = 0.41λ.
Concept Question 2-24: To match an arbitrary load impedance to a lossless transmission line through a matching network, what is the required minimum number of degrees of freedom that the network should provide?
Concept Question 2-25: In the case of the single-stub matching network, what are the two degrees of freedom?
Concept Question 2-26: When a transmission line is matched to a load through a single-stub matching network, no waves are reflected toward the generator.
What happens to the waves reflected by the load and by the shorted stub when they arrive at terminals MM′ in Fig. 2-38?
2-12 TRANSIENTS ON TRANSMISSION LINES 111 Module 2.9 Single-Stub Tuning Instead of inserting a lumped element to match the feedline to the load, this module determines the length of a shorted stub that can accomplish the same goal.
2-12 Transients on Transmission Lines Thus far, our treatment of wave propagation on transmission lines has focused on the analysis of single-frequency, time-harmonic signals under steady-state conditions.
The impedance-matching and Smith chart techniques we developed, while useful for a wide range of applications, are inappropriate for dealing with digital or wideband signals that exist in digital chips, circuits, and computer networks.
For such signals, we need to examine the transient transmission-line response instead.
▶The transient response of a voltage pulse on a transmission line is a time record of its back and forth travel between the sending and receiving ends of the line, taking into account all the multiple reflections (echoes) at both ends. ◀ Let us start by considering the case of a single rectangular pulse of amplitude V0 and duration τ, as shown in Fig. 2-39(a) (page 115). The amplitude of the pulse is zero prior to t = 0, V0 over the interval 0 ≤t ≤τ, and zero afterwards. The pulse 112 TECHNOLOGY BRIEF 4: EM CANCER ZAPPERS Technology Brief 4: EM Cancer Zappers From laser eye surgery to 3-D X-ray imaging, EM sources and sensors have been used as medical diagnostic and treatment tools for many decades. Future advances in information processing and other relevant technologies will undoubtedly lead to greater performance and utility of EM devices, as well as to the introduction of entirely new types of devices. This Technology Brief introduces two recent EM technologies that are still in their infancy, but are fast developing into serious techniques for the surgical treatment of cancer tumors.
Microwave Ablation In medicine, ablation is defined as the “surgical removal of body tissue,” usually through the direct application of chemical or thermal therapies.
▶Microwave ablation applies the same heat-conversion process used in a microwave oven (see TB3), but instead of using microwave energy to cook food, it is used instead to destroy cancerous tumors by exposing them to a focused beam of microwaves. ◀ The technique can be used percutaneously (through the skin), laparoscopically (via an incision), or intraoperatively (open surgical access). Guided by an imaging system, such as a CT scanner or an ultrasound imager, the surgeon can localize the tumor and then insert a thin coaxial transmission line (∼1.5 mm in diameter) directly through the body to position the tip of the transmission line (a probe-like antenna) inside the tumor (Fig. TF4-1). The transmission line is connected to a generator capable of delivering 60 W of power at 915 MHz (Fig. TF4-2). The rise in temperature of the tumor is related to the amount of microwave energy it receives, which is equal to the product of the generator’s power level and the duration of the ablation treatment. Microwave ablation is a promising new technique for the treatment of liver, lung, and adrenal tumors.
Ultrasound transducer Ablation catheter (transmission line) Liver Ultrasound image Figure TF4-1 Microwave ablation for liver cancer treatment.
TECHNOLOGY BRIEF 4: EM CANCER ZAPPERS 113 Figure TF4-2 Photograph of the setup for a percutaneous microwave ablation procedure in which three single microwave applicators are connected to three microwave generators.
High-Power Nanosecond Pulses Bioelectrics is an emerging field focused on the study of how electric fields behave in biological systems. Of particular recent interest is the desire to understand how living cells might respond to the application of extremely short pulses (on the order of nanoseconds (10−9 s), and even as short as picoseconds (10−12 s)) with exceptionally high voltage and current amplitudes.
▶The motivation is to treat cancerous cells by zapping them with high-power pulses. The pulse power is delivered to the cell via a transmission line, as illustrated by the example in Fig. TF4-3. ◀ Note that the pulse is about 200 ns long, and its voltage and current amplitudes are approximately 3,000 V and 60 A, respectively. Thus, the peak power level is about 180,000 W! However, the total energy carried by the pulse is only (1.8 × 105) × (2 × 10−7) = 0.0036 Joules. Despite the low energy content, the very high voltage appears to be very effective in destroying malignant tumors (in mice, so far), with no regrowth.
114 TECHNOLOGY BRIEF 4: EM CANCER ZAPPERS 1 With the switch open, the device is charged up by its connection to the high-voltage source. Closing the switch sets up transient waves.
2 The voltage waves reflect off the ends of the transmission line. The wave near the switch inverts (red)—its polarity changes—when it reflects, because that end is shorted. When the inverted and noninverted waves crash into each other at the load, a pulse of voltage results.
3 When the trailing edges of the waves finally meet, the pulse ends.
Figure TF4-3 High-voltage nanosecond pulse delivered to tumor cells via a transmission line. The cells to be shocked by the pulse sit in a break in one of the transmission-line conductors.
2-12 TRANSIENTS ON TRANSMISSION LINES 115 V(t) V0 τ t (a) Pulse of duration τ (b) V(t) = V1(t) + V2(t) V(t) V1(t) = V0 u(t) V2(t) = −V0 u(t − τ) V0 τ t Figure 2-39 A rectangular pulse V (t) of duration τ can be represented as the sum of two step functions of opposite polarities displaced by τ relative to each other.
can be described mathematically as the sum of two unit step functions: V (t) = V1(t) + V2(t) = V0 u(t) −V0 u(t −τ), (2.147) where the unit step function u(x) is u(x) = 1 for x > 0, 0 for x < 0.
(2.148) The first component, V1(t) = V0 u(t), represents a dc voltage of amplitude V0 that is switched on at t = 0 and retains that value indefinitely, and the second component, V2(t) = −V0 u(t −τ), represents a dc voltage of amplitude −V0 that is switched on at t = τ and remains that way indefinitely. As can be seen from Fig. 2-39(b), the sum V1(t)+V2(t) is equal to V0 for 0 < t < τ and equal to zero for t < 0 and t > τ. This representation of a pulse in terms of two step functions allows us to analyze the transient behavior of the pulse on a transmission line as the superposition of two dc signals.
Hence, if we can develop basic tools for describing the transient behavior of a single step function, we can apply the same tools for each of the two components of the pulse and then add the results to obtain the response to V (t).
2-12.1 Transient Response to a Step Function The circuit shown in Fig. 2-40(a) (page 115) consists of a generator, composed of a dc voltage source Vg and a series resistance Rg, connected to a lossless transmission line of length l and characteristic impedance Z0. The line is terminated in a purely resistive load RL at z = l.
▶Note that whereas in previous sections, z = 0 was defined as the location of the load, now it is more convenient to define it as the location of the source. ◀ Vg I1 + Rg Z0 (a) Transmission-line circuit (b) Equivalent circuit at t = 0+ V1 + t = 0 z = 0 z = l Vg Rg Z0 RL z Transmission line − + − + − + Figure 2-40 At t = 0+, immediately after closing the switch in the circuit in (a), the circuit can be represented by the equivalent circuit in (b).
116 CHAPTER 2 TRANSMISSION LINES (I1 +) (I1 +) I1 + I1 + (I1 + + I1 −) I1 − = −ΓL I1 + I(z, 3T/2) I(z, T/2) (a) V(z) at t = T/2 (b) V(z) at t = 3T/2 (c) V(z) at t = 5T/2 (d) I(z) at t = T/2 (e) I(z) at t = 3T/2 (f) I(z) at t = 5T/2 l/2 0 l l/2 0 l I I z z l/2 0 l (V1 +) V1 + V V(z, T/2) z I1 + (I1 + + I1 − + I2 +) (I1 + + I1 −) I2 + = −ΓG I1 − I(z, 5T/2) I l/2 0 l z l/2 0 l (V1 +) V1 + V (V1 + + V1 −) V1 − = ΓLV1 + V(z, 3T/2) z V1 + V (V1 + + V1 − + V2 +) (V1 + + V1 −) V2 + = ΓGV1 − l/2 0 l V(z, 5T/2) z Figure 2-41 Voltage and current distributions on a lossless transmission line at t = T /2, t = 3T /2, and t = 5T /2, due to a unit step voltage applied to a circuit with Rg = 4Z0 and RL = 2Z0. The corresponding reflection coefficients are L = 1/3 and g = 3/5.
Theswitchbetweenthegeneratorcircuitandthetransmission line is closed at t = 0. The instant the switch is closed, the transmission line appears to the generator circuit as a load with impedance Z0.
This is because, in the absence of a signal on the line, the input impedance of the line is unaffected by the load impedance RL. The circuit representing the initial condition is shown in Fig. 2-40(b). The initial current I + 1 and corresponding initial voltage V + 1 at the sending end of the transmission line are given by I + 1 = Vg Rg + Z0 , (2.149a) V + 1 = I + 1 Z0 = VgZ0 Rg + Z0 .
(2.149b) The combination of V + 1 and I + 1 constitutes a wave that travels along the line with velocity up = 1/ √μϵ , immediately after the switch is closed. The plus-sign superscript denotes the fact that the wave is traveling in the +z direction. The transient response of the wave is shown in Fig. 2-41 at each of three instances in time for a circuit with Rg = 4Z0 and RL = 2Z0. The first response is at time t1 = T /2, where T = l/up is the time it takes the wave to travel the full length of the line. By time t1, the wave has traveled halfway down the line; consequently, the voltage on the first half of the line is equal to V + 1 , while the voltage on the second half is still zero [Fig. 2-41(a)]. At t = T , the wave reaches the load at z = l, and because RL ̸= Z0, the mismatch generates a reflected wave with amplitude V − 1 = LV + 1 , (2.150) where L = RL −Z0 RL + Z0 (2.151) 2-12 TRANSIENTS ON TRANSMISSION LINES 117 is the reflection coefficient of the load. For the specific case illustrated in Fig. 2-41, RL = 2Z0, which leads to L = 1/3.
After this first reflection, the voltage on the line consists of the sum of two waves: the initial wave V + 1 and the reflected wave V − 1 . The voltage on the transmission line at t2 = 3T /2 is shown in Fig. 2-41(b); V (z, 3T /2) equals V + 1 on the first half of the line (0 ≤z < l/2), and (V + 1 + V − 1 ) on the second half (l/2 ≤z ≤l).
At t = 2T , the reflected wave V − 1 arrives at the sending end of the line. If Rg ̸= Z0, the mismatch at the sending end generates a reflection at z = 0 in the form of a wave with voltage amplitude V + 2 given by V + 2 = gV − 1 = gLV + 1 , (2.152) where g = Rg −Z0 Rg + Z0 (2.153) is the reflection coefficient of the generator resistance Rg. For Rg = 4Z0, we have g = 0.6. As time progresses after t = 2T , the wave V + 2 travels down the line toward the load and adds to the previously established voltage on the line. Hence, at t = 5T/2, the total voltage on the first half of the line is V (z, 5T/2) = V + 1 + V − 1 + V + 2 = (1 + L + Lg)V + 1 (0 ≤z < l/2), (2.154) while on the second half of the line the voltage is only V (z, 5T/2) = V + 1 + V − 1 = (1 + L)V + 1 (l/2 ≤z ≤l).
(2.155) The voltage distribution is shown in Fig. 2-41(c).
Sofar, wehaveexaminedthetransientresponseofthevoltage wave V (z, t). The associated transient response of the current I(z, t) is shown in Figs. 2-41(d)–(f).
The current behaves similarly to the voltage V (z, t), except for one important difference.
Whereas at either end of the line the reflected voltage is related to the incident voltage by the reflection coefficient at that end, the reflected current is related to the incident current by the negative of the reflection coefficient.
This property of wave reflection is expressed by Eq. (2.61).
Accordingly, I − 1 = −LI + 1 , (2.156a) I + 2 = −gI − 1 = gLI + 1 , (2.156b) and so on.
▶The multiple-reflection process continues indefinitely, and the ultimate value that V (z, t) reaches as t approaches +∞is the same at all locations on the transmission line. ◀ It is given by V∞=V + 1 +V − 1 +V + 2 +V − 2 +V + 3 +V − 3 +· · · =V + 1 [1+L+Lg+2 Lg+2 L2 g +3 L2 g +· · · ] =V + 1 [(1+L)(1+Lg+2 L2 g +· · · )] =V + 1 (1+L)[1 + x + x2 + · · · ], (2.157) where x = Lg. The series inside the square bracket is the geometric series of the function 1 1 −x = 1 + x + x2 + · · · for |x| < 1.
(2.158) Hence, Eq. (2.157) can be rewritten in the compact form V∞= V + 1 1 + L 1 −Lg .
(2.159) Upon replacing V + 1 , L, and g with Eqs. (2.149b), (2.151), and (2.153), and simplifying the resulting expression, we obtain V∞= VgRL Rg + RL .
(2.160) The voltage V∞is called the steady-state voltage on the line, and its expression is exactly what we should expect on the basis of dc analysis of the circuit in Fig. 2-40(a), wherein we treat the transmission line as simply a connecting wire between the generator circuit and the load. The corresponding steady-state current is I∞= V∞ RL = Vg Rg + RL .
(2.161) 118 CHAPTER 2 TRANSMISSION LINES I(l/4, 4T) T l/4 l/2 3l/4 I1 + −ΓL I1 + Γg ΓL I1 + −Γg ΓL 2 I1 + Γg 2 ΓL 2 I1 + 5T 3T 4T 2T z = 0 t = 0 z = l Γ = −Γg Γ = −ΓL t t (a) Voltage bounce diagram (b) Current bounce diagram (c) Voltage versus time at z = l/4 V1 + ΓLV1 + Γg ΓLV1 + Γg ΓL 2 V1 + Γg 2 ΓL 2 V1 + V(l/4, 4T) (1 + ΓL)V1 + ΓL = 1/3 Γg = 3/5 (1 + ΓL + Γg ΓL)V1 + (1 + ΓL + Γg ΓL + Γg ΓL 2)V1 + (1 + ΓL + Γg ΓL + Γg ΓL 2 + Γg 2 ΓL 2)V1 + T 2T 3T 4T 5T T 4 7T 4 9T 4 15T 4 17T 4 t V1 + V1 + V(l/4, t) 5T 3T 4T 2T z = 0 t = 0 V I T l/4 l/2 3l/4 z = l Γ = Γg Γ = ΓL t t Figure 2-42 Bounce diagrams for (a) voltage and (b) current. In (c), the voltage variation with time at z = l/4 for a circuit with g = 3/5 and L = 1/3 is deduced from the vertical dashed line at l/4 in (a).
2-12.2 Bounce Diagrams Keeping track of the voltage and current waves as they bounce back and forth on the line is a rather tedious process. The bounce diagram is a graphical presentation that allows us to accomplish the same goal, but with relative ease.
The horizontal axes in Figs. 2-42(a) and (b) represent position along the transmission line, while the vertical axes denote time. Figures 2-42(a) and (b) pertain to V (z, t) and I(z, t), respectively.
The bounce diagram in Fig. 2-42(a) consists 2-12 TRANSIENTS ON TRANSMISSION LINES 119 of a zigzag line indicating the progress of the voltage wave on the line. The incident wave V + 1 starts at z = t = 0 and travels in the +z direction until it reaches the load at z = l at time t = T . At the very top of the bounce diagram, the reflection coefficients are indicated by = g at the generator end and by = L at the load end. At the end of the first straight-line segment of the zigzag line, a second line is drawn to represent the reflected voltage wave V − 1 = LV + 1 .
The amplitude of each new straight-line segment equals the product of the amplitude of the preceding straight-line segment and the reflection coefficient at that end of the line. The bounce diagram for the current I(z, t) in Fig. 2-42(b) adheres to the same principle except for the reversal of the signs of L and g at the top of the bounce diagram.
Using the bounce diagram, the total voltage (or current) at any point z1 and time t1 can be determined by drawing a vertical line through point z1, then adding the voltages (or currents) of all the zigzag segments intersected by that line between t = 0 and t = t1. To find the voltage at z = l/4 and T = 4T , for example, we draw a dashed vertical line in Fig. 2-42(a) through z = l/4 and we extend it from t = 0 to t = 4T . The dashed line intersects four line segments. The total voltage at z = l/4 and t = 4T therefore is V (l/4, 4T ) = V + 1 + LV + 1 + gLV + 1 + g2 LV + 1 = V + 1 (1 + L + gL + g2 L).
The time variation of V (z, t) at a specific location z can be obtained by plotting the values of V (z, t) along the (dashed) vertical line passing through z.
Figure 2-42(c) shows the variation of V as a function of time at z = l/4 for a circuit with g = 3/5 and L = 1/3.
Example 2-15: Pulse Propagation The transmission-line circuit of Fig. 2-43(a) is excited by a rectangular pulse of duration τ = 1 ns that starts at t = 0.
Establish the waveform of the voltage response at the load, given that the pulse amplitude is 5 V, the phase velocity is c, and the length of the line is 0.6 m.
Solution: The one-way propagation time is T = l c = 0.6 3 × 108 = 2 ns.
0 5 V 1 ns + _ 150 Ω 12.5 Ω Z0 = 50 Ω ΓL = 0.5 Γg = −0.6 1 ns 2 ns 3 ns 4 ns 5 ns 6 ns 7 ns 8 ns 9 ns 10 ns 11 ns 12 ns 1 ns 2 ns 3 ns 4 ns 5 ns 6 ns 7 ns 8 ns 9 ns 10 ns 11 ns 12 ns z = 0 z = l l/4 l/2 3l/4 t = 0 t t ΓL = 0.5 Γg = −0.6 V1 + = 4 V 2 V 1.2 V 0.6 V 0.36 V 0.18 V −4 V −2 V −1.2 V −0.6 V −0.36 V −0.18 V (a) Pulse circuit 1 2 2 V 0.54 V 4 V 6 V 3 4 5 VL (V) −2 V −4 V −1.8 V 8 9 10 11 t (ns) 12 6 7 (b) Bounce diagram (c) Voltage waveform at the load First step function Second step function Figure 2-43 Example 2-15.
120 CHAPTER 2 TRANSMISSION LINES The reflection coefficients at the load and the sending end are L = RL −Z0 RL + Z0 = 150 −50 150 + 50 = 0.5, g = Rg −Z0 Rg + Z0 = 12.5 −50 12.5 + 50 = −0.6.
By Eq. (2.147), the pulse is treated as the sum of two step functions, one that starts at t = 0 with an amplitude V10 = 5 V and a second one that starts at t = 1 ns with an amplitude V20 = −5 V. Except for the time delay of 1 ns and the sign reversal of all voltage values, the two step functions generate identical bounce diagrams, as shown in Fig. 2-43(b). For the first step function, the initial voltage is given by V + 1 = V01Z0 Rg + Z0 = 5 × 50 12.5 + 50 = 4 V.
Using the information displayed in the bounce diagram, it is straightforward to generate the voltage response shown in Fig. 2-43(c).
Example 2-16: Time-Domain Reflectometer A time-domain reflectometer (TDR) is an instrument used to locate faults on a transmission line. Consider, for example, a long underground or undersea cable that gets damaged at some distance d from the sending end of the line. The damage may alter the electrical properties or the shape of the cable, causing it to exhibit at the fault location an impedance RLf. A TDR sends a step voltage down the line, and by observing the voltage at the sending end as a function of time, it is possible to determine the location of the fault and its severity.
If the voltage waveform shown in Fig. 2-44(a) is seen on an oscilloscope connected to the input of a 75 matched transmission line, determine (a) the generator voltage, (b) the location of the fault, and (c) the fault shunt resistance. The line’s insulating material is Teflon with ϵr = 2.1.
(a) Observed voltage at the sending end (b) The fault at z = d is represented by a fault resistance Rf Drop in level caused by reflection from fault 12 μs 0 6 V 3 V V(0, t) t t = 0 z = 0 z = d Vg Rg = Z0 Rf Z0 Z0 RL = Z0 − + Figure 2-44 Time-domain reflectometer of Example 2-16.
Solution: (a) Since the line is properly matched, Rg = RL = Z0. In Fig. 2-44(b), the fault located a distance d from the sending end is represented by a shunt resistance Rf.
For a matched line, Eq. (2.149b) gives V + 1 = VgZ0 Rg + Z0 = VgZ0 2Z0 = Vg 2 .
According to Fig. 2-44(a), V + 1 = 6 V. Hence, Vg = 2V + 1 = 12 V.
(b) The propagation velocity on the line is up = c √ϵr = 3 × 108 √ 2.1 = 2.07 × 108 m/s.
For a fault at a distance d, the round-trip time delay of the echo is t = 2d up .
2-12 TRANSIENTS ON TRANSMISSION LINES 121 Module 2.10 Transient Response For a lossless line terminated in a resistive load, the module simulates the dynamic response, at any location on the line, to either a step or pulse waveform sent by the generator.
From Fig. 2-44(a), t = 12 μs. Hence, d = t 2 up = 12 × 10−6 2 × 2.07 × 108 = 1, 242 m.
(c) The change in level of V (0, t) shown in Fig. 2-44(a) represents V − 1 . Thus, V − 1 = fV + 1 = −3 V, or f = −3 6 = −0.5, where f is the reflection coefficient due to the fault load RLf that appears at z = d.
From Eq. (2.59), f = RLf −Z0 RLf + Z0 , which leads to RLf = 25 . This fault load is composed of the fault shunt resistance Rf and the characteristic impedance Z0 of the line to the right of the fault: 1 RLf = 1 Rf + 1 Z0 , so the shunt resistance must be 37.5 .
Concept Question 2-27: What is transient analysis used for?
Concept Question 2-28: The transient analysis pre-sented in this section was for a step voltage. How does one use it for analyzing the response to a pulse?
Concept Question 2-29: What is the difference be-tween the bounce diagram for voltage and the bounce diagram for current?
122 CHAPTER 2 TRANSMISSION LINES Chapter 2 Summary Concepts • A transmission line is a two-port network connecting a generator to a load. EM waves traveling on the line may experience ohmic power losses, dispersive effects, and reflections at the generator and load ends of the line.
These transmission-line effects may be ignored if the line length is much shorter than λ.
• TEM transmission lines consist of two conductors that can support the propagation of transverse electromagnetic waves characterized by electric and magnetic fields that are transverse to the direction of propagation.
TEM lines may be represented by a lumped-element model consisting of four line parameters (R′, L′, G′, and C ′) whose values are specified by the specific line geometry, the constitutive parameters of the conductors and of the insulating material between them, and the angular frequency ω.
• Wave propagation on a transmission line, which is represented by the phasor voltage V (z) and associated current ˜ I(z), is governed by the propagation constant of the line, γ = α + jβ, and its characteristic impedance Z0. Both γ and Z0 are specified by ω and the four line parameters.
• If R′ = G′ = 0, the line becomes lossless (α = 0). A losslesslineisgenerallynondispersive, meaningthatthe phasevelocityofawaveisindependentofthefrequency.
• In general, a line supports two waves, an incident wave supplied by the generator and another wave reflected by the load. The sum of the two waves generates a standing-wave pattern with a period of λ/2. The voltage standing-wave ratio S, which is equal to the ratio of the maximum to minimum voltage magnitude on the line, varies between 1 for a matched load (ZL = Z0) to ∞ for a line terminated in an open circuit, a short circuit, or a purely reactive load.
• The input impedance of a line terminated in a short circuit or open circuit is purely reactive.
This property can be used to design equivalent inductors and capacitors.
• The fraction of the incident power delivered to the load by a lossless line is equal to (1 −||2).
• The Smith chart is a useful graphical tool for analyzing transmission-line problems and for designing impedance-matching networks.
• Matching networks are placed between the load and the feed transmission line for the purpose of eliminating reflections toward the generator. A matching network may consist of lumped elements in the form of capacitors and/or inductors, or it may consist of sections of transmission lines with appropriate lengths and terminations.
• Transient analysis of pulses on transmission lines can be performed using a bounce-diagram graphical technique thattracksreflectionsatboth theloadandgeneratorends of the transmission line.
Mathematical and Physical Models TEM Transmission Lines L′C ′ = μϵ G′ C ′ = σ ϵ α = Re(γ ) = Re (R′ + jωL′)(G′ + jωC ′) (Np/m) β = Im(γ ) = Im (R′ + jωL′)(G′ + jωC ′) (rad/m) Z0 = R′ + jωL′ γ = R′ + jωL′ G′ + jωC ′ () = zL −1 zL + 1 CHAPTER 2 SUMMARY 123 Mathematical and Physical Models (continued) Step Function Transient Response V + 1 = VgZ0 Rg + Z0 V∞= VgRL Rg + RL g = Rg −Z0 Rg + Z0 L = RL −Z0 RL + Z0 Lossless Line α = 0 β = ω √ L′C ′ Z0 = L′ C ′ up = 1 √μϵ (m/s) λ = up f = c f 1 √ϵr = λ0 √ϵr dmax = θrλ 4π + nλ 2 dmin = θrλ 4π + (2n + 1)λ 4 S = 1 + || 1 −|| Pav = |V + 0 |2 2Z0 [1 −||2] Important Terms Provide definitions or explain the meaning of the following terms: admittance Y air line attenuation constant α bounce diagram characteristic impedance Z0 coaxial line complex propagation constant γ conductance G current maxima and minima dispersive transmission line distortionless line effective relative permittivity ϵeff guide wavelength λ higher-order transmission lines impedance matching in-phase input impedance Zin load impedance ZL lossless line lumped-element model matched transmission line matching network microstrip line normalized impedance normalized load reactance xL normalized load resistance rL open-circuited line optical fiber parallel-plate line perfect conductor perfect dielectric phase constant β phase opposition phase-shifted reflection coefficient d quarter-wave transformer short-circuited line single-stub matching slotted line Smith chart standing wave standing-wave pattern surface resistance Rs susceptance B SWR circle telegrapher’s equations TEM transmission lines time-average power Pav transient response transmission-line parameters two-wire line unit circle voltage maxima and minima voltage reflection coefficient voltage standing-wave ratio (VSWR or SWR) S wave equations wave impedance Z(d) waveguide WTG and WTL 124 CHAPTER 2 TRANSMISSION LINES PROBLEMS Sections 2-1 to 2-4: Transmission-Line Model 2.1 A transmission line of length l connects a load to a sinusoidal voltage source with an oscillation frequency f .
Assuming that the velocity of wave propagation on the line is c, for which of the following situations is it reasonable to ignore the presence of the transmission line in the solution of the circuit: ∗(a) l = 20 cm, f = 20 kHz (b) l = 50 km, f = 60 Hz ∗(c) l = 20 cm, f = 600 MHz (d) l = 1 mm, f = 100 GHz 2.2 A two-wire copper transmission line is embedded in a dielectric material with ϵr = 2.6 and σ = 2 × 10−6 S/m. Its wires are separated by 3 cm, and their radii are 1 mm each.
(a) Calculate the line parameters R′, L′, G′, and C′ at 2 GHz.
(b) Compare your results with those based on CD Module 2.1.
Include a printout of the screen display.
2.3 Show that the transmission-line model shown in Fig. P2.3 yields the same telegrapher’s equations given by Eqs. (2.14) and (2.16).
Δz R' Δz G' Δz C' Δz 2 L' Δz 2 R' Δz 2 L' Δz 2 i(z, t) i(z + Δz, t) υ(z + Δz, t) υ(z, t) − + − + Figure P2.3 Transmission-line model for Problem 2.3.
∗2.4 A 1 GHz parallel-plate transmission line consists of 1.2 cm wide copper strips separated by a 0.15 cm thick layer of polystyrene.
Appendix B gives μc = μ0 = 4π × 10−7 (H/m) and σc = 5.8 × 107 (S/m) for copper, and ϵr = 2.6 for ∗Answer(s) available in Appendix D.
polystyrene. Use Table 2-1 to determine the line parameters of the transmission line. Assume that μ = μ0 and σ ≈0 for polystyrene.
2.5 For the parallel-plate transmission line of Problem 2.4, the line parameters are given by R′ = 1 /m, L′ = 167 nH/m, G′ = 0, and C′ = 172 pF/m. Find α, β, up, and Z0 at 1 GHz.
2.6 A coaxial line with inner and outer conductor diameters of 0.5 cm and 1 cm, respectively, is filled with an insulating material with ϵr = 4.5 and σ = 10−3 S/m. The conductors are made of copper.
(a) Calculate the line parameters at 1 GHz.
(b) Compare your results with those based on CD Module 2.2.
Include a printout of the screen display.
2.7 Find α, β, up, and Z0 for the two-wire line of Problem 2.2.
Compare results with those based on CD Module 2.1. Include a printout of the screen display.
∗2.8 Find α, β, up, and Z0 for the coaxial line of Problem 2.6. Verify your results by applying CD Module 2.2. Include a printout of the screen display.
Section 2-5: The Lossless Microstrip Line 2.9 A lossless microstrip line uses a 1 mm wide conducting strip over a 1 cm thick substrate with ϵr = 2.5. Determine the line parameters ϵeff, Z0, and β at 10 GHz. Compare your results with those obtained by using CD Module 2.3. Include a printout of the screen display.
∗2.10 Use CD Module 2.3 to design a 100 microstrip transmission line. The substrate thickness is 1.8 mm and its ϵr = 2.3. Select the strip width w, and determine the guide wavelength λ at f = 5 GHz. Include a printout of the screen display.
2.11 A 50 microstrip line uses a 0.6 mm alumina substrate with ϵr = 9. Use CD Module 2.3 to determine the required strip width w. Include a printout of the screen display.
2.12 Generate a plot of Z0 as a function of strip width w, over the range from 0.05 mm to 5 mm, for a microstrip line fabricated on a 0.7 mm thick substrate with ϵr = 9.8.
PROBLEMS 125 Section 2-6: The Lossless Transmission Line: General Considerations 2.13 In addition to not dissipating power, a lossless line has two important features: (1) it is dispersionless (up is independent of frequency); and (2) its characteristic impedance Z0 is purely real. Sometimes, it is not possible to design a transmission line such that R′ ≪ωL′ and G′ ≪ωC ′, but it is possible to choose the dimensions of the line and its material properties so as to satisfy the condition R′C ′ = L′G′ (distortionless line) Such a line is called a distortionless line, because despite the fact that it is not lossless, it nonetheless possesses the previously mentioned features of the lossless line. Show that for a distortionless line, α = R′ C ′ L′ = √ R′G′ , β = ω √ L′C ′ , Z0 = L′ C ′ .
∗2.14 For a distortionless line [see Problem 2.13] with Z0 = 50 , α = 20 (mNp/m), and up = 2.5 × 108 (m/s), find the line parameters and λ at 100 MHz.
2.15 Find α and Z0 of a distortionless line whose R′ = 2 /m and G′ = 2 × 10−4 S/m.
∗2.16 A transmission line operating at 125 MHz has Z0 = 40 , α = 0.02 (Np/m), and β = 0.75 rad/m. Find the line parameters R′, L′, G′, and C ′.
2.17 Using a slotted line, the voltage on a lossless transmission line was found to have a maximum magnitude of 1.5 V and a minimum magnitude of 0.6 V. Find the magnitude of the load’s reflection coefficient.
∗2.18 Polyethylene with ϵr = 2.25 is used as the insulating material in a lossless coaxial line with a characteristic impedanceof50. Theradiusoftheinnerconductoris1.2mm.
(a) What is the radius of the outer conductor?
(b) What is the phase velocity of the line?
2.19 A 50 lossless transmission line is terminated in a load with impedance ZL = (30 −j50) . The wavelength is 8 cm.
Determine: (a) The reflection coefficient at the load.
(b) The standing-wave ratio on the line.
(c) The position of the voltage maximum nearest the load.
(d) The position of the current maximum nearest the load.
(e) Verify quantities in parts (a)–(d) using CD Module 2.4.
Include a printout of the screen display.
2.20 A 300 lossless air transmission line is connected to a complex load composed of a resistor in series with an inductor, as shown in Fig. P2.20. At 5 MHz, determine: (a) , (b) S, (c) location of voltage maximum nearest to the load, and (d) location of current maximum nearest to the load.
L = 0.02 mH Z0 = 300 Ω R = 600 Ω Figure P2.20 Circuit for Problem 2.20.
∗2.21 On a 150 lossless transmission line, the following observations were noted: distance of first voltage minimum from the load = 3 cm; distance of first voltage maximum from the load = 9 cm; S = 3. Find ZL.
2.22 Using a slotted line, the following results were obtained: distance of first minimum from the load = 4 cm; distance of second minimum from the load = 14 cm; voltage standing-wave ratio = 1.5. If the line is lossless and Z0 = 50 , find the load impedance.
∗2.23 A load with impedance ZL = (25 −j50) is to be connected to a lossless transmission line with characteristic impedance Z0, with Z0 chosen such that the standing-wave ratio is the smallest possible. What should Z0 be?
2.24 A 50 lossless line terminated in a purely resistive load has a voltage standing-wave ratio of 3. Find all possible values of ZL.
126 CHAPTER 2 TRANSMISSION LINES 2.25 Apply CD Module 2.4 to generate plots of the voltage standing-wave pattern for a 50 line terminated in a load impedance ZL = (100 −j50) . Set Vg = 1 V, Zg = 50 , ϵr = 2.25, l = 40 cm, and f = 1 GHz. Also determine S, dmax, and dmin.
2.26 A 50 lossless transmission line is connected to a load composed of a 75 resistor in series with a capacitor of unknown capacitance (Fig. P2.26). If at 10 MHz the voltage standing-wave ratio on the line was measured to be 3, determine the capacitance C.
RL = 75 Ω Z0 = 50 Ω C = ?
Figure P2.26 Circuit for Problem 2.26.
Section 2-7: Wave and Input Impedance ∗2.27 At an operating frequency of 300 MHz, a lossless 50 air-spaced transmission line 2.5 m in length is terminated with an impedance ZL = (40 + j20) . Find the input impedance.
2.28 A lossless transmission line of electrical length l = 0.35λ is terminated in a load impedance as shown in Fig. P2.28. Find , S, and Zin. Verify your results using CD Modules 2.4 or 2.5. Include a printout of the screen’s output display.
Zin Z0 = 100 Ω ZL = (60 + j30) Ω l = 0.35λ Figure P2.28 Circuit for Problem 2.28.
2.29 Show that the input impedance of a quarter-wavelength– long lossless line terminated in a short circuit appears as an open circuit.
2.30 Show that at the position where the magnitude of the voltage on the line is a maximum, the input impedance is purely real.
2.31 A voltage generator with vg(t) = 5 cos(2π × 109t) V and internal impedance Zg = 50 is connected to a 50 lossless air-spaced transmission line.
The line length is 5 cm, and the line is terminated in a load with impedance ZL = (100 −j100) . Determine: ∗(a) at the load.
(b) Zin at the input to the transmission line.
(c) The input voltage Vi and input current ˜ Ii.
(d) The quantities in (a)–(c) using CD Modules 2.4 or 2.5.
2.32 A 6 m section of 150 lossless line is driven by a source with vg(t) = 5 cos(8π × 107t −30◦) (V) and Zg = 150 . If the line, which has a relative permittivity ϵr = 2.25, is terminated in a load ZL = (150 −j50) , determine: (a) λ on the line.
∗(b) The reflection coefficient at the load.
(c) The input impedance.
(d) The input voltage Vi.
(e) The time-domain input voltage vi(t).
(f) Quantities in (a) to (d) using CD Modules 2.4 or 2.5.
2.33 Two half-wave dipole antennas, each with an impedance of 75 , are connected in parallel through a pair of transmission lines, and the combination is connected to a feed transmission line, as shown in Fig. P2.33. All lines are 50 and lossless.
∗(a) Calculate Zin1, the input impedance of the antenna-terminated line, at the parallel juncture.
(b) Combine Zin1 and Zin2 in parallel to obtain Z′ L, the effective load impedance of the feedline.
(c) Calculate Zin of the feedline.
PROBLEMS 127 75 Ω (Antenna) 75 Ω (Antenna) 0.3λ 0.2λ 0.2λ Zin1 Zin Zin2 Figure P2.33 Circuit for Problem 2.33.
2.34 A 50 lossless line is terminated in a load impedance ZL = (30 −j20) .
(a) Calculate and S.
(b) It has been proposed that by placing an appropriately selected resistor across the line at a distance dmax from the load (as shown in Fig. P2.34(b)), where dmax is the (a) (b) Z0 = 50 Ω ZL = (30 − j20) Ω Z0 = 50 Ω ZL = (30 − j20) Ω Zi dmax R Figure P2.34 Circuit for Problem 2.34.
distance from the load of a voltage maximum, then it is possible to render Zi = Z0, thereby eliminating reflection back to the end. Show that the proposed approach is valid and find the value of the shunt resistance.
∗2.35 For the lossless transmission line circuit shown in Fig. P2.35, determine the equivalent series lumped-element circuit at 400 MHz at the input to the line. The line has a characteristic impedance of 50 , and the insulating layer has ϵr = 2.25.
Z0 = 50 Ω 75 Ω Zin 1.2 m Figure P2.35 Circuit for Problem 2.35.
Section 2-8: Special Cases of the Lossless Line 2.36 At an operating frequency of 300 MHz, it is desired to use a section of a lossless 50 transmission line terminated in a short circuit to construct an equivalent load with reactance X = 40 . If the phase velocity of the line is 0.75c, what is the shortest possible line length that would exhibit the desired reactance at its input? Verify your result using CD Module 2.5.
∗2.37 A lossless transmission line is terminated in a short circuit. How long (in wavelengths) should the line be for it to appear as an open circuit at its input terminals?
2.38 The input impedance of a 31 cm long lossless transmission line of unknown characteristic impedance was measured at 1 MHz. With the line terminated in a short circuit, the measurement yielded an input impedance equivalent to an inductor with inductance of 0.064 μH, and when the line was open-circuited, the measurement yielded an input impedance equivalent to a capacitor with capacitance of 40 pF. Find Z0 of the line, the phase velocity, and the relative permittivity of the insulating material.
∗2.39 A 75 resistive load is preceded by a λ/4 section of a 50 lossless line, which itself is preceded by another λ/4 section of a 100 line. What is the input impedance?
128 CHAPTER 2 TRANSMISSION LINES Zin Generator 50 Ω λ/2 A B D C 250 V Line 1 λ/2 Line 2 λ/2 Line 3 ZL1 = 75 Ω (Antenna 1) ZL2 = 75 Ω (Antenna 2) − + Figure P2.43 Antenna configuration for Problem 2.43.
Compare your result with that obtained through two successive applications of CD Module 2.5.
2.40 A 100 MHz FM broadcast station uses a 300 transmission line between the transmitter and a tower-mounted half-wave dipole antenna. The antenna impedance is 73 .
You are asked to design a quarter-wave transformer to match the antenna to the line.
(a) Determine the electrical length and characteristic impe-dance of the quarter-wave section.
(b) If the quarter-wave section is a two-wire line with D = 2.5 cm, and the wires are embedded in polystyrene with ϵr = 2.6, determine the physical length of the quarter-wave section and the radius of the two wire conductors.
2.41 A 50 lossless line of length l = 0.375λ connects a 300 MHz generator with Vg = 300 V and Zg = 50 to a load ZL.
Determine the time-domain current through the load for: (a) ZL = (50 −j50) ∗(b) ZL = 50 (c) ZL = 0 (short circuit) For (a), verify your results by deducing the information you need from the output products generated by CD Module 2.4.
Section 2-9: Power Flow on a Lossless Transmission Line 2.42 A generator with Vg = 300 V and Zg = 50 is connected to a load ZL = 75 through a 50 lossless line of length l = 0.15λ.
∗(a) Compute Zin, the input impedance of the line at the generator end.
(b) Compute ˜ Ii and Vi.
(c) Compute the time-average power delivered to the line, Pin = 1 2Re[ Vi ˜ I ∗ i ].
(d) Compute VL, ˜ IL, and the time-average power delivered to the load, PL = 1 2Re[ VL ˜ I ∗ L]. How does Pin compare to PL?
Explain.
(e) Compute the time-average power delivered by the generator, Pg, and the time-average power dissipated in Zg. Is conservation of power satisfied?
2.43 If the two-antenna configuration shown in Fig. P2.43 is connected to a generator with Vg = 250V and Zg = 50 , how much average power is delivered to each antenna?
PROBLEMS 129 ∗2.44 For the circuit shown in Fig. P2.44, calculate the average incident power, the average reflected power, and the average power transmitted into the infinite 100 line. The λ/2 line is lossless and the infinitely long line is slightly lossy. (Hint: The input impedance of an infinitely long line is equal to its characteristic impedance so long as α ̸= 0.) Z0 = 50 Ω Z1 = 100 Ω λ/2 50 Ω 2 V Pav i Pav r Pav t 8 − + Figure P2.44 Circuit for Problem 2.44.
2.45 The circuit shown in Fig. P2.45 consists of a 100 lossless transmission line terminated in a load with ZL = (50 + j100) . If the peak value of the load voltage was measured to be | VL| = 12 V, determine: ∗(a) the time-average power dissipated in the load, (b) the time-average power incident on the line, (c) the time-average power reflected by the load.
Z0 = 100 Ω Rg Vg ~ ZL = (50 + j100) Ω − + Figure P2.45 Circuit for Problem 2.45.
2.46 An antenna with a load impedance ZL = (75 + j25) is connected to a transmitter through a 50 lossless transmission line. If under matched conditions (50 load) the transmitter can deliver 20 W to the load, how much power can it deliver to the antenna? Assume that Zg = Z0.
Section 2-10: The Smith Chart 2.47 Use the Smith chart to find the reflection coefficient corresponding to a load impedance of (a) ZL = 3Z0 ∗(b) ZL = (2 −j2)Z0 (c) ZL = −j2Z0 (d) ZL = 0 (short circuit) 2.48 Repeat Problem 2.47 using CD Module 2.6.
2.49 Use the Smith chart to find the normalized load impedance corresponding to a reflection coefficient of (a) = 0.5 (b) = 0.5∠ 60◦ (c) = −1 (d) = 0.3∠ −30◦ (e) = 0 (f) = j ∗2.50 Use the Smith chart to determine the input impedance Zin of the two-line configuration shown in Fig. P2.50.
2.51 Repeat Problem 2.50 using CD Module 2.6.
∗2.52 On a lossless transmission line terminated in a load ZL = 100 , the standing-wave ratio was measured to be 2.5.
Use the Smith chart to find the two possible values of Z0.
2.53 A lossless 50 transmission line is terminated in a load with ZL = (50 + j25) . Use the Smith chart to find the following: (a) The reflection coefficient .
∗(b) The standing-wave ratio.
(c) The input impedance at 0.35λ from the load.
(d) The input admittance at 0.35λ from the load.
(e) The shortest line length for which the input impedance is purely resistive.
(f) The position of the first voltage maximum from the load.
130 CHAPTER 2 TRANSMISSION LINES Z01 = 100 Ω l1 = 3λ/8 l2 = 5λ/8 Z02 = 50 Ω Zin B C A ZL = (75 − j50) Ω Figure P2.50 Circuit for Problem 2.50.
2.54 Repeat Problem 2.53 using CD Module 2.6.
∗2.55 A lossless 50 transmission line is terminated in a short circuit. Use the Smith chart to determine: (a) The input impedance at a distance 2.3λ from the load.
(b) The distance from the load at which the input admittance is Yin = −j0.04 S.
2.56 Repeat Problem 2.55 using CD Module 2.6.
∗2.57 Use the Smith chart to find yL if zL = 1.5 −j0.7.
2.58 A lossless 100 transmission line 3λ/8 in length is terminated in an unknown impedance. If the input impedance is Zin = −j2.5 , (a) Use the Smith chart to find ZL.
(b) Verify your results using CD Module 2.6.
2.59 A 75 lossless line is 0.6λ long.
If S = 1.8 and θr = −60◦, use the Smith chart to find ||, ZL, and Zin.
2.60 Repeat Problem 2.59 using CD Module 2.6.
∗2.61 Using a slotted line on a 50 air-spaced lossless line, the following measurements were obtained: S = 1.6 and | V |max occurred only at 10 cm and 24 cm from the load. Use the Smith chart to find ZL.
2.62 At an operating frequency of 5 GHz, a 50 lossless coaxial line with insulating material having a relative permittivity ϵr = 2.25 is terminated in an antenna with an impedance ZL = 150 . Use the Smith chart to find Zin. The line length is 30 cm.
Section 2-11: Impedance Matching ∗2.63 A 50 lossless line 0.6λ long is terminated in a load with ZL = (50+j25) . At 0.3λ from the load, a resistor with resistance R = 30 is connected as shown in Fig. P2.63. Use the Smith chart to find Zin.
Zin ZL ZL = (50 + j25) Ω Z0 = 50 Ω Z0 = 50 Ω 30 Ω 0.3λ 0.3λ Figure P2.63 Circuit for Problem 2.63.
2.64 Use CD Module 2.7 to design a quarter-wavelength transformer to match a load with ZL = (100 −j200) to a 50 line.
2.65 Use CD Module 2.7 to design a quarter-wavelength transformer to match a load with ZL = (50 + j10) to a 100 line.
2.66 A 200 transmission line is to be matched to a computer terminal with ZL = (50 −j25) by inserting an appropriate reactance in parallel with the line. If f = 800 MHz and ϵr = 4, determine the location nearest to the load at which inserting: (a) A capacitor can achieve the required matching, and the value of the capacitor.
PROBLEMS 131 (b) An inductor can achieve the required matching, and the value of the inductor.
2.67 Repeat Problem 2.66 using CD Module 2.8.
2.68 A 50 lossless line is to be matched to an antenna with ZL = (75 −j20) using a shorted stub. Use the Smith chart to determine the stub length and distance between the antenna and stub.
∗2.69 Repeat Problem 2.68 for a load with ZL = (100 + j50) .
2.70 Repeat Problem 2.68 using CD Module 2.9.
2.71 Repeat Problem 2.69 using CD Module 2.9.
2.72 Determine Zin of the feed line shown in Fig. P2.72. All lines are lossless with Z0 = 50 .
Z1 = (50 + j50) Ω Z2 = (50 − j50) Ω Zin 0.7λ 0.3λ 0.3λ Z1 Z2 Figure P2.72 Network for Problem 2.72.
∗2.73 Repeat Problem 2.72 for the case where all three transmission lines are λ/4 in length.
2.74 A 25 antenna is connected to a 75 lossless transmission line. Reflections back toward the generator can be eliminated by placing a shunt impedance Z at a distance l from the load (Fig. P2.74). Determine the values of Z and l.
Z0 = 75 Ω l = ?
Z = ?
ZL = 25 Ω B A Figure P2.74 Circuit for Problem 2.74.
Section 2-12: Transients on Transmission Lines 2.75 Generate a bounce diagram for the voltage V (z, t) for a 1 m long lossless line characterized by Z0 = 50 and up = 2c/3 (where c is the velocity of light) if the line is fed by a step voltage applied at t = 0 by a generator circuit with Vg = 60 V and Rg = 100 . The line is terminated in a load RL = 25 .
Use the bounce diagram to plot V (t) at a point midway along the length of the line from t = 0 to t = 25 ns.
2.76 Repeat Problem 2.75 for the current I(z, t) on the line.
2.77 In response to a step voltage, the voltage waveform shown in Fig. P2.77 was observed at the sending end of a lossless transmission line with Rg = 50 , Z0 = 50 , and ϵr = 2.25. Determine the following: (a) The generator voltage.
(b) The length of the line.
(c) The load impedance.
6 μs 0 V(0, t) t 5 V 3 V Figure P2.77 Voltage waveform for Problems 2.77 and 2.79.
132 CHAPTER 2 TRANSMISSION LINES ∗2.78 In response to a step voltage, the voltage waveform shown in Fig. P2.78 was observed at the sending end of a shorted line with Z0 = 50 and ϵr = 4. Determine Vg, Rg, and the line length.
7 μs 14 μs 0 V(0, t) t 12 V 0.75 V 3 V Figure P2.78 Voltage waveform of Problem 2.78.
2.79 Suppose the voltage waveform shown in Fig. P2.77 was observed at the sending end of a 50 transmission line in response to a step voltage introduced by a generator with Vg = 15 V and an unknown series resistance Rg. The line is 1 km in length, its velocity of propagation is 1 × 108 m/s, and it is terminated in a load RL = 100 .
(a) Determine Rg.
(b) Explain why the drop in level of V (0, t) at t = 6 μs cannot be due to reflection from the load.
(c) Determine the shunt resistance Rf and location of the fault responsible for the observed waveform.
2.80 A generator circuit with Vg = 200V and Rg = 25 was used to excite a 75 lossless line with a rectangular pulse of duration τ = 0.4 μs. The line is 200 m long, its up = 2 × 108 m/s, and it is terminated in a load RL = 125 .
(a) Synthesize the voltage pulse exciting the line as the sum of two step functions, Vg1(t) and Vg2(t).
(b) For each voltage step function, generate a bounce diagram for the voltage on the line.
(c) Use the bounce diagrams to plot the total voltage at the sending end of the line.
(d) Confirm the result of part (c) by applying CD Module 2.10.
2.81 For the circuit of Problem 2.80, generate a bounce diagram for the current and plot its time history at the middle of the line.
∗2.82 In response to a step voltage, the voltage waveform shown in Fig. P2.82 was observed at the midpoint of a lossless transmission line with Z0 = 50 and up = 2 × 108 m/s.
Determine: (a) the length of the line, (b) ZL, (c) Rg, and (d) Vg.
t (μs) V(l/2, t) −3 V 12 V 0 3 9 15 21 Figure P2.82 Circuit for Problem 2.82.
C H A P T E R 3 Vector Analysis Chapter Contents Overview, 134 3-1 Basic Laws of Vector Algebra, 134 3-2 Orthogonal Coordinate Systems, 140 3-3 Transformations between Coordinated Systems, 147 TB5 Global Positioning System, 150 3-4 Gradient of a Scalar Field, 154 3-5 Divergence of a Vector Field, 158 3-6 Curl of a Vector Field, 162 TB6 X-Ray Computed Tomography, 164 3-7 Laplacian Operator, 167 Chapter 3 Summary, 169 Problems, 171 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Use vector algebra in Cartesian, cylindrical, and spherical coordinate systems.
2. Transform vectors between the three primary coordinate systems.
3. Calculate the gradient of a scalar function and the divergence and curl of a vector function in any of the three primary coordinate systems.
4. Apply the divergence theorem and Stokes’s theorem.
134 CHAPTER 3 VECTOR ANALYSIS Overview In our examination of wave propagation on a transmission line in Chapter 2, the primary quantities we worked with were voltage, current, impedance, and power. Each of these is a scalar quantity, meaning that it can be completely specified by its magnitude, if it is a positive real number, or by its magnitude and phase angle if it is a negative or a complex number (a negative number has a positive magnitude and a phase angle of π (rad)). This chapter is concerned with vectors. A vector has a magnitude and a direction. The speed of an object is a scalar, whereas its velocity is a vector.
Starting in the next chapter and throughout the succeeding chapters in the book, the primary electromagnetic quantities we deal with are the electric and magnetic fields, E and H. These, and many other related quantities, are vectors. Vector analysis provides the mathematical tools necessary for expressing and manipulating vector quantities in an efficient and convenient manner. To specify a vector in three-dimensional space, it is necessary to specify its components along each of the three directions.
▶Several types of coordinate systems are used in the study of vector quantities, the most common being the Cartesian (or rectangular), cylindrical, and spherical systems.
A particular coordinate system is usually chosen to best suit the geometry of the problem under consideration. ◀ Vector algebra governs the laws of addition, subtraction, and “multiplication” of vectors. The rules of vector algebra and vector representation in each of the aforementioned orthogonal coordinate systems (including vector transformation between them) are two of the three major topics treated in this chapter.
The third topic is vector calculus, which encompasses the laws of differentiation and integration of vectors, the use of special vector operators (gradient, divergence, and curl), and the application of certain theorems that are particularly useful in the study of electromagnetics, most notably the divergence and Stokes’s theorems.
3-1 Basic Laws of Vector Algebra A vector is a mathematical object that resembles an arrow.
Vector A in Fig. 3-1 has magnitude (or length) A = |A| and unit vector ˆ a: A = ˆ a|A| = ˆ aA.
(3.1) a A = aA A 1 ˆ ˆ Figure 3-1 Vector A = ˆ aA has magnitude A = |A| and points in the direction of unit vector ˆ a = A/A.
The unit vector ˆ a has a magnitude of one (|ˆ a| = 1), and points from A’s tail or anchor to its head or tip. From Eq. (3.1), ˆ a = A |A| = A A .
(3.2) In the Cartesian (or rectangular) coordinate system shown in Fig. 3-2(a), the x, y, and z coordinate axes extend along directions of the three mutually perpendicular unit vectors ˆ x, ˆ y, and ˆ z, also called base vectors. The vector A in Fig. 3-2(b) may be decomposed as A = ˆ xAx + ˆ yAy + ˆ zAz, (3.3) where Ax, Ay, and Az are A’s scalar components along the x-, y-, and z axes, respectively. The component Az is equal to the perpendicular projection of A onto the z axis, and similar definitions apply to Ax and Ay. Application of the Pythagorean theorem, first to the right triangle in the x–y plane to express the hypotenuse Ar in terms of Ax and Ay, and then again to the vertical right triangle with sides Ar and Az and hypotenuse A, yields the following expression for the magnitude of A: A = |A| = + A2 x + A2 y + A2 z .
(3.4) Since A is a nonnegative scalar, only the positive root applies.
From Eq. (3.2), the unit vector ˆ a is ˆ a = A A = ˆ xAx + ˆ yAy + ˆ zAz + A2 x + A2 y + A2 z .
(3.5) Occasionally, we use the shorthand notation A = (Ax, Ay, Az) to denote a vector with components Ax, Ay, and Az in a Cartesian coordinate system.
3-1 BASIC LAWS OF VECTOR ALGEBRA 135 (a) Base vectors (b) Components of A z Az Az Ay Ax Ar A x y z x y z y x 1 1 1 2 3 2 3 2 3 ˆ ˆ ˆ Figure 3-2 Cartesian coordinate system: (a) base vectors ˆ x, ˆ y, and ˆ z, and (b) components of vector A.
3-1.1 Equality of Two Vectors Two vectors A and B are equal if they have equal magnitudes and identical unit vectors. Thus, if A = ˆ aA = ˆ xAx + ˆ yAy + ˆ zAz, (3.6a) B = ˆ bB = ˆ xBx + ˆ yBy + ˆ zBz, (3.6b) then A = B if and only if A = B and ˆ a = ˆ b, which requires that Ax = Bx, Ay = By, and Az = Bz.
▶Equality of two vectors does not necessarily imply that they are identical; in Cartesian coordinates, two displaced parallel vectors of equal magnitude and pointing in the same direction are equal, but they are identical only if they lie on top of one another. ◀ 3-1.2 Vector Addition and Subtraction The sum of two vectors A and B is a vector C = ˆ x Cx + ˆ y Cy + ˆ z Cz, given by C = A + B = (ˆ xAx + ˆ yAy + ˆ zAz) + (ˆ xBx + ˆ yBy + ˆ zBz) = ˆ x(Ax + Bx) + ˆ y(Ay + By) + ˆ z(Az + Bz) = ˆ x Cx + ˆ y Cy + ˆ z Cz.
(3.7) ▶Hence, vector addition is commutative: C = A + B = B + A.
(3.8) Graphically, vector addition can be accomplished by either the parallelogram or the head-to-tail rule (Fig. 3-3). Vector C is the diagonal of the parallelogram with sides A and B. With the head-to-tail rule, we may either add A to B or B to A. When A is added to B, it is repositioned so that its tail starts at the tip of B, while keeping its length and direction unchanged. The sum vector C starts at the tail of B and ends at the tip of A.
Subtraction of vector B from vector A is equivalent to the addition of A to negative B. Thus, D = A −B = A + (−B) = ˆ x(Ax −Bx) + ˆ y(Ay −By) + ˆ z(Az −Bz).
(3.9) Graphically, the same rules used for vector addition are also applicable to vector subtraction; the only difference is that the arrowhead of (−B) is drawn on the opposite end of the line segment representing the vector B (i.e., the tail and head are interchanged).
A B C (a) Parallelogram rule A B C (b) Head-to-tail rule Figure 3-3 Vector addition by (a) the parallelogram rule and (b) the head-to-tail rule.
136 CHAPTER 3 VECTOR ANALYSIS z2 y2 z1 y1 x1 x2 x y R1 R2 R12 z P1 = (x1, y1, z1) P2 = (x2, y2, z2) O Figure3-4 DistancevectorR12 = − − − → P1P2 = R2−R1, whereR1 and R2 are the position vectors of points P1 and P2, respectively.
3-1.3 Position and Distance Vectors The position vector of a point P in space is the vector from the origin to P . Assuming points P1 and P2 are at (x1, y1, z1) and (x2, y2, z2) in Fig. 3-4, their position vectors are R1 = − → OP1 = ˆ xx1 + ˆ yy1 + ˆ zz1, (3.10a) R2 = − → OP2 = ˆ xx2 + ˆ yy2 + ˆ zz2, (3.10b) where point O is the origin. The distance vector from P1 to P2 is defined as R12 = − − → P1P2 = R2 −R1 = ˆ x(x2 −x1) + ˆ y(y2 −y1) + ˆ z(z2 −z1), (3.11) and the distance d between P1 and P2 equals the magnitude of R12: d = |R12| = [(x2 −x1)2 + (y2 −y1)2 + (z2 −z1)2]1/2.
(3.12) Note that the first and second subscripts of R12 denote the locations of its tail and head, respectively (Fig. 3-4).
3-1.4 Vector Multiplication Thereexistthreetypesofproductsinvectorcalculus: thesimple product, the scalar (or dot) product, and the vector (or cross) product.
Simple product The multiplication of a vector by a scalar is called a simple product. The product of the vectorA = ˆ aA by a scalar k results in a vector B with magnitude B = kA and direction the same as A. That is, ˆ b = ˆ a. In Cartesian coordinates, B = kA = ˆ akA = ˆ x(kAx) + ˆ y(kAy) + ˆ z(kAz) = ˆ x Bx + ˆ y By + ˆ z Bz.
(3.13) Scalar or dot product The scalar (or dot) product of two co-anchored vectors A and B, denoted A· B and pronounced “A dot B,” is defined geometrically as the product of the magnitude of A and the scalar component of B along A, or vice versa. Thus, A· B = AB cos θAB, (3.14) where θAB is the angle between A and B (Fig. 3-5) measured from the tail of A to the tail of B. Angle θAB is assumed to be in the range 0 ≤θAB ≤180◦. The scalar product of A and B yields a scalar whose magnitude is less than or equal to the products of their magnitudes (equality holds when θAB = 0) and whose sign is positive if 0 < θAB < 90◦and negative if 90◦< θAB < 180◦. When θAB = 90◦,A and B are orthogonal, and their dot product is zero. The quantity A cos θAB is the (a) (b) B θBA θAB A θAB θBA A B Figure 3-5 The angle θAB is the angle between A and B, measured from A to B between vector tails. The dot product is positive if 0 ≤θAB < 90◦, as in (a), and it is negative if 90◦< θAB ≤180◦, as in (b).
3-1 BASIC LAWS OF VECTOR ALGEBRA 137 scalar component of A along B. Similarly B cos θBA is the scalar component of B along A.
The dot product obeys both the commutative and distributive properties of multiplication: A· B = B·A, (3.15a) (commutative property) A·(B + C) = A· B + A· C, (3.15b) (distributive property) The commutative property follows from Eq. (3.14) and the fact that θAB = θBA. The distributive property expresses the fact that the scalar component of the sum of two vectors along a third one equals the sum of their respective scalar components.
The dot product of a vector with itself gives A·A = |A|2 = A2, (3.16) which implies that A = |A| = + √ A·A .
(3.17) Also, θAB can be determined from θAB = cos−1 A· B + √ A·A + √ B· B .
(3.18) Since the base vectors ˆ x, ˆ y, and ˆ z are each orthogonal to the other two, it follows that ˆ x· ˆ x = ˆ y· ˆ y = ˆ z· ˆ z = 1, (3.19a) ˆ x· ˆ y = ˆ y· ˆ z = ˆ z· ˆ x = 0.
(3.19b) If A = (Ax, Ay, Az) and B = (Bx, By, Bz), then A· B = (ˆ xAx + ˆ yAy + ˆ zAz)·(ˆ xBx + ˆ yBy + ˆ zBz).
(3.20) Use of Eqs. (3.19a) and (3.19b) in Eq. (3.20) leads to A· B = AxBx + AyBy + AzBz.
(3.21) Vector or cross product The vector (or cross) product of two vectors A and B, denoted A × × × B and pronounced “A cross B,” yields a vector defined as A × × × B = ˆ n AB sin θAB, (3.22) where ˆ n is a unit vector normal to the plane containing A and B [Fig. 3-6(a)].
The magnitude of the cross product, AB| sin θAB|, equals the area of the parallelogram defined by the two vectors. The direction of ˆ n is governed by the following right-hand rule [Fig. 3-6(b)]: ˆ n points in the direction of the right thumb when the fingers rotate from A to B through the angle θAB. Note that, since ˆ n is perpendicular to the plane containing A and B, A × × × B is perpendicular to both vectors A and B.
(a) Cross product (b) Right-hand rule z y x n B A θAB A × B = n AB sin θAB ˆ ˆ B A A × B Figure 3-6 Cross productA× × ×B points in the direction ˆ n, which is perpendicular to the plane containing A and B and defined by the right-hand rule.
138 CHAPTER 3 VECTOR ANALYSIS The cross product is anticommutative and distribu-tive: A × × × B = −B × × × A (anticommutative).
(3.23a) The anticommutative property follows from the application of the right-hand rule to determine ˆ n. The distributive property follows from the fact that the area of the parallelogram formed by A and (B + C) equals the sum of those formed by (A and B) and (A and C): A × × × (B + C) = A × × × B + A × × × C, (3.23b) (distributive) The cross product of a vector with itself vanishes. That is, A × × × A = 0.
(3.24) From the definition of the cross product given by Eq. (3.22), it is easy to verify that the base vectors ˆ x, ˆ y, and ˆ z of the Cartesian coordinate system obey the following right-hand cyclic relations: ˆ x × × × ˆ y = ˆ z, ˆ y × × × ˆ z = ˆ x, ˆ z × × × ˆ x = ˆ y.
(3.25) Note the cyclic order (xyzxyz . . .). Also, ˆ x × × × ˆ x = ˆ y × × × ˆ y = ˆ z × × × ˆ z = 0.
(3.26) If A = (Ax, Ay, Az) and B = (Bx, By, Bz), then use of Eqs. (3.25) and (3.26) leads to A × × × B = (ˆ xAx + ˆ yAy + ˆ zAz) × × × (ˆ xBx + ˆ yBy + ˆ zBz) = ˆ x(AyBz −AzBy) + ˆ y(AzBx −AxBz) + ˆ z(AxBy −AyBx).
(3.27) The cyclical form of the result given by Eq. (3.27) allows us to express the cross product in the form of a determinant: A × × × B = ˆ x ˆ y ˆ z Ax Ay Az Bx By Bz .
(3.28) Example 3-1: Vectors and Angles In Cartesian coordinates, vector A points from the origin to point P1 = (2, 3, 3), and vector B is directed from P1 to point P2 = (1, −2, 2). Find: (a) vector A, its magnitude A, and unit vector ˆ a, (b) the angle between A and the y axis, (c) vector B, (d) the angle θAB between A and B, and (e) perpendicular distance from the origin to vector B.
Solution: (a) Vector A is given by the position vector of P1 = (2, 3, 3) (Fig. 3-7). Thus, A = ˆ x2 + ˆ y3 + ˆ z3, A = |A| = 22 + 32 + 32 = √ 22 , ˆ a = A A = (ˆ x2 + ˆ y3 + ˆ z3)/ √ 22 .
β θAB A P3 B P2 = (1, –2, 2) P1 = (2, 3, 3) –2 1 1 2 3 3 2 z y x O Figure 3-7 Geometry of Example 3-1.
3-1 BASIC LAWS OF VECTOR ALGEBRA 139 (b) The angle β between A and the y axis is obtained from A· ˆ y = |A||ˆ y| cos β = A cos β, or β = cos−1 A· ˆ y A = cos−1 3 √ 22 = 50.2◦.
(c) B = ˆ x(1 −2) + ˆ y(−2 −3) + ˆ z(2 −3) = −ˆ x −ˆ y5 −ˆ z.
(d) θAB = cos−1 A· B |A||B| = cos−1 (−2 −15 −3) √ 22 √ 27 = 145.1◦.
(e) The perpendicular distance between the origin and vector B is the distance |− → OP3 | shown in Fig. 3-7. From right triangle OP1P3, |− → OP3 | = |A| sin(180◦−θAB) = √ 22 sin(180◦−145.1◦) = 2.68.
Exercise 3-1: Find the distance vector between P1 = (1, 2, 3) and P2 = (−1, −2, 3) in Cartesian coordinates.
Answer: − − → P1P2 = −ˆ x2 −ˆ y4. (See EM.) Exercise 3-2: Find the angle θAB between vectors A and B of Example 3-1 from the cross product between them.
Answer: θAB = 145.1◦. (See EM.) Exercise 3-3: Find the angle between vector B of Example 3-1 and the z axis.
Answer: 101.1◦. (See EM.) Exercise 3-4: VectorsAandBlieinthey-z planeandboth have the same magnitude of 2 (Fig. E3.4). Determine (a) A· B and (b) A × × × B.
y z x 2 2 A B 30◦ Figure E3.4 Answer: (a) A· B = −2; (b) A × × × B = ˆ x 3.46. (See EM.) Exercise 3-5: IfA· B = A· C, does it follow that B = C?
Answer: No. (See EM.) 3-1.5 Scalar and Vector Triple Products When three vectors are multiplied, not all combinations of dot and cross products are meaningful. For example, the product A × × × (B· C) does not make sense because B· C is a scalar, and the cross product of the vector A with a scalar is not defined under the rules of vector algebra. Other than the product of the form A(B· C), the only two meaningful products of three vectors are the scalar triple product and the vector triple product.
Scalar triple product The dot product of a vector with the cross product of two other vectors is called a scalar triple product, so named because the result is a scalar. A scalar triple product obeys the following cyclic order: A·(B × × × C) = B·(C × × × A) = C·(A × × × B).
(3.29) The equalities hold as long as the cyclic order (ABCABC . . .) is preserved.
The scalar triple product of vectors 140 CHAPTER 3 VECTOR ANALYSIS A = (Ax, Ay, Az), B = (Bx, By, Bz), and C = (Cx, Cy, Cz) can be expressed in the form of a 3 × 3 determinant: A·(B × × × C) = Ax Ay Az Bx By Bz Cx Cy Cz .
(3.30) The validity of Eqs. (3.29) and (3.30) can be verified by expanding A, B, and C in component form and carrying out the multiplications.
Vector triple product The vector triple product involves the cross product of a vector with the cross product of two others, such as A × × × (B × × × C).
(3.31) Since each cross product yields a vector, the result of a vector triple product is also a vector. The vector triple product does not obey the associative law. That is, A × × × (B × × × C) ̸= (A × × × B) × × × C, (3.32) which means that it is important to specify which cross multiplicationistobeperformedfirst. Byexpandingthevectors A, B, and C in component form, it can be shown that A × × × (B × × × C) = B(A· C) −C(A· B), (3.33) which is known as the “bac-cab” rule.
Example 3-2: Vector Triple Product Given A = ˆ x −ˆ y + ˆ z2, B = ˆ y + ˆ z, and C = −ˆ x2 + ˆ z3, find (A × × × B) × × × C and compare it with A × × × (B × × × C).
Solution: A × × × B = ˆ x ˆ y ˆ z 1 −1 2 0 1 1 = −ˆ x3 −ˆ y + ˆ z and (A × × × B) × × × C = ˆ x ˆ y ˆ z −3 −1 1 −2 0 3 = −ˆ x3 + ˆ y7 −ˆ z2.
A similar procedure gives A × × × (B × × × C) = ˆ x2 + ˆ y4 + ˆ z. The fact that the results of two vector triple products are different demonstrates the inequality stated in Eq. (3.32).
Concept Question 3-1: When are two vectors equal and when are they identical?
Concept Question 3-2: When is the position vector of a point identical to the distance vector between two points?
Concept Question 3-3: If A· B = 0, what is θAB?
Concept Question 3-4: If A × × × B = 0, what is θAB?
Concept Question 3-5: Is A(B· C) a vector triple product?
Concept Question 3-6: IfA· B = A· C, does it follow that B = C?
3-2 Orthogonal Coordinate Systems A three-dimensional coordinate system allows us to uniquely specify locations of points in space and the magnitudes and directions of vectors. Coordinate systems may be orthogonal or nonorthogonal.
▶ An orthogonal coordinate system is one in which coordinates are measured along locally mutually perpendicular axes. ◀ Nonorthogonal systems are very specialized and seldom used in solving practical problems.
Many orthogonal coordinate systems have been devised, but the most commonly used are • the Cartesian (also called rectangular), • the cylindrical, and • the spherical coordinate system.
Why do we need more than one coordinate system? Whereas a point in space has the same location and an object has the same shape regardless of which coordinate system is 3-2 ORTHOGONAL COORDINATE SYSTEMS 141 used to describe them, the solution of a practical problem can be greatly facilitated by the choice of a coordinate system that best fits the geometry under consideration. The following subsections examine the properties of each of the aforementioned orthogonal systems, and Section 3-3 describes how a point or vector may be transformed from one system to another.
3-2.1 Cartesian Coordinates The Cartesian coordinate system was introduced in Section 3-1 to illustrate the laws of vector algebra. Instead of repeating these laws for the Cartesian system, we summarize them in Table 3-1. Differential calculus involves the use of differential lengths, areas, and volumes.
In Cartesian coordinates a differential length vector (Fig. 3-8) is expressed as dl = ˆ x dlx + ˆ y dly + ˆ z dlz = ˆ x dx + ˆ y dy + ˆ z dz, (3.34) where dlx = dx is a differential length along ˆ x, and similar interpretations apply to dly = dy and dlz = dz.
A differential area vector ds is a vector with magnitude ds equal to the product of two differential lengths (such as dly and dlz), and direction specified by a unit vector along the third direction (such as ˆ x). Thus, for a differential area vector in the y–z plane, dsx = ˆ x dly dlz = ˆ x dy dz (y–z plane), (3.35a) Table 3-1 Summary of vector relations.
Cartesian Cylindrical Spherical Coordinates Coordinates Coordinates Coordinate variables x, y, z r, φ, z R, θ, φ Vector representation A = ˆ xAx + ˆ yAy + ˆ zAz ˆ rAr + ˆ φ φ φAφ + ˆ zAz ˆ RAR + ˆ θ θ θAθ + ˆ φ φ φAφ Magnitude of A |A| = + A2 x + A2 y + A2 z + A2 r + A2 φ + A2 z + A2 R + A2 θ + A2 φ Position vector − → OP1 = ˆ xx1 + ˆ yy1 + ˆ zz1, ˆ rr1 + ˆ zz1, ˆ RR1, for P(x1, y1, z1) for P(r1, φ1, z1) for P(R1, θ1, φ1) Base vectors properties ˆ x· ˆ x = ˆ y· ˆ y = ˆ z· ˆ z = 1 ˆ r· ˆ r = ˆ φ φ φ· ˆ φ φ φ = ˆ z· ˆ z = 1 ˆ R· ˆ R = ˆ θ θ θ· ˆ θ θ θ = ˆ φ φ φ· ˆ φ φ φ = 1 ˆ x· ˆ y = ˆ y· ˆ z = ˆ z· ˆ x = 0 ˆ r· ˆ φ φ φ = ˆ φ φ φ· ˆ z = ˆ z· ˆ r = 0 ˆ R· ˆ θ θ θ = ˆ θ θ θ· ˆ φ φ φ = ˆ φ φ φ· ˆ R = 0 ˆ x × × × ˆ y = ˆ z ˆ r × × × ˆ φ φ φ = ˆ z ˆ R × × × ˆ θ θ θ = ˆ φ φ φ ˆ y × × × ˆ z = ˆ x ˆ φ φ φ × × × ˆ z = ˆ r ˆ θ θ θ × × × ˆ φ φ φ = ˆ R ˆ z × × × ˆ x = ˆ y ˆ z × × × ˆ r = ˆ φ φ φ ˆ φ φ φ × × × ˆ R = ˆ θ θ θ Dot product A· B = AxBx + AyBy + AzBz ArBr + AφBφ + AzBz ARBR + AθBθ + AφBφ Cross product A × × × B = ˆ x ˆ y ˆ z Ax Ay Az Bx By Bz ˆ r ˆ φ φ φ ˆ z Ar Aφ Az Br Bφ Bz ˆ R ˆ θ θ θ ˆ φ φ φ AR Aθ Aφ BR Bθ Bφ Differential length dl = ˆ x dx + ˆ y dy + ˆ z dz ˆ r dr + ˆ φ φ φr dφ + ˆ z dz ˆ R dR + ˆ θ θ θR dθ + ˆ φ φ φR sin θ dφ Differential surface areas dsx = ˆ x dy dz dsy = ˆ y dx dz dsz = ˆ z dx dy dsr = ˆ rr dφ dz dsφ = ˆ φ φ φ dr dz dsz = ˆ zr dr dφ dsR = ˆ RR2 sin θ dθ dφ dsθ = ˆ θ θ θR sin θ dR dφ dsφ = ˆ φ φ φR dR dθ Differential volume dv = dx dy dz r dr dφ dz R2 sin θ dR dθ dφ 142 CHAPTER 3 VECTOR ANALYSIS dsz = z dx dy dsy = y dx dz dsx = x dy dz dx dz dy dv = dx dy dz dz dy dx dl z y x ˆ ˆ ˆ Figure 3-8 Differential length, area, and volume in Cartesian coordinates.
x φ1 φ z r R1 z y O φ = φ1 plane r = r1 cylinder ˆ ˆ ˆ r1 R1 P = (r1, φ1, z1) z = z1 plane z1 Figure 3-9 Point P (r1, φ1, z1) in cylindrical coordinates; r1 is the radial distance from the origin in the x–y plane, φ1 is the angle in the x–y plane measured from the x axis toward the y axis, and z1 is the vertical distance from the x–y plane.
with the subscript on ds denoting its direction. Similarly, dsy = ˆ y dx dz (x–z plane), (3.35b) dsz = ˆ z dx dy (x–y plane).
(3.35c) A differential volume equals the product of all three differential lengths: dv = dx dy dz.
(3.36) 3-2.2 Cylindrical Coordinates The cylindrical coordinate system is useful for solving problems involving structures with cylindrical symmetry, such as calculating the capacitance per unit length of a coaxial transmission line.
In the cylindrical coordinate system, the location of a point in space is defined by three variables, r, φ, and z (Fig. 3-9). The coordinate r is the radial distance in the x–y plane, φ is the azimuth angle measured from the positive x axis, and z is as previously defined in the Cartesian coordinate system. Their ranges are 0 ≤r < ∞, 0 ≤φ < 2π, and −∞< z < ∞. Point P(r1, φ1, z1) in Fig. 3-9 is located 3-2 ORTHOGONAL COORDINATE SYSTEMS 143 at the intersection of three surfaces. These are the cylindrical surface defined by r = r1, the vertical half-plane defined by φ = φ1 (which extends outwardly from the z axis), and the horizontal plane defined by z = z1.
▶The mutually perpendicular base vectors are ˆ r, ˆ φ φ φ, and ˆ z, with ˆ r pointing away from the origin along r, ˆ φ φ φ pointing in a direction tangential to the cylindrical surface, and ˆ z pointing along the vertical. Unlike the Cartesian system, in which the base vectors ˆ x, ˆ y, and ˆ z are independent of the location of P, in the cylindrical system both ˆ r and ˆ φ φ φ are functions of φ. ◀ The base unit vectors obey the following right-hand cyclic relations: ˆ r × × × ˆ φ φ φ = ˆ z, ˆ φ φ φ × × × ˆ z = ˆ r, ˆ z × × × ˆ r = ˆ φ φ φ, (3.37) and like all unit vectors, ˆ r· ˆ r = ˆ φ φ φ· ˆ φ φ φ = ˆ z· ˆ z = 1, and ˆ r × × × ˆ r = ˆ φ φ φ × × × ˆ φ φ φ = ˆ z × × × ˆ z = 0.
In cylindrical coordinates, a vector is expressed as A = ˆ a|A| = ˆ rAr + ˆ φ φ φAφ + ˆ zAz, (3.38) where Ar, Aφ, and Az are the components of A along the ˆ r, ˆ φ φ φ, and ˆ z directions. The magnitude of A is obtained by applying Eq. (3.17), which gives |A| = + √ A·A = + A2 r + A2 φ + A2 z .
(3.39) The position vector − → OP shown in Fig. 3-9 has components along r and z only. Thus, R1 = − → OP = ˆ rr1 + ˆ zz1.
(3.40) The dependence of R1 on φ1 is implicit through the dependence of ˆ r on φ1. Hence, when using Eq. (3.40) to denote the position vector of point P(r1, φ1, z1), it is necessary to specify that ˆ r is at φ1.
dv = r dr dφ dz dr r dφ dsφ = ϕ dr dz dsr = r r dφ dz dz dz φ r dr r dφ z y x O dsz = z r dr dφ ˆ ˆ ˆ Figure 3-10 Differential areas and volume in cylindrical coordinates.
Figure 3-10 shows a differential volume element in cylindrical coordinates. The differential lengths along ˆ r, ˆ φ φ φ, and ˆ z are dlr = dr, dlφ = r dφ, dlz = dz.
(3.41) Note that the differential length along ˆ φ φ φ is r dφ, not just dφ.
The differential length dl in cylindrical coordinates is given by dl = ˆ r dlr + ˆ φ φ φ dlφ + ˆ z dlz = ˆ r dr + ˆ φ φ φr dφ + ˆ z dz. (3.42) As was stated previously for the Cartesian coordinate system, the product of any pair of differential lengths is equal to the magnitude of a vector differential surface area with a surface normal pointing along the direction of the third coordinate.
Thus, dsr = ˆ r dlφ dlz = ˆ rr dφ dz (φ–z cylindrical surface), (3.43a) dsφ = ˆ φ φ φ dlr dlz = ˆ φ φ φ dr dz (r–z plane), (3.43b) dsz = ˆ z dlr dlφ = ˆ zr dr dφ (r–φ plane).
(3.43c) 144 CHAPTER 3 VECTOR ANALYSIS The differential volume is the product of the three differential lengths, dv = dlr dlφ dlz = r dr dφ dz.
(3.44) These properties of the cylindrical coordinate system are summarized in Table 3-1.
Example 3-3: Distance Vector in Cylindrical Coordinates Find an expression for the unit vector of vector A shown in Fig. 3-11 in cylindrical coordinates.
Solution: In triangle OP1P2, − → OP2 = − → OP1 +A.
Hence, A = − → OP2 −− → OP1 = ˆ rr0 −ˆ zh, and ˆ a = A |A| = ˆ rr0 −ˆ zh r2 0 + h2 .
We note that the expression for A is independent of φ0. This implies that all vectors from point P1 to any point on the circle defined by r = r0 in the x–y plane are equal in the cylindrical φ0 r0 P2 = (r0, φ0, 0) P1 = (0, 0, h) O a A x y z h ˆ Figure 3-11 Geometry of Example 3-3.
coordinate system, which is not true. The ambiguity can be resolved by specifying that A passes through a point whose φ = φ0.
Example 3-4: Cylindrical Area Find the area of a cylindrical surface described by r = 5, 30◦≤φ ≤60◦, and 0 ≤z ≤3 (Fig. 3-12).
60° 30° z = 3 r = 5 x y z Figure 3-12 Cylindrical surface of Example 3-4.
Solution: The prescribed surface is shown in Fig. 3-12. Use of Eq. (3.43a) for a surface element with constant r gives S = r 60◦ φ=30◦ dφ 3 z=0 dz = 5φ π/3 π/6 z 3 0 = 5π 2 .
Note that φ had to be converted to radians before evaluating the integration limits.
Exercise 3-6: A circular cylinder of radius r = 5 cm is concentric with the z axis and extends between z = −3 cm and z = 3 cm. Use Eq. (3.44) to find the cylinder’s volume.
Answer: 471.2 cm3. (See EM.) 3-2 ORTHOGONAL COORDINATE SYSTEMS 145 Module 3.1 Vector Addition and Subtraction Display two vectors in rectangular or cylindrical coordinates, and compute their sum and difference.
3-2.3 Spherical Coordinates In the spherical coordinate system, the location of a point in space is uniquely specified by the variables R, θ, and φ (Fig. 3-13).
The range coordinate R, which measures the distance from the origin to the point, describes a sphere of radius R centered at the origin. The zenith angle θ is measured from the positive z axis and it describes a conical surface with its apex at the origin, and the azimuth angle φ is the same as in cylindrical coordinates. The ranges of R, θ, and φ are 0 ≤R < ∞, 0 ≤θ ≤π, and 0 ≤φ < 2π. The base vectors ˆ R, ˆ θ θ θ, and ˆ φ φ φ obey the following right-hand cyclic relations: ˆ R × × × ˆ θ θ θ = ˆ φ φ φ, ˆ θ θ θ × × × ˆ φ φ φ = ˆ R, ˆ φ φ φ × × × ˆ R = ˆ θ θ θ.
(3.45) A vector with components AR, Aθ, and Aφ is written as A = ˆ a|A| = ˆ RAR + ˆ θ θ θAθ + ˆ φ φ φAφ, (3.46) and its magnitude is |A| = + √ A·A = + A2 R + A2 θ + A2 φ .
(3.47) The position vector of point P(R1, θ1, φ1) is simply R1 = − → OP = ˆ RR1, (3.48) while keeping in mind that ˆ R is implicitly dependent on θ1 and φ1.
As shown in Fig. 3-14, the differential lengths along ˆ R, ˆ θ θ θ, and ˆ φ φ φ are dlR = dR, dlθ = R dθ, dlφ = R sin θ dφ.
(3.49) 146 CHAPTER 3 VECTOR ANALYSIS θ1 R1 φ1 R θ y z x θ = θ1 conical surface P = (R1, θ1, φ1) φ φ ˆ ˆ ˆ ˆ Figure 3-13 Point P (R1, θ1, φ1) in spherical coordinates.
dθ R dθ dφ R dR y z x R sin θ dφ dν = R2 sin θ dR dθ dφ θ φ Figure 3-14 Differential volume in spherical coordinates.
Hence, the expressions for the vector differential length dl, thevectordifferentialsurfaceds, andthedifferentialvolumedv are dl = ˆ R dlR + ˆ θ θ θ dlθ + ˆ φ φ φ dlφ = ˆ R dR + ˆ θ θ θR dθ + ˆ φ φ φR sin θ dφ, (3.50a) dsR = ˆ R dlθ dlφ = ˆ RR2 sin θ dθ dφ (3.50b) (θ–φ spherical surface), dsθ = ˆ θ θ θ dlR dlφ = ˆ θ θ θR sin θ dR dφ (3.50c) (R–φ conical surface), dsφ = ˆ φ φ φ dlR dlθ = ˆ φ φ φR dR dθ (R–θ plane), (3.50d) dv = dlR dlθ dlφ = R2 sin θ dR dθ dφ.
(3.50e) These relations are summarized in Table 3-1.
Example 3-5: Surface Area in Spherical Coordinates The spherical strip shown in Fig. 3-15 is a section of a sphere of radius 3 cm. Find the area of the strip.
z y x 3 cm 60o 30o Figure 3-15 Spherical strip of Example 3-5.
3-3 TRANSFORMATIONS BETWEEN COORDINATE SYSTEMS 147 Solution: Use of Eq. (3.50b) for the area of an elemental spherical area with constant radius R gives S = R2 60◦ θ=30◦ sin θ dθ 2π φ=0 dφ = 9(−cos θ) 60◦ 30◦φ 2π 0 (cm2) = 18π(cos 30◦−cos 60◦) = 20.7 cm2.
Example 3-6: Charge in a Sphere A sphere of radius 2 cm contains a volume charge density ρv given by ρv = 4 cos2 θ (C/m3).
Find the total charge Q contained in the sphere.
Solution: Q = v ρv dv = 2π φ=0 π θ=0 2×10−2 R=0 (4 cos2 θ)R2 sin θ dR dθ dφ = 4 2π 0 π 0 R3 3 2×10−2 0 sin θ cos2 θ dθ dφ = 32 3 × 10−6 2π 0 −cos3 θ 3 π 0 dφ = 64 9 × 10−6 2π 0 dφ = 128π 9 × 10−6 = 44.68 (μC).
Note that the limits on R were converted to meters prior to evaluating the integral on R.
3-3 Transformations between Coordinate Systems The position of a given point in space of course does not depend on the choice of coordinate system. That is, its location is the same irrespective of which specific coordinate system is used to represent it. The same is true for vectors. Nevertheless, certain coordinate systems may be more useful than others in solving a given problem, so it is essential that we have the tools to “translate” the problem from one system to another. In this section, we shall establish the relations between the variables (x, y, z) of the Cartesian system, (r, φ, z) of the cylindrical system, and (R, θ, φ) of the spherical system. These relations will then be used to transform expressions for vectors expressed in any one of the three systems into expressions applicable in the other two.
3-3.1 Cartesian to Cylindrical Transformations Point P in Fig. 3-16 has Cartesian coordinates (x, y, z) and cylindrical coordinates (r, φ, z).
Both systems share the coordinate z, and the relations between the other two pairs of coordinates can be obtained from the geometry in Fig. 3-16.
They are r = + x2 + y2, φ = tan−1 y x , (3.51) z x y φ r P(x, y, z) z y = r sin φ x = r cos φ 123 123 Figure 3-16 Interrelationships between Cartesian coordinates (x, y, z) and cylindrical coordinates (r, φ, z).
148 CHAPTER 3 VECTOR ANALYSIS r r x y ϕ ϕ ˆ ϕ ˆ ˆ y x ˆ ˆ −ϕ Figure 3-17 Interrelationships between base vectors (ˆ x, ˆ y) and (ˆ r, ˆ φ φ φ).
and the inverse relations are x = r cos φ, y = r sin φ.
(3.52) Next, with the help of Fig. 3-17, which shows the directions of the unit vectors ˆ x, ˆ y, ˆ r, and ˆ φ φ φ in the x–y plane, we obtain the following relations: ˆ r· ˆ x = cos φ, ˆ r· ˆ y = sin φ, (3.53a) ˆ φ φ φ· ˆ x = −sin φ, ˆ φ φ φ· ˆ y = cos φ.
(3.53b) To express ˆ r in terms of ˆ x and ˆ y, we write ˆ r as ˆ r = ˆ xa + ˆ yb, (3.54) where a and b are unknown transformation coefficients. The dot product ˆ r· ˆ x gives ˆ r· ˆ x = ˆ x· ˆ xa + ˆ y· ˆ xb = a.
(3.55) Comparison of Eq. (3.55) with Eq. (3.53a) yields a = cos φ.
Similarly, application of the dot product ˆ r· ˆ y to Eq. (3.54) gives b = sin φ. Hence, ˆ r = ˆ x cos φ + ˆ y sin φ.
(3.56a) Repetition of the procedure for ˆ φ φ φ leads to ˆ φ φ φ = −ˆ x sin φ + ˆ y cos φ.
(3.56b) The third base vector ˆ z is the same in both coordinate systems.
By solving Eqs. (3.56a) and (3.56b) simultaneously for ˆ x and ˆ y, we obtain the following inverse relations: ˆ x = ˆ r cos φ −ˆ φ φ φ sin φ, (3.57a) ˆ y = ˆ r sin φ + ˆ φ φ φ cos φ.
(3.57b) The relations given by Eqs. (3.56a) to (3.57b) are not only useful for transforming the base vectors (ˆ x, ˆ y) into (ˆ r, ˆ φ φ φ), and vice versa, they also can be used to transform the components of a vector expressed in either coordinate system into its corresponding components expressed in the other system.
For example, a vector A = ˆ xAx + ˆ yAy + ˆ zAz in Cartesian coordinates can be described by A = ˆ rAr + ˆ φ φ φAφ + ˆ zAz in cylindrical coordinates by applying Eqs. (3.56a) and (3.56b).
That is, Ar = Ax cos φ + Ay sin φ, (3.58a) Aφ = −Ax sin φ + Ay cos φ, (3.58b) and, conversely, Ax = Ar cos φ −Aφ sin φ, (3.59a) Ay = Ar sin φ + Aφ cos φ.
(3.59b) The transformation relations given in this and the following two subsections are summarized in Table 3-2.
3-3 TRANSFORMATIONS BETWEEN COORDINATE SYSTEMS 149 Table 3-2 Coordinate transformation relations.
Transformation Coordinate Variables Unit Vectors Vector Components Cartesian to r = + x2 + y2 ˆ r = ˆ x cos φ + ˆ y sin φ Ar = Ax cos φ + Ay sin φ cylindrical φ = tan−1(y/x) ˆ φ φ φ = −ˆ x sin φ + ˆ y cos φ Aφ = −Ax sin φ + Ay cos φ z = z ˆ z = ˆ z Az = Az Cylindrical to x = r cos φ ˆ x = ˆ r cos φ −ˆ φ φ φ sin φ Ax = Ar cos φ −Aφ sin φ Cartesian y = r sin φ ˆ y = ˆ r sin φ + ˆ φ φ φ cos φ Ay = Ar sin φ + Aφ cos φ z = z ˆ z = ˆ z Az = Az Cartesian to R = + x2 + y2 + z2 ˆ R = ˆ x sin θ cos φ AR = Ax sin θ cos φ spherical + ˆ y sin θ sin φ + ˆ z cos θ + Ay sin θ sin φ + Az cos θ θ = tan−1[ + x2 + y2/z] ˆ θ θ θ = ˆ x cos θ cos φ Aθ = Ax cos θ cos φ + ˆ y cos θ sin φ −ˆ z sin θ + Ay cos θ sin φ −Az sin θ φ = tan−1(y/x) ˆ φ φ φ = −ˆ x sin φ + ˆ y cos φ Aφ = −Ax sin φ + Ay cos φ Spherical to x = R sin θ cos φ ˆ x = ˆ R sin θ cos φ Ax = AR sin θ cos φ Cartesian + ˆ θ θ θ cos θ cos φ −ˆ φ φ φ sin φ + Aθ cos θ cos φ −Aφ sin φ y = R sin θ sin φ ˆ y = ˆ R sin θ sin φ Ay = AR sin θ sin φ + ˆ θ θ θ cos θ sin φ + ˆ φ φ φ cos φ + Aθ cos θ sin φ + Aφ cos φ z = R cos θ ˆ z = ˆ R cos θ −ˆ θ θ θ sin θ Az = AR cos θ −Aθ sin θ Cylindrical to R = + r2 + z2 ˆ R = ˆ r sin θ + ˆ z cos θ AR = Ar sin θ + Az cos θ spherical θ = tan−1(r/z) ˆ θ θ θ = ˆ r cos θ −ˆ z sin θ Aθ = Ar cos θ −Az sin θ φ = φ ˆ φ φ φ = ˆ φ φ φ Aφ = Aφ Spherical to r = R sin θ ˆ r = ˆ R sin θ + ˆ θ θ θ cos θ Ar = AR sin θ + Aθ cos θ cylindrical φ = φ ˆ φ φ φ = ˆ φ φ φ Aφ = Aφ z = R cos θ ˆ z = ˆ R cos θ −ˆ θ θ θ sin θ Az = AR cos θ −Aθ sin θ Example 3-7: Cartesian to Cylindrical Transformations Given point P1 = (3, −4, 3) and vector A = ˆ x2 −ˆ y3 + ˆ z4, definedinCartesiancoordinates, expressP1 andAincylindrical coordinates and evaluate A at P1.
Solution: For point P1, x = 3, y = −4, and z = 3. Using Eq. (3.51), we have r = + x2 + y2 = 5, φ = tan−1 y x = −53.1◦= 306.9◦, and z remains unchanged. Hence, P1 = P1(5, 306.9◦, 3) in cylindrical coordinates.
The cylindrical components of vectorA = ˆ rAr + ˆ φ φ φAφ +ˆ zAz can be determined by applying Eqs. (3.58a) and (3.58b): Ar = Ax cos φ + Ay sin φ = 2 cos φ −3 sin φ, Aφ = −Ax sin φ + Ay cos φ = −2 sin φ −3 cos φ, Az = 4.
Hence, A = ˆ r(2 cos φ −3 sin φ) −ˆ φ φ φ(2 sin φ + 3 cos φ) + ˆ z4.
At point P, φ = 306.9◦, which gives A = ˆ r3.60 −ˆ φ φ φ0.20 + ˆ z4.
150 TECHNOLOGY BRIEF 5: GLOBAL POSITIONING SYSTEM Technology Brief 5: Global Positioning System The Global Positioning System (GPS), initially developed in the 1980s by the U.S. Department of Defense as a navigation tool for military use, has evolved into a system with numerous civilian applications, including vehicle tracking, aircraft navigation, map displays in automobiles and hand-held cell phones (Fig. TF5-1), and topographic mapping.
The overall GPS comprises three segments. The space segment consists of 24 satellites (Fig. TF5-2), each circling Earth every 12 hours at an orbital altitude of about 12,000 miles and transmitting continuous coded time signals. All satellite transmitters broadcast coded messages at two specific frequencies: 1.57542 GHz and 1.22760 GHz. The user segment consists of hand-held or vehicle-mounted receivers that determine their own locations by receiving and processing multiple satellite signals. The third segment is a network of five ground stations, distributed around the world, that monitor the satellites and provide them with updates on their precise orbital information.
▶GPS provides a location inaccuracy of about 30 m, both horizontally and vertically, but it can be improved to within 1 m by differential GPS. (See final section.) ◀ Principle of Operation The triangulation technique allows the determination of the location (x0, y0, z0) of any object in 3-D space from knowledge of the distances d1, d2, and d3 between that object and three other independent points in space of known locations (x1, y1, z1) to (x3, y3, z3). In GPS, the distances are established by measuring the times it takes the signals to travel from the satellites to the GPS receivers, and then multiplying them by the speed of light c = 3 × 108 m/s. Time synchronization is achieved by using atomic clocks. The satellites use very precise clocks, accurate to 3 nanoseconds (3×10−9 s), but receivers use less accurate, inexpensive, ordinary quartz clocks. Consequently, the receiver clock may have an unknown time offset error t0 relative to the satellite clocks. To correct for the time error of a GPS receiver, a signal from a fourth satellite is needed.
The GPS receiver of the automobile in Fig.TF5-3 is at distances d1 to d4 from the GPS satellites. Each satellite sends a message identifying its orbital coordinates (x1, y1, z1) for satellite 1, and so on for the other satellites, together with a binary-coded sequence common to all satellites. The GPS receiver generates the same binary sequence (Fig.TF5-3), Figure TF5-1 iPhone map feature.
Figure TF5-2 GPS nominal satellite constellation.
Four satellites in each plane, 20,200 km altitudes, 55◦inclination.
TECHNOLOGY BRIEF 5: GLOBAL POSITIONING SYSTEM 151 SA T 4 (x4, y4, z4) SA T 3 (x3, y3, z3) SA T 2 (x2, y2, z2) SA T 1 (x1, y1, z1) d1 d2 d3 d4 Time delay Receiver Code Satellite Code (x0, y0, z0) Figure TF5-3 Automobile GPS receiver at location (x0, y0, z0).
and by comparing its code with the one received from satellite 1, it determines the time t1 corresponding to travel time over the distance d1. A similar process applies to satellites 2 to 4, leading to four equations: d2 1 = (x1 −x0)2 + (y1 −y0)2 + (z1 −z0)2 = c [(t1 + t0)]2 d2 2 = (x2 −x0)2 + (y2 −y0)2 + (z2 −z0)2 = c [(t2 + t0)]2 d2 3 = (x3 −x0)2 + (y3 −y0)2 + (z3 −z0)2 = c [(t3 + t0)]2 d2 4 = (x4 −x0)2 + (y4 −y0)2 + (z4 −z0)2 = c [(t4 + t0)]2 .
The four satellites report their coordinates (x1, y1, z1) to (x4, y4, z4) to the GPS receiver, and the time delays t1 to t4 are measured directly by the receiver. The unknowns are (x0, y0, z0), the coordinates of the GPS receiver, and the time offset of its clock t0. Simultaneous solution of the four equations provides the desired location information.
Differential GPS The 30 m GPS position inaccuracy is attributed to several factors, including time-delay errors (due to the difference between the speed of light and the actual signal speed in the troposphere) that depend on the receiver’s location on Earth, delays due to signal reflections by tall buildings, and satellites’ locations misreporting errors.
▶Differential GPS, or DGPS, uses a stationary reference receiver at a location with known coordinates. ◀ By calculating the difference between its location on the basis of the GPS estimate and its true location, the reference receiver establishes coordinate correction factors and transmits them to all DGPS receivers in the area. Application of the correction information usually reduces the location inaccuracy down to about 1 m.
152 CHAPTER 3 VECTOR ANALYSIS 3-3.2 Cartesian to Spherical Transformations From Fig. 3-18, we obtain the following relations between the Cartesian coordinates (x, y, z) and the spherical coordinates (R, θ, φ): R = + x2 + y2 + z2 , (3.60a) θ = tan−1 + x2 + y2 z , (3.60b) φ = tan−1 y x .
(3.60c) The converse relations are x = R sin θ cos φ, (3.61a) y = R sin θ sin φ, (3.61b) z = R cos θ.
(3.61c) The unit vector ˆ R lies in the ˆ r–ˆ z plane.
Hence, it can be expressed as a linear combination of ˆ r and ˆ z as follows: ˆ R = ˆ ra + ˆ zb, (3.62) z x y = r sin φ x = r cos φ z = R cos θ y R r z R (π/2 – θ) r φ ˆ φ ˆ r ˆ θ θ ˆ ˆ Figure 3-18 Interrelationships between (x, y, z) and (R, θ, φ).
where a and b are transformation coefficients. Since ˆ r and ˆ z are mutually perpendicular, ˆ R· ˆ r = a, (3.63a) ˆ R· ˆ z = b.
(3.63b) From Fig. 3-18, the angle between ˆ R and ˆ r is the complement of θ and that between ˆ R and ˆ z is θ. Hence, a = ˆ R· ˆ r = sin θ and b = ˆ R· ˆ z = cos θ. Upon inserting these expressions for a and b in Eq. (3.62) and replacing ˆ r with Eq. (3.56a), we have ˆ R = ˆ x sin θ cos φ + ˆ y sin θ sin φ + ˆ z cos θ.
(3.64a) A similar procedure can be followed to obtain the following expression for ˆ θ θ θ: ˆ θ θ θ = ˆ x cos θ cos φ + ˆ y cos θ sin φ −ˆ z sin θ.
(3.64b) Finally ˆ φ φ φ is given by Eq. (3.56b) as ˆ φ φ φ = −ˆ x sin φ + ˆ y cos φ.
(3.64c) Equations (3.64a) through (3.64c) can be solved simultaneously to give the following expressions for (ˆ x, ˆ y, ˆ z) in terms of ( ˆ R, ˆ θ θ θ, ˆ φ φ φ): ˆ x = ˆ R sin θ cos φ + ˆ θ θ θ cos θ cos φ −ˆ φ φ φ sin φ, (3.65a) ˆ y = ˆ R sin θ sin φ + ˆ θ θ θ cos θ sin φ + ˆ φ φ φ cos φ, (3.65b) ˆ z = ˆ R cos θ −ˆ θ θ θ sin θ.
(3.65c) Equations (3.64a) to (3.65c) can also be used to transform Cartesian components (Ax, Ay, Az) of vector A into their spherical counterparts (AR, Aθ, Aφ), and vice versa, by replacing (ˆ x, ˆ y, ˆ z, ˆ R, ˆ θ θ θ, ˆ φ φ φ) with (Ax, Ay, Az, AR, Aθ, Aφ).
3-3 TRANSFORMATION BETWEEN COORDINATE SYSTEMS 153 Example 3-8: Cartesian to Spherical Transformation Express vector A = ˆ x(x + y) + ˆ y(y −x) + ˆ zz in spherical coordinates.
Solution: Using the transformation relation for AR given in Table 3-2, we have AR = Ax sin θ cos φ + Ay sin θ sin φ + Az cos θ = (x + y) sin θ cos φ + (y −x) sin θ sin φ + z cos θ.
Using the expressions for x, y, and z given by Eq. (3.61c), we have AR = (R sin θ cos φ + R sin θ sin φ) sin θ cos φ + (R sin θ sin φ−R sin θ cos φ) sin θ sin φ + R cos2 θ = R sin2 θ (cos2 φ + sin2 φ) + R cos2 θ = R sin2 θ + R cos2 θ = R.
Similarly, Aθ = (x + y) cos θ cos φ + (y −x) cos θ sin φ −z sin θ, Aφ = −(x + y) sin φ + (y −x) cos φ, and following the procedure used with AR, we obtain Aθ = 0, Aφ = −R sin θ.
Hence, A = ˆ RAR + ˆ θ θ θAθ + ˆ φ φ φAφ = ˆ RR −ˆ φ φ φR sin θ.
3-3.3 Cylindrical to Spherical Transformations Transformations between cylindrical and spherical coordinates can be realized by combining the transformation relations of the preceding two subsections. The results are given in Table 3-2.
3-3.4 Distance between Two Points In Cartesian coordinates, the distance d between two points P1 = (x1, y1, z1) and P2 = (x2, y2, z2) is given by Eq. (3.12) as d = |R12| = [(x2 −x1)2 + (y2 −y1)2 + (z2 −z1)2]1/2.
(3.66) Upon using Eq. (3.52) to convert the Cartesian coordinates of P1 and P2 into their cylindrical equivalents, we have d = (r2 cos φ2 −r1 cos φ1)2 + (r2 sin φ2 −r1 sin φ1)2 + (z2 −z1)21/2 = r2 2 +r2 1 −2r1r2 cos(φ2−φ1)+(z2−z1)21/2.
(cylindrical) (3.67) A similar transformation using Eqs. (3.61a-c) leads to an expression for d in terms of the spherical coordinates of P1 and P2: d = R2 2 + R2 1 −2R1R2[cos θ2 cos θ1 + sin θ1 sin θ2 cos(φ2 −φ1)] 1/2.
(3.68) (spherical) Concept Question 3-7: Why do we use more than one coordinate system?
Concept Question 3-8: Why is it that the base vectors (ˆ x, ˆ y, ˆ z) are independent of the location of a point, but ˆ r and ˆ φ φ φ are not?
Concept Question 3-9: What are the cyclic relations for the base vectors in (a) Cartesian coordinates, (b) cylindrical coordinates, and (c) spherical coordinates?
154 CHAPTER 3 VECTOR ANALYSIS Concept Question 3-10: How is the position vector of a point in cylindrical coordinates related to its position vector in spherical coordinates?
Exercise 3-7: Point P = (2 √ 3, π/3, −2) is given in cylindrical coordinates.
Express P in spherical coordinates.
Answer: P = (4, 2π/3, π/3). (See EM.) Exercise 3-8: Transform vector A = ˆ x(x + y) + ˆ y(y −x) + ˆ zz from Cartesian to cylindrical coordinates.
Answer: A = ˆ rr −ˆ φ φ φr + ˆ zz. (See EM.) 3-4 Gradient of a Scalar Field When dealing with a scalar physical quantity whose magnitude depends on a single variable, such as the temperature T as a function of height z, the rate of change of T with height can be described by the derivative dT/dz. However, if T is also a function of x and y, its spatial rate of change becomes more difficult to describe because we now have to deal with three separate variables. The differential change in T along x, y, and z can be described in terms of the partial derivatives of T with respect to the three coordinate variables, but it is not immediately obvious as to how we should combine the three partial derivatives so as to describe the spatial rate of change of T along a specified direction. Furthermore, many of the quantities we deal with in electromagnetics are vectors, and therefore both their magnitudes and directions may vary with spatial position. To this end, we introduce three fundamental operators to describe the differential spatial variations of scalars and vectors; these are the gradient, divergence, and curl operators. The gradient operator applies to scalar fields and is the subject of the present section. The other two operators, which apply to vector fields, are discussed in succeeding sections.
Suppose that T1 = T (x, y, z) is the temperature at point P1 = (x, y, z) in some region of space, and T2 = T (x + dx, y + dy, z + dz) is the temperature at a nearby point P2 = (x + dx, y + dy, z + dz) (Fig. 3-19). The differential distances dx, dy, and dz are the components of the differential distance vector dl. That is, dl = ˆ x dx + ˆ y dy + ˆ z dz.
(3.69) From differential calculus, the temperature difference between points P1 and P2, dT = T2 −T1, is dT = ∂T ∂x dx + ∂T ∂y dy + ∂T ∂z dz.
(3.70) Because dx = ˆ x· dl, dy = ˆ y· dl, and dz = ˆ z· dl, Eq. (3.70) can be rewritten as dT = ˆ x∂T ∂x · dl + ˆ y∂T ∂y · dl + ˆ z∂T ∂z · dl = ˆ x∂T ∂x + ˆ y∂T ∂y + ˆ z∂T ∂z · dl.
(3.71) The vector inside the square brackets in Eq. (3.71) relates the change in temperature dT to a vector change in direction dl.
This vector is called the gradient of T , or grad T for short, and denoted ∇T : ∇T = grad T = ˆ x∂T ∂x + ˆ y∂T ∂y + ˆ z∂T ∂z .
(3.72) Equation (3.71) can then be expressed as dT = ∇T · dl.
(3.73) dl P1 = (x, y, z) P2 = (x + dx, y + dy, z + dz) dx dy dz z y x Figure 3-19 Differential distance vector dl between points P1 and P2.
3-4 GRADIENT OF A SCALAR FIELD 155 The symbol ∇is called the del or gradient operator and is defined as ∇= ˆ x ∂ ∂x + ˆ y ∂ ∂y + ˆ z ∂ ∂z (Cartesian).
(3.74) ▶Whereas the gradient operator itself has no physical meaning, it attains a physical meaning once it operates on a scalar quantity, and the result of the operation is a vector with magnitude equal to the maximum rate of change of the physical quantity per unit distance and pointing in the direction of maximum increase. ◀ With dl = ˆ aldl, where ˆ al is the unit vector of dl, the directional derivative of T along ˆ al is dT dl = ∇T · ˆ al.
(3.75) We can find the difference (T2 −T1), where T1 = T (x1, y1, z1) and T2 = T (x2, y2, z2) are the values of T at points P1 = (x1, y1, z1) and P2 = (x2, y2, z2), not necessarily in-finitesimally close to one another, by integrating both sides of Eq. (3.73). Thus, T2 −T1 = P2 P1 ∇T · dl.
(3.76) Example 3-9: Directional Derivative Find the directional derivative of T = x2 +y2z along direction ˆ x2 + ˆ y3 −ˆ z2 and evaluate it at (1, −1, 2).
Solution: First, we find the gradient of T : ∇T = ˆ x ∂ ∂x + ˆ y ∂ ∂y + ˆ z ∂ ∂z (x2 + y2z) = ˆ x2x + ˆ y2yz + ˆ zy2.
We denote l as the given direction, l = ˆ x2 + ˆ y3 −ˆ z2.
Its unit vector is ˆ al = l |l| = ˆ x2 + ˆ y3 −ˆ z2 √ 22 + 32 + 22 = ˆ x2 + ˆ y3 −ˆ z2 √ 17 .
Application of Eq. (3.75) gives dT dl = ∇T · ˆ al = (ˆ x2x + ˆ y2yz + ˆ zy2)· ˆ x2 + ˆ y3 −ˆ z2 √ 17 = 4x + 6yz −2y2 √ 17 .
At (1, −1, 2), dT dl (1,−1,2) = 4 −12 −2 √ 17 = −10 √ 17 .
3-4.1 Gradient Operator in Cylindrical and Spherical Coordinates Even though Eq.
(3.73) was derived using Cartesian coordinates, it should have counterparts in other coordinate systems.
To convert Eq. (3.72) into cylindrical coordinates (r, φ, z), we start by restating the coordinate relations r = x2 + y2 , tan φ = y x .
(3.77) From differential calculus, ∂T ∂x = ∂T ∂r ∂r ∂x + ∂T ∂φ ∂φ ∂x + ∂T ∂z ∂z ∂x .
(3.78) Since z is orthogonal to x and ∂z/∂x = 0, the last term in Eq. (3.78) vanishes. From the coordinate relations given by Eq. (3.77), it follows that ∂r ∂x = x x2 + y2 = cos φ, (3.79a) ∂φ ∂x = −1 r sin φ.
(3.79b) Hence, ∂T ∂x = cos φ ∂T ∂r −sin φ r ∂T ∂φ .
(3.80) This expression can be used to replace the coefficient of ˆ x in Eq. (3.72), and a similar procedure can be followed to obtain an expression for ∂T/∂y in terms of r and φ. If, in addition, we 156 CHAPTER 3 VECTOR ANALYSIS use the relations ˆ x = ˆ r cos φ−ˆ φ φ φ sin φ and ˆ y = ˆ r sin φ+ ˆ φ φ φ cos φ [from Eqs. (3.57a) and (3.57b)], then Eq. (3.72) becomes ∇T = ˆ r∂T ∂r + ˆ φ φ φ1 r ∂T ∂φ + ˆ z∂T ∂z .
(3.81) Hence, the gradient operator in cylindrical coordinates can be expressed as ˆ r ∂ ∂r + ˆ φ φ φ1 r ∂ ∂φ + ˆ z ∂ ∂z (cylindrical).
(3.82) A similar procedure leads to the following expression for the gradient in spherical coordinates: ∇= ˆ R ∂ ∂R + ˆ θ θ θ 1 R ∂ ∂θ + ˆ φ φ φ 1 R sin θ ∂ ∂φ .
(3.83) (spherical) 3-4.2 Properties of the Gradient Operator For any two scalar functions U and V , the following relations apply: (1) ∇(U + V ) = ∇U + ∇V, (3.84a) (2) ∇(UV ) = U ∇V + V ∇U, (3.84b) (3) ∇V n = nV n−1 ∇V, for any n.
(3.84c) Example 3-10: Calculating the Gradient Find the gradient of each of the following scalar functions and then evaluate it at the given point.
(a) V1 = 24V0 cos (πy/3) sin (2πz/3) at (3, 2, 1) in Carte-sian coordinates, (b) V2 = V0e−2r sin 3φ at (1, π/2, 3) in cylindrical coordi-nates, (c) V3 = V0 (a/R) cos 2θ at (2a, 0, π) in spherical coordi-nates.
Solution: (a) Using Eq. (3.72) for ∇, ∇V1 = ˆ x∂V1 ∂x + ˆ y∂V1 ∂y + ˆ z∂V1 ∂z = −ˆ y8πV0 sin πy 3 sin 2πz 3 + ˆ z16πV0 cos πy 3 cos 2πz 3 = 8πV0 −ˆ y sin πy 3 sin 2πz 3 + ˆ z2 cos πy 3 cos 2πz 3 .
At (3, 2, 1), ∇V1 = 8πV0 −ˆ y sin2 2π 3 + ˆ z2 cos2 2π 3 = πV0 −ˆ y6 + ˆ z4 .
(b) The function V2 is expressed in terms of cylindrical variables. Hence, we need to use Eq. (3.82) for ∇: ∇V2 = ˆ r ∂ ∂r + ˆ φ φ φ1 r ∂ ∂φ + ˆ z ∂ ∂z V0e−2r sin 3φ = −ˆ r2V0e−2r sin 3φ + ˆ φ φ φ(3V0e−2r cos 3φ)/r = −ˆ r2 sin 3φ + ˆ φ φ φ3 cos 3φ r V0e−2r.
At (1, π/2, 3), r = 1 and φ = π/2. Hence, ∇V2 = −ˆ r2 sin 3π 2 + ˆ φ φ φ3 cos 3π 2 V0e−2 = ˆ r2V0e−2 = ˆ r0.27V0.
(c) As V3 is expressed in spherical coordinates, we apply Eq. (3.83) to V3: ∇V3= ˆ R ∂ ∂R + ˆ θ θ θ 1 R ∂ ∂θ + ˆ φ φ φ 1 R sin θ ∂ ∂φ V0 a R cos 2θ =−ˆ RV0a R2 cos 2θ −ˆ θ θ θ2V0a R2 sin 2θ =−[ ˆ R cos 2θ + ˆ θ θ θ2 sin 2θ]V0a R2 .
At (2a, 0, π), R = 2a and θ = 0, which yields ∇V3 = −ˆ RV0 4a .
Exercise 3-9: Given V = x2y + xy2 + xz2, (a) find the gradient of V , and (b) evaluate it at (1, −1, 2).
Answer: (a) ∇V = ˆ x(2xy + y2 + z2) + ˆ y(x2 + 2xy) + ˆ z2xz, (b) ∇V (1,−1,2) = ˆ x3 −ˆ y + ˆ z4. (See EM.) 3-4 GRADIENT OF A SCALAR FIELD 157 Exercise 3-10: Find the directional derivative of V = rz2 cos 2φ along the direction of A = ˆ r2 −ˆ z and evaluate it at (1, π/2, 2).
Answer: (dV/dl) (1,π/2,2) = −4/ √ 5 . (See EM.) Exercise 3-11: The power density radiated by a star [Fig. E3.11(a)] decreases radially as S(R) = S0/R2, where R is the radial distance from the star and S0 is a constant. Recalling that the gradient of a scalar function denotes the maximum rate of change of that function per unit distance and the direction of the gradient is along the direction of maximum increase, generate an arrow representation of ∇S.
Figure E3.11 S (a) (b) S ∆ Answer: ∇S = −ˆ R 2S0/R3 (Fig. 3.11(b)). (See EM.) Exercise 3-12: The graph in Fig. E3.12(a) depicts a gentle change in atmospheric temperature from T1 over the sea to T2 over land.
The temperature profile is described by the function T (x) = T1 + (T2 −T1)/(e−x + 1), where x is measured in kilometers and x = 0 is the sea– land boundary. (a) In which direction does ∇T point and (b) at what value of x is it a maximum?
Figure E3.12 (a) T1 T x T2 Sea Land (b) x Sea Land T ∆ Answer: (a) +ˆ x; (b) at x = 0.
T (x) = T1 + T2 −T1 e−x + 1 , ∇T = ˆ x ∂T ∂x = ˆ x e−x(T2 −T1) (e−x + 1)2 .
(See EM.) 158 CHAPTER 3 VECTOR ANALYSIS Module 3.2 Gradient Select a scalar function f (x, y, z), evaluate its gradient, and display both in an appropriate 2-D plane.
3-5 Divergence of a Vector Field From our brief introduction of Coulomb’s law in Chapter 1, we know that an isolated, positive point charge q induces an electric field E in the space around it, with the direction of E being outward away from the charge. Also, the strength (magnitude) of E is proportional to q and decreases with distance R from the charge as 1/R2. In a graphical presentation, a vector field is usually represented by field lines, as shown in Fig. 3-20. The arrowhead denotes the direction of the field at the point where the field line is drawn, and the length of the line provides a qualitative depiction of the field’s magnitude.
At a surface boundary, flux density is defined as the amount of outward flux crossing a unit surface ds: Flux density of E = E· ds |ds| = E· ˆ n ds ds = E· ˆ n, (3.85) where ˆ n is the normal to ds. The total flux outwardly crossing a closed surface S, such as the enclosed surface of the imaginary sphere outlined in Fig. 3-20, is Total flux = S E· ds.
(3.86) Let us now consider the case of a differential rectangular parallelepiped, such as a cube, whose edges align with the Cartesian axes shown in Fig. 3-21. The edges are of lengths x along x, y along y, and z along z. A vector field E(x, y, z) exists in the region of space containing the parallelepiped, and we wish to determine the flux of E through its total surface S.
Since S includes six faces, we need to sum up the fluxes through all of them, and by definition the flux through any face is the outward flux from the volume v through that face.
Let E be defined as E = ˆ xEx + ˆ yEy + ˆ zEz.
(3.87) 3-5 DIVERGENCE OF A VECTOR FIELD 159 Imaginary spherical surface +q n ˆ E Figure 3-20 Flux lines of the electric field E due to a positive charge q.
ˆ (x, y + Δy, z) (x + Δx, y, z) (x, y, z + Δz) Δy Δz Δx y x z E E E n3 ˆ n2 ˆ n1 ˆ n4 Face 3 Face 1 Face 2 Face 4 (x, y, z) Figure 3-21 Flux lines of a vector field E passing through a differential rectangular parallelepiped of volume v = x y z.
The area of the face marked 1 in Fig. 3-21 is y z, and its unit vector ˆ n1 = −ˆ x. Hence, the outward flux F1 through face 1 is F1 = Face 1 E· ˆ n1 ds = Face 1 (ˆ xEx + ˆ yEy + ˆ zEz)· (−ˆ x) dy dz ≈−Ex(1) y z, (3.88) where Ex(1) is the value of Ex at the center of face 1.
Approximating Ex over face 1 by its value at the center is justified by the assumption that the differential volume under consideration is very small.
Similarly, the flux out of face 2 (with ˆ n2 = ˆ x) is F2 = Ex(2) y z, (3.89) where Ex(2) is the value of Ex at the center of face 2. Over a differential separation x between the centers of faces 1 and 2, Ex(2) is related to Ex(1) by Ex(2) = Ex(1) + ∂Ex ∂x x, (3.90) where we have ignored higher-order terms involving (x)2 and higher powers because their contributions are negligibly small when x is very small. Substituting Eq. (3.90) into Eq. (3.89) gives F2 = Ex(1) + ∂Ex ∂x x y z.
(3.91) The sum of the fluxes out of faces 1 and 2 is obtained by adding Eqs. (3.88) and (3.91), F1 + F2 = ∂Ex ∂x x y z.
(3.92a) 160 CHAPTER 3 VECTOR ANALYSIS Repeating the same procedure to each of the other face pairs leads to F3 + F4 = ∂Ey ∂y x y z, (3.92b) F5 + F6 = ∂Ez ∂z x y z.
(3.92c) The sum of fluxes F1 through F6 gives the total flux through surface S of the parallelepiped: S E· ds = ∂Ex ∂x + ∂Ey ∂y + ∂Ez ∂z x y z = (div E) v, (3.93) where v = x y z and div E is a scalar function called the divergence of E, specified in Cartesian coordinates as div E = ∂Ex ∂x + ∂Ey ∂y + ∂Ez ∂z .
(3.94) ▶By shrinking the volume v to zero, we define the divergence of E at a point as the net outward flux per unit volume over a closed incremental surface. ◀ Thus, from Eq. (3.93), we have div E ≜ lim v→0 S E· ds v , (3.95) where S encloses the elemental volume v.
Instead of denoting the divergence of E by div E, it is common practice to denote it as ∇· E. That is, ∇· E = div E = ∂Ex ∂x + ∂Ey ∂y + ∂Ez ∂z (3.96) for a vector E in Cartesian coordinates.
▶From the definition of the divergence of E given by Eq. (3.95), field E has positive divergence if the net flux out of surface S is positive, which may be “viewed” as if volume v contains a source of field lines.
If the divergence is negative, v may be viewed as containing a sink of field lines because the net flux is into v. For a uniform field E, the same amount of flux enters v as leaves it; hence, its divergence is zero and the field is said to be divergenceless. ◀ The divergence is a differential operator; it always operates on vectors, and the result of its operation is a scalar. This is in contrast with the gradient operator, which always operates on scalars and results in a vector. Expressions for the divergence of a vector in cylindrical and spherical coordinates are provided on the inside back cover of this book.
The divergence operator is distributive. That is, for any pair of vectors E1 and E2, ∇·(E1 + E2) = ∇· E1 + ∇· E2.
(3.97) If ∇· E = 0, the vector field E is called divergenceless.
The result given by Eq. (3.93) for a differential volume v can be extended to relate the volume integral of ∇· E over any volume v to the flux of E through the closed surface S that bounds v. That is, v ∇· E dv = S E· ds.
(3.98) (divergence theorem) This relationship, known as the divergence theorem, is used extensively in electromagnetics.
Example 3-11: Calculating the Divergence Determine the divergence of each of the following vector fields and then evaluate them at the indicated points: (a) E = ˆ x3x2 + ˆ y2z + ˆ zx2z at (2, −2, 0); (b) E = ˆ R(a3 cos θ/R2) −ˆ θ θ θ(a3 sin θ/R2) at (a/2, 0, π).
3-5 DIVERGENCE OF A VECTOR FIELD 161 Solution: (a) ∇· E = ∂Ex ∂x + ∂Ey ∂y + ∂Ez ∂z = ∂ ∂x (3x2) + ∂ ∂y (2z) + ∂ ∂z(x2z) = 6x + 0 + x2 = x2 + 6x.
At (2, −2, 0), ∇· E (2,−2,0) = 16.
(b) From the expression given on the inside of the back cover of the book for the divergence of a vector in spherical coordinates, it follows that ∇· E = 1 R2 ∂ ∂R (R2ER) + 1 R sin θ ∂ ∂θ (Eθ sin θ) + 1 R sin θ ∂Eφ ∂φ = 1 R2 ∂ ∂R (a3 cos θ) + 1 R sin θ ∂ ∂θ −a3 sin2 θ R2 = 0 −2a3 cos θ R3 = −2a3 cos θ R3 .
At R = a/2 and θ = 0, ∇· E (a/2,0,π) = −16.
Exercise 3-13: GivenA = e−2y(ˆ x sin 2x+ˆ y cos 2x), find ∇·A.
Answer: ∇·A = 0. (See EM.) Exercise 3-14: GivenA = ˆ rr cos φ+ ˆ φ φ φr sin φ+ˆ z3z, find ∇·A at (2, 0, 3).
Answer: ∇·A = 6. (See EM.) Exercise 3-15: If E = ˆ RAR in spherical coordinates, calculate the flux of E through a spherical surface of radius a, centered at the origin.
Answer: S E· ds = 4πAa3. (See EM.) Exercise 3-16: Verify the divergence theorem by calculating the volume integral of the divergence of the field E of Exercise 3.15 over the volume bounded by the surface of radius a.
Exercise 3-17: The arrow representation in Fig. E3.17 represents the vector fieldA = ˆ x x −ˆ y y. At a given point in space, A has a positive divergence ∇· A if the net flux flowing outward through the surface of an imaginary infinitesimalvolumecenteredatthatpointispositive, ∇·A is negative if the net flux is into the volume, and ∇·A = 0 if the same amount of flux enters into the volume as leaves it. Determine ∇· A everywhere in the x–y plane.
Figure E3.17 Answer: ∇· A = 0 everywhere. (See EM.) 162 CHAPTER 3 VECTOR ANALYSIS Module 3.3 Divergence Select a vector function f(x, y, z), evaluate its divergence, and display both in an appropriate 2-D plane.
3-6 Curl of a Vector Field So far we have defined and discussed two of the three fundamental operators used in vector analysis, the gradient of a scalar and the divergence of a vector. Now we introduce the curl operator. The curl of a vector field B describes its rotational property, or circulation. The circulation of B is defined as the line integral of B around a closed contour C; Circulation = C B· dl.
(3.99) To gain a physical understanding of this definition, we consider two examples. The first is for a uniform field B = ˆ xB0, whose field lines are as depicted in Fig. 3-22(a). For the rectangular contour abcd shown in the figure, we have Circulation = b a ˆ xB0 · ˆ x dx + c b ˆ xB0 · ˆ y dy + d c ˆ xB0 · ˆ x dx + a d ˆ xB0 · ˆ y dy = B0 x −B0 x = 0, (3.100) where x = b −a = c −d and, because ˆ x· ˆ y = 0, the second and fourth integrals are zero. According to Eq. (3.100), the circulation of a uniform field is zero.
Next, we consider the magnetic flux density B induced by an infinite wire carrying a dc current I. If the current is in free space and it is oriented along the z direction, then, from Eq. (1.13), B = ˆ φ φ φμ0I 2πr , (3.101) 3-6 CURL OF A VECTOR FIELD 163 (a) Uniform field (b) Azimuthal field B a d y x b c Δx Δx Contour C r B Current I z Contour C y x φ ˆ φ Figure 3-22 Circulation is zero for the uniform field in (a), but it is not zero for the azimuthal field in (b).
where μ0 is the permeability of free space and r is the radial distance from the current in the x–y plane. The direction of B is along the azimuth unit vector ˆ φ φ φ. The field lines of B are concentric circles around the current, as shown in Fig. 3-22(b).
For a circular contour C of radius r centered at the origin in the x–y plane, the differential length vector dl = ˆ φ φ φr dφ, and the circulation of B is Circulation = C B· dl = 2π 0 ˆ φ φ φμ0I 2πr · ˆ φ φ φr dφ = μ0I.
(3.102) In this case, the circulation is not zero.
However, had the contour C been in the x–z or y–z planes, dl would not have had a ˆ φ φ φ component, and the integral would have yielded a zero circulation. Clearly, the circulation of B depends on the choice of contour and the direction in which it is traversed. To describe the circulation of a tornado, for example, we would like to choose our contour such that the circulation of the wind fieldismaximum, andwewouldlikethecirculationtohaveboth a magnitude and a direction, with the direction being toward the tornado’s vortex. The curl operator embodies these properties.
The curl of a vector field B, denoted curl B or ∇× × ×B, is defined as ∇× × × B = curl B = lim s→0 1 s ⎡ ⎣ˆ n C B· dl ⎤ ⎦ max .
(3.103) ▶Curl B is the circulation of B per unit area, with the area s of the contour C being oriented such that the circulation is maximum. ◀ The direction of curl B is ˆ n, the unit normal of s, defined according to the right-hand rule: with the four fingers of the right hand following the contour direction dl, the thumb points along ˆ n (Fig. 3-23). When we use the notation ∇× × ×B to denote curl B, it should not be interpreted as the cross product of ∇ and B.
164 TECHNOLOGY BRIEF 6: X-RAY COMPUTED TOMOGRAPHY Technology Brief 6: X-Ray Computed Tomography ▶The word tomography is derived from the Greek words tome, meaning section or slice, and graphia, meaning writing. ◀ Computed tomography, also known as CT scan or CAT scan (for computed axial tomography), refers to a technique capable of generating 3-D images of the X-ray attenuation (absorption) properties of an object. This is in contrast to the traditional, X-ray technique that produces only a 2-D profile of the object (Fig. TF6-1). CT was invented in 1972 by British electrical engineer Godfrey Hounsfeld and independently by Allan Cormack, a South African-born American physicist. The two inventors shared the 1979 Nobel Prize in Physiology or Medicine. Among diagnostic imaging techniques, CT has the decided advantage in having the sensitivity to image body parts on a wide range of densities, from soft tissue to blood vessels and bones.
Principle of Operation In the system shown in Fig.TF6-2, the X-ray source and detector array are contained inside a circular structure through which the patient is moved along a conveyor belt. A CAT scan technician can monitor the reconstructed images to insure that they do not contain artifacts such as streaks or blurry sections caused by movement on the part of the patient during the measurement process.
A CT scanner uses an X-ray source with a narrow slit that generates a fan-beam, wide enough to encompass the extent of the body, but only a few millimeters in thickness [Fig. TF6-3(a)]. Instead of recording the attenuated X-ray beam on film, it is captured by an array of some 700 detectors. The X-ray source and the detector array are mounted on a circular frame that rotates in steps of a fraction of a degree over a full 360◦circle around the patient, each time recording an X-ray attenuation profile from a different angular perspective. Typically, 1,000 such profiles are recorded Figure TF6-1 2-D X-ray image.
Figure TF6-2 CT scanner.
TECHNOLOGY BRIEF 6: X-RAY COMPUTED TOMOGRAPHY 165 (a) CT scanner X-ray source Fan beam of X-rays Detector array Computer and monitor (b) Detector measures integrated attenuation along anatomical path (c) CT image of a normal brain Detector Voxel X-ray source Figure TF6-3 Basic elements of a CT scanner.
per each thin traverse slice of anatomy. In today’s technology, this process is completed in less than 1 second. To image an entire part of the body, such as the chest or head, the process is repeated over multiple slices (layers), which typically takes about 10 seconds to complete.
Image Reconstruction For each anatomical slice, the CT scanner generates on the order of 7 × 105 measurements (1,000 angular orientations × 700 detector channels). Each measurement represents the integrated path attenuation for the narrow beam between the X-ray source and the detector [Fig. TF6-3(b)], and each volume element (voxel) contributes to 1,000 such measurement beams.
▶Commercial CT machines use a technique called filtered back-projection to “reconstruct” an image of the attenuation rate of each voxel in the anatomical slice and, by extension, for each voxel in the entire body organ.
This is accomplished through the application of a sophisticated matrix inversion process. ◀ A sample CT image of the brain is shown in Fig. TF6-3(c).
166 CHAPTER 3 VECTOR ANALYSIS S contour C ds n ds = n ds dl ˆ ˆ Figure 3-23 The direction of the unit vector ˆ n is along the thumb when the other four fingers of the right hand follow dl.
For a vector B specified in Cartesian coordinates as B = ˆ xBx + ˆ yBy + ˆ zBz, (3.104) it can be shown, through a rather lengthy derivation, that Eq. (3.103) leads to ∇× × × B = ˆ x ∂Bz ∂y −∂By ∂z + ˆ y ∂Bx ∂z −∂Bz ∂x + ˆ z ∂By ∂x −∂Bx ∂y = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z Bx By Bz .
(3.105) Expressions for ∇× × ×B are given on the inside back cover of the book for the three orthogonal coordinate systems considered in this chapter.
3-6.1 Vector Identities Involving the Curl For any two vectors A and B and scalar V , (1) ∇× × × (A + B) = ∇× × × A + ∇× × × B, (3.106a) (2) ∇·(∇× × × A) = 0, (3.106b) (3) ∇× × × (∇V ) = 0.
(3.106c) 3-6.2 Stokes’s Theorem ▶Stokes’s theorem converts the surface integral of the curl of a vector over an open surface S into a line integral of the vector along the contour C bounding the surface S. ◀ For the geometry shown in Fig. 3-23, Stokes’s theorem states S (∇× × × B) · ds = C B· dl.
(3.107) (Stokes’s theorem) Its validity follows from the definition of ∇× × × B given by Eq. (3.103). If ∇× × ×B = 0, the field B is said to be conservative or irrotational because its circulation, represented by the right-hand side of Eq. (3.107), is zero, irrespective of the contour chosen.
Example 3-12: Verification of Stokes’s Theorem For vector field B = ˆ z cos φ/r, verify Stokes’s theorem for a segment of a cylindrical surface defined by r = 2, π/3 ≤φ ≤π/2, and 0 ≤z ≤3 (Fig. 3-24).
Solution: Stokes’s theorem states that S (∇× × × B)· ds = C B· dl.
Left-handside: WithBhavingonlyacomponentBz = cos φ/r, use of the expression for ∇× × × B in cylindrical coordinates from the inside back cover of the book gives ∇× × × B = ˆ r 1 r ∂Bz ∂φ −∂Bφ ∂z + ˆ φ φ φ ∂Br ∂z −∂Bz ∂r + ˆ z1 r ∂ ∂r (rBφ) −∂Br ∂φ = ˆ r1 r ∂ ∂φ cos φ r −ˆ φ φ φ ∂ ∂r cos φ r = −ˆ rsin φ r2 + ˆ φ φ φcos φ r2 .
3-7 LAPLACIAN OPERATOR 167 π/3 2 r = 2 z = 3 0 z y x a b c d n = r π/2 ˆ ˆ Figure 3-24 Geometry of Example 3-12.
The integral of ∇× × × B over the specified surface S is S (∇× × × B)· ds = 3 z=0 π/2 φ=π/3 −ˆ rsin φ r2 + ˆ φ φ φcos φ r2 · ˆ rr dφ dz = 3 0 π/2 π/3 −sin φ r dφ dz = −3 2r = −3 4 .
Right-hand side: The surface S is bounded by contour C = abcd shown in Fig. 3-24. The direction of C is chosen so that it is compatible with the surface normal ˆ r by the right-hand rule. Hence, C B· dl = b a Bab · dl + c b Bbc · dl + d c Bcd · dl + a d Bda · dl, where Bab, Bbc, Bcd, and Bda are the field B along segments ab, bc, cd, and da, respectively.
Over segment ab, the dot product of Bab = ˆ z (cos φ) /2 and dl = ˆ φ φ φr dφ is zero, and the same is true for segment cd.
Over segment bc, φ = π/2; hence, Bbc = ˆ z(cos π/2)/2 = 0.
For the last segment, Bda = ˆ z(cos π/3)/2 = ˆ z/4 and dl = ˆ z dz. Hence, C B· dl = a d ˆ z 1 4 · ˆ z dz = 0 3 1 4 dz = −3 4 , which is the same as the result obtained by evaluating the left-hand side of Stokes’s equation.
Exercise 3-18: Find ∇× × × A at (2, 0, 3) in cylindrical coordinates for the vector field A = ˆ r10e−2r cos φ + ˆ z10 sin φ.
Answer: (See EM.) ∇× × × A = ˆ r 10 cos φ r + ˆ z 10e−2r r sin φ (2,0,3) = ˆ r5.
Exercise 3-19: Find ∇× × × A at (3, π/6, 0) in spherical coordinates for the vector field A = ˆ θ θ θ12 sin θ.
Answer: (See EM.) ∇× × × A = ˆ φ φ φ 12 sin θ R (3,π/6,0) = ˆ φ φ φ2.
3-7 Laplacian Operator In later chapters, we sometimes deal with problems involving multiple combinations of operations on scalars and vectors. A 168 CHAPTER 3 VECTOR ANALYSIS Module 3.4 Curl Select a vector f(x, y), evaluate its curl, and display both in the x-y plane.
frequently encountered combination is the divergence of the gradient of a scalar. For a scalar function V defined in Cartesian coordinates, its gradient is ∇V = ˆ x∂V ∂x + ˆ y∂V ∂y + ˆ z∂V ∂z = ˆ xAx + ˆ yAy + ˆ zAz = A, (3.108) where we defined a vector A with components Ax = ∂V/∂x, Ay = ∂V/∂y, and Az = ∂V/∂z. The divergence of ∇V is ∇·(∇V ) = ∇·A = ∂Ax ∂x + ∂Ay ∂y + ∂Az ∂z = ∂2V ∂x2 + ∂2V ∂y2 + ∂2V ∂z2 .
(3.109) For convenience, ∇·(∇V ) is called the Laplacian of V and is denoted by ∇2V (the symbol ∇2 is pronounced “del square”).
That is, ∇2V = ∇· (∇V ) = ∂2V ∂x2 + ∂2V ∂y2 + ∂2V ∂z2 .
(3.110) As we can see from Eq. (3.110), the Laplacian of a scalar function is a scalar. Expressions for ∇2V in cylindrical and spherical coordinates are given on the inside back cover of the book.
The Laplacian of a scalar can be used to define the Laplacian of a vector. For a vector E specified in Cartesian coordinates as E = ˆ xEx + ˆ yEy + ˆ zEz, (3.111) CHAPTER 3 SUMMARY 169 the Laplacian of E is ∇2E = ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 E = ˆ x ∇2Ex + ˆ y ∇2Ey + ˆ z ∇2Ez.
(3.112) Thus, in Cartesian coordinates the Laplacian of a vector is a vector whose components are equal to the Laplacians of the vector components.
Through direct substitution, it can be shown that ∇2E = ∇(∇· E) −∇× × × (∇× × × E).
(3.113) Concept Question 3-11: What do the magnitude and direction of the gradient of a scalar quantity represent?
Concept Question 3-12: Prove the validity of Eq. (3.84c) in Cartesian coordinates.
Concept Question 3-13: What is the physical meaning of the divergence of a vector field?
Concept Question 3-14: If a vector field is solenoidal at a given point in space, does it necessarily follow that the vector field is zero at that point? Explain.
Concept Question 3-15: What is the meaning of the transformation provided by the divergence theorem?
Concept Question 3-16: How is the curl of a vector field at a point related to the circulation of the vector field?
Concept Question 3-17: What is the meaning of the transformation provided by Stokes’s theorem?
Concept Question 3-18: When is a vector field “con-servative”?
Chapter 3 Summary Concepts • Vector algebra governs the laws of addition, subtraction, and multiplication of vectors, and vector calculus encompasses the laws of differentiation and integration of vectors.
• In a right-handed orthogonal coordinate system, the three base vectors are mutually perpendicular to each other at any point in space, and the cyclic relations governing the cross products of the base vectors obey the right-hand rule.
• The dot product of two vectors produces a scalar, whereas the cross product of two vectors produces another vector.
• A vector expressed in a given coordinate system can be expressed in another coordinate system through the use of transformation relations linking the two coordinate systems.
• The fundamental differential functions in vector calculus are the gradient, the divergence, and the curl.
• The gradient of a scalar function is a vector whose magnitude is equal to the maximum rate of increasing change of the scalar function per unit distance, and its direction is along the direction of maximum increase.
• The divergence of a vector field is a measure of the net outward flux per unit volume through a closed surface surrounding the unit volume.
• The divergence theorem transforms the volume integral of the divergence of a vector field into a surface integral of the field’s flux through a closed surface surrounding the volume.
• The curl of a vector field is a measure of the circulation of the vector field per unit area s, with the orientation of s chosen such that the circulation is maximum.
• Stokes’s theorem transforms the surface integral of the curl of a vector field into a line integral of the field over a contour that bounds the surface.
• The Laplacian of a scalar function is defined as the divergence of the gradient of that function.
170 CHAPTER 3 VECTOR ANALYSIS Mathematical and Physical Models Distance between Two Points d = [(x2 −x1)2 + (y2 −y1)2 + (z2 −z1)2]1/2 d = r2 2 +r2 1 −2r1r2 cos(φ2−φ1)+(z2−z1)21/2 d = R2 2 + R2 1 −2R1R2[cos θ2 cos θ1 + sin θ1 sin θ2 cos(φ2 −φ1)] 1/2 Coordinate Systems Table 3-1 Coordinate Transformations Table 3-2 Vector Products A· B = AB cos θAB A × × × B = ˆ n AB sin θAB A·(B × × × C) = B·(C × × × A) = C·(A × × × B) A × × × (B × × × C) = B(A· C) −C(A· B) Divergence Theorem v ∇· E dv = S E· ds Vector Operators ∇T = ˆ x∂T ∂x + ˆ y∂T ∂y + ˆ z∂T ∂z ∇· E = ∂Ex ∂x + ∂Ey ∂y + ∂Ez ∂z ∇× × × B = ˆ x ∂Bz ∂y −∂By ∂z + ˆ y ∂Bx ∂z −∂Bz ∂x + ˆ z ∂By ∂x −∂Bx ∂y ∇2V = ∂2V ∂x2 + ∂2V ∂y2 + ∂2V ∂z2 (see back cover for cylindrical and spherical coordinates) Stokes’s Theorem S (∇× × × B)· ds = C B· dl Important Terms Provide definitions or explain the meaning of the following terms: azimuth angle base vectors Cartesian coordinate system circulation of a vector conservative field cross product curl operator cylindrical coordinate system differential area vector differential length vector differential volume directional derivative distance vector divergenceless divergence operator divergence theorem dot product field lines flux density flux lines gradient operator irrotational field Laplacian operator magnitude orthogonal coordinate system position vector radial distance r range R right-hand rule scalar product scalar quantity simple product solenoidal field spherical coordinate system Stokes’s theorem vector product vector quantity unit vector zenith angle PROBLEMS 171 PROBLEMS Section 3-1: Basic Laws of Vector Algebra ∗3.1 Vector A starts at point (1, −1, −3) and ends at point (2, −1, 0). Find a unit vector in the direction of A.
3.2 Given vectors A = ˆ x2 −ˆ y3 + ˆ z, B = ˆ x2 −ˆ y + ˆ z3, and C = ˆ x4+ ˆ y2−ˆ z2, show that C is perpendicular to bothA and B.
∗3.3 In Cartesian coordinates, the three corners of a triangle are P1 = (0, 4, 4), P2 = (4, −4, 4), and P3 = (2, 2, −4). Find the area of the triangle.
3.4 Given A = ˆ x2 −ˆ y3 + ˆ z1 and B = ˆ xBx + ˆ y2 + ˆ zBz: (a) Find Bx and Bz if A is parallel to B.
(b) Find a relation between Bx and Bz if A is perpendicular to B.
3.5 Given vectors A = ˆ x + ˆ y2 −ˆ z3, B = ˆ x2 −ˆ y4, and C = ˆ y2 −ˆ z4, find the following: ∗(a) A and ˆ a (b) The component of B along C (c) θAC (d) A × × × C ∗(e) A·(B × × × C) (f) A × × × (B × × × C) (g) ˆ x × × × B ∗(h) (A × × × ˆ y)· ˆ z 3.6 Given vectors A = ˆ x2 −ˆ y + ˆ z3 and B = ˆ x3 −ˆ z2, find a vector C whose magnitude is 9 and whose direction is perpendicular to both A and B.
3.7 GivenA = ˆ x(x +2y)−ˆ y(y +3z)+ ˆ z(3x −y), determine a unit vector parallel to A at point P = (1, −1, 2).
3.8 By expansion in Cartesian coordinates, prove: (a) The relation for the scalar triple product given by Eq. (3.29).
(b) The relation for the vector triple product given by Eq. (3.33).
∗Answer(s) available in Appendix D.
∗3.9 Find an expression for the unit vector directed toward the origin from an arbitrary point on the line described by x = 1 and z = −3.
3.10 Find an expression for the unit vector directed toward the point P located on the z axis at a height h above the x–y plane from an arbitrary point Q = (x, y, −5) in the plane z = −5.
∗3.11 Find a unit vector parallel to either direction of the line described by 2x + z = 4.
3.12 Two lines in the x–y plane are described by the following expressions: Line 1 x + 2y = −6.
Line 2 3x + 4y = 8.
Use vector algebra to find the smaller angle between the lines at their intersection point.
∗3.13 A given line is described by x + 2y = 4.
Vector A starts at the origin and ends at point P on the line such that A is orthogonal to the line. Find an expression for A.
3.14 Show that, given two vectors A and B, (a) The vector C defined as the vector component of B in the direction of A is given by C = ˆ a(B · ˆ a) = A(B · A) |A|2 where ˆ a is the unit vector of A.
(b) The vector D defined as the vector component of B perpendicular to A is given by D = B −A(B · A) |A|2 .
172 CHAPTER 3 VECTOR ANALYSIS ∗3.15 A certain plane is described by 2x + 3y + 4z = 16.
Find the unit vector normal to the surface in the direction away from the origin.
3.16 Given B = ˆ x(z −3y) + ˆ y(2x −3z) −ˆ z(x + y), find a unit vector parallel to B at point P = (1, 0, −1).
∗3.17 Find a vector G whose magnitude is 4 and whose direction is perpendicular to both vectors E and F, where E = ˆ x + ˆ y 2 −ˆ z 2 and F = ˆ y 3 −ˆ z 6.
3.18 A given line is described by the equation: y = x −1.
Vector A starts at point P1 = (0, 2) and ends at point P2 on the line, at which A is orthogonal to the line. Find an expression for A.
3.19 Vector field E is given by E = ˆ R 5R cos θ −ˆ θ θ θ12 R sin θ cos φ + ˆ φ φ φ3 sin φ.
Determine the component of E tangential to the spherical surface R = 2 at point P = (2, 30◦, 60◦).
3.20 When sketching or demonstrating the spatial variation of a vector field, we often use arrows, as in Fig. P3.20, wherein the length of the arrow is made to be proportional to the strength of the field and the direction of the arrow is the same as that of the field’s. The sketch shown in Fig. P3.20, which represents the vector field E = ˆ rr, consists of arrows pointing radially away from the origin and their lengths increasing linearly in proportion to their distance away from the origin. Using this arrow representation, sketch each of the following vector fields: (a) E1 = −ˆ xy (b) E2 = ˆ yx (c) E3 = ˆ xx + ˆ yy (d) E4 = ˆ xx + ˆ y2y (e) E5 = ˆ φ φ φr (f) E6 = ˆ r sin φ 3.21 Use arrows to sketch each of the following vector fields: (a) E1 = ˆ xx −ˆ yy (b) E2 = −ˆ φ φ φ (c) E3 = ˆ y (1/x) (d) E4 = ˆ r cos φ Sections 3-2 and 3-3: Coordinate Systems 3.22 Convert the coordinates of the following points from Cartesian to cylindrical and spherical coordinates: ∗(a) P1 = (1, 2, 0) (b) P2 = (0, 0, 2) (c) P3 = (1, 1, 3) ∗(d) P4 = (−2, 2, −2) 3.23 Convert the coordinates of the following points from cylindrical to Cartesian coordinates: (a) P1 = (2, π/4, −3) (b) P2 = (3, 0, −2) (c) P3 = (4, π, 5) 3.24 Convert the coordinates of the following points from spherical to cylindrical coordinates: ∗(a) P1 = (5, 0, 0) (b) P2 = (5, 0, π) (c) P3 = (3, π/2, 0) x y E E E E Figure P3.20 Arrow representation for vector field E = ˆ r r (Problem 3.20).
PROBLEMS 173 3.25 Use the appropriate expression for the differential surface area ds to determine the area of each of the following surfaces: (a) r = 3; 0 ≤φ ≤π/3; −2 ≤z ≤2 (b) 2 ≤r ≤5; π/2 ≤φ ≤π; z = 0 ∗(c) 2 ≤r ≤5; φ = π/4; −2 ≤z ≤2 (d) R = 2; 0 ≤θ ≤π/3; 0 ≤φ ≤π (e) 0 ≤R ≤5; θ = π/3; 0 ≤φ ≤2π Also sketch the outline of each surface.
3.26 Find the volumes described by the following: ∗(a) 2 ≤r ≤5; π/2 ≤φ ≤π; 0 ≤z ≤2 (b) 0 ≤R ≤5; 0 ≤θ ≤π/3; 0 ≤φ ≤2π Also sketch the outline of each volume.
3.27 A section of a sphere is described by 0 ≤R ≤2, 0 ≤θ ≤90◦, and 30◦≤φ ≤90◦. Find the following: (a) The surface area of the spherical section.
(b) The enclosed volume.
Also sketch the outline of the section.
3.28 A vector field is given in cylindrical coordinates by E = ˆ rr cos φ + ˆ φ φ φr sin φ + ˆ zz2.
Point P = (2, π, 3) is located on the surface of the cylinder described by r = 2. At point P, find: (a) The vector component of E perpendicular to the cylinder.
(b) The vector component of E tangential to the cylinder.
3.29 At a given point in space, vectors A and B are given in spherical coordinates by A = ˆ R4 + ˆ θ θ θ2 −ˆ φ φ φ, B = −ˆ R2 + ˆ φ φ φ3.
Find: (a) The scalar component, or projection, of B in the direction of A.
(b) The vector component of B in the direction of A.
(c) The vector component of B perpendicular to A.
∗3.30 Given vectors A = ˆ r(cos φ + 3z) −ˆ φ φ φ(2r + 4 sin φ) + ˆ z(r −2z) B = −ˆ r sin φ + ˆ z cos φ find (a) θAB at (2, π/2, 0) (b) A unit vector perpendicular to both A and B at (2, π/3, 1) 3.31 Find the distance between the following pairs of points: (a) P1 = (1, 2, 3) and P2 = (−2, −3, −2) in Cartesian coordinates.
(b) P3 = (1, π/4, 3) and P4 = (3, π/4, 4) in cylindrical coordinates.
(c) P5 = (4, π/2, 0) and P6 = (3, π, 0) in spherical coordi-nates.
3.32 Determine the distance between the following pairs of points: ∗(a) P1 = (1, 1, 2) and P2 = (0, 2, 3) (b) P3 = (2, π/3, 1) and P4 = (4, π/2, 3) (c) P5 = (3, π, π/2) and P6 = (4, π/2, π) 3.33 Transform the vector A = ˆ R sin2 θ cos φ + ˆ θ θ θ cos2 φ −ˆ φ φ φ sin φ into cylindrical coordinates and then evaluate it at P = (2, π/2, π/2).
3.34 Transform the following vectors into cylindrical coordi-nates and then evaluate them at the indicated points: (a) A = ˆ x(x + y) at P1 = (1, 2, 3) (b) B = ˆ x(y −x) + ˆ y(x −y) at P2 = (1, 0, 2) ∗(c) C = ˆ xy2/(x2 + y2) −ˆ yx2/(x2 + y2) + ˆ z4 at P3 = (1, −1, 2) (d) D = ˆ R sin θ + ˆ θ θ θ cos θ + ˆ φ φ φ cos2 φ at P4 = (2, π/2, π/4) ∗(e) E = ˆ R cos φ + ˆ θ θ θ sin φ + ˆ φ φ φ sin2 θ at P5 = (3, π/2, π) 3.35 Transform the following vectors into spherical coordi-nates and then evaluate them at the indicated points: (a) A = ˆ xy2 + ˆ yxz + ˆ z4 at P1 = (1, −1, 2) (b) B = ˆ y(x2 + y2 + z2) −ˆ z(x2 + y2) at P2 = (−1, 0, 2) ∗(c) C = ˆ r cos φ −ˆ φ φ φ sin φ + ˆ z cos φ sin φ at P3 = (2, π/4, 2) (d) D = ˆ xy2/(x2 + y2) −ˆ yx2/(x2 + y2) + ˆ z4 at P4 = (1, −1, 2) 174 CHAPTER 3 VECTOR ANALYSIS Sections 3-4 to 3-7: Gradient, Divergence, and Curl Operators 3.36 Find the gradient of the following scalar functions: (a) T = 3/(x2 + z2) (b) V = xy2z4 (c) U = z cos φ/(1 + r2) (d) W = e−R sin θ ∗(e) S = 4x2e−z + y3 (f) N = r2 cos2 φ (g) M = R cos θ sin φ 3.37 For each of the following scalar fields, obtain an analytical solution for ∇T and generate a corresponding arrow representation.
(a) T = 10 + x, for −10 ≤x ≤10 ∗(b) T = x2, for −10 ≤x ≤10 (c) T = 100 + xy, for −10 ≤x ≤10 (d) T = x2y2, for −10 ≤x, y ≤10 (e) T = 20 + x + y, for −10 ≤x, y ≤10 (f) T = 1 + sin(πx/3), for −10 ≤x ≤10 ∗(g) T = 1 + cos(πx/3), for −10 ≤x ≤10 (h) T = 15 + r cos φ, for 0 ≤r ≤10 0 ≤φ ≤2π.
(i) T = 15 + r cos2 φ, for 0 ≤r ≤10 0 ≤φ ≤2π.
∗3.38 The gradient of a scalar function T is given by ∇T = ˆ z e−3z.
If T = 10 at z = 0, find T (z).
3.39 Follow a procedure similar to that leading to Eq. (3.82) to derive the expression given by Eq. (3.83) for ∇in spherical coordinates.
∗3.40 For the scalar function V = xy2 −z2, determine its directional derivative along the direction of vector A = (ˆ x −ˆ yz) and then evaluate it at P = (1, −1, 4).
3.41 Evaluate the line integral of E = ˆ x x −ˆ y y along the segment P1 to P2 of the circular path shown in Fig. P3.41.
x y P1 = (0, 3) P2 = (−3, 0) Figure P3.41 Problem 3.41.
3.42 For the scalar function T = 1 2 e−r/5 cos φ, determine its directional derivative along the radial direction ˆ r and then evaluate it at P = (2, π/4, 3).
∗3.43 For the scalar function U = 1 R sin2 θ, determine its directional derivative along the range direction ˆ R and then evaluate it at P = (5, π/4, π/2).
3.44 Each of the following vector fields is displayed in Fig. P3.44 in the form of a vector representation. Determine ∇· A analytically and then compare the result with your expectations on the basis of the displayed pattern.
(a) A = −ˆ x cos x sin y + ˆ y sin x cos y, for −π ≤x, y ≤π Figure P3.44(a) PROBLEMS 175 (b) A = −ˆ x sin 2y + ˆ y cos 2x, for −π ≤x, y ≤π Figure P3.44(b) (c) A = −ˆ x xy + ˆ y y2, for −10 ≤x, y ≤10 Figure P3.44(c) (d) A = −ˆ x cos x + ˆ y sin y, for −π ≤x, y ≤π Figure P3.44(d) (e) A = ˆ x x, for −10 ≤x ≤10 Figure P3.44(e) (f) A = ˆ x xy2, for −10 ≤x, y ≤10 Figure P3.44(f) (g) A = ˆ x xy2 + ˆ y x2y, for −10 ≤x, y ≤10 Figure P3.44(g) 176 CHAPTER 3 VECTOR ANALYSIS (h) A = ˆ x sin πx 10 + ˆ y sin πy 10 , for −10 ≤x, y ≤10 Figure P3.44(h) (i) A = ˆ r r + ˆ φ φ φ r cos φ, for 0 ≤r ≤10 0 ≤φ ≤2π.
Figure P3.44(i) (j) A = ˆ r r2 + ˆ φ φ φ r2 sin φ, for 0 ≤r ≤10 0 ≤φ ≤2π.
Figure P3.44(j) ∗3.45 Vector field E is characterized by the following properties: (a) E points along ˆ R; (b) the magnitude of E is a function of only the distance from the origin; (c) E vanishes at the origin; and (d) ∇· E = 12, everywhere. Find an expression for E that satisfies these properties.
3.46 For the vector field E = ˆ xxz −ˆ yyz2 −ˆ zxy, verify the divergence theorem by computing (a) The total outward flux flowing through the surface of a cube centered at the origin and with sides equal to 2 units each and parallel to the Cartesian axes.
(b) The integral of ∇· E over the cube’s volume.
3.47 For the vector field E = ˆ r10e−r −ˆ z3z, verify the divergence theorem for the cylindrical region enclosed by r = 2, z = 0, and z = 4.
∗3.48 A vector field D = ˆ rr3 exists in the region between two concentriccylindricalsurfacesdefinedbyr = 1andr = 2, with both cylinders extending between z = 0 and z = 5. Verify the divergence theorem by evaluating the following: (a) S D· ds (b) v ∇· D dv 3.49 For the vector field D = ˆ R3R2, evaluate both sides of the divergence theorem for the region enclosed between the spherical shells defined by R = 1 and R = 2.
3.50 For the vector field E = ˆ xxy −ˆ y(x2 + 2y2), calculate (a) C E· dl around the triangular contour shown in Fig. P3.50(a).
(a) 1 1 x y 0 1 1 2 x y (b) 0 Figure P3.50 Contours for (a) Problem 3.50 and (b) Problem 3.51.
PROBLEMS 177 (b) S (∇× × × E)· ds over the area of the triangle.
3.51 Repeat Problem 3.50 for the contour shown in Fig. P3.50(b).
3.52 Verify Stokes’s theorem for the vector field B = (ˆ rr cos φ + ˆ φ φ φ sin φ) by evaluating the following: (a) C B· dl over the semicircular contour shown in Fig. P3.52(a).
(b) S (∇× × × B)· ds over the surface of the semicircle.
(a) 2 –2 2 0 x y (b) x 1 0 2 1 2 y Figure P3.52 Contour paths for (a) Problem 3.52 and (b) Problem 3.53.
3.53 Repeat Problem 3.52 for the contour shown in Fig. P3.52(b).
3.54 Verify Stokes’s theorem for the vector field A = ˆ R cos θ + ˆ φ φ φ sin θ by evaluating it on the hemisphere of unit radius.
3.55 Verify Stokes’s theorem for the vector field B = (ˆ r cos φ + ˆ φ φ φ sin φ) by evaluating: (a) C B · dℓ ℓ ℓover the path comprising a quarter section of a circle, as shown in Fig. P3.55, and (b) S (∇× × × B)· ds over the surface of the quarter section.
x y (0, 3) L1 L3 (−3, 0) Figure P3.55 Problem 3.55.
3.56 Determine if each of the following vector fields is solenoidal, conservative, or both: ∗(a) A = ˆ xx2 −ˆ y2xy (b) B = ˆ xx2 −ˆ yy2 + ˆ z2z (c) C = ˆ r(sin φ)/r2 + ˆ φ φ φ(cos φ)/r2 ∗(d) D = ˆ R/R (e) E = ˆ r 3 − r 1+r + ˆ zz (f) F = (ˆ xy + ˆ yx)/(x2 + y2) (g) G = ˆ x(x2 + z2) −ˆ y(y2 + x2) −ˆ z(y2 + z2) ∗(h) H = ˆ R(Re−R) 3.57 Find the Laplacian of the following scalar functions: (a) V = 4xy2z3 (b) V = xy + yz + zx ∗(c) V = 3/(x2 + y2) (d) V = 5e−r cos φ (e) V = 10e−R sin θ 3.58 Find the Laplacian of the following scalar functions: (a) V1 = 10r3 sin 2φ (b) V2 = (2/R2) cos θ sin φ C H A P T E R 4 Electrostatics Chapter Contents 4-1 Maxwell’s Equations, 179 4-2 Charge and Current Distributions, 180 4-3 Coulomb’s Law, 182 4-4 Gauss’s Law, 187 4-5 Electric Scalar Potential, 189 4-6 Conductors, 195 TB7 Resistive Sensors, 196 4-7 Dielectrics, 201 4-8 Electric Boundary Conditions, 203 4-9 Capacitance, 210 4-10 Electrostatic Potential Energy, 213 TB8 Supercapacitors as Batteries, 214 TB9 Capacitive Sensors, 218 4-11 Image Method, 223 Chapter 4 Summary, 225 Problems, 226 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Evaluate the electric field and electric potential due to any distribution of electric charges.
2. Apply Gauss’s law.
3. Calculate the resistance R of any shaped object, given the electric field at every point in its volume.
4. Describe the operational principles of resistive and capacitive sensors.
5. Calculate the capacitance of two-conductor configura-tions.
4-1 MAXWELL’S EQUATIONS 179 4-1 Maxwell’s Equations The modern theory of electromagnetism is based on a set of four fundamental relations known as Maxwell’s equations: ∇· D = ρv, (4.1a) ∇× × × E = −∂B ∂t , (4.1b) ∇· B = 0, (4.1c) ∇× × × H = J + ∂D ∂t .
(4.1d) Here E and D are the electric field intensity and flux density, interrelated by D = ϵE where ϵ is the electrical permittivity; H and B are magnetic field intensity and flux density, interrelated by B = μH where μ is the magnetic permeability; ρv is the electric charge density per unit volume; and J is the current density per unit area. The fields and fluxes E, D, B, H were introduced in Section 1-3, and ρv and J will be discussed in Section 4-2.
Maxwell’s equations hold in any material, including free space (vacuum). In general, all of the above quantities may depend on spatial location and time t. In the interest of readability, we will not, however, explicitly reference these dependencies [as in E(x, y, z, t)] except when the context calls for it. By formulating these equations, published in a classic treatise in 1873, James Clerk Maxwell established the first unified theory of electricity and magnetism. His equations, deduced from experimental observations reported by Coulomb, Gauss, Amp ere, Faraday, and others, not only encapsulate the connection between the electric field and electric charge and between the magnetic field and electric current, but also capture the bilateral coupling between electric and magnetic fields and fluxes.
Together with some auxiliary relations, Maxwell’s equations comprise the fundamental tenets of electromagnetic theory.
Under static conditions, none of the quantities appearing in Maxwell’s equations are functions of time (i.e., ∂/∂t = 0).
This happens when all charges are permanently fixed in space, or, if they move, they do so at a steady rate so that ρv and J are constant in time. Under these circumstances, the time derivatives of B and D in Eqs. (4.1b) and (4.1d) vanish, and Maxwell’s equations reduce to Electrostatics ∇· D = ρv, (4.2a) ∇× × × E = 0.
(4.2b) Magnetostatics ∇· B = 0, (4.3a) ∇× × × H = J.
(4.3b) Maxwell’s four equations separate into two uncoupled pairs, with the first pair involving only the electric field and flux E and D and the second pair involving only the magnetic field and flux H and B.
▶Electric and magnetic fields become decoupled in the static case. ◀ This allows us to study electricity and magnetism as two distinct and separate phenomena, as long as the spatial distributions of charge and current flow remain constant in time. We refer to the study of electric and magnetic phenomena under static conditions as electrostatics and magnetostatics, respectively.
Electrostatics is the subject of the present chapter, and in Chapter 5 we learn about magnetostatics.
The experience gained through studying electrostatic and magnetostatic phenomena will prove invaluable in tackling the more involved material in subsequent chapters, which deal with time-varying fields, charge densities, and currents.
We study electrostatics not only as a prelude to the study of time-varying fields, but also because it is an important field in its own right. Many electronic devices and systems are based on the principles of electrostatics.
They include x-ray machines, oscilloscopes, ink-jet electrostatic printers, liquid crystal displays, copy machines, micro-electromechanical switches and accelerometers, and many solid-state–based control devices. Electrostatic principles also guide the design of medical diagnostic sensors, such as the electrocardiogram, which records the heart’s pumping pattern, and the electroencephalogram, which records brain activity, as well as the development of numerous industrial applications.
180 CHAPTER 4 ELECTROSTATICS 4-2 Charge and Current Distributions In electromagnetics, we encounter various forms of electric charge distributions.
When put in motion, these charge distributions constitute current distributions.
Charges and currents may be distributed over a volume of space, across a surface, or along a line.
4-2.1 Charge Densities At the atomic scale, the charge distribution in a material is discrete, meaning that charge exists only where electrons and nuclei are and nowhere else.
In electromagnetics, we usually are interested in studying phenomena at a much larger scale, typically three or more orders of magnitude greater than the spacing between adjacent atoms. At such a macroscopic scale, we can disregard the discontinuous nature of the charge distribution and treat the net charge contained in an elemental volume v as if it were uniformly distributed within. Accordingly, we define the volume charge density ρv as ρv = lim v→0 q v = dq dv (C/m3), (4.4) whereq isthechargecontainedinv. Ingeneral, ρv depends on spatial location (x, y, z) and t; thus, ρv = ρv(x, y, z, t).
Physically, ρv represents the average charge per unit volume for a volume v centered at (x, y, z), with v being large enough to contain a large number of atoms, yet small enough to be regarded as a point at the macroscopic scale under consideration. The variation of ρv with spatial location is called its spatial distribution, or simply its distribution. The total charge contained in volume v is Q = v ρv dv (C).
(4.5) In some cases, particularly when dealing with conductors, electric charge may be distributed across the surface of a material, in which case the quantity of interest is the surface charge density ρs, defined as ρs = lim s→0 q s = dq ds (C/m2), (4.6) where q is the charge present across an elemental surface area s. Similarly, if the charge is, for all practical purposes, confined to a line, which need not be straight, we characterize its distribution in terms of the line charge density ρℓ, defined as ρℓ= lim l→0 q l = dq dl (C/m).
(4.7) Example 4-1: Line Charge Distribution Calculate the total charge Q contained in a cylindrical tube oriented along the z axis as shown in Fig. 4-1(a). The line charge density is ρℓ= 2z, where z is the distance in meters from the bottom end of the tube. The tube length is 10 cm.
Solution: The total charge Q is Q = 0.1 0 ρℓdz = 0.1 0 2z dz = z2 0.1 0 = 10−2 C.
(a) Line charge distribution (b) Surface charge distribution Surface charge ρs z x y 3 cm r ϕ z x y Line charge ρl 10 cm Figure 4-1 Charge distributions for Examples 4-1 and 4-2.
4-2 CHARGE AND CURRENT DISTRIBUTIONS 181 Example 4-2: Surface Charge Distribution The circular disk of electric charge shown in Fig. 4-1(b) is characterized by an azimuthally symmetric surface charge density that increases linearly with r from zero at the center to 6 C/m2 at r = 3 cm. Find the total charge present on the disk surface.
Solution: Since ρs is symmetrical with respect to the azimuth angle φ, it depends only on r and is given by ρs = 6r 3 × 10−2 = 2 × 102r (C/m2), where r is in meters. In polar coordinates, an elemental area is ds = r dr dφ, and for the disk shown in Fig. 4-1(b), the limits of integration are from 0 to 2π (rad) for φ and from 0 to 3 × 10−2 m for r. Hence, Q = S ρs ds = 2π φ=0 3×10−2 r=0 (2 × 102r)r dr dφ = 2π × 2 × 102 r3 3 3×10−2 0 = 11.31 (mC).
Exercise 4-1: A square plate residing in the x–y plane is situated in the space defined by −3 m ≤x ≤3 m and −3 m ≤y ≤3 m. Find the total charge on the plate if the surface charge density is ρs = 4y2 (μC/m2).
Answer: Q = 0.432 (mC). (See EM.) Exercise 4-2: Athicksphericalshellcenteredattheorigin extends between R = 2 cm and R = 3 cm. If the volume charge density is ρv = 3R × 10−4 (C/m3), find the total charge contained in the shell.
Answer: Q = 0.61 (nC). (See EM.) Volume charge ρv u ∆l ∆q' = ρvu ∆s' ∆t ∆s' ∆s ρv u ∆q = ρvu • ∆s ∆t = ρvu ∆s ∆t cos θ ∆s = n ∆s (a) (b) θ ˆ Figure 4-2 Charges with velocity u moving through a cross section s′ in (a) and s in (b).
4-2.2 Current Density Consider a tube with volume charge density ρv [Fig. 4-2(a)].
The charges in the tube move with velocity u along the tube axis.
Over a period t, the charges move a distance l = u t.
The amount of charge that crosses the tube’s cross-sectional surface s′ in time t is therefore q′ = ρv v = ρv l s′ = ρvu s′ t.
(4.8) Now consider the more general case where the charges are flowing through a surface s with normal ˆ n not necessarily parallel to u [Fig. 4-2(b)]. In this case, the amount of charge q flowing through s is q = ρvu· s t, (4.9) where s = ˆ n s and the corresponding total current flowing in the tube is I = q t = ρvu· s = J· s, (4.10) where J = ρvu (A/m2) (4.11) 182 CHAPTER 4 ELECTROSTATICS is defined as the current density in ampere per square meter.
Generalizing to an arbitrary surface S, the total current flowing through it is I = S J· ds (A).
(4.12) ▶When a current is due to the actual movement of electrically charged matter, it is called a convection current, and J is called a convection current density. ◀ A wind-driven charged cloud, for example, gives rise to a convection current.
In some cases, the charged matter constituting the convection current consists solely of charged particles, such as the electron beam of a scanning electron microscope or the ion beam of a plasma propulsion system.
When a current is due to the movement of charged particles relative to their host material, J is called a conduction current density. In a metal wire, for example, there are equal amounts of positive charges (in atomic nuclei) and negative charges (in the electron shells of the atoms). None of the positive charges and few of the negative charges can move; only those electrons in the outermost electron shells of the atoms can be pushed from one atom to the next if a voltage is applied across the ends of the wire.
▶This movement of electrons from atom to atom constitutes a conduction current.
The electrons that emerge from the wire are not necessarily the same electrons that entered the wire at the other end. ◀ Conduction current, which is discussed in more detail in Section 4-6, obeys Ohm’s law, whereas convection current does not.
Concept Question 4-1: What happens to Maxwell’s equations under static conditions?
Concept Question 4-2: How is the current density J related to the volume charge density ρv?
Concept Question 4-3: What is the difference between convection and conduction currents?
4-3 Coulomb’s Law One of the primary goals of this chapter is to develop dexterity in applying the expressions for the electric field intensity E and associated electric flux density D induced by a specified distribution of charge.
Our discussion will be limited to electrostatic fields induced by stationary charge densities.
We begin by reviewing the expression for the electric field introduced in Section 1-3.2 on the basis of the results of Coulomb’s experiments on the electrical force between charged bodies. Coulomb’s law, which was first introduced for electrical charges in air and later generalized to material media, implies that: (1) An isolated charge q induces an electric field E at every point in space, and at any specific point P, E is given by E = ˆ R q 4πϵR2 (V/m), (4.13) where ˆ R is a unit vector pointing from q to P (Fig. 4-3), R is the distance between them, and ϵ is the electrical permittivity of the medium containing the observation point P.
(2) In the presence of an electric field E at a given point in space, which may be due to a single charge or a distribution of charges, the force acting on a test charge q′ when placed at P, is F = q′E (N).
(4.14) With F measured in newtons (N) and q′ in coulombs (C), the unit of E is (N/C), which will be shown later in Section 4-5 to be the same as volt per meter (V/m).
R P E R +q ˆ Figure 4-3 Electric-field lines due to a charge q.
4-3 COULOMB’S LAW 183 For a material with electrical permittivity ϵ, the electric field quantities D and E are related by D = ϵE (4.15) with ϵ = ϵrϵ0, (4.16) where ϵ0 = 8.85 × 10−12 ≈(1/36π) × 10−9 (F/m) is the electrical permittivity of free space, and ϵr = ϵ/ϵ0 is called the relative permittivity (or dielectric constant) of the material.
For most materials and under a wide range of conditions, ϵ is independent of both the magnitude and direction of E [as implied by Eq. (4.15)].
▶If ϵ is independent of the magnitude of E, then the material is said to be linear because D and E are related linearly, and if it is independent of the direction of E, the material is said to be isotropic. ◀ Materials usually do not exhibit nonlinear permittivity behavior except when the amplitude of E is very high (at levels approaching dielectric breakdown conditions discussed later in Section 4-7), and anisotropy is present only in certain materials with peculiar crystalline structures. Hence, except for unique materials under very special circumstances, the quantities D and E are effectively redundant; for a material with known ϵ, knowledge of either D or E is sufficient to specify the other in that material.
4-3.1 Electric Field Due to Multiple Point Charges The expression given by Eq. (4.13) for the field E due to a single point charge can be extended to multiple charges. We begin by considering two point charges, q1 and q2 with position vectors R1 and R2 (measured from the origin in Fig. 4-4). The electric field E is to be evaluated at a point P with position vector R.
At P , the electric field E1 due to q1 alone is given by Eq. (4.13) with R, the distance between q1 and P, replaced with |R −R1| and the unit vector ˆ R replaced with (R −R1)/|R −R1|. Thus, E1 = q1(R −R1) 4πϵ|R −R1|3 (V/m).
(4.17a) Similarly, the electric field at P due to q2 alone is E2 = q2(R −R2) 4πϵ|R −R2|3 (V/m).
(4.17b) ▶The electric field obeys the principle of linear superposition. ◀ Hence, the total electric field E at P due to q1 and q2 together is E = E1 + E2 = 1 4πϵ q1(R −R1) |R −R1|3 + q2(R −R2) |R −R2|3 .
(4.18) Generalizing the preceding result to the case of N point charges, the electric field E at point P with position vector R due to charges q1, q2, . . . , qN located at points with position vectors R1, R2, . . . , RN, equals the vector sum of the electric fields induced by all the individual charges, or E = 1 4πϵ N i=1 qi(R −Ri) |R −Ri|3 (V/m).
(4.19) z y x q1 q2 P R – R1 R – R2 R1 E2 E1 E R2 R Figure 4-4 The electric field E at P due to two charges is equal to the vector sum of E1 and E2.
184 CHAPTER 4 ELECTROSTATICS Example 4-3: Electric Field Due to Two Point Charges Two point charges with q1 = 2 × 10−5 C and q2 = −4 × 10−5 C are located in free space at points with Cartesian coordinates (1, 3, −1) and (−3, 1, −2), respectively. Find (a) the electric field E at (3, 1, −2) and (b) the force on a 8 × 10−5 C charge located at that point. All distances are in meters.
Solution: (a) From Eq. (4.18), the electric field E with ϵ = ϵ0 (free space) is E = 1 4πϵ0 q1 (R −R1) |R −R1|3 + q2 (R −R2) |R −R2|3 (V/m).
The vectors R1, R2, and R are R1 = ˆ x + ˆ y3 −ˆ z, R2 = −ˆ x3 + ˆ y −ˆ z2, R = ˆ x3 + ˆ y −ˆ z2.
Hence, E = 1 4πϵ0 2(ˆ x2 −ˆ y2 −ˆ z) 27 −4(ˆ x6) 216 × 10−5 = ˆ x −ˆ y4 −ˆ z2 108πϵ0 × 10−5 (V/m).
(b) The force on q3 is F = q3E = 8 × 10−5 × ˆ x −ˆ y4 −ˆ z2 108πϵ0 × 10−5 = ˆ x2 −ˆ y8 −ˆ z4 27πϵ0 × 10−10 (N).
Exercise 4-3: Four charges of 10 μC each are located in free space at points with Cartesian coordinates (−3, 0, 0), (3, 0, 0), (0, −3, 0), and (0, 3, 0). Find the force on a 20-μC charge located at (0, 0, 4). All distances are in meters.
Answer: F = ˆ z0.23 N. (See EM.) Exercise 4-4: Two identical charges are located on the x axis at x = 3 and x = 7. At what point in space is the net electric field zero?
Answer: At point (5, 0, 0). (See EM.) Exercise 4-5: In a hydrogen atom the electron and proton are separated by an average distance of 5.3 × 10−11 m.
Find the magnitude of the electrical force Fe between the two particles, and compare it with the gravitational force Fg between them.
Answer: Fe = 8.2 × 10−8 N, and Fg = 3.6 × 10−47 N. (See EM.) 4-3.2 Electric Field Due to a Charge Distribution We now extend the results obtained for the field due to discrete point charges to continuous charge distributions. Consider a volume v ′ that contains a distribution of electric charge with volume charge density ρv, which may vary spatially within v ′ (Fig. 4-5). The differential electric field at a point P due to a differential amount of charge dq = ρv dv ′ contained in a differential volume dv ′ is dE = ˆ R′ dq 4πϵR′2 = ˆ R′ ρv dv ′ 4πϵR′2 , (4.20) where R′ is the vector from the differential volume dv ′ to point P. Applying the principle of linear superposition, the P R' dE ρv dv' v' Figure 4-5 Electric field due to a volume charge distribution.
4-3 COULOMB’S LAW 185 total electric field E is obtained by integrating the fields due to all differential charges in v ′. Thus, E = v ′ dE = 1 4πϵ v ′ ˆ R′ ρv dv ′ R′2 .
(4.21a) (volume distribution) It is important to note that, in general, both R′ and ˆ R′ vary as a function of position over the integration volume v ′.
If the charge is distributed across a surface S′ with surface charge density ρs, then dq = ρs ds′, and if it is distributed along a line l′ with a line charge density ρℓ, then dq = ρℓdl′.
Accordingly, the electric fields due to surface and line charge distributions are E = 1 4πϵ S′ ˆ R′ ρs ds′ R′2 , (4.21b) (surface distribution) E = 1 4πϵ l′ ˆ R′ ρℓdl′ R′2 .
(4.21c) (line distribution) Example 4-4: Electric Field of a Ring of Charge A ring of charge of radius b is characterized by a uniform line charge density of positive polarity ρℓ. The ring resides in free space and is positioned in the x–y plane as shown in Fig. 4-6.
Determine the electric field intensity E at a point P = (0, 0, h) along the axis of the ring at a distance h from its center.
Solution: We start by considering the electric field generated by a differential ring segment with cylindrical coordinates (a) (b) φ + + + + + + + + + + + + + b h z P = (0, 0, h) R'1 ρl dE1r dE1 dE1z dφ 1 y x dl = b dφ R'1 R'2 + + + + + + + + + + + + y x dE1 dE2 dE1r dE2r 1 2 dE = dE1 + dE2 φ + π φ z Figure 4-6 Ring of charge with line density ρℓ.
(a) The field dE1 due to infinitesimal segment 1 and (b) the fields dE1 and dE2 due to segments at diametrically opposite locations (Example 4-4).
(b, φ, 0) in Fig. 4-6(a). The segment has length dl = b dφ and contains charge dq = ρℓdl = ρℓb dφ. The distance vector R′ 1 from segment 1 to point P = (0, 0, h) is R′ 1 = −ˆ rb + ˆ zh, 186 CHAPTER 4 ELECTROSTATICS from which it follows that R′ 1 = |R′ 1| = b2 + h2 , ˆ R′ 1 = R′ 1 |R′ 1| = −ˆ rb + ˆ zh √ b2 + h2 .
The electric field at P = (0, 0, h) due to the charge in segment 1 therefore is dE1 = 1 4πϵ0 ˆ R′ 1 ρℓdl R′ 1 2 = ρℓb 4πϵ0 (−ˆ rb + ˆ zh) (b2 + h2)3/2 dφ.
The field dE1 has component dE1r along −ˆ r and compo-nent dE1z along ˆ z.
From symmetry considerations, the field dE2 generated by differential segment 2 in Fig. 4-6(b), which is located diametrically opposite to segment 1, is identical to dE1 except that the ˆ r component of dE2 is opposite that of dE1. Hence, the ˆ r components in the sum cancel and the ˆ z contributions add. The sum of the two contributions is dE = dE1 + dE2 = ˆ z ρℓbh 2πϵ0 dφ (b2 + h2)3/2 .
(4.22) Since for every ring segment in the semicircle defined over the azimuthal range 0 ≤φ ≤π (the right-hand half of the circular ring) there is a corresponding segment located diametrically opposite at (φ + π), we can obtain the total field generated by the ring by integrating Eq. (4.22) over a semicircle as E = ˆ z ρℓbh 2πϵ0(b2 + h2)3/2 π 0 dφ = ˆ z ρℓbh 2ϵ0(b2 + h2)3/2 = ˆ z h 4πϵ0(b2 + h2)3/2 Q, (4.23) where Q = 2πbρℓis the total charge on the ring.
Example 4-5: Electric Field of a Circular Disk of Charge Find the electric field at point P with Cartesian coordinates (0, 0, h) due to a circular disk of radius a and uniform charge density ρs residing in the x–y plane (Fig. 4-7). Also, evaluate E due to an infinite sheet of charge density ρs by letting a →∞.
z P = (0, 0, h) h y x a a r dr dq = 2πρsr dr ρs E Figure 4-7 Circular disk of charge with surface charge density ρs. The electric field at P = (0, 0, h) points along the z direction (Example 4-5).
Solution: Building on the expression obtained in Example 4-4 for the on-axis electric field due to a circular ring of charge, we can determine the field due to the circular disk by treating the disk as a set of concentric rings.
A ring of radius r and width dr has an area ds = 2πr dr and contains charge dq = ρs ds = 2πρsr dr.
Upon using this expression in Eq. (4.23) and also replacing b with r, we obtain the following expression for the field due to the ring: dE = ˆ z h 4πϵ0(r2 + h2)3/2 (2πρsr dr).
The total field at P is obtained by integrating the expression over the limits r = 0 to r = a: E = ˆ z ρsh 2ϵ0 a 0 r dr (r2 + h2)3/2 = ±ˆ z ρs 2ϵ0 1 − |h| √ a2 + h2 , (4.24) with the plus sign for h > 0 (P above the disk) and the minus sign when h < 0 (P below the disk).
4-4 GAUSS’S LAW 187 For an infinite sheet of charge with a = ∞, E = ±ˆ z ρs 2ϵ0 .
(4.25) (infinite sheet of charge) We note that for an infinite sheet of charge E is the same at all points above the x–y plane, and a similar statement applies for points below the x–y plane.
Concept Question 4-4: When characterizing the elec-trical permittivity of a material, what do the terms linear and isotropic mean?
Concept Question 4-5: If the electric field is zero at a given point in space, does this imply the absence of electric charges?
Concept Question 4-6: State the principle of linear superposition as it applies to the electric field due to a distribution of electric charge.
Exercise 4-6: An infinite sheet with uniform surface charge density ρs is located at z = 0 (x–y plane), and another infinite sheet with density −ρs is located at z = 2 m, both in free space. Determine E everywhere.
Answer: E = 0 for z < 0; E = ˆ zρs/ϵ0 for 0 < z < 2 m; and E = 0 for z > 2 m. (See EM.) 4-4 Gauss’s Law In this section, we use Maxwell’s equations to confirm the expressions for the electric field implied by Coulomb’s law, and propose alternative techniques for evaluating electric fields induced by electric charge. To that end, we restate Eq. (4.1a): ∇· D = ρv, (4.26) (differential form of Gauss’s law) which is referred to as the differential form of Gauss’s law. The adjective “differential” refers to the fact that the divergence operation involves spatial derivatives.
As we see shortly, Eq. (4.26) can be converted to an integral form. When solving electromagnetic problems, we often go back and forth between equations in differential and integral form, depending on which of the two happens to be the more applicable or convenient to use. To convert Eq. (4.26) into integral form, we multiply both sides by dv and evaluate their integrals over an arbitrary volume v: v ∇· D dv = v ρv dv = Q.
(4.27) Here, Q is the total charge enclosed in v. The divergence theorem, given by Eq. (3.98), states that the volume integral of the divergence of any vector over a volume v equals the total outward flux of that vector through the surface S enclosing v.
Thus, for the vector D, v ∇· D dv = S D· ds.
(4.28) Comparison of Eq. (4.27) with Eq. (4.28) leads to S D· ds = Q.
(4.29) (integral form of Gauss’s law) ▶The integral form of Gauss’s law is illustrated diagrammatically in Fig. 4-8; for each differential surface element ds, D· ds is the electric field flux flowing outward of v through ds, and the total flux through surface S equals the enclosed charge Q. The surface S is called a Gaussian surface. ◀ The integral form of Gauss’s law can be applied to determine D due to a single isolated point charge q by enclosing the latter with a closed, spherical, Gaussian surface S of arbitrary radius R centered at q (Fig. 4-9). From symmetry considerations and assuming that q is positive, the direction of D must be radially outward along the unit vector ˆ R, and DR, the magnitude of D, must be the same at all points on S. Thus, at any point on S, D = ˆ RDR, (4.30) 188 CHAPTER 4 ELECTROSTATICS Q D • ds Gaussian surface S enclosing volume v Total charge in v v Figure 4-8 The integral form of Gauss’s law states that the outward flux of D through a surface is proportional to the enclosed charge Q.
and ds = ˆ R ds. Applying Gauss’s law gives S D· ds = S ˆ RDR · ˆ R ds = S DR ds = DR(4πR2) = q.
(4.31) Solving for DR and then inserting the result in Eq. (4.30) gives the following expression for the electric field E induced by an isolated point charge in a medium with permittivity ϵ: E = D ϵ = ˆ R q 4πϵR2 (V/m).
(4.32) Gaussian surface q R D ds R ˆ Figure 4-9 Electric field D due to point charge q.
This is identical with Eq. (4.13) obtained from Coulomb’s law; after all, Maxwell’s equations incorporate Coulomb’s law.
For this simple case of an isolated point charge, it does not matter whether Coulomb’s law or Gauss’s law is used to obtain the expression for E.
However, it does matter as to which approach we follow when we deal with multiple point charges or continuous charge distributions. Even though Coulomb’s law can be used to find E for any specified distribution of charge, Gauss’s law is easier to apply than Coulomb’s law, but its utility is limited to symmetrical charge distributions.
▶Gauss’s law, as given by Eq. (4.29), provides a convenient method for determining the flux density D when the charge distribution possesses symmetry properties that allow us to infer the variations of the magnitude and direction of D as a function of spatial location, thereby facilitating the integration of D over a cleverly chosen Gaussian surface. ◀ Because at every point on the surface the direction of ds is alongitsoutwardnormal, onlythenormalcomponentofDatthe surface contributes to the integral in Eq. (4.29). To successfully apply Gauss’s law, the surface S should be chosen such that, from symmetry considerations, across each subsurface of S, D is constant in magnitude and its direction is either normal or purely tangential to the subsurface. These aspects are illustrated in Example 4-6.
Example 4-6: Electric Field of an Infinite Line Charge UseGauss’slawtoobtainanexpressionforEduetoaninfinitely long line with uniform charge density ρℓthat resides along the z axis in free space.
Solution: Since the charge density along the line is uniform, infinite in extent and residing along the z axis, symmetry considerations dictate that D is in the radial ˆ r direction and cannot depend on φ or z. Thus, D = ˆ r Dr. Therefore, we construct a finite cylindrical Gaussian surface of radius r and height h, concentric around the line of charge (Fig. 4-10). The total charge contained within the cylinder is Q = ρℓh. Since D is along ˆ r, the top and bottom surfaces of the cylinder do not contribute to the surface integral on the left-hand side of 4-5 ELECTRIC SCALAR POTENTIAL 189 z r D Gaussian surface uniform line charge ρl h ds Figure 4-10 Gaussian surface around an infinitely long line of charge (Example 4-6).
Eq. (4.29); that is, only the curved surface contributes to the integral. Hence, h z=0 2π φ=0 ˆ rDr · ˆ rr dφ dz = ρℓh or 2πhDrr = ρℓh, which yields E = D ϵ0 = ˆ r Dr ϵ0 = ˆ r ρℓ 2πϵ0r .
(4.33) (infinite line charge) Note that Eq. (4.33) is applicable for any infinite line of charge, regardless of its location and direction, as long as ˆ r is properly defined as the radial distance vector from the line charge to the observation point (i.e., ˆ r is perpendicular to the line of charge).
Concept Question 4-7: Explain Gauss’s law.
Under what circumstances is it useful?
Concept Question 4-8: How should one choose a Gaussian surface?
Exercise 4-7: Two infinite lines, each carrying a uniform charge density ρℓ, reside in free space parallel to the z axis at x = 1 and x = −1. Determine E at an arbitrary point along the y axis.
Answer: E = ˆ yρℓy/ πϵ0(y2 + 1) . (See EM.) Exercise 4-8: A thin spherical shell of radius a carries a uniform surface charge density ρs. Use Gauss’s law to determine E everywhere in free space.
Answer: E = 0 for R < a; E = ˆ Rρsa2/(ϵR2) for R > a. (See EM.) Exercise 4-9: A spherical volume of radius a contains a uniform volume charge density ρv. Use Gauss’s law to determine D for (a) R ≤a and (b) R ≥a.
Answer: (a) D = ˆ RρvR/3, (b) D = ˆ Rρva3/(3R2). (See EM.) 4-5 Electric Scalar Potential The operation of an electric circuit usually is described in terms of the currents flowing through its branches and the voltages at its nodes.
The voltage difference V between two points in a circuit represents the amount of work, or potential energy, required to move a unit charge from one to the other.
▶The term “voltage” is short for “voltage potential” and synonymous with electric potential. ◀ Even though when analyzing a circuit we may not consider the electric fields present in the circuit, it is in fact the existence of these fields that gives rise to voltage differences across circuit elements such as resistors or capacitors. The relationship between the electric field E and the electric potential V is the subject of this section.
4-5.1 Electric Potential as a Function of Electric Field We begin by considering the simple case of a positive charge q in a uniform electric field E = −ˆ yE, in the −y direction (Fig. 4-11). The presence of the field E exerts a force Fe = qE 190 CHAPTER 4 ELECTROSTATICS q y dy x E E E Fext Fe E Figure 4-11 Work done in moving a charge q a distance dy against the electric field E is dW = qE dy.
on the charge in the −y direction. To move the charge along the positive y direction (against the force Fe), we need to provide an external force Fext to counteract Fe, which requires the expenditure of energy.
To move q without acceleration (at constant speed), the net force acting on the charge must be zero, which means that Fext + Fe = 0, or Fext = −Fe = −qE.
(4.34) The work done, or energy expended, in moving any object a vector differential distance dl while exerting a force Fext is dW = Fext · dl = −qE· dl (J).
(4.35) Work, or energy, is measured in joules (J). If the charge is moved a distance dy along ˆ y, then dW = −q(−ˆ yE)· ˆ y dy = qE dy.
(4.36) The differential electric potential energy dW per unit charge is called the differential electric potential (or differential voltage) dV . That is, dV = dW q = −E· dl (J/C or V).
(4.37) The unit of V is the volt (V), with 1 V = 1 J/C, and since V is measured in volts, the electric field is expressed in volts per meter (V/m).
The potential difference corresponding to moving a point charge from point P1 to point P2 (Fig. 4-12) is obtained by integrating Eq. (4.37) along any path between them. That is, P2 P1 dV = − P2 P1 E· dl, (4.38) P1 C3 C2 C1 P2 path 1 path 2 path 3 E E Figure 4-12 In electrostatics, the potential difference between P2 and P1 is the same irrespective of the path used for calculating the line integral of the electric field between them.
or V21 = V2 −V1 = − P2 P1 E· dl, (4.39) where V1 and V2 are the electric potentials at points P1 and P2, respectively. The result of the line integral on the right-hand side of Eq. (4.39) is independent of the specific integration path that connects points P1 and P2. This follows immediately from the law of conservation of energy. To illustrate with an example, consider a particle in Earth’s gravitational field. If the particle is raised from a height h1 above Earth’s surface to height h2, the particle gains potential energy in an amount proportional to (h2 −h1). If, instead, we were to first raise the particle from height h1 to a height h3 greater than h2, thereby giving it potential energy proportional to (h3 −h1), and then let it drop back to height h2 by expending an energy amount proportional to (h3 −h2), its net gain in potential energy would againbeproportionalto(h2−h1). Thesameprincipleappliesto the electric potential energy W and to the potential difference (V2 −V1). The voltage difference between two nodes in an electric circuit has the same value regardless of which path in thecircuitwefollowbetweenthenodes. Moreover, Kirchhoff’s voltage law states that the net voltage drop around a closed loop is zero. If we go from P1 to P2 by path 1 in Fig. 4-12 and then return from P2 to P1 by path 2, the right-hand side of Eq. (4.39) 4-5 ELECTRIC SCALAR POTENTIAL 191 becomes a closed contour and the left-hand side vanishes. In fact, the line integral of the electrostatic field E around any closed contour C is zero: C E· dl = 0 (electrostatics).
(4.40) ▶A vector field whose line integral along any closed path is zero is called a conservative or an irrotational field. Hence, the electrostatic field E is conservative. ◀ As we will see later in Chapter 6, if E is a time-varying function, it is no longer conservative, and its line integral along a closed path is not necessarily zero.
The conservative property of the electrostatic field can be deduced from Maxwell’s second equation, Eq. (4.1b).
If ∂/∂t = 0, then ∇× × × E = 0.
(4.41) If we take the surface integral of ∇× × × E over an open surface S and then apply Stokes’s theorem expressed by Eq. (3.107) to convert the surface integral into a line integral, we obtain S (∇× × × E)· ds = C E· dl = 0, (4.42) where C is a closed contour surrounding S. Thus, Eq. (4.41) is the differential-form equivalent of Eq. (4.40).
We now define what we mean by the electric potential V at a point in space. Before we do so, however, let us revisit our electric-circuit analogue. Just as a node in a circuit cannot be assigned an absolute voltage, a point in space cannot have an absolute electric potential. The voltage of a node in a circuit is measured relative to that of a conveniently chosen reference point to which we have assigned a voltage of zero, which we call ground. The same principle applies to the electric potential V .
Usually (but not always), the reference point is chosen to be at infinity. That is, in Eq. (4.39) we assume that V1 = 0 when P1 is at infinity, and therefore the electric potential V at any point P is V = − P ∞ E· dl (V).
(4.43) 4-5.2 Electric Potential Due to Point Charges The electric field due to a point charge q located at the origin is given by Eq. (4.32) as E = ˆ R q 4πϵR2 (V/m).
(4.44) The field is radially directed and decays quadratically with the distance R from the observer to the charge.
As was stated earlier, the choice of integration path between the end points in Eq. (4.43) is arbitrary.
Hence, we can conveniently choose the path to be along the radial direction ˆ R, in which case dl = ˆ R dR and V = − R ∞ ˆ R q 4πϵR2 · ˆ R dR = q 4πϵR (V).
(4.45) If the charge q is at a location other than the origin, say at position vector R1, then V at observation position vector R becomes V = q 4πϵ|R −R1| (V), (4.46) where |R −R1| is the distance between the observation point and the location of the charge q. The principle of superposition applied previously to the electric field E also applies to the electric potential V .
Hence, for N discrete point charges q1, q2, . . . , qN residing at position vectors R1, R2, . . . , RN, the electric potential is V = 1 4πϵ N i=1 qi |R −Ri| (V).
(4.47) 4-5.3 Electric Potential Due to Continuous Distributions To obtain expressions for the electric potential V due to continuous charge distributions over a volume v ′, across a surface S ′, or along a line l′, we (1) replace qi in Eq. (4.47) with ρv dv ′, ρs ds′, and ρℓdl′, respectively; (2) convert the summation into an integration; and (3) define R′ = |R−Ri| as 192 CHAPTER 4 ELECTROSTATICS the distance between the integration point and the observation point. These steps lead to the following expressions: V = 1 4πϵ v ′ ρv R′ dv ′ (volume distribution), (4.48a) V = 1 4πϵ S′ ρs R′ ds′ (surface distribution), (4.48b) V = 1 4πϵ l′ ρℓ R′ dl′ (line distribution).
(4.48c) 4-5.4 Electric Field as a Function of Electric Potential In Section 4-5.1, we expressed V in terms of a line integral over E.
Now we explore the inverse relationship by re-examining Eq. (4.37): dV = −E· dl.
(4.49) For a scalar function V , Eq. (3.73) gives dV = ∇V · dl, (4.50) where ∇V is the gradient of V . Comparison of Eq. (4.49) with Eq. (4.50) leads to E = −∇V.
(4.51) ▶This differential relationship between V and E allows us to determine E for any charge distribution by first calculating V and then taking the negative gradient of V to find E. ◀ The expressions for V , given by Eqs. (4.47) to (4.48c), involve scalar sums and scalar integrals, and as such are usually much easier to evaluate than the vector sums and integrals in the expressions for E derived in Section 4-3 on the basis of Coulomb’s law. Thus, even though the electric potential approach for finding E is a two-step process, it is conceptually and computationally simpler to apply than the direct method based on Coulomb’s law.
Example 4-7: Electric Field of an Electric Dipole An electric dipole consists of two point charges of equal magnitude but opposite polarity, separated by a distance d [Fig. 4-13(a)]. Determine V and E at any point P, given that P is at a distance R ≫d from the dipole center, and the dipole resides in free space.
Solution: To simplify the derivation, we align the dipole along the z axis and center it at the origin [Fig. 4-13(a)]. For the two charges shown in Fig. 4-13(a), application of Eq. (4.47) gives V = 1 4πϵ0 q R1 + −q R2 = q 4πϵ0 R2 −R1 R1R2 .
(a) Electric dipole (b) Electric-field pattern +q –q R1 R2 R d y x z θ d cos θ P = (R, θ, φ) E Figure 4-13 Electric dipole with dipole moment p = qd (Example 4-7).
4-5 ELECTRIC SCALAR POTENTIAL 193 Since d ≪R, the lines labeled R1 and R2 in Fig. 4-13(a) are approximately parallel to each other, in which case the following approximations apply: R2 −R1 ≈d cos θ, R1R2 ≈R2.
Hence, V = qd cos θ 4πϵ0R2 .
(4.52) To generalize this result to an arbitrarily oriented dipole, note that the numerator of Eq. (4.52) can be expressed as the dot product of qd (where d is the distance vector from −q to +q) and the unit vector ˆ R pointing from the center of the dipole toward the observation point P. That is, qd cos θ = qd· ˆ R = p· ˆ R, (4.53) where p = qd is called the dipole moment. Using Eq. (4.53) in Eq. (4.52) then gives V = p· ˆ R 4πϵ0R2 (electric dipole).
(4.54) In spherical coordinates, Eq. (4.51) is given by E = −∇V = − ˆ R ∂V ∂R + ˆ θ θ θ 1 R ∂V ∂θ + ˆ φ φ φ 1 R sin θ ∂V ∂φ , (4.55) where we have used the expression for ∇V in spherical coordinates given on the inside back cover of the book. Upon takingthe derivativesoftheexpressionforV givenby Eq.(4.52) with respect to R and θ and then substituting the results in Eq. (4.55), we obtain E = qd 4πϵ0R3 ( ˆ R 2 cos θ + ˆ θ θ θ sin θ) (V/m).
(4.56) We stress that the expressions for V and E given by Eqs. (4.54) and (4.56) apply only when R ≫d. To compute V and E at points in the vicinity of the two dipole charges, it is necessary to perform all calculations without resorting to the far-distance approximations that led to Eq. (4.52). Such an exact calculation for E leads to the field pattern shown in Fig. 4-13(b).
4-5.5 Poisson’s Equation With D = ϵE, the differential form of Gauss’s law given by Eq. (4.26) may be cast as ∇· E = ρv ϵ .
(4.57) Inserting Eq. (4.51) in Eq. (4.57) gives ∇· (∇V ) = −ρv ϵ .
(4.58) Given Eq. (3.110) for the Laplacian of a scalar function V , ∇2V = ∇· (∇V ) = ∂2V ∂x2 + ∂2V ∂y2 + ∂2V ∂z2 , (4.59) Eq. (4.58) can be cast in the abbreviated form ∇2V = −ρv ϵ (Poisson’s equation).
(4.60) This is known as Poisson’s equation.
For a volume v ′ containing a volume charge density distribution ρv, the solution for V derived previously and expressed by Eq. (4.48a) as V = 1 4πϵ v ′ ρv R′ dv ′ (4.61) satisfies Eq. (4.60). If the medium under consideration contains no charges, Eq. (4.60) reduces to ∇2V = 0 (Laplace’s equation), (4.62) and it is then referred to as Laplace’s equation. Poisson’s and Laplace’s equations are useful for determining the electrostatic potential V in regions with boundaries on which V is known, such as the region between the plates of a capacitor with a specified voltage difference across it.
Concept Question 4-9: What is a conservative field?
Concept Question 4-10: Why is the electric potential at a point in space always defined relative to the potential at some reference point?
194 CHAPTER 4 ELECTROSTATICS Module 4.1 Fields due to Charges For any group of point charges, this module calculates and displays the electric field E and potential V across a 2-D grid. The user can specify the locations, magnitudes and polarities of the charges.
Concept Question 4-11: Explain why Eq. (4.40) is a mathematical statement of Kirchhoff’s voltage law.
Concept Question 4-12: Why is it usually easier to compute V for a given charge distribution and then find E using E = −∇V than to compute E directly by applying Coulomb’s law?
Concept Question 4-13: What is an electric dipole?
Exercise 4-10: Determine the electric potential at the origin due to four 20 μC charges residing in free space at the corners of a 2 m × 2 m square centered about the origin in the x–y plane.
Answer: V = √ 2 × 10−5/(πϵ0) (V). (See EM.) Exercise 4-11: A spherical shell of radius a has a uniform surface charge density ρs.
Determine (a) the electric potential and (b) the electric field, both at the center of the shell.
Answer: (a) V = ρsa/ϵ (V), (b) E = 0. (See EM.) 4-6 CONDUCTORS 195 4-6 Conductors The electromagnetic constitutive parameters of a material medium are its electrical permittivity ϵ, magnetic permeabil-ity μ, and conductivity σ. A material is said to be homogeneous if its constitutive parameters do not vary from point to point, and isotropic if they are independent of direction. Most materials are isotropic, but some crystals are not. Throughout this book, all materials are assumed to be homogeneous and isotropic.
This section is concerned with σ, Section 4-7 examines ϵ, and discussion of μ is deferred to Chapter 5.
▶The conductivity of a material is a measure of how easily electrons can travel through the material under the influence of an externally applied electric field. ◀ Materials are classified as conductors (metals) or dielectrics (insulators) according to the magnitudes of their conductivities.
A conductor has a large number of loosely attached electrons in the outermost shells of its atoms. In the absence of an external electric field, these free electrons move in random directions and with varying speeds. Their random motion produces zero average current through the conductor.
Upon applying an external electric field, however, the electrons migrate from one atom to the next in the direction opposite that of the external field. Their movement gives rise to a conduction current J = σE (A/m2) (Ohm’s law), (4.63) where σ is the material’s conductivity with units of siemen per meter (S/m).
In yet other materials, called dielectrics, the electrons are tightly bound to the atoms, so much so that it is very difficult to detach them under the influence of an electric field. Consequently, no significant conduction current can flow through them.
▶A perfect dielectric is a material with σ = 0.
In contrast, a perfect conductor is a material with σ = ∞.
Some materials, called superconductors, exhibit such a behavior. ◀ The conductivity σ of most metals is in the range from 106 to 107 S/m, compared with 10−10 to 10−17 S/m for good insulators Table 4-1 Conductivity of some common materials at 20 ◦C.
Material Conductivity, σ (S/m) Conductors Silver 6.2 × 107 Copper 5.8 × 107 Gold 4.1 × 107 Aluminum 3.5 × 107 Iron 107 Mercury 106 Carbon 3 × 104 Semiconductors Pure germanium 2.2 Pure silicon 4.4 × 10−4 Insulators Glass 10−12 Paraffin 10−15 Mica 10−15 Fused quartz 10−17 (Table 4-1). A class of materials called semiconductors allow for conduction currents even though their conductivities are much smaller than those of metals. The conductivity of pure germanium, for example, is 2.2 S/m. Tabulated values of σ at room temperature (20 ◦C) are given in Appendix B for some common materials, and a subset is reproduced in Table 4-1.
▶The conductivity of a material depends on several factors, including temperature and the presence of impurities.
In general, σ of metals increases with decreasing temperature. Most superconductors operate in the neighborhood of absolute zero. ◀ Concept Question 4-14: What are the electromagnetic constitutive parameters of a material?
Concept Question 4-15: What classifies a material as a conductor, a semiconductor, or a dielectric? What is a superconductor?
Concept Question 4-16: What is the conductivity of a perfect dielectric?
196 TECHNOLOGY BRIEF 7: RESISTIVE SENSORS Technology Brief 7: Resistive Sensors An electrical sensor is a device capable of responding to an applied stimulus by generating an electrical signal whose voltage, current, or some other attribute is related to the intensity of the stimulus.
▶The family of possible stimuli encompasses a wide array of physical, chemical, and biological quantities, including temperature, pressure, position, distance, motion, velocity, acceleration, concentration (of a gas or liquid), blood flow, etc. ◀ The sensing process relies on measuring resistance, capacitance, inductance, induced electromotive force (emf), oscillation frequency or time delay, among others. Sensors are integral to the operation of just about every instrument that uses electronic systems, from automobiles and airplanes to computers and cell phones. This Technology Brief covers resistive sensors. Capacitive, inductive, and emf sensors are covered separately (here and in later chapters).
Piezoresistivity According to Eq. (4.70), the resistance of a cylindrical resistor or wire conductor is given by R = l/σA, where l is the cylinder’s length, A is its cross-sectional area, and σ is the conductivity of its material.
Stretching the wire by an applied external force causes l to increase and A to decrease.
Consequently, R increases (Fig. TF7-1).
Conversely, compressing the wire causes R to decrease. The Greek word piezein means to press, from which the term piezoresistivity is derived. This should not be confused with piezoelectricity, which is an emf effect. (See EMF Sensors in Technology Brief 12.) The relationship between the resistance R of a piezoresistor and the applied force F can be modeled by the approximate linear equation R = R0 1 + αF A0 , where R0 is the unstressed resistance (@ F = 0), A0 is the unstressed cross-sectional area of the resistor, and α is the piezoresistive coefficient of the resistor material. The force F is positive if it is causing the resistor to stretch and negative if it is compressing it.
An elastic resistive sensor is well suited for measuring the deformation z of a surface (Fig. TF7-2), which can be related to the pressure applied to the surface; and if z is recorded as a function of time, it is possible to derive the velocity and acceleration of the surface’s motion. To realize high longitudinal piezoresistive sensitivity (the ratio of the normalized change in resistance, R/R0, to the corresponding change in length, l/l0, caused by the applied force), the piezoresistor is often designed as a serpentine-shaped wire [Fig. TF7-3(a)] bonded on a flexible plastic substrate and glued onto the surface whose deformation is to be monitored. Copper and nickel alloys are commonly used for making the sensor wires, although in some applications silicon is used instead [Fig. TF7-3(b)] because it has a very high piezoresistive sensitivity.
▶By connecting the piezoresistor to a Wheatstone bridge circuit (Fig. TF7-4) in which the other three resistors are all identical in value and equal to R0 (the resistance of the piezoresistor when no external force is present), the voltage output becomes directly proportional to the normalized resistance change: R/R0. ◀ TECHNOLOGY BRIEF 7: RESISTIVE SENSORS 197 Compression Stretching Force (N) F F F F R (Ω) F = 0 Figure TF7-1 Piezoresistance varies with applied force.
F = 0 Flat Film Stretched z Figure TF7-2 Piezoresistor films.
Metal wire Ohmic contacts (b) Silicon piezoresistor Silicon piezoresistor (a) Serpentine wire Figure TF7-3 Metal and silicon piezoresistors.
R0 R0 R0 R0 + ΔR V0 V1 Vout V2 V0 + − Flexible resistor Vout = V0 4 R R0 FigureTF7-4 Wheatstone bridge circuit with piezoresis-tor.
198 CHAPTER 4 ELECTROSTATICS 4-6.1 Drift Velocity The drift velocity ue of electrons in a conducting material is related to the externally applied electric field E through ue = −μeE (m/s), (4.64a) where μe is a material property called the electron mobility with units of (m2/V·s). In a semiconductor, current flow is due to the movement of both electrons and holes, and since holes are positive-charge carriers, the hole drift velocity uh is in the same direction as E, uh = μhE (m/s), (4.64b) where μh is the hole mobility. The mobility accounts for the effective mass of a charged particle and the average distance over which the applied electric field can accelerate it before it is stopped by colliding with an atom and then starts accelerating all over again.
From Eq. (4.11), the current density in a medium containing a volume density ρv of charges moving with velocity u is J = ρvu.
In the most general case, the current density consists of a component Je due to electrons and a component Jh due to holes. Thus, the total conduction current density is J = Je + Jh = ρveue + ρvhuh (A/m2), (4.65) where ρve = −Nee and ρvh = Nhe, with Ne and Nh being the number of free electrons and the number of free holes per unit volume, and e = 1.6×10−19 C is the absolute charge of a single hole or electron. Use of Eqs. (4.64a) and (4.64b) gives J = (−ρveμe + ρvhμh)E = σE, (4.66) where the quantity inside the parentheses is defined as the conductivity of the material, σ. Thus, σ = −ρveμe + ρvhμh = (Neμe + Nhμh)e (S/m), (4.67a) (semiconductor) and its unit is siemens per meter (S/m). For a good conductor, Nhμh ≪Neμe, and Eq. (4.67a) reduces to σ = −ρveμe = Neμee (S/m).
(4.67b) (good conductor) ▶In view of Eq. (4.66), in a perfect dielectric with σ = 0, J = 0 regardless of E. Similarly, in a perfect conductor with σ = ∞, E = J/σ = 0 regardless of J. ◀ That is, Perfect dielectric: J = 0, Perfect conductor: E = 0.
Because σ is on the order of 106 S/m for most metals, such as silver, copper, gold, and aluminum (Table 4-1), it is common practice to treat them as perfect conductors and to set E = 0 inside them.
A perfect conductor is an equipotential medium, meaning that the electric potential is the same at every point in the conductor. This property follows from the fact that V21, the voltage difference between two points in the conductor equals the line integral of E between them, as indicated by Eq. (4.39), and since E = 0 everywhere in the perfect conductor, the voltage difference V21 = 0. The fact that the conductor is an equipotential medium, however, does not necessarily imply that the potential difference between the conductor and some other conductor is zero. Each conductor is an equipotential medium, but the presence of different distributions of charges on their two surfaces can generate a potential difference between them.
Example 4-8: Conduction Current in a Copper Wire A2mmdiametercopperwirewithconductivityof5.8×107 S/m and electron mobility of 0.0032 (m2/V·s) is subjected to an electric field of 20 (mV/m). Find (a) the volume charge density of the free electrons, (b) the current density, (c) the current flowing in the wire, (d) the electron drift velocity, and (e) the volume density of the free electrons.
4-6 CONDUCTORS 199 Solution: (a) ρve = −σ μe = −5.8 × 107 0.0032 = −1.81 × 1010 (C/m3).
(b) J = σE = 5.8 × 107 × 20 × 10−3 = 1.16 × 106 (A/m2).
(c) I = JA = J πd2 4 = 1.16 × 106 π × 4 × 10−6 4 = 3.64 A.
(d) ue = −μeE = −0.0032 × 20 × 10−3 = −6.4 × 10−5 m/s.
The minus sign indicates that ue is in the opposite direction of E.
(e) Ne = −ρve e = 1.81 × 1010 1.6 × 10−19 = 1.13 × 1029 electrons/m3.
Exercise 4-12: Determine the density of free electrons in aluminum, given that its conductivity is 3.5 × 107 (S/m) and its electron mobility is 0.0015 (m2/V · s).
Answer: Ne = 1.46 × 1029 electrons/m3. (See EM.) Exercise 4-13: The current flowing through a 100 m long conducting wire of uniform cross section has a density of 3 × 105 (A/m2). Find the voltage drop along the length of the wire if the wire material has a conductivity of 2 × 107 (S/m).
Answer: V = 1.5 V. (See EM.) 4-6.2 Resistance To demonstrate the utility of the point form of Ohm’s law, we apply it to derive an expression for the resistance R of a conductor of length l and uniform cross section A, as shown in Fig. 4-14. The conductor axis is along the x direction and extends between points x1 and x2, with l = x2−x1. A voltage V applied across the conductor terminals establishes an electric field E = ˆ xEx; the direction of E is from the point with higher potential (point 1 in Fig. 4-14) to the point with lower potential (point 2).
The relation between V and Ex is obtained by applying Eq. (4.39): V = V1 −V2 = − x1 x2 E· dl = − x1 x2 ˆ xEx · ˆ x dl = Exl (V).
(4.68) Using Eq. (4.63), the current flowing through the cross section A at x2 is I = A J· ds = A σE· ds = σExA (A).
(4.69) From R = V/I, the ratio of Eq. (4.68) to Eq. (4.69) gives R = l σA ( ).
(4.70) x1 x2 l 1 2 I I A J E + – V y x Figure 4-14 Linear resistor of cross section A and length l connected to a dc voltage source V .
200 CHAPTER 4 ELECTROSTATICS We now generalize our result for R to any resistor of arbitrary shape by noting that the voltage V across the resistor is equal to the line integral of E over a path l between two specified points and the current I is equal to the flux of J through the surface S of the resistor. Thus, R = V I = − l E· dl S J· ds = − l E· dl S σE· ds .
(4.71) The reciprocal of R is called the conductance G, and the unit of G is ( −1), or siemens (S). For the linear resistor, G = 1 R = σA l (S).
(4.72) Example 4-9: Conductance of Coaxial Cable The radii of the inner and outer conductors of a coaxial cable of length l are a and b, respectively (Fig. 4-15). The insulation material has conductivity σ. Obtain an expression for G′, the conductance per unit length of the insulation layer.
Solution: Let I be the total current flowing radially (along ˆ r) from the inner conductor to the outer conductor through the insulation material. At any radial distance r from the axis of – +Vab l E r a b σ Figure 4-15 Coaxial cable of Example 4-9.
the center conductor, the area through which the current flows is A = 2πrl. Hence, J = ˆ r I A = ˆ r I 2πrl , (4.73) and from J = σE, E = ˆ r I 2πσrl .
(4.74) In a resistor, the current flows from higher electric potential to lower potential. Hence, if J is in the ˆ r direction, the inner conductor must be at a potential higher than that at the outer conductor. Accordingly, the voltage difference between the conductors is Vab = − a b E· dl = − a b I 2πσl ˆ r· ˆ r dr r = I 2πσl ln b a .
(4.75) The conductance per unit length is then G′ = G l = 1 Rl = I Vabl = 2πσ ln(b/a) (S/m).
(4.76) 4-6.3 Joule’s Law We now consider the power dissipated in a conducting medium in the presence of an electrostatic field E.
The medium contains free electrons and holes with volume charge densities ρve and ρvh, respectively.
The electron and hole charge contained in an elemental volume v is qe = ρve v and qh = ρvh v, respectively. The electric forces acting on qe and qh are Fe = qeE = ρveE v and Fh = qhE = ρvhE v.
The work (energy) expended by the electric field in moving qe a differential distance le and moving qh a distance lh is W = Fe · le + Fh · lh.
(4.77) Power P, measured in watts (W), is defined as the time rate of change of energy. The power corresponding to W is P = W t = Fe · le t + Fh · lh t = Fe · ue + Fh · uh = (ρveE· ue + ρvhE· uh) v = E· J v, (4.78) 4-7 DIELECTRICS 201 where ue = le/ t and uh = lh/ t are the electron and hole drift velocities, respectively. Equation (4.65) was used in the last step of the derivation leading to Eq. (4.78). For a volumev, the total dissipated power is P = v E· J dv (W) (Joule’s law), (4.79) and in view of Eq. (4.63), P = v σ|E|2 dv (W).
(4.80) Equation (4.79) is a mathematical statement of Joule’s law.
For the resistor example considered earlier, |E| = Ex and its volume isv = lA. Separating the volume integral in Eq. (4.80) into a product of a surface integral over A and a line integral over l, we have P = v σ|E|2 dv = A σEx ds l Ex dl = (σExA)(Exl) = IV (W), (4.81) where use was made of Eq. (4.68) for the voltage V and Eq. (4.69) for the current I.
With V = IR, we obtain the familiar expression P = I 2R (W).
(4.82) Concept Question 4-17: What is the fundamental difference between an insulator, a semiconductor, and a conductor?
Concept Question 4-18: Show that the power dissipated in the coaxial cable of Fig.
4-15 is P = I 2 ln(b/a)/(2πσl).
Exercise 4-14: A 50 m long copper wire has a circular cross section with radius r = 2 cm.
Given that the conductivity of copper is 5.8×107 S/m, determine (a) the resistance R of the wire and (b) the power dissipated in the wire if the voltage across its length is 1.5 mV.
Answer: (a) R = 6.9 × 10−4 , (b) P = 3.3 mW.
(See EM.) Exercise 4-15: Repeat part (b) of Exercise 4.14 by applying Eq. (4.80). (See EM.) 4-7 Dielectrics The fundamental difference between a conductor and a dielectric is that electrons in the outermost atomic shells of a conductor are only weakly tied to atoms and hence can freely migrate through the material, whereas in a dielectric they are strongly bound to the atom.
In the absence of an electric field, the electrons in so-called nonpolar molecules form a symmetrical cloud around the nucleus, with the center of the cloud coinciding with the nucleus [Fig. 4-16(a)]. The electric field generated by the positively charged nucleus attracts and holds the electron cloud around it, and the mutual repulsion of the electron clouds of adjacent atoms shapes its form. When a conductor is subjected to an externally applied electric field, the most loosely bound electrons in each atom can jump from one atom to the next, thereby setting up an electric current. In a dielectric, however, an externally applied electric field E cannot effect mass migration of charges since none are able to move freely. Instead, E will polarize the atoms or molecules in the material by moving the center of the electron cloud away from the nucleus [Fig. 4-16(b)].
The polarized atom or molecule may be represented by an electric dipole consisting of charges +q in the nucleus and −q at the center of the electron cloud [Fig. 4-16(c)].
Each such dipole sets up a small electric field, pointing from the positively charged nucleus to the center of the equally but negatively charged electron cloud. This induced electric field, called a polarization field, generally is weaker than and opposite in directionto,E. Consequently, thenetelectricfieldpresentinthe 202 CHAPTER 4 ELECTROSTATICS (a) External Eext = 0 (b) External Eext ≠ 0 (c) Electric dipole – – – – – – – – – Atom Nucleus Electron – – – – – – – – – Nucleus E E Center of electron cloud d q –q Figure 4-16 In the absence of an external electric field E, the center of the electron cloud is co-located with the center of the nucleus, but when a field is applied, the two centers are separated by a distance d.
dielectric material is smaller than E. At the microscopic level, each dipole exhibits a dipole moment similar to that described in Example 4-7. Within a block of dielectric material subject to a uniform external field, the dipoles align themselves linearly, as shown in Fig. 4-17. Along the upper and lower edges of the material, the dipole arrangement exhibits positive and negative surface charge densities, respectively.
It is important to stress that this description applies to only nonpolar molecules, which do not have permanent dipole moments. Nonpolar molecules become polarized only when an external electric field is applied, and when the field is removed, the molecules return to their original unpolarized state.
In polar materials, such as water, the molecules possess built-in permanent dipole moments that are randomly oriented in the absence of an applied electric field, and owing to their random orientations, the dipoles of polar materials produce no net macroscopic dipole moment (at the macroscopic scale, each point in the material represents a small volume containing thousands of molecules). Under the influence of an applied field, the permanent dipoles tend to align themselves along the direction of the electric field, in a manner similar to that shown in Fig. 4-17 for nonpolar materials.
+ – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – E E E E E Polarized molecule Positive surface charge Negative surface charge Figure 4-17 A dielectric medium polarized by an external electric field E.
4-7.1 Polarization Field Whereas in free space D = ϵ0E, the presence of microscopic dipoles in a dielectric material alters that relationship to D = ϵ0E + P, (4.83) where P, called the electric polarization field, accounts for the polarization properties of the material. The polarization field is produced by the electric field E and depends on the material properties. A dielectric medium is said to be linear if the magnitude of the induced polarization field P is directly proportional to the magnitude of E, and isotropic if P and E are in the same direction. Some crystals allow more polarization to take place along certain directions, such as the crystal axes, than along others. In such anisotropic dielectrics, E and P may have different directions. A medium is said to be homogeneous if its constitutive parameters (ϵ, μ, and σ) are constant throughout the medium. Our present treatment will be limited to media that are linear, isotropic, and homogeneous. For such media P is directly proportional to E and is expressed as P = ϵ0χeE, (4.84) 4-8 ELECTRIC BOUNDARY CONDITIONS 203 where χe is called the electric susceptibility of the material.
Inserting Eq. (4.84) into Eq. (4.83), we have D = ϵ0E + ϵ0χeE = ϵ0(1 + χe)E = ϵE, (4.85) which defines the permittivity ϵ of the material as ϵ = ϵ0(1 + χe).
(4.86) It is often convenient to characterize the permittivity of a material relative to that of free space, ϵ0; this is accommodated by the relative permittivity ϵr = ϵ/ϵ0. Values of ϵr are listed in Table 4-2 for a few common materials, and a longer list is given in Appendix B. In free space ϵr = 1, and for most conductors ϵr ≈1. The dielectric constant of air is approximately 1.0006 at sea level, and decreases toward unity with increasing altitude. Except in some special circumstances, such as when calculating electromagnetic wave refraction (bending) through the atmosphere over long distances, air can be treated as if it were free space.
4-7.2 Dielectric Breakdown The preceding dielectric-polarization model presumes that the magnitude of E does not exceed a certain critical value, known as the dielectric strength Eds of the material, beyond which electrons will detach from the molecules and accelerate through the material in the form of a conduction current. When this happens, sparking can occur, and the dielectric material can sustain permanent damage due to electron collisions with the molecular structure. This abrupt change in behavior is called dielectric breakdown.
▶The dielectric strength Eds is the largest magnitude of E that the material can sustain without breakdown. ◀ Dielectric breakdown can occur in gases, liquids, and solids.
The dielectric strength Eds depends on the material composition, as well as other factors such as temperature and humidity. For air Eds is roughly 3 (MV/m); for glass 25 to 40 (MV/m); and for mica 200 (MV/m) (see Table 4-2).
A charged thundercloud at electric potential V relative to the ground induces an electric field E = V/d in the air beneath it, where d is the height of the cloud base above the ground. If V is sufficiently large so that E exceeds the dielectric strength of air, ionization occurs and a lightning discharge follows. The breakdown voltage Vbr of a parallel-plate capacitor is discussed later in Example 4-11.
Concept Question 4-19: What is a polar material? A nonpolar material?
Concept Question 4-20: Do D and E always point in the same direction? If not, when do they not?
Concept Question 4-21: What happens when dielec-tric breakdown occurs?
4-8 Electric Boundary Conditions A vector field is said to be spatially continuous if it does not exhibit abrupt changes in either magnitude or direction as a function of position.
Even though the electric field may be continuous in adjoining dissimilar media, it may well be discontinuous at the boundary between them. Boundary conditions specify how the components of fields tangential and normal to an interface between two media relate across the interface. Here we derive a general set of boundary conditions for E, D, and J, applicable at the interface between any two dissimilar media, be they two dielectrics or a conductor and a dielectric.
Of course, any of the dielectrics may be free space.
Even though these boundary conditions are derived assuming electrostatic conditions, they remain valid for time-varying electric fields as well. Figure 4-18 shows an interface between medium 1 with permittivity ϵ1 and medium 2 with permittivity ϵ2. In the general case, the interface may contain a surface charge density ρs (unrelated to the dielectric polarization charge density).
To derive the boundary conditions for the tangential components of E and D, we consider the closed rectangular loop abcda shown in Fig. 4-18 and apply the conservative property of the electric field expressed by Eq. (4.40), which states that the line integral of the electrostatic field around a closed path is always zero. By letting h →0, the contributions to the line 204 CHAPTER 4 ELECTROSTATICS Table 4-2 Relative permittivity (dielectric constant) and dielectric strength of common materials.
Material Relative Permittivity, ϵr Dielectric Strength, Eds (MV/m) Air (at sea level) 1.0006 3 Petroleum oil 2.1 12 Polystyrene 2.6 20 Glass 4.5–10 25–40 Quartz 3.8–5 30 Bakelite 5 20 Mica 5.4–6 200 ϵ = ϵrϵ0 and ϵ0 = 8.854 × 10−12 F/m.
Δh 2 Δh 2 E1 E1n E1t E2 E2n E2t } } c b a d Δl ε1 ε2 Medium 1 Medium 2 Δh 2 Δh 2 Δs { { ρs n2 n1 D1n D2n ˆ ˆ ˆ l l1 ˆ l l2 Figure 4-18 Interface between two dielectric media.
integral by segments bc and da vanish. Hence, C E· dl = b a E1 · ˆ ℓ ℓ ℓ1 dl + d c E2 · ˆ ℓ ℓ ℓ2 dl = 0, (4.87) where ˆ ℓ ℓ ℓ1 and ˆ ℓ ℓ ℓ2 are unit vectors along segments ab and cd, and E1 and E2 are the electric fields in media 1 and 2. Next, we decompose E1 and E2 into components tangential and normal to the boundary (Fig. 4-18), E1 = E1t + E1n, (4.88a) E2 = E2t + E2n.
(4.88b) Noting that ˆ ℓ ℓ ℓ1 = −ˆ ℓ ℓ ℓ2, it follows that (E1 −E2)· ˆ ℓ ℓ ℓ1 = 0.
(4.89) In other words, the component of E1 along ˆ ℓ ℓ ℓ1 equals that of E2 along ˆ ℓ ℓ ℓ1, for all ˆ ℓ ℓ ℓ1 tangential to the boundary. Hence, E1t = E2t (V/m).
(4.90) ▶Thus, the tangential component of the electric field is continuous across the boundary between any two media. ◀ Upon decomposing D1 and D2 into tangential and normal components (in the manner of Eq. (4.88)) and noting that 4-8 ELECTRIC BOUNDARY CONDITIONS 205 D1t = ϵ1E1t and D2t = ϵ2E2t, the boundary condition on the tangential component of the electric flux density is D1t ϵ1 = D2t ϵ2 .
(4.91) Next, we apply Gauss’s law, as expressed by Eq. (4.29), to determine boundary conditions on the normal components of E and D. According to Gauss’s law, the total outward flux of D through the three surfaces of the small cylinder shown in Fig. 4-18 must equal the total charge enclosed in the cylinder.
By letting the cylinder’s height h →0, the contribution to the total flux through the side surface goes to zero.
Also, even if each of the two media happens to contain free charge densities, the only charge remaining in the collapsed cylinder is that distributed on the boundary. Thus, Q = ρs s, and S D· ds = top D1 · ˆ n2 ds + bottom D2 · ˆ n1 ds = ρs s, (4.92) where ˆ n1 and ˆ n2 are the outward normal unit vectors of the bottom and top surfaces, respectively.
It is important to remember that the normal unit vector at the surface of any medium is always defined to be in the outward direction away from that medium. Since ˆ n1 = −ˆ n2, Eq. (4.92) simplifies to ˆ n2 ·(D1 −D2) = ρs (C/m2).
(4.93) If D1n and D2n denote as the normal components of D1 and D2 along ˆ n2, we have D1n −D2n = ρs (C/m2).
(4.94) ▶The normal component of D changes abruptly at a charged boundary between two different media in an amount equal to the surface charge density. ◀ The corresponding boundary condition for E is ˆ n2 ·(ϵ1E1 −ϵ2E2) = ρs, (4.95a) or equivalently ϵ1E1n −ϵ2E2n = ρs.
(4.95b) In summary, (1) the conservative property of E, ∇× × × E = 0 C E· dl = 0, (4.96) led to the result that E has a continuous tangential component across a boundary, and (2) the divergence property of D, ∇· D = ρv S D· ds = Q, (4.97) led to the result that the normal component of D changes by ρs across the boundary. A summary of the conditions that apply at the boundary between different types of media is given in Table 4-3.
Example 4-10: Application of Boundary Conditions The x–y plane is a charge-free boundary separating two dielectric media with permittivities ϵ1 and ϵ2, as shown in Fig. 4-19.
If the electric field in medium 1 is E1 = ˆ xE1x + ˆ yE1y + ˆ zE1z, find (a) the electric field E2 in medium 2 and (b) the angles θ1 and θ2.
Solution: (a) Let E2 = ˆ xE2x + ˆ yE2y + ˆ zE2z. Our task is to find the components of E2 in terms of the given components of E1. The normal to the boundary is ˆ z. Hence, the x and y components of the fields are tangential to the boundary and the z components are normal to the boundary. At a charge-206 CHAPTER 4 ELECTROSTATICS Table 4-3 Boundary conditions for the electric fields.
Field Component Any Two Media Medium 1 Dielectric ϵ1 Medium 2 Conductor Tangential E E1t = E2t E1t = E2t = 0 Tangential D D1t/ϵ1 = D2t/ϵ2 D1t = D2t = 0 Normal E ϵ1E1n −ϵ2E2n = ρs E1n = ρs/ϵ1 E2n = 0 Normal D D1n −D2n = ρs D1n = ρs D2n = 0 Notes: (1) ρs is the surface charge density at the boundary; (2) normal components of E1, D1, E2, and D2 are along ˆ n2, the outward normal unit vector of medium 2.
E1z E1t E2t E2z E1 E2 θ2 ε1 ε2 θ1 z x-y plane Figure 4-19 Application of boundary conditions at the interface between two dielectric media (Example 4-10).
free interface, the tangential components of E and the normal components of D are continuous. Consequently, E2x = E1x, E2y = E1y, and D2z = D1z or ϵ2E2z = ϵ1E1z.
Hence, E2 = ˆ xE1x + ˆ yE1y + ˆ z ϵ1 ϵ2 E1z.
(4.98) (b) The tangential components of E1 and E2 are E1t = E2 1x + E2 1y and E2t = E2 2x + E2 2y .
The angles θ1 and θ2 are then given by tan θ1 = E1t E1z = E2 1x + E2 1y E1z , tan θ2 = E2t E2z = E2 2x + E2 2y E2z = E2 1x + E2 1y (ϵ1/ϵ2)E1z , and the two angles are related by tan θ2 tan θ1 = ϵ2 ϵ1 .
(4.99) Exercise 4-16: Find E1 in Fig. 4-19 if E2 = ˆ x2 −ˆ y3 + ˆ z3 (V/m), ϵ1 = 2ϵ0, ϵ2 = 8ϵ0, and the boundary is charge free.
Answer: E1 = ˆ x2 −ˆ y3 + ˆ z12 (V/m). (See EM.) Exercise 4-17: Repeat Exercise 4.16 for a boundary with surface charge density ρs = 3.54 × 10−11 (C/m2).
Answer: E1 = ˆ x2 −ˆ y3 + ˆ z14 (V/m). (See EM.) 4-8 ELECTRIC BOUNDARY CONDITIONS 207 Module 4.2 Charges in Adjacent Dielectrics In two adjoining half-planes with selectable permittivities, the user can place point charges anywhere in space and select their magnitudes and polarities. The module then displays E, V , and the equipotential contours of V .
4-8.1 Dielectric-Conductor Boundary Consider the case when medium 1 is a dielectric and medium 2 is a perfect conductor. Because in a perfect conductor, electric fields and fluxes vanish, it follows that E2 = D2 = 0, which implies that components of E2 and D2 tangential and normal to the interface are zero. Consequently, from Eq. (4.90) and Eq. (4.94), the fields in the dielectric medium, at the boundary with the conductor, satisfy E1t = D1t = 0, (4.100a) D1n = ϵ1E1n = ρs.
(4.100b) These two boundary conditions can be combined into D1 = ϵ1E1 = ˆ nρs, (4.101) (at conductor surface) where ˆ n is a unit vector directed normally outward from the conducting surface.
▶The electric field lines point directly away from the conductor surface when ρs is positive and directly toward the conductor surface when ρs is negative. ◀ Figure 4-20 shows an infinitely long conducting slab placed in a uniform electric field E1. The media above and below the slab have permittivity ϵ1. Because E1 points away from the upper surface, it induces a positive charge density ρs = ϵ1|E1| 208 CHAPTER 4 ELECTROSTATICS E1 E1 E1 E1 E1 E1 Ei Ei Ei + + + + + – – – – – – – – – – – – – – – – + + + + + + + + + + + ρs = ε1E1 −ρs Conducting slab ε1 ε1 Figure 4-20 When a conducting slab is placed in an external electric field E1, charges that accumulate on the conductor surfaces induce an internal electric field Ei = −E1. Consequently, the total field inside the conductor is zero.
on the upper slab surface. On the bottom surface, E1 points toward the surface, and therefore the induced charge density is −ρs. The presence of these surface charges induces an electric field Ei in the conductor, resulting in a total field E = E1 + Ei.
To satisfy the condition that E must be everywhere zero in the conductor, Ei must equal −E1.
If we place a metallic sphere in an electrostatic field (Fig. 4-21), positive and negative charges accumulate on the upper and lower hemispheres, respectively. The presence of the sphere causes the field lines to bend to satisfy the condition expressed by Eq. (4.101); that is, E is always normal to a conductor boundary.
– E0 metal sphere + + + + + + + + + + – – –– – – –– – Figure 4-21 Metal sphere placed in an external electric field E0.
4-8.2 Conductor-Conductor Boundary We now examine the general case of the boundary between two media neither of which is a perfect dielectric or a perfect conductor (Fig. 4-22).
Medium 1 has permittivity ϵ1 and conductivity σ1, medium 2 has ϵ2 and σ2, and the interface between them holds a surface charge density ρs. For the electric fields, Eqs. (4.90) and (4.95b) give E1t = E2t, ϵ1E1n −ϵ2E2n = ρs.
(4.102) Medium 1 ε1, σ1 Medium 2 ε2, σ2 J1n J2n J1t J2t J1 J2 n ˆ Figure 4-22 Boundary between two conducting media.
4-8 ELECTRIC BOUNDARY CONDITIONS 209 Module 4.3 Charges above Conducting Plane When electric charges are placed in a dielectric medium adjoining a conducting plane, some of the conductor’s electric charges move to its surface boundary, thereby satisfying the boundary conditions outlined in Table 4-3. This module displays E and V everywhere and ρs along the dielectric-conductor boundary.
Since we are dealing with conducting media, the electric fields give rise to current densities J1 = σ1E1 and J2 = σ2E2. Hence J1t σ1 = J2t σ2 , ϵ1 J1n σ1 −ϵ2 J2n σ2 = ρs.
(4.103) ThetangentialcurrentcomponentsJ1t andJ2t representcurrents flowing in the two media in a direction parallel to the boundary, and hence there is no transfer of charge between them. This is not the case for the normal components. If J1n ̸= J2n, then a different amount of charge arrives at the boundary than leaves it. Hence, ρs cannot remain constant in time, which violates the condition of electrostatics requiring all fields and charges to remain constant. Consequently, the normal component of J has to be continuous across the boundary between two different media under electrostatic conditions. Upon setting J1n = J2n in Eq. (4.103), we have J1n ϵ1 σ1 −ϵ2 σ2 = ρs (electrostatics).
(4.104) Concept Question 4-22: What are the boundary con-ditions for the electric field at a conductor–dielectric boundary?
Concept Question 4-23: Under electrostatic condi-tions, we require J1n = J2n at the boundary between two conductors. Why?
210 CHAPTER 4 ELECTROSTATICS Module 4.4 Charges near Conducting Sphere This module is similar to Module 4.3, except that now the conducting body is a sphere of selectable size.
4-9 Capacitance When separated by an insulating (dielectric) medium, any two conducting bodies, regardless of their shapes and sizes, form a capacitor.
If a dc voltage source is connected across them (Fig. 4-23) the surfaces of the conductors connected to the positive and negative source terminals accumulate charges +Q and −Q, respectively.
▶When a conductor has excess charge, it distributes the charge onitssurfaceinsuch amannerastomaintainazero electric field everywhere within the conductor, thereby ensuring that the electric potential is the same at every point in the conductor. ◀ The capacitance of a two-conductor configuration is defined as C = Q V (C/V or F), (4.105) – – – – – – – – – – – E Surface S V + – ρs + + + + + + + + + + + +Q Conductor 1 −Q Conductor 2 Figure 4-23 A dc voltage source connected to a capacitor composed of two conducting bodies.
4-9 CAPACITANCE 211 where V is the potential (voltage) difference between the conductors. Capacitance is measured in farads (F), which is equivalent to coulombs per volt (C/V).
The presence of free charges on the conductors’ surfaces gives rise to an electric field E (Fig. 4-23) with field lines originating on the positive charges and terminating on the negative ones. Since the tangential component of E always vanishes at a conductor’s surface, E is always perpendicular to the conducting surfaces. The normal component of E at any point on the surface of either conductor is given by En = ˆ n· E = ρs ϵ , (4.106) (at conductor surface) where ρs is the surface charge density at that point, ˆ n is the outward normal unit vector at the same location, and ϵ is the permittivity of the dielectric medium separating the conductors.
The charge Q is equal to the integral of ρs over surface S (Fig. 4-23): Q = S ρs ds = S ϵ ˆ n· E ds = S ϵE· ds, (4.107) where use was made of Eq. (4.106). The voltage V is related to E by Eq. (4.39): V = V12 = − P1 P2 E· dl, (4.108) where points P1 and P2 are any two arbitrary points on conductors 1 and 2, respectively. Substituting Eqs. (4.107) and (4.108) into Eq. (4.105) gives C = S ϵE· ds − l E· dl (F), (4.109) where l is the integration path from conductor 2 to conductor 1.
To avoid making sign errors when applying Eq. (4.109), it is important to remember that surface S is the +Q surface and P1 is on S. (Alternatively, if you compute C and it comes out negative, just change its sign.) Because E appears in both the numerator and denominator of Eq. (4.109), the value of C obtained for any specific capacitor configuration is always independent of E’s magnitude. In fact, C depends only on the capacitor geometry (sizes, shapes, and relative positions of the two conductors) and the permittivity of the insulating material.
If the material between the conductors is not a perfect dielectric (i.e., if it has a small conductivity σ), then current can flow through the material between the conductors, and the material exhibits a resistance R. The general expression for R for a resistor of arbitrary shape is given by Eq. (4.71): R = − l E· dl S σE· ds ( ).
(4.110) For a medium with uniform σ and ϵ, the product of Eqs. (4.109) and (4.110) gives RC = ϵ σ .
(4.111) This simple relation allows us to find R if C is known, and vice versa.
Example 4-11: Capacitance and Break-down Voltage of Parallel-Plate Capacitor Obtain an expression for the capacitance C of a parallel-plate capacitor consisting of two parallel plates each of surface area A and separated by a distance d. The capacitor is filled with a dielectric material with permittivity ϵ. Also, determine the breakdown voltage if d = 1 cm and the dielectric material is quartz.
Solution: In Fig. 4-24, we place the lower plate of the capacitorinthex–y planeandtheupperplateintheplane z = d.
Because of the applied voltage difference V , charges +Q and −Q accumulate on the top and bottom capacitor plates.
If the plate dimensions are much larger than the separation d, then these charges distribute themselves quasi-uniformly across the plates, giving rise to a quasi-uniform field between them pointing in the −ˆ z direction. In addition, a fringing field will 212 CHAPTER 4 ELECTROSTATICS + + + + + + + – – – – – – – ρs –ρs –Q E E Area A +Q z = d z + + + + + + + + + + – + – – – – – – – – – – ds E E E V Conducting plate Conducting plate Dielectric ε z = 0 Fringing field lines Figure 4-24 A dc voltage source connected to a parallel-plate capacitor (Example 4-11).
exist near the capacitor edges, but its effects may be ignored because the bulk of the electric field exists between the plates.
The charge density on the upper plate is ρs = Q/A. Hence, in the dielectric medium E = −ˆ zE, and from Eq. (4.106), the magnitude of E at the conductor– dielectric boundary is E = ρs/ϵ = Q/ϵA. From Eq. (4.108), the voltage difference is V = − d 0 E· dl = − d 0 (−ˆ zE)· ˆ z dz = Ed, (4.112) and the capacitance is C = Q V = Q Ed = ϵA d , (4.113) where use was made of the relation E = Q/ϵA.
From V = Ed, as given by Eq. (4.112), V = Vbr when E = Eds, the dielectric strength of the material. According to Table 4-2, Eds = 30 (MV/m) for quartz. Hence, the breakdown voltage is Vbr = Edsd = 30 × 106 × 10−2 = 3 × 105 V.
Example 4-12: Capacitance per Unit Length of Coaxial Line Obtain an expression for the capacitance of the coaxial line shown in Fig. 4-25.
Solution: For a given voltage V across the capacitor, charges +Q and −Q accumulate on the surfaces of the outer and inner conductors, respectively. We assume that these charges are uniformly distributed along the length and circumference of the conductors with surface charge density ρ′ s = Q/2πbl on the outer conductor and ρ′′ s = −Q/2πal on the inner one.
Ignoring fringing fields near the ends of the coaxial line, we can construct a cylindrical Gaussian surface in the dielectric in between the conductors, with radius r such that a < r < b.
Symmetry implies that the E-field is identical at all points on this surface, directed radially inward. From Gauss’s law, it follows that the field magnitude equals the absolute value of the total charge enclosed, divided by the surface area. That is, E = −ˆ r Q 2πϵrl .
(4.114) The potential difference V between the outer and inner conductors is V = − b a E· dl = − b a −ˆ r Q 2πϵrl · (ˆ r dr) = Q 2πϵl ln b a .
(4.115) 4-10 ELECTROSTATIC POTENTIAL ENERGY 213 + + + + + + + + + + + + + + + + + + + + + + + + – – – – – – – – – – – + – – – – – – – – – – – – – – + – V + – b a ρl –ρl l E E E E E E Dielectric material ε Outer conductor Inner conductor Figure 4-25 Coaxial capacitor filled with insulating material of permittivity ϵ (Example 4-12).
The capacitance C is then given by C = Q V = 2πϵl ln(b/a) , (4.116) and the capacitance per unit length of the coaxial line is C ′ = C l = 2πϵ ln(b/a) (F/m).
(4.117) Concept Question 4-24: How is the capacitance of a two-conductor structure related to the resistance of the insulating material between the conductors?
Concept Question 4-25: What are fringing fields and when may they be ignored?
4-10 Electrostatic Potential Energy A source connected to a capacitor expends energy in charging up the capacitor. If the capacitor plates are made of a good conductor with effectively zero resistance, and if the dielectric separating the two plates has negligible conductivity, then no real current can flow through the dielectric, and no ohmic losses occur anywhere in the capacitor. Where then does the energy expended in charging up the capacitor go? The energy ends up getting stored in the dielectric medium in the form of electrostatic potential energy. The amount of stored energy We is related to Q, C, and V .
Suppose we were to charge up a capacitor by ramping up the voltage across it from υ = 0 to υ = V . During the process, charge +q accumulates on one conductor, and −q on the other.
In effect, a charge q has been transferred from one of the conductors to the other. The voltage υ across the capacitor is related to q by υ = q C .
(4.118) From the definition of υ, the amount of work dWe required to transfer an additional incremental charge dq from one conductor to the other is dWe = υ dq = q C dq.
(4.119) If we transfer a total charge Q between the conductors of an initially uncharged capacitor, then the total amount of work performed is We = Q 0 q C dq = 1 2 Q2 C (J).
(4.120) Using C = Q/V , where V is the final voltage, We also can be expressed as We = 1 2CV 2 (J).
(4.121) The capacitance of the parallel-plate capacitor discussed in Example 4-11 is given by Eq. (4.113) as C = ϵA/d, where A is the surface area of each of its plates and d is the separation between them.
Also, the voltage V across the capacitor is 214 TECHNOLOGY BRIEF 8: SUPERCAPACITORS AS BATTERIES Technology Brief 8: Supercapacitors as Batteries As recent additions to the language of electronics, the names supercapacitor, ultracapacitor, and nanocapacitor suggest that they represent devices that are somehow different from or superior to traditional capacitors. Are these just fancy names attached to traditional capacitors by manufacturers, or are we talking about a really different type of capacitor?
▶The three aforementioned names refer to variations on an energy storage device known by the technical name electrochemical double-layer capacitor (EDLC), in which energy storage is realized by a hybrid process that incorporates features from both the traditional electrostatic capacitor and the electrochemical voltaic battery. ◀ For the purposes of this Technology Brief, we refer to this relatively new device as a supercapacitor: The battery is far superior to the traditional capacitor with regard to energy storage, but a capacitor can be charged and discharged much more rapidly than a battery. As a hybrid technology, the supercapacitor offers features that are intermediate between those of the battery and the traditional capacitor. The supercapacitor is now used to support a wide range of applications, from motor startups in large engines (trucks, locomotives, submarines, etc.) to flash lights in digital cameras, and its use is rapidly extending into consumer electronics (cell phones, MP3 players, laptop computers) and electric cars (Fig. TF8-2).
Figure TF8-1 Examples of electromechanical double-layer capacitors (EDLC), otherwise known as a superca-pacitor.
Figure TF8-2 Examples of systems that use supercapacitors.
TECHNOLOGY BRIEF 8: SUPERCAPACITORS AS BATTERIES 215 Capacitor Energy Storage Limitations Energy density W ′ is often measured in watts-hours per kg (Wh/kg), with 1 Wh = 3.6 × 103 joules. Thus, the energy capacity of a device is normalized to its mass. For batteries, W ′ extends between about 30 Wh/kg for a lead-acid battery to as high as 150 Wh/kg for a lithium-ion battery. In contrast, W ′ rarely exceeds 0.02 Wh/kg for a traditional capacitor.
Let us examine what limits the value of W ′ for the capacitor by considering a small parallel-plate capacitor with plate area A and separation between plates d. For simplicity, we assign the capacitor a voltage rating of 1 V (maximum anticipated voltage across the capacitor). Our goal is to maximize the energy density W ′. For a parallel-plate capacitor C = ϵA/d, where ϵ is the permittivity of the insulating material. Using Eq. (4.121) leads to W ′ = W m = 1 2m CV 2 = ϵAV 2 2md (J/kg), where m is the mass of the conducting plates and the insulating material contained in the capacitor. To keep the analysis simple, we assume that the plates can be made so thin as to ignore their masses relative to the mass of the insulating material. If the material’s density is ρ (kg/m3), then m = ρAd and W ′ = ϵV 2 2ρd2 (J/kg).
To maximize W ′, we need to select d to be the smallest possible, but we also have to be aware of the constraint associated with dielectric breakdown. To avoid sparking between the capacitor’s two plates, the electric field strength should not exceed Eds, the dielectric strength of the insulating material. Among the various types of materials commonly used in capacitors, mica has one of the highest values of Eds, nearly 2×108 V/m. Breakdown voltage Vbr is related to Eds by Vbr = Edsd, so given that the capacitor is to have a voltage rating of 1 V, let us choose Vbr to be 2 V, thereby allowing a 50% safety margin. With Vbr = 2 V and Eds = 2 × 108 V/m, it follows that d should not be smaller than 10−8 m, or 10 nm. For mica, ϵ ≃6ϵ0 and ρ = 3 × 103 kg/m3. Ignoring for the moment the practical issues associated with building a capacitor with a spacing of only 10 nm between conductors, the expression for energy density leads to W ′ ≃90 J/kg.
Converting W ′ to Wh/kg (by dividing by 3.6 × 103 J/Wh) gives W ′(max) = 2.5 × 10−2 (Wh/kg), for a traditional capacitor at a voltage rating of 1 V.
▶The energy storage capacity of a traditional capacitor is about four orders of magnitude smaller than the energy density capability of a lithium-ion battery. ◀ Energy Storage Comparison The table in the upper part of Fig.TF8-3 displays typical values or ranges of values for each of five attributes commonly used to characterize the performance of energy storage devices. In addition to the energy density W ′, they include the power density P ′, the charge and discharge rates, and the number of charge/discharge cycles that the device can withstand before deteriorating in performance. For most energy storage devices, the discharge rate usually is shorter than the charge rate, but for the purpose of the present discussion we treat them as equal. As a first-order approximation, the discharge rate is related to P ′ and W ′ by T = W ′ P ′ .
216 TECHNOLOGY BRIEF 8: SUPERCAPACITORS AS BATTERIES Energy Storage Devices Feature Traditional Capacitor Supercapacitor Battery Energy density W ′ (Wh/kg) ∼10−2 1 to 10 5 to 150 Power density P ′ (W/kg) 1,000 to 10,000 1,000 to 5,000 10 to 500 Charge and discharge rate T 10−3 sec ∼1 sec to 1 min ∼1 to 5 hrs Cycle life Nc ∞ ∼106 ∼103 1000 Fuel cells Batteries Future developments Supercapacitors Energy density W′ (Wh/kg) Power density P′ (W/kg) 100 10 10 100 10,000 1000 1 0.1 0.01 Traditional capacitors FigureTF8-3 Comparison of energy storage devices.
▶Supercapacitors are capable of storing 100 to 1000 times more energy than a traditional capacitor, but 10 times less than a battery (Fig.TF8-3). On the other hand, supercapacitors can discharge their stored energy in a matter of seconds, compared with hours for a battery. ◀ Moreover, the supercapacitor’s cycle life is on the order of 1 million, compared with only 1000 for a rechargeable battery. Because of these features, the supercapacitor has greatly expanded the scope and use of capacitors in electronic circuits and systems.
Future Developments The upper right-hand corner of the plot in Fig. TF8-3 represents the ideal energy storage device with W ′ ≃100–1000 Wh/kg and P ′ ≃103–104 W/kg. The corresponding discharge rate is T ≃10–100 ms. Current research aims to extend the capabilities of batteries and supercapacitors in the direction of this prized domain of the energy-power space.
4-10 ELECTROSTATIC POTENTIAL ENERGY 217 related to the magnitude of the electric field, E in the dielectric by V = Ed. Using these two expressions in Eq. (4.121) gives We = 1 2 ϵA d (Ed)2 = 1 2 ϵE2(Ad) = 1 2 ϵE2v, (4.122) where v = Ad is the volume of the capacitor. This expression affirms the assertion made at the beginning of this section, namely that the energy expended in charging up the capacitor is being stored in the electric field present in the dielectric material in between the two conductors.
The electrostatic energy density we is defined as the electrostatic potential energy We per unit volume: we = We v = 1 2 ϵE2 (J/m3).
(4.123) Even though this expression was derived for a parallel-plate capacitor, it is equally valid for any dielectric medium containing an electric field E, including vacuum. Furthermore, for any volumev, the total electrostatic potential energy stored in it is We = 1 2 v ϵE2 dv (J).
(4.124) Returning to the parallel-plate capacitor, the oppositely chargedplatesareattractedtoeachotherbyanelectricalforceF.
The force acting on any system of charges may be obtained from energy considerations.
In the discussion that follows, we show how F can be determined from We, the electrostatic energy stored in the system by virtue of the presence of electric charges.
If two conductors comprising a capacitor are allowed to move closer to each other under the influence of the electrical force F by a differential distance dl, while maintaining the charges on the plates constant, then the mechanical work done by the charged capacitor is dW = F· dl.
(4.125) The mechanical work is provided by expending electrostatic energy. Hence, dW equals the loss of energy stored in the dielectric insulating material of the capacitor, or dW = −dWe.
(4.126) From Eq. (3.73), dWe may be written in terms of the gradient of We as dWe = ∇We · dl.
(4.127) In view of Eq. (4.126), comparison of Eqs. (4.125) and (4.127) leads to F = −∇We (N).
(4.128) To apply Eq. (4.128) to the parallel-plate capacitor, we rewrite Eq. (4.120) in the form We = 1 2 Q2 C = Q2z 2ϵA , (4.129) where we replaced d with the variable z, representing the vertical spacing between the conducting plates.
Use of Eq. (4.129) in Eq. (4.128) gives F = −∇We = −ˆ z ∂ ∂z Q2z 2ϵA = −ˆ z Q2 2ϵA , (4.130) and since Q = ϵAE, F can also be expressed as F = −ˆ z ϵAE2 2 .
(4.131) (parallel-plate capacitor) Concept Question 4-26: To bring a charge q from infinity to a given point in space, a certain amount of work W is expended.
Where does the energy corresponding to W go?
Concept Question 4-27: When a voltage source is connected across a capacitor, what is the direction of the electrical force acting on its two conducting surfaces?
Exercise 4-18: Theradiioftheinnerandouterconductors of a coaxial cable are 2 cm and 5 cm, respectively, and the insulating material between them has a relative permittivity of 4.
The charge density on the outer conductor is ρℓ= 10−4 (C/m). Use the expression for E derived in Example 4-12 to calculate the total energy stored in a 20 cm length of the cable.
Answer: We = 4.1 J. (See EM.) 218 TECHNOLOGY BRIEF 9: CAPACITIVE SENSORS Technology Brief 9: Capacitive Sensors To sense is to respond to a stimulus. (See Tech Brief 7 on resistive sensors.) A capacitor can function as a sensor if the stimulus changes the capacitor’s geometry—usually the spacing between its conductive elements—or the effective dielectric properties of the insulating material situated between them. Capacitive sensors are used in a multitude of applications. A few examples follow.
Fluid Gauge The two metal electrodes in [Fig. TF9-1(a)], usually rods or plates, form a capacitor whose capacitance is directly proportional to the permittivity of the material between them. If the fluid section is of height hf and the height of the empty space above it is (h −hf), then the overall capacitance is equivalent to two capacitors in parallel, or C = Cf + Ca = ϵfw hf d + ϵaw (h −hf) d , where w is the electrode plate width, d is the spacing between electrodes, and ϵf and ϵa are the permittivities of the fluid and air, respectively. Rearranging the expression as a linear equation yields C = khf + C0, where the constant coefficient is k = (ϵf −ϵa)w/d and C0 = ϵawh/d is the capacitance of the tank when totally empty.
Using the linear equation, the fluid height can be determined by measuring C with a bridge circuit [Fig. TF9-1(b)].
g Vout C0 (empty tank) C (b) Bridge circuit with 150 kHz ac source R R Air To capacitive bridge circuit C Fluid Tank d h − hf hf w (a) Fluid tank Figure TF9-1 Fluid gauge and associated bridge circuit, with C0 being the capacitance that an empty tank would have and C the capacitance of the tank under test.
TECHNOLOGY BRIEF 9: CAPACITIVE SENSORS 219 Silicon substrate Electrodes Figure TF9-2 Interdigital capacitor used as a humidity sensor.
▶The output voltage Vout assumes a functional form that depends on the source voltage υg, the capacitance C0 of the empty tank, and the unknown fluid height hf. ◀ Humidity Sensor Thin-film metal electrodes shaped in an interdigitized pattern (to enhance the ratio A/d) are fabricated on a silicon substrate (Fig. TF9-2). The spacing between digits is typically on the order of 0.2 μm. The effective permittivity of the material separating the electrodes varies with the relative humidity of the surrounding environment. Hence, the capacitor becomes a humidity sensor.
Pressure Sensor A flexible metal diaphragm separates an oil-filled chamber with reference pressure P0 from a second chamber exposed to the gas or fluid whose pressure P is to be measured by the sensor [Fig. TF9-3(a)]. The membrane is sandwiched, but electrically isolated, between two conductive parallel surfaces, forming two capacitors in series (Fig. TF9-3(b)).
When P > P0, the membrane bends in the direction of the lower plate. Consequently, d1 increases and d2 decreases, and in turn, C1 decreases and C2 increases [Fig. TF9-3(c)]. The converse happens when P < P0. With the use of a capacitance bridge circuit, such as the one in Fig. TF9-1(b), the sensor can be calibrated to measure the pressure P with good precision.
Noncontact Sensors Precision positioning is a critical ingredient in semiconductor device fabrication, as well as in the operation and control of many mechanical systems. Noncontact capacitive sensors are used to sense the position of silicon wafers during the deposition, etching, and cutting processes, without coming in direct contact with the wafers.
220 TECHNOLOGY BRIEF 9: CAPACITIVE SENSORS (a) Pressure sensor (b) C1 = C2 (c) C1 < C2 Fluid Conducting plate Flexible metallic membrane Oil Conducting plate 1 2 3 C1 d1 C2 P P0 d2 Plate Membrane Plate 1 2 C1 C2 d1 d2 3 1 2 3 C1 = C2 P = P0 To bridge circuit Plate Membrane Plate 1 2 P d1 3 P > P0 C1 < C2 1 2 3 To bridge circuit C1 C2 d2 FigureTF9-3 Pressure sensor responds to deflection of metallic membrane.
▶They are also used to sense and control robot arms in equipment manufacturing and to position hard disc drives, photocopier rollers, printing presses, and other similar systems. ◀ TECHNOLOGY BRIEF 9: CAPACITIVE SENSORS 221 Conductive plates Electric field lines Insulator C Figure TF9-4 Concentric-plate capacitor.
External object (a) Adjacent-plates capacitor (b) Perturbation field C0 C ≠ C0 Figure TF9-5 (a) Adjacent-plates capacitor; (b) perturbation field.
The concentric plate capacitor in Fig. TF9-4 consists of two metal plates, sharing the same plane, but electrically isolated from each other by an insulating material. When connected to a voltage source, charges of opposite polarity form on the two plates, resulting in the creation of electric-field lines between them. The same principle applies to the adjacent-plates capacitor in Fig. TF9-5. In both cases, the capacitance is determined by the shapes and sizes of the conductive elements and by the effective permittivity of the dielectric medium containing the electric field lines between them. Often, the capacitor surface is covered by a thin film of nonconductive material, the purpose of which is to keep the plate surfaces clean and dust free.
▶The introduction of an external object into the proximity of the capacitor [Fig. TF9-5(b)] changes the effective permittivity of the medium, perturbs the electric field lines, and modifies the charge distribution on the plates. ◀ This, in turn, changes the value of the capacitance as would be measured by a capacitance meter or bridge circuit.
Hence, the capacitor becomes a proximity sensor, and its sensitivity depends, in part, on how different the permittivity of the external object is from that of the unperturbed medium and on whether it is or is not made of a conductive material.
Fingerprint Imager An interesting extension of noncontact capacitive sensors is the development of a fingerprint imager consisting of a two-dimensional array of capacitive sensor cells, constructed to record an electrical representation of a fingerprint (Fig. TF9-6). Each sensor cell is composed of an adjacent-plates capacitor connected to a capacitance measurement circuit (Fig.TF9-7). The entire surface of the imager is covered by a thin layer of nonconductive oxide. When the finger is placed on the oxide surface, it perturbs the field lines of the individual sensor cells to varying degrees, depending on the distance between the ridges and valleys of the finger’s surface from the sensor cells.
▶Given that the dimensions of an individual sensor are on the order of 65 μm on the side, the imager is capable of recording a fingerprint image at a resolution corresponding to 400 dots per inch or better. ◀ 222 TECHNOLOGY BRIEF 9: CAPACITIVE SENSORS FigureTF9-6 Elements of a fingerprint matching system.
FigureTF9-7 Fingerprint representation.
4-11 IMAGE METHOD 223 4-11 Image Method Consider a point charge Q at a distance d above a horizontally infinite, perfectly conducting plate [Fig. 4-26(a)]. We want to determine V and E at any point in the space above the plate, as well as the surface charge distribution on the plate. Three different methods for finding E have been introduced in this chapter The first method, based on Coulomb’s law, requires knowledge of the magnitudes and locations of all the charges.
In the present case, the charge Q induces an unknown and nonuniform distribution of charge on the plate.
Hence, we cannot utilize Coulomb’s method. The second method, based on Gauss’s law, is equally difficult to use because it is not clear how to construct a Gaussian surface across which E is only tangential or only normal.
The third method is based on evaluating the electric field using E = −∇V after solving Poisson’s or Laplace’s equation for V subject to the available boundary conditions, but it is mathematically involved.
Alternatively, the problem at hand can be solved using image theory.
▶Any given charge configuration above an infinite, perfectly conducting plane is electrically equivalent to the combination of the given charge configuration and its image configuration, with the conducting plane removed. ◀ The image-method equivalent of the charge Q above a conductingplaneisshownintheright-handsectionofFig.4-26.
It consists of the charge Q itself and an image charge −Q at a distance 2d from Q, with nothing else between them.
The electric field due to the two isolated charges can now be easily found at any point (x, y, z) by applying Coulomb’s method, as demonstrated by Example 4-13. By symmetry, the combination of the two charges always produces a potential V = 0 at every point in the plane previously occupied by the conductingsurface. Ifthechargeresidesinthepresenceofmore than one grounded plane, it is necessary to establish its images relative to each of the planes and then to establish images of each of those images against the remaining planes. The process is continued until the condition V = 0 is satisfied everywhere on all grounded planes. The image method applies not only to point charges, but also to distributions of charge, such as the line and volume distributions depicted in Fig. 4-27. Once E has been determined, the charge induced on the plate can be found from ρs = (ˆ n· E)ϵ0, (4.132) where ˆ n is the normal unit vector to the plate [Fig. 4-26(a)].
Example 4-13: Image Method for Charge above Conducting Plane Use image theory to determine E at an arbitrary point P = (x, y, z) in the region z > 0 due to a charge Q in free space at a distance d above a grounded conducting plate residing in the z = 0 plane.
Solution: In Fig. 4-28, charge Q is at (0, 0, d) and its image −Q is at (0, 0, −d). From Eq. (4.19), the electric field at point P(x, y, z) due to the two charges is given by E = 1 4πϵ0 QR1 R3 1 + −QR2 R3 2 = Q 4πϵ0 ˆ xx + ˆ yy + ˆ z(z −d) [x2 + y2 + (z −d)2]3/2 − ˆ xx + ˆ yy + ˆ z(z + d) [x2 + y2 + (z + d)2]3/2 for z ≥0.
Exercise 4-19: Use the result of Example 4-13 to find the surface charge density ρs on the surface of the conducting plane.
Answer: ρs = −Qd/[2π(x2 + y2 + d2)3/2]. (See EM.) Concept Question 4-28: What is the fundamental premise of the image method?
Concept Question 4-29: Given a charge distribution, what are the various approaches described in this chapter for computing the electric field E at a given point in space?
224 CHAPTER 4 ELECTROSTATICS (a) Charge Q above grounded plane (b) Equivalent configuration Electric field lines V = 0 d d Q −Q ε ε + – V = 0 d z Q σ = ∞ ε + n ˆ Figure 4-26 By image theory, a charge Q above a grounded perfectly conducting plane is equivalent to Q and its image −Q with the ground plane removed.
(a) Charge distributions above ground plane (b) Equivalent distributions V = 0 ρl ρv σ = ∞ ε ρl –ρl ρv –ρv ε ε V = 0 Figure 4-27 Charge distributions above a conducting plane and their image-method equivalents.
Q = (0, 0, d) P = (x, y, z) –Q = (0, 0, –d) R1 R2 z = 0 plane z + – Figure 4-28 Application of the image method for finding E at point P (Example 4-13).
CHAPTER 4 SUMMARY 225 Chapter 4 Summary Concepts • Maxwell’s equations are the fundamental tenets of electromagnetic theory.
• Under static conditions, Maxwell’s equations separate into two uncoupled pairs, with one pair pertaining to electrostatics and the other to magnetostatics.
• Coulomb’s law provides an explicit expression for the electric field due to a specified charge distribution.
• Gauss’s law states that the total electric field flux through a closed surface is equal to the net charge enclosed by the surface.
• The electrostatic field E at a point is related to the electric potential V at that point by E = −∇V , with V often being referenced to zero at infinity.
• Because most metals have conductivities on the order of 106 (S/m), they are treated in practice as perfect conductors.
By the same token, insulators with conductivities smaller than 10−10 (S/m) often are treated as perfect dielectrics.
• Boundary conditions at the interface between two materials specify the relations between the normal and tangential components of D, E, and J in one of the materials to the corresponding components in the other.
• The capacitance of a two-conductor body and resistance of the medium between them can be computed from knowledge of the electric field in that medium.
• The electrostatic energy density stored in a dielectric medium is we = 1 2ϵE2 (J/m3).
• When a charge configuration exists above an infinite, perfectly conducting plane, the induced field E is the same as that due to the configuration itself and its image with the conducting plane removed.
Important Terms Provide definitions or explain the meaning of the following terms: boundary conditions capacitance C charge density conductance G conduction current conductivity σ conductor conservative field constitutive parameters convection current Coulomb’s law current density J dielectric breakdown voltage Vbr dielectric material dielectric strength Eds dipole moment p electric dipole electric field intensity E electric flux density D electric potential V electric susceptibility χe electron drift velocity ue electron mobility μe electrostatic energy density we electrostatic potential energy We electrostatics equipotential Gaussian surface Gauss’s law hole drift velocity uh hole mobility μh homogeneous material image method isotropic material Joule’s law Kirchhoff’s voltage law Laplace’s equation linear material Ohm’s law perfect conductor perfect dielectric permittivity ϵ Poisson’s equation polarization vector P relative permittivity ϵr semiconductor static condition superconductor volume, surface, and line charge densities 226 CHAPTER 4 ELECTROSTATICS Mathematical and Physical Models Maxwell’s Equations for Electrostatics Name Differential Form Integral Form Gauss’s law ∇· D = ρv S D· ds = Q Kirchhoff’s law ∇× × × E = 0 C E· dl = 0 Electric Field Current density J = ρvu Poisson’s equation ∇2V = −ρv ϵ Laplace’s equation ∇2V = 0 Resistance R = − l E· dl S σE· ds Boundary conditions Table 4-3 Capacitance C = S ϵE· ds − l E· dl RC relation RC = ϵ σ Energy density we = 1 2ϵE2 Point charge E = ˆ R q 4πϵR2 Many point charges E = 1 4πϵ N i=1 qi(R −Ri) |R −Ri|3 Volume distribution E = 1 4πϵ v ′ ˆ R′ ρv dv ′ R′2 Surface distribution E = 1 4πϵ S′ ˆ R′ ρs ds′ R′2 Line distribution E = 1 4πϵ l′ ˆ R′ ρℓdl′ R′2 Infinite sheet of charge E = ˆ z ρs 2ϵ0 Infinite line of charge E = D ϵ0 = ˆ r Dr ϵ0 = ˆ r ρℓ 2πϵ0r Dipole E = qd 4πϵ0R3 ( ˆ R 2 cos θ + ˆ θ θ θ sin θ) Relation to V E = −∇V PROBLEMS Section 4-2: Charge and Current Distributions ∗4.1 A cube 2 m on a side is located in the first octant in a Cartesian coordinate system, with one of its corners at the origin. Find the total charge contained in the cube if the charge density is given by ρv = xy2e−2z (mC/m3).
∗Answer(s) available in Appendix D.
4.2 Find the total charge contained in a cylindrical volume defined by r ≤2 m and 0 ≤z ≤3 m if ρv = 20rz (mC/m3).
∗4.3 Find the total charge contained in a round-top cone defined by R ≤2 m and 0 ≤θ ≤π/4, given that ρv = 10R2 cos2 θ (mC/m3).
4.4 If the line charge density is given by ρl = 24y2 (mC/m), find the total charge distributed on the y axis from y = −5 to y = 5.
PROBLEMS 227 4.5 Find the total charge on a circular disk defined by r ≤a and z = 0 if: (a) ρs = ρs0 cos φ (C/m2) (b) ρs = ρs0 sin2 φ (C/m2) (c) ρs = ρs0e−r (C/m2) (d) ρs = ρs0e−r sin2 φ (C/m2) where ρs0 is a constant.
4.6 If J = ˆ y4xz (A/m2), find the current I flowing through a square with corners at (0, 0, 0), (2, 0, 0), (2, 0, 2), and (0, 0, 2).
∗4.7 If J = ˆ R5/R (A/m2), find I through the surface R = 5 m.
4.8 An electron beam shaped like a circular cylinder of radius r0 carries a charge density given by ρv = −ρ0 1 + r2 (C/m3) where ρ0 is a positive constant and the beam’s axis is coincident with the z axis.
(a) Determine the total charge contained in length L of the beam.
(b) If the electrons are moving in the +z direction with uniform speed u, determine the magnitude and direction of the current crossing the z-plane.
4.9 A circular beam of charge of radius a consists of electrons moving with a constant speed u along the +z direction. The beam’s axis is coincident with the z axis and the electron charge density is given by ρv = −cr2 (c/m3) where c is a constant and r is the radial distance from the axis of the beam.
∗(a) Determine the charge density per unit length.
(b) Determine the current crossing the z-plane.
4.10 A line of charge of uniform density ρℓoccupies a semicircle of radius b as shown in Fig. P4.10. Use the material presented in Example 4-4 to determine the electric field at the origin.
x y b z ρl Figure P4.10 Problem 4.10.
Section 4-3: Coulomb’s Law ∗4.11 A square with sides of 2 m has a charge of 40 μC at each of its four corners. Determine the electric field at a point 5 m above the center of the square.
4.12 Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x–y plane, with one corner at the origin, another at (2 cm, 0, 0), and the third at (0, 2 cm, 0).
Find the force acting on the charge located at the origin.
∗4.13 Charge q1 = 6 μC is located at (1 cm, 1 cm, 0) and charge q2 is located at (0, 0, 4 cm). What should q2 be so that E at (0, 2 cm, 0) has no y component?
4.14 A line of charge with uniform density ρℓ= 8 (μC/m) exists in air along the z axis between z = 0 and z = 5 cm. Find E at (0,10 cm,0).
4.15 Electric charge is distributed along an arc located in the x–y plane and defined by r = 2 cm and 0 ≤φ ≤π/4. If ρℓ= 5 (μC/m), find E at (0, 0, z) and then evaluate it at: ∗(a) The origin.
(b) z = 5 cm (c) z = −5 cm 4.16 A line of charge with uniform density ρl extends between z = −L/2 and z = L/2 along the z axis. Apply Coulomb’s law to obtain an expression for the electric field at any point P(r, φ, 0) on the x–y plane. Show that your result reduces to the expression given by (4.33) as the length L is extended to infinity.
∗4.17 Repeat Example 4-5 for the circular disk of charge of radius a, but in the present case, assume the surface charge density to vary with r as ρs = ρs0r2 (C/m2) 228 CHAPTER 4 ELECTROSTATICS where ρs0 is a constant.
4.18 Multiple charges at different locations are said to be in equilibrium if the force acting on any one of them is identical in magnitude and direction to the force acting on any of the others.
Suppose we have two negative charges, one located at the origin and carrying charge −9e, and the other located on the positive x axis at a distance d from the first one and carrying charge −36e. Determine the location, polarity, and magnitude of a third charge whose placement would bring the entire system into equilibrium.
Section 4-4: Gauss’s Law 4.19 Three infinite lines of charge, all parallel to the z axis, are located at the three corners of the kite-shaped arrangement shown in Fig. P4.19. If the two right triangles are symmetrical and of equal corresponding sides, show that the electric field is zero at the origin.
y x −2ρl ρl ρl Figure P4.19 Kite-shaped arrangement of line charges for Problem 4.19.
∗4.20 Three infinite lines of charge, ρl1 = 3 (nC/m), ρl2 = −3 (nC/m), and ρl3 = 3 (nC/m), are all parallel to the z axis. If they pass through the respective points (0, −b), (0, 0), and (0, b) in the x–y plane, find the electric field at (a, 0, 0). Evaluate your result for a = 2 cm and b = 1 cm.
4.21 A horizontal strip lying in the x–y plane is of width d in the y direction and infinitely long in the x direction. If the strip is in air and has a uniform charge distribution ρs, use Coulomb’s law to obtain an explicit expression for the electric field at a point P located at a distance h above the centerline of the strip. Extend your result to the special case where d is infinite and compare it with Eq. (4.25).
4.22 Given the electric flux density D = ˆ x2(x + y) + ˆ y(3x −2y) (C/m2) determine (a) ρv by applying Eq. (4.26).
(b) The total charge Q enclosed in a cube 2 m on a side, located in the first octant with three of its sides coincident with the x-, y-, and z axes and one of its corners at the origin.
(c) The total charge Q in the cube, obtained by applying Eq. (4.29).
∗4.23 Repeat Problem 4.22 for D = ˆ xxy3z3 (C/m2).
4.24 Charge Q1 is uniformly distributed over a thin spherical shell of radius a, and charge Q2 is uniformly distributed over a second spherical shell of radius b, with b > a. Apply Gauss’s law to find E in the regions R < a, a < R < b, and R > b.
∗4.25 The electric flux density inside a dielectric sphere of radius a centered at the origin is given by D = ˆ Rρ0R (C/m2) where ρ0 is a constant. Find the total charge inside the sphere.
4.26 In a certain region of space, the charge density is given in cylindrical coordinates by the function: ρv = 5re−r (C/m3) Apply Gauss’s law to find D.
∗4.27 An infinitely long cylindrical shell extending between r = 1 m and r = 3 m contains a uniform charge density ρv0.
Apply Gauss’s law to find D in all regions.
4.28 If the charge density increases linearly with distance from the origin such that ρv = 0 at the origin and ρv = 4 C/m3 at R = 2 m, find the corresponding variation of D.
4.29 A spherical shell with outer radius b surrounds a charge-free cavity of radius a < b (Fig. P4.29). If the shell contains a charge density given by ρv = −ρv0 R2 , a ≤R ≤b, where ρv0 is a positive constant, determine D in all regions.
PROBLEMS 229 b r3 r1 a ρv r2 Figure P4.29 Problem 4.29.
Section 4-5: Electric Potential ∗4.30 A square in the x–y plane in free space has a point charge of +Q at corner (a/2, a/2), the same at corner (a/2, −a/2), and a point charge of −Q at each of the other two corners.
(a) Find the electric potential at any point P along the x axis.
(b) Evaluate V at x = a/2.
4.31 The circular disk of radius a shown in Fig. 4-7 has uniform charge density ρs across its surface.
(a) Obtain an expression for the electric potential V at a point P (0, 0, z) on the z axis.
(b) Use your result to find E and then evaluate it for z = h.
Compare your final expression with (4.24), which was obtained on the basis of Coulomb’s law.
∗4.32 A circular ring of charge of radius a lies in the x–y plane and is centered at the origin. Assume also that the ring is in air and carries a uniform density ρℓ.
(a) Show that the electrical potential at (0, 0, z) is given by V = ρℓa/[2ϵ0(a2 + z2)1/2].
(b) Find the corresponding electric field E.
4.33 Show that the electric potential difference V12 between two points in air at radial distances r1 and r2 from an infinite line of charge with density ρℓalong the z axis is V12 = (ρℓ/2πϵ0) ln(r2/r1).
∗4.34 Find the electric potential V at a location a distance b from the origin in the x–y plane due to a line charge with charge density ρℓand of length l. The line charge is coincident with the z axis and extends from z = −l/2 to z = l/2.
4.35 For the electric dipole shown in Fig. 4-13, d = 1 cm and |E| = 4 (mV/m) at R = 1 m and θ = 0◦. Find E at R = 2 m and θ = 90◦.
4.36 For each of the distributions of the electric potential V shown in Fig. P4.36, sketch the corresponding distribution of E (in all cases, the vertical axis is in volts and the horizontal axis is in meters).
3 3 6 9 12 15 30 −30 4 −4 5 8 11 13 16 V V x (a) (b) x 3 6 9 12 15 4 −4 V (c) x Figure P4.36 Electric potential distributions of Problem 4.36.
230 CHAPTER 4 ELECTROSTATICS ∗4.37 Two infinite lines of charge, both parallel to the z axis, lie in the x–z plane, one with density ρℓand located at x = a and the other with density −ρℓand located at x = −a. Obtain an expression for the electric potential V (x, y) at a point P = (x, y) relative to the potential at the origin.
x y (−a, 0) −ρl ρl (a, 0) P = (x, y) r' r'' Figure P4.37 Problem 4.37.
4.38 Given the electric field E = ˆ R 18 R2 (V/m) find the electric potential of point A with respect to point B where A is at +2 m and B at −4 m, both on the z axis.
∗4.39 An infinitely long line of charge with uniform density ρl = 9 (nC/m) lies in the x–y plane parallel to the y axis at x = 2 m.
Find the potential VAB at point A(3 m, 0, 4 m) in Cartesian coordinates with respect to point B(0, 0, 0) by applying the result of Problem 4.33.
4.40 The x–y plane contains a uniform sheet of charge with ρs1 = 0.2 (nC/m2). A second sheet with ρs2 = −0.2 (nC/m2) occupies the plane z = 6 m. Find VAB, VBC, and VAC for A(0, 0, 6 m), B(0, 0, 0), and C(0, −2 m, 2 m).
Section 4-6: Conductors 4.41 A cylindrical bar of silicon has a radius of 4 mm and a length of 8 cm. If a voltage of 5 V is applied between the ends of the bar and μe = 0.13 (m2/V·s), μh = 0.05 (m2/V·s), Ne = 1.5×1016 electrons/m3, and Nh = Ne, find the following: (a) The conductivity of silicon.
(b) The current I flowing in the bar.
∗(c) The drift velocities ue and uh.
(d) The resistance of the bar.
(e) The power dissipated in the bar.
4.42 Repeat Problem 4.41 for a bar of germanium with μe = 0.4 (m2/V·s), μh = 0.2 (m2/V·s), and Ne = Nh = 2.4 × 1019 electrons or holes/m3.
4.43 A 100 m long conductor of uniform cross-section has a voltagedropof4Vbetweenitsends. Ifthedensityofthecurrent flowing through it is 1.4 × 106 (A/m2), identify the material of the conductor.
4.44 A coaxial resistor of length l consists of two concentric cylinders.
The inner cylinder has radius a and is made of a material with conductivity σ1, and the outer cylinder, extending between r = a and r = b, is made of a material with conductivity σ2. If the two ends of the resistor are capped with conducting plates, show that the resistance between the two ends is R = l/[π(σ1a2 + σ2(b2 −a2))].
∗4.45 Apply the result of Problem 4.44 to find the resistance of a 20 cm long hollow cylinder (Fig. P4.45) made of carbon with σ = 3 × 104 (S/m).
3 cm 2 cm Carbon Figure P4.45 Cross section of hollow cylinder of Problem 4.45.
4.46 A 2 × 10−3 mm thick square sheet of aluminum has 5 cm × 5 cm faces. Find the following: (a) The resistance between opposite edges on a square face.
(b) The resistance between the two square faces.
(See Appendix B for the electrical constants of materials.) 4.47 A cylinder-shaped carbon resistor is 8 cm in length and its circular cross section has a diameter d = 1 mm.
(a) Determine the resistance R.
(b) To reduce its resistance by 40%, the carbon resistor is coated with a layer of copper of thickness t. Use the result of Problem 4.44 to determine t.
PROBLEMS 231 Section 4-8: Boundary Conditions ∗4.48 With reference to Fig.
4-19, find E1 if E2 = ˆ x3 −ˆ y2 + ˆ z2 (V/m), ϵ1 = 2ϵ0, ϵ2 = 18ϵ0, and the boundary has a surface charge density ρs = 3.54 × 10−11 (C/m2).
What angle does E2 make with the z axis?
4.49 An infinitely long cylinder of radius a is surrounded by a dielectric medium that contains no free charges. If the tangential component of the electric field in the region r ≥a is given by Et = −ˆ φ φ φ cos2 φ/r2, find E in that region.
∗4.50 If E = ˆ R150 (V/m) at the surface of a 5-cm conducting sphere centered at the origin, what is the total charge Q on the sphere’s surface?
4.51 Figure P4.51 shows three planar dielectric slabs of equal thickness but with different dielectric constants. If E0 in air makes an angle of 45◦with respect to the z axis, find the angle of E in each of the other layers.
ε0 (air) ε1 = 3ε0 ε2 = 5ε0 ε3 = 7ε0 ε0 (air) 45° z E0 Figure P4.51 Dielectric slabs in Problem 4.51.
Sections 4-9 and 4-10: Capacitance and Electrical Energy 4.52 Determine the force of attraction in a parallel-plate capacitor with A = 5 cm2, d = 2 cm, and ϵr = 4 if the voltage across it is 50 V.
4.53 Dielectric breakdown occurs in a material whenever the magnitude of the field E exceeds the dielectric strength anywhere in that material.
In the coaxial capacitor of Example 4-12, ∗(a) At what value of r is |E| maximum?
(b) What is the breakdown voltage if a = 1 cm, b = 2 cm, and the dielectric material is mica with ϵr = 6?
4.54 An electron with charge Qe = −1.6×10−19 C and mass me = 9.1 × 10−31 kg is injected at a point adjacent to the negatively charged plate in the region between the plates of an air-filled parallel-plate capacitor with separation of 1 cm and rectangular plates each 10 cm2 in area (Fig. P4.54). If the voltage across the capacitor is 10 V, find the following: (a) The force acting on the electron.
(b) The acceleration of the electron.
(c) The time it takes the electron to reach the positively charged plate, assuming that it starts from rest.
Qe 1 cm V0 = 10 V + – Figure P4.54 Electron between charged plates of Problem 4.54.
∗4.55 In a dielectric medium with ϵr = 4, the electric field is given by E = ˆ x(x2 + 2z) + ˆ yx2 −ˆ z(y + z) (V/m) Calculate the electrostatic energy stored in the region −1 m ≤x ≤1 m, 0 ≤y ≤2 m, and 0 ≤z ≤3 m.
4.56 Figure P4.56(a) depicts a capacitor consisting of two parallel, conducting plates separated by a distance d. The space between the plates contains two adjacent dielectrics, one with permittivity ϵ1 and surface area A1 and another with ϵ2 and A2.
232 CHAPTER 4 ELECTROSTATICS (a) (b) ε1 A1 A2 ε2 d + − V C1 C2 V + − Figure P4.56 (a) Capacitor with parallel dielectric section, and (b) equivalent circuit.
The objective of this problem is to show that the capacitance C of the configuration shown in Fig. P4.56(a) is equivalent to two capacitances in parallel, as illustrated in Fig. P4.56(b), with C = C1 + C2 (4.133) where C1 = ϵ1A1 d (4.134) C2 = ϵ2A2 d (4.135) To this end, proceed as follows: (a) Find the electric fields E1 and E2 in the two dielectric layers.
(b) Calculate the energy stored in each section and use the result to calculate C1 and C2.
(c) Use the total energy stored in the capacitor to obtain an expression for C.
Show that (4.133) is indeed a valid result.
4.57 Use the result of Problem 4.56 to determine the capacitance for each of the following configurations: (a) (b) ε1 = 8ε0; ε2 = 4ε0; ε3 = 2ε0 3 cm 5 cm 1 cm 2 cm εr = 2 εr = 4 r1 = 2 mm r2 = 4 mm r3 = 8 mm ε3 2 cm ε2 ε1 Figure P4.57 Dielectric sections for Problems 4.57 and 4.59.
∗(a) Conducting plates are on top and bottom faces of the rectangular structure in Fig. P4.57(a).
(b) Conducting plates are on front and back faces of the structure in Fig. P4.57(a).
PROBLEMS 233 (c) Conducting plates are on top and bottom faces of the cylindrical structure in Fig. P4.57(b).
4.58 The capacitor shown in Fig. P4.58 consists of two parallel dielectric layers. Use energy considerations to show that the equivalent capacitance of the overall capacitor, C, is equal to the series combination of the capacitances of the individual layers, C1 and C2, namely C = C1C2 C1 + C2 (4.136) where C1 = ϵ1 A d1 , C2 = ϵ2 A d2 .
(a) Let V1 and V2 be the electric potentials across the upper and lower dielectrics, respectively. What are the corresponding electric fields E1 and E2? By applying the appropriate boundary condition at the interface between the two dielectrics, obtain explicit expressions for E1 and E2 in terms of ϵ1, ϵ2, V , and the indicated dimensions of the capacitor.
(b) Calculate the energy stored in each of the dielectric layers and then use the sum to obtain an expression for C.
(c) Show that C is given by Eq. (4.136).
(a) (b) V + − C1 C2 + − d1 d2 V A ε1 ε2 Figure P4.58 (a) Capacitor with parallel dielectric layers, and (b) equivalent circuit (Problem 4.58).
4.59 Use the expressions given in Problem 4.58 to determine the capacitance for the configurations in Fig. P4.57(a) when the conducting plates are placed on the right and left faces of the structure.
4.60 A coaxial capacitor consists of two concentric, conducting, cylindrical surfaces, one of radius a and another of radiusb, asshowninFig.P4.60. Theinsulatinglayerseparating the two conducting surfaces is divided equally into two semi-cylindrical sections, one filled with dielectric ϵ1 and the other filled with dielectric ϵ2.
b E − + V l a ε2 ε1 Figure P4.60 Problem 4.60.
(a) Develop an expression for C in terms of the length l and the given quantities.
∗(b) EvaluatethevalueofC fora = 2mm, b = 6mm, ϵr1 = 2, ϵr2 = 4, and l = 4 cm.
Section 4-12: Image Method 4.61 With reference to Fig. P4.61, charge Q is located at a distance d above a grounded half-plane located in the x–y plane and at a distance d from another grounded half-plane in the x–z plane. Use the image method to (a) Establish the magnitudes, polarities, and locations of the images of charge Q with respect to each of the two ground planes (as if each is infinite in extent).
234 CHAPTER 4 ELECTROSTATICS d d z y P = (0, y, z) Q = (0, d, d) Figure P4.61 Charge Q next to two perpendicular, grounded, conducting half-planes.
(b) Find the electric potential and electric field at an arbitrary point P = (0, y, z).
4.62 Conducting wires above a conducting plane carry currents I1 and I2 in the directions shown in Fig. P4.62.
Keepinginmindthatthedirectionofacurrentisdefinedinterms of the movement of positive charges, what are the directions of the image currents corresponding to I1 and I2?
I1 I2 (a) (b) Figure P4.62 Currents above a conducting plane (Problem 4.62).
∗4.63 Use the image method to find the capacitance per unit length of an infinitely long conducting cylinder of radius a situated at a distance d from a parallel conducting plane, as shown in Fig. P4.63.
V = 0 a d Figure P4.63 Conducting cylinder above a conducting plane (Problem 4.63).
C H A P T E R 5 Magnetostatics Chapter Contents Overview, 236 5-1 Magnetic Forces and Torques, 237 5-2 The Biot–Savart Law, 244 5-3 Maxwell’s Magnetostatic Equations, 251 TB10 Electromagnets, 256 5-4 Vector Magnetic Potential, 259 5-5 Magnetic Properties of Materials, 260 5-6 Magnetic Boundary Conditions, 264 5-7 Inductance, 265 TB11 Inductive Sensors, 268 5-8 Magnetic Energy, 271 Chapter 5 Summary, 272 Problems, 274 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Calculate the magnetic force on a current-carrying wire placed in a magnetic field and the torque exerted on a current loop.
2. Apply the Biot–Savart law to calculate the magnetic field due to current distributions.
3. Apply Ampere’s law to configurations with appropriate symmetry.
4. Explain magnetic hysteresis in ferromagnetic materials.
5. Calculate the inductance of a solenoid, a coaxial transmission line, or other configurations.
6. Relate the magnetic energy stored in a region to the magnetic field distribution in that region.
236 CHAPTER 5 MAGNETOSTATICS Overview This chapter on magnetostatics parallels the preceding one on electrostatics.
Stationary charges produce static electric fields, and steady (i.e., non–time-varying) currents produce static magnetic fields. When ∂/∂t = 0, the magnetic fields in a medium with magnetic permeability μ are governed by the second pair of Maxwell’s equations [Eqs. (4.3a,b)]: ∇· B = 0, (5.1a) ∇× × × H = J, (5.1b) where J is the current density. The magnetic flux density B and the magnetic field intensity H are related by B = μH.
(5.2) When examining electric fields in a dielectric medium in Chapter 4, we noted that the relation D = ϵE is valid only when the medium is linear and isotropic. These properties, which hold true for most materials, allow us to treat the permittivity ϵ as a constant, scalar quantity, independent of both the magnitude and the direction of E. A similar statement applies to the relation given by Eq. (5.2). With the exception of ferromagnetic materials, for which the relationship between B and H is nonlinear, most materials are characterized by constant permeabilities.
▶Furthermore, μ = μ0 for most dielectrics and metals (excluding ferromagnetic materials). ◀ The objective of this chapter is to develop an understanding of the relationship between steady currents and the magnetic flux B and field H due to various types of current distributions and in various types of media, and to introduce a number of related quantities, such as the magnetic vector potential A, the magneticenergydensitywm, andtheinductanceofaconducting structure, L.
The parallelism that exists between these magnetostatic quantities and their electrostatic counterparts is elucidated in Table 5-1.
Table 5-1 Attributes of electrostatics and magnetostatics.
Attribute Electrostatics Magnetostatics Sources Stationary charges ρv Steady currents J Fields and fluxes E and D H and B Constitutive parameter(s) ϵ and σ μ Governing equations • Differential form • Integral form ∇· D = ρv ∇× × × E = 0 S D· ds = Q C E· dl = 0 ∇· B = 0 ∇× × × H = J S B· ds = 0 C H· dl = I Potential Scalar V , with Vector A, with E = −∇V B = ∇× × × A Energy density we = 1 2ϵE2 wm = 1 2μH 2 Force on charge q Fe = qE Fm = qu × × × B Circuit element(s) C and R L 5-1 MAGNETIC FORCES AND TORQUES 237 5-1 Magnetic Forces and Torques The electric field E at a point in space was defined as the electric force Fe per unit charge acting on a charged test particle placed at that point. We now define the magnetic flux density B at a point in space in terms of the magnetic force Fm that acts on a charged test particle moving with velocity u through that point.
Experiments revealed that a particle of charge q moving with velocity u in a magnetic field experiences a magnetic force Fm given by Fm = qu × × × B (N).
(5.3) Accordingly, the strength of B is measured in newtons/(C·m/s), also called the tesla (T). For a positively charged particle, the direction of Fm is that of the cross product u × × × B, which is perpendicular to the plane containing u and B and governed by the right-hand rule. If q is negative, the direction of Fm is reversed (Fig. 5-1). The magnitude of Fm is given by Fm = quB sin θ, (5.4) where θ is the angle between u and B. We note that Fm is maximum when u is perpendicular to B (θ = 90◦), and zero when u is parallel to B (θ = 0 or 180◦).
If a charged particle resides in the presence of both an electric field E and a magnetic field B, then the total electromagnetic force acting on it is F = Fe + Fm = qE + qu × × × B = q(E + u × × × B).
(5.5) The force expressed by Eq. (5.5) also is known as the Lorentz force.
Electric and magnetic forces exhibit a number of important differences: 1. Whereas the electric force is always in the direction of the electric field, the magnetic force is always perpendicular to the magnetic field.
2. Whereas the electric force acts on a charged particle whether or not it is moving, the magnetic force acts on it only when it is in motion.
3. Whereas the electric force expends energy in displacing a charged particle, the magnetic force does no work when a particle is displaced.
This last statement requires further elaboration. Because the magnetic force Fm is always perpendicular to u, Fm · u = 0.
(a) (b) θ +q u B Fm = quB sin θ − + F F B u u Figure 5-1 The direction of the magnetic force exerted on a charged particle moving in a magnetic field is (a) perpendicular to both B and u and (b) depends on the charge polarity (positive or negative).
Hence, the work performed when a particle is displaced by a differential distance dl = u dt is dW = Fm · dl = (Fm · u) dt = 0.
(5.6) ▶Since no work is done, a magnetic field cannot change the kinetic energy of a charged particle; the magnetic field can change the direction of motion of a charged particle, but not its speed. ◀ Exercise 5-1: An electron moving in the positive x direction perpendicular to a magnetic field is deflected in the negative z direction. What is the direction of the magnetic field?
Answer: Positive y direction. (See EM.) 238 CHAPTER 5 MAGNETOSTATICS Module 5.1 Electron Motion in Static Fields This module demonstrates the Lorentz force on an electron moving under the influence of an electric field alone, a magnetic field alone, or both acting simultaneously.
Exercise 5-2: A proton moving with a speed of 2 × 106 m/s through a magnetic field with magnetic flux density of 2.5 T experiences a magnetic force of magnitude 4 × 10−13 N. What is the angle between the magnetic field and the proton’s velocity?
Answer: θ = 30◦or 150◦. (See EM.) Exercise 5-3: A charged particle with velocity u is moving in a medium with uniform fields E = ˆ xE and B = ˆ yB. Whatshouldubesothattheparticleexperiences no net force?
Answer: u = ˆ zE/B.
[u may also have an arbitrary y component uy]. (See EM.) 5-1.1 Magnetic Force on a Current-Carrying Conductor A current flowing through a conducting wire consists of charged particles drifting through the material of the wire.
Consequently, when a current-carrying wire is placed in a magnetic field, it experiences a force equal to the sum of the magnetic forces acting on the charged particles moving within it. Consider, for example, the arrangement shown in Fig. 5-2 5-1 MAGNETIC FORCES AND TORQUES 239 z y x I B B B I = 0 (a) I (b) (c) Figure 5-2 When a slightly flexible vertical wire is placed in a magnetic field directed into the page (as denoted by the crosses), it is (a) not deflected when the current through it is zero, (b) deflected to the left when I is upward, and (c) deflected to the right when I is downward.
in which a vertical wire oriented along the z direction is placed in a magnetic field B (produced by a magnet) oriented along the −ˆ x direction (into the page). With no current flowing in the wire, Fm = 0 and the wire maintains its vertical orientation [Fig. 5-2(a)], but when a current is introduced in the wire, the wire deflects to the left (−ˆ y direction) if the current direction is upward (+ˆ z direction), and to the right (+ˆ y direction) if the current direction is downward (−ˆ z direction). The directions of these deflections are in accordance with the cross product given by Eq. (5.3).
To quantify the relationship between Fm and the current I flowing in a wire, let us consider a small segment of the wire of cross-sectional area A and differential length dl, with the direction of dl denoting the direction of the current.
Without loss of generality, we assume that the charge carriers constituting the current I are exclusively electrons, which is always a valid assumption for a good conductor. If the wire contains a free-electron charge density ρve = −Nee, where Ne is the number of moving electrons per unit volume, then the total amount of moving charge contained in an elemental volume of the wire is dQ = ρveA dl = −NeeA dl, (5.7) and the corresponding magnetic force acting on dQ in the presence of a magnetic field B is dFm = dQ ue × × × B = −NeeA dl ue × × × B, (5.8a) whereue isthedriftvelocityoftheelectrons. Sincethedirection ofacurrentisdefinedasthedirectionofflowofpositivecharges, the electron drift velocity ue is parallel to dl, but opposite in direction. Thus, dl ue = −dl ue and Eq. (5.8a) becomes dFm = NeeAue dl × × × B.
(5.8b) From Eqs. (4.11) and (4.12), the current I flowing through a cross-sectional area A due to electrons with density ρve = −Nee, moving with velocity −ue, is I = ρve(−ue)A = (−Nee)(−ue)A = NeeAue. Hence, Eq. (5.8b) may be written in the compact form dFm = I dl × × × B (N).
(5.9) For a closed circuit of contour C carrying a current I, the total magnetic force is Fm = I C dl × × × B (N).
(5.10) If the closed wire shown in Fig. 5-3(a) resides in a uniform external magnetic field B, then B can be taken outside the integral in Eq. (5.10), in which case Fm = I ⎛ ⎝ C dl ⎞ ⎠× × × B = 0.
(5.11) ▶This result, which is a consequence of the fact that the vector sum of the infinitesimal vectors dl over a closed path equals zero, states that the total magnetic force on any closed current loop in a uniform magnetic field is zero. ◀ 240 CHAPTER 5 MAGNETOSTATICS (a) (b) dl B C I dl a b I B Vector l l Figure 5-3 In a uniform magnetic field, (a) the net force on a closed current loop is zero because the integral of the displacement vector dl over a closed contour is zero, and (b) the force on a line segment is proportional to the vector between the end point (Fm = Iℓ ℓ ℓ× B).
In the study of magnetostatics, all currents flow through closed paths. To understand why, consider the curved wire in Fig. 5-3(b) carrying a current I from point a to point b. In doing so, negative charges accumulate at a, and positive ones at b. The time-varying nature of these charges violates the static assumptions underlying Eqs. (5-1a,b).
If we are interested in the magnetic force exerted on a wire segment l [Fig. 5-3(b)] residing in a uniform magnetic field (while realizing that it is part of a closed current loop), we can integrate Eq. (5.9) to obtain Fm = I ⎛ ⎝ ℓ dl ⎞ ⎠× × × B = Iℓ ℓ ℓ× × × B, (5.12) where ℓ ℓ ℓis the vector directed from a to b [Fig. 5-3(b)]. The integral of dl from a to b has the same value irrespective of the path taken between a and b. For a closed loop, points a and b become the same point, in which case ℓ ℓ ℓ= 0 and Fm = 0.
Example 5-1: Force on a Semicircular Conductor The semicircular conductor shown in Fig. 5-4 lies in the x–y plane and carries a current I. The closed circuit is exposed to a uniform magnetic field B = ˆ yB0. Determine (a) the magnetic dφ dl I B r y x φ φ Figure 5-4 Semicircular conductor in a uniform field (Example 5-1).
5-1 MAGNETIC FORCES AND TORQUES 241 force F1 on the straight section of the wire and (b) the force F2 on the curved section.
Solution: (a) To evaluate F1, consider that the straight section of the circuit is of length 2r and its current flows along the +x direction. Application of Eq. (5.12) with ℓ ℓ ℓ= ˆ x 2r gives F1 = ˆ x(2Ir) × × × ˆ yB0 = ˆ z 2IrB0 (N).
The ˆ z direction in Fig. 5-4 is out of the page.
(b) To evaluate F2, consider a segment of differential length dl on the curved part of the circle. The direction of dl is chosen to coincide with the direction of the current. Since dl and B are both in the x–y plane, their cross product dl × × × B points in the negative z direction, and the magnitude of dl × × × B is proportional to sin φ, where φ is the angle between dl and B.
Moreover, the magnitude of dl is dl = r dφ. Hence, F2 = I π φ=0 dl × × × B = −ˆ zI π φ=0 rB0 sin φ dφ = −ˆ z 2IrB0 (N).
The −ˆ z direction of the force acting on the curved part of the conductor is into the page. We note that F2 = −F1, implying that no net force acts on the closed loop, although opposing forces act on its two sections.
Concept Question 5-1: What are the major differences between the electric force Fe and the magnetic force Fm?
Concept Question 5-2: The ends of a 10 cm long wire carrying a constant current I are anchored at two points on the x axis, namely x = 0 and x = 6 cm. If the wire lies in the x–y plane in a magnetic field B = ˆ yB0, which of the following arrangements produces a greater magnetic force on the wire: (a) wire is V-shaped with corners at (0, 0), (3, 4), and (6, 0), (b) wire is an open rectangle with corners at (0, 0), (0, 2), (6, 2), and (6, 0).
Exercise 5-4: A horizontal wire with a mass per unit length of 0.2 kg/m carries a current of 4 A in the +x direction. If the wire is placed in a uniform magnetic flux density B, what should the direction and minimum magnitude of B be in order to magnetically lift the wire vertically upward? [Hint: The acceleration due to gravity is g = −ˆ z 9.8 m/s2.] Answer: B = ˆ y0.49 T. (See EM.) 5-1.2 Magnetic Torque on a Current-Carrying Loop When a force is applied on a rigid body that can pivot about a fixed axis, the body will, in general, react by rotating about that axis. The angular acceleration depends on the cross product of the applied force vector F and the distance vector d, measured from a point on the rotation axis (such that d is perpendicular to the axis) to the point of application of F (Fig. 5-5). The length of d is called the moment arm, and the cross product T = d × × × F (N·m) (5.13) is called the torque. The unit for T is the same as that for work or energy, even though torque does not represent either. The force F applied on the disk shown in Fig. 5-5 lies in the x–y plane and makes an angle θ with d. Hence, T = ˆ zrF sin θ, (5.14) θ F T d z x y Pivot axis Figure 5-5 The force F acting on a circular disk that can pivot along the z axis generates a torque T = d × F that causes the disk to rotate.
242 CHAPTER 5 MAGNETOSTATICS where |d| = r, the radius of the disk, and F = |F|.
From Eq. (5.14) we observe that a torque along the positive z direction corresponds to a tendency for the cylinder to rotate counterclockwise and, conversely, a torque along the −z direction corresponds to clockwise rotation.
▶These directions are governed by the following right-hand rule: when the thumb of the right hand points along the direction of the torque, the four fingers indicate the direction that the torque tries to rotate the body. ◀ We now consider the magnetic torque exerted on a conducting loop under the influence of magnetic forces. We begin with the simple case where the magnetic field B is in the plane of the loop, and then we extend the analysis to the more general case where B makes an angle θ with the surface normal of the loop.
Magnetic Field in the Plane of the Loop The rectangular conducting loop shown in Fig. 5-6(a) is constructed from rigid wire and carries a current I. The loop lies in the x–y plane and is allowed to pivot about the axis shown. Under the influence of an externally generated uniform magnetic field B = ˆ xB0, arms 1 and 3 of the loop are subjected to forces F1 and F3, given by F1 = I(−ˆ yb) × × × (ˆ xB0) = ˆ zIbB0, (5.15a) and F3 = I(ˆ yb) × × × (ˆ xB0) = −ˆ zIbB0.
(5.15b) These results are based on the application of Eq. (5.12). We note that the magnetic forces acting on arms 1 and 3 are in opposite directions, and no magnetic force is exerted on either arm 2 or 4 because B is parallel to the direction of the current flowing in those arms.
A bottom view of the loop, depicted in Fig. 5-6(b), reveals that forces F1 and F3 produce a torque about the origin O, causing the loop to rotate in a clockwise direction. The moment arm is a/2 for both forces, but d1 and d3 are in opposite directions, resulting in a total magnetic torque of T = d1 × × × F1 + d3 × × × F3 = −ˆ x a 2 × × × ˆ zIbB0 + ˆ x a 2 × × × −ˆ zIbB0 = ˆ yIabB0 = ˆ yIAB0, (5.16) (a) (b) I y x b O a Pivot axis B B 2 4 3 1 Loop arm 3 Loop arm 1 z x y a/2 O z −z d1 F1 B F3 d3 Figure 5-6 Rectangular loop pivoted along the y axis: (a) front view and (b) bottom view. The combination of forces F1 and F3 on the loop generates a torque that tends to rotate the loop in a clockwise direction as shown in (b).
where A = ab is the area of the loop. The right-hand rule tells us that the sense of rotation is clockwise. The result given by Eq. (5.16) is valid only when the magnetic field B is parallel to the plane of the loop. As soon as the loop starts to rotate, the torque T decreases, and at the end of one quarter of a complete rotation, the torque becomes zero, as discussed next.
Magnetic Field Perpendicular to the Axis of a Rectangular Loop For the situation represented by Fig. 5-7, where B = ˆ xB0, the field is still perpendicular to the loop’s axis of rotation, but 5-1 MAGNETIC FORCES AND TORQUES 243 (a) (b) x y z F4 F3 B B I F2 F1 n a b 3 Pivot axis 2 1 4 ˆ θ m (magnetic moment) Arm 1 Arm 3 O B F1 F3 n a/2 (a/2) sin θ θ θ ˆ Figure 5-7 Rectangular loop in a uniform magnetic field with flux density B whose direction is perpendicular to the rotation axis of the loop but makes an angle θ with the loop’s surface normal ˆ n.
because its direction may be at any angle θ with respect to the loop’s surface normal ˆ n, we may now have nonzero forces on all four arms of the rectangular loop. However, forces F2 and F4 are equal in magnitude and opposite in direction and are along the rotation axis; hence, the net torque contributed by their combination is zero. The directions of the currents in arms 1 and 3 are always perpendicular to B regardless of the magnitude of θ. Hence, F1 and F3 have the same expressions given previously by Eqs. (5.15a,b), and for 0 ≤θ ≤π/2 their moment arms are of magnitude (a/2) sin θ, as illustrated in Fig. 5-7(b). Consequently, the magnitude of the net torque exerted by the magnetic field about the axis of rotation is the same as that given by Eq. (5.16), but modified by sin θ: T = IAB0 sin θ.
(5.17) According to Eq. (5.17), the torque is maximum when the magnetic field is parallel to the plane of the loop (θ = 90◦) and zero when the field is perpendicular to the plane of the loop (θ = 0). If the loop consists of N turns, each contributing a torque given by Eq. (5.17), then the total torque is T = NIAB0 sin θ.
(5.18) The quantity NIA is called the magnetic moment m of the loop. Now, consider the vector m = ˆ n NIA = ˆ n m (A·m2), (5.19) where ˆ n is the surface normal of the loop and governed by the following right-hand rule: when the four fingers of the right hand advance in the direction of the current I, the direction of the thumb specifies the direction of ˆ n. In terms of m, the torque vector T can be written as T = m × × × B (N·m).
(5.20) Even though the derivation leading to Eq. (5.20) was obtained for B perpendicular to the axis of rotation of a rectangular loop, the expression is valid for any orientation of B and for a loop of any shape.
Concept Question 5-3: How is the direction of the magnetic moment of a loop defined?
Concept Question 5-4: If one of two wires of equal length is formed into a closed square loop and the other into a closed circular loop, and if both wires are carrying equal currents and both loops have their planes parallel to a uniform magnetic field, which loop would experience the greater torque?
244 CHAPTER 5 MAGNETOSTATICS Exercise 5-5: A square coil of 100 turns and 0.5 m long sides is in a region with a uniform magnetic flux density of 0.2 T. If the maximum magnetic torque exerted on the coil is 4 × 10−2 (N·m), what is the current flowing in the coil?
Answer: I = 8 mA. (See EM.) 5-2 The Biot–Savart Law In the preceding section, we elected to use the magnetic flux density B to denote the presence of a magnetic field in a given region of space. We now work with the magnetic field intensity H instead. We do this in part to remind the reader that for most materials the flux and field are linearly related by B = μH, and therefore knowledge of one implies knowledge of the other (assuming that μ is known).
Through his experiments on the deflection of compass needles by current-carrying wires, Hans Oersted established that currents induce magnetic fields that form closed loops around the wires (see Section 1-3.3). Building upon Oersted’s results, Jean Biot and F´ elix Savart arrived at an expression that relates the magnetic field H at any point in space to the current I that generates H. The Biot–Savart law states that the differential magnetic field dH generated by a steady current I flowing through a differential length vector dl is dH = I 4π dl × × × ˆ R R2 (A/m), (5.21) where R = ˆ RR is the distance vector between dl and the observation point P shown in Fig. 5-8.
The SI unit for H is ampere·m/m2 = (A/m). It is important to remember that Eq. (5.21) assumes that dl is along the direction of the current I and the unit vector ˆ R points from the current element to the observation point.
According to Eq. (5.21), dH varies as R−2, which is similar to the distance dependence of the electric field induced by an electric charge. However, unlike the electric field vector E, whosedirectionisalongthedistancevectorRjoiningthecharge to the observation point, the magnetic field H is orthogonal to the plane containing the direction of the current element dl and the distance vector R. At point P in Fig. 5-8, the direction of dH is out of the page, whereas at point P ′ the direction of dH is into the page.
dH P P' R R dl I (dH into the page) dH (dH out of the page) θ ˆ Figure 5-8 Magnetic field dH generated by a current element I dl. The direction of the field induced at point P is opposite to that induced at point P ′.
To determine the total magnetic field H due to a conductor of finite size, we need to sum up the contributions due to all the current elements making up the conductor. Hence, the Biot– Savart law becomes H = I 4π l dl × × × ˆ R R2 (A/m), (5.22) where l is the line path along which I exists.
5-2.1 Magnetic Field Due to Surface and Volume Current Distributions The Biot–Savart law may also be expressed in terms of distributed current sources (Fig. 5-9) such as the volume current density J, measured in (A/m2), or the surface current density Js, measured in (A/m). The surface current density Js applies to currents that flow on the surface of a conductor in the form of a sheet of effectively zero thickness. When current sources are specified in terms of Js over a surface S or in terms of J over a volume v, we can use the equivalence given by I dl Js ds J dv (5.23) 5-2 THE BIOT–SAVART LAW 245 (a) Volume current density J in A/m2 (b) Surface current density Js in A/m J S Js l Figure 5-9 (a) The total current crossing the cross section S of the cylinder is I = S J·ds. (b) The total current flowing across the surface of the conductor is I = l Js dl.
to express the Biot–Savart law as H = 1 4π S Js × × × ˆ R R2 ds, (5.24a) (surface current) H = 1 4π v J × × × ˆ R R2 dv.
(5.24b) (volume current) Example 5-2: Magnetic Field of a Linear Conductor A free-standing linear conductor of length l carries a current I along the z axis as shown in Fig. 5-10. Determine the magnetic flux density B at a point P located at a distance r in the x–y plane.
Solution: From Fig. 5-10, the differential length vector dl = ˆ z dz. Hence, dl × × × ˆ R = dz (ˆ z × × × ˆ R) = ˆ φ φ φ sin ˆ θ θ θ dz, where ˆ φ φ φ is the azimuth direction and θ is the angle between dl and ˆ R.
(a) (b) z I P θ1 θ2 r R2 R1 l I P dl dl dθ r z R R dH into the page l θ ˆ Figure 5-10 Linear conductor of length l carrying a current I.
(a)The field dH at point P due to incremental current element dl.
(b)Limitinganglesθ1 andθ2, eachmeasuredbetweenvectorI dl and the vector connecting the end of the conductor associated with that angle to point P (Example 5-2).
246 CHAPTER 5 MAGNETOSTATICS Module 5.2 Magnetic Fields due to Line Sources You can place z-directed linear currents anywhere in the display plane (x-y plane), select their magnitudes and directions, and then observe the spatial pattern of the induced magnetic flux B(x, y).
Application of Eq. (5.22) gives H = I 4π z=l/2 z=−l/2 dl × × × ˆ R R2 = ˆ φ φ φ I 4π l/2 −l/2 sin θ R2 dz.
(5.25) Both R and θ are dependent on the integration variable z, but the radial distance r is not. For convenience, we convert the integration variable from z to θ by using the transformations R = r csc θ, (5.26a) z = −r cot θ, (5.26b) dz = r csc2 θ dθ.
(5.26c) Upon inserting Eqs. (5.26a) and (5.26c) into Eq. (5.25), we have H = ˆ φ φ φ I 4π θ2 θ1 sin θ r csc2 θ dθ r2 csc2 θ = ˆ φ φ φ I 4πr θ2 θ1 sin θ dθ = ˆ φ φ φ I 4πr (cos θ1 −cos θ2), (5.27) 5-2 THE BIOT–SAVART LAW 247 where θ1 and θ2 are the limiting angles at z = −l/2 and z = l/2, respectively. From the right triangle in Fig. 5-10(b), it follows that cos θ1 = l/2 r2 + (l/2)2 , (5.28a) cos θ2 = −cos θ1 = −l/2 r2 + (l/2)2 .
(5.28b) Hence, B = μ0H = ˆ φ φ φ μ0Il 2πr √ 4r2 + l2 (T).
(5.29) For an infinitely long wire with l ≫r, Eq. (5.29) reduces to B = ˆ φ φ φ μ0I 2πr (infinitely long wire).
(5.30) ▶This is a very important and useful expression to keep in mind. It states that in the neighborhood of a linear conductor carrying a current I, the induced magnetic field forms concentric circles around the wire (Fig. 5-11), and its intensity is directly proportional to I and inversely proportional to the distance r. ◀ I Magnetic field I B B B Figure 5-11 Magnetic field surrounding a long, linear current-carrying conductor.
Example 5-3: Magnetic Field of a Circular Loop A circular loop of radius a carries a steady current I. Determine the magnetic field H at a point on the axis of the loop.
Solution: Let us place the loop in the x–y plane (Fig. 5-12).
Our task is to obtain an expression for H at point P(0, 0, z).
We start by noting that any element dl on the circular loop is perpendicular to the distance vector R, and that all elements around the loop are at the same distance R from P, with R = √ a2 + z2 . From Eq. (5.21), the magnitude of dH due to current element dl is dH = I 4πR2 |dl × × × ˆ R| = I dl 4π(a2 + z2) , (5.31) and the direction of dH is perpendicular to the plane containing R and dl. dH is in the r–z plane (Fig. 5-12), and therefore it has components dHr and dHz. If we consider element dl′, located diametrically opposite to dl, we observe that the z components φ θ θ a φ + π P = (0, 0, z) R dHr dH'r dHz dH'z dH dH' z dl' dl I x y Figure 5-12 Circular loop carrying a current I (Example 5-3).
248 CHAPTER 5 MAGNETOSTATICS of the magnetic fields due to dl and dl′ add because they are in the same direction, but their r components cancel because they are in opposite directions. Hence, the net magnetic field is along z only. That is, dH = ˆ z dHz = ˆ z dH cos θ = ˆ z I cos θ 4π(a2 + z2) dl.
(5.32) For a fixed point P(0, 0, z) on the axis of the loop, all quantities in Eq. (5.32) are constant, except for dl. Hence, integrating Eq. (5.32) over a circle of radius a gives H = ˆ z I cos θ 4π(a2 + z2) dl = ˆ z I cos θ 4π(a2 + z2) (2πa).
(5.33) Upon using the relation cos θ = a/(a2 + z2)1/2, we obtain H = ˆ z Ia2 2(a2 + z2)3/2 (A/m).
(5.34) At the center of the loop (z = 0), Eq. (5.34) reduces to H = ˆ z I 2a (at z = 0), (5.35) and at points very far away from the loop such that z2 ≫a2, Eq. (5.34) simplifies to H = ˆ z Ia2 2|z|3 (at |z| ≫a).
(5.36) 5-2.2 Magnetic Field of a Magnetic Dipole In view of the definition given by Eq. (5.19) for the magnetic moment m of a current loop, a single-turn loop situated in the x–y plane (Fig. 5-12) has magnetic moment m = ˆ zm with m = Iπa2. Consequently, Eq. (5.36) may be expressed as H = ˆ z m 2π|z|3 (at |z| ≫a).
(5.37) This expression applies to a point P far away from the loop and on its axis.
Had we solved for H at any distant point P = (R, θ, φ) in a spherical coordinate system, with R the distance between the center of the loop and point P, we would have obtained the expression H = m 4πR3 ( ˆ R 2 cos θ + ˆ θ θ θ sin θ) (5.38) (for R ≫a).
▶A current loop with dimensions much smaller than the distance between the loop and the observation point is called a magnetic dipole. This is because the pattern of its magnetic field lines is similar to that of a permanent magnet, as well as to the pattern of the electric field lines of the electric dipole (Fig. 5-13). ◀ Concept Question 5-5: Two infinitely long parallel wires carry currents of equal magnitude. What is the resultant magnetic field due to the two wires at a point midway between the wires, compared with the magnetic field due to one of them alone, if the currents are (a) in the same direction and (b) in opposite directions?
Concept Question 5-6: Devisearight-handruleforthe direction of the magnetic field due to a linear current-carrying conductor.
Concept Question 5-7: What is a magnetic dipole?
Describe its magnetic field distribution.
Exercise 5-6: A semi-infinite linear conductor extends between z = 0 and z = ∞along the z axis. If the current I in the conductor flows along the positive z direction, find H at a point in the x–y plane at a radial distance r from the conductor.
Answer: H = ˆ φ φ φ I 4πr (A/m). (See EM.) 5-2 THE BIOT–SAVART LAW 249 (a) Electric dipole (b) Magnetic dipole (c) Bar magnet + − E S N H H I Figure 5-13 Patterns of (a) the electric field of an electric dipole, (b) the magnetic field of a magnetic dipole, and (c) the magnetic field of a bar magnet. Far away from the sources, the field patterns are similar in all three cases.
Module 5.3 Magnetic Field of a Current Loop Examine how the field along the loop axis changes with loop parameters.
250 CHAPTER 5 MAGNETOSTATICS d/2 d x y z d/2 F' 1 F' 2 B1 I1 I2 l Figure 5-14 Magnetic forces on parallel current-carrying conductors.
Exercise 5-7: A wire carrying a current of 4 A is formed into a circular loop. If the magnetic field at the center of the loop is 20 A/m, what is the radius of the loop if the loop has (a) only one turn and (b) 10 turns?
Answer: (a) a = 10 cm, (b) a = 1 m. (See EM.) Exercise 5-8: A wire is formed into a square loop and placed in the x–y plane with its center at the origin and each of its sides parallel to either the x or y axes. Each side is 40 cm in length, and the wire carries a current of 5A whose direction is clockwise when the loop is viewed from above. Calculate the magnetic field at the center of the loop.
Answer: H = −ˆ z 4I √ 2πl = −ˆ z11.25 A/m. (See EM.) 5-2.3 Magnetic Force Between Two Parallel Conductors In Section 5-1.1 we examined the magnetic force Fm that acts on a current-carrying conductor when placed in an external magnetic field. The current in the conductor, however, also generates its own magnetic field.
Hence, if two current-carrying conductors are placed in each other’s vicinity, each will exert a magnetic force on the other.
Let us consider two very long (or effectively infinitely long), straight, free-standing, parallel wires separated by a distance d and carrying currents I1 and I2 in the z direction (Fig. 5-14) at y = −d/2 and y = d/2, respectively. We denote by B1 the magnetic field due to current I1, defined at the location of the wire carrying current I2 and, conversely, by B2 the field due to I2 at the location of the wire carrying current I1. From Eq. (5.30), with I = I1, r = d, and ˆ φ φ φ = −ˆ x at the location of I2, the field B1 is B1 = −ˆ x μ0I1 2πd .
(5.39) The force F2 exerted on a length l of wire I2 due to its presence in field B1 may be obtained by applying Eq. (5.12): F2 = I2lˆ z × × × B1 = I2lˆ z × × × (−ˆ x) μ0I1 2πd = −ˆ y μ0I1I2l 2πd , (5.40) and the corresponding force per unit length is F′ 2 = F2 l = −ˆ y μ0I1I2 2πd .
(5.41) A similar analysis performed for the force per unit length exerted on the wire carrying I1 leads to F′ 1 = ˆ y μ0I1I2 2πd .
(5.42) ▶Thus, two parallel wires carrying currents in the same direction attract each other with equal force.
If the currents are in opposite directions, the wires would repel one another with equal force. ◀ 5-3 MAXWELL’S MAGNETOSTATIC EQUATIONS 251 Module 5.4 Magnetic Force Between Two Parallel Conductors Observe the direction and magnitude of the force exerted on parallel current-carrying wires.
5-3 Maxwell’s Magnetostatic Equations Thus far, we have introduced the Biot–Savart law for finding the magnetic flux density B and field H due to any distribution of electric currents in free space, and we examined how magnetic fields can exert magnetic forces on moving charged particles and current-carrying conductors. We now examine two additional important properties of magnetostatic fields.
5-3.1 Gauss’s Law for Magnetism In Chapter 4 we learned that the net outward flux of the electric flux density D through a closed surface equals the enclosed net charge Q. We referred to this property as Gauss’s law (for electricity), and expressed it mathematically in differential and integral forms as ∇· D = ρv S D· ds = Q.
(5.43) Conversion from differential to integral form was accomplished by applying the divergence theorem to a volume v that is enclosed by a surface S and contains a total charge Q = v ρv dv (Section 4-4).
The magnetostatic counterpart of Eq. (5.43), often called Gauss’s law for magnetism, is ∇· B = 0 S B· ds = 0.
(5.44) The differential form is one of Maxwell’s four equations, and the integral form is obtained with the help of the divergence theorem.
Note that the right-hand side of Gauss’s law for magnetism is zero, reflecting the fact that the magnetic equivalence of an electric point charge does not exist in nature.
252 CHAPTER 5 MAGNETOSTATICS (b) Bar magnet + − E S N H (a) Electric dipole Closed imaginary surface Figure 5-15 Whereas (a) the net electric flux through a closed surface surrounding a charge is not zero, (b) the net magnetic flux through a closed surface surrounding one of the poles of a magnet is zero.
▶The hypothetical magnetic analogue to an electric point charge is called a magnetic monopole. Magnetic monopoles, however, always occur in pairs (that is, as dipoles). ◀ No matter how many times a permanent magnet is subdivided, each new piece will always have a north and a south pole, even if the process were to be continued down to the atomic level.
Consequently, there is no magnetic equivalence to an electric charge q or charge density ρv.
Formally, the name “Gauss’s law” refers to the electric case, even when no specific reference to electricity is indicated. The property described by Eq. (5.44) has been called “the law of nonexistence of isolated monopoles,” “the law of conservation of magnetic flux,” and “Gauss’s law for magnetism,” among others. We prefer the last of the three cited names because it reminds us of the parallelism, as well as the differences, between the electric and magnetic laws of nature.
The difference between Gauss’s law for electricity and its magnetic counterpart can be elucidated in terms of field lines.
Electric field lines originate from positive electric charges and terminate on negative ones. Hence, for the electric field lines of the electric dipole shown in Fig. 5-15(a), the electric flux through a closed surface surrounding one of the charges is nonzero.
In contrast, magnetic field lines always form continuous closed loops.
As we saw in Section 5-2, the magnetic field lines due to currents do not begin or end at any point; this is true for the linear conductor of Fig. 5-11 and the circular loop of Fig. 5-12, as well as for any current distribution.
It is also true for a bar magnet [Fig. 5-15(b)]. Because the magnetic field lines form closed loops, the net magnetic flux through any closed surface surrounding the south pole of the magnet (or through any other closed surface) is always zero, regardless of its shape.
5-3.2 Amp ere’s Law In Chapter 4 we learned that the electrostatic field is conservative, meaning that its line integral along a closed contour always vanishes. This property of the electrostatic field was expressed in differential and integral forms as ∇× × × E = 0 C E· dℓ ℓ ℓ= 0.
(5.45) Conversion of the differential to integral form was accom-plished by applying Stokes’s theorem to a surface S with contour C.
The magnetostatic counterpart of Eq. (5.45), known as Ampere’s law, is ∇× × × H = J C H· dℓ ℓ ℓ= I, (5.46) where I is the total current passing through S. The differential form again is one of Maxwell’s equations, and the integral form is obtained by integrating both sides of Eq. (5.46) over an open surface S, S (∇× × × H)· ds = S J· ds, (5.47) and then invoking Stokes’s theorem with I = J· ds.
▶The sign convention for the direction of the contour path C in Amp ere’s law is taken so that I and H satisfy the right-hand rule defined earlier in connection with the Biot–Savart law. That is, if the direction of I is aligned with the direction of the thumb of the right hand, then the direction of the contour C should be chosen along that of the other four fingers. ◀ In words, Ampere’s circuital law states that the line integral of H around a closed path is equal to the current traversing the surface bounded by that path. To apply Amp ere’s law, the 5-3 MAXWELL’S MAGNETOSTATIC EQUATIONS 253 I C H H I C H H I H C H (a) (b) (c) Figure 5-16 Ampere’s law states that the line integral of H around a closed contour C is equal to the current traversing the surface bounded by the contour. This is true for contours (a) and (b), but the line integral of H is zero for the contour in (c) because the current I (denoted by the symbol⊙) is not enclosed by the contour C.
current must flow through a closed path. By way of illustration, for both configurations shown in Figs. 5-16(a) and (b), the line integral of H is equal to the current I, even though the paths have very different shapes and the magnitude of H is not uniform along the path of configuration (b). By the same token, because path (c) in Fig. 5-16 does not enclose the current I, the line integral of H along it vanishes, even though H is not zero along the path.
When we examined Gauss’s law in Section 4-4, we discovered that in practice its usefulness for calculating the electric flux density D is limited to charge distributions that possess a certain degree of symmetry and that the calculation procedure is subject to the proper choice of a Gaussian surface enclosing the charges. A similar restriction applies toAmp ere’s law: its usefulness is limited to symmetric current distributions that allow the choice of convenient Amperian contours around them, as illustrated by Examples 5-4 to 5-6.
Example 5-4: Magnetic Field of a Long Wire A long (practically infinite) straight wire of radius a carries a steady current I that is uniformly distributed over its cross section. Determine the magnetic field H a distance r from the wire axis for (a) r ≤a (inside the wire) and (b) r ≥a (outside the wire).
Solution: (a) We choose I to be along the +z direction [Fig. 5-17(a)]. To determine H1 = H at a distance r = r1 ≤a, we choose the Amp erian contour C1 to be a circular path of radius r = r1 [Fig. 5-17(b)]. In this case, Ampere’s law takes the form C1 H1 · dl1 = I1, (5.48) where I1 is the fraction of the total current I flowing through C1.
From symmetry, H1 must be constant in magnitude and parallel to the contour at any point along the path. Furthermore, to satisfy the right-hand rule and given that I is along the z direction, H1 must be in the +φ direction. Hence, H1 = ˆ φ φ φH1, dl1 = ˆ φ φ φr1 dφ, and the left-hand side of Eq. (5.48) becomes C1 H1 · dl1 = 2π 0 H1( ˆ φ φ φ· ˆ φ φ φ)r1 dφ = 2πr1H1.
The current I1 flowing through the area enclosed by C1 is equal to the total current I multiplied by the ratio of the area enclosed by C1 to the total cross-sectional area of the wire: I1 = πr2 1 πa2 I = r1 a 2 I.
Equating both sides of Eq. (5.48) and then solving for H1 yields H1 = ˆ φ φ φH1 = ˆ φ φ φ r1 2πa2 I (for r1 ≤a).
(5.49a) (b) For r = r2 ≥a, we choose path C2, which encloses all the current I. Hence, H2 = ˆ φ φ φ H2, dℓ ℓ ℓ2 = ˆ φ φ φ r2 dφ, and C2 H2 · dl2 = 2πr2H2 = I, which yields H2 = ˆ φ φ φH2 = ˆ φ φ φ I 2πr2 (for r2 ≥a).
(5.49b) Ignoring the subscript 2, we observe that Eq. (5.49b) provides the same expression for B = μ0H as Eq. (5.30), which was derived on the basis of the Biot–Savart law.
254 CHAPTER 5 MAGNETOSTATICS (b) Wire cross section (a) Cylindrical wire (c) a H(a) = I 2πa H(r) r H1 H2 C2 I z a C1 r2 r1 Contour C2 for r2 a Contour C1 for r1 a C2 a x y C1 φ1 r1 r2 ˆ φ Figure 5-17 Infinitely long wire of radius a carrying a uniform current I along the +z direction: (a) general configuration showing contours C1 and C2; (b) cross-sectional view; and (c) a plot of H versus r (Example 5-4).
The variation of the magnitude of H as a function of r is plotted in Fig. 5-17(c); H increases linearly between r = 0 and r = a (inside the conductor), and then decreases as 1/r for r > a (outside the conductor).
Exercise 5-9: A current I flows in the inner conductor of a long coaxial cable and returns through the outer conductor. Whatisthemagneticfieldintheregionoutside the coaxial cable and why?
Answer: H = 0 outside the coaxial cable because the net current enclosed by an Amp erian contour enclosing the cable is zero.
Exercise 5-10: The metal niobium becomes a superconductor with zero electrical resistance when it is cooled to below 9 K, but its superconductive behavior ceases when the magnetic flux density at its surface exceeds 0.12 T. Determine the maximum current that a 0.1 mm diameter niobium wire can carry and remain superconductive.
Answer: I = 30 A. (See EM.) Example 5-5: Magnetic Field inside a Toroidal Coil A toroidal coil (also called a torus or toroid) is a doughnut-shaped structure (called the core) wrapped in closely spaced turns of wire (Fig. 5-18). For clarity, we show the turns in the figure as spaced far apart, but in practice they are wound in r I I b a Ampèrian contour H C ˆ φ Figure 5-18 Toroidal coil with inner radius a and outer radius b. The wire loops usually are much more closely spaced than shown in the figure (Example 5-5).
5-3 MAXWELL’S MAGNETOSTATIC EQUATIONS 255 a closely spaced arrangement to form approximately circular loops.
The toroid is used to magnetically couple multiple circuits and to measure the magnetic properties of materials, as illustrated later in Fig. 5-30.
For a toroid with N turns carrying a current I, determine the magnetic field H in each of the following three regions: r < a, a < r < b, and r > b, all in the azimuthal symmetry plane of the toroid.
Solution: From symmetry, it is clear that H is uniform in the azimuthal direction. If we construct a circular Amperian contour with center at the origin and radius r < a, there will be no current flowing through the surface of the contour, and therefore H = 0 for r < a.
Similarly, for an Amp erian contour with radius r > b, the net current flowing through its surface is zero because an equal number of current coils cross the surface in both directions; hence, H = 0 for r > b (region exterior to the toroidal coil).
For the region inside the core, we construct a path of radius r (Fig. 5-18). For each loop, we know from Example 5-3 that the field H at the center of the loop points along the axis of the loop, which in this case is the φ direction, and in view of the direction of the current I shown in Fig. 5-18, the right-hand rule tells us that H must be in the −φ direction. Hence, H = −ˆ φ φ φH. The total current crossing the surface of the contour with radius r is NI and its direction is into the page. According to the right-hand rule associated with Ampere’s law, the current is positive if it crosses the surface of the contour in the direction of the four fingers of the right hand when the thumb is pointing along the direction of the contour C.
Hence, the current through the surface spanned by the contour is −NI. Application of Amp ere’s law then gives C H· dl = 2π 0 (−ˆ φ φ φH)· ˆ φ φ φr dφ = −2πrH = −NI.
Hence, H = NI/(2πr) and H = −ˆ φ φ φH = −ˆ φ φ φ NI 2πr (for a < r < b).
(5.50) Example 5-6: Magnetic Field of an Infinite Current Sheet The x–y plane contains an infinite current sheet with surface current density Js = ˆ xJs (Fig. 5-19). Find the magnetic field H everywhere in space.
y z l w H H Js (out of the page) Ampèrian contour Figure 5-19 A thin current sheet in the x–y plane carrying a surface current density Js = ˆ xJs (Example 5-6).
Solution: From symmetry considerations and the right-hand rule, for z > 0 and z < 0 H must be in the directions shown in the figure. That is, H = −ˆ yH for z > 0, ˆ yH for z < 0.
To evaluate the line integral in Ampere’s law, we choose a rectangular Amp erian path around the sheet, with dimensions l and w (Fig. 5-19). Recalling that Js represents current per unit length along the y direction, the total current crossing the surface of the rectangular loop is I = Jsl.
Hence, applying Ampere’s law over the loop, while noting that H is perpendicular to the paths of length w, we have C H· dl = 2Hl = Jsl, from which we obtain the result H = ⎧ ⎪ ⎨ ⎪ ⎩ −ˆ y Js 2 for z > 0, ˆ y Js 2 for z < 0.
(5.51) 256 TECHNOLOGY BRIEF 10: ELECTROMAGNETS Technology Brief 10: Electromagnets William Sturgeon developed the first practical electromagnet in the 1820s. Today, the principle of the electromagnet is used in motors, relay switches in read/write heads for hard disks and tape drives, loud speakers, magnetic levitation, and many other applications.
Basic Principle Electromagnets can be constructed in various shapes, including the linear solenoid and horseshoe geometries depicted in Fig. TF10-1. In both cases, when an electric current flows through the insulated wire coiled around the central core, it induces a magnetic field with lines resembling those generated by a bar magnet. The strength of the magnetic field is proportional to the current, the number of turns, and the magnetic permeability of the core material.
By using a ferromagnetic core, the field strength can be increased by several orders of magnitude, depending on the purity of the iron material. When subjected to a magnetic field, ferromagnetic materials, such as iron or nickel, get magnetized and act like magnets themselves.
Magnetic Relays A magnetic relay is a switch or circuit breaker that can be activated into the “ON” and “OFF” positions magnetically.
One example is the low-power reed relay used in telephone equipment, which consists of two flat nickel–iron blades separated by a small gap (Fig. TF10-2). The blades are shaped in such a way that in the absence of an external force, they remain apart and unconnected (OFF position). Electrical contact between the blades (ON position) is realized by applying a magnetic field along their length. The field, induced by a current flowing in the wire coiled around the glass envelope, causes the two blades to assume opposite magnetic polarities, thereby forcing them to attract together and close out the gap.
The Doorbell In a doorbell circuit (Fig. TF10-3), the doorbell button is a switch; pushing and holding it down serves to connect the circuit to the household ac source through an appropriate step-down transformer. The current from the source flows (a) Solenoid (b) Horseshoe electromagnet Iron core B Insulated wire Switch N S N S Iron core Magnetic field B B FigureTF10-1 Solenoid and horseshoe magnets.
TECHNOLOGY BRIEF 10: ELECTROMAGNETS 257 N S Glass envelope Electronic circuit FigureTF10-2 Microreed relay (size exaggerated for illustration purposes).
through the electromagnet, via a contact arm with only one end anchored in place (and the other moveable), and onward to the switch. The magnetic field generated by the current flowing in the windings of the electromagnet pulls the unanchored end of the contact arm (which has an iron bar on it) closer in, in the direction of the electromagnet, thereby losing connection with the metal contact and severing current flow in the circuit. With no magnetic field to pull on the contact arm, it snaps back into its earlier position, re-establishing the current in the circuit. This back and forth cycle is repeated many times per second, so long as the doorbell button continues to be pushed down, and with every cycle, the clapper arm attached to the contact arm hits the metal bell and generates a ringing sound.
The Loudspeaker By using a combination of a stationary, permanent magnet, and a moveable electromagnet, the electromagnet/speaker-cone of the loudspeaker (Fig.TF10-4) can be made to move back and forth in response to the electrical signal exciting the electromagnet. The vibrating movement of the cone generates sound waves with the same distribution of frequencies as contained in the spectrum of the electrical signal.
Transformer Button Bell Clapper Contact arm Magnetic field ac source Metal contact Electromagnet FigureTF10-3 Basic elements of a doorbell.
258 TECHNOLOGY BRIEF 10: ELECTROMAGNETS Permanent magnet Cone Audio signal Electrical signal FigureTF10-4 The basic structure of a speaker.
Magnetic Levitation ▶Magnetically levitated trains [Fig. TF10-5(a)], called maglevs for short, can achieve speeds as high as 500 km/hr, primarily because there is no friction between the train and the track. ◀ The train basically floats at a height of 1 or more centimeters above the track, made possible by magnetic levitation [Fig.TF10-5(b)]. The train carries superconducting electromagnets that induce currents in coils built into the guide rails alongside the train. The magnetic interaction between the train’s superconducting electromagnets and the guide-rail coils serves not only to levitate the train, but also to propel it along the track.
(a) (b) (c) S N S N S N N S N S N S N S S N Figure TF10-5 (a) A maglev train, (b) electrodynamic suspension of an SCMaglev train, and (c) electrodynamic maglev propulsion via propulsion coils.
5-4 VECTOR MAGNETIC POTENTIAL 259 Concept Question 5-8: What are the fundamental differences between electric and magnetic fields?
Concept Question 5-9: If the line integral of H over a closed contour is zero, does it follow that H = 0 at every point on the contour? If not, what then does it imply?
Concept Question 5-10: Compare the utility of apply-ing the Biot–Savart law versus applying Amp ere’s law for computing the magnetic field due to current-carrying conductors.
Concept Question 5-11: What is a toroid? What is the magnetic field outside the toroid?
5-4 Vector Magnetic Potential In our treatment of electrostatic fields in Chapter 4, we defined the electrostatic potential V as the line integral of the electric field E, and found that V and E are related by E = −∇V .
This relationship proved useful not only in relating electric field distributions in circuit elements (such as resistors and capacitors) to the voltages across them, but also to determine E for a given charge distribution by first computing V using Eq. (4.48). We now explore a similar approach in connection with the magnetic flux density B.
According to Eq. (5.44), ∇· B = 0. We wish to define B in terms of a magnetic potential with the constraint that such a definition guarantees that the divergence of B is always zero.
This can be realized by taking advantage of the vector identity given by Eq. (3.106b), which states that, for any vector A, ∇·(∇× × × A) = 0.
(5.52) Hence, by introducing the vector magnetic potential A such that B = ∇× × × A (Wb/m2), (5.53) we are guaranteed that ∇· B = 0. The SI unit for B is the tesla (T).
An equivalent unit is webers per square meter (Wb/m2). Consequently, the SI unit for A is (Wb/m).
With B = μH, the differential form of Ampere’s law given by Eq. (5.46) can be written as ∇× × × B = μJ.
(5.54) If we substitute Eq. (5.53) into Eq. (5.54), we obtain ∇× × × (∇× × × A) = μJ.
(5.55) For any vector A, the Laplacian of A obeys the vector identity given by Eq. (3.113), that is, ∇2A = ∇(∇·A) −∇× × × (∇× × × A), (5.56) where, by definition, ∇2A in Cartesian coordinates is ∇2A = ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 A = ˆ x∇2Ax + ˆ y∇2Ay + ˆ z∇2Az.
(5.57) Combining Eq. (5.55) with Eq. (5.56) gives ∇(∇·A) −∇2A = μJ.
(5.58) This equation contains a term involving ∇·A.
It turns out that we have a fair amount of latitude in specifying a value or a mathematical form for ∇·A, without conflicting with the requirement represented by Eq. (5.53). The simplest among these allowed restrictions on A is ∇· A = 0.
(5.59) Using this choice in Eq. (5.58) leads to the vector Poisson’s equation ∇2A = −μJ.
(5.60) Using the definition for ∇2A given by Eq. (5.57), the vector Poisson’s equation can be decomposed into three scalar Poisson’s equations: ∇2Ax = −μJx, (5.61a) ∇2Ay = −μJy, (5.61b) ∇2Az = −μJz.
(5.61c) 260 CHAPTER 5 MAGNETOSTATICS In electrostatics, Poisson’s equation for the scalar potential V is given by Eq. (4.60) as ∇2V = −ρv ϵ , (5.62) and its solution for a volume charge distribution ρv occupying a volume v ′ was given by Eq. (4.61) as V = 1 4πϵ v ′ ρv R′ dv ′.
(5.63) Poisson’s equations for Ax, Ay, and Az are mathematically identical in form to Eq. (5.62). Hence, for a current density J with x component Jx distributed over a volumev ′, the solution for Eq. (5.61a) is Ax = μ 4π v ′ Jx R′ dv ′ (Wb/m).
(5.64) Similar solutions can be written for Ay in terms of Jy and for Az in terms of Jz. The three solutions can be combined into a vector equation: A = μ 4π v ′ J R′ dv ′ (Wb/m).
(5.65) In view of Eq. (5.23), if the current distribution is specified over a surface S′, then J dv ′ should be replaced with Js ds′ and v ′ should be replaced with S′; similarly, for a line distribution, J dv ′ should be replaced with I dl′ and the integration should be performed over the associated path l′.
The vector magnetic potential provides a third approach for computing the magnetic field due to current-carrying conductors, in addition to the methods suggested by the Biot– Savart and Amp ere laws. For a specified current distribution, Eq. (5.65) can be used to find A, and then Eq. (5.53) can be used to find B.
Except for simple current distributions with symmetrical geometries that lend themselves to the application of Ampere’s law, in practice we often use the approaches provided by the Biot–Savart law and the vector magnetic potential, and among these two the latter often is more convenient to apply because it is easier to perform the integration in Eq. (5.65) than that in Eq. (5.22).
The magnetic flux linking a surface S is defined as the total magnetic flux density passing through it, or = S B · ds (Wb).
(5.66) If we insert Eq. (5.53) into Eq. (5.66) and then invoke Stokes’s theorem, we obtain = S (∇× A) · ds = C A · dl (Wb), (5.67) where C is the contour bounding the surface S. Thus, can be determined by either Eq. (5.66) or Eq. (5.67), whichever is easier to integrate for the specific problem under consideration.
5-5 Magnetic Properties of Materials Because of the similarity between the pattern of the magnetic field lines generated by a current loop and those exhibited by a permanent magnet, the loop can be regarded as a magnetic dipole with north and south poles (Section 5-2.2 and Fig. 5-13).
The magnetic moment m of a loop of area A has magnitude m = IA and a direction normal to the plane of the loop (in accordance with the right-hand rule).
Magnetization in a material is due to atomic scale current loops associated with: (1) orbital motions of the electrons and protons around and inside the nucleus and (2) electron spin. The magnetic moment due to proton motion typically is three orders of magnitude smaller than that of the electrons, and therefore the total orbital and spin magnetic moment of an atom is dominated by the sum of the magnetic moments of its electrons.
▶The magnetic behavior of a material is governed by the interaction of the magnetic dipole moments of its atoms with an external magnetic field.
The nature of the behavior depends on the crystalline structure of the material and is used as a basis for classifying materials as diamagnetic, paramagnetic, or ferromagnetic. ◀ The atoms of a diamagnetic material have no permanent magnetic moments.
In contrast, both paramagnetic and ferromagnetic materials have atoms with permanent magnetic dipole moments, albeit with very different organizational structures.
5-5 MAGNETIC PROPERTIES OF MATERIALS 261 5-5.1 Electron Orbital and Spin Magnetic Moments This section presents a semiclassical, intuitive model of the atom, which provides quantitative insight into the origin of electron magnetic moments. An electron with charge of −e moving at constant speed u in a circular orbit of radius r [Fig. 5-20(a)] completes one revolution in time T = 2πr/u.
This circular motion of the electron constitutes a tiny loop with current I given by I = −e T = −eu 2πr .
(5.68) The magnitude of the associated orbital magnetic moment mo is mo = IA = −eu 2πr (πr2) = −eur 2 = − e 2me Le, (5.69) where Le = meur is the angular momentum of the electron and me is its mass. According to quantum physics, the orbital angularmomentumisquantized; specifically, Le isalwayssome integer multiple of ℏ= h/2π, where h is Planck’s constant.
That is, Le = 0, ℏ, 2ℏ, . . . . Consequently, the smallest nonzero magnitude of the orbital magnetic moment of an electron is mo = −eℏ 2me .
(5.70) Despite the fact that all materials contain electrons that exhibit magnetic dipole moments, most are effectively nonmagnetic.
This is because, in the absence of an external magnetic field, the atoms of most materials are oriented randomly, as a result of which they exhibit a zero or very small net magnetic moment.
In addition to the magnetic moment due to its orbital motion, an electron has an intrinsic spin magnetic moment ms due to its spinning motion about its own axis [Fig. 5-20(b)].
The magnitude of ms predicted by quantum theory is ms = −eℏ 2me , (5.71) which is equal to the minimum orbital magnetic moment mo.
The electrons of an atom with an even number of electrons usuallyexistinpairs, withthemembersofapairhavingopposite spin directions, thereby canceling each others’ spin magnetic moments. If the number of electrons is odd, the atom has a net nonzero spin magnetic moment due to its unpaired electron.
r e e mo ms (a) Orbiting electron (b) Spinning electron Figure 5-20 An electron generates (a) an orbital magnetic moment mo as it rotates around the nucleus and (b) a spin magnetic moment ms, as it spins about its own axis.
5-5.2 Magnetic Permeability In Chapter 4, we learned that the relationship D = ϵ0E, between the electric flux and field in free space, is modified to D = ϵ0E+P in a dielectric material. Likewise, the relationship B = μ0H in free space is modified to B = μ0H + μ0M = μ0(H + M), (5.72) where the magnetization vector M is defined as the vector sum of the magnetic dipole moments of the atoms contained in a unit volume of the material. Scale factors aside, the roles and interpretations of B, H, and M in Eq. (5.72) mirror those of D, E, and P in Eq. (4.83). Moreover, just as in most dielectrics P and E are linearly related, in most magnetic materials M = χmH, (5.73) where χm is a dimensionless quantity called the magnetic susceptibility of the material.
For diamagnetic and paramagnetic materials, χm is a (temperature-dependent) constant, resulting in a linear relationship between M and H at a given temperature. This is not the case for ferromagnetic substances; the relationship between M and H not only is nonlinear, but also depends on the “history” of the material, as explained in the next section.
Keeping this fact in mind, we can combine Eqs. (5.72) and (5.73) to get B = μ0(H + χmH) = μ0(1 + χm)H, (5.74) or B = μH, (5.75) 262 CHAPTER 5 MAGNETOSTATICS where μ, the magnetic permeability of the material, relates to χm as μ = μ0(1 + χm) (H/m).
(5.76) Often it is convenient to define the magnetic properties of a material in terms of the relative permeability μr: μr = μ μ0 = 1 + χm.
(5.77) A material usually is classified as diamagnetic, paramagnetic, or ferromagnetic on the basis of the value of its χm (Table 5-2).
Diamagnetic materials have negative susceptibilities whereas paramagnetic materials have positive ones.
However, the absolutemagnitudeofχm isontheorderof10−5 forbothclasses of materials, which for most applications allows us to ignore χm relative to 1 in Eq. (5.77).
▶Thus, μr ≈1 or μ ≈μ0 for diamagnetic and paramagnetic substances, which include dielectric materials and most metals.
In contrast, |μr| ≫1 for ferromagnetic materials; |μr| of purified iron, for example, is on the order of 2 × 105. ◀ Ferromagnetic materials are discussed next.
Exercise 5-11: The magnetic vector M is the vector sum of the magnetic moments of all the atoms contained in a unit volume (1m3). If a certain type of iron with 8.5 × 1028 atoms/m3 contributes one electron per atom to align its spin magnetic moment along the direction of the applied field, find (a) the spin magnetic moment of a single electron, given that me = 9.1 × 10−31 (kg) and ℏ= 1.06 × 10−34 (J·s), and (b) the magnitude of M.
Answer: (a) ms = 9.3 × 10−24 (A·m2), (b) M = 7.9 × 105 (A/m). (See EM.) 5-5.3 Magnetic Hysteresis of Ferromagnetic Materials Ferromagneticmaterials, whichincludeiron, nickel, andcobalt, exhibit unique magnetic properties due to the fact that their magnetic moments tend to readily align along the direction of an external magnetic field. Moreover, such materials remain partially magnetized even after the external field is removed.
Because of these peculiar properties, ferromagnetic materials are used in the fabrication of permanent magnets.
A key to understanding the properties of ferromagnetic materials is the notion of magnetized domains, microscopic regions (on the order of 10−10 m3) within which the magnetic moments of all atoms (typically on the order of 1019 atoms) are permanently aligned with each other. This alignment, which occurs in all ferromagnetic materials, is due to strong coupling forces between the magnetic dipole moments constituting an individual domain. In the absence of an external magnetic field, the domains take on random orientations relative to each other [Fig. 5-21(a)], resulting in zero net magnetization. The domain walls forming the boundaries between adjacent domains consist of thin transition regions. When an unmagnetized sample of a ferromagnetic material is placed in an external magnetic field, the domains partially align with the external field, as illustrated in Fig. 5-21(b). A quantitative understanding of how the domains form and how they behave under the influence of an external magnetic field requires a heavy dose of quantum mechanics, and is outside the scope of the present treatment.
Hence, we confine our discussion to a qualitative description of the magnetization process and its implications.
The magnetization behavior of a ferromagnetic material is described in terms of its B–H magnetization curve, where B and H refer to the amplitudes of the B flux and H field in the material. Suppose that we start with an unmagnetized sample of iron, denoted by point O in Fig. 5-22. When we increase H continuously by, for example, increasing the current passing through a wire wound around the sample, B increases also along the B–H curve from point O to point A1, at which nearly all the domains have become aligned with H. Point A1 represents a saturation condition. If we then decrease H from its value at point A1 back to zero (by reducing the current through the wire), the magnetization curve follows the path from A1 to A2. At point A2, the external field H is zero (owing to the fact that the current through the wire is zero), but the flux density B in the material is not. The magnitude of B at A2 is called the residual flux density Br. The iron material is now magnetized and ready to be used as a permanent magnet owing 5-5 MAGNETIC PROPERTIES OF MATERIALS 263 Table 5-2 Properties of magnetic materials.
Diamagnetism Paramagnetism Ferromagnetism Permanent magnetic No Yes, but weak Yes, and strong dipole moment Primary magnetization Electron orbital Electron spin Magnetized mechanism magnetic moment magnetic moment domains Direction of induced Opposite Same Hysteresis magnetic field [see Fig. 5-22] (relative to external field) Common substances Bismuth, copper, diamond, Aluminum, calcium, Iron, gold, lead, mercury, silver, chromium, magnesium, nickel, silicon niobium, platinum, cobalt tungsten Typical value of χm ≈−10−5 ≈10−5 |χm| ≫1 and hysteretic Typical value of μr ≈1 ≈1 |μr| ≫1 and hysteretic (a) Unmagnetized domains (b) Magnetized domains Figure 5-21 Comparison of (a) unmagnetized and (b) magnetized domains in a ferromagnetic material.
to the fact that a large fraction of its magnetized domains have remained aligned. Reversing the direction of H and increasing its intensity causes B to decrease from Br at point A2 to zero at point A3, and if the intensity of H is increased further while maintaining its direction, the magnetization moves to the saturation condition at point A4. Finally, as H is made to return to zero and is then increased again in the positive direction, the curve follows the path from A4 to A1.
This process is called magnetic hysteresis. Hysteresis means “lag A2 A1 A3 H O B A4 Br Figure 5-22 Typical hysteresis curve for a ferromagnetic material.
264 CHAPTER 5 MAGNETOSTATICS H B H B (a) Hard material (b) Soft material Figure 5-23 Comparison of hysteresis curves for (a) a hard ferromagnetic material and (b) a soft ferromagnetic material.
behind.” The existence of a hysteresis loop implies that the magnetization process in ferromagnetic materials depends not only on the magnetic field H, but also on the magnetic history of the material. The shape and extent of the hysteresis loop depend on the properties of the ferromagnetic material and the peak-to-peak range over which H is made to vary. Hard ferromagnetic materials are characterized by wide hysteresis loops [Fig. 5-23(a)]. They cannot be easily demagnetized by an external magnetic field because they have a large residual magnetization Br. Hard ferromagnetic materials are used in the fabrication of permanent magnets for motors and generators.
Soft ferromagnetic materials have narrow hysteresis loops [Fig. 5-23(b)], and hence can be more easily magnetized and demagnetized. To demagnetize any ferromagnetic material, the materialissubjectedtoseveralhysteresiscycleswhilegradually decreasing the peak-to-peak range of the applied field.
Concept Question 5-12: What are the three types of magnetic materials and what are typical values of their relative permeabilities?
Concept Question 5-13: What causes magnetic hys-teresis in ferromagnetic materials?
Concept Question 5-14: What does a magnetization curve describe?
What is the difference between the magnetization curves of hard and soft ferromagnetic materials?
5-6 Magnetic Boundary Conditions In Chapter 4, we derived a set of boundary conditions that describes how, at the boundary between two dissimilar contiguous media, the electric flux and field D and E in the first medium relate to those in the second medium. We now derive a similar set of boundary conditions for the magnetic flux and field B and H. By applying Gauss’s law to a pill box that straddles the boundary, we determined that the difference between the normal components of the electric flux densities in two media equals the surface charge density ρs. That is, S D· ds = Q D1n −D2n = ρs.
(5.78) By analogy, application of Gauss’s law for magnetism, as expressed by Eq. (5.44), leads to the conclusion that S B· ds = 0 B1n = B2n.
(5.79) ▶Thus the normal component of B is continuous across the boundary between two adjacent media. ◀ Because B1 = μ1H1 and B2 = μ2H2 for linear, isotropic media, the boundary condition for H corresponding to Eq. (5.79) is μ1H1n = μ2H2n.
(5.80) Comparison of Eqs. (5.78) and (5.79) reveals a striking difference between the behavior of the magnetic and electric fluxes across a boundary: whereas the normal component of B is continuous across the boundary, the normal component of D is not (unless ρs = 0). The reverse applies to the tangential components of the electric and magnetic fields E and H: whereas the tangential component of E is continuous across the boundary, the tangential component of H is not (unless the surface current density Js = 0). To obtain the boundary conditionforthetangentialcomponentofH, wefollowthesame basic procedure used in Chapter 4 to establish the boundary condition for the tangential component of E. With reference to Fig. 5-24, we apply Amp ere’s law [Eq. (5.47)] to a closed 5-7 INDUCTANCE 265 ∆h 2 ∆h 2 H1 H1n H1t H2 H2n H2t } } a d c b ∆l μ1 Medium 1 μ2 Medium 2 n2 Js ˆ n ˆ ˆ l l l1 ˆ l l l2 Figure 5-24 Boundary between medium 1 with μ1 and medium 2 with μ2.
rectangular path with sides of lengths l and h, and then let h →0, to obtain C H· dl = b a H1 · ˆ ℓ ℓ ℓ1 dℓ+ d c H2 · ˆ ℓ ℓ ℓ2 dℓ= I, (5.81) where I is the net current crossing the surface of the loop in the direction specified by the right-hand rule (I is in the direction of the thumb when the fingers of the right hand extend in the directionoftheloop C). Aswelet h oftheloopapproachzero, the surface of the loop approaches a thin line of length l. The total current flowing through this thin line is I = Js l, where Js is the magnitude of the component of the surface current density Js normal to the loop. That is, Js = Js · ˆ n, where ˆ n is the normal to the loop. In view of these considerations, Eq. (5.81) becomes (H1 −H2)· ˆ ℓ ℓ ℓ1 l = Js · ˆ n l.
(5.82) The vector ˆ ℓ ℓ ℓ1 can be expressed as ˆ ℓ ℓ ℓ1 = ˆ n × × × ˆ n2, where ˆ n and ˆ n2 are the normals to the loop and to the surface of medium 2 (Fig. 5-24), respectively. Using this relation in Eq. (5.82), and then applying the vector identityA·(B× × ×C) = B·(C× × ×A) leads to ˆ n·[ˆ n2 × × × (H1 −H2)] = Js · ˆ n.
(5.83) Since Eq. (5.83) is valid for any ˆ n, it follows that ˆ n2 × × × (H1 −H2) = Js.
(5.84) This equation implies that the tangential components of H parallel to Js are continuous across the interface, whereas those orthogonal to Js are discontinuous in the amount of Js.
Surface currents can exist only on the surfaces of perfect conductors and superconductors.
Hence, at the interface between media with finite conductivities, Js = 0 and H1t = H2t.
(5.85) Exercise 5-12: With reference to Fig. 5-24, determine the angle between H1 and ˆ n2 = ˆ z if H2 = (ˆ x3 + ˆ z2) (A/m), μr1 = 2, and μr2 = 8, and Js = 0.
Answer: θ = 20.6◦. (See EM.) 5-7 Inductance An inductor is the magnetic analogue of an electric capacitor.
Just as a capacitor can store energy in the electric field in the medium between its conducting surfaces, an inductor can store energy in the magnetic field near its current-carrying conductors. A typical inductor consists of multiple turns of wire helically coiled around a cylindrical core [Fig. 5-25(a)].
Such a structure is called a solenoid. Its core may be air filled or may contain a magnetic material with magnetic permeability μ.
If the wire carries a current I and the turns are closely spaced, the solenoid will produce a relatively uniform magnetic field within its interior with magnetic field lines resembling those of the permanent magnet [Fig. 5-25(b)].
5-7.1 Magnetic Field in a Solenoid As a prelude to our discussion of inductance we derive an expression for the magnetic flux density B in the interior region of a tightly wound solenoid. The solenoid is of length l and radius a, and comprises N turns carrying current I. The number of turns per unit length is n = N/l, and the fact that the turns are tightly wound implies that the pitch of a single turn is small compared with the solenoid’s radius. Even though the turns are slightly helical in shape, we can treat them as circular loops (Fig. 5-26). Let us start by considering the magnetic 266 CHAPTER 5 MAGNETOSTATICS (a) Loosely wound solenoid (b) Tightly wound solenoid N S B N B S Figure 5-25 Magnetic field lines of (a) a loosely wound solenoid and (b) a tightly wound solenoid.
flux density B at point P on the axis of the solenoid.
In Example 5-3, we derived the following expression for the magnetic field H along the axis of a circular loop of radius a, a distance z away from its center: H = ˆ z I ′a2 2(a2 + z2)3/2 , (5.86) where I ′ is the current carried by the loop.
If we treat an incremental length dz of the solenoid as an equivalent loop composed of n dz turns carrying a current I ′ = In dz, then the induced field at point P is dB = μ dH = ˆ z μnIa2 2(a2 + z2)3/2 dz.
(5.87) The total field B at P is obtained by integrating the contributions from the entire length of the solenoid. This is facilitated by expressing the variable z in terms of the angle θ, as seen from P to a point on the solenoid rim. That is, z = a tan θ, (5.88a) a2 + z2 = a2 + a2 tan2 θ = a2 sec2 θ, (5.88b) dz = a sec2 θ dθ.
(5.88c) θ dθ θ2 θ1 dz z l z x P I (out) I (in) a B Figure 5-26 Solenoid cross section showing geometry for calculating H at a point P on the solenoid axis.
Upon substituting the last two expressions in Eq. (5.87) and integrating from θ1 to θ2, we have B = ˆ z μnIa2 2 θ2 θ1 a sec2 θ dθ a3 sec3 θ = ˆ z μnI 2 (sin θ2 −sin θ1).
(5.89) If the solenoid length l is much larger than its radius a, then for points P away from the solenoid’s ends, θ1 ≈−90◦and θ2 ≈90◦, in which case Eq. (5.89) reduces to B ≈ˆ zμnI = ˆ zμNI l (long solenoid with l/a ≫1).
(5.90) Even though Eq. (5.90) was derived for the field B at the midpoint of the solenoid, it is approximately valid everywhere in the solenoid’s interior, except near the ends.
We now return to a discussion of inductance, which includes the notion of self-inductance, representing the magnetic flux linkage of a coil or circuit with itself, and mutual inductance, which involves the magnetic flux linkage in a circuit due to the magnetic field generated by a current in another one. Usually, when the term inductance is used, the intended reference is to self-inductance.
5-7 INDUCTANCE 267 Exercise 5-13: Use Eq. (5.89) to obtain an expression for B at a point on the axis of a very long solenoid but situated at its end points. How does B at the end points compare to B at the midpoint of the solenoid?
Answer: B = ˆ z(μNI/2l) at the end points, which is half as large as B at the midpoint. (See EM.) 5-7.2 Self-Inductance From Eq. (5.66), the magnetic flux linking a surface S is = S B· ds (Wb).
(5.91) In a solenoid characterized by an approximately uniform magnetic field throughout its cross-section given by Eq. (5.90), the flux linking a single loop is = S ˆ z μ N l I · ˆ z ds = μ N l IS, (5.92) where S is the cross-sectional area of the loop. Magnetic flux linkage is defined as the total magnetic flux linking a given circuit or conducting structure. If the structure consists of a single conductor with multiple loops, as in the case of the solenoid, equals the flux linking all loops of the structure.
For a solenoid with N turns, = N = μ N2 l IS (Wb).
(5.93) If, on the other hand, the structure consists of two separate conductors, as in the case of the parallel-wire and coaxial transmission lines shown in Fig. 5-27, the flux linkage associated with a length l of either line refers to the flux through a closed surface between the two conductors, such as the shaded areas in Fig. 5-27. In reality, there is also some magnetic flux that passes through the conductors themselves, but it may be ignored by assuming that currents flow only on the surfaces of the conductors, in which case the magnetic field inside the conductors vanishes.
This assumption is justified by the fact that our interest in calculating is for the purpose of determining the inductance of a given structure, and inductance is of interest primarily in the ac case (i.e., time-varying currents, voltages, and fields). As we will see later in Section 7-5, the current flowing in a conductor under ac (a) Parallel-wire transmission line (b) Coaxial transmission line Radius a z x y I S I d l l a b I S I c Figure 5-27 To compute the inductance per unit length of a two-conductor transmission line, we need to determine the magnetic flux through the area S between the conductors.
conditions is concentrated within a very thin layer on the skin of the conductor.
▶For the parallel-wire transmission line, ac currents flow on the outer surfaces of the wires, and for the coaxial line, the current flows on the outer surface of the inner conductor and on the inner surface of the outer one (the current-carrying surfaces are those adjacent to the electric and magnetic fields present in the region between the conductors). ◀ The self-inductance of any conducting structure is defined as the ratio of the magnetic flux linkage to the current I flowing through the structure: L = I (H).
(5.94) 268 TECHNOLOGY BRIEF 11: INDUCTIVE SENSORS Technology Brief 11: Inductive Sensors Magnetic coupling between different coils forms the basis of several different types of inductive sensors. Applications include the measurement of position and displacement (with submillimeter resolution) in device-fabrication processes, proximity detection of conductive objects, and other related applications.
Linear Variable Differential Transformer (LVDT) ▶An LVDT comprises a primary coil connected to an ac source (typically a sine wave at a frequency in the 1–10 kHz range) and a pair of secondary coils, all sharing a common ferromagnetic core (Fig. TF11-1). ◀ The magnetic core serves to couple the magnetic flux generated by the primary coil into the two secondaries, thereby inducing an output voltage across each of them. The secondary coils are connected in opposition, so that when the core is positioned at the magnetic center of the LVDT, the individual output signals of the secondaries cancel each other out, producing a null output voltage. The core is connected to the outside world via a nonmagnetic push rod.
When the rod moves the core away from the magnetic center, the magnetic fluxes induced in the secondary coils are no longer equal, resulting in a nonzero output voltage. The LVDT is called a “linear” transformer because the amplitude of the output voltage is a linear function of displacement over a wide operating range (Fig. TF11-2).
The cutaway view of the LVDT model in Fig.TF11-3 depicts a configuration in which all three coils—with the primary straddled by the secondaries—are wound around a glass tube that contains the magnetic core and attached rod.
Sample applications are illustrated in Fig. TF11-4.
Eddy-Current Proximity Sensor The transformer principle can be applied to build a proximity sensor in which the output voltage of the secondary coil becomes a sensitive indicator of the presence of a conductive object in its immediate vicinity (Fig. TF11-5).
Vin Primary coil Push rod Secondary coils Vout + _ + _ Ferromagnetic core Figure TF11-1 Linear variable differential transformer (LVDT) circuit.
Distance traveled Phase Amplitude −10 −5 0 5 10 Amplitude and phase output Figure TF11-2 Amplitude and phase responses as a function of the distance by which the magnetic core is moved away from the center position.
TECHNOLOGY BRIEF 11: INDUCTIVE SENSORS 269 Stainless steel housing Rod Secondary coils Primary coil Electronics module Magnetic core Figure TF11-3 Cutaway view of LVDT.
Sagging beam LVDT LVDT Float Figure TF11-4 LVDT for measuring beam deflection and as a fluid-level gauge.
▶When an object is placed in front of the secondary coil, the magnetic field of the coil induces eddy (circular) currents in the object, which generate magnetic fields of their own having a direction that opposes the magnetic field of the secondary coil. ◀ The reduction in magnetic flux causes a drop in output voltage, with the magnitude of the change being dependent on the conductive properties of the object and its distance from the sensor.
Primary coil Sensing coil Eddy currents Conductive object Vin + + Vout _ _ FigureTF11-5 Eddy-current proximity sensor.
270 CHAPTER 5 MAGNETOSTATICS The SI unit for inductance is the henry (H), which is equivalent to webers per ampere (Wb/A).
For a solenoid, use of Eq. (5.93) gives L = μ N2 l S (solenoid), (5.95) and for two-conductor configurations similar to those of Fig. 5-27, L = I = I = 1 I S B· ds.
(5.96) Example 5-7: Inductance of a Coaxial Transmission Line Develop an expression for the inductance per unit length of a coaxial transmission line with inner and outer conductors of radii a and b (Fig. 5-28) and an insulating material of permeability μ.
Solution: The current I in the inner conductor generates a magnetic field B throughout the region between the two conductors. It is given by Eq. (5.30) as B = ˆ φ φ φ μI 2πr , (5.97) where r is the radial distance from the axis of the coaxial line.
Consider a transmission-line segment of length l as shown in Fig. 5-28. Because B is perpendicular to the planar surface S between the conductors, the flux through S is = l b a B dr = l b a μI 2πr dr = μIl 2π ln b a .
(5.98) Using Eq. (5.96), the inductance per unit length of the coaxial transmission line is L′ = L l = lI = μ 2π ln b a .
(5.99) a r μ l b I I I z Outer conductor Inner conductor Outer conductor S Figure 5-28 Cross-sectional view of coaxial transmission line (Example 5-7).
and denote H field out of and into the page, respectively.
I1 B1 S1 S2 C1 C2 N1 turns N2 turns Figure 5-29 Magnetic field lines generated by current I1 in loop 1 linking surface S2 of loop 2.
5-7.3 Mutual Inductance Magnetic coupling between two different conducting structures is described in terms of the mutual inductance between them.
For simplicity, consider the case of two multiturn closed loops with surfaces S1 and S2. Current I1 flows through the first loop 5-8 MAGNETIC ENERGY 271 R V1 I1 V2 + − Figure 5-30 Toroidal coil with two windings used as a transformer.
(Fig. 5-29), and no current flows through the second one. The magnetic field B1 generated by I1 results in a flux 12 through loop 2, given by 12 = S2 B1 · ds, (5.100) and if loop 2 consists of N2 turns all coupled by B1 in exactly the same way, then the total magnetic flux linkage through loop 2 is 12 = N2 12 = N2 S2 B1 · ds.
(5.101) The mutual inductance associated with this magnetic coupling is given by L12 = 12 I1 = N2 I1 S2 B1 · ds (H).
(5.102) Mutual inductance is important in transformers (as discussed in Chapter 6) wherein the windings of two or more circuits share a common magnetic core, as illustrated by the toroidal arrangement shown in Fig. 5-30.
Concept Question 5-15: What is the magnetic field like in the interior of a long solenoid?
Concept Question 5-16: What is the difference be-tween self-inductance and mutual inductance?
Concept Question 5-17: How is the inductance of a solenoid related to its number of turns N?
5-8 Magnetic Energy When we introduced electrostatic energy in Section 4-10, we did so by examining what happens to the energy expended in charging up a capacitor from zero voltage to some final voltage V .
We introduce the concept of magnetic energy by considering an inductor with inductance L connected to a current source. Suppose that we were to increase the current i flowing through the inductor from zero to a final value I. From circuit theory, we know that the instantaneous voltage υ across the inductor is given by υ = L di/dt.
We will derive this relationship from Maxwell’s equations in Chapter 6, thereby justifying the use of the i–υ relationship for the inductor.
Power p equals the product of υ and i, and the time integral of power is work, or energy. Hence, the total energy in joules (J) expended in building up a current I in the inductor is Wm = p dt = iv dt = L I 0 i di = 1 2LI 2 (J).
(5.103) We call this the magnetic energy stored in the inductor.
To justify this association, consider the solenoid inductor.
Its inductance is given by Eq. (5.95) as L = μN2S/l, and the magnitude of the magnetic flux density in its interior is given by Eq. (5.90) as B = μNI/l, implying that I = Bl/(μN). Using these expressions for L and I in Eq. (5.103), we obtain Wm = 1 2 LI 2 = 1 2 μ N2 l S Bl μN 2 = 1 2 B2 μ (lS) = 1 2 μH 2v, (5.104) where v = lS is the volume of the interior of the solenoid and H = B/μ. The expression for Wm suggests that the energy expended in building up the current in the inductor is stored in the magnetic field with magnetic energy density wm, defined as the magnetic energy Wm per unit volume, 272 CHAPTER 5 MAGNETOSTATICS wm = Wm v = 1 2μH 2 (J/m3).
(5.105) ▶Even though this expression was derived for a solenoid, itremainsvalidforanymediumwithamagneticfieldH.◀ Furthermore, for any volume v containing a material with permeability μ (including free space with permeability μ0), the total magnetic energy stored in a magnetic field H is Wm = 1 2 v μH 2 dv (J).
(5.106) Example 5-8: Magnetic Energy in a Coaxial Cable Derive an expression for the magnetic energy stored in a coaxial cable of length l and inner and outer radii a and b. The current flowing through the cable is I and its insulation material has permeability μ.
Solution: From Eq. (5.97), the magnitude of the magnetic field in the insulating material is H = B μ = I 2πr , where r is the radial distance from the center of the inner conductor (Fig. 5-28).
The magnetic energy stored in the coaxial cable therefore is Wm = 1 2 v μH 2 dv = μI 2 8π2 v 1 r2 dv.
Since H is a function of r only, we choose dv to be a cylindrical shell of length l, radius r, and thickness dr along the radial direction. Thus, dv = 2πrl dr and Wm = μI 2 8π2 b a 1 r2 · 2πrl dr = μI 2l 4π ln b a = 1 2 LI 2 (J), with L given by Eq. (5.99).
Chapter 5 Summary Concepts • The magnetic force acting on a charged particle q moving with a velocity u in a region containing a magnetic flux density B is Fm = qu × × × B.
• The total electromagnetic force, known as the Lorentz force, acting on a moving charge in the presence of both electric and magnetic fields is F = q(E + u × × × B).
• Magnetic forces acting on current loops can generate magnetic torques.
• The magnetic field intensity induced by a current element is defined by the Biot–Savart law.
• Gauss’s law for magnetism states that the net magnetic flux flowing out of any closed surface is zero.
• Ampere’s law states that the line integral of H over a closed contour is equal to the net current crossing the surface bounded by the contour.
• The vector magnetic potential A is related to B by B = ∇× × × A.
• Materials are classified as diamagnetic, paramagnetic, or ferromagnetic, depending on their crystalline structure and the behavior under the influence of an external magnetic field.
• Diamagnetic and paramagnetic materials exhibit a linear behavior between B and H, with μ ≈μ0 for both.
• Ferromagnetic materials exhibit a nonlinear hysteretic behavior between B and H and, for some, μ may be as large as 105μ0.
• At the boundary between two different media, the normalcomponentofBiscontinuous, andthetangential components of H are related by H2t −H1t = Js, where Js is the surface current density flowing in a direction orthogonal to H1t and H2t.
• The inductance of a circuit is defined as the ratio of magnetic flux linking the circuit to the current flowing through it.
• Magnetic energy density is given by wm = 1 2μH 2.
CHAPTER 5 SUMMARY 273 Important Terms Provide definitions or explain the meaning of the following terms: Amp ere’s law Amperian contour Biot–Savart law current density (volume) J diamagnetic ferromagnetic Gauss’s law for magnetism hard and soft ferromagnetic materials inductance (self- and mutual) Lorentz force F magnetic dipole magnetic energy Wm magnetic energy density wm magnetic flux magnetic flux density B magnetic flux linkage magnetic force Fm magnetic hysteresis magnetic moment m magnetic potential A magnetic susceptibility χm magnetization curve magnetization vector M magnetized domains moment arm d orbital and spin magnetic moments paramagnetic solenoid surface current density Js toroid toroidal coil torque T vector Poisson’s equation Mathematical and Physical Models Maxwell’s Magnetostatics Equations Gauss’s Law for Magnetism ∇· B = 0 S B· ds = 0 Amp ere’s Law ∇× × × H = J C H· dℓ ℓ ℓ= I Lorentz Force on Charge q F = q(E + u × × × B) Magnetic Force on Wire Fm = I C dl × × × B (N) Magnetic Torque on Loop T = m × × × B (N·m) m = ˆ n NIA (A·m2) Biot–Savart Law H = I 4π l dl × × × ˆ R R2 (A/m) Magnetic Field Infinitely Long Wire B = ˆ φ φ φ μ0I 2πr (Wb/m2) Circular Loop H = ˆ z Ia2 2(a2 + z2)3/2 (A/m) Solenoid B ≈ˆ z μnI = ˆ z μNI l (Wb/m2) Vector Magnetic Potential B = ∇× × × A (Wb/m2) Vector Poisson’s Equation ∇2A = −μJ Inductance L = I = I = 1 I S B· ds (H) Magnetic Energy Density wm = 1 2 μH 2 (J/m3) 274 CHAPTER 5 MAGNETOSTATICS PROBLEMS Section 5-1: Magnetic Forces and Torques ∗5.1 An electron with a speed of 8 × 106 m/s is projected along the positive x direction into a medium containing a uniform magnetic flux density B = (ˆ x4 −ˆ z3) T.
Given that e = 1.6 × 10−19 C and the mass of an electron is me = 9.1 × 10−31 kg, determine the initial acceleration vector of the electron (at the moment it is projected into the medium).
5.2 When a particle with charge q and mass m is introduced into a medium with a uniform field B such that the initial velocity of the particle u is perpendicular to B (Fig. P5.2), the magnetic force exerted on the particle causes it to move in a circle of radius a. By equating Fm to the centripetal force on the particle, determine a in terms of q, m, u, and B.
Fm Fm B Fm + + + + q a q q q u u u P FigureP5.2 Particleofchargeq projectedwithvelocityuintoa medium with a uniform field B perpendicular to u (Problem 5.2).
5.3 The circuit shown in Fig. P5.3 uses two identical springs to support a 10 cm long horizontal wire with a mass of 20 g. In the absence of a magnetic field, the weight of the wire causes the springs to stretch a distance of 0.2 cm each.
When a uniform magnetic field is turned on in the region containing the horizontal wire, the springs are observed to stretch an additional 0.5 cm each. What is the intensity of the magnetic flux density B? The force equation for a spring is F = kd, where k is the spring constant and d is the distance it has been stretched.
∗Answer(s) available in Appendix D.
4 Ω 12 V B 10 cm Springs + − Figure P5.3 Configuration of Problem 5.3.
∗5.4 The rectangular loop shown in Fig. P5.4 consists of 20 closely wrapped turns and is hinged along the z axis. The plane of the loop makes an angle of 30◦with the y axis, and the current in the windings is 0.5 A. What is the magnitude of the torque exerted on the loop in the presence of a uniform field B = ˆ y 2.4 T? When viewed from above, is the expected direction of rotation clockwise or counterclockwise?
30◦ 0.4 m 0.2 m y x z I = 0.5 A 20 turns Figure P5.4 Hinged rectangular loop of Problem 5.4.
5.5 In a cylindrical coordinate system, a 2 m long straight wire carrying a current of 5 A in the positive z direction is located at r = 4 cm, φ = π/2, and −1 m ≤z ≤1 m.
PROBLEMS 275 ∗(a) If B = ˆ r 0.2 cos φ (T), what is the magnetic force acting on the wire?
(b) How much work is required to rotate the wire once about the z axis in the negative φ direction (while maintaining r = 4 cm)?
(c) At what angle φ is the force a maximum?
5.6 A 20-turn rectangular coil with sides l = 30 cm and w = 10 cm is placed in the y–z plane as shown in Fig. P5.6.
(a) If the coil, which carries a current I = 10 A, is in the presence of a magnetic flux density B = 2 × 10−2(ˆ x + ˆ y2) (T), determine the torque acting on the coil.
(b) At what angle φ is the torque zero?
(c) At what angle φ is the torque maximum? Determine its value.
z y x l w I n 20-turn coil φ ˆ Figure P5.6 Rectangular loop of Problem 5.6.
Section 5-2: The Biot–Savart Law ∗5.7 An 8 cm × 12 cm rectangular loop of wire is situated in the x–y plane with the center of the loop at the origin and its long sides parallel to the x axis. The loop has a current of 50 A flowing clockwise (when viewed from above). Determine the magnetic flux density at the center of the loop.
5.8 Use the approach outlined in Example 5-2 to develop an expression for the magnetic field H at an arbitrary point P due to the linear conductor defined by the geometry shown in Fig. P5.8. If the conductor extends between z1 = 3 m and z2 = 7 m and carries a current I = 15 A, find H at P = (2, φ, 0).
I z P = (r, φ, z) P1(z1) P2(z2) θ2 θ1 r Figure P5.8 Current-carrying linear conductor of Problem 5.8.
∗5.9 The loop shown in Fig. P5.9 consists of radial lines and segments of circles whose centers are at point P. Determine the magnetic field H at P in terms of a, b, θ, and I.
θ b a P I Figure P5.9 Configuration of Problem 5.9.
5.10 An infinitely long, thin conducting sheet defined over the space 0 ≤x ≤w and −∞≤y ≤∞is carrying a current with a uniform surface current density Js = ˆ y5 (A/m). Obtain an expression for the magnetic field at point P = (0, 0, z) in Cartesian coordinates.
276 CHAPTER 5 MAGNETOSTATICS ∗5.11 An infinitely long wire carrying a 25 A current in the positive x direction is placed along the x axis in the vicinity of a 20-turn circular loop located in the x–y plane (Fig. P5.11). If the magnetic field at the center of the loop is zero, what is the direction and magnitude of the current flowing in the loop?
1 m d = 2 m I1 x Figure P5.11 Circular loop next to a linear current (Problem 5.11).
5.12 Two infinitely long, parallel wires are carrying 6 A currents in opposite directions. Determine the magnetic flux density at point P in Fig. P5.12.
I2 = 6 A I1 = 6 A 0.5 m 2 m P Figure P5.12 Arrangement for Problem 5.12.
∗5.13 A long, East-West–oriented power cable carrying an unknown current I is at a height of 8 m above the Earth’s surface. If the magnetic flux density recorded by a magnetic-field meter placed at the surface is 15 μT when the current is flowing through the cable and 20 μT when the current is zero, what is the magnitude of I?
5.14 Two parallel, circular loops carrying a current of 40 A each are arranged as shown in Fig. P5.14. The first loop is situated in the x–y plane with its center at the origin, and the second loop’s center is at z = 2 m. If the two loops have the same radius a = 3 m, determine the magnetic field at: (a) z = 0 (b) z = 1 m (c) z = 2 m z = 2 m 0 z y x a a I I Figure P5.14 Parallel circular loops of Problem 5.14.
5.15 A circular loop of radius a carrying current I1 is located in the x–y plane as shown in Fig. P5.15.
In addition, an infinitely long wire carrying current I2 in a direction parallel with the z axis is located at y = y0.
(a) Determine H at P = (0, 0, h).
(b) Evaluate H for a = 3 cm, y0 = 10 cm, h = 4 cm, I1 = 10 A, and I2 = 20 A.
PROBLEMS 277 x y z y0 a n parallel to z P = (0, 0, h) I2 I1 ˆ ˆ Figure P5.15 Problem 5.15.
∗5.16 The long, straight conductor shown in Fig. P5.16 lies in the plane of the rectangular loop at a distance d = 0.1 m. The loop has dimensions a = 0.2 m and b = 0.5 m, and the currents are I1 = 20 A and I2 = 30 A. Determine the net magnetic force acting on the loop.
I2 I1 a = 0.2 m d = 0.1 m b = 0.5 m Figure P5.16 Current loop next to a conducting wire (Problem 5.16).
5.17 In the arrangement shown in Fig. P5.17, each of the two long, parallel conductors carries a current I, is supported by 8 cm long strings, and has a mass per unit length of 1.2 g/cm. Due to the repulsive force acting on the conductors, the angle θ between the supporting strings is 10◦. Determine the magnitude of I and the relative directions of the currents in the two conductors.
F12 F21 θ = 10◦ z x y Figure P5.17 Parallel conductors supported by strings (Problem 5.17).
5.18 An infinitely long, thin conducting sheet of width w along the x direction lies in the x–y plane and carries a current I in the −y direction. Determine the following: ∗(a) The magnetic field at a point P midway between the edges of the sheet and at a height h above it (Fig. P5.18).
(b) The force per unit length exerted on an infinitely long wire passing through point P and parallel to the sheet if the current through the wire is equal in magnitude but opposite in direction to that carried by the sheet.
I I P h w Figure P5.18 A linear current source above a current sheet (Problem 5.18).
5.19 Three long, parallel wires are arranged as shown in Fig. P5.19. Determine the force per unit length acting on the wire carrying I3.
278 CHAPTER 5 MAGNETOSTATICS I1 = 10 A I2 = 10 A I3 = 10 A 2 m 2 m 2 m Figure P5.19 Three parallel wires of Problem 5.19.
∗5.20 A square loop placed as shown in Fig. P5.20 has 2 m sidesandcarriesacurrentI1 = 5A. Ifastraight, longconductor carrying a current I2 = 10A is introduced and placed just above the midpoints of two of the loop’s sides, determine the net force acting on the loop.
z x y a a 1 3 4 2 I1 I2 Figure P5.20 Long wire carrying current I2, just above a square loop carrying I1 (Problem 5.20).
Section 5-3: Maxwell’s Magnetostatic Equations 5.21 Current I flows along the positive z direction in the inner conductor of a long coaxial cable and returns through the outer conductor. The inner conductor has radius a, and the inner and outer radii of the outer conductor are b and c, respectively.
(a) Determine the magnetic field in each of the following regions: 0 ≤r ≤a, a ≤r ≤b, b ≤r ≤c, and r ≥c.
(b) Plot the magnitude of H as a function of r over the range from r = 0 to r = 10 cm, given that I = 10 A, a = 2 cm, b = 4 cm, and c = 5 cm.
5.22 A long cylindrical conductor whose axis is coincident with the z axis has a radius a and carries a current characterized by a current density J = ˆ zJ0/r, where J0 is a constant and r is the radial distance from the cylinder’s axis. Obtain an expression for the magnetic field H for (a) 0 ≤r ≤a (b) r > a 5.23 Repeat Problem 5.22 for a current density J = ˆ zJ0e−r.
∗5.24 In a certain conducting region, the magnetic field is given in cylindrical coordinates by H = ˆ φ φ φ4 r [1 −(1 + 3r)e−3r] Find the current density J.
5.25 A cylindrical conductor whose axis is coincident with the z axis has an internal magnetic field given by H = ˆ φ φ φ 2 r [1 −(4r + 1)e−4r] (A/m) for r ≤a where a is the conductor’s radius. If a = 5 cm, what is the total current flowing in the conductor?
Section 5-4: Vector Magnetic Potential 5.26 With reference to Fig. 5-10: ∗(a) Derive an expression for the vector magnetic potential A at a point P located at a distance r from the wire in the x–y plane.
(b) Derive B from A. Show that your result is identical with the expression given by Eq. (5.29), which was derived by applying the Biot–Savart law.
5.27 In a given region of space, the vector magnetic potential is given by A = ˆ x5 cos πy + ˆ z(2 + sin πx) (Wb/m).
∗(a) Determine B.
(b) Use Eq. (5.66) to calculate the magnetic flux passing through a square loop with 0.25 m long edges if the loop is in the x–y plane, its center is at the origin, and its edges are parallel to the x and y axes.
(c) Calculate again using Eq. (5.67).
PROBLEMS 279 5.28 A uniform current density given by J = ˆ zJ0 (A/m2) gives rise to a vector magnetic potential A = −ˆ z μ0J0 4 (x2 + y2) (Wb/m).
(a) Apply the vector Poisson’s equation to confirm the above statement.
(b) Use the expression for A to find H.
(c) Use the expression for J in conjunction with Ampere’s law to find H. Compare your result with that obtained in part (b).
∗5.29 A thin current element extending between z = −L/2 and z = L/2 carries a current I along +ˆ z through a circular cross-section of radius a.
(a) FindA at a point P located very far from the origin (assume R is so much larger than L that point P may be considered to be at approximately the same distance from every point along the current element).
(b) Determine the corresponding H.
Section 5-5: Magnetic Properties of Materials 5.30 In the model of the hydrogen atom proposed by Bohr in 1913, the electron moves around the nucleus at a speed of 2×106 m/s in a circular orbit of radius 5×10−11 m. What is the magnitude of the magnetic moment generated by the electron’s motion?
∗5.31 Iron contains 8.5 × 1028 atoms/m3. At saturation, the alignment of the electrons’ spin magnetic moments in iron can contribute 1.5 T to the total magnetic flux density B. If the spin magnetic moment of a single electron is 9.27 × 10−24 (A·m2), how many electrons per atom contribute to the saturated field?
Section 5-6: Magnetic Boundary Conditions 5.32 The x–y plane separates two magnetic media with magnetic permeabilities μ1 and μ2 (Fig. P5.32). If there is no surface current at the interface and the magnetic field in medium 1 is H1 = ˆ xH1x + ˆ yH1y + ˆ zH1z find: (a) H2 (b) θ1 and θ2 (c) Evaluate H2, θ1, and θ2 for H1x = 2 (A/m), H1y = 0, H1z = 4 (A/m), μ1 = μ0, and μ2 = 4μ0 θ1 μ1 μ2 H1 z x-y plane Figure P5.32 Adjacent magnetic media (Problem 5.32).
∗5.33 Given that a current sheet with surface current density Js = ˆ x 8 (A/m) exists at y = 0, the interface between two magnetic media, and H1 = ˆ z 11 (A/m) in medium 1 (y > 0), determine H2 in medium 2 (y < 0).
5.34 In Fig. P5.34, the plane defined by x −y = 1 separates medium 1 of permeability μ1 from medium 2 of permeability μ2. If no surface current exists on the boundary and B1 = ˆ x2 + ˆ y3 (T), find B2 and then evaluate your result for μ1 = 5μ2. Hint: Start by deriving the equation for the unit vector normal to the given plane.
y x (1, 0) (0, −1) μ2 Medium 2 Plane x − y = 1 μ1 Medium 1 Figure P5.34 Magnetic media separated by the planex−y = 1 (Problem 5.34).
280 CHAPTER 5 MAGNETOSTATICS ∗5.35 The plane boundary defined by z = 0 separates air from a block of iron. If B1 = ˆ x4 −ˆ y6 + ˆ z8 in air (z ≥0), find B2 in iron (z ≤0), given that μ = 5000μ0 for iron.
5.36 Show that if no surface current densities exist at the parallel interfaces shown in P5.36, the relationship between θ4 and θ1 is independent of μ2.
μ1 μ2 μ3 B3 B2 B1 θ4 θ3 θ2 θ1 Figure P5.36 Three magnetic media with parallel interfaces (Problem 5.36).
Sections 5-7 and 5-8: Inductance and Magnetic Energy ∗5.37 Obtain an expression for the self-inductance per unit length for the parallel wire transmission line of Fig. 5-27(a) in terms of a, d, and μ, where a is the radius of the wires, d is the axis-to-axis distance between the wires, and μ is the permeability of the medium in which they reside.
5.38 A solenoid with a length of 20 cm and a radius of 5 cm consists of 400 turns and carries a current of 12 A. If z = 0 represents the midpoint of the solenoid, generate a plot for |H(z)| as a function of z along the axis of the solenoid for the range −20 cm ≤z ≤20 cm in 1 cm steps.
5.39 In terms of the dc current I, how much magnetic energy is stored in the insulating medium of a 3 m long, air-filled section of a coaxial transmission line, given that the radius of the inner conductor is 5 cm and the inner radius of the outer conductor is 10 cm?
∗5.40 The rectangular loop shown in Fig. P5.40 is coplanar with the long, straight wire carrying the current I = 20 A.
Determine the magnetic flux through the loop.
y x z 5 cm 20 A 20 cm 30 cm Figure P5.40 Loop and wire arrangement for Problem 5.40.
5.41 Determine the mutual inductance between the circular loop and the linear current shown in Fig. P5.41.
a d I1 x y Figure P5.41 Linear conductor with current I1 next to a circular loop of radius a at distance d (Problem 5.41).
C H A P T E R 6 Maxwell’s Equations for Time-Varying Fields Chapter Contents Dynamic Fields, 282 6-1 Faraday’s Law, 282 6-2 Stationary Loop in a Time-Varying Magnetic Field, 284 6-3 The Ideal Transformer, 288 6-4 Moving Conductor in a Static Magnetic Field, 289 TB12 EMF Sensors, 292 6-5 The Electromagnetic Generator, 294 6-6 Moving Conductor in a Time-Varying Magnetic Field, 296 6-7 Displacement Current, 297 6-8 Boundary Conditions for Electromagnetics, 299 6-9 Charge-Current Continuity Relation, 299 6-10 Free-Charge Dissipation in a Conductor, 302 6-11 Electromagnetic Potentials, 302 Chapter 6 Summary, 307 Problems, 308 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Apply Faraday’s law to compute the voltage induced by a stationary coil placed in a time-varying magnetic field or moving in a medium containing a magnetic field.
2. Describe the operation of the electromagnetic generator.
3. Calculate the displacement current associated with a time-varying electric field.
4. Calculate the rate at which charge dissipates in a material with known ϵ and σ.
282 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS Dynamic Fields Electric charges induce electric fields, and electric currents induce magnetic fields.
As long as the charge and current distributions remain constant in time, so will the fields they induce. If the charges and currents vary in time, the electric and magnetic fields vary accordingly. Moreover, the electric and magnetic fields become coupled and travel through space in the form of electromagnetic waves. Examples of such waves include light, x-rays, infrared rays, gamma rays, and radio waves (see Fig. 1-16).
To study time-varying electromagnetic phenomena, we need to consider the entire set of Maxwell’s equations simultaneously.
These equations, first introduced in the opening section of Chapter 4, are given in both differential and integral form in Table 6-1. In the static case (∂/∂t = 0), we use the first pair of Maxwell’s equations to study electric phenomena (Chapter 4) and the second pair to study magnetic phenomena (Chapter 5).
In the dynamic case (∂/∂t ̸= 0), the coupling that exists between the electric and magnetic fields, as expressed by the second and fourth equations in Table 6-1, prevents such decomposition. The first equation represents Gauss’s law for electricity, and it is equally valid for static and dynamic fields. The same is true for the third equation, Gauss’s law for magnetism. By contrast, the second and fourth equations—Faraday’s and Amp ere’s laws—are of a totally different nature. Faraday’s law expresses the fact that a time-varying magnetic field gives rise to an electric field.
Conversely, Ampere’s law states that a time-varying electric field must be accompanied by a magnetic field.
Some statements in this and succeeding chapters contradict conclusions reached in Chapter 4 and 5 as those pertained to the special case of static charges and dc currents. The behavior of dynamic fields reduces to that of static ones when ∂/∂t is set to zero.
We begin this chapter by examining Faraday’s and Amp ere’s laws and some of their practical applications. We then combine Maxwell’s equations to obtain relations among the charge and current sources, ρv and J, the scalar and vector potentials, V and A, and the electromagnetic fields, E, D, H, and B, for the most general time-varying case and for the specific case of sinusoidal-time variations.
6-1 Faraday’s Law The close connection between electricity and magnetism was established by Oersted, who demonstrated that a wire carrying an electric current exerts a force on a compass needle and that the needle always turns so as to point in the ˆ φ φ φ direction when the current is along the ˆ z direction. The force acting on the compass needle is due to the magnetic field produced by the current in the Table 6-1 Maxwell’s equations.
Reference Differential Form Integral Form Gauss’s law ∇· D = ρv S D· ds = Q (6.1) Faraday’s law ∇× × × E = −∂B ∂t C E· dl = − S ∂B ∂t · ds (6.2)∗ No magnetic charges ∇· B = 0 S B· ds = 0 (6.3) (Gauss’s law for magnetism) Ampere’s law ∇× × × H = J + ∂D ∂t C H· dl = S J + ∂D ∂t · ds (6.4) ∗For a stationary surface S.
6-1 FARADAY’S LAW 283 Galvanometer Loop Coil Battery B I I Figure 6-1 The galvanometer (predecessor of the ammeter) shows a deflection whenever the magnetic flux passing through the square loop changes with time.
wire. Following this discovery, Faraday hypothesized that if a current produces a magnetic field, then the converse should also be true: a magnetic field should produce a current in a wire.
To test his hypothesis, he conducted numerous experiments in his laboratory in London over a period of about 10 years, all aimed at making magnetic fields induce currents in wires.
Similar work was being carried out by Henry in Albany, New York. Wires were placed next to permanent magnets or current-carrying loops of all different sizes, but no currents were ever detected. Eventually, these experiments led to the discovery by both Faraday and Henry that: ▶Magnetic fields can produce an electric current in a closed loop, but only if the magnetic flux linking the surface area of the loop changes with time. The key to the induction process is change. ◀ To elucidate the induction process, consider the arrangement shown in Fig. 6-1.
A conducting loop connected to a galvanometer, a sensitive instrument used in the 1800s to detect current flow, is placed next to a conducting coil connected to a battery. The current in the coil produces a magnetic field B whose lines pass through the loop. In Section 5-4, we defined the magnetic flux passing through a loop as the integral of the normal component of the magnetic flux density over the surface area of the loop, S, or = S B· ds (Wb).
(6.5) Under stationary conditions, the dc current in the coil produces a constant magnetic field B, which in turn produces a constant flux through the loop. When the flux is constant, no current is detected by the galvanometer. However, when the battery is disconnected, thereby interrupting the flow of current in the coil, the magnetic field drops to zero, and the consequent change in magnetic flux causes a momentary deflection of the galvanometer needle.
When the battery is reconnected, the galvanometer again exhibits a momentary deflection, but in the opposite direction.
Thus, current is induced in the loop when the magnetic flux changes, and the direction of the current depends on whether the flux increases (when the battery is being connected) or decreases (when the battery is being disconnected). It was further discovered that current can also flow in the loop while the battery is connected to the coil if the loop turns or moves closer to, or away from, the coil.
The physical movement of the loop changes the amount of flux linking its surface S, even though the field B due to the coil has not changed.
A galvanometer is a predecessor of the voltmeter and ammeter. When a galvanometer detects the flow of current through the coil, it means that a voltage has been induced across the galvanometer terminals.
This voltage is called the electromotive force (emf), Vemf, and the process is called electromagnetic induction.
The emf induced in a closed conducting loop of N turns is given by Vemf = −N d dt = −N d dt S B· ds (V).
(6.6) Even though the results leading to Eq. (6.6) were also discovered independently by Henry, Eq. (6.6) is attributed to Faraday and known as Faraday’s law. The significance of the negative sign in Eq. (6.6) is explained in the next section.
We note that the derivative in Eq. (6.6) is a total time derivative that operates on the magnetic field B, as well as the differential surface area ds.
Accordingly, an emf can be generated in a closed conducting loop under any of the following three conditions: 1. A time-varying magnetic field linking a stationary loop; the induced emf is then called the transformer emf, V tr emf.
284 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS 2. A moving loop with a time-varying area (relative to the normal component of B) in a static field B; the induced emf is then called the motional emf, V m emf.
3. A moving loop in a time-varying field B.
The total emf is given by Vemf = V tr emf + V m emf, (6.7) with V m emf = 0 if the loop is stationary [case (1)] and V tr emf = 0 if B is static [case (2)]. For case (3), both terms are important.
Each of the three cases is examined separately in the following sections.
6-2 Stationary Loop in a Time-Varying Magnetic Field The stationary, single-turn, conducting, circular loop with contour C and surface area S shown in Fig. 6-2(a) is exposed to a time-varying magnetic field B(t). As stated earlier, the emf induced when S is stationary and the field is time varying (a) Loop in a changing B field (b) Equivalent circuit 1 R 2 Bind Changing B(t) C I I S Vemf tr R 1 2 Ri Vemf (t) tr Figure 6-2 (a) Stationary circular loop in a changing magnetic field B(t), and (b) its equivalent circuit.
is called the transformer emf and is denoted V tr emf. Since the loop is stationary, d/dt in Eq. (6.6) now operates on B(t) only.
Hence, V tr emf = −N S ∂B ∂t · ds (transformer emf), (6.8) where the full derivative d/dt has been moved inside the integral and changed into a partial derivative ∂/∂t to signify that it operates on B only. The transformer emf is the voltage difference that would appear across the small opening between terminals 1 and 2, even in the absence of the resistor R. That is, V tr emf = V12, where V12 is the open-circuit voltage across the open ends of the loop. Under dc conditions, V tr emf = 0. For the loop shown in Fig. 6-2(a) and the associated definition for V tr emf given by Eq. (6.8), the direction of ds, the loop’s differential surface normal, can be chosen to be either upward or downward.
The two choices are associated with opposite designations of the polarities of terminals 1 and 2 in Fig. 6-2(a).
▶The connection between the direction of ds and the polarity of V tr emf is governed by the following right-hand rule: if ds points along the thumb of the right hand, then the direction of the contour C indicated by the four fingers is such that it always passes across the opening from the positive terminal of V tr emf to the negative terminal. ◀ If the loop has an internal resistance Ri, the circuit in Fig. 6-2(a) can be represented by the equivalent circuit shown in Fig. 6-2(b), in which case the current I flowing through the circuit is given by I = V tr emf R + Ri .
(6.9) For good conductors, Ri usually is very small, and it may be ignored in comparison with practical values of R.
▶The polarity of V tr emf and hence the direction of I is governed by Lenz’s law, which states that the current in the loop is always in a direction that opposes the change of magnetic flux (t) that produced I. ◀ 6-2 STATIONARY LOOP IN A TIME-VARYING MAGNETIC FIELD 285 The current I induces a magnetic field of its own, Bind, with a corresponding flux ind. The direction of Bind is governed by the right-hand rule; if I is in a clockwise direction, then Bind points downward through S and, conversely, if I is in a counterclockwise direction, then Bind points upward through S.
If the original field B(t) is increasing, which means that d/dt > 0, then according to Lenz’s law, I has to be in the direction shown in Fig. 6-2(a) in order for Bind to be in opposition to B(t). Consequently, terminal 2 would be at a higher potential than terminal 1, and V tr emf would have a negative value. However, if B(t) were to remain in the same direction but to decrease in magnitude, then d/dt would become negative, the current would have to reverse direction, and its induced field Bind would be in the same direction as B(t) so as to oppose the change (decrease) of B(t). In that case, V tr emf would be positive.
▶It is important to remember that Bind serves to oppose the change in B(t), and not necessarily B(t) itself. ◀ Despite the presence of the small opening between terminals 1 and 2 of the loop in Fig. 6-2(a), we shall treat the loop as a closed path with contour C. We do this in order to establish the link between B and the electric field E associated with the induced emf, V tr emf. Also, at any point along the loop, the field E is related to the current I flowing through the loop. For contour C, V tr emf is related to E by V tr emf = C E· dl.
(6.10) For N = 1 (a loop with one turn), equating Eqs. (6.8) and (6.10) gives C E· dl = − S ∂B ∂t · ds, (6.11) which is the integral form of Faraday’s law given in Table 6-1.
We should keep in mind that the direction of the contour C and the direction of ds are related by the right-hand rule.
By applying Stokes’s theorem to the left-hand side of Eq. (6.11), we have S (∇× × × E)· ds = − S ∂B ∂t · ds, (6.12) and in order for the two integrals to be equal for all possible choices of S, their integrands must be equal, which gives ∇× × × E = −∂B ∂t (Faraday’s law).
(6.13) ThisdifferentialformofFaraday’slawstatesthatatime-varying magnetic field induces an electric field E whose curl is equal to the negative of the time derivative of B. Even though the derivation leading to Faraday’s law started out by considering the field associated with a physical circuit, Eq. (6.13) applies at any point in space, whether or not a physical circuit exists at that point.
Example 6-1: Inductor in a Changing Magnetic Field An inductor is formed by winding N turns of a thin conducting wire into a circular loop of radius a. The inductor loop is in the x–y plane with its center at the origin, and connected to a resistor R, as shown in Fig. 6-3. In the presence of a magnetic field B = B0(ˆ y2+ ˆ z3) sin ωt, where ω is the angular frequency, find (a) the magnetic flux linking a single turn of the inductor, (b) the transformer emf, given that N = 10, B0 = 0.2 T, a = 10 cm, and ω = 103 rad/s, (c) the polarity of V tr emf at t = 0, and (d) the induced current in the circuit for R = 1 k (assume the wire resistance to be much smaller than R).
Vemf tr 1 R 2 I z y a B B N turns Figure 6-3 Circular loop with N turns in the x–y plane. The magnetic field is B = B0(ˆ y2 + ˆ z3) sin ωt (Example 6-1).
286 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS Solution: (a) The magnetic flux linking each turn of the inductor is = S B· ds = S [B0(ˆ y 2 + ˆ z 3) sin ωt]· ˆ z ds = 3πa2B0 sin ωt.
(b) To find V tr emf, we can apply Eq. (6.8) or we can apply the general expression given by Eq. (6.6) directly. The latter approach gives V tr emf = −N d dt = −d dt (3πNa2B0 sin ωt) = −3πNωa2B0 cos ωt.
For N = 10, a = 0.1 m, ω = 103 rad/s, and B0 = 0.2 T, V tr emf = −188.5 cos 103t (V).
(c) At t = 0, d/dt > 0 and V tr emf = −188.5 V. Since the flux is increasing, the current I must be in the direction shown in Fig. 6-3 in order to satisfy Lenz’s law. Consequently, point 2 is at a higher potential than point 1 and V tr emf = V1 −V2 = −188.5 (V).
(d) The current I is given by I = V2 −V1 R = 188.5 103 cos 103t = 0.19 cos 103t (A).
Exercise 6-1: For the loop shown in Fig. 6-3, what is V tr emf if B = ˆ yB0 cos ωt?
Answer: V tr emf = 0 because B is orthogonal to the loop’s surface normal ds. (See EM.) Exercise 6-2: Suppose that the loop of Example 6-1 is replaced with a 10-turn square loop centered at the origin and having 20 cm sides oriented parallel to the x and y axes. If B = ˆ zB0x2 cos 103t and B0 = 100 T, find the current in the circuit.
Answer: I = −133 sin 103t (mA). (See EM.) Example 6-2: Lenz’s Law Determine voltages V1 and V2 across the 2 and 4 resistors shown in Fig. 6-4. The loop is located in the x–y plane, its area is 4 m2, the magnetic flux density is B = −ˆ z0.3t (T), and the internal resistance of the wire may be ignored.
Solution: The flux flowing through the loop is = S B· ds = S (−ˆ z0.3t)· ˆ z ds = −0.3t × 4 = −1.2t (Wb), and the corresponding transformer emf is V tr emf = −d dt = 1.2 (V).
I y x V1 V2 4 Ω B 2 Ω Area = 4 m2 Figure 6-4 Circuit for Example 6-2.
6-2 STATIONARY LOOP IN A TIME-VARYING MAGNETIC FIELD 287 Module 6.1 Circular Loop inTime-varying Magnetic Field Faraday’s law of induction is demonstrated by simulating the current induced in a loop in response to the change in magnetic flux flowing through it.
Since the magnetic flux through the loop is along the −z direction (into the page) and increases in magnitude with time t, Lenz’s law states that the induced current I should be in a direction such that the magnetic flux density Bind it induces counteracts the direction of change of . Hence, I has to be in the direction shown in the circuit because the corresponding Bind is along the +z direction in the region inside the loop area.
This, in turn, means that V1 and V2 are positive voltages.
The total voltage of 1.2 V is distributed across two resistors in series. Consequently, I = V tr emf R1 + R2 = 1.2 2 + 4 = 0.2 A, and V1 = IR1 = 0.2 × 2 = 0.4 V, V2 = IR2 = 0.2 × 4 = 0.8 V.
Concept Question 6-1: Explain Faraday’s law and the function of Lenz’s law.
Concept Question 6-2: Under what circum-stances is the net voltage around a closed loop equal to zero?
Concept Question 6-3: Suppose the magnetic flux density linking the loop of Fig. 6-4 (Example 6-2) is given by B = −ˆ z 0.3e−t (T). What would the direction of the current be, relative to that shown in Fig. 6-4, for t ≥0?
288 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS 6-3 The Ideal Transformer The transformer shown in Fig. 6-5(a) consists of two coils wound around a common magnetic core. The primary coil has N1 turns and is connected to an ac voltage source V1(t). The secondary coil has N2 turns and is connected to a load resistor RL. In an ideal transformer the core has infinite permeability (μ = ∞), and the magnetic flux is confined within the core.
▶The directions of the currents flowing in the two coils, I1 and I2, are defined such that, when I1 and I2 are both positive, the flux generated by I2 is opposite to that generated by I1.
The transformer gets its name from the fact that it transforms currents, voltages, and impedances between its primary and secondary circuits, and vice versa. ◀ (a) Magnetic core (b) V1(t) V2(t) N1 N2 Ф I1 I2 RL Ф I2 I1 V1(t) V2(t) RL Ф Ф N1 N2 Figure 6-5 In a transformer, the directions of I1 and I2 are such that the flux generated by one of them is opposite to that generated by the other. The direction of the secondary winding in (b) is opposite to that in (a), and so are the direction of I2 and the polarity of V2.
On the primary side of the transformer, the voltage source V1 generates current I1 in the primary coil, which establishes a flux in the magnetic core. The flux and voltage V1 are related by Faraday’s law: V1 = −N1 d dt .
(6.14) A similar relation holds true on the secondary side: V2 = −N2 d dt .
(6.15) The combination of Eqs. (6.14) and (6.15) gives V1 V2 = N1 N2 .
(6.16) In an ideal lossless transformer, all the instantaneous power supplied by the source connected to the primary coil is delivered to the load on the secondary side. Thus, no power is lost in the core, and P1 = P2.
(6.17) Since P1 = I1V1 and P2 = I2V2, and in view of Eq. (6.16), we have I1 I2 = N2 N1 .
(6.18) Thus, whereas the ratio of the voltages given by Eq. (6.16) is proportional to the corresponding turns ratio, the ratio of the currents is equal to the inverse of the turns ratio.
If N1/N2 = 0.1, V2 of the secondary circuit would be 10 times V1 of the primary circuit, but I2 would be only I1/10.
The transformer shown in Fig. 6-5(b) is identical to that in Fig. 6-5(a) except for the direction of the windings of the secondary coil. Because of this change, the direction of I2 and the polarity of V2 in Fig. 6-5(b) are the reverse of those in Fig. 6-5(a).
The voltage and current in the secondary circuit in Fig. 6-5(a) are related by V2 = I2RL. To the input circuit, the transformer may be represented by an equivalent input resistance Rin, as shown in Fig. 6-6, defined as Rin = V1 I1 .
(6.19) 6-4 MOVING CONDUCTOR IN A STATIC MAGNETIC FIELD 289 V1(t) I1(t) Rin Figure 6-6 Equivalent circuit for the primary side of the transformer.
Use of Eqs. (6.16) and (6.18) gives Rin = V2 I2 N1 N2 2 = N1 N2 2 RL.
(6.20) When the load is an impedance ZL and V1 is a sinusoidal source, the phasor-domain equivalent of Eq. (6.20) is Zin = N1 N2 2 ZL.
(6.21) 6-4 Moving Conductor in a Static Magnetic Field Consider a wire of length l moving across a static magnetic field B = ˆ zB0 with constant velocity u, as shown in Fig. 6-7. The conducting wire contains free electrons. From Eq. (5.3), the magnetic force Fm acting on a particle with charge q moving with velocity u in a magnetic field B is Fm = q(u × × × B).
(6.22) This magnetic force is equivalent to the electrical force that would be exerted on the particle by the electric field Em given by Em = Fm q = u × × × B.
(6.23) The field Em generated by the motion of the charged particle is called a motional electric field and is in the direction perpendicular to the plane containing u and B. For the wire shown in Fig. 6-7, Em is along −ˆ y. The magnetic force acting on the (negatively charged) electrons in the wire causes them to drift in the direction of −Em; that is, toward the wire end u u Em 1 2 l Moving wire y x z Magnetic field line (out of the page) B B Figure 6-7 Conducting wire moving with velocity u in a static magnetic field.
labeled 1 in Fig. 6-7. This, in turn, induces a voltage difference between ends 1 and 2, with end 2 being at the higher potential.
The induced voltage is called a motional emf, V m emf, and is defined as the line integral of Em between ends 2 and 1 of the wire, V m emf = V12 = 1 2 Em · dl = 1 2 (u × × × B)· dl.
(6.24) For the conducting wire, u × × × B = ˆ xu × × × ˆ zB0 = −ˆ yuB0 and dl = ˆ y dl. Hence, V m emf = V12 = −uB0l.
(6.25) In general, if any segment of a closed circuit with contour C moves with a velocity u across a static magnetic field B, then the induced motional emf is given by V m emf = C (u × × × B)· dl (motional emf).
(6.26) ▶Only those segments of the circuit that cross magnetic field lines contribute to V m emf. ◀ 290 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS y I R x z Magnetic field B dl u u Vemf m 1 2 4 3 x = 0 l x0 Figure 6-8 Sliding bar with velocity u in a magnetic field that increases linearly with x; that is, B = ˆ zB0x (Example 6-3).
Example 6-3: Sliding Bar The rectangular loop shown in Fig. 6-8 has a constant width l, but its length x0 increases with time as a conducting bar slides with uniform velocity u in a static magnetic field B = ˆ zB0x.
Note that B increases linearly with x. The bar starts from x = 0 at t = 0. Find the motional emf between terminals 1 and 2 and the current I flowing through the resistor R. Assume that the loop resistance Ri ≪R.
Solution: This problem can be solved by using the motional emf expression given by Eq. (6.26) or by applying the general formulaofFaraday’slaw. Wenowshowthatthetwoapproaches yield the same result.
The sliding bar, being the only part of the circuit that crosses the lines of the field B, is the only part of contour 2341 that contributes to V m emf. Hence, at x = x0, for example, V m emf = V12 = V43 = 4 3 (u × × × B)· dl = 4 3 (ˆ xu × × × ˆ zB0x0)· ˆ y dl = −uB0x0l.
The length of the loop is related to u by x0 = ut. Hence, V m emf = −B0u2lt (V).
(6.27) Since B is static, V tr emf = 0 and Vemf = V m emf only. To verify that the same result can be obtained by the general form of Faraday’s law, we evaluate the flux through the surface of the loop. Thus, = S B· ds = S (ˆ zB0x)· ˆ z dx dy = B0l x0 0 x dx = B0lx2 0 2 .
(6.28) Substituting x0 = ut in Eq. (6.28) and then evaluating the negative of the derivative of the flux with respect to time gives Vemf = −d dt = −d dt B0lu2t2 2 = −B0u2lt (V), (6.29) which is identical with Eq. (6.27). Since V12 is negative, the current I = B0u2lt/R flows in the direction shown in Fig. 6-8.
Example 6-4: Moving Loop The rectangular loop shown in Fig. 6-9 is situated in the x–y plane and moves away from the origin with velocity u = ˆ y5 (m/s) in a magnetic field given by B(y) = ˆ z 0.2e−0.1y (T).
6-4 MOVING CONDUCTOR IN A STATIC MAGNETIC FIELD 291 1 u u 4 2 3 I R 0.5 m V12 V43 y1 = 2 m y2 = 2.5 m l = 2 m x z y Figure 6-9 Moving loop of Example 6-4.
If R = 5 , find the current I at the instant that the loop sides are at y1 = 2 m and y2 = 2.5 m. The loop resistance may be ignored.
Solution: Since u× × × B is along ˆ x, voltages are induced across only the sides oriented along ˆ x, namely the sides linking points 1 and 2 and points 3 and 4. Had B been uniform, the induced voltages would have been the same and the net voltage across the resistor would have been zero. In the present case, however, B decreases exponentially with y, thereby assuming a different value over side 1-2 than over side 3-4. Side 1-2 is at y1 = 2 m, and the corresponding magnetic field is B(y1) = ˆ z 0.2e−0.1y1 = ˆ z 0.2e−0.2 (T).
The induced voltage V12 is then given by V12 = 1 2 [u × × × B(y1)] · dl = −l/2 l/2 (ˆ y5 × × × ˆ z0.2e−0.2)· ˆ x dx = −e−0.2l = −2e−0.2 = −1.637 (V).
Similarly, V43 = −u B(y2) l = −5 × 0.2e−0.25 × 2 = −1.558 (V).
Consequently, the current is in the direction shown in the figure and its magnitude is I = V43 −V12 R = 0.079 5 = 15.8 (mA).
u B B B B I = 10 A 10 cm 1 2 30 cm z r B B Wire Metal rod Figure 6-10 Moving rod of Example 6-5.
Example 6-5: Moving Rod Next to a Wire The wire shown in Fig. 6-10 carries a current I = 10 A. A 30 cm long metal rod moves with a constant velocity u = ˆ z5 m/s.
Find V12.
Solution: The current I induces a magnetic field B = ˆ φ φ φ μ0I 2πr , where r is the radial distance from the wire and the direction of ˆ φ φ φ is into the page on the rod side of the wire. The movement of the rod in the presence of the field B induces a motional emf given by V12 = 10 cm 40 cm (u × × × B)· dl = 10 cm 40 cm ˆ z 5 × × × ˆ φ φ φ μ0I 2πr · ˆ r dr = −5μ0I 2π 10 cm 40 cm dr r = −5 × 4π × 10−7 × 10 2π × ln 10 40 = 13.9 (μV).
292 TECHNOLOGY BRIEF 12: EMF SENSORS Technology Brief 12: EMF Sensors An electromotive force (emf) sensor is a device that can generate an induced voltage in response to an external stimulus. Three types of emf sensors are profiled in this technical brief: the piezoelectric transducer, the Faraday magnetic flux sensor, and the thermocouple.
Piezoelectric Transducers ▶Piezoelectricity is the property exhibited by certain crystals, such as quartz, that become electrically polarized when the crystal is subjected to mechanical pressure, thereby inducing a voltage across it. ◀ The crystal consists of polar domains represented by equivalent dipoles (Fig. TF12-1). Under the absence of an external force, the polar domains are randomly oriented throughout the material, but when compressive or tensile (stretching) stress is applied to the crystal, the polar domains align themselves along one of the principal axes of the crystal, leading to a net polarization (electric charge) at the crystal surfaces. Compression and stretching generate voltages of opposite polarity. The piezoelectric effect (piezein means to press or squeeze in Greek) was discovered by the Curie brothers, Pierre and Paul-Jacques, in 1880, and a year later, Lippmann predicted the converse property, namely that, if subjected to an electric field, the crystal would change in shape.
▶The piezoelectric effect is a reversible (bidirectional) electromechanical process; application of force induces a voltage across the crystal, and conversely, application of a voltage changes the shape of the crystal. ◀ Piezoelectric crystals are used in microphones to convert mechanical vibrations (of the crystal surface) caused by acoustic waves into a corresponding electrical signal, and the converse process is used in loudspeakers to convert electrical signals into sound. In addition to having stiffness values comparable to that of steel, some piezoelectric materials exhibit very high sensitivity to the force applied upon them, with excellent linearity over a wide dynamic range.
They can be used to measure surface deformations as small as nanometers (10−9 m), making them particularly attractive as positioning sensors in scanning tunneling microscopes.
As accelerometers, they can measure acceleration levels as low as 10−4 g to as high as 100 g (where g is the acceleration due to gravity). Piezoelectric crystals and ceramics are used in cigarette lighters and gas grills as spark generators, in clocks and electronic circuitry as precision oscillators, in medical ultrasound diagnostic equipment as transducers (Fig. TF12-2), and in numerous other applications.
(a) No force (b) Compressed crystal Vemf > 0 F + _ + _ + _ + _ + _ + _ (c) Stretched crystal F Vemf < 0 + _ + _ + _ + _ + _ + _ Vemf = 0 F = 0 Dipole + _ Figure TF12-1 Response of a piezoelectric crystal to an applied force.
TECHNOLOGY BRIEF 12: EMF SENSORS 293 Wear plate Case Electrodes Piezoelectric element Backing material Epoxy potting Ground wire Signal wire Coaxial cable connector Figure TF12-2 The ultrasonic transducer uses piezoelectric crystals.
Vemf Conducting loop Magnet x l u + _ N S Figure TF12-3 In a Faraday accelerometer, the induced emf is directly proportional to the velocity of the loop (into and out of the magnet’s cavity).
Faraday Magnetic Flux Sensor According to Faraday’s law [Eq. (6.6)], the emf voltage induced across the terminals of a conducting loop is directly proportional to the time rate of change of the magnetic flux passing through the loop. For the configuration in Fig.TF12-3, Vemf = −uB0l, where u = dx/dt is the velocity of the loop (into or out of the magnet’s cavity), with the direction of u defined as positive when the loop is moving inward into the cavity, B0 is the magnetic field of the magnet, and l is the loop width. With B0 and l being constant, the variation of Vemf(t) with time t becomes a direct indicator of the time variation of u(t). The time derivative of u(t) provides the acceleration a(t).
Thermocouple In 1821, Thomas Seebeck discovered that when a junction made of two different conducting materials, such as bismuth and copper, is heated, it generates a thermally induced emf, which we now call the Seebeck potential VS (Fig. TF12-4). When connected to a resistor, a current flows through the resistor, given by I = VS/R.
This feature was advanced by A. C. Becquerel in 1826 as a means to measure the unknown temperature T2 of a junction relative to a temperature T1 of a (cold) reference junction. Today, such a generator of thermoelectricity is called a thermocouple. Initially, an ice bath was used to maintain T1 at 0◦C, but in today’s temperature sensor designs, an artificial cold junction is used instead. The artificial junction is an electric circuit that generates a potential equal to that expected from a reference junction at temperature T1.
Vs R I + _ T1 Cold reference junction T2 Copper Bismuth Measurement junction FigureTF12-4 Principle of the thermocouple.
294 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS Concept Question 6-4: Suppose that no friction is involved in sliding the conducting bar of Fig. 6-8 and that the horizontal arms of the circuit are very long.
Hence, if the bar is given an initial push, it should continue moving at a constant velocity, and its movement generates electrical energy in the form of an induced emf, indefinitely. Is this a valid argument? If not, why not? Can we generate electrical energy without having to supply an equal amount of energy by other means?
Concept Question 6-5: Isthecurrentflowingintherod of Fig. 6-10 a steady current? Examine the force on a charge q at ends 1 and 2 and compare.
Exercise 6-3: For the moving loop of Fig. 6-9, find I when the loop sides are at y1 = 4 m and y2 = 4.5 m.
Also, reverse the direction of motion such that u = −ˆ y5 (m/s).
Answer: I = −13 (mA). (See EM.) Exercise 6-4: Suppose that we turn the loop of Fig. 6-9 so that its surface is parallel to the x–z plane. What would I be in that case?
Answer: I = 0. (See EM.) 6-5 The Electromagnetic Generator The electromagnetic generator is the converse of the electromagnetic motor. The principles of operation of both instruments may be explained with the help of Fig. 6-11. A permanent magnet is used to produce a static magnetic field B in the slot between its two poles. When a current is passed through the conducting loop, as depicted in Fig. 6-11(a), the current flows in opposite directions in segments 1–2 and 3–4 of the loop. The induced magnetic forces on the two segments are also opposite, resulting in a torque that causes the loop to rotate about its axis. Thus, in a motor, electrical energy supplied by a (b) ac generator (a) ac motor R I I Vemf m N Magnet 1 2 3 4 Axis of rotation ω V(t) R I I S B N Magnet 1 2 3 4 Axis of rotation ω S B Figure 6-11 Principles of the ac motor and the ac generator.
In (a) the magnetic torque on the wires causes the loop to rotate, and in (b) the rotating loop generates an emf.
voltage source is converted into mechanical energy in the form of a rotating loop, which can be coupled to pulleys, gears, or other movable objects.
If, instead of passing a current through the loop to make it turn, the loop is made to rotate by an external force, the movement of the loop in the magnetic field produces a motional emf, V m emf, as shown in Fig. 6-11(b). Hence, the motor has become a generator, and mechanical energy is being converted into electrical energy.
6-5 THE ELECTROMAGNETIC GENERATOR 295 dl l w y z x n 1 4 3 2 B Vemf m Brushes Slip rings Loop surface normal ˆ α ω Figure 6-12 A loop rotating in a magnetic field induces an emf.
Let us examine the operation of the electromagnetic generator in more detail using the coordinate system shown in Fig. 6-12. The magnetic field is B = ˆ zB0, (6.30) and the axis of rotation of the conducting loop is along the x axis. Segments 1–2 and 3–4 of the loop are of length l each, and both cross the magnetic flux lines as the loop rotates. The other two segments are each of width w, and neither crosses the B lines when the loop rotates. Hence, only segments 1–2 and 3–4 contribute to the generation of the motional emf, V m emf.
As the loop rotates with an angular velocity ω about its own axis, segment 1–2 moves with velocity u given by u = ˆ nω w 2 , (6.31) where ˆ n, the surface normal to the loop, makes an angle α with the z axis. Hence, ˆ n × × × ˆ z = ˆ x sin α.
(6.32) Segment 3–4 moves with velocity −u.
Application of Eq. (6.26), consistent with our choice of ˆ n, gives V m emf = V14 = 1 2 (u × × × B)· dl + 3 4 (u × × × B)· dl = l/2 −l/2 ˆ nωw 2 × × × ˆ zB0 · ˆ x dx + −l/2 l/2 −ˆ nωw 2 × × × ˆ zB0 · ˆ x dx.
(6.33) Using Eq. (6.32) in Eq. (6.33), we obtain the result V m emf = wlωB0 sin α = AωB0 sin α, (6.34) where A = wl is the surface area of the loop. The angle α is related to ω by α = ωt + C0, (6.35) where C0 is a constant determined by initial conditions. For example, if α = 0 at t = 0, then C0 = 0. In general, V m emf = AωB0 sin(ωt + C0) (V).
(6.36) This same result can also be obtained by applying the general form of Faraday’s law given by Eq. (6.6). The flux linking the surface of the loop is = S B· ds = S ˆ zB0 · ˆ n ds = B0A cos α = B0A cos(ωt + C0), (6.37) and Vemf = −d dt = −d dt [B0A cos(ωt + C0)] = AωB0 sin(ωt + C0), (6.38) which is identical with the result given by Eq. (6.36).
296 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS Module 6.2 Rotating Wire Loop in Constant Magnetic Field The principle of the electromagnetic generator is demonstrated by a rectangular loop rotating in the presence of a magnetic field.
▶The voltage induced by the rotating loop is sinusoidal in time with an angular frequency ω equal to that of the rotating loop, and its amplitude is equal to the product of the surface area of the loop, the magnitude of the magnetic field generated by the magnet, and the angular frequency ω. ◀ Concept Question 6-6: Contrast the operation of an ac motor with that of an ac generator.
Concept Question 6-7: The rotating loop of Fig. 6-12 had a single turn. What would be the emf generated by a loop with 10 turns?
Concept Question 6-8: The magnetic flux linking the loop shown in Fig. 6-12 is maximum when α = 0 (loop in x–y plane), and yet according to Eq. (6.34), the induced emf is zero when α = 0. Conversely, when α = 90◦, the flux linking the loop is zero, but V m emf is at a maximum. Is this consistent with your expectations? Why?
6-6 Moving Conductor in a Time-Varying Magnetic Field For the general case of a single-turn conducting loop moving in a time-varying magnetic field, the induced emf is the sum of a transformer component and a motional component. Thus, the 6-7 DISPLACEMENT CURRENT 297 sum of Eqs. (6.8) and (6.26) gives Vemf = V tr emf + V m emf = C E· dl = − S ∂B ∂t · ds + C (u × × × B)· dl.
(6.39) Vemf is also given by the general expression of Faraday’s law: Vemf = −d dt = −d dt S B· ds (total emf).
(6.40) In fact, it can be shown mathematically that the right-hand side of Eq. (6.39) is equivalent to the right-hand side of Eq. (6.40).
For a particular problem, the choice between using Eq. (6.39) or Eq. (6.40) is usually made on the basis of which is the easier to apply. In either case, for an N-turn loop, the right-hand sides of Eqs. (6.39) and (6.40) should be multiplied by N.
Example 6-6: Electromagnetic Generator Find the induced voltage when the rotating loop of the electromagnetic generator of Section 6-5 is in a magnetic field B = ˆ zB0 cos ωt. Assume that α = 0 at t = 0.
Solution: The flux is given by Eq. (6.37) with B0 replaced with B0 cos ωt. Thus, = B0A cos2 ωt, and Vemf = −∂ ∂t = −∂ ∂t (B0A cos2 ωt) = 2B0Aω cos ωt sin ωt = B0Aω sin 2ωt.
6-7 Displacement Current Amp ere’s law in differential form is given by ∇× × × H = J + ∂D ∂t (Ampere’s law).
(6.41) Integrating both sides of Eq. (6.41) over an arbitrary open surface S with contour C, we have S (∇× × × H)· ds = S J· ds + S ∂D ∂t · ds.
(6.42) The surface integral of J equals the conduction current Ic flowing through S, and the surface integral of ∇× × × H can be converted into a line integral of H over the contour C bounding C by invoking Stokes’s theorem. Hence, C H· dl = Ic + S ∂D ∂t · ds.
(6.43) (Amp ere’s law) The second term on the right-hand side of Eq. (6.43) of course has the same unit (amperes) as the current Ic, and because it is proportional to the time derivative of the electric flux density D, which is also called the electric displacement, it is called the displacement current Id. That is, Id = S Jd · ds = S ∂D ∂t · ds, (6.44) where Jd = ∂D/∂t represents a displacement current density.
In view of Eq. (6.44), C H· dl = Ic + Id = I, (6.45) where I is the total current. In electrostatics, ∂D/∂t = 0 and therefore Id = 0 and I = Ic.
The concept of displacement current was first introduced in 1873 by James Clerk Maxwell when he formulated the unified theory of electricity and magnetism under time-varying conditions.
298 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS E + + + + + + + – – – – – – – I2d Vs(t) I1 = I1c y Imaginary surface S2 Imaginary surface S1 Figure 6-13 The displacement current I2d in the insulating material of the capacitor is equal to the conducting current I1c in the wire.
The parallel-plate capacitor is commonly used as an example to illustrate the physical meaning of the displacement current Id. The simple circuit shown in Fig. 6-13 consists of a capacitor and an ac source with voltage Vs(t) given by Vs(t) = V0 cos ωt (V).
(6.46) According to Eq. (6.45), the total current flowing through any surface consists, in general, of a conduction current Ic and a displacement current Id.
Let us find Ic and Id through each of the following two imaginary surfaces: (1) the cross section of the conducting wire, S1, and (2) the cross section of the capacitor S2 (Fig. 6-13).
We denote the conduction and displacement currents in the wire as I1c and I1d and those through the capacitor as I2c and I2d.
In the perfectly conducting wire, D = E = 0; hence, Eq. (6.44) gives I1d = 0. As for I1c, we know from circuit theory that it is related to the voltage across the capacitor VC by I1c = C dVC dt = C d dt (V0 cos ωt) = −CV0ω sin ωt, (6.47) where we used the fact that VC = Vs(t). With I1d = 0, the total current in the wire is simply I1 = I1c = −CV0ω sin ωt.
In the perfect dielectric with permittivity ϵ between the capacitor plates, σ = 0. Hence, I2c = 0 because no conduction current exists there.
To determine I2d, we need to apply Eq. (6.44).
From Example 4-11, the electric field E in the dielectric spacing is related to the voltage Vc across its plates by E = ˆ y Vc d = ˆ y V0 d cos ωt, (6.48) where d is the spacing between the plates and ˆ y is the direction from the higher-potential plate toward the lower-potential plate at t = 0. The displacement current I2d is obtained by applying Eq. (6.44) with ds = ˆ y ds: I2d = S ∂D ∂t · ds = A ∂ ∂t ˆ y ϵV0 d cos ωt ·(ˆ y ds) = −ϵA d V0ω sin ωt = −CV0ω sin ωt, (6.49) where we used the relation C = ϵA/d for the capacitance of the parallel-plate capacitor with plate area A. The expression for I2d in the dielectric region between the conducting plates is identical with that given by Eq. (6.47) for the conduction current I1c in the wire. The fact that these two currents are equal ensures the continuity of total current flow through the circuit.
▶Even though the displacement current does not transport free charges, it nonetheless behaves like a real current. ◀ In the capacitor example, we treated the wire as a perfect conductor, and we assumed that the space between the capacitor plates was filled with a perfect dielectric. If the wire has a finite conductivity σw, then D in the wire would not be zero, and thereforethecurrentI1 wouldconsistofaconductioncurrentI1c 6-8 BOUNDARY CONDITIONS FOR ELECTROMAGNETICS 299 as well as a displacement current I1d; that is, I1 = I1c +I1d. By the same token, if the dielectric spacing material has a nonzero conductivity σd, then free charges would flow between the two plates, and I2c would not be zero. In that case, the total current flowing through the capacitor would be I2 = I2c + I2d.
No matter the circumstances, the total capacitor current remains equal to the total current in the wire. That is, I1 = I2.
Example 6-7: Displacement Current Density The conduction current flowing through a wire with conductiv-ity σ = 2 × 107 S/m and relative permittivity ϵr = 1 is given by Ic = 2 sin ωt (mA). If ω = 109 rad/s, find the displacement current.
Solution: The conduction current Ic = JA = σEA, where A is the cross section of the wire. Hence, E = Ic σA = 2 × 10−3 sin ωt 2 × 107A = 1 × 10−10 A sin ωt (V/m).
Application of Eq. (6.44), with D = ϵE, leads to Id = JdA = ϵA ∂E ∂t = ϵA ∂ ∂t 1 × 10−10 A sin ωt = ϵω × 10−10 cos ωt = 0.885 × 10−12 cos ωt (A), where we used ω = 109 rad/s and ϵ = ϵ0 = 8.85×10−12 F/m.
Note that Ic and Id are in phase quadrature (90◦phase shift between them). Also, Id is about nine orders of magnitude smaller than Ic, which is why the displacement current usually is ignored in good conductors.
Exercise 6-5: A poor conductor is characterized by a conductivity σ = 100 (S/m) and permittivity ϵ = 4ϵ0.
At what angular frequency ω is the amplitude of the conduction current density J equal to the amplitude of the displacement current density Jd?
Answer: ω = 2.82 × 1012 (rad/s). (See EM.) 6-8 Boundary Conditions for Electromagnetics In Chapters 4 and 5 we applied the integral form of Maxwell’s equations under static conditions to obtain boundary conditions applicable to the tangential and normal components of E, D, B, and H on interfaces between contiguous media (Section 4-8 for E and D and in Section 5-6 for B and H). In the dynamic case, Maxwell’s equations (Table 6-1) include two new terms not accounted for in electrostatics and magnetostatics, namely, ∂B/∂t in Faraday’s law and ∂D/∂t in Ampere’s law.
▶ Nevertheless, the boundary conditions derived previously for electrostatic and magnetostatic fields remain valid for time-varying fields as well. ◀ This is because, if we were to apply the procedures outlined in the above-referenced sections for time-varying fields, we would find that the combination of the aforementioned terms vanish as the areas of the rectangular loops in Figs. 4-18 and 5-24 are made to approach zero.
The combined set of electromagnetic boundary conditions is summarized in Table 6-2.
Concept Question 6-9: When conduction current flows through a material, a certain number of charges enter the material on one end and an equal number leave on the other end. What’s the situation like for the displacement current through a perfect dielectric?
Concept Question 6-10: Verify that the integral form of Amp ere’s law given by Eq. (6.43) leads to the boundary condition that the tangential component of H is continuous across the boundary between two dielectric media.
6-9 Charge-Current Continuity Relation Under static conditions, the charge density ρv and the current density J at a given point in a material are totally independent of one another.
This is no longer true in the time-varying case.
To show the connection between ρv and J, we start by considering an arbitrary volume v bounded by a closed surface S (Fig. 6-14).
The net positive charge contained in v is Q.
Since, according to the law of conservation of 300 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS Table 6-2 Boundary conditions for the electric and magnetic fields.
Field Components General Form Medium 1 Dielectric Medium 2 Dielectric Medium 1 Dielectric Medium 2 Conductor Tangential E ˆ n2 × × × (E1 −E2) = 0 E1t = E2t E1t = E2t = 0 Normal D ˆ n2 ·(D1 −D2) = ρs D1n −D2n = ρs D1n = ρs D2n = 0 Tangential H ˆ n2 × × × (H1 −H2) = Js H1t = H2t H1t = Js H2t = 0 Normal B ˆ n2 · (B1 −B2) = 0 B1n = B2n B1n = B2n = 0 Notes: (1) ρs is the surface charge density at the boundary; (2) Js is the surface current density at the boundary; (3) normal componentsofallfieldsarealong ˆ n2, theoutwardunitvectorofmedium2; (4)E1t = E2t impliesthatthetangentialcomponents are equal in magnitude and parallel in direction; (5) direction of Js is orthogonal to (H1 −H2).
Module 6.3 Displacement Current Observe the displacement current through a parallel plate capacitor.
electric charge (Section 1-3.2), charge can neither be created nor destroyed, the only way Q can increase is as a result of a net inward flow of positive charge into the volume v. By the same token, for Q to decrease there has to be a net outward flow of charge fromv. The inward and outward flow of charge constitute currents flowing across the surface S into and out of v, respectively.
We define I as the net current flowing across S out of v. Accordingly, I is equal to the negative rate of change of Q: I = −dQ dt = −d dt v ρv dv, (6.50) where ρv is the volume charge density in v. According to Eq. (4.12), the current I is also defined as the outward flux of 6-9 CHARGE-CURRENT CONTINUITY RELATION 301 J J J J Charge density ρv S encloses ν ν Figure 6-14 The total current flowing out of a volume v is equal to the flux of the current density J through the surface S, which in turn is equal to the rate of decrease of the charge enclosed in v.
the current density J through the surface S. Hence, S J· ds = −d dt v ρv dv.
(6.51) By applying the divergence theorem given by Eq. (3.98), we can convert the surface integral of J into a volume integral of its divergence ∇· J, which then gives S J· ds = v ∇· J dv = −d dt v ρv dv.
(6.52) For a stationary volume v, the time derivative operates on ρv only. Hence, we can move it inside the integral and express it as a partial derivative of ρv: v ∇· J dv = − v ∂ρv ∂t dv.
(6.53) In order for the volume integrals on both sides of Eq. (6.53) to be equal for any volume v, their integrands have to be equal at every point within v. Hence, ∇· J = −∂ρv ∂t , (6.54) which is known as the charge-current continuity relation, or simply the charge continuity equation.
If the volume charge density within an elemental volume v (such as a small cylinder) is not a function of time (i.e., ∂ρv/∂t = 0), it means that the net current flowing out of v is zero or, equivalently, that the current flowing into v is equal to the current flowing out of it. In this case, Eq. (6.54) implies ∇· J = 0, (6.55) and its integral-form equivalent [from Eq. (6.51)] is S J· ds = 0 (Kirchhoff’s current law).
(6.56) Let us examine the meaning of Eq. (6.56) by considering a junction (or node) connecting two or more branches in an electric circuit.
No matter how small, the junction has a volume v enclosed by a surface S. The junction shown in Fig. 6-15 has been drawn as a cube, and its dimensions have been artificially enlarged to facilitate the present discussion.
The junction has six faces (surfaces), which collectively constitute the surface S associated with the closed-surface integration given by Eq. (6.56). For each face, the integration represents the current flowing out through that face.
Thus, Eq. (6.56) can be cast as i Ii = 0 (Kirchhoff’s current law), (6.57) I1 I2 I3 Figure 6-15 Kirchhoff’s current law states that the algebraic sum of all the currents flowing out of a junction is zero.
302 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS where Ii is the current flowing outward through the ith face.
For the junction of Fig. 6-15, Eq. (6.57) translates into (I1 + I2 + I3) = 0. In its general form, Eq. (6.57) is an expression of Kirchhoff’s current law, which states that in an electric circuit the sum of all the currents flowing out of a junction is zero.
6-10 Free-Charge Dissipation in a Conductor We stated earlier that current flow in a conductor is realized by the movement of loosely attached electrons under the influence of an externally applied electric field. These electrons, however, are not excess charges; their charge is balanced by an equal amount of positive charge in the atoms’ nuclei. In other words, the conductor material is electrically neutral, and the net charge density in the conductor is zero (ρv = 0). What happens then if an excess free charge q is introduced at some interior point in a conductor? The excess charge gives rise to an electric field, which forces the charges of the host material nearest to the excess charge to rearrange their locations, which in turn cause other charges to move, and so on. The process continues until neutrality is reestablished in the conductor material and a charge equal to q resides on the conductor’s surface.
How fast does the excess charge dissipate? To answer this question, let us introduce a volume charge density ρvo at the interior of a conductor and then find out the rate at which it decays down to zero. From Eq. (6.54), the continuity equation is given by ∇· J = −∂ρv ∂t .
(6.58) In a conductor, the point form of Ohm’s law, given by Eq. (4.63), states that J = σE. Hence, σ∇· E = −∂ρv ∂t .
(6.59) Next, we use Eq. (6.1), ∇· E = ρv/ϵ, to obtain the partial differential equation ∂ρv ∂t + σ ϵ ρv = 0.
(6.60) Given that ρv = ρvo at t = 0, the solution of Eq. (6.60) is ρv(t) = ρvoe−(σ/ϵ)t = ρvoe−t/τr (C/m3), (6.61) where τr = ϵ/σ is called the relaxation time constant. We see from Eq. (6.61) that the initial excess charge ρvo decays exponentially at a rate τr.
At t = τr, the initial charge ρvo will have decayed to 1/e ≈37% of its initial value, and at t = 3τr, it will have decayed to e−3 ≈5% of its initial value at t = 0. For copper, with ϵ ≈ϵ0 = 8.854 × 10−12 F/m and σ = 5.8 × 107 S/m, τr = 1.53 × 10−19 s. Thus, the charge dissipation process in a conductor is extremely fast. In contrast, the decay rate is very slow in a good insulator. For a material like mica with ϵ = 6ϵ0 and σ = 10−15 S/m, τr = 5.31×104 s, or approximately 14.8 hours.
Concept Question 6-11: Explain how the charge con-tinuity equation leads to Kirchhoff’s current law.
Concept Question 6-12: How long is the relaxation time constant for charge dissipation in a perfect conductor? In a perfect dielectric?
Exercise 6-6: Determine (a) the relaxation time constant and (b) the time it takes for a charge density to decay to 1% of its initial value in quartz, given that ϵr = 5 and σ = 10−17 S/m.
Answer: (a) τr = 51.2 days, (b) 236 days. (See EM.) 6-11 Electromagnetic Potentials Our discussion of Faraday’s and Ampere’s laws revealed two aspects of the link between time-varying electric and magnetic fields. We now examine the implications of this interconnection on the electric scalar potential V and the vector magnetic potential A.
In the static case, Faraday’s law reduces to ∇× × × E = 0 (static case), (6.62) which states that the electrostatic field E is conservative.
According to the rules of vector calculus, if a vector field E is conservative, it can be expressed as the gradient of a scalar.
Hence, in Chapter 4 we defined E as E = −∇V (electrostatics).
(6.63) 6-11 ELECTROMAGNETIC POTENTIALS 303 In the dynamic case, Faraday’s law is ∇× × × E = −∂B ∂t .
(6.64) In view of the relation B = ∇× × ×A, Eq. (6.64) can be expressed as ∇× × × E = −∂ ∂t (∇× × × A), (6.65) which can be rewritten as ∇× × × E + ∂A ∂t = 0 (dynamic case).
(6.66) Let us for the moment define E′ = E + ∂A ∂t .
(6.67) Using this definition, Eq. (6.66) becomes ∇× × × E′ = 0.
(6.68) Following the same logic that led to Eq. (6.63) from Eq. (6.62), we define E′ = −∇V.
(6.69) Upon substituting Eq. (6.67) for E′ in Eq. (6.69) and then solving for E, we have E = −∇V −∂A ∂t (dynamic case).
(6.70) Equation (6.70) reduces to Eq. (6.63) in the static case.
When the scalar potential V and the vector potential A are known, E can be obtained from Eq. (6.70), and B can be obtained from B = ∇× × × A.
(6.71) Next we examine the relations between the potentials, V andA, and their sources, the charge and current distributions ρv and J, in the time-varying case.
6-11.1 Retarded Potentials Consider the situation depicted in Fig. 6-16.
A charge distribution ρv exists over a volume v ′ embedded in a perfect dielectric with permittivity ϵ.
Were this a static charge distribution, then from Eq. (4.48a), the electric potential V (R) at an observation point in space specified by the position vector R would be V (R) = 1 4πϵ v ′ ρv(Ri) R′ dv ′, (6.72) where Ri denotes the position vector of an elemental volume v ′ containing charge density ρv(Ri), and R′ = |R−Ri| is the distance between v ′ and the observation point. If the charge Ri Charge distribution ρv R' R V(R) z x y ∆ν' ν' Figure 6-16 Electric potential V (R) due to a charge distribution ρv over a volume v ′.
304 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS distribution is time-varying, we may be tempted to rewrite Eq. (6.72) for the dynamic case as V (R, t) = 1 4πϵ v ′ ρv(Ri, t) R′ dv ′, (6.73) but such a form does not account for “reaction time.” If V1 is the potential due to a certain distribution ρv1, and if ρv1 were to suddenly change to ρv2, it will take a finite amount of time before V1 a distance R′ away changes to V2. In other words, V (R, t) cannot change instantaneously. The delay time is equal to t′ = R′/up, where up is the velocity of propagation in the medium between the charge distribution and the observation point. Thus, V (R, t) at time t corresponds to ρv at an earlier time, that is, (t −t′). Hence, Eq. (6.73) should be rewritten as V (R, t) = 1 4πϵ v ′ ρv(Ri, t −R′/up) R′ dv ′ (V), (6.74) and V (R, t) is appropriately called the retarded scalar potential. If the propagation medium is vacuum, up is equal to the velocity of light c.
Similarly, the retarded vector potential A(R, t) is related to the distribution of current density J by A(R, t) = μ 4π v ′ J(Ri, t −R′/up) R′ dv ′ (Wb/m).
(6.75) This expression is obtained by extending the expression for the magnetostatic vector potential A(R) given by Eq. (5.65) to the time-varying case.
6-11.2 Time-Harmonic Potentials The expressions given by Eqs. (6.74) and (6.75) for the retarded scalar and vector potentials are valid under both static and dynamic conditions and for any type of time dependence of the source functions ρv and J. Because V and A depend linearly on ρv and J, and as E and B depend linearly on V and A, the relationships interconnecting all these quantities obey the rules of linear systems. When analyzing linear systems, we can take advantage of sinusoidal-time functions to determine the system’s response to a source with arbitrary time dependence.
As was noted in Section 1-7, if the time dependence is described by a (nonsinusoidal) periodic time function, it can always be expanded into a Fourier series of sinusoidal components, and if the time function is nonperiodic, it can be represented by a Fourier integral. In either case, if the response of the linear system is known for all steady-state sinusoidal excitations, the principle of superposition can be used to determine its response to an excitation with arbitrary time dependence.
Thus, the sinusoidal response of the system constitutes a fundamental building block that can be used to determine the response due to a source described by an arbitrary function of time. The term time-harmonic is often used in this context as a synonym for “steady-state sinusoidal time-dependent.” In this subsection, we derive expressions for the scalar and vector potentials due to time-harmonic sources. Suppose that ρv(Ri, t) is a sinusoidal-time function with angular frequency ω, given by ρv(Ri, t) = ρv(Ri) cos(ωt + φ).
(6.76) Phasor analysis, which was first introduced in Section 1-7 and then used extensively in Chapter 2 to study wave propagation on transmission lines, is a useful tool for analyzing time-harmonic scenarios. A time harmonic charge distribution ρv(Ri, t) is related to its phasor ˜ ρv(Ri) as ρv(Ri, t) = Re ˜ ρv(Ri) ejωt , (6.77) Comparison of Eqs. (6.76) and (6.77) shows that in the present case ˜ ρv(Ri) = ρv(Ri) ejφ.
6-11 ELECTROMAGNETIC POTENTIALS 305 Next, we express the retarded charge density ρv(Ri, t −R′/up) in phasor form by replacing t with (t −R′/up) in Eq. (6.77): ρv(Ri, t −R′/up) = Re ˜ ρv(Ri) ejω(t−R ′/up) = Re ˜ ρv(Ri) e−jωR ′/upejωt = Re ˜ ρv(Ri) e−jkR ′ejωt , (6.78) where k = ω up (6.79) is called the wavenumber or phase constant of the propagation medium.
(In general, the phase constant is denoted by the symbol “β”, but for lossless dielectric media, it is commonly denoted by the symbol “k” and called the wavenumber.) Similarly, we define the phasor V (R) of the time function V (R, t) according to V (R, t) = Re V (R) ejωt .
(6.80) Using Eqs. (6.78) and (6.80) in Eq. (6.74) gives Re V (R) ejωt = Re ⎡ ⎣1 4πϵ v ′ ˜ ρv(Ri) e−jkR ′ R′ ejωt dv ′ ⎤ ⎦.
(6.81) By equating the quantities inside the square brackets on both sides of Eq. (6.81) and cancelling the common ejωt factor, we obtain the phasor-domain expression V (R) = 1 4πϵ v ′ ˜ ρv(Ri) e−jkR ′ R′ dv ′ (V).
(6.82) For any given charge distribution, Eq. (6.82) can be used to compute V (R), and then the resultant expression can be used in Eq. (6.80) to find V (R, t).
Similarly, the expression for A(R, t) given by Eq. (6.75) can be transformed into A(R, t) = Re A(R) ejωt (6.83) with A(R) = μ 4π v ′ J(Ri) e−jkR ′ R′ dv ′, (6.84) where J(Ri) is the phasor function corresponding to J(Ri, t).
The magnetic field phasor H corresponding to A is given by H = 1 μ ∇× × × A.
(6.85) Recalling that differentiation in the time domain is equivalent to multiplication by jω in the phasor domain, in a nonconducting medium (J = 0), Amp ere’s law given by Eq. (6.41) becomes ∇× × × H = jωϵ E or E = 1 jωϵ ∇× × × H.
(6.86) Hence, given a time-harmonic current-density distribution with phasor J, Eqs. (6.84) to (6.86) can be used successively to determine both E and H. The phasor vectors E and H also are related by the phasor form of Faraday’s law: ∇× × × E = −jωμ H or H = − 1 jωμ ∇× × × E.
(6.87) 306 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS Example 6-8: Relating E to H In a nonconducting medium with ϵ = 16ϵ0 and μ = μ0, the electric field intensity of an electromagnetic wave is E(z, t) = ˆ x 10 sin(1010t −kz) (V/m).
(6.88) Determine the associated magnetic field intensity H and find the value of k.
Solution: We begin by finding the phasor E(z) of E(z, t).
Since E(z, t) is given as a sine function and phasors are defined in this book with reference to the cosine function, we rewrite Eq. (6.88) as E(z, t) = ˆ x 10 cos(1010t −kz −π/2) (V/m) = Re E(z) ejωt , (6.89) with ω = 1010 (rad/s) and E(z) = ˆ x 10e−jkze−jπ/2 = −ˆ xj10e−jkz.
(6.90) To find both H(z) and k, we will perform a “circle”: we will use the given expression for E(z) in Faraday’s law to find H(z); then we will use H(z) in Ampere’s law to find E(z), which we will then compare with the original expression for E(z); and the comparison will yield the value of k. Application of Eq. (6.87) gives H(z) = − 1 jωμ ∇× × × E = − 1 jωμ ˆ x ˆ y ˆ z ∂/∂x ∂/∂y ∂/∂z −j10e−jkz 0 0 = − 1 jωμ ˆ y ∂ ∂z(−j10e−jkz) = −ˆ yj 10k ωμ e−jkz.
(6.91) So far, we have used Eq. (6.90) for E(z) to find H(z), but k remains unknown. To find k, we use H(z) in Eq. (6.86) to find E(z): E(z) = 1 jωϵ ∇× × × H = 1 jωϵ −ˆ x ∂ ∂z −j 10k ωμ e−jkz = −ˆ xj 10k2 ω2μϵ e−jkz.
(6.92) Equating Eqs. (6.90) and (6.92) leads to k2 = ω2μϵ, or k = ω√μϵ = 4ω√μ0ϵ0 = 4ω c = 4 × 1010 3 × 108 = 133 (rad/m).
(6.93) With k known, the instantaneous magnetic field intensity is then given by H(z, t) = Re H(z) ejωt = Re −ˆ yj 10k ωμ e−jkzejωt = ˆ y 0.11 sin(1010t −133z) (A/m).
(6.94) We note that k has the same expression as the phase constant of a lossless transmission line [Eq. (2.49)].
Exercise 6-7: The magnetic field intensity of an electromagnetic wave propagating in a lossless medium with ϵ = 9ϵ0 and μ = μ0 is H(z, t) = ˆ x 0.3 cos(108t −kz + π/4) (A/m).
Find E(z, t) and k.
Answer: E(z, t) = −ˆ y 37.7 cos(108t −z + π/4) (V/m); k = 1 (rad/m). (See EM.) CHAPTER 6 SUMMARY 307 Chapter 6 Summary Concepts • Faraday’s law states that a voltage is induced across the terminals of a loop if the magnetic flux linking its surface changes with time.
• In an ideal transformer, the ratios of the primary to secondary voltages, currents, and impedances are governed by the turns ratio.
• Displacement current accounts for the “apparent” flow of charges through a dielectric. In reality, charges of opposite polarity accumulate along the two ends of a dielectric, giving the appearance of current flow through it.
• Boundary conditions for the electromagnetic fields at the interface between two different media are the same for both static and dynamic conditions.
• The charge continuity equation is a mathematical statement of the law of conservation of electric charge.
• Excess charges in the interior of a good conductor dissipate very quickly; through a rearrangement process, the excess charge is transferred to the surface of the conductor.
• In the dynamic case, the electric field E is related to both the scalar electric potential V and the magnetic vector potential A.
• The retarded scalar and vector potentials at a given observation point take into account the finite time required for propagation between their sources, the charge and current distributions, and the location of the observation point.
Mathematical and Physical Models Faraday’s Law Vemf = −d dt = −d dt S B· ds = V tr emf + V m emf Transformer V tr emf = −N S ∂B ∂t · ds (N loops) Motional V m emf = C (u × × × B)· dl Charge-Current Continuity ∇· J = −∂ρv ∂t EM Potentials E = −∇V −∂A ∂t B = ∇× × × A Current Density Conduction Jc = σ E Displacement Jd = ∂D ∂t Conductor Charge Dissipation ρv(t) = ρvoe−(σ/ϵ)t = ρvoe−t/τr 308 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS Important Terms Provide definitions or explain the meaning of the following terms: charge continuity equation charge dissipation displacement current Id electromagnetic induction electromotive force Vemf Faraday’s law Kirchhoff’s current law Lenz’s law motional emf V m emf relaxation time constant retarded potential transformer emf V tr emf wavenumber k PROBLEMS Sections 6-1 to 6-6: Faraday’s Law and its Applications ∗6.1 The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?
R2 R1 t = 0 t = t1 I Figure P6.1 Loops of Problem 6.1.
6.2 The loop in Fig. P6.2 is in the x–y plane and B = ˆ zB0 sin ωt with B0 positive. What is the direction of I ( ˆ φ φ φ or −ˆ φ φ φ) at: (a) t = 0 (b) ωt = π/4 (c) ωt = π/2 ∗Answer(s) available in Appendix D.
R Vemf z y x I Figure P6.2 Loop of Problem 6.2.
6.3 A coil consists of 100 turns of wire wrapped around a square frame of sides 0.25 m. The coil is centered at the origin with each of its sides parallel to the x- or y axis. Find the induced emf across the open-circuited ends of the coil if the magnetic field is given by ∗(a) B = ˆ z 20e−3t (T) (b) B = ˆ z 20 cos x cos 103t (T) (c) B = ˆ z 20 cos x sin 2y cos 103t (T) 6.4 A stationary conducting loop with an internal resistance of 0.5 is placed in a time-varying magnetic field. When the loop is closed, a current of 5 A flows through it. What will the current be if the loop is opened to create a small gap and a 2 resistor is connected across its open ends?
∗6.5 A circular-loop TV antenna with 0.02 m2 area is in the presence of a uniform-amplitude 300 Mhz signal. When oriented for maximum response, the loop develops an emf with a peak value of 30 (mV). What is the peak magnitude of B of the incident wave?
6.6 The square loop shown in Fig. P6.6 is coplanar with a long, straight wire carrying a current I(t) = 5 cos(2π × 104t) (A).
PROBLEMS 309 y x z 5 cm I(t) 10 cm 10 cm Figure P6.6 Loop coplanar with long wire (Problem 6.6).
(a) Determine the emf induced across a small gap created in the loop.
(b) Determine the direction and magnitude of the current that would flow through a 4 resistor connected across the gap. The loop has an internal resistance of 1 .
∗6.7 The rectangular conducting loop shown in Fig. P6.7 rotates at 6,000 revolutions per minute in a uniform magnetic flux density given by B = ˆ y 50 (mT).
Determine the current induced in the loop if its internal resistance is 0.5 .
6.8 The transformer shown in Fig. P6.8 consists of a long wire coincident with the z axis carrying a current I = I0 cos ωt, coupling magnetic energy to a toroidal coil situated in the x–y plane and centered at the origin. The toroidal core uses iron material with relative permeability μr, around which 100 turns of a tightly wound coil serves to induce a voltage Vemf, as shown in the figure.
(a) Develop an expression for Vemf.
(b) Calculate Vemf for f = 60 Hz, μr = 4000, a = 5 cm, b = 6 cm, c = 2 cm, and I0 = 50 A.
y x z B B φ(t) 3 cm 2 cm ω Figure P6.7 Rotating loop in a magnetic field (Problem 6.7).
a b x y z I Vemf c N Iron core with μr Figure P6.8 Problem 6.8.
6.9 A rectangular conducting loop 5 cm×10 cm with a small air gap in one of its sides is spinning at 7200 revolutions per minute. IfthefieldBisnormaltotheloopaxisanditsmagnitude is 6 × 10−6 T, what is the peak voltage induced across the air gap?
∗6.10 A 50 cm long metal rod rotates about the z axis at 90 revolutions per minute, with end 1 fixed at the origin as shown in Fig. P6.10. Determine the induced emf V12 if B = ˆ z 2×10−4 T.
310 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS 1 2 x y z B Figure P6.10 Rotating rod of Problem 6.10.
6.11 The loop shown in P6.11 moves away from a wire carrying a current I1 = 10 A at a constant velocity u = ˆ y7.5 (m/s). If R = 10 and the direction of I2 is as defined in the figure, find I2 as a function of y0, the distance between the wire and the loop. Ignore the internal resistance of the loop.
u u I1 = 10 A I2 20 cm 10 cm R z R y0 Figure P6.11 Moving loop of Problem 6.11.
∗6.12 The electromagnetic generator shown in Fig. 6-12 is connected to an electric bulb with a resistance of 150 . If the loop area is 0.1 m2 and it rotates at 3,600 revolutions per minute in a uniform magnetic flux density B0 = 0.4 T, determine the amplitude of the current generated in the light bulb.
6.13 The circular, conducting, disk shown in Fig. P6.13 lies in the x–y plane and rotates with uniform angular velocity ω about the z axis. The disk is of radius a and is present in a uniform magnetic flux density B = ˆ zB0. Obtain an expression for the emf induced at the rim relative to the center of the disk.
z y x a V ω Figure P6.13 Rotating circular disk in a magnetic field (Problem 6.13).
Section 6-7: Displacement Current 6.14 The plates of a parallel-plate capacitor have areas of 10 cm2 each and are separated by 2 cm. The capacitor is filled with a dielectric material with ϵ = 4ϵ0, and the voltage across it is given by V (t) = 30 cos 2π×106t (V). Find the displacement current.
∗6.15 A coaxial capacitor of length l = 6 cm uses an insulating dielectric material with ϵr = 9. The radii of the cylindrical conductors are 0.5 cm and 1 cm. If the voltage applied across the capacitor is V (t) = 50 sin(120πt) (V), what is the displacement current?
6.16 The parallel-plate capacitor shown in Fig. P6.16 is filled with a lossy dielectric material of relative permittivity ϵr and conductivity σ. The separation between the plates is d and each plate is of area A. The capacitor is connected to a time-varying voltage source V (t).
(a) Obtain an expression for Ic, the conduction current flowing between the plates inside the capacitor, in terms of the given quantities.
PROBLEMS 311 V(t) I A d ε, σ Figure P6.16 Parallel-plate capacitor containing a lossy dielectric material (Problem 6.16).
(b) Obtain an expression for Id, the displacement current flowing inside the capacitor.
(c) Based on your expressions for parts (a) and (b), give an equivalent-circuit representation for the capacitor.
(d) Evaluate the values of the circuit elements for A = 4 cm2, d = 0.5 cm, ϵr = 4, σ = 2.5 (S/m), and V (t) = 10 cos(3π × 103t) (V).
∗6.17 In wet soil, characterized by σ = 10−2 (S/m), μr = 1, and ϵr = 36, at what frequency is the conduction current density equal in magnitude to the displacement current density?
6.18 An electromagnetic wave propagating in seawater has an electric field with a time variation given by E = ˆ zE0 cos ωt.
If the permittivity of water is 81ϵ0 and its conductivity is 4 (S/m), find the ratio of the magnitudes of the conduction current density to displacement current density at each of the following frequencies: (a) 1 kHz ∗(b) 1 MHz (c) 1 GHz (d) 100 GHz Sections 6-9 and 6-10: Continuity Equation and Charge Dissipation 6.19 At t = 0, charge density ρv0 was introduced into the interior of a material with a relative permittivity ϵr = 9. If at t = 1 μs the charge density has dissipated down to 10−3ρv0, what is the conductivity of the material?
∗6.20 If the current density in a conducting medium is given by J(x, y, z; t) = (ˆ xz −ˆ y4y2 + ˆ z2x) cos ωt determine the corresponding charge distribution ρv(x, y, z; t).
6.21 In a certain medium, the direction of current density J points in the radial direction in cylindrical coordinates and its magnitude is independent of both φ and z. Determine J, given that the charge density in the medium is ρv = ρ0r cos ωt (C/m3).
6.22 If we were to characterize how good a material is as an insulator by its resistance to dissipating charge, which of the following two materials is the better insulator?
Dry Soil: ϵr = 2.5, σ = 10−4 (S/m) Fresh Water: ϵr = 80, σ = 10−3 (S/m) Sections 6-11: Electromagnetic Potentials 6.23 The electric field of an electromagnetic wave propagat-ing in air is given by E(z, t) = ˆ x4 cos(6 × 108t −2z) + ˆ y3 sin(6 × 108t −2z) (V/m).
Find the associated magnetic field H(z, t).
∗6.24 The magnetic field in a dielectric material with ϵ = 4ϵ0, μ = μ0, and σ = 0 is given by H(y, t) = ˆ x5 cos(2π × 107t + ky) (A/m).
Find k and the associated electric field E.
6.25 Given an electric field E = ˆ xE0 sin ay cos(ωt −kz), where E0, a, ω, and k are constants, find H.
312 CHAPTER 6 MAXWELL’S EQUATIONS FOR TIME-VARYING FIELDS ∗6.26 The electric field radiated by a short dipole antenna is given in spherical coordinates by E(R, θ; t) = ˆ θ θ θ 2 × 10−2 R sin θ cos(6π × 108t −2πR) (V/m).
Find H(R, θ; t).
6.27 A Hertzian dipole is a short conducting wire carrying an approximately constant current over its length l. If such a dipole is placed along the z axis with its midpoint at the origin, and if the current flowing through it is i(t) = I0 cos ωt, find the following: (a) The retarded vector potential A(R, θ, φ) at an observation point Q(R, θ, φ) in a spherical coordinate system.
(b) The magnetic field phasor H(R, θ, φ).
Assume l to be sufficiently small so that the observation point is approximately equidistant to all points on the dipole; that is, assume R′ ≈R.
6.28 In free space, the magnetic field is given by H = ˆ φ φ φ 36 r cos(6 × 109t −kz) (mA/m).
∗(a) Determine k.
(b) Determine E.
(c) Determine Jd.
6.29 The magnetic field in a given dielectric medium is given by H = ˆ y 6 cos 2z sin(2 × 107t −0.1x) (A/m), where x and z are in meters. Determine: (a) E, (b) the displacement current density Jd, and (c) the charge density ρv.
C H A P T E R 7 Plane-Wave Propagation Chapter Contents Unbounded EM Waves, 314 7-1 Time-Harmonic Fields, 315 7-2 Plane-Wave Propagation in Lossless Media, 316 TB13 RFID Systems, 322 7-3 Wave Polarization, 324 7-4 Plane-Wave Propagation in Lossy Media, 331 TB14 Liquid Crystal Display (LCD), 336 7-5 Current Flow in a Good Conductor, 339 7-6 Electromagnetic Power Density, 343 Chapter 7 Summary, 346 Problems, 348 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Describe mathematically the electric and magnetic fields of TEM waves.
2. Describe the polarization properties of an EM wave.
3. Relate the propagation parameters of a wave to the constitutive parameters of the medium.
4. Characterize the flow of current in conductors and use it to calculate the resistance of a coaxial cable.
5. Calculate the rate of power carried by an EM wave, in both lossless and lossy media.
314 CHAPTER 7 PLANE-WAVE PROPAGATION Unbounded EM Waves It was established in Chapter 6 that a time-varying electric field produces a magnetic field and, conversely, a time-varying magnetic field produces an electric field. This cyclic pattern often results in electromagnetic (EM) waves propagating through free space and in material media.
When a wave propagatesthroughahomogeneousmediumwithoutinteracting with obstacles or material interfaces, it is said to be unbounded.
Light waves emitted by the sun and radio transmissions by antennas are good examples. Unbounded waves may propagate in both lossless and lossy media.
Waves propagating in a lossless medium (e.g., air and perfect dielectrics) are similar to thoseonalosslesstransmissionlineinthattheydonotattenuate.
When propagating in a lossy medium (material with nonzero conductivity, such as water), part of the power carried by an EM wave gets converted into heat. A wave produced by a localized source, such as an antenna, expands outwardly in the form of a spherical wave, as depicted in Fig. 7-1(a).
Even though an antenna may radiate more energy along some directions than along others, the spherical wave travels at the same speed in all directions.
To an observer very far away from the source, however, the wavefront of the spherical wave appears approximately planar, as if it were part of a uniform plane wave with identical properties at all points in the plane tangent to the wavefront [Fig. 7-1(b)]. Plane-waves are easily described using a Cartesian coordinate system, which is mathematically easier to work with than the spherical coordinate system needed to describe spherical waves.
When a wave propagates along a material structure, it is said to be guided. Earth’s surface and ionosphere constitute parallel boundaries of a natural structure capable of guiding short-wave radio transmissions in the HF band† (3 to 30 MHz); indeed, the ionosphere is a good reflector at these frequencies, thereby allowing the waves to zigzag between the two boundaries (Fig. 7-2).
When we discussed wave propagation on a transmission line in Chapter 2, we dealt with voltages and currents. For a transmission-line circuit such as that shown in Fig. 7-3, the ac voltage source excites an incident wave that travels down the coaxial line toward the load, and unless the load is matched to the line, part (or all) of the incident wave is reflected back toward the generator. At any point on the line, the instantaneous total voltage v(z, t) is the sum of the incident and reflected waves, both of which vary sinusoidally with time. Associated with the voltage difference between the inner and outer conductors of the coaxial line is a radial electric †See Fig. 1-17.
(a) Spherical wave (b) Plane-wave approximation Radiating antenna Spherical wavefront Uniform plane wave Aperture Observer Figure 7-1 Waves radiated by an EM source, such as a light bulb or an antenna, have spherical wavefronts, as in (a); to a distant observer, however, the wavefront across the observer’s aperture appears approximately planar, as in (b).
field E(z, t) that exists in the dielectric material between the conductors, and since v(z, t) varies sinusoidally with time, so doesE(z, t). Furthermore, thecurrentflowingthroughtheinner conductor induces an azimuthal magnetic field H(z, t) in the dielectric material surrounding it. These coupled fields, E(z, t) and H(z, t), constitute an electromagnetic wave. Thus, we can model wave propagation on a transmission line either in terms of the voltages across the line and the currents in its conductors, 7-1 TIME-HARMONIC FIELDS 315 Transmitter Earth's surface Ionosphere Figure 7-2 The atmospheric layer bounded by the ionosphere at the top and Earth’s surface at the bottom forms a guiding structure for the propagation of radio waves in the HF band.
Vg Rg RL H H H H E E Figure 7-3 A guided electromagnetic wave traveling in a coaxial transmission line consists of time-varying electric and magnetic fields in the dielectric medium between the inner and outer conductors.
or in terms of the electric and magnetic fields in the dielectric medium between the conductors.
In this chapter we focus our attention on wave propagation in unbounded media. Unbounded waves have many practical applications in science and engineering.
We consider both lossless and lossy media.
Even though strictly speaking uniform plane waves cannot exist, we study them in this chapter to develop a physical understanding of wave propagation in lossless and lossy media. In Chapter 8 we examine how waves, both planar and spherical, are reflected by, and transmitted through, boundaries between dissimilar media. The processes of radiation and reception of waves by antennas are treated in Chapter 9.
7-1 Time-Harmonic Fields Time-varying electric and magnetic fields (E, D, B, and H) and their sources (the charge density ρv and current density J) generally depend on the spatial coordinates (x, y, z) and the time variable t.
However, if their time variation is sinusoidal with angular frequency ω, then these quantities can be represented by a phasor that depends on (x, y, z) only. The vectorphasor E(x, y, z)andtheinstantaneousfieldE(x, y, z; t) it describes are related as E(x, y, z; t) = Re E(x, y, z) ejωt .
(7.1) Similar definitions apply to D, B, and H, as well as to ρv and J.
For a linear, isotropic, and homogeneous medium with electrical permittivity ϵ, magnetic permeability μ, and conductivity σ, Maxwell’s equations (6.1) to (6.4) assume the following form in the phasor domain: ∇· E = ˜ ρv/ϵ, (7.2a) ∇× × × E = −jωμ H, (7.2b) ∇· H = 0, (7.2c) ∇× × × H = J + jωϵ E.
(7.2d) To derive these equations we used D = ϵE and B = μH, and the fact that for time-harmonic quantities, differentiation in the time domain corresponds to multiplication by jω in the phasor domain. These equations are the starting point for the subject matter treated in this chapter.
7-1.1 Complex Permittivity In a medium with conductivity σ, the conduction current density J is related to E by J = σ E. Assuming no other current flows in the medium, Eq. (7.2d) may be written as ∇× × × H = J+jωϵ E = (σ +jωϵ) E = jω ϵ −j σ ω E. (7.3) By defining the complex permittivity ϵc as ϵc = ϵ −j σ ω , (7.4) Eq. (7.3) can be rewritten as ∇× × × H = jωϵc E.
(7.5) Taking the divergence of both sides of Eq. (7.5), and recalling that the divergence of the curl of any vector field vanishes (i.e., 316 CHAPTER 7 PLANE-WAVE PROPAGATION ∇· ∇× × × H = 0), it follows that ∇·(jωϵc E) = 0, or ∇· E = 0.
Comparing this with Eq. (7.2a) implies that ˜ ρv = 0.
Upon replacing Eq. (7.2d) with Eq. (7.5) and setting ˜ ρv = 0 in Eq. (7.2a), Maxwell’s equations become ∇· E = 0, (7.6a) ∇× × × E = −jωμ H, (7.6b) ∇· H = 0, (7.6c) ∇× × × H = jωϵc E.
(7.6d) The complex permittivity ϵc given by Eq. (7.4) is often written in terms of a real part ϵ′ and an imaginary part ϵ′′. Thus, ϵc = ϵ −j σ ω = ϵ′ −jϵ′′, (7.7) with ϵ′ = ϵ, (7.8a) ϵ′′ = σ ω .
(7.8b) For a lossless medium with σ = 0, it follows that ϵ′′ = 0 and ϵc = ϵ′ = ϵ.
7-1.2 Wave Equations Next, we derive wave equations for E and H and then solve them to obtain explicit expressions for E and H as a function of the spatial variables (x, y, z). To that end, we start by taking the curl of both sides of Eq. (7.6b) to get ∇× × × (∇× × × E) = −jωμ(∇× × × H).
(7.9) Upon substituting Eq. (7.6d) into Eq. (7.9) we obtain ∇× × × (∇× × × E) = −jωμ(jωϵc E) = ω2μϵc E.
(7.10) From Eq. (3.113), we know that the curl of the curl of E is ∇× × × (∇× × × E) = ∇(∇· E) −∇2 E, (7.11) where ∇2 E is the Laplacian of E, which in Cartesian coordinates is given by ∇2 E = ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 E.
(7.12) In view of Eq. (7.6a), the use of Eq. (7.11) in Eq. (7.10) gives ∇2 E + ω2μϵc E = 0, (7.13) which is known as the homogeneous wave equation for E. By defining the propagation constant γ as γ 2 = −ω2μϵc, (7.14) Eq. (7.13) can be written as ∇2 E −γ 2 E = 0 (wave equation for E).
(7.15) To derive Eq. (7.15), we took the curl of both sides of Eq. (7.6b) and then we used Eq. (7.6d) to eliminate H and obtain an equation in E only. If we reverse the process, that is, if we start by taking the curl of both sides of Eq. (7.6d) and then use Eq. (7.6b) to eliminate E, we obtain a wave equation for H: ∇2 H −γ 2 H = 0 (wave equation for H).
(7.16) Since the wave equations for E and H are of the same form, so are their solutions.
7-2 Plane-Wave Propagation in Lossless Media The properties of an electromagnetic wave, such as its phase velocity up and wavelength λ, depend on the angular frequency ω and the medium’s three constitutive parameters: ϵ, μ, and σ. If the medium is nonconducting (σ = 0), the wave does not suffer any attenuation as it travels and hence the medium is said to be lossless. Because in a lossless medium ϵc = ϵ, Eq. (7.14) becomes γ 2 = −ω2μϵ.
(7.17) For lossless media, it is customary to define the wavenumber k as k = ω√μϵ .
(7.18) In view of Eq. (7.17), γ 2 = −k2 and Eq. (7.15) becomes ∇2 E + k2 E = 0.
(7.19) 7-2 PLANE-WAVE PROPAGATION IN LOSSLESS MEDIA 317 7-2.1 Uniform Plane Waves For an electric field phasor defined in Cartesian coordinates as E = ˆ x Ex + ˆ y Ey + ˆ z Ez, (7.20) substitution of Eq. (7.12) into Eq. (7.19) gives ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 (ˆ x Ex + ˆ y Ey + ˆ z Ez) + k2(ˆ x Ex + ˆ y Ey + ˆ z Ez) = 0.
(7.21) To satisfy Eq. (7.21), each vector component on the left-hand side of the equation must vanish. Hence, ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 + k2 Ex = 0, (7.22) and similar expressions apply to Ey and Ez.
▶A uniform plane wave is characterized by electric and magnetic fields that have uniform properties at all points across an infinite plane. ◀ If this happens to be the x–y plane, then E and H do not vary with x or y. Hence, ∂ Ex/∂x = 0 and ∂ Ex/∂y = 0, and Eq. (7.22) reduces to d2 Ex dz2 + k2 Ex = 0.
(7.23) Similar expressions apply to Ey, Hx, and Hy. The remaining components of E and H are zero; that is, Ez = Hz = 0. To show that Ez = 0, let us consider the z component of Eq. (7.6d), ˆ z ∂ Hy ∂x −∂ Hx ∂y = ˆ zjωϵ Ez.
(7.24) Since ∂ Hy/∂x = ∂ Hx/∂y = 0, it follows that Ez = 0.
A similar examination involving Eq. (7.6b) reveals that Hz = 0.
▶This means that a plane wave has no electric-field or magnetic-field components along its direction of propagation. ◀ For the phasor quantity Ex, the general solution of the ordinary differential equation given by Eq. (7.23) is Ex(z) = E+ x (z) + E− x (z) = E+ x0e−jkz + E− x0ejkz, (7.25) where E+ x0 and E− x0 are constants to be determined from boundary conditions.
The solution given by Eq. (7.25) is similar in form to the solution for the phasor voltage V (z) given by Eq. (2.54a) for the lossless transmission line. The first term in Eq. (7.25), containing the negative exponential e−jkz, represents a wave with amplitude E+ x0 traveling in the +z direction. Likewise, the second term (with ejkz) represents a wave with amplitude E− x0 traveling in the −z direction.
Assume for the time being that E only has a component along x (i.e., Ey = 0) and that Ex is associated with a wave traveling in the +z direction only (i.e., E− x0 = 0). Under these conditions, E(z) = ˆ x E+ x (z) = ˆ xE+ x0e−jkz.
(7.26) To find the magnetic field H associated with this wave, we apply Eq. (7.6b) with Ey = Ez = 0: ∇× × × E = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z E+ x (z) 0 0 = −jωμ(ˆ x Hx + ˆ y Hy + ˆ z Hz).
(7.27) For a uniform plane wave traveling in the +z direction, ∂E+ x (z)/∂x = ∂E+ x (z)/∂y = 0.
Hence, Eq. (7.27) gives Hx = 0, (7.28a) Hy = 1 −jωμ ∂ E+ x (z) ∂z , (7.28b) Hz = 1 −jωμ ∂E+ x (z) ∂y = 0.
(7.28c) Use of Eq. (7.26) in Eq. (7.28b) gives Hy(z) = k ωμE+ x0e−jkz = H + y0e−jkz, (7.29) 318 CHAPTER 7 PLANE-WAVE PROPAGATION x z y H E k ˆ Figure 7-4 A transverse electromagnetic (TEM) wave propagating in the direction ˆ k = ˆ z. For all TEM waves, ˆ k is parallel to E × × × H.
where H + y0 is the amplitude of Hy(z) and is given by H + y0 = k ωμE+ x0.
(7.30) For a wave traveling from the source toward the load on a transmission line, the amplitudes of its voltage and current phasors, V + 0 and I + 0 , are related by the characteristic impedance of the line, Z0. A similar connection exists between the electric and magnetic fields of an electromagnetic wave. The intrinsic impedance of a lossless medium is defined as η = ωμ k = ωμ ω√μϵ = μ ϵ ( ), (7.31) where we used the expression for k given by Eq. (7.18).
In view of Eq. (7.31), the electric and magnetic fields of a +z-propagating plane wave with E field along ˆ x are: E(z) = ˆ x E+ x (z) = ˆ xE+ x0e−jkz, (7.32a) H(z) = ˆ y E+ x (z) η = ˆ yE+ x0 η e−jkz.
(7.32b) ▶The electric and magnetic fields of a plane wave are perpendicular to each other, and both are perpendicular to the direction of wave travel (Fig. 7-4). These attributes qualify the wave as a transverse electromagnetic (TEM). ◀ Other examples of TEM waves include waves traveling on coaxial transmission lines (E is along ˆ r, H is along ˆ φ φ φ, and the direction of travel is along ˆ z) and spherical waves radiated by antennas.
In the general case, E+ x0 is a complex quantity with magnitude |E+ x0| and phase angle φ+. That is, E+ x0 = |E+ x0|ejφ+.
(7.33) The instantaneous electric and magnetic fields therefore are E(z, t) = Re E(z) ejωt = ˆ x|E+ x0| cos(ωt −kz + φ+) (V/m), (7.34a) and H(z, t) = Re H(z) ejωt = ˆ y|E+ x0| η cos(ωt −kz + φ+) (A/m).
(7.34b) Because E(z, t) and H(z, t) exhibit the same functional dependence on z and t, they are said to be in phase; when the amplitude of one of them reaches a maximum, the amplitude of the other does so too. The fact that E and H are in phase is characteristic of waves propagating in lossless media.
From the material on wave motion presented in Section 1-4, we deduce that the phase velocity of the wave is up = ω k = ω ω√μϵ = 1 √μϵ (m/s), (7.35) and its wavelength is λ = 2π k = up f (m).
(7.36) In vacuum, ϵ = ϵ0 and μ = μ0, and the phase velocity up and the intrinsic impedance η given by Eq. (7.31) are up = c = 1 √μ0ϵ0 = 3 × 108 (m/s), (7.37) η = η0 = μ0 ϵ0 = 377 ( ) ≈120π ( ), (7.38) 7-2 PLANE-WAVE PROPAGATION IN LOSSLESS MEDIA 319 where c is the velocity of light and η0 is called the intrinsic impedance of free space.
Example 7-1: EM Plane Wave in Air This example is analogous to the “Sound Wave in Water” problem given by Example 1-1.
The electric field of a 1 MHz plane wave traveling in the +z direction in air points along the x direction. If this field reaches a peak value of 1.2π (mV/m) at t = 0 and z = 50 m, obtain expressions for E(z, t) and H(z, t) and then plot them as a function of z at t = 0.
Solution: At f = 1 MHz, the wavelength in air is λ = c f = 3 × 108 1 × 106 = 300 m, and the corresponding wavenumber is k = (2π/300) (rad/m).
The general expression for an x-directed electric field traveling in the +z direction is given by Eq. (7.34a) as E(z, t) = ˆ x|E+ x0| cos(ωt −kz + φ+) = ˆ x 1.2π cos 2π × 106t −2πz 300 + φ+ (mV/m).
The field E(z, t) is maximum when the argument of the cosine functionequalszerooramultipleof2π. Att = 0 and z = 50 m, this condition yields −2π × 50 300 + φ+ = 0 or φ+ = π 3 .
Hence, E(z, t) = ˆ x 1.2π cos 2π × 106t −2πz 300 + π 3 (mV/m), and from Eq. (7.34b) we have H(z, t) = ˆ y E(z, t) η0 = ˆ y 10 cos 2π × 106t −2πz 300 + π 3 (μA/m), where we have used the approximation η0 ≈120π ( ).
z 0 x E E 1.2π (mV/m) y H H 10 (μA/m) λ Figure 7-5 Spatial variations of E and H at t = 0 for the plane wave of Example 7-1.
At t = 0, E(z, 0) = ˆ x 1.2π cos 2πz 300 −π 3 (mV/m), H(z, 0) = ˆ y 10 cos 2πz 300 −π 3 (μA/m).
Plots of E(z, 0) and H(z, 0) as a function of z are shown in Fig. 7-5.
7-2.2 General Relation between E and H It can be shown that, for any uniform plane wave traveling in an arbitrary direction denoted by the unit vector ˆ k, the electric and magnetic field phasors E and H are related as H = 1 η ˆ k × × × E, (7.39a) E = −η ˆ k × × × H.
(7.39b) 320 CHAPTER 7 PLANE-WAVE PROPAGATION ▶The following right-hand rule applies: when we rotate the four fingers of the right hand from the direction of E toward that of H, the thumb points in the direction of wave travel, ˆ k. ◀ The relations given by Eqs. (7.39a) and (b) are valid not only for lossless media but for lossy ones as well. As we see later in Section 7-4, the expression for η of a lossy medium is different from that given by Eq. (7.31). As long as the expression used for η is appropriate for the medium in which the wave is traveling, the relations given by Eqs. (7.39a) and (b) always hold.
(a) Wave propagating along +z with E along ˆ x Let us apply Eq. (7.39a) to the wave given by Eq. (7.32a). The direction of propagation ˆ k = ˆ z and E = ˆ x E+ x (z). Hence, H = 1 η ˆ k × × × E = 1 η(ˆ z × × × ˆ x) E+ x (z) = ˆ y E+ x (z) η , (7.40) which is the same as the result given by Eq. (7.32b). For a wave traveling in the −z direction with electric field given by E = ˆ x E− x (z) = ˆ xE− x0ejkz, (7.41) application of Eq. (7.39a) gives H = 1 η(−ˆ z × × × ˆ x) E− x (z) = −ˆ y E− x (z) η = −ˆ y E− x0 η ejkz.
(7.42) Hence, in this case, H points in the negative y direction.
(b) Wave propagating along +z with E along ˆ x and ˆ y In general, a uniform plane wave traveling in the +z direction may have both x and y components, in which case E is given by E = ˆ x E+ x (z) + ˆ y E+ y (z), (7.43a) and the associated magnetic field is H = ˆ x H + x (z) + ˆ y H + y (z).
(7.43b) Application of Eq. (7.39a) gives H = 1 η ˆ z × × × E = −ˆ x E+ y (z) η + ˆ y E+ x (z) η .
(7.44) H E y z x Hy + Hx + Ex + Ey + Figure 7-6 The wave (E, H) is equivalent to the sum of two waves, one with fields (E+ x , H + y ) and another with (E+ y , H + x ), with both traveling in the +z direction.
By equating Eq. (7.43b) to Eq. (7.44), we have H + x (z) = − E+ y (z) η , H + y (z) = E+ x (z) η .
(7.45) These results are illustrated in Fig. 7-6.
The wave may be considered the sum of two waves, one with electric and magneticcomponents(E+ x , H + y ), andanotherwithcomponents (E+ y , H + x ). In general, a TEM wave may have an electric field in any direction in the plane orthogonal to the direction of wave travel, and the associated magnetic field is also in the same plane and its direction is dictated by Eq. (7.39a).
Concept Question 7-1: What is a uniform plane wave?
Describe its properties, both physically and mathematically. Under what conditions is it appropriate to treat a spherical wave as a plane wave?
Concept Question 7-2: Since E and H are governed by wave equations of the same form [Eqs. (7.15) and (7.16)], does it follow that E = H? Explain.
7-2 PLANE-WAVE PROPAGATION IN LOSSLESS MEDIA 321 Module 7.1 Linking E to H Select the directions and magnitudes of E and H and observe the resultant wave vector.
Concept Question 7-3: If a TEM wave is traveling in the ˆ y direction, can its electric field have components along ˆ x, ˆ y, and ˆ z? Explain.
Exercise 7-1: A 10 MHz uniform plane wave is traveling in a nonmagnetic medium with μ = μ0 and ϵr = 9.
Find (a) the phase velocity, (b) the wavenumber, (c) the wavelength in the medium, and (d) the intrinsic impedance of the medium.
Answer: (a) up = 1 × 108 m/s, (b) k = 0.2π rad/m, (c) λ = 10 m, (d) η = 125.67 . (See EM.) Exercise 7-2: The electric field phasor of a uniform plane wave traveling in a lossless medium with an intrinsic impedance of 188.5 is given by E = ˆ z 10e−j4πy (mV/m).
Determine (a) the associated magnetic field phasor and (b) the instantaneous expression for E(y, t) if the medium is nonmagnetic (μ = μ0).
Answer: (a) H = ˆ x 53e−j4πy (μA/m), (b) E(y, t) = ˆ z 10 cos(6π × 108t −4πy) (mV/m).
(See EM.) 322 TECHNOLOGY BRIEF 13: RFID SYSTEMS Technology Brief 13: RFID Systems In 1973, two separate patents were issued in the United States for Radio Frequency Identification (RFID) concepts. The first, granted to Mario Cardullo, was for an active RFID tag with rewritable memory. An active tag has a power source (such as a battery) of its own, whereas a passive RFID tag does not. The second patent was granted to Charles Walton, who proposed the use of a passive tag for keyless entry (unlocking a door without a key). Shortly thereafter a passive RFID tag was developed for tracking cattle (Fig.TF13-1), and then the technology rapidly expanded into many commercial enterprises, from tracking vehicles and consumer products to supply chain management and automobile anti-theft systems.
RFID System Overview In an RFID system, communication occurs between a reader—which actually is a transceiver—and a tag (Fig. TF13-2). When interrogated by the reader, a tag responds with information about its identity, as well as other relevant information depending on the specific application.
▶The tag is, in essence, a transponder commanded by the reader. ◀ The functionality and associated capabilities of the RFID tag depend on two important attributes: (a) whether the tag is of the active or passive type, and (b) the tag’s operating frequency. Usually the RFID tag remains dormant (asleep) until activated by an electromagnetic signal radiated by the reader’s antenna. The magnetic field of the EM signal induces a current in the coil contained in the tag’s circuit (Fig. TF13-3). For a passive tag, the induced current has to be sufficient to generate the power necessary to activate the chip as well as to transmit the response to the reader.
▶Passive RFID systems are limited to short read ranges (between reader and tag) on the order of 30 cm to 3 m, depending on the system’s frequency band (as noted in Table TT13-1). ◀ The obvious advantage of active RFID systems is that they can operate over greater distances and do not require reception of a signal from the reader’s antenna to get activated. However, active tags are significantly more expensive to fabricate than their passive cousins.
RFID Frequency Bands Table TT13-1 provides a comparison among the four frequency bands commonly used for RFID systems. Generally speaking, the higher-frequency tags can operate over longer read ranges and can carry higher data rates, but they are more expensive to fabricate.
FigureTF13-1 Passive RFID tags were developed in the 1970s for tracking cows.
TECHNOLOGY BRIEF 13: RFID SYSTEMS 323 Tag reader The reader forwards the data it received from the RFID tag to a database that can then match the tag’s identifying serial number to an authorized account and debit that account.
Once activated by the signal from the tag reader (which acts as both a transmitter and a receiver), the RFID tag responds by transmitting the identifying serial number programmed into its electronic chip. FigureTF13-2 How an RFID system works is illustrated through this EZ-Pass example. The UHF RFID shown is courtesy of Prof. C. F. Huang of Tatung University, Taiwan.
RFID reader Antenna Tag Chip Antenna FigureTF13-3 Simplified diagram for how the RFID reader communicates with the tag. At the two lower carrier frequencies commonly used for RFID communication, namely 125 kHz and 13.56 MHz, coil inductors act as magnetic antennas. In systems designed to operate at higher frequencies (900 MHz and 2.54 GHz), dipole antennas are used instead.
TableTT13-1 Comparison of RFID frequency bands.
Band LF HF UHF Microwave RFID frequency 125–134 kHz 13.56 MHz 865–956 MHz 2.45 GHz Read range ≤0.5 m ≤1.5 m ≤5 m ≤10 m Data rate 1 kbit/s 25 kbit/s 30 kbit/s 100 kbit/s Typical • Animal ID • Smart cards • Supply chain • Vehicle toll collection applications • Automobile key/antitheft • Article surveillance management • Railroad car monitoring • Access control • Airline baggage tracking • Logistics • Library book tracking 324 CHAPTER 7 PLANE-WAVE PROPAGATION Module 7.2 PlaneWave Observe a plane wave propagating along the z direction; note the temporal and spatial variations of E and H, and examine how the wave properties change as a function of the values selected for the wave parameters—frequency and E field amplitude and phase—and the medium’s constitutive parameters (ϵ, μ, σ).
Exercise 7-3: If the magnetic field phasor of a plane wave traveling in a medium with intrinsic impedance η = 100 is given by H = (ˆ y 10 + ˆ z 20)e−j4x (mA/m), find the associated electric field phasor.
Answer: E = (−ˆ z + ˆ y 2)e−j4x (V/m). (See EM.) Exercise 7-4: Repeat Exercise 7-3 for a magnetic field given by H = ˆ y(10e−j3x −20ej3x) (mA/m).
Answer: E = −ˆ z(e−j3x + 2ej3x) (V/m). (See EM.) 7-3 Wave Polarization ▶The polarization of a uniform plane wave describes the locus traced by the tip of the E vector (in the plane orthogonal to the direction of propagation) at a given point in space as a function of time. ◀ In the most general case, the locus of the tip of E is an ellipse, and the wave is said to be elliptically polarized. Under certain conditions, the ellipse may degenerate into a circle or a straight line, in which case the polarization state is called circular or linear, respectively.
It was shown in Section 7-2 that the z components of the electric and magnetic fields of a z-propagating plane wave are both zero. Hence, in the most general case, the electric field phasor E(z) of a +z-propagating plane wave may consist of an x component, ˆ x Ex(z), and a y component, ˆ y Ey(z), or E(z) = ˆ x Ex(z) + ˆ y Ey(z), (7.46) with Ex(z) = Ex0e−jkz, (7.47a) Ey(z) = Ey0e−jkz, (7.47b) where Ex0 and Ey0 are the amplitudes of Ex(z) and Ey(z), respectively. For the sake of simplicity, the plus sign superscript 7-3 WAVE POLARIZATION 325 has been suppressed; the negative sign in e−jkz is sufficient to remind us that the wave is traveling in the positive z direction.
The two amplitudes Ex0 and Ey0 are, in general, complex quantities, eachcharacterizedbyamagnitudeandaphaseangle.
The phase of a wave is defined relative to a reference state, such as z = 0 and t = 0 or any other combination of z and t. As will become clear from the discussion that follows, the polarization of the wave described by Eqs. (7.46) and (7.47) depends on the phase of Ey0 relative to that of Ex0, but not on the absolute phases of Ex0 and Ey0. Hence, for convenience, we assign Ex0 a phase of zero and denote the phase of Ey0, relative to that of Ex0, as δ. Thus, δ is the phase difference between the y and x components of E. Accordingly, we define Ex0 and Ey0 as Ex0 = ax, (7.48a) Ey0 = ayejδ, (7.48b) where ax = |Ex0| ≥0 and ay = |Ey0| ≥0 are the magnitudes of Ex0 and Ey0, respectively. Thus, by definition, ax and ay maynotassumenegativevalues. UsingEqs.(7.48a)and(7.48b) in Eqs. (7.47a) and (7.47b), the total electric field phasor is E(z) = (ˆ xax + ˆ yayejδ)e−jkz, (7.49) and the corresponding instantaneous field is E(z, t) = Re E(z) ejωt = ˆ xax cos(ωt −kz) + ˆ yay cos(ωt −kz + δ).
(7.50) When characterizing an electric field at a given point in space, two of its attributes that are of particular interest are its magnitude and direction. The magnitude of E(z, t) is |E(z, t)| = [E2 x(z, t) + E2 y(z, t)]1/2 = [a2 x cos2(ωt −kz) + a2 y cos2(ωt −kz + δ)]1/2.
(7.51) The electric field E(z, t) has components along the x and y directions. At a specific position z, the direction of E(z, t) is characterized by its inclination angle ψ, defined with respect to the x axis and given by ψ(z, t) = tan−1 Ey(z, t) Ex(z, t) .
(7.52) In the general case, both the intensity of E(z, t) and its direction are functions of z and t. Next, we examine some special cases.
7-3.1 Linear Polarization ▶A wave is said to be linearly polarized if for a fixed z, the tip of E(z, t) traces a straight line segment as a function of time. This happens when Ex(z, t) and Ey(z, t) are in phase (i.e., δ = 0) or out of phase (δ = π). ◀ Under these conditions Eq. (7.50) simplifies to E(0, t) = (ˆ xax + ˆ yay) cos(ωt −kz) (in phase), (7.53a) E(0, t) = (ˆ xax −ˆ yay) cos(ωt −kz) (out of phase).
(7.53b) Let us examine the out-of-phase case. The field’s magnitude is |E(z, t)| = [a2 x + a2 y]1/2| cos(ωt −kz)|, (7.54a) and the inclination angle is ψ = tan−1 −ay ax (out of phase).
(7.54b) We note that ψ is independent of both z and t. Figure 7-7 displays the line segment traced by the tip of E at z = 0 over a half of a cycle. The trace would be the same at any other value of z as well. At z = 0 and t = 0, |E(0, 0)| = [a2 x + a2 y]1/2.
The length of the vector representing E(0, t) decreases to zero at ωt = π/2. The vector then reverses direction and increases in magnitude to [a2 x + a2 y]1/2 in the second quadrant of the x–y plane at ωt = π. Since ψ is independent of both z and t, E(z, t) maintains a direction along the line making an angle ψ with the x axis, while oscillating back and forth across the origin.
If ay = 0, then ψ = 0◦or 180◦, and the wave is x-polarized; conversely, if ax = 0, then ψ = 90◦or −90◦, and the wave is y-polarized.
326 CHAPTER 7 PLANE-WAVE PROPAGATION ay Ey Ex ax −ax −ay ωt = π ωt = 0 y x z E ψ Figure 7-7 Linearly polarized wave traveling in the +z direction (out of the page).
7-3.2 Circular Polarization We now consider the special case when the magnitudes of the x and y components of E(z) are equal, and the phase difference δ = ±π/2. For reasons that become evident shortly, the wave polarization is called left-hand circular when δ = π/2, and right-hand circular when δ = −π/2.
Left-hand circular (LHC) polarization For ax = ay = a and δ = π/2, Eqs. (7.49) and (7.50) become E(z) = (ˆ xa + ˆ yaejπ/2)e−jkz = a(ˆ x + j ˆ y)e−jkz, (7.55a) E(z, t) = Re E(z) ejωt = ˆ xa cos(ωt −kz) + ˆ ya cos(ωt −kz + π/2) = ˆ xa cos(ωt −kz) −ˆ ya sin(ωt −kz).
(7.55b) The corresponding field magnitude and inclination angle are |E(z, t)| = E2 x(z, t) + E2 y(z, t) 1/2 = [a2 cos2(ωt −kz) + a2 sin2(ωt −kz)]1/2 = a (7.56a) and ψ(z, t) = tan−1 Ey(z, t) Ex(z, t) = tan−1 −a sin(ωt −kz) a cos(ωt −kz) = −(ωt −kz). (7.56b) We observe that the magnitude of E is independent of both z and t, whereas ψ depends on both variables.
These functional dependencies are the converse of those for the linear polarization case.
At z = 0, Eq. (7.56b) gives ψ = −ωt; the negative sign implies that the inclination angle decreases as time increases.
As illustrated in Fig. 7-8(a), the tip of E(t) traces a circle in the x–y plane and rotates in a clockwise direction as a function of time (when viewing the wave approaching). Such a wave is called left-hand circularly polarized because, when the thumb of the left hand points along the direction of propagation (the z direction in this case), the other four fingers point in the direction of rotation of E.
Right-hand circular (RHC) polarization For ax = ay = a and δ = −π/2, we have |E(z, t)| = a, ψ = (ωt −kz).
(7.57) The trace of E(0, t) as a function of t is shown in Fig. 7-8(b).
For RHC polarization, the fingers of the right hand point in the direction of rotation of E when the thumb is along the propagation direction.
Figure 7-9 depicts a right-hand circularly polarized wave radiated by a helical antenna.
7-3 WAVE POLARIZATION 327 (a) LHC polarization (b) RHC polarization z ψ ω a y z x a E ψ ω a y z z x a E Figure 7-8 Circularly polarized plane waves propagating in the +z direction (out of the page).
▶Polarization handedness is defined in terms of the rotation of E as a function of time in a fixed plane orthogonal to the direction of propagation, which is opposite of the direction of rotation of E as a function of distance at a fixed point in time. ◀ Example 7-2: RHC-Polarized Wave An RHC-polarized plane wave with electric field magnitude of 3 (mV/m) is traveling in the +y direction in a dielectric x y z Right sense of rotation in plane Left screw sense in space Transmitting antenna E Figure 7-9 Right-hand circularly polarized wave radiated by a helical antenna.
medium with ϵ = 4ϵ0, μ = μ0, and σ = 0. If the frequency is 100 MHz, obtain expressions for E(y, t) and H(y, t).
Solution: Since the wave is traveling in the +y direction, its field must have components along the x and z directions. The rotation of E(y, t) is depicted in Fig. 7-10, where ˆ y is out of the ω x y z E Figure 7-10 Right-hand circularly polarized wave of Example 7-2.
328 CHAPTER 7 PLANE-WAVE PROPAGATION page. By comparison with the RHC-polarized wave shown in Fig. 7-8(b), we assign the z component of E(y) a phase angle of zero and the x component a phase shift of δ = −π/2. Both components have magnitudes of a = 3 (mV/m). Hence, E(y) = ˆ x Ex + ˆ z Ez = ˆ xae−jπ/2e−jky + ˆ zae−jky = (−ˆ xj + ˆ z)3e−jky (mV/m), and application of (7.39a) gives H(y) = 1 η ˆ y × × × E(y) = 1 η ˆ y × × × (−ˆ xj + ˆ z)3e−jky = 3 η(ˆ zj + ˆ x)e−jky (mA/m).
With ω = 2πf = 2π × 108 (rad/s), the wavenumber k is k = ω√ϵr c = 2π × 108√ 4 3 × 108 = 4 3π (rad/m), and the intrinsic impedance η is η = η0 √ϵr ≈120π √ 4 = 60π ( ).
The instantaneous fields E(y, t) and H(y, t) are E(y, t) = Re E(y) ejωt = Re (−ˆ xj + ˆ z)3e−jkyejωt = 3[ˆ x sin(ωt −ky) + ˆ z cos(ωt −ky)] (mV/m) and H(y, t) = Re H(y) ejωt = Re 3 η(ˆ zj + ˆ x)e−jkyejωt = 1 20π [ˆ x cos(ωt −ky) −ˆ z sin(ωt −ky)] (mA/m).
7-3.3 Elliptical Polarization Plane waves that are not linearly or circularly polarized are elliptically polarized. That is, the tip of E(z, t) traces an ellipse in the plane perpendicular to the direction of propagation. The shape of the ellipse and the field’s handedness (left-hand or right-hand) are determined by the values of the ratio (ay/ax) and the phase difference δ.
The polarization ellipse shown in Fig. 7-11 has its major axis with length aξ along the ξ direction and its minor axis with length aη along the η direction. The rotation angle γ is defined as the angle between the major axis of the ellipse and a reference direction, chosen here to be the x axis, with γ being bounded within the range −π/2 ≤γ ≤π/2. The shape of the ellipse and its handedness are characterized by the ellipticity angle χ, defined as tan χ = ±aη aξ = ± 1 R , (7.58) ψ0 γ ξ η x y z χ aξ aη ax ay Major axis Minor axis Polarization ellipse Ellipticity angle Rotation angle Figure 7-11 Polarization ellipse in the x–y plane, with the wave traveling in the z direction (out of the page).
7-3 WAVE POLARIZATION 329 with the plus sign corresponding to left-handed rotation and the minus sign corresponding to right-handed rotation. The limits for χ are −π/4 ≤χ ≤π/4. The quantity R = aξ/aη is called the axial ratio of the polarization ellipse, and it varies between 1 for circular polarization and ∞for linear polarization. The polarization angles γ and χ are related to the wave parameters ax, ay, and δ by† tan 2γ = (tan 2ψ0) cos δ (−π/2 ≤γ ≤π/2), (7.59a) sin 2χ = (sin 2ψ0) sin δ (−π/4 ≤χ ≤π/4), (7.59b) where ψ0 is an auxiliary angle defined by tan ψ0 = ay ax 0 ≤ψ0 ≤π 2 .
(7.60) Sketches of the polarization ellipse are shown in Fig. 7-12 for various combinations of the angles (γ, χ). The ellipse reduces to a circle for χ = ±45◦and to a line for χ = 0.
▶Positive values of χ, corresponding to sin δ > 0, are associated with left-handed rotation, and negative values of χ, corresponding to sin δ < 0, are associated with right-handed rotation. ◀ Since the magnitudes ax and ay are, by definition, nonnegative numbers, the ratio ay/ax may vary between zero for an x-polarized linear polarization and ∞for a y-polarized linear polarization. Consequently, the angle ψ0 is limited to the range 0 ≤ψ0 ≤90◦. Application of Eq. (7.59a) leads to two possible solutions for the value of γ , both of which fall within the defined range from −π/2 to π/2. The correct choice is governed by the following rule: γ > 0 if cos δ > 0, γ < 0 if cos δ < 0.
†From M. Born and E. Wolf, Principles of Optics, New York: Macmillan, 1965, p. 27.
▶In summary, the sign of the rotation angle γ is the same as the sign of cos δ and the sign of the ellipticity angle χ is the same as the sign of sin δ. ◀ Example 7-3: Polarization State Determine the polarization state of a plane wave with electric field E(z, t) = ˆ x 3 cos(ωt −kz + 30◦) −ˆ y 4 sin(ωt −kz + 45◦) (mV/m).
Solution: We begin by converting the second term to a cosine reference, E = ˆ x 3 cos(ωt −kz + 30◦) −ˆ y 4 cos(ωt −kz + 45◦−90◦) = ˆ x 3 cos(ωt −kz + 30◦) −ˆ y 4 cos(ωt −kz −45◦).
The corresponding field phasor E(z) is E(z) = ˆ x 3e−jkzej30◦−ˆ y 4e−jkze−j45◦ = ˆ x 3e−jkzej30◦+ ˆ y 4e−jkze−j45◦ej180◦ = ˆ x 3e−jkzej30◦+ ˆ y 4e−jkzej135◦, where we have replaced the negative sign of the second term with ej180◦in order to have positive amplitudes for both terms, thereby allowing us to use the definitions given in Section 7-3.3.
According to the expression for E(z), the phase angles of the x and y components are δx = 30◦and δy = 135◦, giving a phase difference δ = δy −δx = 135◦−30◦= 105◦. The auxiliary angle ψ0 is obtained from ψ0 = tan−1 ay ax = tan−1 4 3 = 53.1◦.
From Eq. (7.59a), tan 2γ = (tan 2ψ0) cos δ = tan 106.2◦cos 105◦= 0.89, 330 CHAPTER 7 PLANE-WAVE PROPAGATION 45◦ 45◦ 22.5◦ 0◦ 0◦ −22.5◦ −45◦ −45◦ −90◦ 90◦ Left circular polarization Left elliptical polarization Linear polarization Right elliptical polarization Right circular polarization χ γ Figure 7-12 Polarization states for various combinations of the polarization angles (γ, χ) for a wave traveling out of the page.
which gives two solutions for γ , namely γ = 20.8◦and γ = −69.2◦. Since cos δ < 0, the correct value of γ is −69.2◦.
From Eq. (7.59b), sin 2χ = (sin 2ψ0) sin δ = sin 106.2◦sin 105◦ = 0.93 or χ = 34.0◦.
The magnitude of χ indicates that the wave is elliptically polarized and its positive polarity specifies its rotation as left handed.
Concept Question 7-4: An elliptically polarized wave is characterized by amplitudes ax and ay and by the phase difference δ. If ax and ay are both nonzero, what should δ be in order for the polarization state to reduce to linear polarization?
Concept Question 7-5: Which of the following two descriptions defines an RHC-polarized wave: A wave incident upon an observer is RHC-polarized if its electric field appears to the observer to rotate in a counterclockwise direction (a) as a function of time in a fixed plane perpendicular to the direction of wave travel or (b) as a function of travel distance at a fixed time t?
Exercise 7-5: The electric field of a plane wave is given by E(z, t) = ˆ x 3 cos(ωt −kz) + ˆ y 4 cos(ωt −kz) (V/m).
Determine (a) the polarization state, (b) the modulus of E, and (c) the auxiliary angle.
Answer: (a) Linear, (b) |E| = 5 cos(ωt −kz) (V/m), (c) ψ0 = 53.1◦. (See EM.) 7-4 PLANE-WAVE PROPAGATION IN LOSSY MEDIA 331 Module 7.3 Polarization I Upon specifying the amplitudes and phases of the x and y components of E, the user can observe the trace of E in the x–y plane.
Exercise 7-6: If the electric field phasor of aTEM wave is given by E = (ˆ y −ˆ zj)e−jkx, determine the polarization state.
Answer: RHC polarization. (See EM.) 7-4 Plane-Wave Propagation in Lossy Media To examine wave propagation in a lossy (conducting) medium, we return to the wave equation given by Eq. (7.15), ∇2 E −γ 2 E = 0, (7.61) with γ 2 = −ω2μϵc = −ω2μ(ϵ′ −jϵ′′), (7.62) where ϵ′ = ϵ and ϵ′′ = σ/ω. Since γ is complex, we express it as γ = α + jβ, (7.63) where α is the medium’s attenuation constant and β its phase constant. By replacing γ with (α+jβ) in Eq. (7.62), we obtain (α + jβ)2 = (α2 −β2) + j2αβ = −ω2μϵ′ + jω2μϵ′′.
(7.64) 332 CHAPTER 7 PLANE-WAVE PROPAGATION Module 7.4 Polarization II Upon specifying the amplitudes and phases of the x and y components of E, the user can observe the 3-D profile of the E vector over a specified length span.
The rules of complex algebra require the real and imaginary parts on one side of an equation to equal, respectively, the real and imaginary parts on the other side. Hence, α2 −β2 = −ω2μϵ′, (7.65a) 2αβ = ω2μϵ′′.
(7.65b) Solving these two equations for α and β gives α = ω ⎧ ⎨ ⎩ μϵ′ 2 ⎡ ⎣ 1 + ϵ′′ ϵ′ 2 −1 ⎤ ⎦ ⎫ ⎬ ⎭ 1/2 (Np/m), (7.66a) β = ω ⎧ ⎨ ⎩ μϵ′ 2 ⎡ ⎣ 1 + ϵ′′ ϵ′ 2 + 1 ⎤ ⎦ ⎫ ⎬ ⎭ 1/2 (rad/m).
(7.66b) For a uniform plane wave with electric field E = ˆ x Ex(z) traveling along the z direction, the wave equation given by Eq. (7.61) reduces to d2 Ex(z) dz2 −γ 2 Ex(z) = 0.
(7.67) The general solution of the wave equation given by Eq. (7.67) comprises two waves, one traveling in the +z direction and another traveling in the −z direction. Assuming only the former is present, the solution of the wave equation leads to E(z) = ˆ x Ex(z) = ˆ xEx0e−γ z = ˆ xEx0e−αze−jβz.
(7.68) The associated magnetic field H can be determined by applying Eq. (7.2b): ∇× × × E = −jωμ H, or using Eq. (7.39a): H = (ˆ k × × × E)/ηc, where ηc is the intrinsic impedance of the lossy medium. Both approaches give H(z) = ˆ y Hy(z) = ˆ y Ex(z) ηc = ˆ y Ex0 ηc e−αze−jβz, (7.69) 7-4 PLANE-WAVE PROPAGATION IN LOSSY MEDIA 333 e−αz e−1 |Ex(z)| ~ δs z 1 |Ex0| Figure 7-13 Attenuation of the magnitude of Ex(z) with distance z.
The skin depth δs is the value of z at which | Ex(z)|/|Ex0| = e−1, or z = δs = 1/α.
where ηc = μ ϵc = μ ϵ′ 1 −j ϵ′′ ϵ′ −1/2 ( ).
(7.70) We noted earlier that in a lossless medium, E(z, t) is in phase with H(z, t). This property no longer holds true in a lossy medium because ηc is complex. This fact is demonstrated in Example 7-4.
From Eq. (7.68), the magnitude of Ex(z) is given by | Ex(z)| = |Ex0e−αze−jβz| = |Ex0|e−αz, (7.71) which decreases exponentially with z at a rate dictated by the attenuation constant α.
Since Hy = Ex/ηc, the magnitude of Hy also decreases as e−αz. As the field attenuates, part of the energy carried by the electromagnetic wave is converted into heat due to conduction in the medium. As the wave travels through a distance z = δs with δs = 1 α (m), (7.72) the wave magnitude decreases by a factor of e−1 ≈0.37 (Fig. 7-13). At depth z = 3δs, the field magnitude is less than 5% of its initial value, and at z = 5δs, it is less than 1%.
▶This distance δs, called the skin depth of the medium, characterizes how deep an electromagnetic wave can penetrate into a conducting medium. ◀ In a perfect dielectric, σ = 0 and ϵ′′ = 0; use of Eq. (7.66a) yields α = 0 and therefore δs = ∞. Thus, in free space, a plane wave can propagate indefinitely with no loss in magnitude. On the other extreme, in a perfect conductor, σ = ∞and use of Eq. (7.66a) leads to α = ∞and hence δs = 0. If the outer conductorofacoaxialcableisdesignedtobeseveralskindepths thick, it prevents energy inside the cable from leaking outward and shields against penetration of electromagnetic energy from external sources into the cable.
The expressions given by Eqs. (7.66a), (7.66b), and (7.70) for α, β, and ηc are valid for any linear, isotropic, and homogeneous medium. For a perfect dielectric (σ = 0), these expressions reduce to those for the lossless case (Section 7-2), wherein α = 0, β = k = ω√μϵ, and ηc = η.
For a lossy medium, the ratio ϵ′′/ϵ′ = σ/ωϵ, which appears in all these expressions, plays an important role in classifying how lossy the medium is. When ϵ′′/ϵ′ ≪1, the medium is considered a low-loss dielectric, and when ϵ′′/ϵ′ ≫1, it is considered a good conductor. In practice, the medium may be regarded as a low-loss dielectric if ϵ′′/ϵ′ < 10−2, as a good conductor if ϵ′′/ϵ′ > 102, and as a quasi-conductor if 10−2 ≤ϵ′′/ϵ′ ≤102.
For low-loss dielectrics and good conductors, the expressions given by Eq. (7.66) can be significantly simplified, as shown next.
7-4.1 Low-Loss Dielectric From Eq. (7.62), the general expression for γ is γ = jω μϵ′ 1 −j ϵ′′ ϵ′ 1/2 .
(7.73) For |x| ≪1, the function (1 −x)1/2 can be approximated by the first two terms of its binomial series; that is, (1 −x)1/2 ≈ 1 −x/2. By applying this approximation to Eq. (7.73) for a low-loss dielectric with x = jϵ′′/ϵ′ and ϵ′′/ϵ′ ≪1, we obtain γ ≈jω μϵ′ 1 −j ϵ′′ 2ϵ′ .
(7.74) The real and imaginary parts of Eq. (7.74) are 334 CHAPTER 7 PLANE-WAVE PROPAGATION α ≈ωϵ′′ 2 μ ϵ′ = σ 2 μ ϵ (Np/m), (7.75a) β ≈ω μϵ′ = ω√μϵ (rad/m).
(7.75b) (low-loss medium) We note that the expression for β is the same as that for the wavenumber k of a lossless medium. Applying the binomial approximation (1 −x)−1/2 ≈(1 + x/2) to Eq. (7.70) leads to ηc ≈ μ ϵ′ 1 + j ϵ′′ 2ϵ′ = μ ϵ 1 + j σ 2ωϵ .
(7.76a) In practice, because ϵ′′/ϵ′ = σ/ωϵ < 10−2, the second term in Eq. (7.76a) often is ignored. Thus, ηc ≈ μ ϵ , (7.76b) which is the same as Eq. (7.31) for the lossless case.
7-4.2 Good Conductor When ϵ′′/ϵ′ > 100, Eqs. (7.66a), (7.66b), and (7.70) can be approximated as α ≈ω μϵ′′ 2 = ω μσ 2ω = πf μσ (Np/m), (7.77a) β = α ≈ πf μσ (rad/m), (7.77b) ηc ≈ j μ ϵ′′ = (1 + j) πf μ σ = (1 + j)α σ ( ).
(good conductor) (7.77c) In Eq. (7.77c), we used the relation given by Eq. (1.53): √j = (1 + j)/ √ 2. For a perfect conductor with σ = ∞, these expressionsyieldα = β = ∞, andηc = 0. Aperfectconductor is equivalent to a short circuit in a transmission line equivalent.
Expressions for the propagation parameters in various types of media are summarized in Table 7-1.
Example 7-4: Plane Wave in Seawater A uniform plane wave is traveling in seawater. Assume that the x–y plane resides just below the sea surface and the wave travels in the +z direction into the water.
The constitutive parameters of seawater are ϵr = 80, μr = 1, and σ = 4 S/m.
If the magnetic field at z = 0 is H(0, t) = ˆ y 100 cos(2π × 103t + 15◦) (mA/m), (a) obtain expressions for E(z, t) and H(z, t), and (b) determine the depth at which the magnitude of E is 1% of its value at z = 0.
Solution: (a) Since H is along ˆ y and the propagation direction is ˆ z, E must be along ˆ x. Hence, the general expressions for the phasor fields are E(z) = ˆ xEx0e−αze−jβz, (7.78a) H(z) = ˆ y Ex0 ηc e−αze−jβz.
(7.78b) To determine α, β, and ηc for seawater, we begin by evaluating the ratio ϵ′′/ϵ′. From the argument of the cosine function of H(0, t), we deduce that ω = 2π × 103 (rad/s), and therefore f = 1 kHz. Hence, ϵ′′ ϵ′ = σ ωϵ = σ ωϵrϵ0 = 4 2π × 103 × 80 × (10−9/36π) = 9 × 105.
This qualifies seawater as a good conductor at 1 kHz and allows us to use the good-conductor expressions given in Table 7-1: α = πf μσ = π × 103 × 4π × 10−7 × 4 = 0.126 (Np/m), (7.79a) β = α = 0.126 (rad/m), (7.79b) ηc = (1 + j)α σ = ( √ 2 ejπ/4)0.126 4 = 0.044ejπ/4 ( ).
(7.79c) As no explicit information has been given about the electric field amplitude Ex0, we should assume it to be complex; that 7-4 PLANE-WAVE PROPAGATION IN LOSSY MEDIA 335 Table 7-1 Expressions for α, β, ηc, up, and λ for various types of media.
Lossless Low-loss Good Any Medium Medium Medium Conductor Units (σ = 0) (ϵ′′/ϵ′ ≪1) (ϵ′′/ϵ′ ≫1) α = ω ⎡ ⎣μϵ′ 2 ⎡ ⎣ 1 + ϵ′′ ϵ′ 2 −1 ⎤ ⎦ ⎤ ⎦ 1/2 0 σ 2 μ ϵ πf μσ (Np/m) β = ω ⎡ ⎣μϵ′ 2 ⎡ ⎣ 1 + ϵ′′ ϵ′ 2 + 1 ⎤ ⎦ ⎤ ⎦ 1/2 ω√μϵ ω√μϵ πf μσ (rad/m) ηc = μ ϵ′ 1 −j ϵ′′ ϵ′ −1/2 μ ϵ μ ϵ (1 + j) α σ ( ) up = ω/β 1/√μϵ 1/√μϵ √4πf/μσ (m/s) λ = 2π/β = up/f up/f up/f up/f (m) Notes: ϵ′ = ϵ; ϵ′′ = σ/ω; in free space, ϵ = ϵ0, μ = μ0; in practice, a material is considered a low-loss medium if ϵ′′/ϵ′ = σ/ωϵ < 0.01 and a good conducting medium if ϵ′′/ϵ′ > 100.
is, Ex0 = |Ex0|ejφ0. The wave’s instantaneous electric and magnetic fields are given by E(z, t) = Re ˆ x|Ex0|ejφ0e−αze−jβzejωt = ˆ x|Ex0|e−0.126z cos(2π × 103t −0.126z + φ0) (V/m), (7.80a) H(z, t) = Re ˆ y |Ex0|ejφ0 0.044ejπ/4 e−αze−jβzejωt = ˆ y22.5|Ex0|e−0.126z cos(2π × 103t −0.126z + φ0 −45◦) (A/m).
(7.80b) At z = 0, H(0, t) = ˆ y 22.5|Ex0| cos(2π × 103t + φ0 −45◦) (A/m).
(7.81) By comparing Eq. (7.81) with the expression given in the problem statement, H(0, t) = ˆ y 100 cos(2π × 103t + 15◦) (mA/m), we deduce that 22.5|Ex0| = 100 × 10−3 or |Ex0| = 4.44 (mV/m), and φ0 −45◦= 15◦ or φ0 = 60◦.
Hence, the final expressions for E(z, t) and H(z, t) are E(z, t) = ˆ x 4.44e−0.126z cos(2π × 103t −0.126z + 60◦) (mV/m), (7.82a) H(z, t) = ˆ y 100e−0.126z cos(2π × 103t −0.126z + 15◦) (mA/m).
(7.82b) 336 TECHNOLOGY BRIEF 14: LIQUID CRYSTAL DISPLAY (LCD) Technology Brief 14: Liquid Crystal Display (LCD) LCDs are used in digital clocks, cellular phones, desktop and laptop computers, and some televisions and other electronic systems. They offer a decided advantage over former display technologies, such as cathode ray tubes, in that they are much lighter and thinner and consume a lot less power to operate. LCD technology relies on special electrical and optical properties of a class of materials known as liquid crystals, first discovered in the 1880s by botanist Friedrich Reinitzer.
Physical Principle ▶Liquid crystals are neither a pure solid nor a pure liquid, but rather a hybrid of both. ◀ One particular variety of interest is the twisted nematic liquid crystal whose rod-shaped molecules have a natural tendency to assume a twisted spiral structure when the material is sandwiched between finely grooved glass substrates with orthogonal orientations (Fig. TF14-1). Note that the molecules in contact with the grooved surfaces align themselves in parallel along the grooves, from a y orientation at the entrance substrate into an x orientation at the x x-polarized light x x-oriented exit substrate y-oriented entrance substrate Rod-shaped molecules y Only y-polarized component can pass through polarizing filter x-polarized component of incident light y-oriented polarizing filter x-oriented polarizing filter Orthogonal groove orientations Unpolarized light Figure TF14-1 The rod-shaped molecules of a liquid crystal sandwiched between grooved substrates with orthogonal orientations cause the electric field of the light passing through it to rotate by 90◦.
TECHNOLOGY BRIEF 14: LIQUID CRYSTAL DISPLAY (LCD) 337 V (a) ON state (switch open) Polarizing filter Molecule of liquid crystal V Dark pixel 5 μm (b) OFF state (switch closed) + _ Bright pixel Liquid crystal + _ FigureTF14-2 Single-pixel LCD.
exit substrate. The molecular spiral causes the crystal to behave like a wave polarizer: unpolarized light incident upon the entrance substrate follows the orientation of the spiral, emerging through the exit substrate with its polarization (direction of electric field) parallel to the groove’s direction, which in Fig. TF14-1 is along the x direction. Thus, of the x and y components of the incident light, only the y component is allowed to pass through the y-polarized filter, but as a consequence of the spiral action facilitated by the liquid crystal’s molecules, the light that emerges from the LCD structure is x-polarized.
LCD Structure A single-pixel LCD structure is shown in Fig. TF14-2 for the OFF and ON states, with OFF corresponding to a bright-looking pixel and ON to a dark-looking pixel.
▶The sandwiched liquid-crystal layer (typically on the order of 5 microns in thickness, or 1/20 of the width of a human hair) is straddled by a pair of optical filters with orthogonal polarizations. ◀ When no voltage is applied across the crystal layer [Fig. TF14-2(a)], incoming unpolarized light gets polarized as it passes through the entrance polarizer, then rotates by 90◦as it follows the molecular spiral, and finally emerges from the exit polarizer, giving the exited surface a bright appearance. A useful feature of nematic liquid crystals is that their spiral untwists [Fig.TF14-2(b)] under the influence of an electric field (induced by a voltage difference across the layer).
The degree of untwisting depends on the strength of the electric field. With no spiral to rotate the wave polarization as the light travels through the crystal, the light polarization becomes orthogonal to that of the exit polarizer, allowing no light to pass through it. Hence, the pixel exhibits a dark appearance.
338 TECHNOLOGY BRIEF 14: LIQUID CRYSTAL DISPLAY (LCD) LCD display Liquid crystal Unpolarized light Exit polarizer Entrance polarizer 2-D pixel array Molecular spiral 678 FigureTF14-3 2-D LCD array.
▶By extending the concept to a two-dimensional array of pixels and devising a scheme to control the voltage across each pixel individually (usually by using a thin-film transistor), a complete image can be displayed as illustrated in Fig. TF14-3. For color displays, each pixel is made up of three subpixels with complementary color filters (red, green, and blue). ◀ FigureTF14-4 LCD display.
7-5 CURRENT FLOW IN A GOOD CONDUCTOR 339 Module 7.5 Wave Attenuation Observe the profile of a plane wave propagating in a lossy medium. Determine the skin depth, the propagation parameters, and the intrinsic impedance of the medium.
(b) The depth at which the amplitude of E has decreased to 1% of its initial value at z = 0 is obtained from 0.01 = e−0.126z or z = ln(0.01) −0.126 = 36.55 m ≈37 m.
Exercise 7-7: The constitutive parameters of copper are μ = μ0 = 4π × 10−7 (H/m), ϵ = ϵ0 ≈(1/36π) × 10−9 (F/m), and σ = 5.8 × 107 (S/m).
Assuming that these parameters are frequency independent, over what frequency range of the electromagnetic spectrum (see Fig. 1-16) is copper a good conductor?
Answer: f < 1.04 × 1016 Hz, which includes the radio, infrared, visible, and part of the ultraviolet regions of the EM spectrum. (See EM.) Exercise 7-8: Over what frequency range may dry soil, with ϵr = 3, μr = 1, and σ = 10−4 (S/m), be regarded as a low-loss dielectric?
Answer: f > 60 MHz. (See EM.) Exercise 7-9: For a wave traveling in a medium with a skin depth δs, what is the amplitude of E at a distance of 3δs compared with its initial value?
Answer: e−3 ≈0.05 or 5%. (See EM.) 7-5 Current Flow in a Good Conductor When a dc voltage source is connected across the ends of a conducting wire, the current flowing through the wire is uniformly distributed over its cross section. That is, the current density J is the same along the axis of the wire and along its outer perimeter [Fig. 7-14(a)]. This is not true in the ac case. As we will see shortly, a time-varying current density is maximum along the perimeter of the wire and decreases exponentially as 340 CHAPTER 7 PLANE-WAVE PROPAGATION (a) dc case (b) ac case J R V I J I R V(t) Figure 7-14 Current density J in a conducting wire is (a) uniform across its cross section in the dc case, but (b) in the ac case, J is highest along the wire’s perimeter.
a function of distance toward the axis of the wire [Fig. 7-14(b)].
In fact, at very high frequencies most of the current flows in a thin layer near the wire surface, and if the wire material is a perfect conductor, the current flows entirely on the surface of the wire.
Before analyzing a wire with circular cross section, let us consider the simpler geometry of a semi-infinite conducting solid, as shown in Fig. 7-15(a). The solid’s planar interface with a perfect dielectric is the x–y plane. If at z = 0−(just above the surface), an x-polarized electric field with E = ˆ xE0 exists in the dielectric, a similarly polarized field is induced in the conducting medium and propagates as a plane wave along the +z direction. As a consequence of the boundary condition mandating continuity of the tangential component of E across the boundary between any two contiguous media, the electric field at z = 0+ (just below the boundary) is E(0) = ˆ xE0 also.
The EM fields at any depth z in the conductor are then given by E(z) = ˆ xE0e−αze−jβz, (7.83a) H(z) = ˆ y E0 ηc e−αze−jβz.
(7.83b) J0 E0 H0 Jx(z) ~ x z (b) Equivalent J0 over skin depth δs (a) Exponentially decaying Jx(z) ~ J0 l δs z w ∞ Figure 7-15 Exponential decay of current density Jx(z) with z in a solid conductor. The total current flowing through (a) a section of width w extending between z = 0 and z = ∞is equivalent to (b) a constant current density J0 flowing through a section of depth δs.
From J = σE, the current flows in the x direction, and its density is J(z) = ˆ x Jx(z), (7.84) with Jx(z) = σE0e−αze−jβz = J0e−αze−jβz, (7.85) where J0 = σE0 is the amplitude of the current density at the surface. In terms of the skin depth δs = 1/α defined by Eq. (7.72) and using the fact that in a good conductor α = β as expressed by Eq. (7.77b), Eq. (7.85) can be written as Jx(z) = J0e−(1+j)z/δs (A/m2).
(7.86) 7-5 CURRENT FLOW IN A GOOD CONDUCTOR 341 The current flowing through a rectangular strip of width w along the y direction and extending between zero and ∞in the z direction is ˜ I = w ∞ 0 Jx(z) dz = w ∞ 0 J0e−(1+j)z/δs dz = J0wδs (1 + j) (A).
(7.87) The numerator of Eq. (7.87) is reminiscent of a uniform current density J0 flowing through a thin surface of width w and depth δs. Because Jx(z) decreases exponentially with depth z, a conductor of finite thickness d can be considered electrically equivalent to one of infinite depth as long as d exceeds a few skin depths. Indeed, if d = 3δs [instead of ∞in the integral of Eq. (7.87)], the error incurred in using the result on the right-hand side of Eq. (7.87) is less than 5%; and if d = 5δs, the error is less than 1%.
The voltage across a length l at the surface [Fig. 7-15(b)] is given by V = E0l = J0 σ l.
(7.88) Hence, the impedance of a slab of width w, length l, and depth d = ∞(or, in practice, d > 5δs) is Z = V ˜ I = 1 + j σδs l w ( ).
(7.89) It is customary to represent Z as Z = Zs l w , (7.90) where Zs, the internal or surface impedance of the conductor, is defined as the impedance Z for a length l = 1 m and a width w = 1 m. Thus, Zs = 1 + j σδs ( ).
(7.91) Since the reactive part of Zs is positive, Zs can be defined as Zs = Rs + jωLs δs 2πa (a) Coaxial cable (b) Equivalent inner conductor Dielectric Dielectric 2a 2b Inner conductor Outer conductor Figure 7-16 The inner conductor of the coaxial cable in (a) is represented in (b) by a planar conductor of width 2πa and depth δs, as if its skin has been cut along its length on the bottom side and then unfurled into a planar geometry.
with Rs = 1 σδs = πf μ σ ( ), (7.92a) Ls = 1 ωσδs = 1 2 μ πf σ (H), (7.92b) where we used the relation δs = 1/α ≈1/ √πf μσ given by Eq. (7.77a).
In terms of the surface resistance Rs, the ac resistance of a slab of width w and length l is R = Rs l w = l σδsw ( ).
(7.93) The expression for the ac resistance R is equivalent to the dc resistance of a plane conductor of length l and cross section A = δsw.
The results obtained for the planar conductor can be extended to the coaxial cable shown in Fig. 7-16(a). If the conductors are made of copper with σ = 5.8 × 107 S/m, the skin depth at 1 MHz is δs = 1/ √πf μσ = 0.066 mm, and since δs varies as 1/ √f , it becomes smaller at higher frequencies. As long as the inner conductor’s radius a is greater than 5δs, or 0.33 mm at 1 MHz, its “depth” may be regarded as infinite. A similar criterion applies to the thickness of the outer conductor. To 342 CHAPTER 7 PLANE-WAVE PROPAGATION Module 7.6 Current in a Conductor Module displays exponential decay of current density in a conductor.
compute the resistance of the inner conductor, note that the current is concentrated near its outer surface and approximately equivalent to a uniform current flowing through a thin layer of depth δs and circumference 2πa.
In other words, the inner conductor’s resistance is nearly the same as that of a planar conductor of depth δs and width w = 2πa, as shown in Fig. 7-16(b). The corresponding resistance per unit length is obtained by setting w = 2πa and dividing by l in Eq. (7.93): R′ 1 = R l = Rs 2πa ( /m).
(7.94) Similarly, for the outer conductor, the current is concentrated within a thin layer of depth δs on the inside surface of the conductor adjacent to the insulating medium between the two conductors, which is where the EM fields exist. The resistance per unit length for the outer conductor with radius b is R′ 2 = Rs 2πb ( /m), (7.95) and the coaxial cable’s total ac resistance per unit length is R′ = R′ 1 + R′ 2 = Rs 2π 1 a + 1 b ( /m).
(7.96) This expression was used in Chapter 2 for characterizing the resistance per unit length of a coaxial transmission line.
Concept Question 7-6: How does β of a low-loss dielectric medium compare to that of a lossless medium?
Concept Question 7-7: In a good conductor, does the phase of H lead or lag that of E and by how much?
Concept Question 7-8: Attenuation means that a wave loses energy as it propagates in a lossy medium. What happens to the lost energy?
Concept Question 7-9: Is a conducting medium dis-persive or dispersionless? Explain.
7-6 ELECTROMAGNETIC POWER DENSITY 343 Concept Question 7-10: Compare the flow of current through a wire in the dc and ac cases.
Compare the corresponding dc and ac resistances of the wire.
7-6 Electromagnetic Power Density This section deals with the flow of power carried by an electromagnetic wave. For any wave with an electric field E and magnetic field H, the Poynting vector S is defined as S = E × × × H (W/m2).
(7.97) The unit of S is (V/m) × (A/m) = (W/m2), and the direction of S is along the wave’s direction of propagation. Thus, S represents the power per unit area (or power density) carried by the wave.
If the wave is incident upon an aperture of area A with outward surface unit vector ˆ n as shown in Fig. 7-17, then the total power that flows through or is intercepted by the aperture is P = A S· ˆ ndA (W).
(7.98) For a uniform plane wave propagating in a direction ˆ k that makes an angle θ with ˆ n, P = SA cos θ, where S = |S|.
Except for the fact that the units of S are per unit area, Eq. (7.97) is the vector analogue of the scalar expression for the instantaneous power P(z, t) flowing through a transmission line, P(z, t) = v(z, t) i(z, t), (7.99) where v(z, t) and i(z, t) are the instantaneous voltage and current on the line.
Since both E and H are functions of time, so is the Poynting vector S. In practice, however, the quantity of greater interest S A n k θ ˆ ˆ Figure 7-17 EM power flow through an aperture.
is the average power density of the wave, Sav, which is the time-average value of S: Sav = 1 2 Re E × × × H∗ (W/m2).
(7.100) This expression may be regarded as the electromagnetic equivalent of Eq. (2.107) for the time-average power carried by a transmission line, namely Pav(z) = 1 2 Re V (z) ˜ I ∗(z) , (7.101) where V (z) and ˜ I(z) are the phasors corresponding to v(z, t) and i(z, t), respectively.
7-6.1 Plane Wave in a Lossless Medium Recall that the general expression for the electric field of a uniform plane wave with arbitrary polarization traveling in the +z direction is E(z) = ˆ x Ex(z) + ˆ y Ey(z) = (ˆ x Ex0 + ˆ y Ey0)e−jkz, (7.102) where, in the general case, Ex0 and Ey0 may be complex quantities. The magnitude of E is | E| = ( E· E∗)1/2 = [|Ex0|2 + |Ey0|2]1/2.
(7.103) The phasor magnetic field associated with E is obtained by applying Eq. (7.39a): H(z) = (ˆ x Hx + ˆ y Hy)e−jkz = 1 η ˆ z × × × E = 1 η (−ˆ x Ey0 + ˆ y Ex0)e−jkz.
(7.104) 344 CHAPTER 7 PLANE-WAVE PROPAGATION The wave can be considered as the sum of two waves, one comprising fields ( Ex, Hy) and another comprising fields ( Ey, Hx). Use of Eqs. (7.102) and (7.104) in Eq. (7.100) leads to Sav = ˆ z 1 2η(|Ex0|2 + |Ey0|2) = ˆ z | E|2 2η (W/m2), (lossless medium) (7.105) which states that power flows in the z direction with average power density equal to the sum of the average power densities of the ( Ex, Hy) and ( Ey, Hx) waves. Note that, because Sav depends only on η and | E|, waves characterized by different polarizations carry the same amount of average power as long as their electric fields have the same magnitudes.
Example 7-5: Solar Power If solar illumination is characterized by a power density of 1 kW/m2 on Earth’s surface, find (a) the total power radiated by the sun, (b) the total power intercepted by Earth, and (c) the electric field of the power density incident upon Earth’s surface, assuming that all the solar illumination is at a single frequency.
The radius of Earth’s orbit around the sun, Rs, is approximately 1.5 × 108 km, and Earth’s mean radius Re is 6,380 km.
Solution: (a) Assuming that the sun radiates isotropically (equally in all directions), the total power it radiates is Sav Asph, where Asph is the area of a spherical shell of radius Rs [Fig. 7-18(a)]. Thus, Psun = Sav(4πR2 s ) = 1 × 103 × 4π × (1.5 × 1011)2 = 2.8 × 1026 W.
(b) With reference to Fig. 7-18(b), the power intercepted by Earth’s cross section Ae = πR2 e is Pint = Sav(πR2 e ) = 1 × 103 × π × (6.38 × 106)2 = 1.28 × 1017 W.
(c) The power density Sav is related to the magnitude of the electric field | E| = E0 by Sav = E2 0 2η0 , Sun S S S S S S Rs Area of spherical surface Asph = 4πRs 2 Earth (a) Radiated solar power (b) Earth intercepted power S Sun Ae = πRe 2 Earth Figure 7-18 Solar radiation intercepted by (a) a spherical surface of radius Rs, and (b) Earth’s surface (Example 7-5).
where η0 = 377 ( ) for air. Hence, E0 = 2η0Sav = 2 × 377 × 103 = 870 (V/m).
7-6.2 Plane Wave in a Lossy Medium The expressions given by Eqs. (7.68) and (7.69) characterize the electric and magnetic fields of an x-polarized plane wave propagating along the z direction in a lossy medium with propagation constant γ = α + jβ. By extending these 7-6 ELECTROMAGNETIC POWER DENSITY 345 expressions to the more general case of a wave with components along both x and y, we have E(z) = ˆ x Ex(z) + ˆ y Ey(z) = (ˆ x Ex0 + ˆ y Ey0)e−αze−jβz, (7.106a) H(z) = 1 ηc (−ˆ x Ey0 + ˆ y Ex0)e−αze−jβz, (7.106b) where ηc is the intrinsic impedance of the lossy medium.
Application of Eq. (7.100) gives Sav(z) = 1 2 Re E × × × H∗ = ˆ z(|Ex0|2 + |Ey0|2) 2 e−2αz Re 1 η∗ c .
(7.107) By expressing ηc in polar form as ηc = |ηc|ejθη, (7.108) Eq. (7.107) can be rewritten as Sav(z) = ˆ z | E(0)|2 2|ηc| e−2αz cos θη (W/m2), (7.109) (lossy medium) where | E(0)|2 = [|Ex0|2 +|Ey0|2]1/2 is the magnitude of E(z) at z = 0.
▶Whereas the fields E(z) and H(z) decay with z as e−αz, the power density Sav decreases as e−2αz. ◀ When a wave propagates through a distance z = δs = 1/α, the magnitudes of its electric and magnetic fields decrease to e−1 ≈37% of their initial values, and its average power density decreases to e−2 ≈14% of its initial value.
7-6.3 Decibel Scale for Power Ratios The unit for power P is watts (W). In many engineering problems, the quantity of interest is the ratio of two power levels, P1 and P2, such as the incident and reflected powers on a transmission line, and often the ratio P1/P2 may vary over several orders of magnitude. The decibel (dB) scale is logarithmic, thereby providing a convenient representation of the power ratio, particularly when numerical values of P1/P2 are plotted against some variable of interest. If G = P1 P2 , (7.110) then G [dB] = 10 log G = 10 log P1 P2 (dB).
(7.111) Table 7-2 provides a comparison between values of G and the corresponding values of G [dB].
Even though decibels are defined for power ratios, they can sometimes be used to represent other quantities. For example, if P1 = V 2 1 /R is the power dissipated in a resistor R with voltage V1 across it at time t1, and P2 = V 2 2 /R is the power dissipated in the same resistor at time t2, then G [dB] = 10 log P1 P2 = 10 log V 2 1 /R V 2 2 /R = 20 log V1 V2 = 20 log(g) = g [dB], (7.112) where g = V1/V2 is the voltage ratio. Note that for voltage (or current) ratios the scale factor is 20 rather than 10, which results in G [dB] = g [dB].
Table 7-2 Power ratios in natural numbers and in decibels.
G G [dB] 10x 10x dB 4 6 dB 2 3 dB 1 0 dB 0.5 −3 dB 0.25 −6 dB 0.1 −10 dB 10−3 −30 dB 346 CHAPTER 7 PLANE-WAVE PROPAGATION The attenuation rate, representing the rate of decrease of the magnitude of Sav(z) as a function of propagation distance, is defined as A = 10 log Sav(z) Sav(0) = 10 log(e−2αz) = −20αz log e = −8.68αz = −α [dB/m] z (dB), (7.113) where α [dB/m] = 8.68α (Np/m).
(7.114) Wealsonotethat, sinceSav(z)isdirectlyproportionalto|E(z)|2, A = 10 log |E(z)|2 |E(0)|2 = 20 log |E(z)| |E(0)| (dB).
(7.115) Example 7-6: Power Received by a Submarine Antenna A submarine at a depth of 200 m below the sea surface uses a wire antenna to receive signal transmissions at 1 kHz.
Determine the power density incident upon the submarine antenna due to the EM wave of Example 7-4.
Solution: From Example 7-4, | E(0)| = |Ex0| = 4.44 (mV/m), α = 0.126 (Np/m), and ηc = 0.044∠ 45◦( ). Appli-cation of Eq. (7.109) gives Sav(z) = ˆ z |E0|2 2|ηc| e−2αz cos θη = ˆ z (4.44 × 10−3)2 2 × 0.044 e−0.252z cos 45◦ = ˆ z 0.16e−0.252z (mW/m2).
At z = 200 m, the incident power density is Sav = ˆ z (0.16 × 10−3e−0.252×200) = 2.1 × 10−26 (W/m2).
Exercise 7-10: Convert the following values of the power ratio G to decibels: (a) 2.3, (b) 4 × 103, (c) 3 × 10−2.
Answer: (a) 3.6 dB, (b) 36 dB, (c) −15.2 dB. (See EM.) Exercise 7-11: Find the voltage ratio g corresponding to the following decibel values of the power ratio G: (a) 23 dB, (b) −14 dB, (c) −3.6 dB.
Answer: (a) 14.13, (b) 0.2, (c) 0.66. (See EM.) Chapter 7 Summary Concepts • A spherical wave radiated by a source becomes approximately a uniform plane wave at large distances from the source.
• The electric and magnetic fields of a transverse electromagnetic (TEM) wave are orthogonal to each other, and both are perpendicular to the direction of wave travel.
• The magnitudes of the electric and magnetic fields of a TEM wave are related by the intrinsic impedance of the medium.
• Wave polarization describes the shape of the locus of the tip of the E vector at a given point in space as a function of time. The polarization state, which may be linear, circular, or elliptical, is governed by the ratio of the magnitudes of and the difference in phase between the two orthogonal components of the electric field vector.
• Media are classified as lossless, low-loss, quasi-conducting, or good-conducting on the basis of the ratio ϵ′′/ϵ′ = σ/ωϵ.
• Unlike the dc case, wherein the current flowing through a wire is distributed uniformly across its cross section, in the ac case most of the current is concentrated along the outer perimeter of the wire.
• Power density carried by a plane EM wave traveling in an unbounded medium is akin to the power carried by the voltage/current wave on a transmission line.
CHAPTER 7 SUMMARY 347 Mathematical and Physical Models Complex Permittivity ϵc = ϵ′ −jϵ′′ ϵ′ = ϵ ϵ′′ = σ ω Lossless Medium k = ω√μϵ η = μ ϵ ( ) up = ω k = 1 √μϵ (m/s) λ = 2π k = up f (m) Wave Polarization H = 1 η ˆ k × × × E E = −η ˆ k × × × H Maxwell’s Equations for Time-Harmonic Fields ∇· E = 0 ∇× × × E = −jωμ H ∇· H = 0 ∇× × × H = jωϵc E Lossy Medium α = ω ⎧ ⎨ ⎩ μϵ′ 2 ⎡ ⎣ 1 + ϵ′′ ϵ′ 2 −1 ⎤ ⎦ ⎫ ⎬ ⎭ 1/2 (Np/m) β = ω ⎧ ⎨ ⎩ μϵ′ 2 ⎡ ⎣ 1 + ϵ′′ ϵ′ 2 + 1 ⎤ ⎦ ⎫ ⎬ ⎭ 1/2 (rad/m) ηc = μ ϵc = μ ϵ′ 1 −j ϵ′′ ϵ′ −1/2 ( ) δs = 1 α (m) Power Density Sav = 1 2 Re E × × × H∗ (W/m2) Important Terms Provide definitions or explain the meaning of the following terms: attenuation constant α attenuation rate A auxiliary angle ψ0 average power density Sav axial ratio circular polarization complex permittivity ϵc dc and ac resistances elliptical polarization ellipticity angle χ good conductor guided wave homogeneous wave equation in phase inclination angle internal or surface impedance intrinsic impedance η LHC and RHC polarizations linear polarization lossy medium low-loss dielectric out of phase phase constant β phase velocity polarization state Poynting vector S propagation constant γ quasi-conductor rotation angle γ skin depth δs spherical wave surface resistance Rs TEM wave unbounded unbounded wave uniform plane wave wave polarization wavefront wavenumber k 348 CHAPTER 7 PLANE-WAVE PROPAGATION PROBLEMS Section 7-2: Plane-Wave Propagation in Lossless Media 7.1 Themagneticfieldofawavepropagatingthroughacertain nonmagnetic material is given by H = ˆ z 30 cos(108t −0.5y) (mA/m).
Find the following: ∗(a) The direction of wave propagation.
(b) The phase velocity.
∗(c) The wavelength in the material.
(d) The relative permittivity of the material.
(e) The electric field phasor.
7.2 Write general expressions for the electric and magnetic fields of a 1 GHz sinusoidal plane wave traveling in the +y direction in a lossless nonmagnetic medium with relative permittivity ϵr = 9. The electric field is polarized along the x direction, its peak value is 6 V/m, and its intensity is 4 V/m at t = 0 and y = 2 cm.
7.3 The electric field phasor of a uniform plane wave is given by E = ˆ y 10ej0.2z (V/m). If the phase velocity of the wave is 1.5 × 108 m/s and the relative permeability of the medium is μr = 2.4, find the following: ∗(a) The wavelength.
(b) The frequency f of the wave.
(c) The relative permittivity of the medium.
(d) The magnetic field H(z, t).
7.4 The electric field of a plane wave propagating in a nonmagnetic material is given by E = [ˆ y 3 sin(π × 107t −0.2πx) + ˆ z 4 cos(π × 107t −0.2πx)] (V/m).
Determine (a) The wavelength.
(b) ϵr.
(c) H.
∗Answer(s) available in Appendix D.
∗7.5 A wave radiated by a source in air is incident upon a soil surface, whereupon a part of the wave is transmitted into the soil medium. If the wavelength of the wave is 60 cm in air and 20cminthesoilmedium, whatisthesoil’srelativepermittivity?
Assume the soil to be a very low-loss medium.
7.6 The electric field of a plane wave propagating in a lossless, nonmagnetic, dielectric material with ϵr = 2.56 is given by E = ˆ y 20 cos(6π × 109t −kz) (V/m).
Determine: (a) f , up, λ, k, and η.
(b) The magnetic field H.
7.7 The magnetic field of a plane wave propagating in a nonmagnetic material is given by H = ˆ x 60 cos(2π × 107t + 0.1πy) ˆ z 30 cos(2π × 107t + 0.1πy) (mA/m).
Determine ∗(a) The wavelength.
(b) ϵr.
(c) E.
7.8 A 60 MHz plane wave traveling in the −x direction in dry soil with relative permittivity ϵr = 4 has an electric field polarized along the z direction.
Assuming dry soil to be approximately lossless, and given that the magnetic field has a peak value of 10 (mA/m) and that its value was measured to be 7 (mA/m) at t = 0 and x = −0.75 m, develop complete expressions for the wave’s electric and magnetic fields.
Section 7-3: Wave Polarization ∗7.9 An RHC-polarized wave with a modulus of 2 (V/m) is traveling in free space in the negative z direction. Write the expression for the wave’s electric field vector, given that the wavelength is 6 cm.
7.10 For a wave characterized by the electric field E(z, t) = ˆ x ax cos(ωt −kz) + ˆ y ay cos(ωt −kz + δ) identify the polarization state, determine the polarization angles (γ, χ), and sketch the locus of E(0, t) for each of the following cases: PROBLEMS 349 (a) ax = 3 V/m, ay = 4 V/m, and δ = 0 (b) ax = 3 V/m, ay = 4 V/m, and δ = 180◦ (c) ax = 3 V/m, ay = 3 V/m, and δ = 45◦ (d) ax = 3 V/m, ay = 4 V/m, and δ = −135◦ 7.11 The electric field of a uniform plane wave propagating in free space is given by E = (ˆ x + j ˆ y)30e−jπz/6 (V/m).
Specify the modulus and direction of the electric field intensity at the z = 0 plane at t = 0, 5, and 10 ns.
∗7.12 The magnetic field of a uniform plane wave propagating in a dielectric medium with ϵr = 36 is given by H = 30(ˆ y + j ˆ z)e−jπx/6 (mA/m).
Specify the modulus and direction of the electric field intensity at the x = 0 plane at t = 0 and 5 ns.
7.13 A linearly polarized plane wave of the form E = ˆ x axe−jkz can be expressed as the sum of an RHC-polarized wave with magnitude aR, and an LHC-polarized wave with magnitude aL.
Prove this statement by finding expressions for aR and aL in terms of ax.
∗7.14 The electric field of an elliptically polarized plane wave is given by E(z, t) = [−ˆ x 10 sin(ωt −kz −60◦) + ˆ y 30 cos(ωt −kz)] (V/m).
Determine the following: (a) The polarization angles (γ, χ).
(b) The direction of rotation.
7.15 Compare the polarization states of each of the following pairs of plane waves: (a) Wave 1: E1 = ˆ x 2 cos(ωt −kz) + ˆ y 2 sin(ωt −kz).
Wave 2: E2 = ˆ x 2 cos(ωt + kz) + ˆ y 2 sin(ωt + kz).
(b) Wave 1: E1 = ˆ x 2 cos(ωt −kz) −ˆ y 2 sin(ωt −kz).
Wave 2: E2 = ˆ x 2 cos(ωt + kz) −ˆ y 2 sin(ωt + kz).
7.16 Plot the locus of E(0, t) for a plane wave with E(z, t) = ˆ x sin(ωt + kz) + ˆ y 2 cos(ωt + kz).
Determine the polarization state from your plot.
Section 7-4: Plane-Wave Propagation in Lossy Media 7.17 For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc: ∗(a) Glass with μr = 1, ϵr = 5, and σ = 10−12 S/m at 10 GHz.
(b) Animal tissue with μr = 1, ϵr = 12, and σ = 0.3 S/m at 100 MHz.
(a) (c)] Wood with μr = 1, ϵr = 3, and σ = 10−4 S/m at 1 kHz.
7.18 Dry soil is characterized by ϵr = 2.5, μr = 1, and σ = 10−4 (S/m).
At each of the following frequencies, determine if dry soil may be considered a good conductor, a quasi-conductor, or a low-loss dielectric, and then calculate α, β, λ, μp, and ηc: (a) 60 Hz (b) 1 kHz (c) 1 MHz (d) 1 GHz ∗7.19 In a medium characterized by ϵr = 9, μr = 1, and σ = 0.1 S/m, determine the phase angle by which the magnetic field leads the electric field at 100 MHz.
7.20 Generate a plot for the skin depth δs versus frequency for seawater for the range from 1 kHz to 10 GHz (use log-log scales). The constitutive parameters of seawater are μr = 1, ϵr = 80, and σ = 4 S/m.
∗7.21 Ignoring reflection at the air–soil boundary, if the amplitude of a 3 GHz incident wave is 10 V/m at the surface of a wet soil medium, at what depth will it be down to 1 mV/m?
Wet soil is characterized by μr = 1, ϵr = 9, and σ = 5 × 10−4 S/m.
7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surface, at what depth will it be down to 1 μV/m? Water has μr = 1, and at 1 GHz, ϵr = 80 and σ = 1 S/m.
∗7.23 The skin depth of a certain nonmagnetic conducting material is 3 μm at 5 GHz. Determine the phase velocity in the material.
7.24 Based on wave attenuation and reflection measurements conducted at 1 MHz, it was determined that the intrinsic impedance of a certain medium is 28.1∠ 45◦( ) and the skin depth is 2 m. Determine the following: 350 CHAPTER 7 PLANE-WAVE PROPAGATION (a) The conductivity of the material.
(b) The wavelength in the medium.
(c) The phase velocity.
∗7.25 The electric field of a plane wave propagating in a nonmagnetic medium is given by E = ˆ z 25e−30x cos(2π × 109t −40x) (V/m).
Obtain the corresponding expression for H.
7.26 The magnetic field of a plane wave propagating in a nonmagnetic medium is given by H = ˆ y 60e−10z cos(2π × 108t −12z) (mA/m).
Obtain the corresponding expression for E.
7.27 At 2 GHz, the conductivity of meat is on the order of 1 (S/m). When a material is placed inside a microwave oven and the field is activated, the presence of the electromagnetic fields in the conducting material causes energy dissipation in the material in the form of heat.
(a) Develop an expression for the time-average power per mm3 dissipated in a material of conductivity σ if the peak electric field in the material is E0.
(b) Evaluatetheresultforan electricfield E0 = 4×104 (V/m).
Section 7-5: Current Flow in Conductors 7.28 In a nonmagnetic, lossy, dielectric medium, a 300 MHz plane wave is characterized by the magnetic field phasor ˜ H = (ˆ x −j4ˆ z)e−2ye−j9y (A/m).
Obtain time-domain expressions for the electric and magnetic field vectors.
∗7.29 A rectangular copper block is 30 cm in height (along z).
In response to a wave incident upon the block from above, a current is induced in the block in the positive x direction.
Determine the ratio of the ac resistance of the block to its dc resistance at 1 kHz. The relevant properties of copper are given in Appendix B.
7.30 Repeat Problem 7.29 at 10 MHz.
7.31 The inner and outer conductors of a coaxial cable have radii of 0.5 cm and 1 cm, respectively. The conductors are made of copper with ϵr = 1, μr = 1, and σ = 5.8×107 S/m, and the outer conductor is 0.5 mm thick. At 10 MHz: (a) Are the conductors thick enough to be considered infinitely thick as far as the flow of current through them is concerned?
(b) Determine the surface resistance Rs.
(c) Determine the ac resistance per unit length of the cable.
7.32 Repeat Problem 7.31 at 1 GHz.
Section 7-6: Electromagnetic Power Density ∗7.33 The magnetic field of a plane wave traveling in air is given by H = ˆ x 50 sin(2π × 107t −ky) (mA/m). Determine the average power density carried by the wave.
7.34 A wave traveling in a nonmagnetic medium with ϵr = 9 is characterized by an electric field given by E = [ˆ y 3 cos(π × 107t + kx) −ˆ z 2 cos(π × 107t + kx)] (V/m).
Determine the direction of wave travel and average power density carried by the wave.
7.35 The electric-field phasor of a uniform plane wave traveling downward in water is given by E = ˆ x 5e−0.2ze−j0.2z (V/m).
where ˆ z is the downward direction and z = 0 is the water surface. If σ = 4 S/m, (a) Obtain an expression for the average power density.
(b) Determine the attenuation rate.
∗(c) Determine the depth at which the power density has been reduced by 40 dB.
7.36 The amplitudes of an elliptically polarized plane wave traveling in a lossless, nonmagnetic medium with ϵr = 4 are Hy0 = 3 (mA/m) and Hz0 = 4 (mA/m). Determine the average power flowing through an aperture in the y–z plane if its area is 20 m2.
∗7.37 A wave traveling in a lossless, nonmagnetic medium has an electric field amplitude of 24.56 V/m and an average power density of 2.4 W/m2. Determine the phase velocity of the wave.
7.38 At microwave frequencies, the power density considered safeforhumanexposureis1(mW/cm2). Aradarradiatesawave with an electric field amplitude E that decays with distance as PROBLEMS 351 E(R) = (3, 000/R) (V/m), where R is the distance in meters.
What is the radius of the unsafe region?
7.39 Consider the imaginary rectangular box shown in Fig. P7.39.
(a) Determine the net power flux P(t) entering the box due to a plane wave in air given by E = ˆ x E0 cos(ωt −ky) (V/m).
∗(b) Determine the net time-average power entering the box.
a b c x y z Figure P7.39 Imaginary rectangular box of Problems 7.39 and 7.40.
7.40 Repeat Problem 7.39 for a wave traveling in a lossy medium in which E = ˆ x 100e−20y cos(2π × 109t −40y) (V/m), H = −ˆ z 0.64e−20y · cos(2π × 109t −40y −36.85◦) (A/m).
The box has dimensions a = 1 cm, b = 2 cm, and c = 0.5 cm.
7.41 Given a wave with E = ˆ x E0 cos(ωt −kz): ∗(a) Calculate the time-average electric energy density (we)av = 1 T T 0 we dt = 1 2T T 0 ϵE2 dt.
(b) Calculate the time-average magnetic energy density (wm)av = 1 T T 0 wm dt = 1 2T T 0 μH 2 dt.
(c) Show that (we)av = (wm)av.
7.42 A team of scientists is designing a radar as a probe for measuring the depth of the ice layer over the antarctic land mass.
In order to measure a detectable echo due to the reflection by the ice-rock boundary, the thickness of the ice sheet should not exceed three skin depths. If ϵ′ r = 3 and ϵ′′ r = 10−2 for ice and if the maximum anticipated ice thickness in the area under exploration is 1.2 km, what frequency range is useable with the radar?
C H A P T E R 8 Wave Reflection and Transmission Chapter Contents EM Waves at Boundaries, 353 8-1 Wave Reflection and Transmission at Normal Incidence, 353 8-2 Snell’s Laws, 362 8-3 Fiber Optics, 365 8-4 Wave Reflection and Transmission at Oblique Incidence, 367 TB15 Lasers, 368 8-5 Reflectivity and Transmissivity, 376 8-6 Waveguides, 380 TB16 Bar-Code Readers, 382 8-7 General Relations for E and H, 383 8-8 TM Modes in Rectangular Waveguide, 384 8-9 TE Modes in Rectangular Waveguide, 388 8-10 Propagation Velocities, 388 8-11 Cavity Resonators, 392 Chapter 8 Summary, 395 Problems, 397 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Characterize the reflection and transmission behavior of plane waves incident upon plane boundaries, for both normal and oblique incidence.
2. Calculate the transmission properties of optical fibers.
3. Characterize wave propagation in a rectangular wave-guide.
4. Determine the behavior of resonant modes inside a rectangular cavity.
8-1 WAVE REFLECTION AND TRANSMISSION AT NORMAL INCIDENCE 353 EM Waves at Boundaries Figure 8-1 depicts the propagation path traveled by a signal transmitted by a shipboard antenna and received by an antenna on a submerged submarine.
Starting from the transmitter (denoted Tx in Fig. 8-1), the signal travels along a transmission line to the transmitting antenna. The relationship between the transmitter(generator)outputpower, Pt, andthepowersupplied to the antenna is governed by the transmission-line equations of Chapter 2. If the transmission line is approximately lossless and properly matched to the transmitting antenna, then all of Pt is delivered to the antenna.
If the antenna itself is lossless too, it will convert all of the power Pt in the guided wave provided by the transmission line into a spherical wave radiated outward into space. The radiation process is the subject of Chapter 9.
From point 1, denoting the location of the shipboard antenna, to point 2, denoting the point of incidence of the wave onto the water’s surface, the signal’s behavior is governed by the equations characterizing wave propagation in lossless media, covered in Chapter 7. As the wave impinges upon the air–water boundary, part of it is reflected by the surface while another part gets transmitted across the boundary into the water.
The transmitted wave is refracted, wherein its propagation direction moves closer toward the vertical, compared with that of the incident wave.
Reflection and transmission processes are treated in this chapter.
Wave travel from point 3, representing a point just below the water Tx Pt 1 Rx Pr 2 3 4 Receiver antenna Transmitter antenna Air Water Figure 8-1 Signal path between a shipboard transmitter (Tx) and a submarine receiver (Rx).
surface, to point 4, which denotes the location of the submarine antenna, is subject to the laws of wave propagation in lossy media, also treated in Chapter 7. Finally, some of the power carried by the wave traveling in water towards the submarine is intercepted by the receiving antenna. The received power, Pr, is then delivered to the receiver via a transmission line. The receiving properties of antennas are covered in Chapter 9. In summary, then, each wave-related aspect of the transmission process depicted in Fig. 8-1, starting with the transmitter and ending with the receiver, is treated in this book.
This chapter begins by examining the reflection and transmission properties of plane waves incident upon planar boundaries and concludes with sections on waveguides and cavityresonators. Applicationsdiscussedalongthewayinclude fiber and laser optics.
8-1 Wave Reflection and Transmission at Normal Incidence We know from Chapter 2 that, when a guided wave encounters a junction between two transmission lines with different characteristic impedances, the incident wave is partly reflected back toward the source and partly transmitted across the junction onto the other line. The same happens to a uniform planewavewhenitencountersaboundarybetweentwomaterial half-spaces with different characteristic impedances. In fact, the situation depicted in Fig. 8-2(b) has an exact analogue in the transmission-line configuration of Fig. 8-2(a). The boundary conditions governing the relationships between the electric and magnetic fields in Fig. 8-2(b) map one to one onto those we developed in Chapter 2 for the voltages and currents on the transmission line.
For convenience, we divide our treatment of wave reflection by, and transmission through, planar boundaries into two parts: in this section we confine our discussion to the normal-incidence case depicted in Fig. 8-3(a), and in Sections 8-2 to 8-4 we examine the more general oblique-incidence case depicted in Fig. 8-3(b). We will show the basis for the analogy between the transmission-line and plane-wave configurations so that we may use transmission-line equivalent models, tools (e.g., Smith chart), and techniques (e.g., quarter-wavelength matching) to expeditiously solve plane wave problems.
Before proceeding, however, we should explain the notion of rays and wavefronts and the relationship between them, as both are used throughout this chapter to represent electromagnetic waves.
A ray is a line representing the direction of flow 354 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION (b) Boundary between different media Transmission line 1 Transmission line 2 Incident wave Reflected wave Incident plane wave Transmitted plane wave Reflected plane wave Transmitted wave Z01 Z02 Medium 1 η1 Medium 2 η2 z = 0 z = 0 (a) Boundary between transmission lines Figure 8-2 Discontinuity between two different transmission lines is analogous to that between two dissimilar media.
Incident wave Transmitted wave Transmitted wave Reflected wave Incident wave Reflected wave Medium 1 η1 Medium 2 η2 Medium 1 η1 Medium 2 η2 (a) Normal incidence (b) Ray representation of oblique incidence θr θi θt Medium 1 η1 Medium 2 η2 (c) Wavefront representation of oblique incidence θr θt θi Figure 8-3 Ray representation of wave reflection and transmission at (a) normal incidence and (b) oblique incidence, and (c) wavefront representation of oblique incidence.
of electromagnetic energy carried by a wave, and therefore it is parallel to the propagation unit vector ˆ k. A wavefront is a surface across which the phase of a wave is constant; it is perpendicular to the wavevector ˆ k.
Hence, rays are perpendicular to wavefronts. The ray representation of wave incidence, reflection, and transmission shown in Fig. 8-3(b) is equivalent to the wavefront representation depicted in Fig. 8-3(c). The two representations are complementary; the ray representation is easier to use in graphical illustrations, whereas the wavefront representation provides greater physical insight into what happens to a wave when it encounters a discontinuous boundary.
8-1.1 Boundary between Lossless Media A planar boundary located at z = 0 [Fig. 8-4(a)] separates two lossless, homogeneous, dielectric media.
Medium 1 has permittivity ϵ1 and permeability μ1 and fills the half-space z ≤0.
Medium 2 has permittivity ϵ2 and permeability μ2 and fills the half-space z ≥0.
An x-polarized plane wave with electric and magnetic fields (Ei, Hi) propagates in medium 1 along direction ˆ ki = ˆ z toward medium 2.
Reflection and transmission at the boundary at z = 0 result in a reflected wave, with electric and magnetic fields (Er, Hr), traveling along direction 8-1 WAVE REFLECTION AND TRANSMISSION AT NORMAL INCIDENCE 355 Hr Er kr Hi Ei ki Ht Et kt x z y Medium 1 (ε1, μ1) Medium 2 (ε2, μ2) z = 0 z = 0 (a) Boundary between dielectric media (b) Transmission-line analogue ˆ ˆ ˆ Infinite line Z01 Z02 Figure 8-4 The two dielectric media separated by the x–y plane in (a) can be represented by the transmission-line analogue in (b).
ˆ kr = −ˆ z in medium 1, and a transmitted wave, with electric and magnetic fields (Et, Ht), traveling along direction ˆ kt = ˆ z in medium 2.
On the basis of the formulations developed in Sections 7-2 and 7-3 for plane waves, the three waves are described in phasor form by: Incident Wave E i(z) = ˆ xEi 0e−jk1z, (8.1a) Hi(z) = ˆ z × × × E i(z) η1 = ˆ y Ei 0 η1 e−jk1z.
(8.1b) Reflected Wave E r(z) = ˆ xEr 0ejk1z, (8.2a) Hr(z) = (−ˆ z) × × × E r(z) η1 = −ˆ y Er 0 η1 ejk1z.
(8.2b) Transmitted Wave E t(z) = ˆ xEt 0e−jk2z, (8.3a) Ht(z) = ˆ z × × × E t(z) η2 = ˆ y Et 0 η2 e−jk2z.
(8.3b) The quantities Ei 0, Er 0, and Et 0 are, respectively, the amplitudes of the incident, reflected, and transmitted electric fields at z = 0 (the boundary between the two media). The wavenumber and intrinsic impedance of medium 1 are k1 = ω√μ1ϵ1 and η1 = √μ1/ϵ1 , and those for medium 2 are k2 = ω√μ2ϵ2 and η2 = √μ2/ϵ2 .
The amplitude Ei 0 is imposed by the source responsible for generating the incident wave, and therefore is assumed known.
Our goal is to relate Er 0 and Et 0 to Ei 0. We do so by applying boundary conditions for the total electric and magnetic fields at z = 0. According to Table 6-2, the tangential component of the total electric field is always continuous across a boundary between two contiguous media, and in the absence of current sources at the boundary, the same is true for the total magnetic field. In the present case, the electric and magnetic fields of the incident, reflected, and transmitted waves are all tangential to the boundary.
The total electric field E1(z) in medium 1 is the sum of the electric fields of the incident and reflected waves, and a similar statement applies to the magnetic field H1(z). Hence, Medium 1 E1(z) = E i(z) + E r(z) = ˆ x(Ei 0e−jk1z + Er 0ejk1z), (8.4a) H1(z) = Hi(z) + Hr(z) = ˆ y 1 η1 (Ei 0e−jk1z −Er 0ejk1z).
(8.4b) With only the transmitted wave present in medium 2, the total fields are Medium 2 E2(z) = E t(z) = ˆ xEt 0e−jk2z, (8.5a) H2(z) = Ht(z) = ˆ y Et 0 η2 e−jk2z.
(8.5b) 356 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION At the boundary (z = 0), the tangential components of the electric and magnetic fields are continuous. Hence, E1(0) = E2(0) or Ei 0 + Er 0 = Et 0, (8.6a) H1(0) = H2(0) or Ei 0 η1 −Er 0 η1 = Et 0 η2 .
(8.6b) Solving these equations for Er 0 and Et 0 in terms of Ei 0 gives Er 0 = η2 −η1 η2 + η1 Ei 0 = Ei 0, (8.7a) Et 0 = 2η2 η2 + η1 Ei 0 = τEi 0, (8.7b) where = Er 0 Ei 0 = η2 −η1 η2 + η1 (normal incidence), (8.8a) τ = Et 0 Ei 0 = 2η2 η2 + η1 (normal incidence).
(8.8b) The quantities and τ are called the reflection and transmission coefficients.
For lossless dielectric media, η1 and η2 are real; consequently, both and τ are real also. As we see in Section 8-1.4, the expressions given by Eqs. (8.8a) and (8.8b) are equally applicable when the media are conductive, even though in that case η1 and η2 may be complex, and hence and τ may be complex as well. From Eqs. (8.8a) and (8.8b), it is easily shown that and τ are interrelated as τ = 1 + (normal incidence).
(8.9) For nonmagnetic media, η1 = η0 √ϵr1 , η2 = η0 √ϵr2 , where η0 is the intrinsic impedance of free space, in which case Eq. (8.8a) may be expressed as = √ϵr1 −√ϵr2 √ϵr1 + √ϵr2 (nonmagnetic media).
(8.10) 8-1.2 Transmission-Line Analogue The transmission-line configuration shown in Fig. 8-4(b) consists of a lossless transmission line with characteristic impedance Z01, connected at z = 0 to an infinitely long lossless transmission line with characteristic impedance Z02. The input impedance of an infinitely long line is equal to its characteristic impedance. Hence, at z = 0, the voltage reflection coefficient (looking toward the boundary from the vantage point of the first line) is = Z02 −Z01 Z02 + Z01 , which is identical in form to Eq. (8.8a). The analogy between plane waves and waves on transmission lines does not end here.
To demonstrate the analogy further, equations pertinent to the two cases are summarized in Table 8-1. Comparison of the two columns shows that there is a one-to-one correspondence between the transmission-line quantities ( V , ˜ I, β, Z0) and the plane-wave quantities ( E, H, k, η).
▶This correspondence allows us to use the techniques developed in Chapter 2, including the Smith-chart method for calculating impedance transformations, to solve plane-wave propagation problems. ◀ The simultaneous presence of incident and reflected waves in medium 1 [Fig. 8-4(a)] gives rise to a standing-wave pattern.
By analogy with the transmission-line case, the standing-wave ratio in medium 1 is defined as S = | E1|max | E1|min = 1 + || 1 −|| .
(8.15) ▶If the two media have equal impedances (η1 = η2), then = 0 and S = 1, and if medium 2 is a perfect conductor with η2 = 0 (which is equivalent to a short-circuited transmission line), then = −1 and S = ∞. ◀ The distance from the boundary to where the magnitude of the electric field intensity in medium 1 is a maximum, denoted lmax, is described by the same expression as that given by Eq. (2.70) for the voltage maxima on a transmission line, namely 8-1 WAVE REFLECTION AND TRANSMISSION AT NORMAL INCIDENCE 357 Table 8-1 Analogy between plane-wave equations for normal incidence and transmission-line equations, both under lossless conditions.
Plane Wave [Fig. 8-4(a)] Transmission Line [Fig. 8-4(b)] E1(z) = ˆ xEi 0(e−jk1z + ejk1z) (8.11a) V1(z) = V + 0 (e−jβ1z + ejβ1z) (8.11b) H1(z) = ˆ y Ei 0 η1 (e−jk1z −ejk1z) (8.12a) ˜ I1(z) = V + 0 Z01 (e−jβ1z −ejβ1z) (8.12b) E2(z) = ˆ xτEi 0e−jk2z (8.13a) V2(z) = τV + 0 e−jβ2z (8.13b) H2(z) = ˆ yτ Ei 0 η2 e−jk2z (8.14a) ˜ I2(z) = τ V + 0 Z02 e−jβ2z (8.14b) = (η2 −η1)/(η2 + η1) = (Z02 −Z01)/(Z02 + Z01) τ = 1 + τ = 1 + k1 = ω√μ1ϵ1 , k2 = ω√μ2ϵ2 β1 = ω√μ1ϵ1 , β2 = ω√μ2ϵ2 η1 = √μ1/ϵ1 , η2 = √μ2/ϵ2 Z01 and Z02 depend on transmission-line parameters −z = lmax = θr + 2nπ 2k1 = θrλ1 4π + nλ1 2 , (8.16) n = 1, 2, . . . , if θr < 0, n = 0, 1, 2, . . . , if θr ≥0, where λ1 = 2π/k1 and θr is the phase angle of (i.e., = ||ejθr, and θr is bounded in the range −π < θr ≤π).
The expression for lmax is valid not only when the two media are lossless dielectrics, but also when medium 1 is a low-loss dielectric. Moreover, medium 2 may be either a dielectric or a conductor. When both media are lossless dielectrics, θr = 0 if η2 > η1 and θr = π if η2 < η1.
The spacing between adjacent maxima is λ1/2, and the spacing between a maximum and the nearest minimum is λ1/4.
The electric-field minima occur at lmin = lmax + λ1/4, if lmax < λ1/4, lmax −λ1/4, if lmax ≥λ1/4.
(8.17) 8-1.3 Power Flow in Lossless Media Medium 1 in Fig. 8-4(a) is host to the incident and reflected waves, which together comprise the total electric and magnetic fields E1(z) and H1(z) given by Eqs. (8.11a) and (8.12a) of Table 8-1. Using Eq. (7.100), the net average power density flowing in medium 1 is Sav1(z) = 1 2Re[ E1(z) × × × H∗ 1(z)] = 1 2Re ˆ xEi 0(e−jk1z + ejk1z) × × × ˆ y Ei∗ 0 η1 (ejk1z −∗e−jk1z) = ˆ z |Ei 0|2 2η1 (1 −||2), (8.18) which is analogous to Eq. (2.106) for the lossless transmission-line case.
The first and second terms inside the bracket in Eq. (8.18) represent the average power density of the incident and reflected waves, respectively. Thus, Sav1 = Si av + Sr av, (8.19a) 358 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION with Si av = ˆ z |Ei 0|2 2η1 , (8.19b) Sr av = −ˆ z||2 |Ei 0|2 2η1 = −||2Si av.
(8.19c) Even though is purely real when both media are lossless dielectrics, we chose to treat it as complex, thereby providing in Eq. (8.19c) an expression that is also valid when medium 2 is conducting.
The average power density of the transmitted wave in medium 2 is Sav2(z) = 1 2Re[ E2(z) × × × H∗ 2(z)] = 1 2Re ˆ xτEi 0e−jk2z × × × ˆ yτ ∗Ei∗ 0 η2 ejk2z = ˆ z|τ|2 |Ei 0|2 2η2 .
(8.20) Through the use of Eqs. (8.8a) and (8.8b), it can be easily shown that for lossless media τ 2 η2 = 1 −2 η1 (lossless media), (8.21) which leads to Sav1 = Sav2.
This result is expected from considerations of power conservation.
Example 8-1: Radar Radome Design A 10 GHz aircraft radar uses a narrow-beam scanning antenna mounted on a gimbal behind a dielectric radome, as shown in Fig. 8-5. Even though the radome shape is far from planar, it is approximately planar over the narrow extent of the radar beam.
If the radome material is a lossless dielectric with ϵr = 9 and μr = 1, choose its thickness d such that the radome appears transparent to the radar beam. Structural integrity requires d to be greater than 2.3 cm.
Solution: Figure 8-6(a) shows a small section of the radome on an expanded scale. The incident wave can be approximated Radar Antenna Dielectric radome Antenna beam d Figure8-5 Antennabeam“looking”throughanaircraftradome of thickness d (Example 8-1).
Incident wave Transmitted wave Medium 1 (air) η0 Line 1 Line 2 Z01 = η0 Z02 = ηr Zin ZL = η0 Medium 3 (air) η0 Medium 2 ηr Radome z = −d z = 0 z = −d z = 0 (b) (a) Figure 8-6 (a) Planar section of the radome of Fig. 8-5 at an expanded scale and (b) its transmission-line equivalent model (Example 8-1).
8-1 WAVE REFLECTION AND TRANSMISSION AT NORMAL INCIDENCE 359 as a plane wave propagating in medium 1 (air) with intrinsic impedance η0. Medium 2 (the radome) is of thickness d and intrinsic impedance ηr, and medium 3 (air) is semi-infinite with intrinsic impedance η0. Figure 8-6(b) shows an equivalent transmission-line model with z = 0 selected to coincide with the outside surface of the radome, and the load impedance ZL = η0 represents the input impedance of the semi-infinite air medium to the right of the radome.
For the radome to “appear” transparent to the incident wave, the reflection coefficient must be zero at z = −d, thereby guaranteeing total transmission of the incident power into medium 3. Since ZL = η0 in Fig. 8-6(b), no reflection takes place at z = −d if Zin = η0, which can be realized by choosing d = nλ2/2 [see Section 2-8.4], where λ2 is the wavelength in medium 2 and n is a positive integer. At 10 GHz, the wavelength in air is λ0 = c/f = 3 cm, while in the radome material it is λ2 = λ0 √ϵr = 3 cm 3 = 1 cm.
Hence, by choosing d = 5λ2/2 = 2.5 cm, the radome is both nonreflecting and structurally stable.
Example 8-2: Yellow Light Incident upon a Glass Surface A beam of yellow light with wavelength 0.6 μm is normally incident in air upon a glass surface.
Assume the glass is sufficiently thick as to ignore its back surface. If the surface is situated in the plane z = 0 and the relative permittivity of glass is 2.25, determine: (a) the locations of the electric field maxima in medium 1 (air), (b) the standing-wave ratio, and (c) the fraction of the incident power transmitted into the glass medium.
Solution: (a) We begin by determining the values of η1, η2, and : η1 = μ1 ϵ1 = μ0 ϵ0 ≈120π ( ), η2 = μ2 ϵ2 = μ0 ϵ0 · 1 √ϵr ≈120π √ 2.25 = 80π ( ), = η2 −η1 η2 + η1 = 80π −120π 80π + 120π = −0.2.
Hence, || = 0.2 and θr = π. From Eq. (8.16), the electric-field magnitude is maximum at lmax = θrλ1 4π + nλ1 2 = λ1 4 + n λ1 2 (n = 0, 1, 2, . . .) with λ1 = 0.6 μm.
(b) S = 1 + || 1 −|| = 1 + 0.2 1 −0.2 = 1.5.
(c) The fraction of the incident power transmitted into the glass medium is equal to the ratio of the transmitted power density, given by Eq. (8.20), to the incident power density, Si av = |Ei 0|2/2η1: Sav2 Si av = τ 2 |Ei 0|2 2η2 |Ei 0|2 2η1 = τ 2 η1 η2 .
In view of Eq. (8.21), Sav2 Si av = 1 −||2 = 1 −(0.2)2 = 0.96, or 96%.
8-1.4 Boundary between Lossy Media In Section 8-1.1 we considered a plane wave in a lossless medium incident normally on a planar boundary of another lossless medium. We now generalize our expressions to lossy media. In a medium with constitutive parameters (ϵ, μ, σ), the propagation constant γ = α+jβ and the intrinsic impedance ηc are both complex. General expressions for α, β, and ηc are given by Eqs. (7.66a), (7.66b), and (7.70), respectively, and approximate expressions are given in Table 7-2 for the special cases of low-loss media and good conductors.
If media 1 and 2 have constitutive parameters (ϵ1, μ1, σ1) and (ϵ2, μ2, σ2) (Fig. 8-7), then expressions for the electric and magnetic fields in media 1 and 2 can be obtained fromEqs. (8.11) through (8.14) of Table 8-1 by replacing jk with γ and η with ηc. Thus, Medium 1 E1(z) = ˆ xEi 0(e−γ1z + eγ1z), (8.22a) H1(z) = ˆ y Ei 0 ηc1 (e−γ1z −eγ1z), (8.22b) 360 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION Hr Er kr Hi Ei ki Ht Et kt x z y Medium 1 (ε1, μ1, σ1) Medium 2 (ε2, μ2, σ2) z = 0 z = 0 (a) Boundary between dielectric media (b) Transmission-line analogue Infinite line Z01 = ηc1 Z02 = ηc2 ˆ ˆ ˆ ηc1 ηc2 Figure 8-7 Normal incidence at a planar boundary between two lossy media.
Medium 2 E2(z) = ˆ xτEi 0e−γ2z, (8.23a) H2(z) = ˆ yτ Ei 0 ηc2 e−γ2z.
(8.23b) Here, γ1 = α1 + jβ1, γ2 = α2 + jβ2, and = ηc2 −ηc1 ηc2 + ηc1 , (8.24a) τ = 1 + = 2ηc2 ηc2 + ηc1 .
(8.24b) Because ηc1 and ηc2 are, in general, complex, and τ may be complex as well.
Example 8-3: Normal Incidence on a Metal Surface A 1 GHz x-polarized plane wave traveling in the +z direction is incident from air upon a copper surface. The air-to-copper interface is at z = 0, and copper has ϵr = 1, μr = 1, and σ = 5.8 × 107 S/m. If the amplitude of the electric field of the incident wave is 12 (mV/m), obtain expressions for the instantaneous electric and magnetic fields in the air medium.
Assume the metal surface to be several skin depths deep.
Solution: In medium 1 (air), α = 0, β = k1 = ω c = 2π × 109 3 × 108 = 20π 3 (rad/m), η1 = η0 = 377 ( ), λ = 2π k1 = 0.3 m.
At f = 1 GHz, copper is an excellent conductor because ϵ′′ ϵ′ = σ ωϵrϵ0 = 5.8 × 107 2π × 109 × (10−9/36π) = 1 × 109 ≫1.
Use of Eq. (7.77c) gives ηc2 = (1 + j) πf μ σ = (1 + j) π × 109 × 4π × 10−7 5.8 × 107 1/2 = 8.25(1 + j) (m ).
Since ηc2 is so small compared to η0 = 377 ( ) for air, the copper surface acts, in effect, like a short circuit. Hence, = ηc2 −η0 ηc2 + η0 ≈−1.
Upon setting = −1 in Eqs. (8.11) and (8.12) of Table 8-1, we obtain E1(z) = ˆ xEi 0(e−jk1z −ejk1z) = −ˆ xj2Ei 0 sin k1z, (8.25a) H1(z) = ˆ y Ei 0 η1 (e−jk1z + ejk1z) = ˆ y2 Ei 0 η1 cos k1z.
(8.25b) 8-1 WAVE REFLECTION AND TRANSMISSION AT NORMAL INCIDENCE 361 With Ei 0 = 12 (mV/m), the instantaneous fields associated with these phasors are E1(z, t) = Re[ E1(z) ejωt] = ˆ x 2Ei 0 sin k1z sin ωt = ˆ x 24 sin(20πz/3) sin(2π × 109t) (mV/m), H1(z, t) = Re[ H1(z) ejωt] = ˆ y 2 Ei 0 η1 cos k1z cos ωt = ˆ y 64 cos(20πz/3) cos(2π × 109t) (μA/m).
Plots of the magnitude of E1(z, t) and H1(z, t) are shown in Fig. 8-8 as a function of negative z for various values of ωt.
The wave patterns exhibit a repetition period of λ/2, and E and H are in phase quadrature (90◦phase shift) in both space and time. This behavior is identical with that for voltage and current waves on a shorted transmission line.
Concept Question 8-1: What boundary conditions were used in the derivations of the expressions for and τ?
Concept Question 8-2: In the radar radome design of Example 8-1, all the incident energy in medium 1 ends up getting transmitted into medium 3, and vice versa. Does this imply that no reflections take place within medium 2?
Explain.
Concept Question 8-3: Explain on the basis of bound-ary conditions why it is necessary that = −1 at the boundary between a dielectric and a perfect conductor.
Exercise 8-1: To eliminate reflections of normally incident plane waves, a dielectric slab of thickness d and relativepermittivityϵr2 istobeinsertedbetweentwosemi-infinite media with relative permittivities ϵr1 = 1 and ϵr3 = 16.
Use the quarter-wave transformer technique to select d and ϵr2. Assume f = 3 GHz.
Answer: ϵr2 = 4 and d = (1.25 + 2.5n) (cm), with n = 0, 1, 2, . . . . (See EM.) 0 64 (μA/m) –64 (μA/m) 0 24 (mV/m) Conductor Conductor –24 (mV/m) ωt = 3π/2 ωt = π/2 ωt = π ωt = 0 ωt = 5π/4 ωt = π/4 ωt = 0 E1(z, t) H1(z, t) –z –z –λ 4 –3λ 4 –λ 2 –λ Figure 8-8 Wave patterns for fields E1(z, t) and H1(z, t) of Example 8-3.
Exercise 8-2: Express the normal-incidence reflection coefficient at the boundary between two nonmagnetic, conducting media in terms of their complex permittivities.
Answer: For incidence in medium 1 (ϵ1, μ0, σ1) onto medium 2 (ϵ2, μ0, σ2), = √ϵc1 −√ϵc2 √ϵc1 + √ϵc2 , with ϵc1 = (ϵ1−jσ1/ω) and ϵc2 = (ϵ2−jσ2/ω). (See EM.) Exercise 8-3: Obtain expressions for the average power densities in media 1 and 2 for the fields described by 362 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION Module 8.1 Normal Incidence on Perfect Conductor Observe the standing wave pattern created by the combination of a wave incident normally onto the plane surface of a conductor and its reflection.
Eqs. (8.22a) through (8.23b), assuming medium 1 is slightly lossy with ηc1 approximately real.
Answer: (See EM.) Sav1 = ˆ z |Ei 0|2 2ηc1 e−2α1z −||2e2α1z , Sav2 = ˆ z|τ|2 |Ei 0|2 2 e−2α2z Re 1 η∗ c2 .
8-2 Snell’s Laws In the preceding sections we examined reflection and transmission of plane waves that are normally incident upon a planar interface between two different media.
We now consider the oblique-incidence case depicted in Fig. 8-9, and for simplicity we assume all media to be lossless. The z = 0 plane forms the boundary between media 1 and 2 with constitutive parameters (ϵ1, μ1) and (ϵ2, μ2), respectively. The two lines in Fig. 8-9 with direction ˆ ki represent rays drawn normal to the wavefront of the incident wave, and those along directions ˆ kr and ˆ kt are similarly associated with the reflected and transmitted waves. The angles of incidence, reflection, and transmission (or refraction), defined with respect to the normal to the boundary (the z axis), are θi, θr, and θt, respectively. These three angles are interrelated by Snell’s laws, which we derive shortly by considering the propagation of the wavefronts of the three waves. Rays of the incident wave intersect the boundary at O and O′. Here AiO represents a constant-phase wavefront of the incident wave. Likewise, ArO′ and AtO′ are constant-phase wavefronts of the reflected and transmitted waves, respectively (Fig. 8-9).
The incident and reflected waves propagate in medium 1 with the same phase velocity up1 = 1/√μ1ϵ1, while the transmitted wave in medium 2 propagates with a velocity up2 = 1/√μ2ϵ2. The time it takes for the incident wave to travel from Ai to O′ is the same as the time it takes for the reflected wave to travel from O to Ar, and also the time it takes the transmitted wave to travel from O to At. Since time equals 8-2 SNELL’S LAWS 363 Ar At θr θt θi Ai O O' x z Transmitted wave Reflected wave Incident wave Medium 1 (ε1, μ1) Medium 2 (ε2, μ2) ki ˆ ki ˆ kt ˆ kt ˆ kr ˆ kr ˆ Figure 8-9 Wave reflection and refraction at a planar boundary between different media.
distance divided by velocity, it follows that AiO′ up1 = OAr up1 = OAt up2 .
(8.26) From the geometries of the three right triangles in Fig. 8-9, we deduce that AiO′ = OO′ sin θi, (8.27a) OAr = OO′ sin θr, (8.27b) OAt = OO′ sin θt.
(8.27c) Use of these expressions in Eq. (8.26) leads to θi = θr (Snell’s law of reflection), (8.28a) sin θt sin θi = up2 up1 = μ1ϵ1 μ2ϵ2 (8.28b) (Snell’s law of refraction).
▶Snell’s law of reflection states that the angle of reflection equals the angle of incidence, and Snell’s law of refraction provides a relation between sin θt and sin θi in terms of the ratio of the phase velocities. ◀ The index of refraction of a medium, n, is defined as the ratio of the phase velocity in free space (i.e., the speed of light c) to the phase velocity in the medium. Thus, n = c up = μϵ μ0ϵ0 = √μrϵr (index of refraction).
(8.29) In view of Eq. (8.29), Eq. (8.28b) may be rewritten as sin θt sin θi = n1 n2 = μr1ϵr1 μr2ϵr2 .
(8.30) For nonmagnetic materials, μr1 = μr2 = 1, in which case sin θt sin θi = n1 n2 = ϵr1 ϵr2 = η2 η1 (for μ1 = μ2).
(8.31) Usually, materials with higher densities have higher permittiv-ities. Air, with μr = ϵr = 1, has an index of refraction n0 = 1.
Since for nonmagnetic materials n = √ϵr, a material is often referred to as more dense than another material if it has a greater index of refraction.
At normal incidence (θi = 0), Eq. (8.31) gives θt = 0, as expected.
At oblique incidence θt < θi when n2 > n1 and θt > θi when n2 < n1.
▶If a wave is incident on a more dense medium [Fig. 8-10(a)], the transmitted wave refracts inwardly (toward the z axis), and the opposite is true if a wave is incident on a less dense medium [Fig. 8-10(b)]. ◀ A case of particular interest is when θt = π/2, as shown in Fig. 8-10(c); in this case, the refracted wave flows along the surface and no energy is transmitted into medium 2. The value 364 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION (a) n1 < n2 (b) n1 > n2 (c) n1 > n2 and θi = θc θr θt θi n1 n2 θr θt θi n1 n2 θt > θi Inward refraction Outward refraction n1 n2 No transmission θr θi θt = 90° Figure 8-10 Snell’s laws state that θr = θi and sin θt = (n1/n2) sin θi. Refraction is (a) inward if n1 < n2 and (b) outward if n1 > n2; and (c) the refraction angle is 90◦if n1 > n2 and θi is equal to or greater than the critical angle θc = sin−1(n2/n1).
of the angle of incidence θi corresponding to θt = π/2 is called the critical angle θc and is obtained from Eq. (8.30) as sin θc = n2 n1 sin θt θt=π/2 = n2 n1 (8.32a) = ϵr2 ϵr1 (for μ1 = μ2).
(8.32b) (critical angle) If θi exceeds θc, the incident wave is totally reflected, and the refracted wave becomes a nonuniform surface wave that travels along the boundary between the two media. This wave behavior is called total internal reflection.
n1 θ1 θ2 θ2 n2 n3 = n1 θ3 = θ1 Figure 8-11 The exit angle θ3 is equal to the incidence angle θ1 if the dielectric slab has parallel boundaries and is surrounded by media with the same index of refraction on both sides (Example 8-4).
Example 8-4: Light Beam Passing through a Slab A dielectric slab with index of refraction n2 is surrounded by a medium with index of refraction n1, as shown in Fig. 8-11. If θi < θc, show that the emerging beam is parallel to the incident beam.
Solution: At the slab’s upper surface, Snell’s law gives sin θ2 = n1 n2 sin θ1 (8.33) and, similarly, at the slab’s lower surface, sin θ3 = n2 n3 sin θ2 = n2 n1 sin θ2.
(8.34) Substituting Eq. (8.33) into Eq. (8.34) gives sin θ3 = n2 n1 n1 n2 sin θ1 = sin θ1.
Hence, θ3 = θ1. The slab displaces the beam’s position, but the beam’s direction remains unchanged.
Exercise 8-4: In the visible part of the electromagnetic spectrum, the index of refraction of water is 1.33. What is the critical angle for light waves generated by an upward-looking underwater light source?
Answer: θc = 48.8◦. (See EM.) 8-3 FIBER OPTICS 365 (a) Optical fiber (b) Successive internal reflections θ2 θi θ3 n0 n0 nc nf nc Fiber core Cladding θi Acceptance cone Figure 8-12 Waves can be guided along optical fibers as long as the reflection angles exceed the critical angle for total internal reflection.
Exercise 8-5: IfthelightsourceofExercise8-4issituated at a depth of 1 m below the water surface and if its beam is isotropic (radiates in all directions), how large a circle would it illuminate when observed from above?
Answer: Circle’s diameter = 2.28 m. (See EM.) 8-3 Fiber Optics By successive total internal reflections, as illustrated in Fig. 8-12(a), light can be guided through thin dielectric rods made of glass or transparent plastic, known as optical fibers.
Because the light is confined to traveling within the rod, the only loss in power is due to reflections at the sending and receiving ends of the fiber and absorption by the fiber material (because it is not a perfect dielectric). Optical fibers are useful for the transmission of wide-band signals as well as many imaging applications.
An optical fiber usually consists of a cylindrical fiber core with an index of refraction nf, surrounded by another cylinder of lower index of refraction, nc, called the cladding [Fig. 8-12(b)].
The cladding layer serves to optically isolate the fiber when a large number of fibers are packed in close proximity, thereby avoiding leakage of light from one fiber into another. To ensure total internal reflection, the incident angle θ3 in the fiber core must be equal to, or greater than, the critical angle θc for a wave in the fiber medium (with nf) incident upon the cladding medium (with nc). From Eq. (8.32a), we have sin θc = nc nf .
(8.35) To meet the total reflection requirement θ3 ≥θc, it is necessary that sin θ3 ≥nc/nf. The angle θ2 is the complement of angle θ3; hence cos θ2 = sin θ3. The necessary condition therefore may be written as cos θ2 ≥nc nf .
(8.36) Moreover, θ2 is related to the incidence angle on the face of the fiber, θi, by Snell’s law: sin θ2 = n0 nf sin θi, (8.37) where n0 is the index of refraction of the medium surrounding the fiber (n0 = 1 for air and n0 = 1.33 if the fiber is in water), or cos θ2 = 1 − n0 nf 2 sin2 θi 1/2 .
(8.38) Using Eq. (8.38) on the left-hand side of Eq. (8.36) and then solving for sin θi gives sin θi ≤1 n0 (n2 f −n2 c)1/2.
(8.39) The acceptance angle θa is defined as the maximum value of θi for which the condition of total internal reflection remains satisfied: sin θa = 1 n0 (n2 f −n2 c)1/2.
(8.40) The angle θa is equal to half the angle of the acceptance cone of the fiber. Any ray of light incident upon the face of the core fiber at an incidence angle within the acceptance cone can propagate down the core. This means that there can be a large number of ray paths, called modes, by which light energy can travel in the core. Rays characterized by large angles θi travel longer paths 366 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION T Core Cladding High-order mode Axial mode Low-order mode τ T τi Figure 8-13 Distortion of rectangular pulses caused by modal dispersion in optical fibers.
than rays that propagate along the axis of the fiber, as illustrated by the three modes shown in Fig. 8-13. Consequently, different modes have different transit times between the two ends of the fiber. This property of optical fibers is called modal dispersion and has the undesirable effect of changing the shape of pulses used for the transmission of digital data. When a rectangular pulse of light incident upon the face of the fiber gets broken up into many modes and the different modes do not arrive at the other end of the fiber at the same time, the pulse gets distorted, both in shape and length. In the example shown in Fig. 8-13, the narrowrectangularpulsesattheinputsideoftheopticalfiberare of width τi separated by a time duration T . After propagating through the fiber core, modal dispersion causes the pulses to look more like spread-out sine waves with spread-out temporal width τ. If the output pulses spread out so much that τ > T , the output signals will smear out, making it impossible to decipher the transmitted message from the output signal.
Hence, to ensure that the transmitted pulses remain distinguishable at the output side of the fiber, it is necessary that τ be shorter than T . As a safety margin, it is common practice to design the transmission system such that T ≥2τ.
The spread-out width τ is equal to the time delay t between the arrival of the slowest ray and the fastest ray. The slowest ray is the one traveling the longest distance and corresponds to the ray incident upon the input face of the fiber at the acceptance angle θa. From the geometry in Fig. 8-12(b) and Eq. (8.36), this ray corresponds to cos θ2 = nc/nf. For an optical fiber of length l, the length of the path traveled by such a ray is lmax = l cos θ2 = l nf nc , (8.41) and its travel time in the fiber at velocity up = c/nf is tmax = lmax up = ln2 f cnc .
(8.42) The minimum time of travel is realized by the axial ray and is given by tmin = l up = l c nf.
(8.43) The total time delay is therefore τ = t = tmax −tmin = lnf c nf −1 nc (s).
(8.44) As we stated before, to retrieve the desired information from the transmitted signals, it is advisable that T , the interpulse period of the input train of pulses, be no shorter than 2τ. This, in turn, means that the data rate (in bits per second), or equivalently the number of pulses per second, that can be transmitted through the fiber is limited to fp = 1 T = 1 2τ = cnc 2lnf(nf −nc) (bits/s).
(8.45) Example 8-5: Transmission Data Rate on Optical Fibers A 1 km long optical fiber (in air) is made of a fiber core with an index of refraction of 1.52 and a cladding with an index of refraction of 1.49. Determine (a) the acceptance angle θa, and (b) the maximum usable data rate of signals that can be transmitted through the fiber.
Solution: (a) From Eq. (8.40), sin θa = 1 n0 (n2 f −n2 c)1/2 = [(1.52)2 −(1.49)2]1/2 = 0.3, which corresponds to θa = 17.5◦.
8-4 WAVE REFLECTION AND TRANSMISSION AT OBLIQUE INCIDENCE 367 Module 8.2 Multimode Step-Index Optical Fiber Choose the indices of refraction on the fibre core and cladding and then observe the zigzag pattern of the wave propagation inside the fiber.
(b) From Eq. (8.45), fp = cnc 2lnf(nf −nc) = 3 × 108 × 1.49 2 × 103 × 1.52(1.52 −1.49) = 4.9 (Mb/s).
Exercise 8-6: If the index of refraction of the cladding material in Example 8-5 is increased to 1.50, what would be the new maximum usable data rate?
Answer: 7.4 (Mb/s). (See EM.) 8-4 Wave Reflection and Transmission at Oblique Incidence In this section we develop a rigorous theory of reflection and refraction of plane waves obliquely incident upon planar boundaries between different media. Our treatment parallels that in Section 8-1 for the normal-incidence case and goes beyond that in Section 8-2 on Snell’s laws, which yielded information on only the angles of reflection and refraction.
For normal incidence, the reflection and transmission coefficients and τ at a boundary between two media are independent of the polarization of the incident wave, as both the electric and magnetic fields of a normally incident plane wave are tangential to the boundary regardless of the wave polarization. This is not the case for obliquely incident waves travelling at an angle θi ̸= 0 with respect to the normal to the interface.
▶The plane of incidence is defined as the plane containing the normal to the boundary and the direction of propagation of the incident wave. ◀ 368 TECHNOLOGY BRIEF 15: LASERS Technology Brief 15: Lasers Lasers are used in CD and DVD players, bar-code readers, eye surgery, and multitudes of other systems and applications (Fig. TF15-1).
▶A laser—acronym for Light Amplification by Stimulated Emission of Radiation—is a source of monochromatic (single wavelength), coherent (uniform wavefront), narrow-beam light. ◀ This is in contrast with other sources of light (such as the sun or a light bulb) which usually encompass waves of many different wavelengths with random phase (incoherent). A laser source generating microwaves is called a maser. The first maser was built in 1953 by Charles Townes and the first laser was constructed in 1960 by Theodore Maiman.
Basic Principles Despite its complex quantum-mechanical structure, an atom can be conveniently modeled as a nucleus (containing protons and neutrons) surrounded by a cloud of electrons. Associated with the atom or molecule of any given material is a specific set of quantized (discrete) energy states (orbits) that the electrons can occupy. Supply of energy (in the form of heat, exposure to intense light, or other means) by an external source can cause an electron to move from a lower-energy state to a higher energy (excited) state. Exciting the atoms is called pumping because it leads to increasing the population of electrons in higher states [Fig. TF15-2(a)]. Spontaneous emission of a photon (light energy) occurs when the electron in the excited state moves to a lower state [Fig.TF15-2(b)], and stimulated emission [Fig. TF15-2(c)] happens when an emitted photon “entices” an electron in an excited state of another atom to move to a lower state, thereby emitting a second photon of identical energy, wavelength, and wavefront (phase).
FigureTF15-1 A few examples of laser applications.
TECHNOLOGY BRIEF 15: LASERS 369 (a) Pumping electron to excited state (b) Spontaneous emission (c) Stimulated emission Photon Photon Original photon Stimulated photon Nucleus Electron Orbit of excited state Orbit of ground state Incident energy or photon FigureTF15-2 Electron excitation and photon emission.
Principle of Operation ▶Highly amplified stimulated emission is called lasing. ◀ The lasing medium can be solid, liquid, or gas. Laser operation is illustrated in Fig.TF15-3 for a ruby crystal surrounded by a flash tube (similar to a camera flash). A perfectly reflecting mirror is placed on one end of the crystal and a partially reflecting mirror on the other end. Light from the flash tube excites the atoms; some undergo spontaneous emission, generating photons that cause others to undergo stimulated emission; photons moving along the axis of the crystal bounce back and forth between the mirrors, causing additional stimulated emission (i.e., amplification), with only a fraction of the photons exiting through the partially reflecting mirror.
▶Because all of the stimulated photons are identical, the light wave generated by the laser is of a single wavelength. ◀ Wavelength (Color) of Emitted Light The atom of any given material has unique energy states. The difference in energy between the excited high-energy state and the stable lower-energy state determines the wavelength of the emitted photons (EM wave). Through proper choice of lasing material, monochromatic waves can be generated with wavelengths in the ultraviolet, visible, infrared or microwave bands.
Amplifying medium Laser light Perfectly reflecting mirror Partially reflecting mirror Excitation energy (e.g., flash tube) FigureTF15-3 Laser schematic.
370 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION A wave of arbitrary polarization may be described as the superposition of two orthogonally polarized waves, one with its electric field parallel to the plane of incidence (parallel H|| t H|| i E|| i E|| r E|| t Hr Ht Hi Ei Er Et θr θt θi x z y z = 0 Medium 1 (ε1, μ1) Medium 2 (ε2, μ2) (a) Perpendicular polarization kr ki kt ˆ ˆ ˆ T T T T T T θr θt θi x z y z = 0 Medium 1 (ε1, μ1) Medium 2 (ε2, μ2) (b) Parallel polarization kr ki kt ˆ ˆ ˆ H|| r Figure 8-14 The plane of incidence is the plane containing the direction of wave travel, ˆ ki, and the surface normal to the boundary. In the present case the plane of incidence containing ˆ ki and ˆ z coincides with the plane of the paper.
A wave is (a) perpendicularly polarized when its electric field vector is perpendicular to the plane of incidence and (b) parallel polarized when its electric field vector lies in the plane of incidence.
polarization) and the other with its electric field perpendicular to the plane of incidence (perpendicular polarization). These two polarization configurations are shown in Fig. 8-14, in which the plane of incidence is coincident with the x–z plane.
Polarization with E perpendicular to the plane of incidence is also called transverse electric (TE) polarization because E is perpendicular to the plane of incidence, and that with E parallel to the plane of incidence is called transverse magnetic (TM) polarization because in that case it is the magnetic field that is perpendicular to the plane of incidence.
For the general case of a wave with an arbitrary polarization, it is common practice to decompose the incident wave (Ei, Hi) into a perpendicularly polarized component (Ei ⊥, Hi ⊥) and a parallel polarized component (Ei ∥, Hi ∥).
Then, after determining the reflected waves (Er ⊥, Hr ⊥) and (Er ∥, Hr ∥) due to the two incident components, the reflected waves are added together to give the total reflected wave (Er, Hr) corresponding to the original incident wave. A similar process can be used to determine the total transmitted wave (Et, Ht).
8-4.1 Perpendicular Polarization Figure 8-15 shows a perpendicularly polarized incident plane wave propagating along the xi direction in dielectric medium 1.
The electric field phasor Ei ⊥points along the y direction, and the associated magnetic field phasor Hi ⊥is along the yi axis.
The directions of Ei ⊥and Hi ⊥are such that Ei ⊥× × × Hi ⊥points along the propagation direction ˆ xi. The electric and magnetic fields of such a plane wave are given by Ei ⊥= ˆ yEi ⊥0e−jk1xi, (8.46a) Hi ⊥= ˆ yi Ei ⊥0 η1 e−jk1xi, (8.46b) where Ei ⊥0 is the amplitude of the electric field phasor at xi = 0, and k1 = ω√μ1ϵ1 and η1 = √μ1/ϵ1 are the wave number and intrinsic impedance of medium 1. From Fig. 8-15, the distance xi and the unit vector ˆ yi may be expressed in terms of the (x, y, z) global coordinate system as xi = x sin θi + z cos θi, (8.47a) ˆ yi = −ˆ x cos θi + ˆ z sin θi.
(8.47b) Substituting Eqs. (8.47a) and (8.47b) into Eqs. (8.46a) and (8.46b) gives 8-4 WAVE REFLECTION AND TRANSMISSION AT OBLIQUE INCIDENCE 371 θt θr θi z yt –x Er H x r z yr xr xr xt xi x z y yi –x θi θr θt xi xt x z xt θt Inset C x z xi θi Inset A x –z xr θr Inset B x z z = 0 Medium 1 (ε1, μ1) Medium 2 (ε2, μ2) xr = x sin θr – z cos θr xi = x sin θi + z cos θi xt = x sin θt + z cos θt yr = x cos θr + z sin θr ˆ ˆ ˆ yt = –x cos θt + z sin θt ˆ ˆ ˆ yi = –x cos θi + z sin θi ˆ ˆ ˆ T T Ei T Et T Hi T Hr T Ht T H x i T H z i T H z t T H x t T H z r T Figure 8-15 Perpendicularly polarized plane wave incident at an angle θi upon a planar boundary.
Incident Wave Ei ⊥= ˆ yEi ⊥0e−jk1(x sin θi+z cos θi), (8.48a) Hi ⊥= (−ˆ x cos θi + ˆ z sin θi) × Ei ⊥0 η1 e−jk1(x sin θi+z cos θi).
(8.48b) With the aid of the directional relationships given in Fig. 8-15 for the reflected and transmitted waves, these fields are given by Reflected Wave Er ⊥= ˆ yEr ⊥0e−jk1xr = ˆ yEr ⊥0e−jk1(x sin θr−z cos θr), (8.49a) Hr ⊥= ˆ yr Er ⊥0 η1 e−jk1xr = (ˆ x cos θr + ˆ z sin θr) × Er ⊥0 η1 e−jk1(x sin θr−z cos θr), (8.49b) Transmitted Wave Et ⊥= ˆ yEt ⊥0e−jk2xt = ˆ yEt ⊥0e−jk2(x sin θt+z cos θt), (8.49c) Ht ⊥= ˆ yt Et ⊥0 η2 e−jk2xt = (−ˆ x cos θt + ˆ z sin θt) × Et ⊥0 η2 e−jk2(x sin θt+z cos θt), (8.49d) where θr and θt are the reflection and transmission angles shown in Fig. 8-15, and k2 and η2 are the wavenumber and intrinsic impedance of medium 2.
Our goal is to describe the reflected and transmitted fields in terms of the parameters that characterize the incident wave, namely the incidence angle θi and the amplitude Ei ⊥0. The four expressions given by Eqs. (8.49a) through (8.49d) contain four unknowns: Er ⊥0, Et ⊥0, θr, and θt. Even though angles θr and θt are related to θi by Snell’s laws (Eqs. (8.28a) and (8.28b)), here we choose to treat them as unknown for the time being, because we intend 372 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION to show that Snell’s laws can also be derived by applying field boundary conditions at z = 0.
The total electric field in medium 1 is the sum of the incident and reflected electric fields: E1 ⊥= Ei ⊥+ Er ⊥; and a similar statement holds true for the total magnetic field in medium 1: H1 ⊥= Hi ⊥+ Hr ⊥.
Boundary conditions state that the tangential components of E and H must each be continuous across the boundary between the two media. Field components tangential to the boundary extend along ˆ x and ˆ y. Since the electric fields in media 1 and 2 have ˆ y components only, the boundary condition for E is ( Ei ⊥y + Er ⊥y) z=0 = Et ⊥y z=0 .
(8.50) Upon using Eqs. (8.48a), (8.49a), and (8.49c) in Eq. (8.50) and then setting z = 0, we have Ei ⊥0e−jk1x sin θi + Er ⊥0e−jk1x sin θr = Et ⊥0e−jk2x sin θt. (8.51) Since the magnetic fields in media 1 and 2 have no ˆ y components, the boundary condition for H is ( H i ⊥x + H r ⊥x) z=0 = H t ⊥x z=0 , (8.52) or −Ei ⊥0 η1 cos θi e−jk1x sin θi + Er ⊥0 η1 cos θr e−jk1x sin θr = −Et ⊥0 η2 cos θt e−jk2x sin θt.
(8.53) To satisfy Eqs. (8.51) and (8.53) for all possible values of x (i.e., all along the boundary), it follows that the arguments of all three exponentials must be equal. That is, k1 sin θi = k1 sin θr = k2 sin θt, (8.54) which is known as the phase-matching condition. The first equality in Eq. (8.54) leads to θr = θi (Snell’s law of reflection), (8.55) while the second equality leads to sin θt sin θi = k1 k2 = ω√μ1ϵ1 ω√μ2ϵ2 = n1 n2 .
(8.56) (Snell’s law of refraction) The results expressed by Eqs. (8.55) and (8.56) are identical with those derived previously in Section 8-2 through consideration of the ray path traversed by the incident, reflected, and transmitted wavefronts.
In view of Eq. (8.54), the boundary conditions given by Eqs. (8.51) and (8.53) reduce to Ei ⊥0 + Er ⊥0 = Et ⊥0, (8.57a) cos θi η1 (−Ei ⊥0 + Er ⊥0) = −cos θt η2 Et ⊥0.
(8.57b) These two equations can be solved simultaneously to yield the following expressions for the reflection and transmission coefficients in the perpendicular polarization case: ⊥= Er ⊥0 Ei ⊥0 = η2 cos θi −η1 cos θt η2 cos θi + η1 cos θt , (8.58a) τ⊥= Et ⊥0 Ei ⊥0 = 2η2 cos θi η2 cos θi + η1 cos θt .
(8.58b) These two coefficients, which formally are known as the Fresnel reflection and transmission coefficients for perpendicular polarization, are related by τ⊥= 1 + ⊥.
(8.59) If medium 2 is a perfect conductor (η2 = 0), Eqs. (8.58a) and (8.58b) reduce to ⊥= −1 and τ⊥= 0, respectively, which means that the incident wave is totally reflected by the conducting medium.
For nonmagnetic dielectrics with μ1 = μ2 = μ0 and with the help of Eq. (8.56), the expression for ⊥can be written as ⊥= cos θi − (ϵ2/ϵ1) −sin2 θi cos θi + (ϵ2/ϵ1) −sin2 θi (8.60) (for μ1 = μ2).
Since (ϵ2/ϵ1) = (n2/n1)2, this expression can also be written in terms of the indices of refraction n1 and n2.
8-4 WAVE REFLECTION AND TRANSMISSION AT OBLIQUE INCIDENCE 373 Example 8-6: Wave Incident Obliquely on a Soil Surface Using the coordinate system of Fig. 8-15, a plane wave radiated by a distant antenna is incident in air upon a plane soil surface located at z = 0. The electric field of the incident wave is given by Ei = ˆ y100 cos(ωt −πx −1.73πz) (V/m), (8.61) and the soil medium may be assumed to be a lossless dielectric with a relative permittivity of 4.
(a) Determine k1, k2, and the incidence angle θi.
(b) Obtain expressions for the total electric fields in air and in the soil.
(c) Determine the average power density carried by the wave traveling in soil.
Solution: (a) We begin by converting Eq. (8.61) into phasor form, akin to the expression given by Eq. (8.46a): Ei = ˆ y100e−jπx−j1.73πz = ˆ y100e−jk1xi (V/m), (8.62) where xi is the axis along which the wave is traveling, and k1xi = πx + 1.73πz.
(8.63) Using Eq. (8.47a), we have k1xi = k1x sin θi + k1z cos θi.
(8.64) Hence, k1 sin θi = π, k1 cos θi = 1.73π, which together give k1 = π2 + (1.73π)2 = 2π (rad/m), θi = tan−1 π 1.73π = 30◦.
The wavelength in medium 1 (air) is λ1 = 2π k1 = 1 m, and the wavelength in medium 2 (soil) is λ2 = λ1 √ϵr2 = 1 √ 4 = 0.5 m.
The corresponding wave number in medium 2 is k2 = 2π λ2 = 4π (rad/m).
Since Ei is along ˆ y, it is perpendicularly polarized (ˆ y is perpendicular to the plane of incidence containing the surface normal ˆ z and the propagation direction ˆ xi).
(b) Given that θi = 30◦, the transmission angle θt is obtained with the help of Eq. (8.56): sin θt = k1 k2 sin θi = 2π 4π sin 30◦= 0.25 or θt = 14.5◦.
With ϵ1 = ϵ0 and ϵ2 = ϵr2ϵ0 = 4ϵ0, the reflection and transmission coefficients for perpendicular polarization are determined with the help of Eqs. (8.59) and (8.60), ⊥= cos θi − (ϵ2/ϵ1) −sin2 θi cos θi + (ϵ2/ϵ1) −sin2 θi = −0.38, τ⊥= 1 + ⊥= 0.62.
Using Eqs. (8.48a) and (8.49a) with Ei ⊥0 = 100 V/m and θi = θr, the total electric field in medium 1 is E1 ⊥= Ei ⊥+ Er ⊥ = ˆ yEi ⊥0e−jk1(x sin θi+z cos θi) + ˆ yEi ⊥0e−jk1(x sin θi−z cos θi) = ˆ y100e−j(πx+1.73πz) −ˆ y38e−j(πx−1.73πz), 374 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION and the corresponding instantaneous electric field in medium 1 is E1 ⊥(x, z, t) = Re E1 ⊥ejωt = ˆ y[100 cos(ωt −πx −1.73πz) −38 cos(ωt −πx + 1.73πz)] (V/m).
In medium 2, using Eq. (8.49c) with Et ⊥0 = τ⊥Ei ⊥0 gives Et ⊥= ˆ yτEi ⊥0e−jk2(x sin θt+z cos θt) = ˆ y62e−j(πx+3.87πz) and, correspondingly, Et ⊥(x, z, t) = Re Et ⊥ejωt = ˆ y62 cos(ωt −πx −3.87πz) (V/m).
(c) In medium 2, η2 = η0/√ϵr2 ≈120π/ √ 4 = 60π ( ), and the average power density carried by the wave is S t av = |Et ⊥0|2 2η2 = (62)2 2 × 60π = 10.2 (W/m2).
8-4.2 Parallel Polarization If we interchange the roles played by E and H in the perpendicular polarization scenario covered in the preceding subsection, while keeping in mind the requirement that E × × × H must point in the direction of propagation for each of the incident, reflected, and transmitted waves, we end up with the parallel polarization scenario shown in Fig. 8-16.
Now the electric fields lie in the plane of incidence, while the associated magnetic fields are perpendicular to the plane of incidence.
With reference to the directions indicated in Fig. 8-16, the fields of the incident, reflected, and transmitted waves are given by Incident Wave Ei ∥= ˆ yiEi ∥0e−jk1xi = (ˆ x cos θi −ˆ z sin θi)Ei ∥0e−jk1(x sin θi+z cos θi), (8.65a) Hi ∥= ˆ y Ei ∥0 η1 e−jk1xi = ˆ y Ei ∥0 η1 e−jk1(x sin θi+z cos θi), (8.65b) θt θr θi yt –z H|| r E|| r z yr xr xr xt xi y yi –z θi θr θt xt x z Medium 1 (ε1, μ1) Medium 2 (ε2, μ2) xr = x sin θr – z cos θr yr = x cos θr + z sin θr ˆ ˆ ˆ xt = x sin θt + z cos θt yt = x cos θt – z sin θt ˆ ˆ ˆ xi = x sin θi + z cos θi yi = x cos θi – z sin θi ˆ ˆ ˆ E||z r E||x r H|| t E|| t E||z i E||z t E||x t H|| i E|| i E||x i Figure 8-16 Parallel-polarized plane wave incident at an angle θi upon a planar boundary.
Reflected Wave Er ∥= ˆ yrEr ∥0e−jk1xr = (ˆ x cos θr + ˆ z sin θr)Er ∥0e−jk1(x sin θr−z cos θr), (8.65c) Hr ∥= −ˆ y Er ∥0 η1 e−jk1xr = −ˆ y Er ∥0 η1 e−jk1(x sin θr−z cos θr), (8.65d) Transmitted Wave Et ∥= ˆ ytEt ∥0e−jk2xt = (ˆ x cos θt −ˆ z sin θt)Et ∥0e−jk2(x sin θt+z cos θt), (8.65e) Ht ∥= ˆ y Et ∥0 η2 e−jk2xt = ˆ y Et ∥0 η2 e−jk2(x sin θt+z cos θt). (8.65f) 8-4 WAVE REFLECTION AND TRANSMISSION AT OBLIQUE INCIDENCE 375 By matching the tangential components of E and H in both media at z = 0, we again obtain the relations defining Snell’s laws, as well as the following expressions for the Fresnel reflection and transmission coefficients for parallel polarization: ∥= Er ∥0 Ei ∥0 = η2 cos θt −η1 cos θi η2 cos θt + η1 cos θi , (8.66a) τ∥= Et ∥0 Ei ∥0 = 2η2 cos θi η2 cos θt + η1 cos θi .
(8.66b) The preceding expressions can be shown to yield the relation τ∥= (1 + ∥) cos θi cos θt .
(8.67) We noted earlier in connection with the perpendicular-polarization case that, when the second medium is a perfect conductor with η2 = 0, the incident wave gets totally reflected at the boundary. The same is true for the parallel polarization case; setting η2 = 0 in Eqs. (8.66a) and (8.66b) gives ∥= −1 and τ∥= 0.
For nonmagnetic materials, Eq. (8.66a) becomes ∥= −(ϵ2/ϵ1) cos θi + (ϵ2/ϵ1) −sin2 θi (ϵ2/ϵ1) cos θi + (ϵ2/ϵ1) −sin2 θi (8.68) (for μ1 = μ2).
To illustrate the angular variations of the magnitudes of ⊥ and ∥, Fig. 8-17 shows plots for waves incident in air onto three different types of dielectric surfaces: dry soil (ϵr = 3), wet soil (ϵr = 25), and water (ϵr = 81). For each of the surfaces, (1) ⊥= ∥at normal incidence (θi = 0), as expected, (2)|⊥| = |∥| = 1atgrazingincidence(θi = 90◦), and(3)∥ goes to zero at an angle called the Brewster angle in Fig. 8-17.
Had the materials been magnetic too (μ1 ̸= μ2), it would have been possible for ⊥to vanish at some angle as well. However, for nonmagnetic materials, the Brewster angle exists only for parallel polarization, and its value depends on the ratio (ϵ2/ϵ1), as we see shortly.
▶ At the Brewster angle, the parallel-polarized component of the incident wave is totally transmitted into medium 2. ◀ 0 1 90 80 70 60 50 40 30 20 10 0.2 0.4 0.6 0.8 Water (εr = 81) Wet soil (εr = 25) Dry soil (εr = 3) |Г | |Г| || (θB dry soil) (θB wet soil) (θB water) Incidence angle θi (degrees) |Г| || or |Г | T T Figure 8-17 Plots for |⊥| and |∥| as a function of θi for a dry-soil surface, a wet-soil surface, and a water surface. For each surface, |∥| = 0 at the Brewster angle.
8-4.3 Brewster Angle The Brewster angle θB is defined as the incidence angle θi at which the Fresnel reflection coefficient = 0.
Perpendicular polarization For perpendicular polarization, the Brewster angle θB⊥can be obtained by setting the numerator of the expression for ⊥, given by Eq. (8.58a), equal to zero. This happens when η2 cos θi = η1 cos θt.
(8.69) By (1) squaring both sides of Eq. (8.69), (2) using Eq. (8.56), (3) solving for θi, and then denoting θi as θB⊥, we obtain sin θB⊥= 1 −(μ1ϵ2/μ2ϵ1) 1 −(μ1/μ2)2 .
(8.70) Because the denominator of Eq. (8.70) goes to zero when μ1 = μ2, θB⊥does not exist for nonmagnetic materials.
376 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION Parallel polarization For parallel polarization, the Brewster angle θB∥at which ∥= 0 can be found by setting the numerator of ∥, Eq. (8.66a), equal to zero. The result is identical to Eq. (8.70), but with μ and ϵ interchanged. That is, sin θB∥= 1 −(ϵ1μ2/ϵ2μ1) 1 −(ϵ1/ϵ2)2 .
(8.71) For nonmagnetic materials, θB∥= sin−1 1 1 + (ϵ1/ϵ2) = tan−1 ϵ2 ϵ1 (for μ1 = μ2).
(8.72) The Brewster angle is also called the polarizing angle. This is because, if a wave composed of both perpendicular and parallel polarization components is incident upon a nonmagnetic surface at the Brewster angle θB∥, the parallel polarized component is totally transmitted into the second medium, and onlytheperpendicularlypolarizedcomponentisreflectedbythe surface. Natural light, including sunlight and light generated by most manufactured sources, is unpolarized because the direction of the electric field of the light waves varies randomly in angle over the plane perpendicular to the direction of propagation. Thus, on average half of the intensity of natural light is perpendicularly polarized and the other half is parallel polarized. When unpolarized light is incident upon a surface at the Brewster angle, the reflected wave is strictly perpendicularly polarized. Hence, the surface acts as a polarizer.
Concept Question 8-4: Can total internal reflection take place for a wave incident from medium 1 (with n1) onto medium 2 (with n2) when n2 > n1?
Concept Question 8-5: What is the difference between the boundary conditions applied in Section 8-1.1 for normal incidence and those applied in Section 8-4.1 for oblique incidence with perpendicular polarization?
Concept Question 8-6: Why is the Brewster angle also called the polarizing angle?
Concept Question 8-7: At the boundary, the vector sum of the tangential components of the incident and reflected electric fields has to equal the tangential component of the transmitted electric field. For ϵr1 = 1 and ϵr2 = 16, determine the Brewster angle and then verify the validity of the preceding statement by sketching to scale the tangential components of the three electric fields at the Brewster angle.
Exercise 8-7: A wave in air is incident upon a soil surface at θi = 50◦. If soil has ϵr = 4 and μr = 1, determine ⊥, τ⊥, ∥, and τ∥.
Answer: ⊥= −0.48, τ⊥= 0.52, ∥= −0.16, τ∥= 0.58. (See EM.) Exercise 8-8: Determine the Brewster angle for the boundary of Exercise 8.7.
Answer: θB = 63.4◦. (See EM.) Exercise 8-9: Show that the incident, reflected, and transmitted electric and magnetic fields given by Eqs. (8.65a) through (8.65f) all have the same exponential phase function along the x direction.
Answer: With the help of Eqs. (8.55) and (8.56), all six fields are shown to vary as e−jk1x sin θi. (See EM.) 8-5 Reflectivity and Transmissivity The reflection and transmission coefficients derived earlier are ratios of the reflected and transmitted electric field amplitudes to the amplitude of the incident electric field. We now examine power ratios, starting with the perpendicular polarization case.
Figure 8-18 shows a circular beam of electromagnetic energy incident upon the boundary between two contiguous, lossless media. The area of the spot illuminated by the beam is A, and the incident, reflected, and transmitted beams have electric-field amplitudes Ei ⊥0, Er ⊥0, and Et ⊥0, respectively.
The average power densities carried by the incident, reflected, and 8-5 REFLECTIVITY AND TRANSMISSIVITY 377 Pi Pr Pt A cos θi A cos θr A cos θt θi θr θt Medium 2 (ε2, μ2) Medium 1 (ε1, μ1) A Figure 8-18 Reflection and transmission of an incident circular beam illuminating a spot of size A on the interface.
transmitted beams are Si ⊥= |Ei ⊥0|2 2η1 , (8.73a) Sr ⊥= |Er ⊥0|2 2η1 , (8.73b) St ⊥= |Et ⊥0|2 2η2 , (8.73c) where η1 and η2 are the intrinsic impedances of media 1 and 2, respectively. The cross-sectional areas of the incident, reflected, and transmitted beams are Ai = A cos θi, (8.74a) Ar = A cos θr, (8.74b) At = A cos θt, (8.74c) and the corresponding average powers carried by the beams are P i ⊥= Si ⊥Ai = |Ei ⊥0|2 2η1 A cos θi, (8.75a) P r ⊥= Sr ⊥Ar = |Er ⊥0|2 2η1 A cos θr, (8.75b) P t ⊥= St ⊥At = |Et ⊥0|2 2η2 A cos θt.
(8.75c) The reflectivity R (also called reflectance in optics) is defined as the ratio of the reflected to the incident power. The reflectivity for perpendicular polarization is then R⊥= P r ⊥ P i ⊥ = |Er ⊥0|2 cos θr |Ei ⊥0|2 cos θi = Er ⊥0 Ei ⊥0 2 , (8.76) where we used the fact that θr = θi, in accordance with Snell’s law of reflection. The ratio of the reflected to incident electric field amplitudes, |Er ⊥0/Ei ⊥0|, is equal to the magnitude of the reflection coefficient ⊥. Hence, R⊥= |⊥|2, (8.77) and, similarly, for parallel polarization R∥= P r ∥ P i ∥ = |∥|2.
(8.78) The transmissivity T (or transmittance in optics) is defined as the ratio of the transmitted power to incident power: T⊥= P t ⊥ P i ⊥ = |Et ⊥0|2 |Ei ⊥0|2 η1 η2 A cos θt A cos θi = |τ⊥|2 η1 cos θt η2 cos θi , (8.79a) T∥= P t ∥ P i ∥ = |τ∥|2 η1 cos θt η2 cos θi .
(8.79b) 378 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION ▶The incident, reflected, and transmitted waves do not have to obey any such laws as conservation of electric field, conservation of magnetic field, or conservation of power density, but they do have to obey the law of conservation of power. ◀ In fact, in many cases the transmitted electric field is larger than the incident electric field. Conservation of power requires that the incident power equals the sum of the reflected and transmitted powers. That is, for perpendicular polarization, P i ⊥= P r ⊥+ P t ⊥, (8.80) or |Ei ⊥0|2 2η1 A cos θi = |Er ⊥0|2 2η1 A cos θr + |Et ⊥0|2 2η2 A cos θt.
(8.81) Use of Eqs. (8.76), (8.79a), and (8.79b) leads to R⊥+ T⊥= 1, (8.82a) R∥+ T∥= 1, (8.82b) or |⊥|2 + |τ⊥|2 η1 cos θt η2 cos θi = 1, (8.83a) |∥|2 + |τ∥|2 η1 cos θt η2 cos θi = 1.
(8.83b) Figure 8-19 shows plots for (R∥, T∥) as a function of θi for an air–glass interface. Note that the sum of R∥and T∥is always equal to 1, as mandated by Eq. (8.82b). We also note that, at the Brewster angle θB, R∥= 0 and T∥= 1.
Table 8-2 provides a summary of the general expressions for , τ, R, and T for both normal and oblique incidence.
Table 8-2 Expressions for , τ, R, and T for wave incidence from a medium with intrinsic impedance η1 onto a medium with intrinsic impedance η2. Angles θi and θt are the angles of incidence and transmission, respectively.
Normal Incidence Perpendicular Parallel Property θi = θt = 0 Polarization Polarization Reflection coefficient = η2 −η1 η2 + η1 ⊥= η2 cos θi −η1 cos θt η2 cos θi + η1 cos θt ∥= η2 cos θt −η1 cos θi η2 cos θt + η1 cos θi Transmission coefficient τ = 2η2 η2 + η1 τ⊥= 2η2 cos θi η2 cos θi + η1 cos θt τ∥= 2η2 cos θi η2 cos θt + η1 cos θi Relation of to τ τ = 1 + τ⊥= 1 + ⊥ τ∥= (1 + ∥) cos θi cos θt Reflectivity R = ||2 R⊥= |⊥|2 R∥= |∥|2 Transmissivity T = |τ|2 η1 η2 T⊥= |τ⊥|2 η1 cos θt η2 cos θi T∥= |τ∥|2 η1 cos θt η2 cos θi Relation of R to T T = 1 −R T⊥= 1 −R⊥ T∥= 1 −R∥ Notes: (1) sin θt = √μ1ϵ1/μ2ϵ2 sin θi; (2) η1 = √μ1/ϵ1; (3) η2 = √μ2/ϵ2; (4) for nonmagnetic media, η2/η1 = n1/n2.
8-5 REFLECTIVITY AND TRANSMISSIVITY 379 Module 8.3 Oblique Incidence Upon specifying the frequency, polarization, and incidence angle of a plane wave incident upon a planar boundary between two lossless media, this module displays vector information and plots of the reflection and transmission coefficients as a function of incidence angle.
θi Air Glass n = 1.5 0 0 30 θi (degrees) 60 90 0.5 1 θB R| | T| | Reflectivity R| | and transmissivity T| | Figure 8-19 Angular plots for (R∥, T∥) for an air–glass interface.
Example 8-7: Beam of Light A 5 W beam of light with circular cross section is incident in air upon the plane boundary of a dielectric medium with index of refraction of 5. If the angle of incidence is 60◦and the incident wave is parallel polarized, determine the transmission angle and the powers contained in the reflected and transmitted beams.
Solution: From Eq. (8.56), sin θt = n1 n2 sin θi = 1 5 sin 60◦= 0.17 or θt = 10◦.
With ϵ2/ϵ1 = n2 2/n2 1 = (5)2 = 25, the reflection coefficient for parallel polarization follows from Eq. (8.68) as ∥= −(ϵ2/ϵ1) cos θi + (ϵ2/ϵ1) −sin2 θi (ϵ2/ϵ1) cos θi + (ϵ2/ϵ1) −sin2 θi = −25 cos 60◦+ 25 −sin2 60◦ 25 cos 60◦+ 25 −sin2 60◦ = −0.435.
380 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION Module 8.4 Oblique Incidence in Lossy Medium This module extends the capabilities of Module 8.1 to situations in which medium 2 is lossy.
The reflected and transmitted powers therefore are P r ∥= P i ∥|∥|2 = 5(0.435)2 = 0.95 W, P t ∥= P i ∥−P r ∥= 5 −0.95 = 4.05 W.
8-6 Waveguides Earlier in Chapter 2, we considered two families of transmission lines, namely those that support transverse-electromagnetic (TEM) modes and those that do not.
Transmission lines belonging to the TEM family (Fig. 2-4), including coaxial, two-wire, and parallel-plate lines, support E and H fields that are orthogonal to the direction of propagation.
Fields supported by lines in the other group, often called higher-order transmission lines, may have E or H orthogonal to the direction of propagation ˆ k, but not both simultaneously. Thus, at least one component of E or H is along ˆ k.
▶If E is transverse to ˆ k but H is not, we call it a transverse electric (TE) mode, and if H is transverse to ˆ k but E is not, we call it a transverse magnetic (TM) mode. ◀ Among all higher-order transmission lines, the two most commonly used are the optical fiber and the metal waveguide.
As noted in Section 8-3, a wave is guided along an optical fiber through successive zigzags by taking advantage of total internal reflection at the boundary between the (inner) core and the (outer) cladding [Fig. 8-20(a)]. Another way to achieve internal reflection at the core’s boundary is to have its surface coated by a conducting material. Under the proper conditions, on which we shall elaborate later, a wave excited in the interior of a hollow conducting pipe, such as the circular or rectangular waveguides shown in Figs. 8-20(b) and (c), undergoes a process similar to that of successive internal reflection in an optical fiber, resulting in propagation down the pipe. Most waveguide applications call for air-filled guides, but in some cases, the waveguide may be filled with a dielectric material so as to alter its propagation velocity or impedance, or it may be vacuum-8-6 WAVEGUIDES 381 (a) Optical fiber (b) Circular waveguide (c) Rectangular waveguide Metal Hollow or dielectric-filled θ2 θ3 n0 n0 nf θi Fiber core Cladding nc nc θi Metal Hollow or dielectric-filled Figure 8-20 Wave travel by successive reflections in (a) an optical fiber, (b) a circular metal waveguide, and (c) a rectangular metal waveguide.
pumped to eliminate air molecules so as to prevent voltage breakdown, thereby increasing its power-handling capabilities.
Figure 8-21 illustrates how a coaxial cable can be connected to a rectangular waveguide. With its outer conductor connected to the metallic waveguide enclosure, the coaxial cable’s inner conductor protrudes through a tiny hole into the waveguide’s (a) Coax-to-waveguide coupler (b) Cross-sectional view at x = a/2 y = b y = 0 Electric field EM wave z Waveguide 0 Coaxial line x b a z y Probe Figure 8-21 The inner conductor of a coaxial cable can excite an EM wave in the waveguide.
interior (without touching the conducting surface).
Time-varying electric field lines extending between the protruding inner conductor and the inside surface of the guide provide the excitation necessary to transfer a signal from the coaxial line to the guide. Conversely, the center conductor can act like a probe, coupling a signal from the waveguide to the coaxial cable.
For guided transmission at frequencies below 30 GHz, the coaxial cable is by far the most widely used transmission line.
At higher frequencies, however, the coaxial cable has a number of limitations: (a) in order for it to propagate only TEM modes, the cable’s inner and outer conductors have to be reduced in size to satisfy a certain size-to-wavelength requirement, making it more difficult to fabricate; (b) the smaller cross section reduces the cable’s power-handling capacity (limited by dielectric breakdown); and (c) the attenuation due to dielectric 382 TECHNOLOGY BRIEF 16: BAR-CODE READERS Technology Brief 16: Bar-Code Readers A bar code consists of a sequence of parallel bars of certain widths, usually printed in black against a white background, configured to represent a particular binary code of information about a product and its manufacturer. Laser scanners can read the code and transfer the information to a computer, a cash register, or a display screen. For both stationary scanners built into checkout counters at grocery stores and handheld units that can be pointed at the bar-coded object like a gun, the basic operation of a bar-code reader is the same.
Basic Operation The scanner uses a laser beam of light pointed at a multifaceted rotating mirror, spinning at a high speed on the order of 6,000 revolutions per minute (Fig. TF16-1). The rotating mirror creates a fan beam to illuminate the bar code on the object. Moreover, by exposing the laser light to its many facets, it deflects the beam into many different directions, allowing the object to be scanned over a wide range of positions and orientations. The goal is to have one of those directions be such that the beam reflected by the bar code ends up traveling in the direction of, and captured by, the light detector (sensor), which then reads the coded sequence (white bars reflect laser light and black ones do not) and converts it into a binary sequence of ones and zeros (Fig.TF16-2). To eliminate interference by ambient light, a glass filter is used as shown in Fig. TF16-1 to block out all light except for a narrow wavelength band centered at the wavelength of the laser light.
Bar code Central store computer Cash register Sensor Rotating mirror (6,000 rpm) Glass filter FigureTF16-1 Elements of a bar-code reader.
Bar code Electrical signal Digital code FigureTF16-2 Bar code contained in reflected laser beam.
8-7 GENERAL RELATIONS FOR E AND H 383 losses increases with frequency. For all of these reasons, metal waveguides have been used as an alternative to coaxial lines for many radar and communication applications that operate at frequencies in the 5–100 GHz range, particularly those requiring the transmission of high levels of radio-frequency (RF) power.
Even though waveguides with circular and elliptical cross sections have been used in some microwave systems, the rectangular shape has been the more prevalent geometry.
8-7 General Relations for E and H The purpose of the next two sections is to derive expressions for E and H for the TE and TM modes in a rectangular waveguide, and to examine their wave properties. We choose the coordinate system shown in Fig. 8-22, in which propagation occurs along ˆ z.
For TE modes, the electric field is transverse to the direction of propagation. Hence, E may have components along ˆ x and ˆ y, but not along ˆ z. In contrast, H has a ˆ z-directed component and may have components along either ˆ x or ˆ y, or both. The converse is true for TM modes.
Our solution procedure consists of four steps: (1) Maxwell’s equations are manipulated to develop general expressions for the phasor-domain transverse field components Ex, Ey, Hx, and Hy in terms of Ez and Hz.
When specialized to the TE case, these expressions become functions of Hz only, and the converse is true for the TM case.
(2) The homogeneous wave equations given by Eqs. (7.15) and (7.16) are solved to obtain valid solutions for Ez (TM case) and Hz (TE case) in a waveguide.
0 z b y a x Figure 8-22 Waveguide coordinate system.
(3) The expressions derived in step 1 are then used to find Ex, Ey, Hx, and Hy.
(4) The solution obtained in step 3 are analyzed to determine the phase velocity and other properties of the TE and TM waves.
The intent of the present section is to realize the stated goals of step 1. We begin with a general form for the E and H fields in the phasor domain: E = ˆ x Ex + ˆ y Ey + ˆ z Ez, (8.84a) H = ˆ x Hx + ˆ y Hy + ˆ z Hz.
(8.84b) In general, all six components of E and H may depend on (x, y, z), and while we do not yet know how they functionally depend on (x, y), our prior experience suggests that E and H of a wave traveling along the +z direction should exhibit a dependence on z of the form e−jβz, where β is a yet-to-be-determined phase constant. Hence, we adopt the form Ex(x, y, z) = ex(x, y) e−jβz, (8.85) where ex(x, y) describes the dependence of Ex(x, y, z) on (x, y) only. The form of Eq. (8.85) can be used for all other components of E and H as well. Thus, E = (ˆ x ex + ˆ y ey + ˆ z ez)e−jβz, (8.86a) H = (ˆ x hx + ˆ y hy + ˆ z hz)e−jβz.
(8.86b) The notation is intended to clarify that, in contrast to E and H, which vary with (x, y, z), the lower-case e and h vary with (x, y) only.
In a lossless, source-free medium (such as the inside of a waveguide) characterized by permittivity ϵ and permeability μ (and conductivity σ = 0), Maxwell’s curl equations are given by Eqs. (7.2b and d) with J = 0, ∇× E = −jωμ H, (8.87a) ∇× H = jωϵ E.
(8.87b) Upon inserting Eqs. (8.86a and b) into Eqs. (8.87a and b), and recalling that each of the curl equations actually consists of 384 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION three separate equations—one for each of the unit vectors ˆ x, ˆ y, and ˆ z, we obtain the following relationships: ∂ ez ∂y + jβ ey = −jωμ hx, (8.88a) −jβ ex −∂ ez ∂x = −jωμ hy, (8.88b) ∂ ey ∂x −∂ ex ∂y = −jωμ hz, (8.88c) ∂ hz ∂y + jβ hy = jωϵ ex, (8.88d) −jβ hx −∂ hz ∂x = jωϵ ey, (8.88e) ∂ hy ∂x −∂ hx ∂y = jωϵ ez.
(8.88f) Equations (8.88a–f) incorporate the fact that differentiation with respect to z is equivalent to multiplication by −jβ.
By manipulating these equations algebraically, we can obtain expressions for the x and y components of E and H in terms of their z components, namely Ex = −j k2 c β ∂ Ez ∂x + ωμ ∂ Hz ∂y , (8.89a) Ey = j k2 c −β ∂ Ez ∂y + ωμ ∂ Hz ∂x , (8.89b) Hx = j k2 c ωϵ ∂ Ez ∂y −β ∂ Hz ∂x , (8.89c) Hy = −j k2 c ωϵ ∂ Ez ∂x + β ∂ Hz ∂y .
(8.89d) Here k2 c = k2 −β2 = ω2μϵ −β2, (8.90) and k is the unbounded-medium wavenumber defined earlier as k = ω√μϵ .
(8.91) For reasons that become clear later (in Section 8-8), the constant kc is called the cutoff wavenumber. In view of Eqs. (8.89a–d), the x and y components of E and H can now be found readily, so long as we have mathematical expressions for Ez and Hz.
For the TE mode, Ez = 0, so all we need to know is Hz, and the converse is true for the TM case.
8-8 TM Modes in Rectangular Waveguide In the preceding section we developed expressions for Ex, Ey, Hx, and Hy, all in terms of Ez and Hz. Since Hz = 0 for the TM mode, our task reduces to obtaining a valid solution for Ez.
Our starting point is the homogeneous wave equation for E.
For a lossless medium characterized by an unbounded-medium wavenumber k, the wave equation is given by Eq. (7.19) as ∇2 E + k2 E = 0.
(8.92) To satisfy Eq. (8.92), each of its ˆ x, ˆ y, and ˆ z components has to be satisfied independently. Its ˆ z component is given by: ∂2 Ez ∂x2 + ∂2 Ez ∂y2 + ∂2 Ez ∂z2 + k2 Ez = 0.
(8.93) By adopting the mathematical form given by Eq. (8.85), namely Ez(x, y, z) = ez(x, y) e−jβz, (8.94) Eq. (8.93) reduces to ∂2 ez ∂x2 + ∂2 ez ∂y2 + k2 c ez = 0, (8.95) where k2 c is as defined by Eq. (8.90).
The form of the partial differential equation (separate, uncoupled derivatives with respect to x and y) allows us to assume a product solution of the form ez(x, y) = X(x) Y(y).
(8.96) Substituting Eq. (8.96) into Eq. (8.95), followed with dividing all terms by X(x) Y(y), leads to: 1 X d2X dx2 + 1 Y d2Y dy2 + k2 c = 0.
(8.97) To satisfy Eq. (8.97), each of the first two terms has to equal a constant. Hence, we define separation constants kx and ky such that d2X dx2 + k2 xX = 0, (8.98a) d2Y dy2 + k2 yY = 0, (8.98b) 8-8 TM MODES IN RECTANGULAR WAVEGUIDE 385 and k2 c = k2 x + k2 y.
(8.99) Before proposing solutions for Eqs. (8.98a and b), we should consider the constraints that the solutions must meet.
The electric field Ez is parallel to all four walls of the waveguide.
Since E = 0 in the conducting walls, the boundary conditions require Ez in the waveguide cavity to go to zero as x approaches 0 and a, and as y approaches 0 and b (Fig. 8-22). To satisfy these boundary conditions, sinusoidal solutions are chosen for X(x) and Y(y) as follows: ez = X(x) Y(y) = (A cos kxx + B sin kxx)(C cos kyy + D sin kyy).
(8.100) These forms for X(x) and Y(y) definitely satisfy the differential equations given by Eqs. (8.98a and b). The boundary conditions for ez are: ez = 0, at x = 0 and a, (8.101a) ez = 0, at y = 0 and b.
(8.101b) Satisfying ez = 0 at x = 0 requires that we set A = 0, and similarly, satisfying ez = 0 at y = 0 requires C = 0. Satisfying ez = 0 at x = a requires kx = mπ a , m = 1, 2, 3, . . .
(8.102a) and similarly, satisfying ez = 0 at y = b requires ky = nπ b , n = 1, 2, 3, . . .
(8.102b) Consequently, Ez = eze−jβz = E0 sin mπx a sin nπy b e−jβz, (8.103) where E0 = BD is the amplitude of the wave in the guide.
Keeping in mind that Hz = 0 for the TM mode, the transverse components of E and H can now be obtained by applying Eq. (8.103) to (8.89a–d), Ex = −jβ k2 c mπ a E0 cos mπx a sin nπy b e−jβz, (8.104a) Ey = −jβ k2 c nπ b E0 sin mπx a cos nπy b e−jβz, (8.104b) Hx = jωϵ k2 c nπ b E0 sin mπx a cos nπy b e−jβz, (8.104c) Hy = −jωϵ k2 c mπ a E0 cos mπx a sin nπy b e−jβz.
(8.104d) Each combination of the integers m and n represents a viable solution, or a mode, denoted TMmn. Associated with each mn mode are specific field distributions for the region inside the guide. Figure 8-23 depicts the E and H field lines for the TM11 mode across two different cross sections of the guide.
According to Eqs. (8.103) and (8.104e), a rectangular waveguide with cross section (a × b) can support the propagation of waves with many different, but discrete, field configurations specified by the integers m and n. The only quantity in the fields’expressions that we have yet to determine is the propagation constant β, contained in the exponential e−jβz. BycombiningEqs.(8.90), (8.99), and(8.102), weobtain the following expression for β: β = k2 −k2 c = ω2μϵ − mπ a 2 − nπ b 2 .
(8.105) (TE and TM) Even though the expression for β was derived for TM modes, it is equally applicable to TE modes.
The exponential e−jβz describes a wave traveling in the +z direction, provided that β is real, which corresponds to k > kc. If k < kc, β becomes imaginary: β = −jα with α real, in which case e−jβz = e−αz, yielding evanescent waves characterized by amplitudes that decay rapidly with z due to the attenuation function e−αz. Corresponding to each mode 386 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION (c) Field lines for side view y b z H field into page E field H field out of page 0 (b) Field lines for front view 0 x y b a H field E field (a) Cross-sectional planes Front view Side view 0 x y z a b Figure 8-23 TM11 electric and magnetic field lines across two cross-sectional planes.
(m, n), there is a cutoff frequency fmn at which β = 0. By setting β = 0 in Eq. (8.105) and then solving for f , we have fmn = up0 2 m a 2 + n b 2 , (8.106) (TE and TM) where up0 = 1/√μϵ is the phase velocity of a TEM wave in an unbounded medium with constitutive parameters ϵ and μ.
▶A wave, in a given mode, can propagate through the guide only if its frequency f > fmn, as only then β = real. ◀ The mode with the lowest cutoff frequency is known as the dominant mode. The dominant mode is TM11 among TM modes and TE10 among TE modes (whose solution is given in Section 8-8). Whereas a value of zero for m or n is allowed for TEmodes, itisnotforTMmodes(becauseifeithermorniszero, Ez in Eq. (8.103) becomes zero and all other field components vanish as well).
By combining Eqs. (8.105) and (8.106), we can express β in terms of fmn, β = ω up0 1 − fmn f 2 (TE and TM).
(8.107) The phase velocity of a TE or TM wave in a waveguide is up = ω β = up0 1 −(fmn/f )2 .
(8.108) (TE and TM) The transverse electric field consists of components Ex and Ey, given by Eqs. (8.104a and b). For a wave traveling in the +z direction, the magnetic field associated with Ex is Hy [according to the right hand rule given by Eq. (7.39a)], and similarly, the magnetic field associated with Ey is − Hx. The 8-8 TM MODES IN RECTANGULAR WAVEGUIDE 387 ratios, obtained by employing Eq. (8.104e), constitute the wave impedance in the guide, ZTM = Ex Hy = − Ey Hx = βη k = η 1 − fmn f 2 , (8.109) where η = √μ/ϵ is the intrinsic impedance of the dielectric material filling the guide.
Example 8-8: Mode Properties A TM wave propagating in a dielectric-filled waveguide of unknown permittivity has a magnetic field with y component given by Hy = 6 cos(25πx) sin(100πy) × sin(1.5π × 1010t −109πz) (mA/m).
If the guide dimensions are a = 2b = 4 cm, determine: (a) the mode numbers, (b) the relative permittivity of the material in the guide, (c) the phase velocity, and (d) obtain an expression for Ex.
Solution: (a) By comparison with the expression for Hy given by Eq. (8.104d), we deduce that the argument of x is (mπ/a) and the argument of y is (nπ/b). Hence, 25π = mπ 4 × 10−2 , 100π = nπ 2 × 10−2 , which yield m = 1 and n = 2. Therefore, the mode is TM12.
(b) The second sine function in the expression for Hy represents sin(ωt −βz), which means that ω = 1.5π × 1010 (rad/s), or f = 7.5 GHz, β = 109π (rad/m).
By rewriting Eq. (8.105) so as to obtain an expression for ϵr = ϵ/ϵ0 in terms of the other quantities, we have ϵr = c2 ω2 β2 + mπ a 2 + nπ b 2 , where c is the speed of light. Inserting the available values, we obtain ϵr = (3 × 108)2 (1.5π × 1010)2 · (109π)2 + π 4 × 10−2 2 + 2π 2 × 10−2 2 = 9.
(c) up = ω β = 1.5π × 1010 109π = 1.38 × 108 m/s, which is slower than the speed of light. However, as explained later in Section 8-10, the phase velocity in a waveguide may exceed c, but the velocity with which energy is carried down the guide is the group velocity ug, which is never greater than c.
(d) From Eq. (8.109), ZTM = η 1 −(f12/f )2 Application of Eq. (8.106) yields f12 = 5.15 GHz for the TM12 mode. Using that in the expression for ZTM, in addition to f = 7.5 GHz and η = √μ/ϵ = (√μ0/ϵ0)/√ϵr = 377/ √ 9 = 125.67 , gives ZTM = 91.3 .
Hence, Ex = ZTMHy = 91.3 × 6 cos(25πx) sin(100πy) × sin(1.5π × 1010t −109πz) (mV/m) = 0.55 cos(25πx) sin(100πy) × sin(1.5π × 1010t −109πz) (V/m).
Concept Question 8-8: What are the primary limita-tions of coaxial cables at frequencies higher than 30 GHz?
Concept Question 8-9: Can a TE mode have a zero magnetic field along the direction of propagation?
Concept Question 8-10: What is the rationale for choosing a solution for ez that involves sine and cosine functions?
388 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION Concept Question 8-11: What is an evanescent wave?
Exercise 8-10: For a square waveguide with a = b, what is the value of the ratio Ex/ Ey for the TM11 mode?
Answer: tan(πy/a)/ tan(πx/a).
Exercise 8-11: What is the cutoff frequency for the dominant TM mode in a waveguide filled with a material with ϵr = 4?
The waveguide dimensions are a = 2b = 5 cm.
Answer: For TM11, f11 = 3.35 GHz.
Exercise 8-12: What is the magnitude of the phase velocity of a TE or TM mode at f = fmn?
Answer: up = ∞! (See explanation in Section 8-10.) 8-9 TE Modes in Rectangular Waveguide In the TM case, for which the wave has no magnetic field component along the z direction (i.e., Hz = 0), we started our treatment in the preceding section by obtaining a solution for Ez, and then we used it to derive expressions for the tangential components of E and H. For the TE case, the same basic procedure can be applied, except for reversing the roles of Ez and Hz. Such a process leads to: Ex = jωμ k2 c nπ b H0 cos mπx a sin nπy b e−jβz, (8.110a) Ey = −jωμ k2 c mπ a H0 sin mπx a cos nπy b e−jβz, (8.110b) Hx = jβ k2 c mπ a H0 sin mπx a cos nπy b e−jβz, (8.110c) Hy = jβ k2 c nπ b H0 cos mπx a sin nπy b e−jβz, (8.110d) Hz = H0 cos mπx a cos nπy b e−jβz, (8.110e) and, of course, Ez = 0.
The expressions for fmn, β, and up given earlier by Eqs. (8.106), (8.107), and (8.108) remain unchanged.
▶Because not all the fields vanish if m or n assume a value of zero, the lowest order TE mode is TE10 if a > b, or TE01 if a < b. It is customary to assign a to be the longer dimension, in which case the TE10 mode is the de facto dominant mode. ◀ Another difference between the TE and TM modes relates to the expression for the wave impedance. For TE, ZTE = Ex Hy = − Ey Hx = η 1 −(fmn/f )2 .
(8.111) A summary of the expressions for the various wave attributes of TE and TM modes is given in Table 8-3. By way of reference, corresponding expressions for the TEM mode on a coaxial transmission line are included as well.
Example 8-9: Cutoff Frequencies For a hollow rectangular waveguide with dimensions a = 3 cm and b = 2 cm, determine the cutoff frequencies for all modes, up to 20 GHz. Over what frequency range will the guide support the propagation of a single dominant mode?
Solution: A hollow guide has μ = μ0 and ϵ = ϵ0. Hence, up0 = 1/√μ0ϵ0 = c.
Application of Eq. (8.106) gives the cutoff frequencies shown in Fig. 8-24, which start at 5 GHz for the TE10 mode. To avoid all other modes, the frequency of operation should be restricted to the 5–7.5 GHz range.
8-10 Propagation Velocities When a wave is used to carry a message through a medium or along a transmission line, information is encoded into the wave’s amplitude, frequency, or phase. A simple example is shown in Fig. 8-25, in which a high-frequency sinusoidal wave of frequency f is amplitude-modulated by a low-frequency Gaussian pulse. The waveform in (b) is the result of multiplying the Gaussian pulse shape in (a) by the carrier waveform.
By Fourier analysis, the waveform in (b) is equivalent to the superposition of a group of sinusoidal waves with specific 8-10 PROPAGATION VELOCITIES 389 Table 8-3 Wave properties for TE and TM modes in a rectangular waveguide with dimensions a × b, filled with a dielectric material with constitutive parameters ϵ and μ. The TEM case, shown for reference, pertains to plane-wave propagation in an unbounded medium.
Rectangular Waveguides Plane Wave TE Modes TM Modes TEM Mode Ex = jωμ k2 c nπ b H0 cos mπx a sin nπy b e−jβz Ex = −jβ k2 c mπ a E0 cos mπx a sin nπy b e−jβz Ex = Ex0e−jβz Ey = −jωμ k2 c mπ a H0 sin mπx a cos nπy b e−jβz Ey = −jβ k2 c nπ b E0 sin mπx a cos nπy b e−jβz Ey = Ey0e−jβz Ez = 0 Ez = E0 sin mπx a sin nπy b e−jβz Ez = 0 Hx = − Ey/ZTE Hx = − Ey/ZTM Hx = − Ey/η Hy = Ex/ZTE Hy = Ex/ZTM Hy = Ex/η Hz = H0 cos mπx a cos nπy b e−jβz Hz = 0 Hz = 0 ZTE = η/ 1 −(fc/f )2 ZTM = η 1 −(fc/f )2 η = √μ/ϵ Properties Common to TE and TM Modes fc = up0 2 m a 2 + n b 2 fc = not applicable β = k 1 −(fc/f )2 k = ω√μϵ up = ω β = up0/ 1 −(fc/f )2 up0 = 1/√μϵ fmn (GHz) TE10 TE01 TE20 TE11 TM11 TM21 TM12 TM31 TE21 TE30 TE31 TE12 TE02 TM22 TM22 0 5 10 15 20 Figure 8-24 Cutoff frequencies for TE and TM modes in a hollow rectangular waveguide with a = 3 cm and b = 2 cm (Example 8-9).
amplitudes and frequencies. Exact equivalence may require a large, or infinite, number of frequency components, but in practice, it is often possible to represent the modulated waveform, to a fairly high degree of fidelity, with a wave group that extends over a relatively narrow bandwidth surrounding the high-frequency carrier f .
The velocity with which the envelope—or equivalently the wave group—travels through the medium is called the group velocity ug.
As such, ug is the velocity of the energy carried by the wave-group, and of the information encoded in it. Depending on whether or not the propagation medium is dispersive, ug may or may not be equal to the phase velocity up. In Section 2-1.1, we described a “dispersive transmission line as one on which the phase velocity is not a constant as a function of frequency,” a consequence of which is that the shape of a pulse transmitted through it gets progressively distorted as it moves down the line.
A rectangular waveguide constitutes a dispersive transmission line because the phase velocity of aTE orTM mode propagating through it is a strong function of frequency [per Eq. (8.108)], 390 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION (a) (b) Gaussian pulse High-frequency carrier Amplitude-modulated waveform Figure 8-25 The amplitude-modulated high-frequency waveform in (b) is the product of the Gaussian-shaped pulse with the sinusoidal high-frequency carrier in (a).
particularly at frequencies close to the cutoff frequency fmn.
As we see shortly, if f ≫fmn, the TE and TM modes become approximately TEM in character, not only in terms of the directional arrangement of the electric and magnetic fields, but also in terms of the frequency dependence of the phase velocity.
We now examine up and ug in more detail.
The phase velocity, defined as the velocity of the sinusoidal pattern of the wave, is given by up = ω β , (8.112) while the group velocity ug is given by ug = 1 dβ/dω .
(8.113) Even though we will not derive Eq. (8.113) in this book, it is nevertheless important that we understand its properties for TE and TM modes in a metal waveguide. Using the expression for β given by Eq. (8.107), f21 f11 f01 f10 TE21 and TM21 TE11 and TM11 TE01 TE10 TEM ω (rad/s) β (rad/m) Figure 8-26 ω-β diagram for TE and TM modes in a hollow rectangular waveguide. The straight line pertains to propagation in an unbounded medium or on a TEM transmission line.
ug = 1 dβ/dω = up0 1 −(fmn/f )2 , (8.114) where, as before, up0 is the phase velocity in an unbounded dielectric medium.
In view of Eq. (8.108) for the phase velocity up, upug = u2 p0.
(8.115) Above cutoff (f > fmn), up ≥up0, and ug ≤up0. As f →∞, ormorepreciselyas(fmn/f ) →0,TEandTMmodesapproach the TEM case, for which up = ug = up0.
A useful graphical tool for describing the propagation properties of a medium or transmission line is the ω-β diagram. In Fig. 8-26, the straight line starting at the origin represents the ω-β relationship for a TEM wave propagating in an unbounded medium (or on a TEM transmission line). The TEM line provides a reference to which the ω-β curves of the TE/TM modes can be compared. At a given location on the ω-β line or curve, the ratio of the value of ω to that of β defines up = ω/β, whereas it is the slope dω/dβ of the curve at that 8-10 PROPAGATION VELOCITIES 391 point that defines the group velocity ug. For the TEM line, the ratio and the slope have identical values (hence, up = ug), and the line starts at ω = 0. In contrast, the curve for each of the indicated TE/TM modes starts at a cutoff frequency specific to that mode, below which the waveguide cannot support the propagation of a wave in that mode.
At frequencies close to cutoff, up and ug assume very different values; in fact, at cutoff up = ∞and ug = 0. On the other end of the frequency spectrum, at frequencies much higher than fmn, the ω-β curves of the TE/TM modes approach the TEM line. We should note that for TE and TM modes, up may easily exceed the speed of light, but ug will not, and since it is ug that represents the actual transport of energy, Einstein’s assertion that there is an upper bound on the speed of physical phenomena, is not violated.
So far, we have described the fields in the guide, but we have yet to interpret them in terms of plane waves that zigzag along the guide through successive reflections. To do just that, consider the simple case of a TE10 mode.
For m = 1 and n = 0, the only nonzero component of the electric field given by Eq. (8.110) is Ey, Ey = −j ωμ k2 c π a H0 sin πx a e−jβz.
(8.116) Using the identity sin θ = (ejθ −e−jθ)/2j for any argument θ, we obtain Ey = ωμπH0 2k2 c a (e−jπx/a −ejπx/a)e−jβz = E′ 0(e−jβ(z+πx/βa) −e−jβ(z−πx/βa)) = E′ 0(e−jβz′ −e−jβz′′), (8.117) where we have consolidated the quantities multiplying the two exponential terms into the constant E′ 0. The first exponential term represents a wave with propagation constant β traveling in the z′ direction, where z′ = z + πx βa , (8.118a) and the second term represents a wave travelling in the z′′ direction, with z′′ = z −πx βa .
(8.118b) From the diagram shown in Fig. 8-27(a), it is evident that the z′ direction is at an angle θ′ relative to z and the z′′ direction is at an angle θ′′ = −θ′. This means that the electric field Ey (and its associated magnetic field H) of the TE10 mode is composed of two TEM waves, as shown in Fig. 8-27(b), both traveling in the (a) z' and z'' propagation directions From Eq. (8.118a), Hence, From Eq. (8.118b), Hence, x z .
.
.
.
a x H H E E z z' z'' (b) TEM waves Figure 8-27 The TE10 mode can be constructed as the sum of two TEM waves.
+z direction by zigzagging between the opposite walls of the waveguide. Along the zigzag directions (z′ and z′′), the phase velocity of the individual wave components is up0, but the phase velocity of the combination of the two waves along z is up.
Example 8-10: Zigzag Angle For the TE10 mode, express the zigzag angle θ′ in terms of the ratio (f/f10), and then evaluate it at f = f10 and for f ≫f10.
Solution: From Fig. 8-27, θ′ 10 = tan−1 π β10a , 392 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION where the subscript 10 has been added as a reminder that the expression applies to the TE10 mode specifically.
For m = 1 and n = 0, Eq. (8.106) reduces to f10 = up0/2a. After replacing β with the expression given by Eq. (8.107) and replacing a with up0/2f10, we obtain θ′ = tan−1 1 (f/f10)2 −1 .
At f = f10, θ′ = 90◦, whichmeansthatthewavebouncesback and forth at normal incidence between the two side walls of the waveguide, making no progress in the z direction. At the other end of the frequency spectrum, when f ≫f10, θ′ approaches 0 and the wave becomes TEM-like as it travels straight down the guide.
Concept Question 8-12: For TE waves, the dominant mode is TE10, but for TM the dominant mode is TM11.
Why is it not TM10?
Concept Question 8-13: Why is it acceptable for up to exceed the speed of light c, but not so for ug?
Exercise 8-13: What do the wave impedances for TE and TM look like as f approaches fmn?
Answer: At f = fmn, ZTE looks like an open circuit, and ZTM looks like a short circuit.
Exercise 8-14: What are the values for (a) up, (b) ug, and (c) the zigzag angle θ′ at f = 2f10 for a TE10 mode in a hollow waveguide?
Answer: (a) up = 1.15c, (b) ug = 0.87c, (c) θ′ = 30◦.
8-11 Cavity Resonators A rectangular waveguide has metal walls on four sides. When the two remaining sides are terminated with conducting walls, the waveguide becomes a cavity.
By designing cavities to resonate at specific frequencies, they can be used as circuit elements in microwave oscillators, amplifiers, and bandpass filters.
ThecavityshowninFig.8-28(a), withdimensions(a×b×d), is connected to two coaxial cables that feed and extract signals into and from the cavity via input and output probes. As a bandpass filter, the function of a resonant cavity is to block all spectral components of the input signal except for those with frequencies that fall within a narrow band surrounding a specific center frequency f0, the cavity’s resonant frequency.
Comparison of the spectrum in Fig. 8-28(b), which describes the range of frequencies that might be contained in a typical input signal, with the narrow output spectrum in Fig. 8-28(c) demonstrates the filtering action imparted by the cavity.
In a rectangular waveguide, the fields constitute standing waves along the x and y directions, and a propagating wave along ˆ z. The terms TE and TM were defined relative to the propagation direction; TE meant that E was entirely transverse to ˆ z, and TM meant that H had no component along ˆ z.
In a cavity, there is no unique propagation direction, as no fields propagate. Instead, standing waves exist along all three directions. Hence, the terms TE and TM need to be modified by defining the fields relative to one of the three rectangular axes. For the sake of consistency, we will continue to define the transverse direction to be any direction contained in the plane whose normal is ˆ z.
The TE mode in the rectangular waveguide consists of a single propagating wave whose Hz component is given by Eq. (8.110e) as Hz = H0 cos mπx a cos nπy b e−jβz, (8.119) where the phase factor e−jβz signifies propagation along +ˆ z.
Because the cavity has conducting walls at both z = 0 and z = d, it will contain two such waves, one with amplitude H0 traveling along +ˆ z, and another with amplitude H − 0 traveling along −ˆ z. Hence, Hz = (H0e−jβz + H − 0 ejβz) cos mπx a cos nπy b .
(8.120) Boundary conditions require the normal component of H to be zero at a conducting boundary. Consequently, Hz must be zero at z = 0 and z = d. To satisfy these conditions, it is necessary that H − 0 = −H0 and βd = pπ, with p = 1, 2, 3, . . ., in which case Eq. (8.120) becomes Hz = −2jH0 cos mπx a cos nπy b sin pπz d . (8.121) Giventhat Ez = 0 fortheTEmodes, alloftheothercomponents of E and H can be derived readily through the application of the relationships given by Eq. (8.89). A similar procedure can also be used to characterize cavity modes for the TM case.
8-11 CAVITY RESONATORS 393 Module 8.5 Rectangular Waveguide When givin the waveguide dimensions, the frequency f , and the mode type (TE or TM) and number, this module provides information about the wave impedance, cutoff frequency, and other wave attributes.
It also displays the electric and magnetic field distributions inside the guide.
8-11.1 Resonant Frequency The consequence of the quantization condition imposed on β, namely β = pπ/d with p assuming only integer values, is that for any specific set of integer values of (m, n, p), the wave inside the cavity can exist at only a single resonant frequency, fmnp, whose value has to satisfy Eq. (8.105). The resulting expression for fmnp is fmnp = up0 2 m a 2 + n b 2 + p d 2 .
(8.122) For TE, the indices m and n start at 0, but p starts at 1. The exact opposite applies to TM. By way of an example, the resonant frequency for a TE101 mode in a hollow cavity with dimensions a = 2 cm, b = 3 cm, and d = 4 cm is f101 = 8.38 GHz.
8-11.2 Quality Factor In the ideal case, if a group of frequencies is introduced into the cavity to excite a certain TE or TM mode, only the frequency component at exactly fmnp of that mode will survive, and all others will attenuate. If a probe is used to couple a sample of the resonant wave out of the cavity, the output signal will be a monochromatic sinusoidal wave at fmnp. In practice, the cavity exhibits a frequency response similar to that shown in Fig. 8-28(c), which is very narrow, but not a perfect spike.
The bandwidth f of the cavity is defined as the frequency range between the two frequencies (on either side of fmnp) at which the amplitude is 1/ √ 2 of the maximum amplitude (at fmnp). The normalized bandwidth, defined as f/fmnp, is approximately equal to the reciprocal of the quality factor Q of the cavity, Q ≈fmnp f .
(8.123) 394 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION (a) Resonant cavity (b) Input spectrum (c) Output spectrum f0 ∆f f f0 f 0 x b a z y d Hollow or dielectric-filled resonant cavity Output signal Input signal Figure 8-28 A resonant cavity supports a very narrow bandwidth around its resonant frequency f0.
▶The quality factor is defined in terms of the ratio of the energystoredinthecavityvolumetotheenergydissipated in the cavity walls through conduction. ◀ For an ideal cavity with perfectly conducting walls, no energy loss is incurred, as a result of which Q is infinite and f ≈0.
Metals have very high, but not infinite, conductivities, so a real cavity with metal walls stores most of the energy coupled into it in its volume, but it also loses some of it to heat conduction.
A typical value for Q is on the order of 10,000, which is much higher than can be realized with lumped RLC circuits.
Example 8-11: Q of a Resonant Cavity The quality factor for a hollow resonant cavity operating in the TE101 mode is Q = 1 δs abd(a2 + d2) [a3(d + 2b) + d3(a + 2b)] , (8.124) where δs = 1/πfmnpμ0σc is the skin depth and σc is the conductivity of the conducting walls. Design a cubic cavity with a TE101 resonant frequency of 12.6 GHz and evaluate its bandwidth. The cavity walls are made of copper.
Solution: For a = b = d, m = 1, n = 0, p = 1, and up0 = c = 3 × 108 m/s, Eq. (8.122) simplifies to f101 = 3 √ 2 × 108 2a (Hz), which, for f101 = 12.6 GHz, gives a = 1.68 cm.
At f101 = 12.6 GHz, the skin depth for copper (with σc = 5.8 × 107 S/m) is δs = 1 [πf101μ0σc]1/2 = 1 [π × 12.6 × 109 × 4π × 10−7 × 5.8 × 107]1/2 = 5.89 × 10−7 m.
Upon setting a = b = d in Eq. (8.124), the expression for Q of a cubic cavity becomes Q = a 3δs = 1.68 × 10−2 3 × 5.89 × 10−7 ≈9, 500.
Hence, the cavity bandwidth is f ≈f101 Q ≈12.6 × 109 9, 500 ≈1.3 MHz.
CHAPTER 8 SUMMARY 395 Chapter 8 Summary Concepts • The relations describing the reflection and transmission behavior of a plane EM wave at the boundary between two different media are the consequence of satisfying the conditions of continuity of the tangential components of E and H across the boundary.
• Snell’s laws state that θi = θr and sin θt = (n1/n2) sin θi.
For media such that n2 < n1, the incident wave is reflected totally by the boundary when θi ≥θc, where θc is the critical angle given by θc = sin−1(n2/n1).
• By successive multiple reflections, light can be guided through optical fibers. The maximum data rate of digital pulses that can be transmitted along optical fibers is dictated by modal dispersion.
• At the Brewster angle for a given polarization, the incident wave is transmitted totally across the boundary.
Fornonmagneticmaterials, theBrewsterangleexistsfor parallel polarization only.
• Any plane wave incident on a plane boundary can be synthesized as the sum of a perpendicularly polarized wave and a parallel polarized wave.
• Transmission-line equivalent models can be used to characterize wave propagation and reflection by and transmission through boundaries between different media.
• Waves can travel through a metal waveguide in the form of transverse electric (TE) and transverse magnetic (TM) modes. For each mode, the waveguide has a cutoff frequency below which a wave cannot propagate.
• A cavity resonator can support standing waves at specific resonant frequencies.
Important Terms Provide definitions or explain the meaning of the following terms: ω-β diagram acceptance angle θa angles of incidence, reflection, and transmission Brewster angle θB cladding critical angle θc cutoff frequency fmn cutoff wavenumber kc dominant mode evanescent wave fiber core grazing incidence group velocity ug index of refraction n modal dispersion modes optical fibers parallel polarization perpendicular polarization phase-matching condition plane of incidence polarizing angle quality factor Q reflection coefficient reflectivity (reflectance) R refraction angle resonant cavity resonant frequency Snell’s laws standing-wave ratio S surface wave total internal reflection transmission coefficient τ transmissivity (transmittance) T transverse electric (TE) polarization transverse magnetic (TM) polarization unbounded-medium wavenumber unpolarized wavefront 396 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION Mathematical and Physical Models Normal Incidence = Er 0 Ei 0 = η2 −η1 η2 + η1 τ = Et 0 Ei 0 = 2η2 η2 + η1 τ = 1 + = √ϵr1 −√ϵr2 √ϵr1 + √ϵr2 (if μ1 = μ2) Snell’s Laws θi = θr sin θt sin θi = up2 up1 = μ1ϵ1 μ2ϵ2 Oblique Incidence Perpendicular Polarization ⊥= Er ⊥0 Ei ⊥0 = η2 cos θi −η1 cos θt η2 cos θi + η1 cos θt τ⊥= Et ⊥0 Ei ⊥0 = 2η2 cos θi η2 cos θi + η1 cos θt Parallel Polarization ∥= Er ∥0 Ei ∥0 = η2 cos θt −η1 cos θi η2 cos θt + η1 cos θi τ∥= Et ∥0 Ei ∥0 = 2η2 cos θi η2 cos θt + η1 cos θi Brewster Angle θB∥= sin−1 1 1 + (ϵ1/ϵ2) = tan−1 ϵ2 ϵ1 Waveguides β = ω2μϵ − mπ a 2 − nπ b 2 fmn = up0 2 m a 2 + n b 2 up = ω β = up0 1 −(fmn/f )2 upug = u2 p0 ZTE = η 1 −(fmn/f )2 ZTM = η 1 − fmn f 2 Resonant Cavity fmnp = up0 2 m a 2 + n b 2 + p d 2 Q ≈fmnp f PROBLEMS 397 PROBLEMS Section 8-1: Wave Reflection and Transmission at Normal Incidence ∗8.1 A plane wave in air with an electric field amplitude of 20 V/m is incident normally upon the surface of a lossless, nonmagnetic medium with ϵr = 25. Determine the following: (a) The reflection and transmission coefficients.
(b) The standing-wave ratio in the air medium.
(c) The average power densities of the incident, reflected, and transmitted waves.
8.2 A plane wave traveling in medium 1 with ϵr1 = 2.25 is normally incident upon medium 2 with ϵr2 = 4. Both media are made of nonmagnetic, nonconducting materials. If the electric field of the incident wave is given by Ei = ˆ y8 cos(6π × 109t −30πx) (V/m).
(a) Obtain time-domain expressions for the electric and magnetic fields in each of the two media.
(b) Determine the average power densities of the incident, reflected, and transmitted waves.
8.3 A plane wave traveling in a medium with ϵr1 = 9 is normally incident upon a second medium with ϵr2 = 4. Both media are made of nonmagnetic, nonconducting materials. If the magnetic field of the incident plane wave is given by Hi = ˆ z 2 cos(2π × 109t −ky) (A/m).
(a) Obtain time-domain expressions for the electric and magnetic fields in each of the two media.
∗(b) Determine the average power densities of the incident, reflected, and transmitted waves.
8.4 A 200 MHz, left-hand circularly polarized plane wave with an electric field modulus of 5 V/m is normally incident in air upon a dielectric medium with ϵr = 4 and occupies the region defined by z ≥0.
(a) Write an expression for the electric field phasor of the incident wave, given that the field is a positive maximum at z = 0 and t = 0.
(b) Calculate the reflection and transmission coefficients.
∗Answer(s) available in Appendix D.
(c) Write expressions for the electric field phasors of the reflected wave, the transmitted wave, and the total field in the region z ≤0.
(d) Determine the percentages of the incident average power reflected by the boundary and transmitted into the second medium.
8.5 Repeat Problem 8.4, but replace the dielectric medium with a poor conductor characterized by ϵr = 2.25, μr = 1, and σ = 10−4 S/m.
8.6 A 50 MHz plane wave with electric field amplitude of 50 V/m is normally incident in air onto a semi-infinite, perfect dielectric medium with ϵr = 36. Determine the following: ∗(a) (b) The average power densities of the incident and reflected waves.
(c) The distance in the air medium from the boundary to the nearest minimum of the electric field intensity, |E|.
∗8.7 What is the maximum amplitude of the total electric field in the air medium of Problem 8.6, and at what nearest distance from the boundary does it occur?
8.8 Repeat Problem 8.6, but replace the dielectric medium with a conductor with ϵr = 1, μr = 1, and σ = 2.78 × 10−3 S/m.
∗8.9 The three regions shown in Fig. P8.9 contain perfect dielectrics. For a wave in medium 1, incident normally upon the boundary at z = −d, what combination of ϵr2 and d produces Medium 2 εr2 Medium 3 εr3 Medium 1 εr1 z = −d z = 0 z d Figure P8.9 Dielectric layers for Problems 8.9 to 8.11.
398 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION no reflection? Express your answers in terms of ϵr1, ϵr3, and the oscillation frequency of the wave, f .
8.10 For the configuration shown in Fig.
P8.9, use transmission-line equations (or the Smith chart) to calculate the input impedance at z = −d for ϵr1 = 1, ϵr2 = 9, ϵr3 = 4, d = 1.2 m, and f = 50 MHz. Also determine the fraction of the incident average power density reflected by the structure.
Assume all media are lossless and nonmagnetic.
∗8.11 Repeat Problem 8.10, but interchange ϵr1 and ϵr3.
8.12 Orange light of wavelength 0.61 μm in air enters a block of glass with ϵr = 1.44. What color would it appear to a sensor embedded in the glass? The wavelength ranges of colors are violet (0.39 to 0.45 μm), blue (0.45 to 0.49 μm), green (0.49 to 0.58 μm), yellow (0.58 to 0.60 μm), orange (0.60 to 0.62 μm), and red (0.62 to 0.78 μm).
∗8.13 A plane wave of unknown frequency is normally incident in air upon the surface of a perfect conductor. Using an electric-fieldmeter, itwasdeterminedthatthetotalelectricfieldintheair medium is always zero when measured at a distance of 2 m from the conductor surface. Moreover, no such nulls were observed at distances closer to the conductor. What is the frequency of the incident wave?
8.14 Consider a thin film of soap in air under illumination by yellow light with λ = 0.6 μm in vacuum. If the film is treated as a planar dielectric slab with ϵr = 1.72, surrounded on both sides byair, whatfilmthicknesswouldproducestrongreflection of the yellow light at normal incidence?
∗8.15 A 5 MHz plane wave with electric field amplitude of 10 (V/m) is normally incident in air onto the plane surface of a semi-infinite conducting material with ϵr = 4, μr = 1, and σ = 100 (S/m). Determine the average power dissipated (lost) per unit cross-sectional area in a 2 mm penetration of the conducting medium.
8.16 A 0.5 MHz antenna carried by an airplane flying over the ocean surface generates a wave that approaches the water surface in the form of a normally incident plane wave with an electric-field amplitude of 3,000 (V/m).
Seawater is characterized by ϵr = 72, μr = 1, and σ = 4 (S/m). The plane is trying to communicate a message to a submarine submerged at a depth d below the water surface. If the submarine’s receiver requires a minimum signal amplitude of 0.01 (μV/m), what is the maximum depth d to which successful communication is still possible?
Sections 8-2 and 8-3: Snell’s Laws and Fiber Optics ∗8.17 A light ray is incident on a prism in air at an angle θ as shown in Fig. P8.17. The ray is refracted at the first surface and again at the second surface. In terms of the apex angle φ of the prism and its index of refraction n, determine the smallest value of θ for which the ray will emerge from the other side.
Find this minimum θ for n = 1.4 and φ = 60◦.
θ n φ Surface 1 Surface 2 Figure P8.17 Prism of Problem 8.17.
8.18 For some types of glass, the index of refraction varies with wavelength. A prism made of a material with n = 1.71 −4 30 λ0 (λ0 in μm), where λ0 is the wavelength in vacuum, was used to disperse white light as shown in Fig. P8.18. The white light is incident at an angle of 50◦, the wavelength λ0 of red light is 0.7 μm, and that of violet light is 0.4 μm. Determine the angular dispersion in degrees.
60° 50° Red Green Violet Angular dispersion Figure P8.18 Prism of Problem 8.18.
PROBLEMS 399 ∗8.19 The two prisms in Fig. P8.19 are made of glass with n = 1.5. What fraction of the power density carried by the ray incident upon the top prism emerges from the bottom prism?
Neglect multiple internal reflections.
45° 45° 45° 45° 90° 90° Figure P8.19 Periscope prisms of Problem 8.19.
8.20 A parallel-polarized plane wave is incident from air at an angle θi = 30◦onto a pair of dielectric layers as shown in Fig. P8.20.
(a) Determine the angles of transmission θ2, θ3, and θ4.
(b) Determine the lateral distance d.
d 5 cm 5 cm θi θ2 θ3 θ4 Air Air μr = 1 εr = 6.25 μr = 1 εr = 2.25 Figure P8.20 Problem P8.20.
8.21 A light ray incident at 45◦passes through two dielectric materials with the indices of refraction and thicknesses given in Fig. P8.21. If the ray strikes the surface of the first dielectric at a height of 2 cm, at what height will it strike the screen?
45° 2 cm Screen n = 1 n = 1 3 cm 4 cm 5 cm n = 1.3 n = 1.5 Figure P8.21 Light incident on a screen through a multilayered dielectric (Problem 8.21).
∗8.22 FigureP8.22depictsabeakercontainingablockofglass onthebottomandwateroverit. Theglassblockcontainsasmall 60° 10 cm Water n = 1.33 Glass n = 1.6 Air bubble Apparent position of air bubble 6.81 cm Figure P8.22 Apparent position of the air bubble in Problem 8.22.
400 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION air bubble at an unknown depth below the water surface. When viewed from above at an angle of 60◦, the air bubble appears at a depth of 6.81 cm. What is the true depth of the air bubble?
8.23 A glass semicylinder with n = 1.5 is positioned such that its flat face is horizontal, as shown in Fig. P8.23, and its horizontal surface supports a drop of oil, as also shown. When light is directed radially toward the oil, total internal reflection occurs if θ exceeds 53◦. What is the index of refraction of the oil?
θ nglass noil Oil drop Figure P8.23 Oil drop on the flat surface of a glass semicylinder (Problem 8.23).
∗8.24 A penny lies at the bottom of a water fountain at a depth of 30 cm. Determine the diameter of a piece of paper which, if placed to float on the surface of the water directly above the penny, would totally obscure the penny from view. Treat the penny as a point and assume that n = 1.33 for water.
8.25 Suppose that the optical fiber of Example 8-5 is submerged in water (with n = 1.33) instead of air. Determine θa and fp in that case.
∗8.26 Equation (8.45) was derived for the case where the light incident upon the sending end of the optical fiber extends over the entire acceptance cone shown in Fig. 8-12(b). Suppose the incident light is constrained to a narrower range extending between normal incidence and θ′, where θ′ < θa.
(a) Obtain an expression for the maximum data rate fp in terms of θ′.
(b) Evaluate fp for the fiber of Example 8-5 when θ′ = 5◦.
Sections 8-4 and 8-5: Reflection and Transmission at Oblique Incidence 8.27 A plane wave in air with Ei = ˆ y 20e−j(3x+4z) (V/m) is incident upon the planar surface of a dielectric material, with ϵr = 4, occupying the half-space z ≥0. Determine: (a) The polarization of the incident wave.
∗(b) The angle of incidence.
(c) The time-domain expressions for the reflected electric and magnetic fields.
(d) The time-domain expressions for the transmitted electric and magnetic fields.
(e) The average power density carried by the wave in the dielectric medium.
8.28 Repeat Problem 8.27 for a wave in air with Hi = ˆ y 2 × 10−2e−j(8x+6z) (A/m) incident upon the planar boundary of a dielectric medium (z ≥0) with ϵr = 9.
8.29 A plane wave in air with E i = (ˆ x 9 −ˆ y 4 −ˆ z 6)e−j(2x+3z) (V/m) is incident upon the planar surface of a dielectric material, with ϵr = 2.25, occupying the half-space z ≥0. Determine ∗(a) The incidence angle θi.
(b) The frequency of the wave.
(c) The field E r of the reflected wave.
(d) The field E t of the wave transmitted into the dielectric medium.
(e) The average power density carried by the wave into the dielectric medium.
PROBLEMS 401 8.30 Natural light is randomly polarized, which means that, on average, half the light energy is polarized along any given direction (in the plane orthogonal to the direction of propagation) and the other half of the energy is polarized along the direction orthogonal to the first polarization direction.
Hence, when treating natural light incident upon a planar boundary, we can consider half of its energy to be in the form of parallel-polarized waves and the other half as perpendicularly polarized waves. Determine the fraction of the incident power reflected by the planar surface of a piece of glass with n = 1.5 when illuminated by natural light at 70◦.
∗8.31 A parallel-polarized plane wave is incident from air onto a dielectric medium with ϵr = 9 at the Brewster angle. What is the refraction angle?
8.32 A perpendicularly polarized wave in air is obliquely incident upon a planar glass–air interface at an incidence angle of 30◦. The wave frequency is 600 THz (1 THz = 1012 Hz), which corresponds to green light, and the index of refraction of the glass is 1.6. If the electric field amplitude of the incident wave is 50 V/m, determine the following: (a) The reflection and transmission coefficients.
(b) The instantaneous expressions for E and H in the glass medium.
8.33 Show that the reflection coefficient ⊥can be written in the following form: ⊥= sin(θt −θi) sin(θt + θi) .
8.34 Show that for nonmagnetic media, the reflection coefficient ∥can be written in the following form: ∥= tan(θt −θi) tan(θt + θi) .
∗8.35 A parallel-polarized beam of light with an electric field amplitude of 10 (V/m) is incident in air on polystyrene with μr = 1 and ϵr = 2.6.
If the incidence angle at the air– polystyrene planar boundary is 50◦, determine the following: (a) The reflectivity and transmissivity.
(b) The power carried by the incident, reflected, and transmitted beams if the spot on the boundary illuminated by the incident beam is 1 m2 in area.
8.36 A 50 MHz right-hand circularly polarized plane wave with an electric field modulus of 30 V/m is normally incident in air upon a dielectric medium with ϵr = 9 and occupying the region defined by z ≥0.
(a) Write an expression for the electric field phasor of the incident wave, given that the field is a positive maximum at z = 0 and t = 0.
(b) Calculate the reflection and transmission coefficients.
(c) Write expressions for the electric field phasors of the reflected wave, the transmitted wave, and the total field in the region z ≤0.
(d) Determine the percentages of the incident average power reflected by the boundary and transmitted into the second medium.
8.37 Consider a flat 5 mm thick slab of glass with ϵr = 2.56.
∗(a) If a beam of green light (λ0 = 0.52 μm) is normally incident upon one of the sides of the slab, what percentage of the incident power is reflected back by the glass?
(b) To eliminate reflections, it is desired to add a thin layer of antireflection coating material on each side of the glass. If you are at liberty to specify the thickness of the antireflection material as well as its relative permittivity, what would these specifications be?
Sections 8-6 to 8-11: Waveguides and Resonators 8.38 Derive Eq. (8.89b).
∗8.39 A hollow rectangular waveguide is to be used to transmit signals at a carrier frequency of 6 GHz. Choose its dimensions so that the cutoff frequency of the dominant TE mode is lower than the carrier by 25% and that of the next mode is at least 25% higher than the carrier.
8.40 A TE wave propagating in a dielectric-filled waveguide of unknown permittivity has dimensions a = 5 cm and b = 3 cm.
If the x component of its electric field is given by Ex = −36 cos(40πx) sin(100πy) · sin(2.4π × 1010t −52.9πz), (V/m) determine: (a) the mode number, (b) ϵr of the material in the guide, (c) the cutoff frequency, and (d) the expression for Hy.
402 CHAPTER 8 WAVE REFLECTION AND TRANSMISSION ∗8.41 A waveguide filled with a material whose ϵr = 2.25 has dimensions a = 2 cm and b = 1.4 cm. If the guide is to transmit 10.5 GHz signals, what possible modes can be used for the transmission?
8.42 For a rectangular waveguide operating in theTE10 mode, obtain expressions for the surface charge density ρs and surface current density Js on each of the four walls of the guide.
∗8.43 A waveguide, with dimensions a = 1 cm and b = 0.7 cm, is to be used at 20 GHz.
Determine the wave impedance for the dominant mode when (a) the guide is empty, and (b) the guide is filled with polyethylene (whose ϵr = 2.25).
8.44 A narrow rectangular pulse superimposed on a carrier with a frequency of 9.5 GHz was used to excite all possible modes in a hollow guide with a = 3 cm and b = 2.0 cm. If the guide is 100 m in length, how long will it take each of the excited modes to arrive at the receiving end?
∗8.45 If the zigzag angle θ′ is 25◦for the TE10 mode, what would it be for the TE20 mode?
8.46 Measurement of the TE101 frequency response of an air-filled cubic cavity revealed that its Q is 4802. If its volume is 64 mm3, what material are its sides made of? (Hint: See Appendix B.) 8.47 A hollow cavity made of aluminum has dimensions a = 4 cm and d = 3 cm. Calculate Q of the TE101 mode for ∗(a) b = 2 cm, and (b) b = 3 cm.
C H A P T E R 9 Radiation and Antennas Chapter Contents Overview, 404 9-1 The Hertzian Dipole, 406 9-2 Antenna Radiation Characteristics, 411 9-3 Half-Wave Dipole Antenna, 417 9-4 Dipole of Arbitrary Length, 420 9-5 Effective Area of a Receiving Antenna, 422 TB17 Health Risks of EM Fields, 424 9-6 Friis Transmission Formula, 427 9-7 Radiation by Large-Aperture Antennas, 429 9-8 Rectangular Aperture with Uniform Aperture Distribution, 432 9-9 Antenna Arrays, 435 9-10 N-Element Array with Uniform Phase Distribution, 442 9-11 Electronic Scanning of Arrays, 444 Chapter 9 Summary, 450 Problems, 452 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Calculatetheelectricandmagneticfieldsofwavesradiated by a dipole antenna.
2. Characterize the radiation of an antenna in terms of its radiation pattern, directivity, beamwidth, and radiation resistance.
3. Apply the Friis transmission formula to a free-space communication system.
4. Calculatetheelectricandmagneticfieldsofwavesradiated by aperture antennas.
5. Calculate the radiation pattern of multi-element antenna arrays.
404 CHAPTER 9 RADIATION AND ANTENNAS Overview An antenna is a transducer that converts a guided wave propagatingonatransmissionlineintoanelectromagneticwave propagating in an unbounded medium (usually free space), or vice versa. Figure 9-1 shows how a wave is launched by a hornlike antenna, with the horn acting as the transition segment between the waveguide and free space.
Antennas are made in various shapes and sizes (Fig. 9-2) and are used in radio and television broadcasting and reception, radio-wave communication systems, cellular telephones, radar systems, and anticollision automobile sensors, among many other applications.
The radiation and impedance properties of an antenna are governed by its shape, size, and material properties. The dimensions of an antenna are usually measured in units of λ of the wave it is launching or receiving; a 1 m long Electric field lines of radiated wave Wave launched into free space Antenna Transition region Transmission line Guided EM wave Generator Incident wave Antenna Transition region Transmission line Guided EM wave Detector or receiver Rec (a) Transmission mode (b) Reception mode Figure 9-1 Antenna as a transducer between a guided electromagnetic wave and a free-space wave, for both transmission and reception.
dipole antenna operating at a wavelength λ = 2 m exhibits the same properties as a 1 cm long dipole operating at λ = 2 cm.
Hence, in most of our discussions in this chapter, we refer to antenna dimensions in wavelength units.
Reciprocity The directional function characterizing the relative distribution of power radiated by an antenna is known as the antenna radiation pattern, or simply the antenna pattern. An isotropic antenna is a hypothetical antenna that radiates equally in all directions, and it is often used as a reference radiator when describing the radiation properties of real antennas.
▶Most antennas are reciprocal devices, exhibiting the same radiation pattern for transmission as for reception. ◀ Reciprocity means that, if in the transmission mode a given antennatransmitsindirectionA100timesthepowerittransmits in direction B, then when used in the reception mode it is 100 times more sensitive to electromagnetic radiation incident from direction A than from B.
All the antennas shown in Fig. 9-2 obey the reciprocity law, but not all antennas are reciprocal devices. Reciprocity may not hold for some solid-state antennas composed of nonlinear semiconductors or ferrite materials. Such nonreciprocal antennas are beyond the scope of this chapter, and hence reciprocity is assumed throughout. The reciprocity property is very convenient because it allows us to compute the radiation pattern of an antenna in the transmission mode, evenwhentheantennaisintendedtooperateasareceiver.
To fully characterize an antenna, one needs to study its radiation properties and impedance. The radiation properties include its directional radiation pattern and the associated polarization state of the radiated wave when the antenna is used in the transmission mode, also called the antenna polarization.
▶Being a reciprocal device, an antenna, when operating in the receiving mode, can extract from an incident wave only that component of the wave whose electric field matches the antenna polarization state. ◀ The second aspect, the antenna impedance, pertains to the transfer of power from a generator to the antenna when the antenna is used as a transmitter and, conversely, the transfer of power from the antenna to a load when the antenna is used 405 Phase shifters Feed point Radiating strip Coaxial feed Dielectric substrate Ground metal plane (a) Thin dipole (b) Biconical dipole (c) Loop Circular plate reflector (d) Helix (e) Log-periodic (f) Parabolic dish reflector (h) Microstrip (i) Antenna array (g) Horn Figure 9-2 Various types of antennas.
as a receiver, as will be discussed later in Section 9-5.
It should be noted that throughout our discussions in this chapter it will be assumed that the antenna is properly matched to the transmission line connected to its terminals, thereby avoiding reflections and their associated problems.
Radiation sources Radiation sources fall into two categories: currents and aperture fields. The dipole and loop antennas [Fig. 9-2(a) and (c)] are examples of current sources; the time-varying currents flowing in the conducting wires give rise to the radiated electromagnetic fields.
A horn antenna [Fig. 9-2(g)] is an example of the second group because the electric and magnetic fields across the horn’s aperture serve as the sources of the radiated fields. The aperture fields are themselves induced by time-varying currents on the surfaces of the horn’s walls, and therefore ultimately all radiation is due to time-varying currents. The choice of currents or apertures as the sources is merely a computational convenience arising from the structure of the antenna. We will examine the radiation processes associated with both types of sources.
Far-field region The wave radiated by a point source is spherical in nature, with the wavefront expanding outward at a rate equal to the phase velocity up (or the velocity of light c if the medium is free space). If R, the distance between the transmitting antenna and the receiving antenna, is sufficiently large such that the wavefront across the receiving aperture may be considered planar (Fig. 9-3), then the receiving aperture is said to be in the far-field (or far-zone) region of the transmitting point source.
This region is of particular significance because for most applications, the location of the observation point is indeed in the far-field region of the antenna.
The far-field plane-wave approximation allows the use of certain mathematical approximations that simplify the computation of the radiated field and, conversely, provide convenient techniques for synthesizing the appropriate antenna structure that would give rise to the desired far-field antenna pattern.
Antenna arrays When multiple antennas operate together, the combination is called an antenna array [Fig. 9-2(i)], and the array as a whole 406 CHAPTER 9 RADIATION AND ANTENNAS Source Transmitting antenna Spherical wave Receiving antenna Plane-wave approximation R Figure 9-3 Far-field plane-wave approximation.
behaves as if it were a single antenna.
By controlling the magnitude and phase of the signal feeding each antenna, it is possible to shape the radiation pattern of the array and to electronically steer the direction of the beam electronically.
These topics are treated in Sections 9-9 to 9-11.
9-1 The Hertzian Dipole By regarding a linear antenna as consisting of a large number of infinitesimally short conducting elements, each of which is so short that current may be considered uniform over its length, the field of the entire antenna may be obtained by integrating the fields from all these differential antennas, with the proper magnitudes and phases taken into account.
We shall first examine the radiation properties of such a differential antenna, known as a Hertzian dipole, and then in Section 9-3 we will extend the results to compute the fields radiated by a half-wave dipole, which is commonly used as a standard antenna for many applications.
▶A Hertzian dipole is a thin, linear conductor whose length l is very short compared with the wavelength λ; l should not exceed λ/50. ◀ The wire, oriented along the z direction in Fig. 9-4, carries a sinusoidally varying current given by i(t) = I0 cos ωt = Re[I0ejωt] (A), (9.1) where I0 is the current amplitude. From Eq. (9.1), the phasor current ˜ I = I0. Even though the current has to go to zero at the i(t) i(t) l x z R R' y θ φ Q = (R, θ, φ) Figure 9-4 Short dipole placed at the origin of a spherical coordinate system.
two ends of the dipole, we shall treat it as constant across its entire length.
The customary approach for finding the electric and magnetic fields at a point Q in space (Fig. 9-4) due to radiation by a current source is through the retarded vector potential A.
From Eq. (6.84), the phasor retarded vector potential A(R) at a distance vector R from a volumev ′ containing a phasor current distribution J is given by A(R) = μ0 4π v ′ Je−jkR′ R′ dv ′, (9.2) where μ0 is the magnetic permeability of free space (because the observation point is in air) and k = ω/c = 2π/λ is the wavenumber.
For the dipole, the current density is simply J = ˆ z(I0/s), where s is the cross-sectional area of the dipole wire. Also, dv ′ = s dz and the limits of integration are from z = −l/2 to z = l/2. In Fig. 9-4, the distance R′ between the observation point and a given point along the dipole is not the same as the distance to its center, R, but because we are dealing with a very short dipole, we can set R′ ≈R. Hence, A = μ0 4π e−jkR R l/2 −l/2 ˆ zI0 dz = ˆ z μ0 4π I0l e−jkR R , (9.3) 9-1 THE HERTZIAN DIPOLE 407 θ = 90° θ = 0° θ = 180° θ = 90° φ = 0° φ = 270° θ = 90° φ = 90° x y R z S θ Direction (θ, φ) Radiation source φ Figure 9-5 Spherical coordinate system.
▶The function (e−jkR/R) is called the spherical propagation factor. It accounts for the 1/R decay of the magnitude with distance as well as the phase change represented by e−jkR. ◀ The direction of A is the same as that of the current (z direction).
Because our objective is to characterize the directional character of the radiated power at a fixed distance R from the antenna, antenna pattern plots are presented in a spherical coordinate system (Fig. 9-5). Its variables, R, θ, and φ, are called the range, zenith angle, and azimuth angle, respectively.
To that end, we need to write A in terms of its spherical coordinate components, which is realized (with the help of Eq. (3.65c)) by expressing ˆ z in terms of spherical coordinates: ˆ z = ˆ R cos θ −ˆ θ θ θ sin θ.
(9.4) Upon substituting Eq. (9.4) into Eq. (9.3), we obtain A = ( ˆ R cos θ −ˆ θ θ θ sin θ) μ0I0l 4π e−jkR R = ˆ R AR + ˆ θ θ θ Aθ + ˆ φ φ φ Aφ, (9.5) with AR = μ0I0l 4π cos θ e−jkR R , (9.6a) Aθ = −μ0I0l 4π sin θ e−jkR R , (9.6b) Aφ = 0.
With the spherical components of A known, the next step is straightforward; we simply apply the free-space relationships given by Eqs. (6.85) and (6.86), H = 1 μ0 ∇× × × A, (9.7a) E = 1 jωϵ0 ∇× × × H, (9.7b) to obtain the expressions Hφ = I0lk2 4π e−jkR j kR + 1 (kR)2 sin θ, (9.8a) ER = 2I0lk2 4π η0e−jkR 1 (kR)2 − j (kR)3 cos θ, (9.8b) Eθ = I0lk2 4π η0e−jkR j kR + 1 (kR)2 − j (kR)3 sin θ, (9.8c) where η0 = √μ0/ϵ0 ≃120π ( ) is the intrinsic impedance of free space. The remaining components ( HR, Hθ, and Eφ) are everywhere zero. Figure 9-6 depicts the electric field lines of the wave radiated by the short dipole.
408 CHAPTER 9 RADIATION AND ANTENNAS Dipole axis Broadside direction 2λ 3λ 4λ λ Figure 9-6 Electric field lines surrounding an oscillating dipole at a given instant.
9-1.1 Far-Field Approximation As was stated earlier, in most antenna applications we are primarily interested in the radiation pattern of the antenna at great distances from the source. For the electric dipole, this corresponds to distances R such that R ≫λ or, equivalently, kR = 2πR/λ ≫1. This condition allows us to neglect the terms varying as 1/(kR)2 and 1/(kR)3 in Eqs. (9.8a) to (9.8c) in favor of the terms varying as 1/kR, which yields the far-field expressions Eθ = jI0lkη0 4π e−jkR R sin θ (V/m), (9.9a) Hφ = Eθ η0 (A/m), (9.9b) and ER is negligible. At the observation point Q (Fig. 9-4), the wave now appears similar to a uniform plane wave with its electric and magnetic fields in phase, related by the intrinsic impedance of the medium η0, and their directions orthogonal to each other and to the direction of propagation ( ˆ R). Both 9-1 THE HERTZIAN DIPOLE 409 fields are proportional to sin θ and independent of φ (which is expected from symmetry considerations).
9-1.2 Power Density Given E and H, the time-average Poynting vector of the radiated wave, which is also called the power density, can be obtained by applying Eq. (7.100); that is, Sav = 1 2Re E × × × H∗ (W/m2).
(9.10) For the short dipole, use of Eqs. (9.9a) and (9.9b) yields Sav = ˆ R S(R, θ), (9.11) with S(R, θ) = η0k2I 2 0 l2 32π2R2 sin2 θ = S0 sin2 θ (W/m2).
(9.12) The directional pattern of any antenna is described in terms of the normalized radiation intensity F(θ, φ), defined as the ratio of the power density S(R, θ, φ) at a specified range R to Smax, the maximum value of S(R, θ, φ) at the same range, F(θ, φ) = S(R, θ, φ) Smax (dimensionless).
(9.13) For the Hertzian dipole, the sin2 θ dependence in Eq. (9.12) indicates that the radiation is maximum in the broadside direction (θ = π/2), corresponding to the azimuth plane, and is given by Smax = S0 = η0k2I 2 0 l2 32π2R2 = 15πI 2 0 R2 l λ 2 (W/m2), (9.14) where use was made of the relations k = 2π/λ and η0 ≈120π.
We observe that Smax is directly proportional to I 2 0 and l2 (with l measured in wavelengths), and that it decreases with distance as 1/R2.
(a) Elevation pattern (b) Azimuth pattern Dipole 0 φ x y 1 1 F(φ) Dipole 1 z 1 0 0.5 β = 90° θ = 90° (broadside) θ1 = 45° θ2 = 135° θ F(θ) Figure 9-7 Radiation patterns of a short dipole.
From the definition of the normalized radiation intensity given by Eq. (9.13), it follows that F(θ, φ) = F(θ) = sin2 θ.
(9.15) Plots of F(θ) are shown in Fig. 9-7 in both the elevation plane (the θ plane) and the azimuth plane (φ plane).
▶No energy is radiated by the short dipole along the direction of the dipole axis, and maximum radiation (F = 1) occurs in the broadside direction (θ = 90◦).
Since F(θ) is independent of φ, the pattern is doughnut-shaped in θ–φ space. ◀ 410 CHAPTER 9 RADIATION AND ANTENNAS Module 9.1 Hertzian Dipole (l ≪λ) For a short dipole oriented along the z axis, this module displays the field distributions for E and H in both the horizontal and vertical planes. It can also animate the radiation process and current flow through the dipole.
Concept Question 9-1: What does it mean to say that most antennas are reciprocal devices?
Concept Question 9-2: What is the radiated wave like in the far-field region of the antenna?
Concept Question 9-3: In a Hertzian dipole, what is the underlying assumption about the current flowing through the wire?
Concept Question 9-4: Outline the basic steps used to relate the current in a wire to the radiated power density.
Exercise 9-1: A 1 m long dipole is excited by a 5 MHz current with an amplitude of 5 A. At a distance of 2 km, what is the power density radiated by the antenna along its broadside direction?
Answer: S0 = 8.2 × 10−8 W/m2. (See EM.) 9-2 ANTENNA RADIATION CHARACTERISTICS 411 θ φ R y x z R sin θ dφ dA = R2 sin θ dθ dφ = R2 dΩ R dθ R dφ Azimuth plane Elevation plane Figure 9-8 Definition of solid angle d = sin θ dθ dφ.
9-2 Antenna Radiation Characteristics Anantennapatterndescribesthefar-fielddirectionalproperties of an antenna when measured at a fixed distance from the antenna. In general, the antenna pattern is a three-dimensional plot that displays the strength of the radiated field or power density as a function of direction, with direction being specified by the zenith angle θ and the azimuth angle φ.
▶By virtue of reciprocity, a receiving antenna has the same directional antenna pattern as the pattern that it exhibits when operated in the transmission mode. ◀ Consider a transmitting antenna placed at the origin of the observation sphere shown in Fig. 9-8. The differential power radiated by the antenna through an elemental area dA is dPrad = Sav · dA = Sav · ˆ R dA = S dA (W), (9.16) where S is the radial component of the time-average Poynting vector Sav. In the far-field region of any antenna, Sav is always in the radial direction. In a spherical coordinate system, dA = R2 sin θ dθ dφ, (9.17) and the solid angle d associated with dA, defined as the subtended area divided by R2, is given by d = dA R2 = sin θ dθ dφ (sr).
(9.18) Note that, whereas a planar angle is measured in radians and the angular measure of a complete circle is 2π (rad), a solid angle is measured in steradians (sr), and the angular measure for a spherical surface is = (4πR2)/R2 = 4π (sr). The solid angle of a hemisphere is 2π (sr).
Using the relation dA = R2 d , dPrad can be rewritten as dPrad = R2 S(R, θ, φ) d .
(9.19) The total power radiated by an antenna through a spherical surfaceatafixeddistanceR isobtained byintegratingEq.(9.19) over that surface: Prad = R2 2π φ=0 π θ=0 S(R, θ, φ) sin θ dθ dφ = R2Smax 2π φ=0 π θ=0 F(θ, φ) sin θ dθ dφ = R2Smax 4π F(θ, φ) d (W), (9.20) where F(θ, φ) is the normalized radiation intensity defined by Eq. (9.13). The 4π symbol under the integral sign is used as an abbreviation for the indicated limits on θ and φ. Formally, Prad is called the total radiated power.
9-2.1 Antenna Pattern Each specific combination of the zenith angle θ and the azimuth angle φ denotes a specific direction in the spherical coordinate system of Fig. 9-8. The normalized radiation intensity F(θ, φ) 412 CHAPTER 9 RADIATION AND ANTENNAS −2 −1 0 1 2 92 91 90 89 88 −30 −25 −20 −15 −10 −5 0 Azimuth angle φ (degrees) Zenith angle θ (degrees) Normalized radiation intensity (dB) Figure 9-9 Three-dimensional pattern of a narrow-beam antenna.
characterizes the directional pattern of the energy radiated by an antenna, and a plot of F(θ, φ) as a function of both θ and φ constitutes a three-dimensional pattern, an example of which is shown in Fig. 9-9.
Often, it is of interest to characterize the variation of F(θ, φ) in the form of two-dimensional plots in specific planes in the spherical coordinate system. The two planes most commonly specified for this purpose are the elevation and azimuth planes.
The elevation plane, also called the θ plane, is a plane corresponding to a constant value of φ. For example, φ = 0 defines the x–z plane and φ = 90◦defines the y–z plane, both of which are elevation planes (Fig. 9-8). A plot of F(θ, φ) versus θ in either of these planes constitutes a two-dimensional pattern in the elevation plane. This is not to imply, however, that the elevation-plane pattern is necessarily the same in all elevation planes.
The azimuth plane, also called the φ plane, is specified by θ = 90◦and corresponds to the x–y plane. The elevation and azimuth planes are often called the two principal planes of the spherical coordinate system.
Some antennas exhibit highly directive patterns with narrow beams, in which case it is often convenient to plot the antenna pattern on a decibel scale by expressing F in decibels: F (dB) = 10 log F.
As an example, the antenna pattern shown in Fig. 9-10(a) is plotted on a decibel scale in polar coordinates, with intensity as the radial variable. This format permits a convenient visual interpretation of the directional distribution of the radiation lobes.
Another format commonly used for inspecting the pattern of a narrow-beam antenna is the rectangular display shown in Fig. 9-10(b), which permits the pattern to be easily expanded by changing the scale of the horizontal axis. These plots represent the variation in only one plane in the observation sphere, the φ = 0 plane. Unless the pattern is symmetrical in φ, additional patterns are required to define the overall variation of F(θ, φ) with θ and φ.
Strictly speaking, the polar angle θ is always positive, being defined over the range from 0◦(z direction) to 180◦(−z direction), and yet the θ axis in Fig. 9-10(b) is shown to have both positive and negative values. This is not a contradiction, but rather a different form of plotting antenna patterns. The right-hand half of the plot represents the variation of F (dB) with θ as θ is increased in a clockwise direction in the x–z plane [see inset in Fig. 9-10(b)], corresponding to φ = 0, whereas the left-hand half of the plot represents the variation of F (dB) with θ as θ is increased in a counterclockwise direction at φ = 180◦. Thus, a negative θ value simply denotes that the direction (θ, φ) is in the left-hand half of the x–z plane.
The pattern shown in Fig. 9-10(a) indicates that the antenna is fairly directive, since most of the energy is radiated through a narrow sector called the main lobe.
In addition to the main lobe, the pattern exhibits several side lobes and back lobes as well.
For most applications, these extra lobes are considered undesirable because they represent wasted energy for transmitting antennas and potential interference directions for receiving antennas.
9-2.2 Beam Dimensions For an antenna with a single main lobe, the pattern solid angle p describes the equivalent width of the main lobe of the 9-2 ANTENNA RADIATION CHARACTERISTICS 413 (a) Polar diagram (b) Rectangular plot 0 10 −10 −20 −30 10 20 20 30 40 50 60 70 80 90 100 110 120 130 30 40 50 60 70 80 90 100 110 120 130 140 140 180 170 170 160 160 150 150 −40 Normalized radiation intensity, dB Main lobe First side lobe Minor lobes Back lobes Zenith angle θ (degrees) 0 −5 −3 −10 −15 −20 −25 −30 −35 −50 −40 −30 −20 −10 0 10 20 30 40 50 θ1 θ2 β1/2 βnull Zenith angle θ (degrees) Normalized radiation intensity F(θ), dB φ = 180° θ = 180° θ = 90° θ = 0° θ φ = 0° z x Figure 9-10 Representative plots of the normalized radiation pattern of a microwave antenna in (a) polar form and (b) rectangular form.
antenna pattern (Fig. 9-11). It is defined as the integral of the normalized radiation intensity F(θ, φ) over a sphere: p = 4π F(θ, φ) d (sr).
(9.21) ▶For an isotropic antenna with F(θ, φ) = 1 in all directions, p = 4π (sr). ◀ The pattern solid angle characterizes the directional properties of the three-dimensional radiation pattern.
To characterize the width of the main lobe in a given plane, the term used is beamwidth. The half-power beamwidth, or simply the beamwidth β, is defined as the angular width of the main lobe between the two angles at which the magnitude of F(θ, φ) is equal to half of its peak value (or −3 dB on a decibel scale). For example, for the pattern displayed in Fig. 9-10(b), β is given by β = θ2 −θ1, (9.22) where θ1 and θ2 are the half-power angles at which F(θ, 0) = 0.5 (with θ2 denoting the larger value and θ1 denoting the smaller one, as shown in the figure). If the pattern is symmetrical and the peak value of F(θ, φ) is at θ = 0, then β = 2θ2. For the short-dipole pattern shown earlier in Fig. 9-7(a), F(θ) is maximum at θ = 90◦, θ2 is at 135◦, and θ1 is at 45◦. Hence, β = 135◦−45◦= 90◦. The beamwidth β 414 CHAPTER 9 RADIATION AND ANTENNAS (a) Actual pattern (b) Equivalent solid angle F(θ, φ) 1 Ωp F = 1 within the cone 1 Figure 9-11 The pattern solid angle p defines an equivalent cone over which all the radiation of the actual antenna is concentrated with uniform intensity equal to the maximum of the actual pattern.
is also known as the 3 dB beamwidth. In addition to the half-power beamwidth, other beam dimensions may be of interest for certain applications, such as the null beamwidth βnull, which is the angular width between the first nulls on the two sides of the peak [Fig. 9-10(b)].
9-2.3 Antenna Directivity The directivity D of an antenna is defined as the ratio of its maximum normalized radiation intensity, Fmax (which by definition is equal to 1), to the average value of F(θ, φ) over all directions (4π space): D = Fmax Fav = 1 1 4π 4π F(θ, φ) d = 4π p (dimensionless).
(9.23) Here p is the pattern solid angle defined by Eq. (9.21). Thus, the narrower p of an antenna pattern is, the greater is the directivity.
For an isotropic antenna, p = 4π; hence, its directivity Diso = 1.
By using Eq. (9.20) in Eq. (9.23), D can be expressed as D = 4πR2Smax Prad = Smax Sav , (9.24) βxz z y x 0 dB βyz Figure 9-12 The solid angle of a unidirectional radiation pattern is approximately equal to the product of the half-power beamwidths in the two principal planes; that is, p ≈βxzβyz.
where Sav = Prad/(4πR2) is the average value of the radiated power density and is equal to the total power radiated by the antenna, Prad, dividedbythesurfaceareaofasphereofradius R.
▶Since Sav = Siso, where Siso is the power density radiated by an isotropic antenna, D represents the ratio of the maximum power density radiated by the antenna to the power density radiated by an isotropic antenna, both measured at the same range R and excited by the same amount of input power. ◀ Usually, D is expressed in decibels:† D (dB) = 10 log D.
For an antenna with a single main lobe pointing in the z direction as shown in Fig. 9-12, p may be approximated as the product of the half-power beamwidths βxz and βyz (in radians): p ≈βxzβyz, (9.25) †A note of caution: Even though we often express certain dimensionless quantities in decibels, we should always convert their decibel values to natural values before using them in the relations given in this chapter.
9-2 ANTENNA RADIATION CHARACTERISTICS 415 and therefore D = 4π p ≈ 4π βxzβyz (single main lobe).
(9.26) Although approximate, this relation provides a useful method for estimating the antenna directivity from measurements of the beamwidths in the two orthogonal planes whose intersection is the axis of the main lobe.
Example 9-1: Antenna Radiation Properties Determine (a) the direction of maximum radiation, (b) pattern solid angle, (c) directivity, and (d) half-power beamwidth in the y–z plane for an antenna that radiates only into the upper hemisphere with normalized radiation intensity given by F(θ, φ) = cos2 θ.
Solution: The statement that the antenna radiates through only the upper hemisphere is equivalent to F(θ, φ) = F(θ) = ⎧ ⎨ ⎩ cos2 θ for 0 ≤θ ≤π/2 and 0 ≤φ ≤2π, 0 elsewhere.
(a) The function F(θ) = cos2 θ is independent of φ and is maximum when θ = 0◦. A polar plot of F(θ) is shown in Fig. 9-13.
(b) From Eq. (9.21), the pattern solid angle p is given by p = 4π F(θ, φ) d = 2π φ=0 ⎡ ⎣ π/2 θ=0 cos2 θ sin θ dθ ⎤ ⎦dφ = 2π φ=0 −cos3 θ 3 π/2 0 dφ = 2π 0 1 3 dφ = 2π 3 (sr).
90° –45° 0.5 0.5 45° 1 y x F(θ) = cos2 θ z Figure 9-13 Polar plot of F(θ) = cos2 θ.
(c) Application of Eq. (9.23) gives D = 4π p = 4π 3 2π = 6, which corresponds to D (dB) = 10 log 6 = 7.78 dB.
(d) The half-power beamwidth β is obtained by setting F(θ) = 0.5. That is, F(θ) = cos2 θ = 0.5, which gives the half-power angles θ1 = −45◦and θ2 = 45◦.
Hence, β = θ2 −θ1 = 90◦.
Example 9-2: Directivity of a Hertzian Dipole Calculate the directivity of a Hertzian dipole.
Solution: Application of Eq. (9.23) with F(θ) = sin2 θ [from Eq. (9.15)] gives D = 4π 4π F(θ, φ) sin θ dθ dφ = 4π 2π φ=0 π θ=0 sin3 θ dθ dφ = 4π 8π/3 = 1.5 or, equivalently, 1.76 dB.
416 CHAPTER 9 RADIATION AND ANTENNAS 9-2.4 Antenna Gain Of the total power Pt (transmitter power) supplied to the antenna, a part, Prad, is radiated out into space, and the remainder, Ploss, is dissipated as heat in the antenna structure.
The radiation efficiency ξ is defined as the ratio of Prad to Pt: ξ = Prad Pt (dimensionless).
(9.27) The gain of an antenna is defined as G = 4πR2Smax Pt , (9.28) which is similar in form to the expression given by Eq. (9.24) for the directivity D except that it is referenced to the input power supplied to the antenna, Pt, rather than to the radiated power Prad. In view of Eq. (9.27), G = ξ D (dimensionless).
(9.29) ▶The gain accounts for ohmic losses in the antenna material, whereas the directivity does not. For a lossless antenna, ξ = 1, and G = D. ◀ 9-2.5 Radiation Resistance To a transmission line connected between a generator supplying power Pt on one end and an antenna on the other end, the antenna is merely a load with input impedance Zin. If the line is lossless and properly matched to the antenna, all of Pt is transferred to the antenna. In general, Zin consists of a resistive component Rin and a reactive component Xin: Zin = Rin + jXin.
(9.30) The resistive component is defined as equivalent to a resistor Rin that would consume an average power Pt when the amplitude of the ac current flowing through it is I0, Pt = 1 2 I 2 0 Rin.
(9.31) Since Pt = Prad + Ploss, it follows that Rin can be defined as the sum of a radiation resistance Rrad and a loss resistance Rloss, Rin = Rrad + Rloss, (9.32) with Prad = 1 2 I 2 0 Rrad, (9.33a) Ploss = 1 2 I 2 0 Rloss, (9.33b) where I0 is the amplitude of the sinusoidal current exciting the antenna. As defined earlier, the radiation efficiency is the ratio of Prad to Pt, or ξ = Prad Pt = Prad Prad + Ploss = Rrad Rrad + Rloss .
(9.34) The radiation resistance Rrad can be calculated by integrating the far-field power density over a sphere to obtain Prad and then equating the result to Eq. (9.33a).
Example 9-3: Radiation Resistance and Efficiency of a Hertzian Dipole A 4 cm long center-fed dipole is used as an antenna at 75 MHz. The antenna wire is made of copper and has a radius a = 0.4 mm. From Eqs. (7.92a) and (7.94), the loss resistance of a circular wire of length l is given by Rloss = l 2πa πf μc σc , (9.35) where μc and σc are the magnetic permeability and conductivity of the wire, respectively. Calculate the radiation resistance and the radiation efficiency of the dipole antenna.
Solution: At 75 MHz, λ = c f = 3 × 108 7.5 × 107 = 4 m.
The length to wavelength ratio is l/λ = 4 cm/4 m = 10−2.
Hence, this is a short dipole. From Eq. (9.24), Prad = 4πR2 D Smax.
(9.36) 9-3 HALF-WAVE DIPOLE ANTENNA 417 For the Hertzian dipole, Smax is given by Eq. (9.14), and from Example 9-2 we established that D = 1.5. Hence, Prad = 4πR2 1.5 × 15πI 2 0 R2 l λ 2 = 40π2I 2 0 l λ 2 .
(9.37) Equating this result to Eq. (9.33a) and then solving for the radiation resistance Rrad leads to Rrad = 80π2(l/λ)2 ( ) (short dipole).
(9.38) For l/λ = 10−2, Rrad = 0.08 .
Next, we determine the loss resistance Rloss.
For copper, Appendix B gives μc ≈μ0 = 4π × 10−7 H/m and σc = 5.8 × 107 S/m. Hence, Rloss = l 2πa πf μc σc = 4 × 10−2 2π × 4 × 10−4 π × 75 × 106 × 4π × 10−7 5.8 × 107 1/2 = 0.036 .
Therefore, the radiation efficiency is ξ = Rrad Rrad + Rloss = 0.08 0.08 + 0.036 = 0.69.
Thus, the dipole is 69% efficient.
Concept Question 9-5: What does the pattern solid angle represent?
Concept Question 9-6: What is the magnitude of the directivity of an isotropic antenna?
Concept Question 9-7: What physical and material properties affect the radiation efficiency of a fixed-length Hertzian dipole antenna?
Exercise 9-2: An antenna has a conical radiation pattern with a normalized radiation intensity F(θ) = 1 for θ between 0◦and 45◦and zero for θ between 45◦and 180◦.
The pattern is independent of the azimuth angle φ. Find (a) the pattern solid angle and (b) the directivity.
Answer: (a) p = 1.84 sr, (b) D = 6.83 or, equivalently, 8.3 dB. (See EM.) Exercise 9-3: The maximum power density radiated by a short dipole at a distance of 1 km is 60 (nW/m2). If I0 = 10 A, find the radiation resistance.
Answer: Rrad = 10 m . (See EM.) 9-3 Half-Wave Dipole Antenna In Section 9-1 we developed expressions for the electric and magnetic fields radiated by a Hertzian dipole of length l ≪λ. We now use these expressions as building blocks to obtain expressions for the fields radiated by a half-wave dipole antenna, so named because its length l = λ/2. As shown in Fig. 9-14, the half-wave dipole consists of a thin wire fed at its center by a generator connected to the antenna terminals via a transmission line. The current flowing through the wire has a symmetrical distribution with respect to the center of the dipole, and the current is zero at its ends. Mathematically, i(t) is given by i(t) = I0 cos ωt cos kz = Re I0 cos kz ejωt , (9.39a) whose phasor is I(z) = I0 cos kz, −λ/4 ≤z ≤λ/4 , (9.39b) 418 CHAPTER 9 RADIATION AND ANTENNAS l = λ/2 Transmission line Dipole antenna Current distribution I(z) = I0 cos kz i(t) i(t) (a) (b) l = λ/2 θ θs dz z z = l/2 z = –l/2 z cos θ s z R Q = (R, θ, φ) Figure 9-14 Center-fed half-wave dipole.
and k = 2π/λ. Equation (9.9a) gives an expression for Eθ, the far field radiated by a Hertzian dipole of length l when excited by a current I0.
Let us adapt that expression to an infinitesimal dipole segment of length dz, excited by a current I(z) and located at a distance s from the observation point Q [Fig. 9-14(b)]. Thus, d Eθ(z) = jkη0 4π I(z) dz e−jks s sin θs, (9.40a) and the associated magnetic field is d Hφ(z) = d Eθ(z) η0 .
(9.40b) The far field due to radiation by the entire antenna is obtained by integrating the fields from all of the Hertzian dipoles making up the antenna: Eθ = λ/4 z=−λ/4 d Eθ.
(9.41) Before we calculate this integral, we make the following two approximations. The first relates to the magnitude part of the spherical propagation factor, 1/s. In Fig. 9-14(b), the distance s between the current element and the observation point Q is considered so large in comparison with the length of the dipole that the difference between s and R may be neglected in terms of its effect on 1/s. Hence, we may set 1/s ≈1/R, and by the same argument we set θs ≈θ. The error between s and R is a maximum when the observation point is along the z axis and it is equal to λ/4 (corresponding to half of the antenna length). If R ≫λ, this error will have an insignificant effect on 1/s. The second approximation is associated with the phase factor e−jks. An error in distance corresponds to an error in phase k = (2π/λ)(λ/4) = π/2. As a rule of thumb, a phase error greater than π/8 is considered unacceptable because it may lead to a significant error in the computed value of the field Eθ. Hence, the approximation s ≈R is too crude for the phase factor and cannot be used. A more tolerable option is to use the parallel-ray approximation given by s ≈R −z cos θ, (9.42) as illustrated in Fig. 9-14(b).
Substituting Eq. (9.42) for s in the phase factor of Eq. (9.40a) andreplacings withR andθs withθ elsewhereintheexpression, we obtain d Eθ = jkη0 4π I(z) dz e−jkR R sin θ ejkz cos θ.
(9.43) After (1) inserting Eq. (9.43) into Eq. (9.41), (2) using the expression for I(z) given by Eq. (9.39b), and (3) carrying out the integration, the following expressions are obtained: Eθ = j 60I0 cos[(π/2) cos θ] sin θ e−jkR R , (9.44a) Hφ = Eθ η0 .
(9.44b) 9-3 HALF-WAVE DIPOLE ANTENNA 419 The corresponding time-average power density is S(R, θ) = | Eθ|2 2η0 = 15I 2 0 πR2 cos2[(π/2) cos θ] sin2 θ = S0 cos2[(π/2) cos θ] sin2 θ (W/m2).
(9.45) Examination of Eq. (9.45) reveals that S(R, θ) is maximum at θ = π/2, and its value is Smax = S0 = 15I 2 0 πR2 .
Hence, the normalized radiation intensity is F(θ) = S(R, θ) S0 = cos[(π/2) cos θ] sin θ 2 .
(9.46) The radiation pattern of the half-wave dipole exhibits roughly the same doughnut-like shape shown earlier in Fig. 9-7 for the short dipole. Its directivity is slightly larger (1.64 compared with 1.5 for the short dipole), but its radiation resistance is 73 (as shown later in Section 9-3.2), which is orders of magnitude larger than that of a short dipole.
9-3.1 Directivity of λ/2 Dipole To evaluate both the directivity D and the radiation resistance Rrad of the half-wave dipole, we first need to calculate the total radiated power Prad by applying Eq. (9.20): Prad = R2 4π S(R, θ) d = 15I 2 0 π 2π 0 π 0 cos[(π/2) cos θ] sin θ 2 sin θ dθ dφ.
(9.47) The integration over φ is equal to 2π, and numerical evaluation of the integration over θ gives the value 1.22. Consequently, Prad = 36.6 I 2 0 (W).
(9.48) From Eq. (9.45), we found that Smax = 15I 2 0 /(πR2). Using this in Eq. (9.24) gives the following result for the directivity D of the half-wave dipole: D = 4πR2Smax Prad = 4πR2 36.6I 2 0 15I 2 0 πR2 = 1.64 (9.49) or, equivalently, 2.15 dB.
9-3.2 Radiation Resistance of λ/2 Dipole From Eq. (9.33a), Rrad = 2Prad I 2 0 = 2 × 36.6I 2 0 I 2 0 ≈73 .
(9.50) As was noted earlier in Example 9-3, because the radiation resistance of a Hertzian dipole is comparable in magnitude to that of its loss resistance Rloss, its radiation efficiency ξ is rather small. For the 4 cm long dipole of Example 9-3, Rrad = 0.08 (at 75 MHz) and Rloss = 0.036 . If we keep the frequency the same and increase the length of the dipole to 2 m (λ = 4 m at f = 75 MHz), Rrad becomes 73 and Rloss increases to 1.8 .
The radiation efficiency increases from 69% for the short dipole to 98% for the half-wave dipole. More significant is the fact that it is practically impossible to match a transmission line to an antenna with a resistance on the order of 0.1 , while it is quite easy to do so when Rrad = 73 .
Moreover, since Rloss ≪Rrad for the half-wave dipole, Rin ≈Rrad and Eq. (9.30) becomes Zin ≈Rrad + jXin.
(9.51) Deriving an expression for Xin for the half-wave dipole is fairly complicated and beyond the scope of this book. However, it is significant to note that Xin is a strong function of l/λ, and that it decreasesfrom42 atl/λ = 0.5tozeroatl/λ = 0.48, whereas Rrad remains approximately unchanged. Hence, by reducing the length of the half-wave dipole by 4%, Zin becomes purely real and equal to 73 , thereby making it possible to match the dipole to a 75 transmission line without resorting to the use of a matching network.
420 CHAPTER 9 RADIATION AND ANTENNAS λ/4 I I I Image Conducting plane (a) (b) Figure 9-15 A quarter-wave monopole above a conducting plane is equivalent to a full half-wave dipole in free space.
9-3.3 Quarter-Wave Monopole Antenna ▶When placed over a conducting ground plane, a quarter-wave monopole antenna excited by a source at its base [Fig. 9-15(a)] exhibits the same radiation pattern in the region above the ground plane as a half-wave dipole in free space. ◀ This is because, from image theory (Section 4-11), the conducting plane can be replaced with the image of the λ/4 monopole, as illustrated in Fig. 9-15(b).
Thus, the λ/4 monopole radiates an electric field identical to that given by Eq. (9.44a), and its normalized radiation intensity is given by Eq. (9.46); but the radiation is limited to the upper half-space defined by 0 ≤θ ≤π/2. Hence, a monopole radiates only half as much power as the dipole. Consequently, for a λ/4 monopole, Prad = 18.3I 2 0 and its radiation resistance is Rrad = 36.5 .
The approach used with the quarter-wave monopole is also valid for any vertical wire antenna placed above a conducting plane, including a Hertzian monopole.
Concept Question 9-8: What is the physical length of a half-wave dipole operating at (a) 1 MHz (in the AM broadcast band), (b) 100 MHz (FM broadcast band), and (c) 10 GHz (microwave band)?
Concept Question 9-9: How does the radiation pattern of a half-wave dipole compare with that of a Hertzian dipole? How do their directivities, radiation resistances, and radiation efficiencies compare?
Concept Question 9-10: How does the radiation effi-ciency of a quarter-wave monopole compare with that of a half-wave dipole, assuming that both are made of the same material and have the same cross section?
Exercise 9-4: For the half-wave dipole antenna, evaluate F(θ) versus θ to determine the half-power beamwidth in the elevation plane (the plane containing the dipole axis).
Answer: β = 78◦. (See EM.) Exercise 9-5: If the maximum power density radiated by a half-wave dipole is 50 μW/m2 at a range of 1 km, what is the current amplitude I0?
Answer: I0 = 3.24 A. (See EM.) 9-4 Dipole of Arbitrary Length So far, we examined the radiation properties of the Hertzian and half-wave dipoles. We now consider the more general case of a linear dipole of arbitrary length l, relative to λ. For a center-fed dipole, as depicted in Fig. 9-16, the currents flowing through its two halves are symmetrical and must go to zero at 9-4 DIPOLE OF ARBITRARY LENGTH 421 (a) l = λ/2 (b) l = λ (c) l = 3λ/2 I(z) ~ I(z) ~ I(z) ~ Figure 9-16 Current distribution for three center-fed dipoles.
its ends. Hence, the current phasor ˜ I(z) can be expressed as a sine function with an argument that goes to zero at z = ±l/2: ˜ I(z) = I0 sin [k (l/2 −z)] , for 0 ≤z ≤l/2, I0 sin [k (l/2 + z)] , for −l/2 ≤z < 0, (9.52) where I0 is the current amplitude. The procedure for calculating theelectricandmagneticfieldsandtheassociatedpowerdensity of the wave radiated by such an antenna is basically the same as that used previously in connection with the half-wave dipole antenna. The only difference is the current distribution ˜ I(z). If we insert the expression for ˜ I(z) given by Eq. (9.52) into Eq. (9.43), we obtain the following expression for the differential electric field d Eθ of the wave radiated by an elemental length dz at location z along the dipole: d Eθ = jkη0I0 4π e−jkR R sin θ ejkz cos θ dz × sin [k (l/2 −z)] for 0 ≤z ≤l/2, sin [k (l/2 + z)] for −l/2 ≤z < 0.
(9.53) The total field radiated by the dipole is Eθ = l/2 −l/2 d Eθ = l/2 0 d Eθ + 0 −l/2 d Eθ = jkη0I0 4π e−jkR R sin θ × ⎧ ⎪ ⎨ ⎪ ⎩ l/2 0 ejkz cos θ sin[k(l/2 −z)] dz + 0 −l/2 ejkz cos θ sin[k(l/2 + z)] dz ⎫ ⎪ ⎬ ⎪ ⎭ .
(9.54) If we apply Euler’s identity to express ejkz cos θ as [cos(kz cos θ) + j sin(kz cos θ)], we can integrate the two integrals and obtain the result Eθ = j60I0 e−jkR R · cos kl 2 cos θ −cos kl 2 sin θ .
(9.55) The corresponding time-average power density radiated by the dipole antenna is given by S(θ) = | Eθ|2 2η0 = 15I 2 0 πR2 cos πl λ cos θ −cos πl λ sin θ 2 , (9.56) where we have used the relations η0 ≈120π ( ) and k = 2π/λ. For l = λ/2, Eq. (9.56) reduces to the expression given by Eq. (9.45) for the half-wave dipole.
Plots of the normalized radiation intensity, F(θ) = S(R, θ)/Smax, are shown in Fig. 9-17 for dipoles of lengths λ/2, λ, and 3λ/2. The dipoles with l = λ/2 and l = λ have similar radiation patterns with maxima along θ = 90◦, but the half-power beamwidth of 422 CHAPTER 9 RADIATION AND ANTENNAS z x-y plane 0.5 1 z x-y plane 0.5 1 β=78° z x-y plane (a) l = λ/2 (b) l = λ (c) l = 3λ/2 β=47° Figure 9-17 Radiation patterns of dipoles with lengths of λ/2, λ, and 3λ/2.
the wavelength-long dipole is narrower than that of the half-wave dipole, and Smax = 60I 2 0 /(πR2) for the wavelength-long dipole, which is four times that for the half-wave dipole. The pattern of the dipole with length l = 3λ/2 exhibits a structure with multiple lobes, and its direction of maximum radiation is not along θ = 90◦.
9-5 Effective Area of a Receiving Antenna So far, antennas have been treated as directional radiators of energy. Now, we examine the reverse process, namely how a Module 9.2 Linear Dipole Antenna For a linear antenna of any specified length (in units of λ), this module displays the current distribution along the antenna and the far-field radiation patterns in the horizontal and elevation planes.
It also calculates the total power radiated by the antenna, the radiation resistance, and the antenna directivity.
receiving antenna extracts energy from an incident wave and delivers it to a load. The ability of an antenna to capture energy from an incident wave of power density Si (W/m2) and convert it into an intercepted power Pint (W) for delivery to a matched load is characterized by the effective area Ae: Ae = Pint Si (m2).
(9.57) Other commonly used names for Ae include effective aperture andreceivingcrosssection. Theantennareceivingprocessmay be modeled in terms of a Th´ evenin equivalent circuit (Fig. 9-18) consisting of a voltage Voc in series with the antenna input impedance Zin. Here, Voc is the open-circuit voltage induced by the incident wave at the antenna terminals and ZL is the impedance of the load connected to the antenna (representing a receiver or some other circuit). In general, both Zin and ZL are complex: Zin = Rrad + jXin, (9.58a) ZL = RL + jXL, (9.58b) where Rrad denotes the radiation resistance of the antenna (assuming Rloss ≪Rrad). To maximize power transfer to the load, the load impedance must be chosen such that ZL = Z∗ in, or RL = Rrad and XL = −Xin. In that case, the circuit reduces to a source Voc connected across a resistance equal to 2Rrad.
Since Voc is a sinusoidal voltage phasor, the time-average power delivered to the load is PL = 1 2| IL|2Rrad = 1 2 | Voc| 2Rrad 2 Rrad = | Voc|2 8Rrad , (9.59) 9-5 EFFECTIVE AREA OF A RECEIVING ANTENNA 423 Module 9.3 Detailed Analysis of Linear Antenna This module complements Module 9.2 by offering extensive information about the specified linear antenna, including its directivity and plots of its current and field distributions.
424 TECHNOLOGY BRIEF 17: HEALTH RISKS OF EM FIELDS Technology Brief 17: Health Risks of EM Fields Can the use of cell phones cause cancer? Does exposure to the electromagnetic fields (EMFs) associated with power lines pose health risks to humans? Are we endangered by EMFs generated by home appliances, telephones, electrical wiring, and the myriad of electronic gadgets we use every day (Fig. TF17-1)? Despite reports in some of the popular media alleging a causative relationship between low-level EMFs and many diseases, according to reports issued by governmental and professional boards in the U.S. and Europe, the answer is: ▶NO, we are not at risk, so long as manufacturers adhere to the approved governmental standards for maximum permissible exposure (MPE) levels. With regard to cell phones, the official reports caution that their conclusions are limited to phone use of less than 15 years, since data for longer-term use is not yet available. ◀ Physiological Effects of EMFs The energy carried by a photon with an EM frequency f is given by E = hf , where h is Planck’s constant. The mode of interaction between a photon passing through a material and the material’s atoms or molecules is very much dependent on f . If f is greater than about 1015 Hz (which falls in the ultraviolet (UV) band of the EM spectrum), the photon’s energy is sufficient to free an electron and remove it completely, thereby ionizing the affected atom or molecule. Consequently, the energy carried by such EM waves is called ionizing radiation, in contrast with non-ionizing radiation (Fig.TF17-2) whose photons may be able to cause an electron to move to a higher energy level, but not eject it from its host atom or molecule.
Assessing health risks associated with exposure to EMFs is complicated by the number of variables involved, which include: (1) the frequency f , (2) the intensities of the electric and magnetic fields, (3) the exposure duration, whether FigureTF17-1 Electromagnetic fields are emitted by power lines, cell phones, TV towers, and many other electronic circuits and devices.
TECHNOLOGY BRIEF 17: HEALTH RISKS OF EM FIELDS 425 Non-ionizing Ionizing 0.1 Hz Induces low currents Extremely low frequency Microwave X-rays Gamma rays Frequency Radio Infrared Visible light Ultraviolet Induces high currents Excites electrons Damages DNA 1 MHz 1 THz 1 EHz FigureTF17-2 Different types of electromagnetic radiation.
continuous or discontinuous, and whether pulsed or uniform, and (4) the specific part of the body that is getting exposed. We know that intense laser illumination can cause corneal burn, high-level X-rays can damage living tissue and cause cancer and, in fact, any form of EM energy can be dangerous if the exposure level and/or duration exceed certain safety limits. Governmental and professional safety boards are tasked with establishing maximum permissible exposure (MPE) levels that protect human beings against adverse health effects associated with EMFs. In the U.S., the relevant standards are IEEE Std C95.6 (dated 2002), which addresses EM fields in the 1 Hz to 3 kHz range, and IEEE Std 95.1 (dated 2005), which deals with the frequency range from 3 kHz to 300 GHz. On the European side of the Atlantic, responsibility for establishing MPE levels resides with the Scientific Committee on Emerging and Newly Identified Health Risks (SCENIHR) of the European Commission.
▶At frequencies below 100 kHz, the goal is to minimize adverse effects of exposure to electric fields that can cause electrostimulation of nerve and muscle cells. Above 5 MHz, the main concern is excessive tissue heating, and in the transition region of 100 kHz to 5 MHz, safety standards are designed to protect against both electrostimulation and excessive heating. ◀ Frequency Range 0 ≤f ≤3 kHz: The plots in Fig. TF17-3 display the values of MPE for electric and magnetic fields over the frequency range below 3 kHz. According to IEEE Std C95.6, it is sufficient to demonstrate compliance with the MPE levels for either the electric field E or the magnetic field H. According to the plot for H, exposure at 60 Hz should not exceed 720 A/m. The magnetic field due to power lines is typically in the range of 2–6 A/m underneath the lines, which is at least two orders of magnitude smaller than the established safe level for H.
Frequency Range 3 kHz ≤f ≤300 GHz: At frequencies below 500 MHz, MPE is specified in terms of the electric and magnetic field strengths of the EM energy (Fig.TF17-4). From 100 MHz to 300 GHz (and beyond), MPE is specified in terms of the product of E and H, namely the power density S. Cell phones operate in the 1–2 GHz band; the specified MPE is 1 W/m2, or equivalently 0.1 mW/cm2.
Bottom Line We are constantly bombarded by EM energy, from solar illumination to blackbody radiation emitted by all matter. Our bodies absorb, reflect, and emit EM energy all the time. Living organisms, including humans, require exposure to EM radiation to survive, but excessive exposure can cause adverse effects. The term excessive exposure connotes a complicated set of relationships among such variables as field strength, exposure duration and mode (continuous, 426 TECHNOLOGY BRIEF 17: HEALTH RISKS OF EM FIELDS 0.1 E-field H-field MPE Other tissue MPE Frequency (Hz) Electric Field E (V/m) Magnetic Field H (A/m) E-field Brain MPE 1 10–3 10–2 10–1 102 103 104 105 106 1 10 10 100 1000 3000 FigureTF17-3 Maximum permissible exposure (MPE) levels for E and H over the frequency range from 0.1 Hz to 3 kHz.
pulsed, etc.), body part, etc.
The emission standards established by the Federal Communications Commission in the U.S. and similar governmental bodies in other countries are based on a combination of epidemiological studies, experimental observations, and theoretical understanding of how EM energy interacts with biological material.
Generally speaking, the maximum permissible exposure levels specified by these standards are typically two orders of magnitude lower than the levels known to cause adverse effects, but in view of the multiplicity of variables involved, there is no guarantee that adhering to the standards will avoid health risks absolutely. The bottom line is: use common sense!
Frequency (MHz) Power Density S (W/m2) Magnetic Field H (A/m) Electric Field E (V/m) 1 614 V/m 27.5 V/m 2 W/m2 100 W/m2 163 A/m 0.0729 A/m Electric field E Magnetic field H Power density 10–2 10–1 102 103 104 105 1 0.1 0.01 10 10 100 1000 10000 1 0.1 10 100 1000 10000 0.01 Figure TF17-4 MPE levels for the frequency range from 10 kHz to 300 GHz.
9-6 FRIIS TRANSMISSION FORMULA 427 where IL = Voc/(2Rrad) is the phasor current flowing through the circuit. Since the antenna is lossless, all the intercepted power Pint ends up in the load resistance RL. Hence, Pint = PL = | Voc|2 8Rrad .
(9.60) For an incident wave with electric field Ei parallel to the antenna polarization direction, the power density carried by the wave is Si = | Ei|2 2η0 = | Ei|2 240π .
(9.61) The ratio of the results provided by Eqs. (9.60) and (9.61) gives Ae = Pint Si = 30π| Voc|2 Rrad| Ei|2 .
(9.62) The open-circuit voltage Voc induced in the receiving antenna is due to the incident field Ei, but the relation between them depends on the specific antenna under consideration. By way of illustration, let us consider the case of the short-dipole antenna of Section 9-1. Because the length l of the short dipole is small compared with λ, the current induced by the incident field is uniform across its length, and the open-circuit voltage is simply Voc = Eil. Noting that Rrad = 80π2(l/λ)2 for the short dipole [see Eq. (9.38)] and using Voc = Eil, Eq. (9.62) simplifies to Ae = 3λ2 8π (m2) (short dipole).
(9.63) In Example 9-2 it was shown that for the Hertzian dipole the directivity D = 1.5. In terms of D, Eq. (9.63) can be rewritten in the form Ae = λ2D 4π (m2) (any antenna).
(9.64) ▶Despite the fact that the relation between Ae and D given by Eq. (9.64) was derived for a Hertzian dipole, it can be shown that it is also valid for any antenna under matched-impedance conditions. ◀ (a) Receiving antenna (b) Equivalent circuit ZL Incident wave Antenna Load Zin = Rrad + jXin ZL = RL + jXL Load Antenna equivalent circuit Voc ~ Figure 9-18 Receiving antenna represented by an equivalent circuit.
Exercise 9-6: The effective area of an antenna is 9 m2.
What is its directivity in decibels at 3 GHz?
Answer: D = 40.53 dB. (See EM.) Exercise 9-7: At 100 MHz, the pattern solid angle of an antenna is 1.3 sr. Find (a) the antenna directivity D and (b) its effective area Ae.
Answer: (a) D = 9.67, (b) Ae = 6.92 m2. (See EM.) 9-6 Friis Transmission Formula The two antennas shown in Fig. 9-19 are part of a free-space communication link, with the separation between the antennas, R, being large enough for each to be in the far-field region of the other.
The transmitting and receiving antennas have effective areas At and Ar and radiation efficiencies ξt and ξr, respectively. Our objective is to find a relationship between Pt, the power supplied to the transmitting antenna, and Prec, the power delivered to the receiver. As always, we assume that both antennas are impedance-matched to their respective 428 CHAPTER 9 RADIATION AND ANTENNAS Pt Prec Prad At Ar Pint R Transmitting antenna Receiving antenna Figure 9-19 Transmitter–receiver configuration.
transmission lines. Initially, we consider the case where the two antennas are oriented such that the peak of the radiation pattern of each antenna points in the direction of the other.
We start by treating the transmitting antenna as a lossless isotropic radiator.
The power density incident upon the receiving antenna at a distance R from an isotropic transmitting antenna is simply equal to the transmitter power Pt divided by the surface area of a sphere of radius R: Siso = Pt 4πR2 .
(9.65) The real transmitting antenna is neither lossless nor isotropic.
Hence, the power density Sr due to the real antenna is Sr = GtSiso = ξtDtSiso = ξtDtPt 4πR2 .
(9.66) Through the gain Gt = ξtDt, ξt accounts for the fact that only part of the power Pt supplied to the antenna is radiated out into space, and Dt accounts for the directivity of the transmitting antenna (in the direction of the receiving antenna). Moreover, by Eq. (9.64), Dt is related to At by Dt = 4πAt/λ2. Hence, Eq. (9.66) becomes Sr = ξtAtPt λ2R2 .
(9.67) On the receiving-antenna side, the power intercepted by the receiving antenna is equal to the product of the incident power density Sr and the effective area Ar: Pint = SrAr = ξtAtArPt λ2R2 .
(9.68) The power delivered to the receiver, Prec, is equal to the intercepted power Pint multiplied by the radiation efficiency of the receiving antenna, ξ r. Hence, Prec = ξrPint, which leads to the result Prec Pt = ξtξrAtAr λ2R2 = GtGr λ 4πR 2 .
(9.69) ▶This relation is known as the Friis transmission formula, and Prec/Pt is called the power transfer ratio. ◀ If the two antennas are not oriented in the direction of maximum power transfer, Eq. (9.69) assumes the general form Prec Pt = GtGr λ 4πR 2 Ft(θt, φt) Fr(θr, φr), (9.70) where Ft(θt, φt) is the normalized radiation intensity of the transmitting antenna at angles (θ t, φt) corresponding to the directionofthereceivingantenna(asseenbytheantennapattern of the transmitting antenna), and a similar definition applies to Fr(θr, φr) for the receiving antenna.
Example 9-4: Satellite Communication System A 6 GHz direct-broadcast TV satellite system transmits 100 W through a 2 m diameter parabolic dish antenna from a distance of approximately 40,000 km above Earth’s surface.
Each TV channel occupies a bandwidth of 5 MHz.
Due to electromagnetic noise picked up by the antenna as well as noise generated by the receiver electronics, a home TV receiver has a noise level given by Pn = KTsysB (W), (9.71) where Tsys [measured in kelvins (K)] is a figure of merit called the system noise temperature that characterizes the noise performance of the receiver–antenna combination, K is Boltzmann’s constant [1.38×10−23 (J/K)], and B is the receiver bandwidth in Hz.
The signal-to-noise ratio Sn (which should not be confused with the power density S) is defined as the ratio of Prec to Pn: Sn = Prec/Pn (dimensionless).
(9.72) 9-7 RADIATION BY LARGE-APERTURE ANTENNAS 429 For a receiver with Tsys = 580 K, what minimum diameter of a parabolic dish receiving antenna is required for high-qualityTV reception with Sn = 40 dB? The satellite and ground receiving antennas may be assumed lossless, and their effective areas may be assumed equal to their physical apertures.
Solution: The following quantities are given: Pt = 100 W, f = 6 GHz = 6 × 109 Hz, Sn = 104, Transmit antenna diameter dt = 2 m, Tsys = 580 K, R = 40, 000 km = 4 × 107 m, B = 5 MHz = 5 × 106 Hz.
The wavelength λ = c/f = 5 × 10−2 m, and the area of the transmitting satellite antenna is At = (πd2 t /4) = π (m2). From Eq. (9.71), the receiver noise power is Pn = KTsysB = 1.38 × 10−23 × 580 × 5 × 106 = 4 × 10−14 W.
Using Eq. (9.69) with ξt = ξr = 1, Prec = PtAtAr λ2R2 = 100πAr (5 × 10−2)2(4 × 107)2 = 7.85 × 10−11Ar.
The area of the receiving antenna, Ar, can now be determined by equating the ratio Prec/Pn to Sn = 104: 104 = 7.85 × 10−11Ar 4 × 10−14 , which yields the value Ar = 5.1 m2. The required minimum diameter is dr = √4Ar/π = 2.55 m.
Exercise 9-8: If the operating frequency of the communication system described in Example 9-4 is doubled to 12 GHz, what would then be the minimum required diameter of a home receiving TV antenna?
Answer: dr = 1.27 m. (See EM.) Exercise 9-9: A 3 GHz microwave link consists of two identical antennas each with a gain of 30 dB. Determine the received power, given that the transmitter output power is 1 kW and the two antennas are 10 km apart.
Answer: Prec = 6.33 × 10−4 W. (See EM.) Exercise 9-10: The effective area of a parabolic dish antenna is approximately equal to its physical aperture.
If its directivity is 30 dB at 10 GHz, what is its effective area? If the frequency is increased to 30 GHz, what will be its new directivity?
Answer: Ae = 0.07 m2, D = 39.44 dB. (See EM.) 9-7 Radiation by Large-Aperture Antennas For wire antennas, the sources of radiation are the infinitesimal current elements comprising the current distribution along the wire, and the total radiated field at a given point in space is equal to the sum, or integral, of the fields radiated by all the elements.
A parallel scenario applies to aperture antennas, except that now the source of radiation is the electric-field distribution across the aperture. Consider the horn antenna shown in Fig. 9-20. It is connected to a source through a coaxial transmission line, with the outer conductor of the line connected to the metal body of the horn and the inner conductor made to protrude, through a small hole, partially into the throat end of the horn. The protruding conductor acts as a monopole antenna, generating waves that radiate outwardly toward the horn’s aperture. The electric field of the wave arriving at the aperture, which may z Q R Observation sphere θ xa Ea(xa, ya) ya Figure 9-20 A horn antenna with aperture field distribution Ea(xa, ya).
430 CHAPTER 9 RADIATION AND ANTENNAS vary as a function of xa and ya over the horn’s aperture, is called the electric-field aperture distribution or illumination, Ea(xa, ya). Inside the horn, wave propagation is guided by the horn’s geometry; but as the wave transitions from a guided wave into an unbounded wave, every point of its wavefront serves as a source of spherical secondary wavelets. The aperture may then be represented as a distribution of isotropic radiators. At a distant point Q, the combination of all the waves arriving from all of these radiators constitutes the total wave that would be observed by a receiver placed at that point.
The radiation process described for the horn antenna is equally applicable to any aperture upon which an electromagnetic wave is incident. For example, if a light source is used to illuminate an opening in an opaque screen through a collimating lens, as shown in Fig. 9-21(a), the opening becomes a source of secondary spherical wavelets, much like the aperture of the horn antenna.
In the case of the parabolic reflector shown in Fig. 9-21(b), it can be described in terms of an imaginary aperture representing the electric-field distribution across a plane in front of the reflector.
Two types of mathematical formulations are available for computing the electromagnetic fields of waves radiated by apertures.
The first is a scalar formulation based on Kirchhoff’s work and the second is a vector formulation based on Maxwell’s equations. In this section, we limit our presentation to the scalar diffraction technique, in part because of its inherent simplicity and also because it is applicable across a wide range of practical applications.
▶The key requirement for the validity of the scalar formulation is that the antenna aperture be at least several wavelengths long along each of its principal dimensions. ◀ A distinctive feature of such an antenna is its high directivity and correspondingly narrow beam, which makes it attractive for radar and free-space microwave communication systems.
The frequency range commonly used for such applications is the 1- to 30 GHz microwave band. Because the corresponding wavelength range is 30 to 1 cm, respectively, it is quite practical to construct and use antennas (in this frequency range) with aperture dimensions that are many wavelengths in size.
The xa–ya plane in Fig. 9-22, denoted plane A, contains an aperture with an electric field distribution Ea(xa, ya). For the sake of convenience, the opening has been chosen to be rectangular in shape, with dimensions lx along xa and (b) Parabolic reflector antenna (a) Opening in an opaque screen xa lx ly Ea(xa, ya) ya Collimating lens Imaginary aperture d Figure 9-21 Radiation by apertures: (a) an opening in an opaquescreenilluminatedbyalightsourcethroughacollimating lens and (b) a parabolic dish reflector illuminated by a small horn antenna.
ly along ya, even though the formulation we are about to discussisgeneralenoughtoaccommodateanytwo-dimensional aperture distribution, including those associated with circular and elliptical apertures.
At a distance z from the aperture plane A in Fig. 9-22, we have an observation plane O with axes (x, y). The two planes have parallel axes and are separated by a distance z. Moreover, z is sufficiently large that any point Q 9-7 RADIATION BY LARGE-APERTURE ANTENNAS 431 Aperture illumination Aperture plane A Observation plane O ya xa lx ly dya dxa y z s R x Q θ φ Figure 9-22 Radiation by an aperture in the xa–ya plane at z = 0.
in the observation plane is in the far-field region of the aperture.
To satisfy the far-field condition, it is necessary that R ≥2d2/λ (far-field range), (9.73) where d is the longest linear dimension of the radiating aperture.
The position of observation point Q is specified by the range R between the center of the aperture and point Q and by the angles θ and φ (Fig. 9-22), which jointly define the direction of the observation point relative to the coordinate system of the aperture. In our treatment of the dipole antenna, we oriented the dipole along the z axis and we called θ the zenith angle. In the present context, the z axis is orthogonal to the plane containing the antenna aperture. Also, θ usually is called the elevation angle.
The electric field phasor of the wave incident upon point Q is denoted E(R, θ, φ). Kirchhoff’s scalar diffraction theory provides the following relationship between the radiated field E(R, θ, φ) and the aperture illumination Ea(xa, ya): E(R, θ, φ) = j λ e−jkR R h(θ, φ), (9.74) where h(θ, φ) = ∞ −∞ Ea(xa, ya) · exp [jk sin θ(xa cos φ + ya sin φ)] dxa dya.
(9.75) We shall refer to h(θ, φ) as the form factor of E(R, θ, φ). Its integral is written with infinite limits, with the understanding that Ea(xa, ya) is identically zero outside the aperture. The spherical propagation factor (e−jkR/R) accounts for wave propagation between the center of the aperture and the 432 CHAPTER 9 RADIATION AND ANTENNAS observation point, and h(θ, φ) represents an integration of the exciting field Ea(xa, ya) over the extent of the aperture, taking into account [through the exponential function in Eq. (9.75)] the approximate deviation in distance between R and s, where s is the distance to any point (xa, ya) in the aperture plane (see Fig. 9-22).
▶In Kirchhoff’s scalar formulation, the polarization direction of the radiated field E(R, θ, φ) is the same as that of the aperture field Ea(xa, ya). ◀ Also, the power density of the radiated wave is given by S(R, θ, φ) = | E(R, θ, φ)|2 2η0 = | h(θ, φ)|2 2η0λ2R2 .
(9.76) 9-8 Rectangular Aperture with Uniform Aperture Distribution To illustrate the scalar diffraction technique, consider a rectangular aperture of height lx and width ly, both at least a few wavelengths long. The aperture is excited by a uniform field distribution (i.e., constant value) given by Ea(xa, ya) = ⎧ ⎨ ⎩ E0 for −lx/2 ≤xa ≤lx/2 and −ly/2 ≤ya ≤ly/2, 0 otherwise.
(9.77) To keep the mathematics simple, let us confine our examination to the radiation pattern at a fixed range R in the x–z plane, which corresponds to φ = 0. In this case, Eq. (9.75) simplifies to h(θ) = ly/2 ya=−ly/2 lx/2 xa=−lx/2 E0 exp[jkxa sin θ] dxa dya. (9.78) In preparation for performing the integration in Eq. (9.78), we introduce the intermediate variable u defined as u = k sin θ = 2π sin θ λ .
(9.79) Hence, h(θ) = E0 lx/2 −lx/2 ejuxa dxa · ly/2 −ly/2 dya = E0 ejulx/2 −e−julx/2 ju · ly = 2E0ly u ejulx/2 −e−julx/2 2j = 2E0ly u sin(ulx/2).
(9.80) Upon replacing u with its defining expression, we have h(θ) = 2E0ly 2π λ sin θ sin(πlx sin θ/λ) = E0lxly sin(πlx sin θ/λ) πlx sin θ/λ = E0Ap sinc(πlx sin θ/λ), (9.81) where Ap = lxly is the physical area of the aperture. Also, we used the standard definition of the sinc function, which, for any argument t, is defined as sinc t = sin t t .
(9.82) Using Eq. (9.76), we obtain the following expression for the power density at the observation point: S(R, θ) = S0 sinc2(πlx sin θ/λ) (x–z plane), (9.83) where S0 = E2 0A2 p/(2η0λ2R2).
▶The sinc function is maximum when its argument is zero; sinc(0) = 1. ◀ 9-8 RECTANGULAR APERTURE WITH UNIFORM APERTURE DISTRIBUTION 433 0 dB –5 –3 –10 –15 –20 –30 –1 1 2 3 –2 –3 –25 F(γ) γ = (lx/λ) sin θ βxz –13.2 dB z R θ ya ly lx xa Q = (R, θ) Figure 9-23 Normalized radiation pattern of a uniformly illuminated rectangular aperture in the x–z plane (φ = 0).
This occurs when θ = 0.
Hence, at a fixed range R, Smax = S(θ = 0) = S0. The normalized radiation intensity is then given by F(θ) = S(R, θ) Smax = sinc2(πlx sin θ/λ) = sinc2(πγ ) (x–z plane).
(9.84) Figure 9-23 shows F(θ) plotted (on a decibel scale) as a function of the intermediate variable γ = (lx/λ) sin θ.
The pattern exhibits nulls at nonzero integer values of γ .
9-8.1 Beamwidth The normalized radiation intensity F(θ) is symmetrical in the x–z plane, and its maximum is along the boresight direction (θ = 0, in this case). Its half-power beamwidth βxz = θ2 −θ1, where θ1 and θ2 are the values of θ at which F(θ, 0) = 0.5 (or −3 dB on a decibel scale), as shown in Fig. 9-23. Since the pattern is symmetrical with respect to θ = 0, θ1 = −θ2 and βxz = 2θ2. The angle θ2 can be obtained from a solution of F(θ2) = sinc2(πlx sin θ/λ) = 0.5.
(9.85) From tabulated values of the sinc function, it is found that Eq. (9.85) yields the result πlx λ sin θ2 = 1.39, (9.86) or sin θ2 = 0.44 λ lx .
(9.87) Because λ/lx ≪1 (a fundamental condition of scalar diffraction theory is that the aperture dimensions be much larger than the wavelength λ), θ2 is a small angle, in which case we can use the approximation sin θ2 ≈θ2. Hence, βxz = 2θ2 ≈2 sin θ2 = 0.88 λ lx (rad).
(9.88a) A similar solution for the y–z plane (φ = π/2) gives βyz = 0.88 λ ly (rad).
(9.88b) ▶The uniform aperture distribution ( Ea = E0 across the aperture) gives a far-field pattern with the narrowest possible beamwidth. ◀ The first sidelobe level is 13.2 dB below the peak value (see Fig. 9-23), which is equivalent to 4.8% of the peak value. If the intended application calls for a pattern with a lower sidelobe level (to avoid interference with signals from sources along directions outside the main beam of the antenna pattern), this can be accomplished by using a tapered aperture distribution, onethatisamaximumatthecenter of theapertureanddecreases toward the edges.
▶A tapered distribution provides a pattern with lower side lobes, but the main lobe becomes wider. ◀ 434 CHAPTER 9 RADIATION AND ANTENNAS (a) Pencil beam (b) Fan beam Sidelobes Boresight λ λ d β ≈ βxz ≈ d ly lx lx λ βyz ≈ly Figure 9-24 Radiation patterns of (a) a circular reflector and (b) a cylindrical reflector (side lobes not shown).
The steeper the taper, the lower are the side lobes and the wider is the main lobe. In general, the beamwidth in a given plane, say the x–z plane, is given by βxz = kx λ lx , (9.89) where kx is a constant related to the steepness of the taper. For a uniform distribution with no taper, kx = 0.88, and for a highly tapered distribution, kx ≈2. In the typical case, kx ≈1.
To illustrate the relationship between the antenna dimensions and the corresponding beam shape, we show in Fig. 9-24 the radiation patterns of a circular reflector and a cylindrical reflector.
The circular reflector has a circularly symmetric pattern, whereas the pattern of the cylindrical reflector has a narrow beam in the azimuth plane corresponding to its long dimension and a wide beam in the elevation plane corresponding to its narrow dimension.
For a circularly symmetric antenna pattern, the beamwidth β is related to the diameter d by the approximate relation β ≈λ/d.
9-8.2 Directivity and Effective Area In Section 9-2.3, we derived an approximate expression [Eq. (9.26)] for the antenna directivity D in terms of the half-power beamwidths βxz and βyz for antennas characterized by a single major lobe whose boresight is along the z direction: D ≈ 4π βxzβyz .
(9.90) Ifweusetheapproximaterelationsβxz ≈λ/lx andβyz ≈λ/ly, we obtain D ≈4πlxly λ2 = 4πAp λ2 .
(9.91) For any antenna, its directivity is related to its effective area Ae by Eq. (9.64): D = 4πAe λ2 .
(9.92) ▶For aperture antennas, their effective apertures are approximately equal to their physical apertures; that is, Ae ≈Ap. ◀ Exercise 9-11: Verify that Eq. (9.86) is a solution of Eq. (9.85) by calculating sinc2 t for t = 1.39.
Exercise 9-12: With its boresight direction along z, a square aperture was observed to have half-power beamwidths of 3◦in both the x–z and y–z planes.
Determine its directivity in decibels.
Answer: D = 4,583.66 = 36.61 dB. (See EM.) Exercise 9-13: What condition must be satisfied in order to use scalar diffraction to compute the field radiated by an aperture antenna? Can we use it to compute the directional pattern of the eye’s pupil (d ≃0.2 cm) in the visible part of the spectrum (λ = 0.35 to 0.7 μm)? What would the beamwidth of the eye’s directional pattern be at λ = 0.5 μm?
Answer: β ≈λ/d = 2.5 × 10−4 rad = 0.86′ (arc minute, with 60′ = 1◦). (See EM.) 9-9 ANTENNA ARRAYS 435 Module 9.4 Large Parabolic Reflector For any specified reflector diameter d (such that d ≥2λ) and illumination taper factor α, this module displays the pattern of the radiated field and computes the associated beamwidth and directivity.
9-9 Antenna Arrays AM broadcast services operate in the 535 to 1605 kHz band.
The antennas they use are vertical dipoles mounted along tall towers.
The antennas range in height from λ/6 to 5λ/8, depending on the operating characteristics desired and other considerations. Their physical heights vary from 46 m (150 ft) to 274 m (900 ft); the wavelength at 1 MHz, approximately in the middle of the AM band, is 300 m. Because the field radiated by a single dipole is uniform in the horizontal plane (as discussed in Sections 9-1 and 9-3), it is not possible to direct the horizontal pattern along specific directions of interest, unless twoormoreantennatowersareusedsimultaneously. Directions of interest may include cities serviced by the AM station, and directionstoavoidmayincludeareasservicedbyanotherstation operating at the same frequency (thereby avoiding interference effects). When two or more antennas are used together, the combination is called an antenna array.
The AM broadcast antenna array is only one example of the many antenna arrays used in communication systems and radar applications. Antenna arrays provide the antenna designer the flexibility to obtain high directivity, narrow beams, low side lobes, steerable beams, and shaped antenna patterns starting from very simple antenna elements. Figure 9-25 shows a very large radar system consisting of a transmitter array composed of 5,184 individual dipole antenna elements and a receiver array composed of 4,660 elements. The radar system, part of the Space Surveillance Network operated by the U.S. Air Force, operates at 442 MHz and transmits a combined peak power of 30 MW!
Although an array need not consist of similar radiating elements, most arrays actually use identical elements, such as dipoles, slots, horn antennas, or parabolic dishes.
The 436 CHAPTER 9 RADIATION AND ANTENNAS 9 stories high 6 stories high Figure 9-25 The AN/FPS-85 Phased Array Radar Facility in the Florida panhandle, near the city of Freeport. A several-mile no-fly zone surrounds the radar installation as a safety concern for electroexplosive devices, such as ejection seats and munitions, carried on military aircraft.
antenna elements composing an array may be arranged in various configurations, but the most common are the linear one-dimensional configuration—wherein the elements are arranged along a straight line—and the two-dimensional lattice configuration in which the elements sit on a planar grid. The desired shape of the far-field radiation pattern of the array can be synthesized by controlling the relative amplitudes of the array elements’ excitations.
▶Also, through the use of electronically controlled solid-state phase shifters, the beam direction of the antenna array can be steered electronically by controlling the relative phases of the array elements. ◀ This flexibility of the array antenna has led to numerous applications, including electronic steering and multiple-beam generation.
The purpose of this and the next two sections is to introduce the reader to the basic principles of array theory and design techniques used in shaping the antenna pattern and steering the main lobe. The presentation is confined to the one-dimensional linear array with equal spacing between adjacent elements.
A linear array of N identical radiators is arranged along the z axis as shown in Fig. 9-26. The radiators are fed by a common oscillator through a branching network. In each branch, an attenuator (or amplifier) and phase shifter are inserted in series to control the amplitude and phase of the signal feeding the antenna element in that branch.
In the far-field region of any radiating element, the element electric-field intensity Ee(R, θ, φ) may be expressed as a product of two functions, the spherical propagation factor e−jkR/R, which accounts for the dependence on the range R, and fe(θ, φ), which accounts for the directional dependence of the element’s electric field. Thus, for an isolated element, the radiated field is Ee(R, θ, φ) = e−jkR R fe(θ, φ), (9.93) and the corresponding power density Se is Se(R, θ, φ) = 1 2η0 | Ee(R, θ, φ)|2 = 1 2η0R2 | fe(θ, φ)|2.
(9.94) 9-9 ANTENNA ARRAYS 437 ψN–1 aN–1 ψN–2 aN–2 ψi ai ψ1 a1 ψ0 a0 Phase shifters Amplifiers (or attenuators) Antenna elements (a) Array elements with individual amplitude and phase control (b) Array geometry relative to observation point Element N – 1 Element N – 2 Element 1 Element 0 Element i z y d d θ RN–1 R0 Ri Q = (R0, θ, φ) (N – 1)d Figure 9-26 Linear-array configuration and geometry.
Hence, for the array shown in Fig. 9-26(b), the far-zone field due to element i at range Ri from observation point Q is Ei(Ri, θ, φ) = Ai e−jkRi Ri fe(θ, φ), (9.95) where Ai = aiejψi is a complex feeding coefficient represent-ing the amplitude ai and phase ψi of the excitation giving rise to Ei, relative to a reference excitation. In practice, the excitation of one of the elements is used as reference. Note that Ri and Ai may be different for different elements in the array, but fe(θ, φ) is the same for all of them because they are all identical and hence exhibit identical directional patterns.
The total field at the observation point Q(R0, θ, φ) is the sum of the fields due to the N elements: E(R0, θ, φ) = N−1 !
i=0 Ei(Ri, θ, φ) = N−1 !
i=0 Ai e−jkRi Ri fe(θ, φ), (9.96) where R0 denotes the range of Q from the center of the coordinate system, chosen to be at the location of the zeroth element. To satisfy the far-field condition given by Eq. (9.73) for an array of length l = (N −1)d, where d is the interelement spacing, the range R0 should be sufficiently large to satisfy R0 ≥2l2 λ = 2(N −1)2d2 λ .
(9.97) This condition allows us to ignore differences in the distances from Q to the individual elements as far as the magnitudes of the radiated fields are concerned. Thus, we can set Ri = R0 in the denominator in Eq. (9.96) for all i. With regard to the phase part of the propagation factor, we can use the parallel-ray approximation given by Ri ≈R0 −zi cos θ = R0 −id cos θ, (9.98) where zi = id is the distance between the ith element and the zeroth element (Fig. 9-27).
Employing these two approximations in Eq. (9.96) leads to E(R0, θ, φ) = fe(θ, φ) e−jkR0 R0 · N−1 !
i=0 Aiejikd cos θ , (9.99) and the corresponding array-antenna power density is given by S(R0, θ, φ) = 1 2η0 | E(R0, θ, φ)|2 = 1 2η0R2 0 | fe(θ, φ)|2 " " " " " N−1 !
i=0 Aiejikd cos θ " " " " " 2 = Se(R0, θ, φ) " " " " " N−1 !
i=0 Aiejikd cos θ " " " " " 2 , (9.100) 438 CHAPTER 9 RADIATION AND ANTENNAS Element 0 z y d id θ R0 Element 1 R1 RN–1 zN–1 zN–2 zi Element i Element N − 1 Element N − 2 Q } id cos θ ≈ Figure 9-27 The rays between the elements and a faraway observation point are approximately parallel lines. Hence, the distance Ri ≈R0 −id cos θ.
where use was made of Eq. (9.94). This expression is a product of two factors.
The first factor, Se(R0, θ, φ), is the power density of the energy radiated by an individual element, and the second, called the array factor, is a function of the positions of the individual elements and their feeding coefficients, but not a function of the specific type of radiators used.
▶The array factor represents the far-field radiation intensity of the N elements, had the elements been isotropic radiators. ◀ Denoting the array factor by Fa(θ) = " " " " " N−1 !
i=0 Aiejikd cos θ " " " " " 2 , (9.101) the power density of the antenna array is then written as S(R0, θ, φ) = Se(R0, θ, φ) Fa(θ).
(9.102) This equation demonstrates the pattern multiplication prin-ciple. It allows us to find the far-field power density of the antenna array by first computing the far-field power pattern with the array elements replaced with isotropic radiators, which yields the array factor Fa(θ), and then multiplying the result by Se(R0, θ, φ), the power density for a single element (which is the same for all elements).
ThefeedingcoefficientAi is, ingeneral, acomplexamplitude consisting of an amplitude factor ai and a phase factor ψi: Ai = aiejψi.
(9.103) Insertion of Eq. (9.103) into Eq. (9.101) leads to Fa(θ) = " " " " " N−1 !
i=0 aiejψiejikd cos θ " " " " " 2 .
(9.104) The array factor is governed by two input functions: the array amplitude distribution given by the ai’s and the array phase distribution given by the ψi’s.
▶The amplitude distribution serves to control the shape of the array radiation pattern, while the phase distribution can be used to steer its direction. ◀ Example 9-5: Array of Two Vertical Dipoles An AM radio station uses two vertically oriented half-wave dipoles separated by a distance of λ/2, as shown in Fig. 9-28(a).
The vector from the location of the first dipole to the location of the second dipole points toward the east. The two dipoles are fed with equal-amplitude excitations, and the dipole farther east is excited with a phase shift of −π/2 relative to the other one. Find and plot the antenna pattern of the antenna array in the horizontal plane.
Solution: The array factor given by Eq. (9.104) was derived for radiators arranged along the z axis. To keep the coordinate system the same, we choose the easterly direction to be the z axis as shown in Fig. 9-28(b), and we place the first dipole at z = −λ/4 and the second at z = λ/4. A dipole radiates uniformly in the plane perpendicular to its axis, which in this case is the horizontal plane. Hence, Se = S0 for all angles θ in Fig. 9-28(b), where S0 is the maximum value of the power 9-9 ANTENNA ARRAYS 439 phase shifter λ/2 λ/2 a0 = 1 ψ0 = 0 a1 = 1 ψ1 = −π/2 (a) Dipole array (c) Horizontal-plane pattern (b) Observation plane z y F(θ) θ x z y (North) (East) (South) θ Figure 9-28 Two half-wave dipole array of Example 9-5.
density radiated by each dipole individually. Consequently, the power density radiated by the two-dipole array is S(R, θ) = S0 Fa(θ).
For two elements separated by d = λ/2 and excited with equal amplitudes (a0 = a1 = 1) and with phase angles ψ0 = 0 and ψ1 = −π/2, Eq. (9.104) becomes Fa(θ) = " " " " " 1 !
i=0 aiejψiejikd cos θ " " " " " 2 = " " "1 + e−jπ/2ej(2π/λ)(λ/2) cos θ" " " 2 = " " "1 + ej(π cos θ−π/2)" " " 2 .
A function of the form |1 + ejx|2 can be evaluated by factoring out ejx/2 from both terms: |1 + ejx|2 = |ejx/2(e−jx/2 + ejx/2)|2 = |ejx/2|2 |e−jx/2 + ejx/2|2 = |ejx/2|2 " " " "2 [e−jx/2 + ejx/2] 2 " " " " 2 .
The absolute value of ejx/2 is 1, and we recognize the function inside the square bracket as cos(x/2). Hence, |1 + ejx|2 = 4 cos2 x 2 .
Applying this result to the expression for Fa(θ), we have Fa(θ) = 4 cos2 π 2 cos θ −π 4 .
The power density radiated by the array is then S(R, θ) = S0Fa(θ) = 4S0 cos2 π 2 cos θ −π 4 .
This function has a maximum value Smax = 4S0, and it occurs when the argument of the cosine function is equal to zero. Thus, π 2 cos θ −π 4 = 0, which leads to the solution: θ = 60◦.
Upon normalizing S(R, θ) by its maximum value, we obtain the normalized radiation intensity given by F(θ) = S(R, θ) Smax = cos2 π 2 cos θ −π 4 .
The pattern of F(θ) is shown in Fig. 9-28(c).
440 CHAPTER 9 RADIATION AND ANTENNAS Module 9.5 Two-dipole Array Given two vertical dipoles, the user can specify their individual lengths and current maxima, as well as the distance between them and the phase difference between their current excitations. The module generates plots of the field and power patterns in the far-zone and calculates the maximum directivity and total radiated power.
9-9 ANTENNA ARRAYS 441 Module 9.6 Detailed Analysis of Two-Dipole Ar-ray This module extends the display and computational capabilities of Module 9.5 by offering plots for individual components of E and H at any range from the antenna, including the near-field.
Example 9-6: Pattern Synthesis In Example 9-5, we were given the array parameters a0, a1, ψ0, ψ1, and d, and we were then asked to determine the pattern of the two-element dipole array. We now consider the reverse process; given specifications on the desired pattern, we specify the array parameters to meet those specifications.
Given two vertical dipoles, as depicted in Fig. 9-28(b), specify the array parameters such that the array exhibits maximum radiation toward the east and no radiation toward the north or south.
Solution: From Example 9-5, we established that because eachdipoleradiatesequallyalongalldirectionsinthey–z plane, the radiation pattern of the two-dipole array in that plane is governed solely by the array factor Fa(θ). The shape of the pattern of the array factor depends on three parameters: the amplitude ratio a1/a0, the phase difference ψ1 −ψ0, and the spacing d [Fig. 9-29(a)]. For convenience, we choose a0 = 1 and ψ0 = 0. Accordingly, Eq. (9.101) becomes Fa(θ) = " " " " " 1 !
i=0 aiejψiejikd cos θ " " " " " 2 = |1 + a1ejψ1ej(2πd/λ) cos θ|2.
Next, weconsiderthespecificationthatFa beequaltozerowhen θ = 90◦[north and south directions in Fig. 9-29(a)]. For any observation point on the y axis, the ranges R0 and R1 shown in Fig. 9-29(a) are equal, which means that the propagation phases associated with the time travel of the waves radiated by the two dipoles to that point are identical. Hence, to satisfy the stated condition, we need to choose a1 = a0 and ψ1 = ±π.
With these choices, the signals radiated by the two dipoles have equal amplitudes and opposite phases, thereby interfering destructively. This conclusion can be ascertained by evaluating the array factor at θ = 90◦, with a0 = a1 = 1 and ψ1 = ±π: Fa(θ = 90◦) = |1 + 1e±jπ|2 = |1 −1| = 0.
(b) Array pattern (a) Array arrangement z y θ F(θ) –y z (East) (North) R0 a0 = 1 ψ0 = 0 a1 ψ1 d R1 Figure 9-29 (a) Two vertical dipoles separated by a distance d along the z axis; (b) normalized array pattern in the y–z plane for a0 = a1 = 1, ψ1 = ψ0 = −π, and d = λ/2.
The two values of ψ1, namely π and −π, lead to the same solution for the value of the spacing d to meet the specification that the array radiation pattern is maximum toward the east, corresponding to θ = 0◦. Let us choose ψ1 = −π and examine the array factor at θ = 0◦: Fa(θ = 0) = |1 + e−jπej2πd/λ|2 = |1 + ej(−π+2πd/λ)|2.
442 CHAPTER 9 RADIATION AND ANTENNAS For Fa(θ = 0) to be a maximum, we require the phase angle of the second term to be zero or a multiple of 2π. That is, −π + 2πd λ = 2nπ, or d = (2n + 1)λ 2 , n = 0, 1, 2, . . .
In summary, the two-dipole array will meet the given specifi-cations if a0 = a1, ψ1 −ψ0 = −π, and d = (2n + 1)λ/2.
For d = λ/2, the array factor is Fa(θ) = |1 + e−jπejπ cos θ|2 = |1 −ejπ cos θ|2 = " " " " "2je−j(π/2) cos θ ej(π/2) cos θ −e−j(π/2) cos θ 2j " " " " " 2 = 4 sin2 π 2 cos θ .
The array factor has a maximum value of 4, which is the maximum level attainable from a two-element array with unit amplitudes. The directions along which Fa(θ) is a maximum are those corresponding to θ = 0 (east) and θ = 180◦(west), as shown in Fig. 9-29(b).
Exercise 9-14: Derive an expression for the array factor of a two-element array excited in phase with a0 = 1 and a1 = 3. The elements are positioned along the z axis and are separated by λ/2.
Answer: Fa(θ) = [10 + 6 cos(π cos θ)]. (See EM.) Exercise 9-15: An equally spaced N-element array arranged along the z axis is fed with equal amplitudes and phases; that is, Ai = 1 for i = 0, 1, . . . , (N −1).
What is the magnitude of the array factor in the broadside direction?
Answer: Fa(θ = 90◦) = N2. (See EM.) 9-10 N-Element Array with Uniform Phase Distribution We now consider an array of N elements with equal spacing d and equal-phase excitations; that is, ψi = ψ0 for i = 1, 2, . . . , (N −1). Such an array of in-phase elements is sometimes referred to as a broadside array because the main beam of the radiation pattern of its array factor is always in the direction broadside to the array axis. From Eq. (9.104), the array factor is given by Fa(θ) = " " " " "ejψ0 N−1 !
i=0 aiejikd cos θ " " " " " 2 = |ejψ0|2 " " " " " N−1 !
i=0 aiejikd cos θ " " " " " 2 = " " " " " N−1 !
i=0 aiejikd cos θ " " " " " 2 .
(9.105) The phase difference between the fields radiated by adjacent elements is γ = kd cos θ = 2πd λ cos θ.
(9.106) In terms of γ , Eq. (9.105) takes the compact form Fa(γ ) = " " " " " N−1 !
i=0 aiejiγ " " " " " 2 (uniform phase).
(9.107) For a uniform amplitude distribution with ai = 1 for i = 0, 1, . . . , (N −1), Eq. (9.107) becomes Fa(γ ) = |1 + ejγ + ej2γ + · · · + ej(N−1)γ |2.
(9.108) 9-10 N-ELEMENT ARRAY WITH UNIFORM PHASE DISTRIBUTION 443 This geometric series can be rewritten in a more compact form by applying the following recipe. First, we define Fa(γ ) = |fa(γ )|2, (9.109) with fa(γ ) = [1 + ejγ + ej2γ + · · · + ej(N−1)γ ].
(9.110) Next, we multiply fa(γ ) by ejγ to obtain fa(γ ) ejγ = (ejγ + ej2γ + · · · + ejNγ ).
(9.111) Subtracting Eq. (9.111) from Eq. (9.110) gives fa(γ ) (1 −ejγ ) = 1 −ejNγ , (9.112) which, in turn, gives fa(γ ) = 1 −ejNγ 1 −ejγ = ejNγ/2 ejγ/2 (e−jNγ/2 −ejNγ/2) (e−jγ/2 −ejγ/2) = ej(N−1)γ/2 sin(Nγ/2) sin(γ/2) .
(9.113) After multiplying fa(γ ) by its complex conjugate, we obtain the result: Fa(γ ) = sin2(Nγ/2) sin2(γ/2) .
(9.114) (uniform amplitude and phase) From Eq. (9.108), Fa(γ ) is maximum when all terms are 1, which occurs when γ = 0 (or equivalently, θ = π/2).
Moreover, Fa(0) = N2. Hence, the normalized array factor is given by Fan(γ ) = Fa(γ ) Fa,max nn (9.115) = sin2(Nγ/2) N2 sin2(γ/2) = sin2 Nπd λ cos θ N2 sin2 πd λ cos θ .
(9.116) 0 dB –30 –20 –10 Fan(θ) θ N = 6 d = λ/2 d z 17.2° 1 1 1 1 1 1 –3 dB Broadside (θ = 90°) Figure 9-30 Normalized array pattern of a uniformly excited six-element array with interelement spacing d = λ/2.
A polar plot of Fan(θ) is shown in Fig. 9-30 for N = 6 and d = λ/2.
The reader is reminded that this is a plot of the radiation pattern of the array factor alone; the pattern for the antenna array is equal to the product of this pattern and that of a single element, as discussed earlier in connection with the pattern multiplication principle.
Example 9-7: Multiple-Beam Array Obtain an expression for the array factor of a two-element array with equal excitation and a separation d = 7λ/2, and then plot the array pattern.
Solution: The array factor of a two-element array (N = 2) 444 CHAPTER 9 RADIATION AND ANTENNAS 0 dB –30 –20 –10 Fan(θ) θ z 1 1 d = 7λ/2 8.2° –3 dB Broadside (θ = 90°) Figure 9-31 Normalized array pattern of a two-element array with spacing d = 7λ/2.
with equal excitation (a0 = a1 = 1) is given by Fa(γ ) = " " " " " 1 !
i=0 aiejiγ " " " " " 2 = |1 + ejγ |2, = |ejγ/2(e−jγ/2 + ejγ/2)|2 = |ejγ/2|2 |e−jγ/2 + ejγ/2|2 = 4 cos2(γ/2), where γ = (2πd/λ) cos θ.
The normalized array pattern, shown in Fig. 9-31, consists of seven beams, all with the same peak value, but not the same angular width. The number of beams in the angular range between θ = 0 and θ = π is equal to the separation between the array elements, d, measured in units of λ/2.
9-11 Electronic Scanning of Arrays The discussion in the preceding section was concerned with uniform-phase arrays, in which the phases of the feeding coefficients, ψ0 to ψN−1, are all equal. In this section, we examine the use of phase delay between adjacent elements as a tool to electronically steer the direction of the array-antenna beam from broadside at θ = 90◦to any desired angle θ0. In addition to eliminating the need to mechanically steer an antenna to change its beam’s direction, electronic steering allows beam scanning at very fast rates.
▶Electronic steering is achieved by applying a linear phase distribution across the array: ψ0 = 0, ψ1 = −δ, ψ2 = −2δ, etc. ◀ As shown in Fig. 9-32, the phase of the ith element, relative to that of the zeroth element, is ψi = −iδ, (9.117) where δ is the incremental phase delay between adjacent elements. Use of Eq. (9.117) in Eq. (9.104) leads to Fa(θ) = " " " " " N−1 !
i=0 aie−jiδejikd cos θ " " " " " 2 = " " " " " N−1 !
i=0 aieji(kd cos θ−δ) " " " " " 2 = " " " " " N−1 !
i=0 aieji γ ′ " " " " " 2 = Fa(γ ′), (9.118) where we introduced a new variable given by γ ′ = kd cos θ −δ.
(9.119) For reasons that become clear later, we define the phase shift δ in terms of an angle θ0, which we call the scan angle, as follows: δ = kd cos θ0.
(9.120) Hence, γ ′ becomes γ ′ = kd(cos θ −cos θ0).
(9.121) 9-11 ELECTRONIC SCANNING OF ARRAYS 445 –(N – 1)δ –iδ –2δ –δ N – 1 N – 2 i 2 1 0 θ R0 Q y z –(N – 2)δ Figure 9-32 The application of linear phase.
The array factor given by Eq. (9.118) has the same functional form as the array factor developed earlier for the uniform-phase array [see Eq. (9.107)], except that γ is replaced with γ ′. Hence: ▶Regardless of the amplitude distribution across an array, its array factor Fa(γ ′) when excited by a linear-phase distribution can be obtained from Fa(γ ), the expression developed for the array assuming a uniform-phase distribution, by replacing γ with γ ′. ◀ If the amplitude distribution is symmetrical with respect to the array center, the array factor Fa(γ ′) is maximum when its argument γ ′ = 0.
When the phase is uniform (δ = 0), this condition corresponds to the direction θ = 90◦, which is why the uniform-phase arrangement is called a broadside array.
According to Eq. (9.121), in a linearly phased array, γ ′ = 0 when θ = θ0. Thus, by applying linear phase across the array, the array pattern is shifted along the cos θ axis by an amount cos θ0, and the direction of maximum radiation is steered from the broadside direction (θ = 90◦) to the direction θ = θ0. To steer the beam all the way to the end-fire direction (θ = 0), the incremental phase shift δ should be equal to kd radians.
9-11.1 Uniform-Amplitude Excitation To illustrate the process with an example, consider the case of the N-element array excited by a uniform-amplitude distribution. Its normalized array factor is given by Eq. (9.116).
Upon replacing γ with γ ′, we have Fan(γ ′) = sin2(Nγ ′/2) N2 sin2(γ ′/2) , (9.122) with γ ′ as defined by Eq. (9.121). For an array with N = 10 and d = λ/2, plots of the main lobe of Fan(θ) are shown in Fig. 9-33 for θ0 = 0◦, 45◦, and 90◦. We note that the half-power beamwidth increases as the array beam is steered from broadside to end fire.
9-11.2 Array Feeding According to the foregoing discussion, to steer the antenna beam to an angle θ0, two conditions must be met: (1) the phase distribution must be linear across the array, and (2) the magnitude of the incremental phase delay δ must satisfy Eq. (9.120). The combination of these two conditions provides the necessary tilting of the beam from θ = 90◦(broadside) to θ = θ0.
This can be accomplished by controlling the excitation of each radiating element individually through the use of electronically controlled phase shifters. Alternatively, a technique known as frequency scanning can be used to provide control of the phases of all the elements simultaneously.
Figure 9-34 shows an example of a simple feeding arrangement employed in frequency scanning arrays. A common feed point is connected to the radiating elements through transmission lines of varying lengths. Relative to the zeroth element, the path between the common feed point and a radiating element is longer by l for the first element, by 2l for the second, and by 3l for the third. Thus, the path length for the ith element is li = il + l0, (9.123) where l0 is the path length of the zeroth element. Waves of frequency f propagating through a transmission line of length li are characterized by a phase factor e−jβli, where β = 2πf/up is the phase constant of the line and up is its phase velocity.
446 CHAPTER 9 RADIATION AND ANTENNAS 0 dB –30 –20 –10 Fan(θ) θ N = 10 d = λ/2 10.2° 48.7° End-fire θ0 = 0° θ0 = 45° 15.5° –3 dB Broadside (θ = 90°) Figure 9-33 Normalized array pattern of a 10-element array with λ/2 spacing between adjacent elements. All elements are excited with equal amplitude. Through the application of linear phase across the array, the main beam can be steered from the broadside direction (θ0 = 90◦) to any scan angle θ0. Equiphase excitation corresponds to θ0 = 90◦.
Hence, the incremental phase delay of the ith element, relative to the phase of the zeroth element, is ψi(f ) = −β(li −l0) = −2π up f (li −l0) = −2πi up f l.
(9.124) Suppose that at a given reference frequency f0 we choose the incremental length l such that l = n0up f0 , (9.125) where n0 is a specific positive integer. In this case, the phase delay ψ1(f0) becomes ψ1(f0) = −2π f0l up = −2n0π (9.126) and, similarly, ψ2(f0) = −4n0π and ψ3(f0) = −6n0π. That is, at f0 all the elements have equal phase (within multiples of 2π) and the array radiates in the broadside direction. If f is changed to f0 + f , the new phase shift of the first element relative to the zeroth element is ψ1(f0 + f ) = −2π up (f0 + f )l = −2πf0l up − 2πl up f = −2n0π −2n0π
f f0 = −2n0π −δ, (9.127) where use was made of Eq. (9.125) and δ is defined as δ = 2n0π
f f0 .
(9.128) Similarly, ψ2(f0 + f ) = 2ψ1 and ψ3(f0 + f ) = 3ψ1.
Ignoring the factor of 2π and its multiples (since they exercise no influence on the relative phases of the radiated fields), we see that the incremental phase shifts are directly proportional to the fractional frequency deviation (
f/f0). Thus, in an array with N elements, controlling f provides a direct control of δ, whichinturncontrolsthescanangleθ0 accordingtoEq.(9.120).
Equating Eq. (9.120) to Eq. (9.128) and then solving for cos θ0 leads to cos θ0 = 2n0π kd
f f0 (9.129) As f is changed from f0 to f0 + f , k = 2π/λ = 2πf/c also changes with frequency. However, if f/f0 is small, we may treat k as a constant equal to 2πf0/c; the error in cos θ0 resulting from the use of this approximation in Eq. (9.129) is on the order of f/f0.
Example 9-8: Electronic Steering Design a steerable six-element array with the following specifications: 9-11 ELECTRONIC SCANNING OF ARRAYS 447 Module 9.7 N-Element Array This module displays the far-field patterns of an array of N identical, equally spaced antennas, with N being a selectable integer between 1 and 6. Two types of antennas can be simulated: λ/2-dipoles and parabolic reflectors. The module provides visual examples of the pattern multiplication principle.
1. All elements are excited with equal amplitudes.
2. At f0 = 10 GHz, the array radiates in the broadside direction, and the interelement spacing d = λ0/2, where λ0 = c/f0 = 3 cm.
3. The array pattern is to be electronically steerable in the elevation plane over the angular range extending between θ0 = 30◦and θ0 = 150◦.
4. The antenna array is fed by a voltage-controlled oscillator whose frequency can be varied over the range from 9.5 to 10.5 GHz.
5. The array uses a feeding arrangement of the type shown in Fig. 9-34, and the transmission lines have a phase velocity up = 0.8c.
Solution: The array is to be steerable from θ0 = 30◦to θ0 = 150◦ (Fig.
9-35).
For θ0 = 30◦ and kd = (2π/λ0)(λ0/2) = π, Eq. (9.129) gives 0.87 = 2n0
f f0 .
(9.130) We are given that f0 = 10 GHz and the oscillator frequency can be varied between (f0 −0.5 GHz) and (f0 + 0.5 GHz). Thus, fmax = 0.5 GHz. To satisfy Eq. (9.130), we need to choose n0 such that f is as close as possible to, but not larger than, fmax. Solving Eq. (9.130) for n0 with f = fmax gives n0 = 0.87 2 f0 fmax = 8.7.
448 CHAPTER 9 RADIATION AND ANTENNAS l1 = l0 + l l2 = l0 + 2l l3 = l0 + 3l l0 l Figure 9-34 An example of a feeding arrangement for frequency-scanned arrays.
z θ0 = 30◦ θ0 = 150◦ θ0 = 90◦ (broadside) d Figure 9-35 Steerable six-element array (Example 9-8).
Since n0 is not an integer, we need to modify its value by rounding it upward to the next whole-integer value. Hence, we set n0 = 9.
Application of Eq. (9.125) specifies the magnitude of the incremental length l: l = n0up f0 = 9 × 0.8 × 3 × 108 1010 = 21.6 cm.
In summary, with N = 6 and kd = π, Eq. (9.122) becomes: Fan(γ ′) = sin2(3γ ′) 36 sin2(γ ′/2) , with γ ′ = kd(cos θ −cos θ0) = π(cos θ −cos θ0), and cos θ0 = 2n0π kd
f f0 = 18 f −10 GHz 10 GHz .
(9.131) The shape of the array pattern is similar to that shown in Fig. 9-30, and its main-beam direction is along θ = θ0.
For f = f0 = 10 GHz, θ0 = 90◦(broadside direction); for f = 10.48 GHz, θ0 = 30◦; and for f = 9.52 GHz, θ0 = 150◦.
For any other value of θ0 between 30◦and 150◦, Eq. (9.131) provides the means for calculating the required value of the oscillator frequency f .
Concept Question 9-11: Why are antenna arrays use-ful? Give examples of typical applications.
Concept Question 9-12: Explain how the pattern multiplication principle is used to compute the radiation pattern of an antenna array.
Concept Question 9-13: For a linear array, what roles do the array amplitudes and phases play?
Concept Question 9-14: Explain how electronic beam steering is accomplished.
Concept Question 9-15: Why is frequency scanning an attractive technique for steering the beam of an antenna array?
9-11 ELECTRONIC SCANNING OF ARRAYS 449 Module 9.8 Uniform Dipole Array For an array of up to 50 identical vertical dipoles of selectable length and current maximum, excited with incremental phase delay δ between adjacent elements, the module displays the elevation and azimuthal patterns of the array. By varying δ, the array pattern can be steered in the horizontal plane.
450 CHAPTER 9 RADIATION AND ANTENNAS Chapter 9 Summary Concepts • An antenna is a transducer between a guided wave propagating on a transmission line and an EM wave propagating in an unbounded medium, or vice versa.
• Except for some solid-state antennas composed of non-linear semiconductors or ferrite materials, antennas are reciprocal devices; they exhibit the same radiation patterns for transmission as for reception.
• In the far-field region of an antenna, the radiated energy is approximately a plane wave.
• The electric field radiated by current antennas, such as wires, is equal to the sum of the electric fields radiated by all the Hertzian dipoles making up the antenna.
• The radiation resistance Rrad of a half-wave dipole is 73 , which can be easily matched to a transmission line.
• The directional properties of an antenna are described by its radiation pattern, directivity, pattern solid angle, and half-power beamwidth.
• The Friis transmission formula relates the power received by an antenna to that transmitted by another antenna at a specified distance away.
• The far-zone electric field radiated by a large aperture (measured in wavelengths) is related to the field distribution across the aperture by Kirchhoff’s scalar diffraction theory.
A uniform aperture distribution produces a far-field pattern with the narrowest possible beamwidth.
• By controlling the amplitudes and phases of the individual elements of an antenna array, it is possible to shape the antenna pattern and to steer the direction of the beam electronically.
• The pattern of an array of identical elements is equal to the product of the array factor and the antenna pattern of an individual antenna element.
Important Terms Provide definitions or explain the meaning of the following terms: 3 dB beamwidth antenna antenna array antenna directivity D antenna gain G antenna input impedance antenna pattern antenna polarization aperture distribution array distribution array factor Fa(θ, φ) azimuth angle beamwidth β broadside direction effective area (effective aperture) Ae electronic steering elevation and azimuth planes elevation angle end-fire direction far-field (or far-zone) region feeding coefficient frequency scanning Friis transmission formula half-power beamwidth isotropic antenna linear phase distribution loss resistance Rloss null beamwidth pattern multiplication principle pattern solid angle p power density S(R, θ, φ) Poynting vector principal planes radiation efficiency ξ radiation intensity (normalized) F(θ, φ) radiation lobes radiation pattern radiation resistance Rrad reciprocal scan angle short dipole (Hertzian dipole) signal-to-noise ratio Sn solid angle spherical propagation factor steradian system noise temperature Tsys tapered aperture distribution zenith angle CHAPTER 9 SUMMARY 451 Mathematical and Physical Models Antenna Properties Pattern solid angle p = 4π F(θ, φ) d Effective area Ae = λ2D 4π Directivity D = 4π p Far-field distance R > 2d2 λ Gain G = ξD, ξ = Prad Prad + Ploss Short Dipole (l ≪λ) λ/2 Dipole Eθ = jI0lkη0 4π e−jkR R sin θ Eθ = j 60I0 cos[(π/2) cos θ] sin θ e−jkR R Hφ = Eθ η0 Hφ = Eθ η0 S(R, θ) = η0k2I 2 0 l2 32π2R2 sin2 θ S(R, θ) = 15I 2 0 πR2 cos2[(π/2) cos θ] sin2 θ D = 1.5 D = 1.64 β = 90◦ β = 78◦ Rrad = 80π2(l/λ)2 Rrad ≈73 Friis Transmission Formula Prec Pt = GtGr λ 4πR 2 Ft(θt, φt) Fr(θr, φr) Antenna Arrays Multiplication Principle S(R0, θ, φ) = Se(R0, θ, φ) Fa(θ) Uniform Phase Fa(γ ) = " " " " " N−1 !
i=0 aiejiγ " " " " " 2 , with γ = kd cos θ = 2πd λ cos θ Linear Phase Fa(θ) = " " " " " N−1 !
i=0 aieji γ ′ " " " " " 2 , with γ ′ = kd cos θ −δ Rectangular Aperture (Uniform) S(R, θ) = S0 sinc2(πlx sin θ/λ), x-z plane S(R, θ) = S0 sinc2(πly sin θ/λ), y-z plane βxz = 0.88 λ lx , βyz = 0.88 λ ly D = 4πAe λ2 ≈4πAp λ2 452 CHAPTER 9 RADIATION AND ANTENNAS PROBLEMS Sections 9-1 and 9-2: Hertizan Dipole and Antenna Radiation Characteristics ∗9.1 A center-fed Hertzian dipole is excited by a current I0 = 20 A.
If the dipole is λ/50 in length, determine the maximum radiated power density at a distance of 1 km.
9.2 A 50 cm long center-fed dipole directed along the z direction and located at the origin is excited by a 1 MHz source.
If the current amplitude is I0 = 10 A, determine: (a) The power density radiated at 2 km along the broadside of the antenna pattern.
(b) The fraction of the total power radiated within the sector between θ = 85◦and θ = 95◦?
9.3 A 1 m long dipole is excited by a 1 MHz current with an amplitude of 12 A. What is the average power density radiated by the dipole at a distance of 5 km in a direction that is 45◦ from the dipole axis?
∗9.4 Determine the following: (a) The direction of maximum radiation.
(b) Directivity.
(c) Beam solid angle.
(d) Half-power beamwidth in the x–z plane.
for an antenna whose normalized radiation intensity is given by F(θ, φ) = 1 for 0 ≤θ ≤60◦and 0 ≤φ ≤2π 0 elsewhere.
Suggestion: Sketch the pattern prior to calculating the desired quantities.
9.5 Repeat Problem 9.4 for an antenna with F(θ, φ) = ⎧ ⎨ ⎩ sin2 θ cos2 φ for 0 ≤θ ≤π and −π/2 ≤φ ≤π/2 0 elsewhere 9.6 A 2 m long center-fed dipole antenna operates in the AM broadcast band at 1 MHz. The dipole is made of copper wire with a radius of 1 mm.
(a) Determine the radiation efficiency of the antenna.
∗(b) What is the antenna gain in decibels?
∗Answer(s) available in Appendix D.
(c) What antenna current is required so that the antenna will radiate 80 W, and how much power will the generator have to supply to the antenna?
9.7 Repeat Problem 9.6 for a 20 cm long antenna operating at 5 MHz.
9.8 Determine the frequency dependence of the radiation efficiency of the short dipole, and plot it over the range from 600 kHz to 60 MHz. The dipole is made of copper, its length is 10 cm, and its circular cross section has a radius of 1 mm.
∗9.9 An antenna with a pattern solid angle of 1.5 (sr) radiates 60W of power. At a range of 1 km, what is the maximum power density radiated by the antenna?
9.10 An antenna with a radiation efficiency of 90% has a directivity of 7.0 dB. What is its gain in decibels?
∗9.11 The radiation pattern of a circular parabolic-reflector antenna consists of a circular major lobe with a half-power beamwidth of 3◦and a few minor lobes. Ignoring the minor lobes, obtain an estimate for the antenna directivity in dB.
9.12 The normalized radiation intensity of a certain antenna is given by F(θ) = exp(−20θ2) for 0 ≤θ ≤π where θ is in radians. Determine: (a) The half-power beamwidth.
(b) The pattern solid angle.
(c) The antenna directivity.
Sections 9-3 and 9-4: Dipole Antennas 9.13 Repeat Problem 9.6 for a 1 m long half-wave dipole that operates in the FM/TV broadcast band at 150 MHz.
∗9.14 Assuming the loss resistance of a half-wave dipole antenna to be negligibly small and ignoring the reactance component of its antenna impedance, calculate the standing-wave ratio on a 50 transmission line connected to the dipole antenna.
9.15 A 50 cm long dipole is excited by a sinusoidally varying current with an amplitude I0 = 5 A.
Determine the total radiated power if the oscillating frequency is: (a) 1 MHz, (b) 300 MHz.
PROBLEMS 453 9.16 For a short dipole with length l such that l ≪λ, instead of treating the current ˜ I(z) as constant along the dipole, as was done in Section 9-1, a more realistic approximation that ensures the current goes to zero at the dipole ends is to describe ˜ I(z) by the triangular function ˜ I(z) = I0(1 −2z/l) for 0 ≤z ≤l/2 I0(1 + 2z/l) for −l/2 ≤z ≤0 as shown in Fig. P9.16.
Use this current distribution to determine the following: ∗(a) The far-field E(R, θ, φ).
(b) The power density S(R, θ, φ).
(c) The directivity D.
(d) The radiation resistance Rrad.
l I0 I(z) ~ Figure P9.16 Triangular current distribution on a short dipole (Problem 9.16).
9.17 For a dipole antenna of length l = 3λ/2, ∗(a) Determine the directions of maximum radiation.
(b) Obtain an expression for Smax.
(c) Generate a plot of the normalized radiation pattern F(θ).
(d) Compare your pattern with that shown in Fig. 9-17(c).
9.18 For a dipole antenna of length l = λ/4, (a) Determine the directions of maximum radiation.
(b) Obtain an expression for Smax.
(c) Generate a plot of the normalized radiation pattern F(θ).
9.19 Repeat parts (a)–(c) of Problem 9.17 for a dipole of length l = 3λ/4.
∗9.20 Repeat parts (a)–(c) of Problem 9.17 for a dipole of length l = λ.
9.21 A car antenna is a vertical monopole over a conducting surface.
Repeat Problem 9.6 for a 1 m long car antenna operating at 1 MHz. The antenna wire is made of aluminum with μc = μ0 and σc = 3.5×107 S/m, and its diameter is 1 cm.
Sections 9-5 and 9-6: Effective Area and Friis Formula 9.22 Determine the effective area of a half-wave dipole antenna at 100 MHz, and compare it with its physical cross-section if the wire diameter is 2 cm.
∗9.23 A 3 GHz line-of-sight microwave communication link consists of two lossless parabolic dish antennas, each 1 m in diameter.
If the receive antenna requires 10 nW of receive power for good reception and the distance between the antennas is 40 km, how much power should be transmitted?
9.24 Ahalf-wavedipoleTVbroadcastantennatransmits1kW at 50 MHz. What is the power received by a home television antenna with 3 dB gain if located at a distance of 30 km?
∗9.25 A 150 MHz communication link consists of two vertical half-wave dipole antennas separated by 2 km. The antennas are lossless, the signal occupies a bandwidth of 3 MHz, the system noise temperature of the receiver is 600 K, and the desired signal-to-noise ratio is 17 dB. What transmitter power is required?
9.26 Consider the communication system shown in Fig. P9.26, with all components properly matched.
If Pt = 10 W and f = 6 GHz: (a) What is the power density at the receiving antenna (assuming proper alignment of antennas)?
(b) What is the received power?
(c) If Tsys = 1,000 K and the receiver bandwidth is 20 MHz, what is the signal-to-noise ratio in decibels?
Tx Pt Gt = 20 dB Gr = 23 dB Rx Prec 20 km Figure P9.26 Communication system of Problem 9.26.
9.27 The configuration shown in Fig. P9.27 depicts two vertically oriented half-wave dipole antennas pointed towards each other, with both positioned on 100 m tall towers separated 454 CHAPTER 9 RADIATION AND ANTENNAS 5 km θi Direct Reflected h = 100 m 100 m Figure P9.27 Problem 9.27.
by a distance of 5 km. If the transit antenna is driven by a 50 MHz current with amplitude I0 = 2 A, determine: ∗(a) The power received by the receive antenna in the absence of the surface. (Assume both antennas to be lossless.) (b) The power received by the receive antenna after incorporating reflection by the ground surface, assuming the surface to be flat and to have ϵr = 9 and conductivity σ = 10−3 (S/m).
9.28 Fig. P9.28 depicts a half-wave dipole connected to a generator through a matched transmission line. The directivity d Figure P9.28 Problem 9.28.
of the dipole can be modified by placing a reflecting rod a distance d behind the dipole. What would its reflectivity in the forward direction be if: (a) d = λ/4, (b) d = λ/2.
9.29 The configuration shown in Fig. P9.29 depicts a satellite repeater with two antennas, one pointed towards the antenna of ground station 1 and the other towards the antenna of ground station 2. All antennas are parabolic dishes, antennas A1 and A4 are each 4 m in diameter, antennas A2 and A3 are each 2 m in diameter, and the distance between the satellite and each of the ground stations is 40,000 km. Upon receiving the signal by its antenna A2, the satellite transponder boosts the power gain by 80 dB and then retransmits the signal to A4. The system operates at 10 GHz with Pt = 1 kW. Determine the received power Pr. Assume all antennas to be lossless.
Sections 9-7 and 9-8: Radiation by Apertures ∗9.30 A uniformly illuminated aperture is of length lx = 20λ.
Determine the beamwidth between first nulls in the x–z plane.
9.31 The 10 dB beamwidth is the beam size between the angles at which F(θ) is 10 dB below its peak value. Determine the 10 dB beamwidth in the x–z plane for a uniformly illuminated aperture with length lx = 10λ.
∗9.32 A uniformly illuminated rectangular aperture situated in the x–y plane is 2 m high (along x) and 1 m wide (along y). If f = 10 GHz, determine the following: PROBLEMS 455 G Station 1 Satellite repeater Pt Pr A1 A4 Station 2 A3 A2 Figure P9.29 Satellite repeater system.
(a) The beamwidths of the radiation pattern in the elevation plane (x–z plane) and the azimuth plane (y–z plane).
(b) The antenna directivity D in decibels.
9.33 An antenna with a circular aperture has a circular beam with a beamwidth of 3◦at 20 GHz.
(a) What is the antenna directivity in dB?
(b) If the antenna area is doubled, what will be the new directivity and new beamwidth?
(c) If the aperture is kept the same as in (a), but the frequency is doubled to 40 GHz, what will the directivity and beamwidth become then?
9.34 Compare directivity Dant of a 1 m diameter antenna aperture operating at 10 GHz with directivity Deye of the eye’s pupil operating in the middle of the visible spectrum at λ = 0.5 μm. Treat the pupil as a circular aperture with a diameter of 4 mm.
∗9.35 A 94 GHz automobile collision-avoidance radar uses a rectangular-aperture antenna placed above the car’s bumper. If the antenna is 1 m in length and 10 cm in height, determine the following: (a) Its elevation and azimuth beamwidths.
(b) The horizontal extent of the beam at a distance of 300 m.
9.36 A microwave telescope consisting of a very sensitive receiver connected to a 100 m parabolic-dish antenna is used to measure the energy radiated by astronomical objects at 20 GHz.
If the antenna beam is directed toward the moon and the moon extends over a planar angle of 0.5◦from Earth, what fraction of the moon’s cross-section will be occupied by the beam?
Sections 9-9 through 9-11: Antenna Arrays 9.37 A two-element array consisting of two isotropic antennas separated by a distance d along the z axis is placed in a coordinate system whose z axis points eastward and whose x axis points toward the zenith. If a0 and a1 are the amplitudes of the excitations of the antennas at z = 0 and at z = d, respectively, and if δ is the phase of the excitation of the antenna at z = d relative to that of the other antenna, find the array factor and plot the pattern in the x–z plane for the following: ∗(a) a0 = a1 = 1, δ = π/4, and d = λ/2 (b) a0 = 1, a1 = 2, δ = 0, and d = λ (c) a0 = a1 = 1, δ = −π/2, and d = λ/2 (d) a0 = 1, a1 = 2, δ = π/4, and d = λ/2 (e) a0 = 1, a1 = 2, δ = π/2, and d = λ/4 9.38 If the antennas in part (a) of Problem 9.37 are parallel, vertical, Hertzian dipoles with axes along the x direction, determine the normalized radiation intensity in the x–z plane and plot it.
∗9.39 Consider the two-element dipole array of Fig. 9-29(a). If the two dipoles are excited with identical feeding coefficients (a0 = a1 = 1 and ψ0 = ψ1 = 0), choose (d/λ) such that the array factor has a maximum at θ = 45◦.
9.40 Choose (d/λ) so that the array pattern of the array of Problem 9.39 has a null, rather than a maximum, at θ = 45◦.
9.41 Find and plot the normalized array factor and determine the half-power beamwidth for a five-element linear array excited with equal phase and a uniform amplitude distribution.
The interelement spacing is 3λ/4.
456 CHAPTER 9 RADIATION AND ANTENNAS 9.42 Repeat Problem 9.41 but change the excitation to tapered amplitude distribution such that the amplitude of the central element is 1, the amplitudes of the next adjacent elements are both 0.5, and those of the outer elements are both 0.25.
9.43 Repeat Problem 9.41 for a nine-element array.
∗9.44 A five-element equally spaced linear array with d = λ/2 is excited with uniform phase and an amplitude distribution given by the binomial distribution ai = (N −1)!
i!(N −i −1)! , i = 0, 1, . . . , (N −1), where N is the number of elements. Develop an expression for the array factor.
9.45 A three-element linear array of isotropic sources aligned along the z axis has an interelement spacing of λ/4 (Fig. P9.45).
The amplitude excitation of the center element is twice that of the bottom and top elements, and the phases are −π/2 for the bottom element and π/2 for the top element, relative to that of the center element. Determine the array factor and plot it in the elevation plane.
1 –π/2 λ/4 λ/4 z 2 0 1 π/2 Figure P9.45 Three-element array of Problem 9.48.
∗9.46 An eight-element linear array with λ/2 spacing is excited withequalamplitudes. Tosteerthemainbeamtoadirection60◦ below the broadside direction, what should be the incremental phase delay between adjacent elements?
Also, give the expression for the array factor and plot the pattern.
9.47 A linear array arranged along the z axis consists of 12 equally spaced elements with d = λ/2. Choose an appropriate incremental phase delay δ so as to steer the main beam to a direction 30◦above the broadside direction.
Provide an expression for the array factor of the steered antenna and plot the pattern. From the pattern, estimate the beamwidth.
C H A P T E R 10 Satellite Communication Systems and Radar Sensors Chapter Contents Application Examples, 458 10-1 Satellite Communication Systems, 458 10-2 Satellite Transponders, 460 10-3 Communication-Link Power Budget, 462 10-4 Antenna Beams, 463 10-5 Radar Sensors, 464 10-6 Target Detection, 467 10-7 Doppler Radar, 469 10-8 Monopulse Radar, 470 Chapter 10 Summary, 473 Problems, 474 Objectives Upon learning the material presented in this chapter, you should be able to: 1. Describe the basic operation of satellite transponders.
2. Calculate the power budget for a communication link.
3. Describe how radar attains spatial and angular resolutions, calculate the maximum detectable range, and explain the tradeoff between the probabilities of detection and false alarm.
4. Calculate the Doppler frequency shift observed by a radar.
5. Describe the monopulse-radar technique.
458 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS Application Examples This concluding chapter presents overviews of satellite communication systems and radar sensors, with emphasis on their electromagnetic-related aspects.
10-1 Satellite Communication Systems Today’s world is connected by a vast communication network that provides a wide array of voice, data, and video services to both fixed and mobile terminals (Fig. 10-1). The viability and effectiveness of the network are attributed in large measure to the use of orbiting satellite systems that function as relay stations with wide area coverage of Earth’s surface. From a geostationary orbit at 35,786 km above the equator, a satellite can view over one-third of Earth’s surface and can connect any pair of points within its coverage (Fig. 10-2). The history of communication satellite engineering dates back to the late 1950s when the U.S. navy used the moon as a passive reflector to relay low-data-rate communications between Washington, D.C., and Hawaii.
The first major development involving artificial Earth satellites took place in October of 1957 when the Soviet Union launched Sputnik I and used it for 21 days to transmit (one-way) telemetry information to a ground receiving station.
This was followed by another telemetry satellite, Explorer I, launched by the United States in January 1958.
An important development took place in December of that year when the United States launched the Score satellite and used it to broadcast President Eisenhower’s Christmas message, Land mobile Ship Aircraft Satellite Land network Figure 10-1 Elements of a satellite communication network.
N S 35,786 km Geostationary orbit Equator (a) Geostationary satellite orbit (b) Worldwide coverage by three satellites spaced 120◦ apart 17.4◦ 17.4◦ 17.4◦ N Figure 10-2 Orbits of geostationary satellites.
marking the first instance of two-way voice communication via an artificial satellite.
These achievements were followed by a flurry of space activity, leading to the development of operational commu-nication satellites by many countries for both commercial and governmental services.
This section describes satellite communications links with emphasis on transmitter–receiver power calculations, propagation aspects, frequency allocations, and antenna design considerations.
A satellite is said to be in a geostationary orbit around Earth when it is in a circular orbit in a plane identical with Earth’s equatorial plane at an altitude where the orbital period is identical with Earth’s rotational period, thereby appearing stationary relative to Earth’s surface. A satellite of mass Ms in circular orbit around Earth (Fig. 10-3) is subject to two forces, the attractive gravitational force Fg and the repelling centrifugal 10-1 SATELLITE COMMUNICATION SYSTEMS 459 Re = 6,378 km at equator R0 = Re + h = 42,164 km h = 35,786 km Ms 17.4° us = 11,070 (km/hr) Geostationary orbit Satellite Me Re Re Re Earth Maximum distance = 41,679 km Figure 10-3 Satellite of mass ms in orbit around Earth. For the orbit to be geostationary, the distance R0 between the satellite and Earth’s center should be 42,164 km. At the equator, this corresponds to an altitude of 35,786 km above Earth’s surface.
force Fc. The magnitudes of these two forces are given by Fg = GMsMe R2 0 , (10.1) Fc = Msu2 s R0 = Msω2R0, (10.2) where G = 6.67×10−11 N·m2/kg2 istheuniversalgravitational constant, Me = 5.98 × 1024 kg is Earth’s mass, R0 is the distance between the satellite and the center of Earth, and us is the satellite velocity. For a rotating object, us = ωR0, where ω is its angular velocity. In order for the satellite to remain in orbit, the two opposing forces acting on it have to be equal in magnitude, or G MsMe R2 0 = Msω2R0, (10.3) which yields a solution for R0 given by R0 = GMe ω2 1/3 .
(10.4) To remain stationary with respect to Earth’s surface, the satellite’s angular velocity has to be the same as that of Earth’s own angular velocity around its own axis. Thus, ω = 2π T , (10.5) where T is the period of one sidereal day in seconds. A sidereal day, which takes into account Earth’s rotation around the sun, is equalto23hours, 56minutes, and4.1seconds. UsingEq.(10.5) in Eq. (10.4) gives R0 = GMeT 2 4π2 1/3 , (10.6) anduponusingthenumericalvaluesforT , Me, andG,weobtain the result R0 = 42,164 km. Subtracting 6,378 km for Earth’s mean radius at the equator gives an altitude of h = 35, 786 km above Earth’s surface.
From a geostationary orbit, Earth subtends an angle of 17.4◦, covering an arc of about 18,000 km along the equator, which corresponds to a longitude angle of about 160◦. With three equally spaced satellites in geostationary orbit over Earth’s equator, it is possible to achieve complete global coverage of the entire equatorial plane, with significant overlap between the beams of the three satellites. As far as coverage toward the poles, a global beam can reach Earth stations up to 81◦of latitude on either side of the equator.
Not all satellite communication systems use spacecraft that are in geostationary orbits.
Indeed, because of transmitter power limitations or other considerations, it is sometimes necessary to operate from much lower altitudes, in which case the satellite is placed in a highly elliptical orbit (to satisfy Kepler’s law) such that for part of the orbit (near its perigee) it is at a range of only a few hundred kilometers from Earth’s surface. Whereas only three geostationary satellites are needed to provide near-global coverage of Earth’s surface, a much 460 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS larger number is needed when the satellites operate from highly elliptical orbits. A good example of the latter is the Global Positioning System (GPS) described in Technology Brief 5.
10-2 Satellite Transponders A communication satellite functions as a distant repeater; it receives uplink signals from Earth stations, processes the signals, and then downlinks (retransmits) them to their intended Earth destinations.
The International Telecommunication Union has allocated specific bands for satellite communications (Table 10-1).
Of these, the bands used by the majority of U.S. commercial satellites for domestic communications are the 4/6 GHz band (3.7 to 4.2 GHz downlink and 5.925 to 6.425 GHz uplink) and the 12/14 GHz band (11.7 to 12.2 GHz downlink and 14.0 to 14.5 GHz uplink).
Each uplink and downlink segment has been allocated 500 MHz of bandwidth. By using different frequency bands for Earth-to-satellite uplink segments and for satellite-to-Earth downlink segments, the same antennas can be used for both functions while simultaneously guarding against interference between the two signals.
The downlink segment commonly uses a Table 10-1 Communications satellite frequency allocations.
Downlink Uplink Frequency Frequency Use (MHz) (MHz) Fixed Service Commercial 3,700–4,200 5,925–6,425 (C-band) Military (X-band) 7,250–7,750 7,900–8,400 Commercial (K-band) Domestic (USA) 11,700–12,200 14,000–14,500 International 10,950–11,200 27,500–31,000 Mobile Service Maritime 1,535–1,542.5 1,635–1,644 Aeronautical 1,543.5–1,558.8 1,645–1,660 Broadcast Service 2,500–2,535 2,655–2,690 11,700–12,750 Telemetry, Tracking, and Command 137–138, 401–402, 1,525–1,540 lower-frequency carrier than the uplink segment, because lower frequencies suffer lower attenuation by Earth’s atmosphere, thereby easing the requirement on satellite output power.
We shall use the 4/6 GHz band as a model to discuss the satellite-repeater operation, while keeping in mind that the functional configuration of the repeater is basically the same regardless of which specific communication band is used.
Figure 10-4 shows a generalized block diagram of a typical 12-channel repeater.
The path of each channel—from the point of reception by the antenna, transfer through the repeater, and final retransmission through the antenna—is called a transponder. The available 500 MHz bandwidth is allocated to 12 channels (transponders) of 36 MHz bandwidth per channel and4 MHzseparationbetweenchannels. Thebasicfunctionsof a transponder are: (a) isolation of neighboring radio frequency (RF) channels, (b) frequency translation, and (c) amplification.
With frequency-division multiple access (FDMA)—one of the schemes commonly used for information transmission— each transponder can accommodate thousands of individual telephone channels within its 36 MHz of bandwidth (telephone speech signals require a minimum bandwidth of 3 kHz, so frequency spacing is nominally 4 kHz per telephone channel), several TV channels (each requiring a bandwidth of 6 MHz), millions of bits of digital data, or combinations of all three.
When the same antenna is used for both transmission and reception, a duplexer is used to perform the signal separation.
Many types of duplexers are available, but among the simplest to understand is the circulator shown in Fig. 10-5. A circulator is a three-port device that uses a ferrite material placed in a magnetic field induced by a permanent magnet to achieve power flow from ports 1 to 2, 2 to 3, and 3 to 1, but not in the reverse directions. With the antenna connected to port 1, the received signal is channeled only to port 2; if port 2 is properly matched to the band-pass filter, no part of the received signal is reflected from port 2 to 3. Similarly, the transmitted signal connected to port 3 is channeled by the circulator to port 1 for transmission by the antenna.
Following the duplexer shown in Fig. 10-4, the received signal passes through a receiver band-pass filter that ensures isolation of the received signal from the transmitted signal. The receiver filter covers the bandwidth from 5.925 to 6.425 GHz, which encompasses the cumulative bandwidths of all 12 channels; the first received channel extends from 5,927 to 5,963 MHz, the second one from 5,967 to 6,003 MHz, and so on until the twelfth channel, which covers the range from 6,367 to 6,403 MHz. Tracing the signal path, the next subsystem is the wideband receiver, which consists of three elements: 10-2 SATELLITE TRANSPONDERS 461 Duplexer Receiver bandpass filter 5,900−6,425 MHz Bandpass filter 4,142−4,178 MHz Bandpass filter 3,702−3,738 MHz Antenna Received signal Transmitted signal LNA HPA HPA Amp mixer ft fr f0 f0 = 10,105 MHz ft = 3,702−4,178 MHz fr = 5,927–6,403 MHz Input multiplexer 1 2 3 .
.
.
12 Output multiplexer 1 2 3 .
.
.
12 Local oscillator Wide-band receiver Figure 10-4 Elements of a 12-channel (transponder) communications system.
Received signal Transmitted signal Antenna Circulator Signal to be transmitted Receiver 1 2 3 From output multiplexer Figure 10-5 Basic operation of a ferrite circulator.
a low-noise wideband amplifier, a frequency translator, and an output amplifier.
The frequency translator consists of a stable local oscillator, which generates a signal at frequency f0 = 10,105 MHz, connected to a nonlinear microwave mixer.
The mixer serves to convert the frequency fr of the received signal (which covers the range from 5,927 to 6,403 MHz) to a lower-frequency signal ft = f0 −fr. Thus, the lower end of the received signal frequency band gets converted from 5,927 to 4,178 MHz and the upper end gets converted from 6,403 to 3,702 MHz. This translation results in 12 channels with new frequency ranges, but whose signals carry the same information (modulation) that was present in the received signals. In principle, the receiver output signal can now be further amplified and then channeled to the antenna through the duplexer for transmission back to Earth. Instead, the receiver output signal is separated into the 12 transponder channels through a multiplexer followed by a bank of narrow band-pass filters, each covering the bandwidth of one transponder channel.
Each of the 12 channels is amplified by its own high-power amplifier (HPA), and then the 12 channels are combined by another multiplexer that feeds the combined spectrum into the 462 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS From output multiplexer (ch 1-12) From output multiplexer (ch 13-24) To receiver (ch 1-12) To receiver (ch 13-24) RHC LHC Duplexer Duplexer Figure 10-6 Polarization diversity is used to increase the number of channels from 12 to 24.
duplexer. This channel separation and recombination process is used as a safety measure against losing all 12 channels should a high-power amplifier experience total failure or degradation in performance.
The information-carrying capacity of a satellite repeater can be doubled from 12 to 24 channels over the same 500 MHz bandwidth by using polarization diversity. Instead of transmitting one channel of information over channel 1 (5,927 to 5,963 MHz), for example, the ground station transmits to the satellite two signals carrying different information and covering the same frequency band, but with different antenna polarization configurations, such as right-hand circular (RHC) and left-hand circular (LHC) polarizations.
The satellite antenna is equipped with a feed arrangement that can receive each of the two circular polarization signals individually with negligible interference between them. Two duplexers are used in this case, one connected to the RHC polarization feed and another connected to the LHC polarization feed, as illustrated in Fig. 10-6.
10-3 Communication-Link Power Budget The uplink and downlink segments of a satellite communication link (Fig. 10-7) are each governed by the Friis transmission formula (Section 9-6), which states that the power Pr received by an antenna with gain Gr due to the transmission of power Pt by an antenna with gain Gt at a range R is given by Pr = PtGtGr λ 4πR 2 .
(10.7) Free-space and atmospheric losses Receiver antenna gain Gsr Satellite Transmitter antenna gain Gst Earth station Earth station Free-space and atmospheric losses Pt Pri Uplink Downlink Figure 10-7 Satellite transponder.
This expression applies to a lossless medium, such as free space. To account for attenuation by clouds and rain in Earth’s atmosphere (when present along the propagation path), as well as absorption by certain atmospheric gases (primarily oxygen and water vapor), we rewrite Eq. (10.7) as Pri = ϒ(θ) Pr = ϒ(θ) PtGtGr λ 4πR 2 .
(10.8) Now, Pri represents the input power at the receiver with atmospheric losses taken into account, and ϒ(θ) is the one-way transmissivity of the atmosphere at zenith angle θ. In addition to its dependence on θ, ϒ(θ) is a function of the frequency of the communication link and the rain-rate conditions along the propagation path. At frequencies below 10 GHz, which include the 4/6 GHz band allocated for satellite communication, absorption by atmospheric gases is very small, as is attenuation due to clouds and rain. Consequently, the magnitude of ϒ(θ) is typically on the order of 0.5 to 1 for most conditions.
A transmissivity of 0.5 means that twice as much power needs to be transmitted (compared to the free-space case) in order to receive a specified power level.
Among the various sources of atmospheric attenuation, the most serious is rainfall, and its attenuation coefficient increases rapidly with increasing frequency. Consequently, atmospheric attenuation assumes greater importance with regard to transmitter power 10-4 ANTENNA BEAMS 463 requirements as the communication-system frequency is increased toward higher bands in the microwave region.
The noise appearing at the receiver output, Pno, consists of three contributions: (1) noise internally generated by the receiver electronics, (2) noise picked up by the antenna due to external sources, including emission by the atmosphere, and (3) noise due to thermal emission by the antenna material.
The combination of all noise sources can be represented by an equivalent system noise temperature, Tsys, defined such that Pno = GrecKTsysB, (10.9) whereK isBoltzmann’sconstantandGrec andB arethereceiver power gain and bandwidth. This output noise level is the same as would appear at the output of a noise-free receiver with input noise level Pni = Pno Grec = KTsysB.
(10.10) The signal-to-noise ratio is defined as the ratio of the signal power to the noise power at the input of an equivalent noise-free receiver. Hence, Sn = Pri Pni = ϒ(θ) PtGtGr KTsysB λ 4πR 2 .
(10.11) The performance of a communication system is governed by two sets of issues.
The first encompasses the signal-processing techniques used to encode, modulate, combine, and transmit the signal at the transmitter end and to receive, separate, demodulate, and decode the signal at the receiver end. The second set encompasses the gains and losses in the communication link, and they are represented by the signal-to-noise ratio Sn. For a given set of signal-processing techniques, Sn determines the quality of the received signal, such as the bit error rate in digital data transmission and sound and picture quality in audio and video transmissions. Very high quality signal transmission requires very high values of Sn; in broadcast-quality television by satellite, some systems are designed to provide values of Sn exceeding 50 dB (or a factor of 105).
The performance of a satellite link depends on the composite performance of the uplink and downlink segments. If either segment performs poorly, the composite performance will be poor, regardless of how good the performance of the other segment is.
10-4 Antenna Beams Whereas most Earth-station antennas are designed to provide highly directive beams (to avoid interference effects), the satellite antenna system is designed to produce beams tailored to match the areas served by the satellite. For global coverage, beamwidths of 17.4◦are required. In contrast, for transmission to and reception from a small area, beamwidths on the order of 1◦or less may be needed (Fig. 10-8).
An antenna with a beamwidth β of 1◦would produce a spot beam on Earth covering an area approximately 630 km in diameter.
Beam size has a direct connection to antenna gain and, in turn, to transmitter power requirements. Antenna gain G (a) Zone coverage (b) Multi-spot beams Individual spot areas or footprints Individual transmitters and horn antennas Figure 10-8 Spot and multibeam satellite antenna systems for coverage of defined areas on Earth’s surface.
464 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS is related to the directivity D by G = ξD, where ξ is the radiation efficiency, and D is related to the beamwidth β by the approximate expression given by Eq. (9.26). For a circular beam, G = ξ 4π β2 , (10.12) where β is in radians.
For a lossless antenna (ξ = 1), a global beam with β = 17.4◦(= 0.3 rad) corresponds to a gain G = 136, or 21.3 dB. A narrow 1◦beam, on the other hand, corresponds to an antenna gain of 41,253, or 46.2 dB.
To accommodate the various communication functions associated with satellite systems, four main types of antennas are used†: 1. Dipoles and helices at VHF and UHF for telemetry, tracking, and command functions; 2. Horns and relatively small parabolic dishes (with diameters on the order of a few centimeters) for producing wide-angle beams for global coverage; 3. Parabolic dishes fed by one or more horns to provide a beam for zone coverage [Fig. 10-8(a)] or multiple spot beams [Fig. 10-8(b)]; 4. Antenna arrays consisting of many individual radiating elements for producing multispot beams and for beam steering and scanning.
Concept Question 10-1: What are the advantages and disadvantages of elliptical satellite orbits in comparison to the geostationary orbit?
Concept Question 10-2: Why do satellite communica-tion systems use different frequencies for the uplink and downlink segments?
Which segment uses the higher frequency and why?
Concept Question 10-3: How does the use of antenna polarization increase the number of channels carried by the communication system?
Concept Question 10-4: What are the sources of noise that contribute to the total system noise temperature of a receiver?
†R. G. Meadows and A. J. Parsons, Satellite Communications, Hutchinson Publishers, London, 1989.
10-5 Radar Sensors The term radar is a contracted form of the phrase radio detection and ranging, which conveys some, but not all, of the features of a modern radar system.
Historically, radar systems were first developed and used at radio frequencies, including the microwave band, but we now also have light radars, or lidars, that operate at optical wavelengths.
Over the years, the name radar has lost its original meaning and has come to signify any active electromagnetic sensor that uses its own source to illuminate a region of space and then measure the echoes generated by reflecting objects contained in that region. In addition to detecting the presence of a reflecting object and determining its range by measuring the time delay of short-duration pulses transmitted by the radar, a radar is also capable of specifying the position of the target and its radial velocity. Measurement of the radial velocity of a moving object is realized by measuring the Doppler frequency shift produced by the object. Also, the strength and shape of the reflected pulse carry information about the shape and material properties of the reflecting object.
Radar is used for a wide range of civilian and military applications, including air traffic control, aircraft navigation, law enforcement, control and guidance of weapon systems, remote sensing of Earth’s environment, weather observation, astronomy, and collision avoidance for automobiles.
The frequency bands used for the various types of radar applications extend from the megahertz region to frequencies as high as 225 GHz.
10-5.1 Basic Operation of a Radar System The block diagram shown in Fig. 10-9 contains the basic functional elements of a pulse radar system. The synchronizer– modulator unit serves to synchronize the operation of the transmitter and the videoprocessor–display unit by generating a train of direct-current (dc) narrow-duration, evenly spaced pulses. These pulses, which are supplied to both the transmitter and the videoprocessor–display unit, specify the times at which radar pulses are transmitted. The transmitter contains a high-power radio-frequency (RF) oscillator with an on/off control voltage actuated by the pulses supplied by the synchronizer– modulator unit.
Hence, the transmitter generates pulses of RF energy equal in duration and spacing to the dc pulses generated by the synchronizer–modulator unit. Each pulse is supplied to the antenna through a duplexer, which allows the antenna to be shared between the transmitter and the receiver.
10-5 RADAR SENSORS 465 Duplexer Servo Transmitter Receiver Video processor/display Synchronizer/modulator unit Figure 10-9 Basic block diagram of a radar system.
The duplexer, which often is called the transmitter/receiver (T/R) switch, connects the transmitter to the antenna for the duration of the pulse, and then connects the antenna to the receiver for the remaining period until the start of a new pulse.
Some duplexers, however, are passive devices that perform the sharing and isolation functions continuously. The circulator shown in Fig. 10-5 is an example of a passive duplexer. After transmission by the antenna, a portion of the transmitted signal is intercepted by a reflecting object (often called a target) and scattered in many directions.
The energy reradiated by the target back toward the radar is collected by the antenna and delivered to the receiver, which processes the signal to detect the presence of the target and to extract information on its location and velocity. The receiver converts the reflected RF signals into lower-frequency video signals and supplies them to the videoprocessor–display unit, which displays the extracted information in a format suitable for the intended application.
The servo unit positions the orientation of the antenna beam in response to control signals provided by either an operator, a control unit with preset functions, or a control unit commanded by another system. The control unit of an air-traffic-control radar, for example, commands the servo to rotate the antenna in azimuth continuously. In contrast, the radar antenna placed in the nose of an aircraft is made to scan back and forth over only a specified angular sector.
10-5.2 Unambiguous Range The collective features of the energy transmitted by a radar are called the signal waveform. For a pulse radar, these features τ Tp = 1/fp RF frequency f Pulse waveform Figure 10-10 A pulse radar transmits a continuous train of RF pulses at a repetition frequency fp.
include (1) the carrier frequency f , (2) the pulse length τ, (3) the pulse repetition frequency fp (number of pulses per second), or equivalently the interpulse period Tp = 1/fp, and (4) the modulation (if any) within the pulses. Three of these features are illustrated in Fig. 10-10. Modulation, which refers to control of the amplitude, frequency, or phase of the signal, is beyond the level of the present treatment.
The range to a target is determined by measuring the time delay T taken by the pulse to travel to the target and back. For a target at range R, T = 2R c , (10.13) where c = 3 × 108 m/s is the speed of light, and the factor 2 accounts for the two-way propagation. The maximum target range that a radar can measure unambiguously, called the 466 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS unambiguous range Ru, is determined by the interpulse period Tp and is given by Ru = cTp 2 = c 2fp .
(10.14) The range Ru corresponds to the maximum range that a target can have such that its echo is received before the transmission of the next pulse. If Tp is too short, an echo signal due to a given pulse might arrive after the transmission of the next pulse, in which case the target would appear to be at a much shorter range than it actually is.
According to Eq. (10.14), if a radar is to be used to detect targets that are as far away as 100 km, for example, then fp should be less than 1.5 kHz, and the higher the pulse repetition frequency (PRF), the shorter is the unambiguous range Ru. Consideration of Ru alone suggests selecting a low PRF, but other considerations suggest selecting a very high PRF. As we will see later in Section 10-6, the signal-to-noise ratio of the radar receiver is directly proportional to fp, and hence it would be advantageous to select a PRF as high as possible. Moreover, in addition to determining the maximum unambiguous range Ru, the PRF also determines the maximum Doppler frequency (and hence the target’s maximum radial velocity) that the radar can measure unambiguously.
If the requirements on maximum range and velocity cannot be met by the same PRF, then some compromise may be necessary.
Alternatively, it is possible to use a multiple-PRF radar system that transmits a few pulses at one PRF followed by another series of pulses at another PRF, and then the two sets of received pulses are processed together to remove the ambiguities that would have been present with either PRF alone.
10-5.3 Range and Angular Resolutions Consider a radar observing two targets located at ranges R1 and R2, as shown in Fig. 10-11. Let t = 0 denote the time corresponding to the start of the transmitted pulse. The pulse length is τ. The return due to target 1 will arrive at T1 = 2R1/c and will have a length τ (assuming that the pulse length in space is much greater than the radial extent of the target). Similarly, the return due to target 2 will arrive at T2 = 2R2/c. The two targets are resolvable as distinct targets so long as T2 ≥T1 + τ or, equivalently, 2R2 c ≥2R1 c + τ.
(10.15) Radar Antenna beam R1 R2 Figure 10-11 Radar beam viewing two targets at ranges R1 and R2.
The range resolution of the radar, R, is defined as the minimum spacing between two targets necessary to avoid overlap between the echoes from the two targets.
From Eq. (10.15), this occurs when R = R2 −R1 = cτ/2.
(10.16) Some radars are capable of transmitting pulses as short as 1 ns in duration or even shorter. For τ = 1 ns, R = 15 cm.
The basic angular resolution of a radar system is determined by its antenna beamwidth β, as shown in Fig. 10-12.
The corresponding azimuth resolution x at a range R is given by x = βR, (10.17) where β is in radians. In some cases, special techniques are used to improve the angular resolution down to a fraction of the beamwidth. One example is the monopulse tracking radar described in Section 10-8.
Beamwidth β R ∆x = βR Figure 10-12 The azimuth resolution x at a range R is equal to βR.
10-6 TARGET DETECTION 467 Mean noise level Threshold detection level 1 Threshold detection level 2 Target 2 Target 1 False alarm Pr2 Grec Pr1 Grec Pni Grec Prmin(2) Grec Prmin(2) Grec Time Figure 10-13 The output of a radar receiver as a function of time.
10-6 Target Detection Target detection by radar is governed by two factors: (1) the signal energy received by the radar receiver due to reflection of part of the transmitted energy by the target, and (2) the noise energy generated by the receiver. Figure 10-13, which depicts the output response of a radar receiver as a function of time, shows the signals due to two targets displayed against the noise contributed by external sources as well as by the devices making up the receiver. The random variations exhibited by the noise may at times make it difficult to distinguish the signal reflected by the target from a noise spike. In Fig. 10-13, the mean noise-power level at the receiver output is denoted by Pno = GrecPni, where Grec is the receiver gain and Pni is the noise level referred to the receiver’s input terminals. The power levels Pr1 and Pr2 represent the echoes of the two targets observed by the radar.
Because of the random nature of noise, it is necessary to set a threshold level, Prmin, for detection. For threshold level 1 indi-cated in Fig. 10-13, the radar will produce the presence of both targets, but it will also detect a false alarm. The chance of this occurring is called the false-alarm probability. On the other hand, if the threshold level is raised to level 2 to avoid the false alarm, the radar will not detect the presence of the first target. A radar’s ability to detect the presence of a target is characterized by a detection probability. The setting of the threshold signal level relative to the mean noise level is thus made on the basis of a compromise that weighs both probabilities.
To keep the noise level at a minimum, the receiver is designed such that its bandwidth B is barely wide enough to pass most of theenergycontainedinthereceivedpulse. Suchadesign, called a matched filter, requires that B be equal to the reciprocal of the pulse length τ (i.e., B = 1/τ). Hence, for a matched-filter receiver, Eq. (10.10) becomes Pni = KTsysB = KTsys τ .
(10.18) The signal power received by the radar, Pr, is related to the transmitted power level, Pt, through the radar equation. We will first derive the radar equation for the general case of a bistatic radar configuration in which the transmitter and receiver are not necessarily at the same location, and then we will specialize the results to the monostatic radar case wherein the transmitter and receiver are colocated. In Fig. 10-14, the target is at range Rt from the transmitter and at range Rr from the receiver. The power density illuminating the target is given by St = Pt 4πR2 t Gt (W/m2), (10.19) where (Pt/4πR2 t ) represents the power density that would have been radiated by an isotropic radiator, and Gt is the gain of the transmitting antenna in the direction of the target. The target is characterized by a radar cross section (RCS) σt (m2), defined such that the power intercepted and then reradiated by the target is Prer = Stσt = PtGtσt 4πR2 t (W).
(10.20) 468 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS Rt Rr RCS σt Transmitter Receiver Figure 10-14 Bistatic radar system viewing a target with radar cross section (RCS) σt.
This reradiated power spreads out over a spherical surface, resulting in a power density Sr incident upon the receiving radar antenna. Hence, Sr = Prer 4πR2 r = PtGtσt (4πRtRr)2 (W/m2).
(10.21) With an effective area Ar and radiation efficiency ξr, the receiving radar antenna intercepts and delivers (to the receiver) power Pr given by Pr = ξrArSr = PtGtξrArσt (4πRtRr)2 = PtGtGrλ2σt (4π)3R2 t R2 r , (10.22) where we have used Eqs. (9.29) and (9.64) to relate the effective area of the receiving antenna, Ar, to its gain Gr. For a monostatic antenna that uses the same antenna for the transmit and receive functions, Gt = Gr = G and Rt = Rr = R. Hence, Pr = PtG2λ2σt (4π)3R4 (radar equation).
(10.23) Unlike the one-way communication system for which the dependence on R is as 1/R2, the range dependence given by the radar equation goes as 1/R4, the product of two one-way propagation processes.
The detection process may be based on the echo from a single pulse or on the addition (integration) of echoes from several pulses. We will consider only the single-pulse case here. A target is said to be detectable if its echo signal power Pr exceeds Prmin, the threshold detection level indicated in Fig. 10-13. The maximum detectable range Rmax is the range beyond which the target cannot be detected, corresponding to the range at which Pr = Prmin in Eq. (10.23). Thus, Rmax = PtG2λ2σt (4π)3Prmin 1/4 .
(10.24) The signal-to-noise ratio is equal to the ratio of the received signal power Pr to the mean input noise power Pni given by Eq. (10.18): Sn = Pr Pni = Prτ KTsys , (10.25) and the minimum signal-to-noise ratio Smin corresponds to when Pr = Prmin: Smin = Prminτ KTsys .
(10.26) Use of Eq. (10.26) in Eq. (10.24) gives Rmax = PtτG2λ2σt (4π)3KTsysSmin 1/4 .
(10.27) The product Ptτ is equal to the energy of the transmitted pulse. Hence, according to Eq. (10.27), it is the energy of the transmitted pulse rather than the transmitter power level alone that determines the maximum detectable range. A high-power narrow pulse and an equal-energy, low-power long pulse will yield the same radar performance as far as maximum detectable range is concerned. However, the range-resolution capability 10-7 DOPPLER RADAR 469 (a) Stationary source (b) Moving source λ λ λ λ u (wave moving in direction opposite to that of the source) (wave moving in the same direction as the source) Figure 10-15 A wave radiated from a point source when (a) stationary and (b) moving. The wave is compressed in the direction of motion, spread out in the opposite direction, and unaffected in the direction normal to motion.
of the long pulse is much poorer than that of the short pulse [see Eq. (10.16)].
The maximum detectable range Rmax can also be increased by improving the signal-to-noise ratio.
This can be accomplished by integrating the echoes from multiple pulses in order to increase the total amount of energy received from the target.
The number of pulses available for integration over a specified integration time is proportional to the PRF.
Hence, from the standpoint of maximizing target detection, it is advantageous to use as high a PRF as allowed by other considerations.
10-7 Doppler Radar The Doppler effect is a shift in the frequency of a wave caused by the motion of the transmitting source, the reflecting object, or the receiving system. As illustrated in Fig. 10-15, a wave radiated by a stationary isotropic point source forms equally spaced concentric circles as a function of time travel from the source. In contrast, a wave radiated by a moving source is compressed in the direction of motion and is spread out in the opposite direction.
Compressing a wave shortens its wavelength, which is equivalent to increasing its frequency.
Conversely, spreading it out decreases its frequency.
The change in frequency is called the Doppler frequency shift fd.
That is, if ft is the frequency of the wave radiated by the moving source, thenthefrequencyfr ofthewavethatwouldbeobserved by a stationary receiver is fr = ft + fd.
(10.28) The magnitude and sign of fd depend on the direction of the velocity vector relative to the direction of the range vector connecting the source to the receiver.
Consider a source transmitting an electromagnetic wave with frequency ft (Fig. 10-16). At a distance R from the source, the electric field of the radiated wave is given by E(R) = E0ej(ωtt−kR) = E0ejφ, (10.29) where E0 is the wave’s magnitude, ωt = 2πft, and k = 2π/λt, where λt is the wavelength of the transmitted wave.
The magnitude depends on the distance R and the gain of the source antenna, but it is not of concern as far as the Doppler effect is concerned. The quantity φ = ωtt −kR = 2πftt −2π λt R (10.30) is the phase of the radiated wave relative to its phase at R = 0 and reference time t = 0. If the source is moving toward the receiver, as in Fig. 10-16, or vice versa, at a radial velocity ur, then R = R0 −urt, (10.31) where R0 is the distance between the source and the receiver at t = 0. Hence, φ = 2πftt −2π λt (R0 −urt).
(10.32) This is the phase of the signal detected by the receiver. The frequency of a wave is defined as the time derivative of the phase φ divided by 2π. Thus, fr = 1 2π dφ dt = ft + ur λt .
(10.33) 470 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS Transmitter Receiver Transmitter moving with velocity u Stationary receiver ur Figure 10-16 Transmitter with radial velocity ur approaching a stationary receiver.
Comparison of Eq. (10.33) with Eq. (10.28) leads to fd = ur/λt.
For radar, the Doppler shift happens twice, once for the wave from the radar to the target and again for the wave reflected by the target back to the radar.
Hence, fd = 2ur/λt.
The dependence of fd on direction is given by the dot product of the velocity and range unit vectors, which leads to fd = −2ur λt = −2u λt cos θ, (10.34) where ur is the radial velocity component of u and θ is the angle between the range vector and the velocity vector (Fig. 10-17), with the direction of the range vector defined to be from the radar to the target. For a receding target (relative to the radar), 0 ≤θ ≤90◦, and for an approaching target, 90◦≤θ ≤180◦.
10-8 Monopulse Radar On the basis of information extracted from the echo due to a single pulse, a monopulse radar can track the direction of a target with an angular accuracy equal to a fraction of its antenna beamwidth.
To track a target in both elevation and azimuth, a monopulse radar uses an antenna (such as a parabolic dish), with four separate small horns at its focal point (Fig. 10-18). Monopulse systems are of two types. The first is called amplitude-comparison monopulse because the tracking information is extracted from the amplitudes of the echoes received by the four horns, and the second is called phase-comparison monopulse because it relies on the phases of the received signals. We shall limit our present discussion to the amplitude-comparison scheme.
Individually, each horn would produce its own beam, with the four beams pointing in slightly different directions.
Figure 10-19 shows the beams of two adjacent horns. The basic principle of the amplitude-comparison monopulse is to measure the amplitudes of the echo signals received through the two beams and then apply the difference between them to repoint the (a) (b) θ u θ u Velocity vector Range vector Figure 10-17 The Doppler frequency shift is negative for a receding target (0 ≤θ ≤90◦), as in (a), and positive for an approaching target (90◦≤θ ≤180◦), as in (b).
antenna boresight direction toward the target. Using computer-controlled phase shifters, the phasing network shown in Fig. 10-18 can combine the signal delivered to the four-element horn array by the transmitter or by the echo signals received by them in different ways. Upon transmission, the network excites all four feeds in phase, thereby producing a single main beam called the sum beam. The phasing network uses special microwave devices that allow it to provide the desired 10-8 MONOPULSE RADAR 471 1 2 3 4 Transmit Elevation Azimuth Sum 1 2 3 4 (a) (b) Radar Phasing network Figure 10-18 Antenna feeding arrangement for an amplitude-comparison monopulse radar: (a) feed horns and (b) connection to phasing network.
Error Beam 1 Beam 2 Figure 10-19 A target observed by two overlapping beams of a monopulse radar.
functionality during both the transmit and receive modes. Its equivalent functionality is described by the circuits shown in Fig. 10-20. During the receive period, the phasing network uses power dividers, power combiners, and phase shifters so as to generate three different output channels. One of these is the sum channel, corresponding to adding all four horns in phase, anditsradiationpatternisdepictedinFig.10-21(a). Thesecond channel, called the elevation-difference channel, is obtained by first adding the outputs of the top-right and top-left horns [Fig. 10-20(b)], then adding the outputs of the bottom-right and bottom-left horns, and then subtracting the second sum from the first. The subtraction process is accomplished by adding a 180◦ phase shifter in the path of the second sum before adding it to the first sum. The beam pattern of the elevation-difference channel is shown in Fig. 10-21(b). If the observed target is centered be-tween the two elevation beams, the receiver echoes will have the same strength for both beams, thereby producing a zero output from the elevation-difference channel. If it is not, the amplitude of the elevation-difference channel will be proportional to the angular deviation of the target from the boresight direction, and its sign will denote the direction of the deviation. The third channel (not shown in Fig. 10-20) is the azimuth-difference channel, and it is accomplished through a similar process that generates a beam corresponding to the difference between the sum of the two right horns and the sum of the two left horns.
In practice, the output of the difference channel is multiplied by the output of the sum channel to increase the strength of the difference signal and to provide a phase reference for extracting the sign of the angle. This product, called the angle error signal, is displayed in Fig. 10-21(c) as a function of the angle error. The error signal activates a servo-control system to repo-sition the antenna direction. By applying a similar procedure along the azimuth direction using the product of the azimuth-difference channel and the sum channel, a monopulse radar pro-vides automatic tracking in both directions. The range to the tar-get is obtained by measuring the round-trip delay of the signal.
Concept Question 10-5: How is the PRF related to unambiguous range?
Concept Question 10-6: Explain how the false-alarm probability and the detection probability are related to the noise level of the receiver.
472 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS Top right Top left ÷ Bottom right Bottom left Top right Top left Bottom right Bottom left ÷ ÷ Transmitter + + ÷ ÷ 180° + Sum channel x + Elevation-difference channel Angle error signal (a) Transmit mode (b) Receiver mode for elevation Figure 10-20 Functionality of the phasing network in (a) the transmit mode and (b) the receive mode for the elevation-difference channel.
Concept Question 10-7: In terms of the geometry shown in Fig. 10-17, when is the Doppler shift a maximum?
Concept Question 10-8: What is the principle of the monopulse radar?
Angle Error-signal voltage (c) Angle error signal (a) Sum pattern (b) Elevation-difference pattern Figure 10-21 Monopulse antenna (a) sum pattern, (b) elevation-difference pattern, and (c) angle error signal.
CHAPTER 10 SUMMARY 473 Chapter 10 Summary Concepts • Three equally spaced satellites in geostationary orbit can provide coverage of most of Earth’s surface.
• The use of polarization diversity makes it possible to double the number of channels per unit bandwidth carried by a satellite repeater.
• A satellite antenna system is designed to produce beams tailored to match the areas served by the satellite.
Antenna arrays are particularly suitable for this purpose.
• A radar is an electromagnetic sensor that illuminates a region of space and then measures the echoes due to reflecting objects. From the echoes, information can be extracted about the range of a target, its radial velocity, direction of motion, and other characteristics.
• Due to the random nature of receiver noise, target detection is a statistical process characterized by detection and false-alarm probabilities.
• A moving object produces a Doppler frequency shift proportional to the radial velocity of the object (relative to the radar) and inversely proportional to λ.
• A monopulse radar uses multiple beams to track the direction of a target, with an angular accuracy equal to a fraction of its antenna beamwidth.
Mathematical and Physical Models Satellite Communication Systems Radius of geostationary orbit R0 = GMeT 2 4π2 1/3 Received power Pri = ϒ(θ) Pr = ϒ(θ) PtGtGr λ 4πR 2 Noise power Pni = KTsysB Signal-to-noise ratio Sn = Pri Pni = ϒ(θ) PtGtGr KTsysB λ 4πR 2 Radar Sensors Unambiguous range Ru = cTp 2 = c 2fp Range resolution R = R2 −R1 = cτ/2 Azimuth resolution x = βR Radar equation Pr = PtG2λ2σt (4π)3R4 Doppler frequency shift fd = −2ur λt = −2u λt cos θ 474 CHAPTER 10 SATELLITE COMMUNICATION SYSTEMS AND RADAR SENSORS Important Terms Provide definitions or explain the meaning of the following terms: atmospheric transmissivity ϒ azimuth resolution bistatic radar circulator detection probability Doppler frequency shift fd duplexer Explorer I false-alarm probability FDMA geostationary orbit interpulse period Tp lidar matched filter maximum detectable range Rmax monopulse radar monostatic radar multiplexer polarization diversity pulse length τ pulse repetition frequency (PRF) fp radar radar cross section σt radar equation radial velocity ur range resolution Score signal-to-noise ratio Sputnik I sum and difference channels synchronizer system noise temperature threshold detection level transponder unambiguous range Ru uplink and downlink PROBLEMS Sections 10-1 to 10-4: Satellite Communication Systems ∗10.1 A remote sensing satellite is in circular orbit around Earth at an altitude of 1,100 km above Earth’s surface. What is its orbital period?
10.2 A transponder with a bandwidth of 400 MHz uses polarization diversity. If the bandwidth allocated to transmit a single telephone channel is 4 kHz, how many telephone channels can be carried by the transponder?
∗10.3 Repeat Problem 10.2 for TV channels, each requiring a bandwidth of 6 MHz.
10.4 A geostationary satellite is at a distance of 40,000 km from a ground receiving station.
The satellite transmitting antenna is a circular aperture with a 1 m diameter, and the ground station uses a parabolic dish antenna with an effective diameter of 20 cm. If the satellite transmits 1 kW of power at 12 GHz and the ground receiver is characterized by a system noise temperature of 1,000 K, what would be the signal-to-noise ratio of a received TV signal with a bandwidth of 6 MHz? The antennas and the atmosphere may be assumed lossless.
∗Answer(s) available in Appendix D.
Sections 10-5 to 10-8: Radar Sensors ∗10.5 A collision-avoidance automotive radar is designed to detect the presence of vehicles up to a range of 0.5 km. What is the maximum usable PRF?
10.6 A 10 GHz weather radar uses a 15 cm diameter lossless antenna. At a distance of 1 km, what are the dimensions of the volume resolvable by the radar if the pulse length is 1 μs?
∗10.7 A radar system is characterized by the following parameters: Pt = 1 kW, τ = 0.1 μs, G = 30 dB, λ = 3 cm, and Tsys = 1,500 K. The radar cross section of a car is typically 5 m2. How far away can the car be and remain detectable by the radar with a minimum signal-to-noise ratio of 13 dB?
10.8 A 3 cm wavelength radar is located at the origin of an x–y coordinate system. A car located at x = 100 m and y = 200 m is heading east (x direction) at a speed of 120 km/hr. What is the Doppler frequency measured by the radar?
A P P E N D I X A Symbols, Quantities, and Units Symbol Quantity SI Unit Abbreviation A Magnetic potential (vector) webers/meter Wb/m B Susceptance siemens S B Magnetic flux density teslas or webers/meter2 T or W/m2 C Capacitance farads F D Directivity (antenna) (dimensionless) — D Electric flux density coulombs/meter2 C/m2 d Moment arm meters m E Electric field intensity volts/meter V/m Eds Dielectric strength volts/meter V/m F Radiation intensity (normalized) (dimensionless) — F Force newtons N f Frequency hertz Hz fd Doppler frequency hertz Hz fmn Cutoff frequency hertz Hz G Conductance siemens S G Gain (power) (dimensionless) — H Magnetic field intensity amperes/meter A/m I Current amperes A J Current density (volume) amperes/meter2 A/m2 Js Current density (surface) amperes/meter A/m k Wavenumber radians/meter rad/m kc Cutoff wavenumber radians/second rad/s L Inductance henrys H l Length meters m 476 APPENDIX A SYMBOLS, QUANTITIES, AND UNITS Symbol Quantity SI Unit Abbreviation M, m Mass kilograms kg M Magnetization vector amperes/meter A/m m Magnetic dipole moment ampere-meters2 A·m2 n Index of refraction (dimensionless) — P Power watts W P Electric polarization vector coulombs/meter2 C/m2 p Pressure newtons/meter2 N/m2 p Electric dipole moment coulomb-meters C·m Q Quality factor (dimensionless) — Q, q Charge coulombs C R Reflectivity (reflectance) (dimensionless) — R Resistance ohms R Range meters m r Radial distance meters m S Standing-wave ratio (dimensionless) — S Poynting vector watts/meter2 W/m2 Sav Power density watts/meter2 W/m2 T Temperature kelvins K T Transmissivity (transmittance) (dimensionless) — T Torque newton-meters N·m t Time seconds s T period seconds s u Velocity meters/second m/s ug Group velocity meters/second m/s up Phase velocity meters/second m/s V Electric potential volts V V Voltage volts V Vbv Voltage breakdown volts V Vemf Electromotive force (emf) volts V W Energy (work) joules J w Energy density joules/meter3 J/m3 X Reactance ohms Y Admittance siemens S Z Impedance ohms α Attenuation constant nepers/meter Np/m β Beamwidth degrees ◦ β Phase constant (wavenumber) radians/meter rad/m Reflection coefficient (dimensionless) — γ Propagation constant meters−1 m−1 δs Skin depth meters m ϵ, ϵ0 Permittivity farads/meter F/m ϵr Relative permittivity (dimensionless) — η Impedance ohms λ Wavelength meters m APPENDIX A SYMBOLS, QUANTITIES, AND UNITS 477 Symbol Quantity SI Unit Abbreviation μ, μ0 Permeability henrys/meter H/m μr Relative permeability (dimensionless) — μe, μh Mobility (electron, hole) meters2/volt·second m2/V·s ρl Charge density (linear) coulombs/meter C/m ρs Charge density (surface) coulombs/meter2 C/m2 ρv Charge density (volume) coulombs/meter3 C/m3 σ Conductivity siemens/meter S/m σt Radar cross section meters2 m2 τ Transmission coefficient (dimensionless) — τ Pulse length seconds s ϒ Atmospheric transmissivity (dimensionless) — Magnetic flux webers Wb ψ ψ ψ Gravitational field newtons/kilogram N/kg χe Electric susceptibility (dimensionless) — χm Magnetic susceptibility (dimensionless) — Solid angle steradians sr ω Angular frequency radians/second rad/s ω Angular velocity radians/second rad/s This page intentionally left blank A P P E N D I X B Material Constants of Some Common Materials Table B-1 RELATIVE PERMITTIVITY ϵr OF COMMON MATERIALSa ϵ = ϵrϵ0 and ϵ0 = 8.854 × 10−12 F/m.
Material Relative Permittivity, ϵr Material Relative Permittivity, ϵr Vacuum 1 Dry soil 2.5–3.5 Air (at sea level) 1.0006 Plexiglass 3.4 Styrofoam 1.03 Glass 4.5–10 Teflon 2.1 Quartz 3.8–5 Petroleum oil 2.1 Bakelite 5 Wood (dry) 1.5–4 Porcelain 5.7 Paraffin 2.2 Formica 6 Polyethylene 2.25 Mica 5.4–6 Polystyrene 2.6 Ammonia 22 Paper 2–4 Seawater 72–80 Rubber 2.2–4.1 Distilled water 81 aThese are low-frequency values at room temperature (20◦C).
Note: For most metals, ϵr ≃1.
480 APPENDIX B MATERIAL CONSTANTS OF SOME COMMON MATERIALS Table B-2 CONDUCTIVITY σ OF SOME COMMON MATERIALSa Material Conductivity, σ (S/m) Material Conductivity, σ (S/m) Conductors Semiconductors Silver 6.2 × 107 Pure germanium 2.2 Copper 5.8 × 107 Pure silicon 4.4 × 10−4 Gold 4.1 × 107 Insulators Aluminum 3.5 × 107 Wet soil ∼10−2 Tungsten 1.8 × 107 Fresh water ∼10−3 Zinc 1.7 × 107 Distilled water ∼10−4 Brass 1.5 × 107 Dry soil ∼10−4 Iron 107 Glass 10−12 Bronze 107 Hard rubber 10−15 Tin 9 × 106 Paraffin 10−15 Lead 5 × 106 Mica 10−15 Mercury 106 Fused quartz 10−17 Carbon 3 × 104 Wax 10−17 Seawater 4 Animal body (average) 0.3 (poor cond.) aThese are low-frequency values at room temperature (20◦C).
APPENDIX B MATERIAL CONSTANTS OF SOME COMMON MATERIALS 481 Table B-3 RELATIVE PERMEABILITY μr OF SOME COMMON MATERIALSa μ = μrμ0 and μ0 = 4π × 10−7 H/m.
Relative Material Permeability, μr Diamagnetic Bismuth 0.99983 ≃1 Gold 0.99996 ≃1 Mercury 0.99997 ≃1 Silver 0.99998 ≃1 Copper 0.99999 ≃1 Water 0.99999 ≃1 Paramagnetic Air 1.000004 ≃1 Aluminum 1.00002 ≃1 Tungsten 1.00008 ≃1 Titanium 1.0002 ≃1 Platinum 1.0003 ≃1 Ferromagnetic (nonlinear) Cobalt 250 Nickel 600 Mild steel 2,000 Iron (pure) 4,000–5,000 Silicon iron 7,000 Mumetal ∼100, 000 Purified iron ∼200, 000 aThese are typical values; actual values depend on material variety.
Note: Except for ferromagnetic materials, μr ≃1 for all dielectrics and conductors.
This page intentionally left blank A P P E N D I X C Mathematical Formulas Trigonometric Relations sin(x ± y) = sin x cos y ± cos x sin y cos(x ± y) = cos x cos y ∓sin x sin y 2 sin x sin y = cos(x −y) −cos(x + y) 2 sin x cos y = sin(x + y) + sin(x −y) 2 cos x cos y = cos(x + y) + cos(x −y) sin 2x = 2 sin x cos x cos 2x = 1 −2 sin2 x sin x + sin y = 2 sin x + y 2 cos x −y 2 sin x −sin y = 2 cos x + y 2 sin x −y 2 cos x + cos y = 2 cos x + y 2 cos x −y 2 cos x −cos y = −2 sin x + y 2 sin x −y 2 cos(x ± 90◦) = ∓sin x cos(−x) = cos x sin(x ± 90◦) = ± cos x sin(−x) = −sin x ejx = cos x + j sin x (Euler’s identity) sin x = ejx −e−jx 2j cos x = ejx + e−jx 2 484 APPENDIX C MATHEMATICAL FORMULAS Approximations for Small Quantities For |x| ≪1, (1 ± x)n ≃1 ± nx (1 ± x)2 ≃1 ± 2x √ 1 ± x ≃1 ± x 2 1 √1 ± x ≃1 ∓x 2 ex = 1 + x + x2 2! + · · · ≃1 + x ln(1 + x) ≃x sin x = x −x3 3! + x5 5! + · · · ≃x cos x = 1 −x2 2! + x4 4! + · · · ≃1 −x2 2 lim x→0 sin x x = 1 A P P E N D I X D Answers to Selected Problems Chapter 1 1.1 p(x, t) = 32.36 cos(4π × 103t −12.12πx + 36◦) (N/m2) 1.3 10 cm 1.6 up = 0.83 (m/s); λ = 10.47 m 1.8 (a) y1(x, t) is traveling in positive x direction. y2(x, t) is traveling in negative x direction.
1.10 T = 2.5 s; up = 0.56 m/s; λ = 1.4 m 1.12 y2(t) lags y1(t) by 54◦.
1.14 α = 2 × 10−3 (Np/m) 1.16 (b) z2 = √ 3 ej3π/4 1.17 (c) z1z2 = 18ej109.4◦ 1.19 (c) |z|2, 1.20 (d) t = 0; s = 6 ej30◦ 1.22 ln(z) = 1.76 −j1.03 1.25 vc(t) = 15.57 cos(2π × 103t −81.5◦) V 1.26 (d) ˜ I = −2ej3π/4 = 2e−jπej3π/4 = 2e−jπ/4 A 1.27 (d) i(t) = 3.61 cos(ωt + 146.31◦) A Chapter 2 2.1 (a) l/λ = 1.33 × 10−5; transmission line may be ignored.
(c) l/λ = 0.4; transmission line effects should be included.
2.4 R′ = 1.38 (/m), L′ = 1.57 × 10−7 (H/m), G′ = 0, C′ = 1.84 × 10−10 (F/m) 2.8 α = 0.109 Np/m; β = 44.5 rad/m; Z0 = (19.6 + j0.030) ; up = 1.41 × 108 m/s 486 APPENDIX D ANSWERS TO SELECTED PROBLEMS 2.10 w = 0.613 mm, λ = 0.044 m 2.14 R′ = 1 (/m); L′ = 200 (nH/m); G′ = 400 (μS/m); C ′ = 80 (pF/m); λ = 2.5 m 2.16 R′ = 0.6 /m, L′ = 38.2 nH/m, G′ = 0.5 mS/m, C′ = 23.9 pF/m 2.18 (a) b = 4.2 mm (b) up = 2 × 108 m/s 2.21 ZL = (90 −j120) 2.23 Z0 = 55.9 2.27 Zin = (40 + j20) 2.31 (a) = 0.62e−j29.7◦ 2.32 (b) = 0.16 e−j80.54◦.
2.33 (a) Zin1 = (35.20 −j8.62) 2.35 L = 8.3 × 10−9 H 2.37 l = λ/4 + nλ/2 2.39 Zin = 1002 33.33 = 300 2.41 (b) iL(t) = 3 cos(6π × 108t −135◦) (A) 2.42 (a) Zin = (41.25 −j16.35) 2.44 P i av = 10.0 mW; P r av = −1.1 mW; P t av = 8.9 mW 2.45 (a) Pav = 0.29 W 2.47 (b) = 0.62 exp −29.7◦ 2.50 Zin = (66 −j125) 2.52 Z01 = 40 ; Z02 = 250 2.53 (b) S = 1.64 2.55 (a) Zin = −j154 (b) 0.074λ + (nλ/2), n = 0, 1, 2, . . .
2.57 Refer to Fig. P2.57. The point Z represents 1.5 −j0.7.
The reciprocal of point Z is at point Y, which is at 0.55 + j0.26.
2.61 ZL = (41 −j19.5) 2.63 Zin = (95 −j70) 2.69 First solution: Stub at d = 0.199λ from antenna and stub length l = 0.125λ. Second solution: d = 0.375λ from antenna and stub length l = 0.375λ.
2.73 Zin = 100 2.78 Vg = 19.2 V; Rg = 30 ; l = 525 m 2.82 (a) l = 1200 m (b) ZL = 0 (c) Rg = 1 + g 1 −g Z0 = 1 + 0.25 1 −0.25 50 = 83.3 (d) Vg = 32 V Chapter 3 3.1 ˆ a = ˆ x 0.32 + ˆ z 0.95 3.3 Area = 36 3.5 (a) A = √ 14 ; ˆ aA = (ˆ x + ˆ y2 −ˆ z3)/ √ 14 (e) A· (B × × × C) = 20 (h) (A × × × ˆ y)· ˆ z = 1 3.9 ˆ a = A |A| = −ˆ x −ˆ yy −ˆ z 2 5 + y2 3.11 ˆ a = (ˆ x 2 −ˆ z 4)/ √ 20 3.13 A = ˆ x 0.8 + ˆ y 1.6 3.15 ˆ c = ˆ x 0.37 + ˆ y 0.56 + ˆ z 0.74 3.17 G = ± −ˆ x 8 3 + ˆ y 8 3 + ˆ z 4 3 3.22 (a) P1 = (2.24, 63.4◦, 0) in cylindrical; P1 = (2.24, 90◦, 63.4◦) in spherical (d) P4 = (2.83, 135◦, −2) in cylindrical; P4 = (3.46, 125.3◦, 135◦) in spherical 3.24 (a) P1 = (0, 0, 5) 3.25 (c) A = 12 3.26 (a) V = 21π/2 3.30 (a) θAB = 90◦ APPENDIX D ANSWERS TO SELECTED PROBLEMS 487 (b) ±(ˆ r 0.487 + ˆ φ φ φ 0.228 + ˆ z 0.843) 3.32 (a) d = √ 3 3.34 (c) ⃗ C (P3) = ˆ r0.707 + ˆ z4 (e) E(P5) = −ˆ r + ˆ φ φ φ 3.35 (c) C(P3) = ˆ R0.854 + ˆ θ θ θ0.146 −ˆ φ φ φ0.707 3.36 (e) ∇S = ˆ x8xe−z + ˆ y3y2 −ˆ z4x2e−z 3.37 (b) ∇T = ˆ x 2x (g) ∇T = −ˆ x 2π 6 sin πx 3 3.38 T (z) = 10 + (1 −e−3z)/3 3.40 dV dl (1,−1,4) = 2.18 3.43 dU/dl = −0.02 3.45 E = ˆ R4R 3.48 (a) D· ds = 150π (b) ∇· D dV = 150π 3.56 (a) A is solenoidal, but not conservative.
(d) D is conservative, but not solenoidal.
(h) H is conservative, but not solenoidal.
3.57 (c) ∇2 3 x2 + y2 = 12 x2 + y22 Chapter 4 4.1 Q = 2.62 (mC) 4.3 Q = 86.65 (mC) 4.7 I = 314.2 A 4.9 (a) ρl = −πca4 2 (C/m) 4.11 E = ˆ z 51.2 kV/m 4.13 q2 ≈−94.69 (μC) 4.15 (a) E = −ˆ x 1.6 −ˆ y 0.66 (MV/m) 4.17 E = ˆ z (ρs0h/2ϵ0) √ a2 + h2 + h2/ √ a2 + h2 −2h 4.20 E = −ˆ y ρl πϵ0R1 R1 R2 + ˆ y ρl πϵ0R2 = 0 4.23 (a) ρv = y3z3 (b) Q = 32 (C) (c) Q = 32 (C) 4.25 Q = 4πρ0a3 (C) 4.27 D = ˆ r ρv0(r2 −1) 2r , 1 ≤r ≤3 m D = ˆ r Dr = ˆ r 4ρv0 r , r ≥3 m 4.30 R1 = a 2, R3 = a √ 5 2 , V = 0.55Q πϵ0a 4.32 (b) E = ˆ z(ρla/2ϵ0)[z/(a2 + z2)3/2] (V/m) 4.34 V (b) = (ρl/4πϵ) × ln l + √ l2 + 4b2 −l + √ l2 + 4b2 (V) 4.37 V = ρl 2πϵ0 ln a (x −a)2 + y2 −ln a (x + a)2 + y2 4.39 VAB = −117.09 V 4.41 (c) ue = −8.125E/|E| (m/s); uh = 3.125E/|E| (m/s) 4.45 R = 4.2 (m) 4.48 θ = 61◦ 4.50 Q = 3πϵ0 2 (C) 4.53 (a) |E| is maximum at r = a.
4.55 We = 4.62 × 10−9 (J) 4.57 (a) C = 3.1 pF 4.60 (b) C = 6.07 pF 4.63 C′ = πϵ0 ln[(2d/a) −1] (C/m) 488 APPENDIX D ANSWERS TO SELECTED PROBLEMS Chapter 5 5.1 a = −ˆ y4.22 × 1018 (m/s2) 5.4 ⃗ T = −ˆ z1.66 (N·m); clockwise 5.5 (a) F = 0 5.7 B = −ˆ z0.6 (mT) 5.9 H = ˆ zIθ (b −a) 4πab 5.11 I2 = 2aI1 2πNd = 1 × 25 π × 20 × 2 = 0.2 A 5.13 I = 200 A 5.16 ⃗ F = −ˆ x0.4 (mN) 5.18 (a) H(0, 0, h) = −ˆ x I πw tan−1 w 2h (A/m) 5.20 F = ˆ y 4 × 10−5 N 5.24 J = ˆ z 36e−3r A/m2 5.26 (a) ⃗ A = ˆ z μ0I 4π ln ℓ+ √ ℓ2 + 4r2 −ℓ+ √ ℓ2 + 4r2 5.27 (a) B = ˆ z5π sin πy −ˆ yπ cos πx (T) 5.29 (a) A = ˆ zμ0IL/(4/piR) (b) H = (IL/4π)[(−ˆ xy + ˆ yx)/(x2 + y2 + z2)3/2] 5.31 ne = 1.5 electrons/atom 5.33 H2 = ˆ z 3 5.35 ⃗ B2 = ˆ x20000 −ˆ y30000 + ˆ z8 5.37 L′ = (μ/π) ln[(d −a)/a] (H) 5.40 = 1.66 × 10−6 (Wb) Chapter 6 6.1 At t = 0, current in top loop is momentarily clockwise.
At t = t1, current in top loop is momentarily counterclockwise.
6.3 (a) Vemf = 375e−3t (V) 6.5 B0 = 0.8 (nA/m) 6.7 Iind = 37.7 sin(200πt) mA 6.10 V12 = −236 (μV) 6.12 I = 0.1 (A) 6.15 I = 0.82 cos(120πt) (μA) 6.17 f = 5 MHz 6.18 (b) 888 6.20 ρv = (8y/ω) sin ωt + C0, where C0 is a constant of integration.
6.24 k = (4π/30) rad/m; E = −ˆ z941 cos(2π × 107t + 4πy/30) (V/m) 6.26 H(R, θ; t) = ˆ φ φ φ (53/R) sin θ cos(6π × 108t −2πR) (μA/m) 6.28 (a) k = 20 (rad/m) Chapter 7 7.1 (a) Positive y-direction (c) λ = 12.6 m 7.3 (a) λ = 31.42 m 7.5 ϵr = 9 7.7 (a) λ = 10 m 7.9 E = ˆ x √ 2 cos(ωt + kz) −ˆ y √ 2 sin(ωt + kz) (V/m) APPENDIX D ANSWERS TO SELECTED PROBLEMS 489 7.12 At x = 0 and t = 0, E = −ˆ z1.885 (V/m).
At x = 0 and t = 5 ns, ωt = 0.13 rad and E = −1.885(ˆ y0.13 + ˆ z0.99) (V/m) 7.14 (a) γ = 73.5◦and χ = −8.73◦ (b) Right-hand elliptically polarized 7.17 (a) Low-loss dielectric. α = 8.42 × 10−11 Np/m, β = 468.3 rad/m, λ = 1.34 cm, up = 1.34 × 108 m/s, ηc ≈168.5 7.19 H lags E by 31.72◦ 7.21 z = 287.82 m 7.23 up = 9.42 × 104 (m/s) 7.25 H = −ˆ y0.16 e−30x cos(2π × 109t −40x −36.85◦) (A/m) 7.29 (Rac/Rdc) = 143.55 7.33 Sav = ˆ y0.48 (W/m2) 7.35 (c) z = 23.03 m 7.37 up = 1 × 108 (m/s) 7.39 (b) Pav = 0 7.41 (a) (we)av = ϵE2 0 4 Chapter 8 8.1 (a) = −0.67; τ = 0.33 (b) S = 5 (c) Si av = 0.52 (W/m2); Sr av = 0.24 (W/m2); St av = 0.28 (W/m2) 8.3 (b) Si av = ˆ y 251.34, Sr av = ˆ y 10.05, St av = ˆ y 241.29 (W/m2) 8.6 (a) = −0.71 8.7 | E1|max = 85.5 (V/m); lmax = 1.5 m 8.9 ϵr2 = √ϵr1ϵr3 ; d = c/[4f (ϵr1ϵr3)1/4] 8.11 Zin(−d) = 0.43η0∠ −51.7◦ ||2 = 0.24 8.13 f = 75 MHz 8.15 P ′ = (3.3 × 10−3)2 102 2 × 1.14 [1 −e−2×44.43×2×10−3] = 1.01 × 10−4 (W/m2) 8.17 θmin = 20.4◦ 8.19 St Si = 0.85 8.22 d = 15 cm 8.24 d = 68.42 cm 8.26 fp = 59.88 (Mb/s) 8.27 (b) θi = 36.87◦ 8.29 (a) θi = 33.7◦ 8.31 θt = 18.44◦ 8.35 (a) R = 6.4 × 10−3; T = 0.9936 (b) P i = 85 mW; P r = 0.55 mW; P t = 84.45 mW 8.37 (a) 9.4% 8.39 a = 3.33 cm; b = 2 cm 8.41 Any one of the first four modes.
8.43 570 (empty); 290 (filled) 8.45 θ′ 20 = 57.7◦ 8.47 (a) Q = 8367 Chapter 9 9.1 Smax = 7.6 (μW/m2) 9.4 (a) Direction of maximum radiation is a circular cone 120◦wide, centered around the +z axis.
(b) D = 4 = 6 dB (c) p = π (sr) = 3.14 (sr) (d) β = 120◦ 9.6 (b) G = −3.5 dB 9.9 Smax = 4 × 10−5 (W/m2) 9.11 D = 36.61 dB 490 APPENDIX D ANSWERS TO SELECTED PROBLEMS 9.14 S = 1.46 9.16 (a) E(R, θ, φ) = ˆ θ θ θ Eθ = ˆ θ θ θj I0lkη0 8π e−jkR R sin θ (V/m) 9.17 (a) θmax1 = 42.6◦, θmax2 = 137.4◦ 9.20 (a) θmax1 = 90◦, θmax2 = 270◦ (b) Smax = 60I 2 0 πR2 (c) F(θ) = 1 4 cos (π cos θ) + 1 sin θ 2 9.23 Pt = 259 (mW) 9.25 Pt = 75 (μW) 9.27 (a) Prec = 3.6 × 10−6 W 9.30 βnull = 5.73◦ 9.32 D = 45.6 dB 9.35 (a) βe = 1.8◦; βa = 0.18◦ (b) y = βaR = 0.96 m 9.37 (a) Fa(θ) = 4 cos2 π 8 (4 cos θ + 1) 9.39 d/λ = 1.414 9.44 Fa(θ) = [6 + 8 cos(π cos θ) + 2 cos(2π cos θ)]2 9.46 δ = −2.72 (rad) = −155.9◦ Chapter 10 10.1 T = 82.97 minutes 10.3 133.3 ≈133 channels 10.5 (fp)max = 300 kHz 10.7 Rmax = 4.84 km Bibliography The following list of books, arranged alphabetically by the last name of the first author, provides references for further reading.
Electromagnetics Balanis, C.A., Advanced Engineering Electromagnetics, John Wiley & Sons, Hoboken, NJ, 1989.
Cheng, D.K., Fundamentals of Engineering Electromagnetics, Addison Wesley, Reading, MA, 1993.
Hayt, W.H., Jr. and J.A. Buck, Engineering Electromagnetics, 7th ed., McGraw-Hill, New York, 2005.
Iskander, M.F., Electromagnetic Fields & Waves, Prentice Hall, Upper Saddle River, NJ, 2000.
King, R.W.P. and S. Prasad, Fundamental Electromagnetic Theory and Applications, Prentice Hall, Englewood Cliffs, NJ, 1986.
Ramo, S., J.R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 3rd ed., John Wiley & Sons, Hoboken, NJ, 1994.
Rao, N.N., Elements of Engineering Electromagnetics, Prentice Hall, Upper Saddle River, NJ, 2004.
Shen, L.C. and J.A. Kong, Applied Electromagnetism, 3rd ed., PWS Engineering, Boston, MA, 1995.
Antennas and Radiowave Propagation Balanis, C.A., Antenna Theory: Analysis and Design, John Wiley & Sons, Hoboken, NJ, 2005.
Ishimaru, A., Electromagnetic Wave Propagation, Radiation, and Scattering, Prentice Hall, Upper Saddle River, NJ, 1991.
Stutzman, W.L. and G.A. Thiele, Antenna Theory and Design, John Wiley & Sons, Hoboken, NJ, 1997.
492 Optical Engineering Bohren, C.F. and D.R. Huffman, Absorption and Scattering of Light by Small Particles, John Wiley & Sons, Hoboken, NJ, 1998.
Born, M. and E. Wolf, Principles of Optics, 7th ed., Pergamon Press, New York, 1999.
Hecht, E., Optics, Addison-Wesley, Reading, MA, 2001.
Smith, W.J., Modern Optical Engineering, SPIE Press, Bellingham, WA, 2007.
Walker, B.H., Optical Engineering Fundamentals, SPIE Press, Bellingham, WA, 2009.
Microwave Engineering Freeman, J.C., Fundamentals of Microwave Transmission Lines, John Wiley & Sons, Hoboken, NJ, 1996.
Pozar, D.M., Microwave Engineering, Addison-Wesley, Reading, MA, 2004.
Richharia, M., Satellite Communication Systems, McGraw-Hill, New York, 1999.
Scott, A.W., Understanding Microwaves, John Wiley & Sons, Hoboken, NJ, 2005.
Skolnik, M.I., Introductionto RadarSystems, 3rded., McGraw-Hill, New York, 2002.
Stimson, G.W., Introduction to Airborne Radar, Hughes Aircraft Company, El Segundo, California, 200l.
Index 3-dB beamwidth, 414 A Abacus, 8 Ablation, 10, 112 ac motor, 3, 5 ac resistance R, 341 Acceptance angle θa, 365 Adding machine, 8 Admittance Y, 96 Alternating current (ac), 5 AM radio, 6 Amp ere, Andr´ e-Marie, 4 Amp` ere’s law, 252–255, 273 Amplitude-comparison monopulse radar, 470 Amplitude modulation (AM), 6 Analog computer, 8 Angle error signal, 471 Angle of incidence θi, 363 Angle of reflection θr, 363 Angle of transmission θt, 363 Angular frequency ω, 25, 58 Angular velocity ω, 25 Antennas, 404–449, 463–464 aperture, 429 rectangular, 432–434 scalar formulation 430 vector formulation 430 arrays, 435–442 linear phase, 446 pattern multiplication principle, 438 scanning, 444–449 uniform phase, 442–443 broadside direction, 409 directivity D, 414, 434 effective area, 434 far-field (far-zone) region, 405, 408–409 gain, 416–417 half-wave dipole, 417–422 input impedance, 404 isotropic, 404, 413 large aperture, 429–435 multiplication principle, 438 normalized radiation intensity, 409 pattern solid angle p, 412 patterns, 404, 411 beam dimensions, 412 beamwidth β, 413–414 directivity D, 414–415 polarization, 404 receiving, 422–427 reciprocal, 404 types, 464 arrays, 464 dipoles, 464 helices, 464 493 494 INDEX horns, 464 parabolic dishes, 464 Antenna radiation pattern, 404 Arithmometer, 8 Armstrong, Edwin, 6, 7 ARPANET, 7 Array factor Fa(θ), 438 array amplitude distribution, 438 array phase distribution, 438 Atmospheric transmissivity ϒ, 462 Attenuation constant α, 57, 331 Average power Sav, 343 Average power density Sav, 343 Auxiliary angle ψ0, 329 Axial ratio R, 329 Azimuth angle φ, 407 Azimuth-difference channel, 471 Azimuth plane (φ-plane), 412 Azimuth resolution x, 466 B bac-cab rule, 139 Backus, John, 8 Band gap energy, 39 Bar-code readers, 382–383 Bardeen, John, 7 Base vector, 134 BASIC, 8 Beam dimensions, 412 Beamwidth β, 413, 414, 433–434 Becquerel, Alexandre-Edmond, 38, 293 Bell, Alexander, 6 Berliner, Emil, 6 Berners–Lee, Tim, 9 Bhatia, Sabeer, 9 Bioelectrics, 113 Biot, Jean-Baptiste, 4, 16 Biot–Savart law, 4, 16, 244–251, 273 current distributions, 244–248 surface current density Js, 244 volume current density J, 244 volume distributions, 244–248 Bistatic radar, 467 Bounce diagram, 118 Boundary conditions, 203–210 Brattain, Walter, 7 Braun, Karl, 6 Brewster (polarizing) angle, 375–376, 396 Broadside array, 442 Broadside direction, 409 Bush, Vannevar, 8 C Capacitance C, 210–213 capacitor, 210 of a coaxial line, 212 of a parallel-plate capacitor, 211–213 Capacitive sensors, 196, 218–222 Capacitor, 4, as batteries, 214–216 electrochemical double-layer (EDLC), 214 Cardullo, Mario, 322 Carrier frequency f , 465 Cartesian coordinate system x, y, z 141, 142 CAT (CT) scan, 164 Cathode ray tube (CRT), 6 Cavity resonators, 392–394, 396 Cell phone, 7 Charge continuity equation, 301, 307 Charge dissipation, 302 Charge distribution, 180–181, 184 surface distribution, 185 Circular polarization, 324, 326–328 Circulation, 162 Circulator, 460 Cladding, 365 Coaxial line, 51 Complex conjugate, 34 Complex feeding coefficient Ai, 437 Complex numbers, 32–36 complex conjugate, 34 Euler’s identity, 32, 43 polar form, 32 properties, 34 rectangular form, 32 rectangular-polar relations, 32, 43 Complex permittivity ϵc, 315 Compressive stress, 292 Conductance G, 96 Conductivity σ, 8, 18, 198, 477 Conductors, 195–201 conduction current, 195 conduction current density J, 195 conductivity, 198, 477 equipotential medium, 198 resistance, 199–200 semiconductors, 195, 198 Conservative (irrotational) field, 166, 191 INDEX 495 Constitutive parameters, 195 Convection current, 182 Conversion efficiency, 38 Coordinate systems, 140–154 Cartesian x, y, z 141, 142 cylindrical r, φ, z, 140, 142–145 spherical R, θ, φ, 140, 145–147 Coplanar waveguide, 51 Cormack, Allan, 164 Coulomb (C), 13 Coulomb, Charles-Augustin de, 3, 4, 13 Coulomb’s law, 13, 182–187 charge distribution, 184 circular disk of charge, 186 infinite sheet of charge, 187 line distribution, 185 relative permittivity (dielectric constant) ϵr, 183 ring of charge, 185 surface distribution, 185 two-point charges, 184 volume distribution, 185 Critical angle θc, 364 Cross (vector) product, 138–139 CT (CAT) scan, 164 Curie, Paul-Jacques, 292 Curie, Pierre, 292 Curl operator, 162, 163 Current density, 195, 244, 297 Cutoff frequency fmn, 386 Cutoff wavenumber kc, 384 Cylindrical coordinate system r, φ, z, 140, 142–145 D dc motor, 3 De Forest, Lee, 6 Deep Blue, 9 Del (gradient operator) ∇, 155 Detection, 467–469 maximum detectable range Rmax, 468 threshold detection level Prmin, 468 Diamagnetic, 260 Dielectric constant (relative permittivity) ϵr, 15, 183, 202, 479 Dielectrics, 195, 201–203 anisotropic, 202 breakdown, 203–203 breakdown voltage Vbr, 203 electric polarization field P, 202 electric susceptibility χe, 203 homogeneous, 202 isotropic, 202 linear, 202 nonpolar, 201 perfect, 195, 198 permanent dipole moments, 202 polar materials, 201 polarization, 201 strength Eds, 203 tables, 204, 479 Difference channel, 471 Digital computer, 8 Dimensions, 11 Dipole, 14, 82, 192, 248, 252 electric, 14, 82, 192 half-wave, 417–422, 451 Hertzian, 406–409 linear, 420–422 moment, 193 short, 427, 451 vertical, 435 Direct current (dc), 3 Directional derivative dT /dl, 155 Directivity D, 414, 434 Dispersive, 50 Displacement current Id, 297–299 Displacement current density Jd, 297 Distance vector, 136 Divergence operator, 158–162 Divergence theorem, 159 Dominant mode, 386 Doppler frequency shift fd, 464, 469 Doppler radar, 469–470 Dot (scalar) product, 136–137 Downlink, 460 Drift velocity ue, 198 du Fay, Charles Fran¸ cois, 3, 4 Duplexer (T/R switch), 460, 465 E e electron charge, 13 Echo satellite, 7 Eckert, J. Presper, 8 Edison, Thomas, 6, 20 Effective aperture, 422, See also Effective area Effective area Ae, 422 Einstein, Albert, 3, 5, 38 Electric, 3, 4 496 INDEX Electric charge, 3, 4, 13–14 law of conservation of electric charge, 14 principle of linear superposition, 14 Electric dipole, 14, 82, 192 moment, 193 Electric-field aperture distribution Ea(xa, ya), 430 Electric field intensity E, 14, 179 Electric field phasor E, 319 Electric fields, 13–15, 179, 183–187 dipole, 14, 192 e charge, 13 polarization, 14, 201 Electric flux density D, 15, 179 Electric generator, 3 Electric potential V , 189 Electric scalar potential, 189–194 as a function of electric field, 189–191 due to continuous distributions, 191 due to point charges, 191, 223 electric dipole, 192 Kirchhoff’s voltage law, 190 Laplace’s equation, 193 line distribution, 191 Poisson’s equation, 193 potential energy, 189 Electric susceptibility χe, 203 Electric typewriter, 6 Electrical force Fe, 13 Electrical permittivity ϵ, 13, 66, 183–184, 203 of free space ϵ0, 13 Electrical sensors, 196 capacitive, 196 emf, 196 inductive, 196 resistive, 196–197 Electromagnetic (EM) force, 12, 237 Electromagnetic (EM) spectrum, 30–32 gamma rays, 30, 32 infrared, 30, 32 microwave band, 32, 32 EHF, 32 millimeter-wave band, 32 SHF, 32 UHF, 32 monochromatic, 30 properties, 30 radio spectrum, 30, 32, 32 ultraviolet, 32, 32 visible, 32, 32 X-rays, 30, 32 Electromagnetic generator, 294–296 Electromagnetic induction, 283 Electromagnetic telegraph, 6 Electromagnetic waves, 5, 82, 353–394 Electromagnets, 256–258 ferromagnetic core, 256 horseshoe, 256 loudspeaker, 257–258 magnetic levitation, 258 magnetically levitated trains (maglevs), 258–258 reed relay, 256 step-down transformer, 256 switch, 256 Electromotive force (emf) Vemf, 5, 283 Electron, 3, 5, 13 Electronic beeper, 7 Electronic steering, 436 EM, 3 Electrostatics, 17, 179 Elevation angle (θ-plane), 412 Elevation-difference channel, 471 Elevation plane (θ-plane), 412 Elliptical polarization, 324, 328–330 Ellipticity angle χ, 328 Emf sensor, 196 End-fire direction, 445 Engelbart, Douglas, 9 ENIAC, 8 Equipotential, 198 Euler’s identity, 32, 43 Evanescent wave, 385 Explorer I satellite, 458 F Faraday, Michael, 3, 5, 283 Faraday’s law, 282–284, 307 motional emf, 289, 307 transformer emf, 284, 307 Far-field (far-zone) region, 405 approximation, 408–409 power density, 409 False alarm probability, 467 Feeding coefficient Ai, 437 Felt, Dorr, 8 Ferromagnetic, 260, 262–264 Fessenden, Reginald, 6 Fiber, 7, 51, 365 Fiber optics, 365–367 INDEX 497 Field lines, 158 Floppy disk, 8 Fluorescence, 20 Fluorescent bulb, 20-23 Flux density, 158 Flux sensor, 293 FORTRAN, 8 Franklin, Benjamin, 3, 4 Free space, 13 velocity of light c, 16 magnetic permeability μ0, 16 electric permittivity ϵ, 13 Frequency, 25 Frequency-division multiple access (FDMA), 460 Frequency modulation (FM), 7 Frequency scanning, 445–449 Friis transmission formula, 427–429, 462 Fundamental forces electromagnetic, 12, 179 nuclear, 12 weak-interaction, 12 gravitational, 12 G Gamma rays, 30, 32 Gauss, Carl Friedrich, 5 Gauss’s law, 5, 187–189 differential form, 187 of infinite line charge, 189 integral form, 187 Gaussian surface, 187 Gauss’s law for magnetism, 251, 252, 273 Geostationary orbit, 458 Gilbert, William, 3, 4 Global Positioning System (GPS), 150–151 Grad (gradient) ∇T , 155 Gradient operator, 155–158 Gravitational force, 12 gravitational field ψ ψ ψ, 12 Grazing incidence, 375 Group velocity ug, 387 H Half-power angle, 413 Half-power beamwidth, 413 Half-wave dipole, 417–422 Henry, Joseph, 3, 5, 283 Hertz, Heinrich, 3, 5, 6, 25 Hertzian dipole, 406–409 High-power amplifier, 461 Hoff, Ted, 9 Hole drift velocity uh, 198 Hole mobility μh, 198 Homogeneous material, 195 Homogeneous wave equation, 316 Horn antenna, 405 Hotmail, 9 Hounsfield, Godfrey, 164 Humidity sensor, 219 I Illumination Ea(xa, ya), 430 Image method, 223–224 Imaginary part Im , 32 Impedance, 49, 58, 66, 68, 75, 76, 93 Impedance matching, 101–110 lumped element matching, 102–108 matching points, 107 network, 102 shunt stub, 108 single-stub matching, 108–111 stub, 108 Impulse period Tp, 465 In-phase, 69 Incandescence, 20 Incandescent bulb, 20–23 Inclination angle ψ, 325 Incremental phase delay δ, 446 Index of refraction, 363 Inductance, 5, 265–271, 273 of a coaxial line, 267 mutual, 266, 270–271 self, 266, 267 solenoid, 265 Inductive sensors, 196, 268–269 eddy-current proximity sensor, 268 ferromagnetic core, 268 linear variable differential transformer (LVDT), 268 proximity detection, 268 Infrared rays, 30, 32 In-phase, 69 Input impedance Zin, 416 Integrated circuit (IC), 7 Intercepted power Pint, 422 Internal (surface) impedance Zs, 341 International System of Units (SI), 11 Internet, 7, 9 498 INDEX Intrinsic impedance η, 318 Isotropic, 195 Isotropic antenna, 404, 413 Isotropic material, 195 J Java, 9 Joule’s law, 201 K Kapany, Narinder, 7 Kemeny, John, 8 Kilby, Jack, 7 Kirchhoff’s laws 49, current, 49, 301, 302 voltage, 49, 190 Kurtz, Thomas, 8 L Laplace’s equation, 193 Laplacian operator, 167–169 Lasers, 368–369 Law of conservation of electric charge, 14 LED bulb, 20-23 LED lighting, 20–23 Left-hand circular (LHC) polarization, 326 Leibniz, Gottfried von, 8 Lenz’s law, 285, 286–287 Leyden Jar, 3 Lidars, 464 Light emitting diode (LED), 22 Lightning rod, 4 Line charge, 180 Line charge density ρℓ, 180 Linear phase distribution, 444 Liquid crystal display (LCD), 2, 336–338 Liquid crystals, 2 Logarithm, 8 Lorentz force, 237, 273 Loss resistance Rloss, 416 Lossless media, 358, 376–380 Lossy media, 28 Loudspeaker, 257–258 Low-loss dielectric, 333 Luminous efficacy (LE), 23 M Macintosh, 9 Maiman, Theodore, 368 Maglevs, 258–258 Magnetic dipole, 248 Magnetic energy Wm, 271–272 Magnetic field intensity H, 16, 236 Magnetic field phasor H, 319 Magnetic field, 244–250 between two parallel conductors, 250–251 in a solenoid, 265 inside a toroidal coil, 254–255 of a circular loop, 247–248, 273 of a linear conductor, 244–247 of a long wire, 253–254, 273 of a magnetic dipole, 248 of an infinite current sheet, 255 Magnetic flux , 260 Magnetic flux density B, 15, 236 Magnetic flux linkage , 267 Magnetic force Fm, 16, 236–241 Magnetic hysteresis, 262 Magnetic levitation, 258 Magnetic moment m, 261–262 Magnetic monopole, 252 Magnetic permeability μ, 16, 262 Magnetic potential A, 259–260 Magnetic properties of materials, 260–264 Magnetic sound recorder, 6 Magnetic susceptibility χm, 261 Magnetic torque, 241–244 Magnetite, 3, 15 Magnetization vector M, Magnetized domains, 262 Magnetron tube, 83 Magnus, 4 Marconi, Guglielmo, 6 Mars Pathfinder, 7 Maser, 368 Matched filter, 467 Matched line, 71, 85 Maximum detectable range Rmax, 468 Maxwell, James Clerk, 3, 5, 179 Maxwell’s equations, 251–255, 273, 282 Mauchley, John, 8 Microprocessor, 9 Microstrip line, 51 Microwave band, 32, 32 Mobility μe, 198 INDEX 499 Modal dispersion, 366 Mode, 365, 386 Modem, 8 Moment, 193, 202, 261–262 Monochromatic, 30, 368 Monopulse radar, 470–472, 473 amplitude-comparison monopulse, 470 phase-comparison monopulse, 470 Monostatic radar, 467 Morse, Samuel, 5, 6 Motional emf V m emf, 284, 289, 307 MS-DOS, 9 Multiple-beam generation, 436 Multiple-PRF, 466 Multiplexer, 461 N n-type layer, 38 Nakama, Yoshiro, 8 Nanocapacitor, 214 Napier, John, 8 Negative electric charge, 3 Neutrons, 13 Newton, Isaac, 4 Noise power, 468, 473 Normal incidence, 356, 396 Normalized load impedance zL, 68 Normalized load reactance xL, 90 Normalized load resistance rL, 90 Notation, 11 Noyce, Robert, 7 Nuclear force, 12 Null beamwidth, 414 O Oblique incidence, 362–364, 396 Oersted, Hans Christian, 4, 15, 282 Ohm, Georg Simon, 5 Ohm’s law, 5, 195 Optical fiber, 7, 51, 365–367 Orbital magnetic moment, 261–262 P p–n junction, 38 p-type layer, 38 Pager, 7 Parallel-plate transmission line, 51 Parallel polarization, 374–376 Paramagnetic, 260 Pascal, Blaise, 8 Pattern multiplication principle, 438 Pattern solid angle p, 413 Perfect conductor, 195, 198 Perfect dielectric, 195, 198 Permittivity ϵ, 183, 203, 477 Perpendicular polarization, 370–374 Phase, 24 Phase constant β, 57, 331 Phase constant (wavenumber) k, 305 Phase lag, 26 Phase lead, 26 Phase-matching condition, 372 Phase velocity (propagation velocity) up, 318 Phasor representation, 11 Phasors 36–43 Photoelectric effect, 3, 5, 38 Photovoltaic (PV), 38 Photovoltaic effect, 38 Piezein, 196, 292 Piezoelectric transducer, 292 Piezoresistivity, 196–197 Planck, Max, 3 Plane-wave propagation, 313–346 attenuation rate A, 346 circular polarization, 324, 326–328 left-hand circular (LHC), 326 right-hand circular (RHC), 326–328 complex permittivity ϵc, 315 imaginary part ϵ′′, 316 real part ϵ′, 316 elliptical polarization, 324, 328–330 auxiliary angle ψ0, 329 axial ratio R, 329 ellipticity angle χ, 328 rotation angle γ , 328 electromagnetic power density, 343 linear polarization, 324, 325–326 lossy medium, 314, 331–339 attenuation constant α, 331 skin depth δs, 333 low-loss dielectric, 333 Pocket calculator, 9 Poisson’s equation, 193 Polarization, 14, 324, 370 parallel polarization, 370, 374–376 perpendicular polarization, 370–374 500 INDEX transverse electric (TE) polarization, 370 transverse magnetic (TM) polarization, 370 unpolarized, 376 Polarization diversity, 462 Polarization field P, 202 Polarization state, 324 Position vector, 136 Potential energy We, 213, 217 Poulsen, Valdemar, 6 Power density S(R, θ, φ), 409 Power transfer ratio Prec/Pt, 428 Poynting vector (power density) S, 343, 409 Pressure sensor, 219 Principle of linear superposition, 14 Principal planes, 412 Propagation constant γ , 316 Propagation velocity (phase velocity) up, 25 Pulse code modulation (PCM), 7 Pulse length τ, 465 Pulse repetition frequency (PRF) fp, 465 Q Quality factor Q, 393 Quarter-wavelength transformer, 84 Quasi-conductor, 333 R Radar (radio detection and ranging), 7, 467–469 azimuth resolution x, 466 cross-section, 467 bistatic, 467 detection, 467–469 Doppler, 469–470 monopulse, 470–472, 473 monostatic, 467 multiple-PRF, 466 operation, 464 pulse, 465 range, 465 range resolution R, 466 unambiguous range Ru, 466 Radar cross-section, 467 Radar equation, 468 Radial distance, 16, 142, 464 Radial velocity ur, 464 Radiation efficiency ξ, 416 Radiation intensity, 409 Radiation pattern, 404 Radiation resistance Rrad, 416 Radio frequency identification (RFID) systems, 322–323 Radio telegraphy, 6 Radio waves, 6, 32, Radius of geostationary orbit, 459, 473 Range R, 145 Range resolution R, 466 RC relation, 211, 226 Real part Re, 32 Received power, 463, 473 Receiving cross section, 422, See also Effective area Rectangular aperture, 432–434 Rectangular waveguide, 51 Reeves, H. A., 7 Reflection coefficient, 66–68 Reflectivity R, 377–380 Refraction angle, 363 Reinitzer, Friedrich, 336 Relaxation time constant τr, 302 Resistive sensor, 196–197 Resonant frequency f0, 392, 393–394 Retarded potentials, 303–304 Right-hand circular (RHC) polarization, 326–328 R¨ ontgen, Wilhelm, 3, 5 Rotation angle γ , 328 S Satellite, 458–469 antennas, 463–464 elliptical orbit, 459 geostationary, 458 transponders, 460–462 Savart, F´ elix, 4, 16 Scalar (dot) product, 136–137 Scalar quantity, 11 Scan angle δ, 446 Score satellite, 458 Seebeck, Thomas, 293 Seebeck potential Vs, 293 Semiconductor, 195, 198 Sensors, 196 capacitive, 196, 218–222 emf, 196, 292–293 inductive, 196, 268–269 resistive, 196–197 Shockley, William, 7 Signal-to-noise ratio Sn, 428, 468, 473 Signal waveform, 465 Skin depth δs, 333 INDEX 501 Smith chart, 52, 88–101 admittance Y, 96 admittance transformation, 96–100 angle of reflected coefficient, 91 characteristic admittance Y0, 96 conductance G, 96 constant-SWR (-||) circle, 93 matching points, 107 normalized admittance y, 96 normalized conductance g, 96 normalized susceptance b, 96 normalized load admittance yL, 96 normalized load impedance zL, 90 normalized load reactance xL, 90 normalized load resistance rL, 90 normalized wave impedance z(d), 92 parametric equations, 89–91 phase-shifted coefficient d, 92 standing-wave ratio (SWR), 93–95 susceptance B, 96 unit circle, 90 voltage maxima | V |max, 93–96 voltage minima | V |min, 93–96 wavelengths toward generator (WTG), 93 wavelengths toward load (WTL), 93 Smith, Jack, 9 Smith, P.H., 88 Snell’s laws, 362–364 of reflection, 363, 372, 396 of refraction, 363, 372, 396 Solar cell, 38 Solenoid, 256 Solid angle d, 411 Spherical propagation factor (e−jkR/R), 407 Spherical wave, 314 Spin magnetic moment, 261 Spontaneous emission, 368 Sputnik I satellite, 458 Standing wave, 59, 70–75 first voltage maximum, 72 first voltage minimum, 72 in-phase, 69 interference, 71 minimum value, 71 maximum value, 71 pattern, 71, 83 phase-opposition, 71 properties, 85 voltage standing wave ratio [(VSWR) or (SWR)] S, 72 Static conditions, 179 Steradians (sr), 411 Stimulated emission, 368 Stokes’s theorem, 166–167 Strip line, 51 Sturgeon, William, 6, 7, 256 Sum channel, 471 Sun beam, 470 Supercapacitor, 214 Superconductor, 198 Superheterodyne radio receiver, 6 Surface charge density ρs, 180 Surface current density Js, 244 Surface (internal) impedance Zs, 341 Surface resistance Rs, 341 SWR (standing-wave ratio), 93–95 Synchronizer–modulator, 464 System noise temperature Tsys, 428, 463 T Tapered aperture distribution, 433 Telegraph, 5 Telephone, 6 Television (TV), 7 TEM (transverse electromagnetic), 51–52 Tensile stress, 292 Tesla, Nikola, 3, 5, 16 Thales of Miletus, 3, 4 Thermocouple, 292, 293 Thomas de Colmar, Charles Xavier, 8 Thompson, Joseph, 3, 5 Threshold detection level Prmin, 468 Tomography, 164 Toroidal coil, 254–255 Torque, 241–244 Total internal reflection, 364 Townes, Charles, 368 Transformer emf V tr emf, 283 Transient response, 111–115 Transistor, 7 Transmission coefficients τ, 356 Transmission lines, 48–121 admittance Y, 96 air line, 55, 59 bounce diagram, 118 characteristic impedance Z0, 58 characteristic parameters, 67 coaxial line, 51, 53, 61 502 INDEX complex propagation constant γ , 57 attenuation constant α, 57 phase constant β, 57 conductance G, 96 current maxima and minima, 72 definition, 49 dispersive transmission line, 52 distortionless line, 52 effective relative permittivity ϵeff, 62 load impedance ZL, 66 guide wavelength λ, 59 input impedance Zin, 76, 93 input reactance Xin, 79 input resistance Rin, 79 lossless line, 65–75 lossless microstrip line, 60–65 lumped-element model, 52–53 matched load, 68 matching network, 102 microstrip line, 51, 60–65 nondispersive, 66 open-circuited line, 81 parallel-plate line, 51 parameters, 52–53 phase-shifted coefficient d, 92 power loss, 50 power flow, 86–88 quarter-wavelength transformer, 84 slotted line, 74 Smith chart, 52, 88–101 standing wave, 59, 70–75 Transmission lines (continued) standing wave pattern, 71, 83 SWR circle, 93 TEM (transverse electromagnetic) transmission lines, 51–52 transient response, 111–115 transmission line parameters, 52 capacitance C′, 53 conductance G′, 53 inductance L′, 53 resistance R′, 52 voltage maxima | V |max, 93–96 voltage minima | V |min, 93–96 voltage reflection coefficient , 66–68 voltage standing wave ratio [(VSWR) or (SWR)] S, 72 wave impedance Z(d), 75–78 Transmissivity ϒ(θ), 377–380, 462 Transmitter/receiver (T/R) switch, 465 Transponder, 460–462 Transverse electric (TE), 370 Transverse electric (TE) polarization, 370 Transverse electromagnetic (TEM) wave, 318 Transverse magnetic (TM), 370 Transverse magnetic (TM) polarization, 370 Travelling waves, 18–32, See also Waves Triode tube, 6 Two-wire line, 51 U Ultracapacitor, 214 Ultraviolet rays, 31, 32 Unambiguous range Ru, 466 Uniform field, 162–163 Uniform field distribution, 432 Units, 11 Unit vectors, 11, 134 Uplink, 460 V van Musschenbroek, Pieter, 4 Vector analysis, 133–169 transformations between coordinate systems, 147–154 Vector magnetic potential, 259–260, 273 Vector Poisson’s equation, 259, 273 Vector (cross) product, 138–139 Vector quantities, 11 Velocity of light in free space c, 16 Video processor/display, 464 Visible light, 32, 32 Volta, Alessandro, 3, 4 VSWR (voltage standing wave ratio) S, 72. See also SWR Volume charge density ρv, 180 Volume current density J, 244 W Walton, Charles, 322 Watson-Watt, Robert, 7 Wave polarization, 324 circular, 324, 326–328 elliptically, 324, 328–330 electric field phasor E, 325 inclination angle ψ, 325 linear, 324, 325–326 INDEX 503 Wave polarizer, 337 Wavefront, 314 Waveguides 380–383, 396 Wavelength, 25, 31 Wavenumber (phase constant) k, 305, 316 Waves, 18–32 Weak-interaction force, 12 White light, 4 Wireless transmission, 6 World Wide Web (WWW), 9 X X-rays, 3, 5, 30, 32 Z Zenith angle θ, 145, 407 Zuse, Konrad, 8 Zworykin, Vladimir, 7 ω-β diagram, 390 This page intentionally left blank F U N D A M E NT A L P HY S I C A L C O N ST A NT S CONSTANT SYMBOL VALUE speed of light in vacuum c 2.998 × 108 ≈3 × 108 m/s gravitational constant G 6.67 × 10−11 N·m2/kg2 Boltzmann’s constant K 1.38 × 10−23 J/K elementary charge e 1.60 × 10−19 C permittivity of free space ε0 8.85 × 10−12 ≈ 1 36π × 10−9 F/m permeability of free space μ0 4π × 10−7 H/m electron mass me 9.11 × 10−31 kg proton mass mp 1.67 × 10−27 kg Planck’s constant h 6.63 × 10−34 J·s intrinsic impedance of free space η0 376.7 ≈120π F U N D A M E NT A L S I U N IT S DIMENSION UNIT SYMBOL Length meter m Mass kilogram kg Time second s Electric current ampere A Temperature kelvin K Amount of substance mole mol Luminous Intensity candela cd M U LT I P L E & S U B M U LT I P L E P R E F I X E S PREFIX SYMBOL MAGNITUDE PREFIX SYMBOL MAGNITUDE exa E 1018 milli m 10−3 peta P 1015 micro μ 10−6 tera T 1012 nano n 10−9 giga G 109 pico p 10−12 mega M 106 femto f 10−15 kilo k 103 atto a 10−18 Book Website: www.pearsonhighered.com/ulaby GRADIENT, DIVERGENCE, CURL, & LAPLACIAN OPERATORS CARTESIAN (RECTANGULAR) COORDINATES (x, y, z) ∇V = ˆ x∂V ∂x + ˆ y∂V ∂y + ˆ z∂V ∂z ∇· A = ∂Ax ∂x + ∂Ay ∂y + ∂Az ∂z ∇× A = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z Ax Ay Az = ˆ x ∂Az ∂y −∂Ay ∂z + ˆ y ∂Ax ∂z −∂Az ∂x + ˆ z ∂Ay ∂x −∂Ax ∂y ∇2V = ∂2V ∂x2 + ∂2V ∂y2 + ∂2V ∂z2 CYLINDRICAL COORDINATES ( r , φ , z ) ∇V = ˆ r∂V ∂r + ˆ φ 1 r ∂V ∂φ + ˆ z∂V ∂z ∇· A = 1 r ∂ ∂r (rAr) + 1 r ∂Aφ ∂φ + ∂Az ∂z ∇× A = 1 r ˆ r ˆ φr ˆ z ∂ ∂r ∂ ∂φ ∂ ∂z Ar rAφ Az = ˆ r 1 r ∂Az ∂φ −∂Aφ ∂z + ˆ φ ∂Ar ∂z −∂Az ∂r + ˆ z1 r ∂ ∂r (rAφ) −∂Ar ∂φ ∇2V = 1 r ∂ ∂r r ∂V ∂r + 1 r2 ∂2V ∂φ2 + ∂2V ∂z2 SPHERICAL COORDINATES ( R , θ , φ ) ∇V = ˆ R∂V ∂R + ˆ θ 1 R ∂V ∂θ + ˆ φ 1 R sin θ ∂V ∂φ ∇· A = 1 R2 ∂ ∂R (R2AR) + 1 R sin θ ∂ ∂θ (Aθ sin θ) + 1 R sin θ ∂Aφ ∂φ ∇× A = 1 R2 sin θ ˆ R ˆ θR ˆ φR sin θ ∂ ∂R ∂ ∂θ ∂ ∂φ AR RAθ (R sin θ)Aφ = ˆ R 1 R sin θ ∂ ∂θ (Aφ sin θ) −∂Aθ ∂φ + ˆ θ 1 R 1 sin θ ∂AR ∂φ −∂ ∂R (RAφ) + ˆ φ 1 R ∂ ∂R (RAθ) −∂AR ∂θ ∇2V = 1 R2 ∂ ∂R R2 ∂V ∂R + 1 R2 sin θ ∂ ∂θ sin θ ∂V ∂θ + 1 R2 sin2 θ ∂2V ∂φ2 S O M E U S E F U L V E CT O R I D E NT IT I E S A · B = AB cos θAB Scalar (or dot) product A × B = ˆ nAB sin θAB Vector (or cross) product, ˆ n normal to plane containing A and B A · (B × C) = B · (C × A) = C · (A × B) A × (B × C) = B(A · C) −C(A × B) ∇(U + V ) = ∇U + ∇V ∇(UV ) = U∇V + V ∇U ∇· (A + B) = ∇· A + ∇· B ∇· (UA) = U∇· A + A · ∇U ∇× (UA) = U∇× A + ∇U × A ∇× (A + B) = ∇× A + ∇× B ∇· (A × B) = B · (∇× A) −A · (∇× B) ∇· (∇× A) = 0 ∇× ∇V = 0 ∇· ∇V = ∇2V ∇× ∇× A = ∇(∇· A) −∇2A V (∇· A) dV = S A · ds Divergence theorem (S encloses V) S (∇× A) · ds = C A · dl Stokes’s theorem (S bounded by C) |
10270 | https://www.geeksforgeeks.org/maths/find-the-exact-value-of-sin-150-degrees/ | Find the exact value of sin 150 degrees
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Trigonometry is a discipline of mathematics that studies the relationships between the lengths of the sides and angles of a right-angled triangle. Trigonometric functions, also known as goniometric functions, angle functions, or circular functions, are functions that establish the relationship between an angle to the ratio of two of the sides of a right-angled triangle. The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant.
Angles defined by the ratios of trigonometric functions are known as trigonometry angles. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360°.
As given in the above figure in a right-angled triangle:
Hypotenuse: The side opposite to the right angle is the hypotenuse, It is the longest side in a right-angled triangle and opposite to the 90° angle.
Base: The side on which angle C lies is known as the base.
Perpendicular: It is the side opposite to angle C in consideration.
Trigonometric Functions
Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,
sine: It is defined as the ratio of perpendicular and hypotenuse and It is represented as sin θ
cosine: It is defined as the ratio of base and hypotenuse and it is represented as cos θ
tangent: It is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ
cosecant: It is the reciprocal of sin θ and is represented as cosec θ.
secant: It is the reciprocal of cos θ and is represented as sec θ.
cotangent: It is the reciprocal of tan θ and is represented as cot θ.
Trigonometric Identities of Complementary and Supplementary Angles
Complementary Angles: Pair of angles whose sum is equal to 90°
Supplementary Angles: Pair of angles whose sum is equal to 180°
Identities of Complementary angles
sin (90° - θ) = cos θ
cos (90° - θ) = sin θ
tan (90° - θ) = cot θ
cot (90° - θ) = tan θ
sec (90° - θ) = cosec θ
cosec (90° - θ) = sec θ
Identities of supplementary angles
sin (180° - θ) = sin θ
cos (180° - θ) = - cos θ
tan (180° - θ) = - tan θ
cot (180° - θ) = - cot θ
sec (180° - θ) = - sec θ
cosec (180° - θ) = - cosec θ
Find the exact value of sin 150 degrees.
Solution:
Here sin is positive only in the 1st and 2nd Quadrant.
150° lies in the 2nd Quadrant.
Therefore
sin (180° – θ) = sin θ
sin (150°) = sin (180° – 30°)
sin (150°) = sin (30°)
sin (150°) = 1/2
So the exact value of sin 150° is 1/2
Similar Questions
Question 1: Find the value of sin 135°.
Solution:
Since, we know that sin is positive in the 1st and 2nd Quadrant,
here, 135° lies in the 2nd Quadrant, then
By the Trigonometric Identity of Supplementary Angles,
We know that sin (180° - θ) = sin θ
Hence,
sin 135° = sin(180° - 45°)
= sin 45° {As given by Identity}
= 1/√2
Question 2: What is the Exact value of cos 150°?
Solution:
Here cos is positive only in 1st and 4th Quadrant.
150° lies in 2nd Quadrant.
Therefore cos(180° - θ) = - cos θ
cos(150°) = cos(180° – 30°)
cos(150°) = -cos(30°)
cos (150°) = -√3/2 { as per the trigonometry value table }
So the exact value of cos 150° is -√3/2
A
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10271 | https://www.youtube.com/watch?v=OuPLDo35Q5o | Explain The Relation Between GCD And LCM / Maths Arithmetic
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The product of the two given numbers should always be equal to the product of their LCM and GCD. The same has been explained here with some solved examples.
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Transcript:
Introduction [Music] hey guys welcome to W Academy and here in this presentation we are going to discuss the relation between greatest common devisor and least common multiple Concept so in our previous video we have already discussed what is a greatest common devisor and a least common multiple and in this video we are going to discuss their relation so now let us consider two numbers say A and B and their LCM and gcd as L and G respectively so now let us find the product of the given numbers that is a into B which should be equal to the product of LCM and gcd which can be represented as L into G so thus the relation between greatest common divisor and the least common multiple will be the product of the given numbers will be equal to product of their LCM and gcd so now let me Example explain this concept by using an example over here so now let us consider two numbers say 30 and 48 and now find their relation so first let us find the greatest common divisor of the given numbers so for that reason let us divide the larger number 48 with the smaller number 30 and here 30 1 time is 30 and by subtracting we get the reminder here 18 and again dividing this number with the reminder 18 we get here 30 divided by 18 and here we know that 18 1 time is 18 and by subtracting we get 12 and then repeating the process of dividing this number with 12 we get 18 ided by 12 and here 12 1 time is 12 and then by subtracting we get the reminder here 6 and now repeating the process again 12 is divided by 6 and then 6 2 is 12 and by subtracting we get the remainer here zero so here the last nonzero reminder is six so we can say that the greatest common divisor is six and now let us find the least common multiple of the given numbers so for that sake let us divide the given numbers with two here and here 2 15 is 30 and 2 24 is 48 and now again dividing these numbers by three here 3 5 is 15 and again 3 8 is 24 and now multiplying all the factors here we get 2 into 3 into 5 into 8 which is equal to and their product is 240 so now let us find the product of their least common multiple and the greatest common factor which can be represented as 6 into 240 and hence we get the product as 1,440 and now let us consider the product of the given numbers which can be represented as 30 into 48 and their product is equal to 1, 440 so hence by observing both the values are equal so we can say that product of their least common multiple and the greatest common divisor so this was a basic introduction on relation between least common multiple and greatest common deviser hope you understood the concept do like and subscribe to we Academy thanks for watching have a nice day [Music] |
10272 | https://bestpractice.bmj.com/topics/en-us/399 | When viewing this topic in a different language, you may notice some differences in the way the content is structured, but it still reflects the latest evidence-based guidance.
Lymphogranuloma venereum
Epidemiology
Etiology
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History and exam
Tests
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Criteria
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Approach
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Emerging
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Patient information
Summary
Primary manifestation of lymphogranuloma venereum infection is painless penile or vulvar inflammation and ulceration at the site of inoculation; often not noticed by the patient.
Secondary stage typically occurs weeks after development of the primary lesion; presents as painful, unilateral, inguinal or femoral lymphadenopathy (often referred to as "inguinal syndrome").
Proctocolitis has emerged as a more typical presentation in men who have sex with men (particularly those who are HIV-positive).
Chronic inflammation can lead to scarring and fibrosis causing lymphedema of the genitals, or formation of strictures and fistulae if anorectal involvement.
Identification of Chlamydia trachomatis from the swab of a genital ulcer or aspiration of a bubo is definitive diagnosis.
Doxycycline is the preferred first-line treatment; macrolides are an alternative treatment option (e.g., in pregnant or lactating women, or patients with allergies to tetracyclines).
Large buboes may be aspirated, but incision and drainage or surgical excision of buboes may complicate healing.
Definition
Lymphogranuloma venereum (LGV) is a STI caused by Chlamydia trachomatis genovars/serovars L1, L2, or L3 (collectively termed the "LGV biovar"), which are endemic to the tropics, but now emerging in developed regions. Infection occurs through contact with mucous membranes or abrasions in the skin of the genital region.
History and exam
Key diagnostic factors
Other diagnostic factors
Risk factors
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1st tests to order
Tests to consider
Emerging tests
Treatment algorithm
asymptomatic adolescent and adult patients who have been exposed to lymphogranuloma venereum (LGV)
all stages of lymphogranuloma venereum (LGV)
Contributors
Authors
Benjamin D. Lorenz, MD
Assistant Professor
Division of Hospital Medicine
MedStar Georgetown University Hospital
Washington
DC
Disclosures
BDL declares that he has no competing interests.
Acknowledgements
Dr Benjamin D. Lorenz would like to gratefully acknowledge Dr Mettassebia Kanno, a previous contributor to this topic.
Disclosures
MK declares that she has no competing interests.
Peer reviewers
Cees van Nieuwkoop, MD
Department of General Internal Medicine
Leiden University Medical Centre
Leiden
The Netherlands
Disclosures
CvN declares that he has no competing interests.
David Chelmow, MD
Chair
Department of Obstetrics and Gynecology
Virginia Commonwealth University
Richmond
VA
Disclosures
DC declares that he has no competing interests.
References
Key articles
de Vries HJC, de Barbeyrac B, de Vrieze NHN, et al. 2019 European guideline on the management of lymphogranuloma venereum. J Eur Acad Dermatol Venereol. 2019 Jun 26;33(10):1821-8.Full text
Abstract
Centers for Disease Control and Prevention. Lymphogranuloma venereum among men who have sex with men - Netherlands, 2003-2004. MMWR Morb Mortal Wkly Rep. 2004 Oct 29;53(42):985-8.Full text
Abstract
Van der Bij AK, Spaargaren J, Morre SA, et al. Diagnostic and clinical implications of anorectal lymphogranuloma venereum in men who have sex with men: a retrospective case-control study. Clin Infect Dis. 2006 Jan 15;42(2):186-94.Full text
Abstract
Stamm WE. Lymphogranuloma venereum. In: Holmes KK, Sparling PF, Stamm WE, et al., eds. Sexually transmitted diseases. 4th ed. New York, NY: McGraw Hill; 2007:595-606.
Workowski KA, Bachmann LH, Chan PA, et al. Sexually transmitted infections treatment guidelines, 2021. MMWR Recomm Rep. 2021 Jul 23;70(4):1-187.Full text
Abstract
Mabey D, Peeling RW. Lymphogranuloma venereum. Sex Transm Infect. 2002 Apr;78(2):90-2.Full text
Abstract
Thorsteinsson SB. Lymphogranuloma venereum: review of clinical manifestations, epidemiology, diagnosis, and treatment. Scand J Infect Dis Suppl. 1982;32:127-31. Abstract
Annamuthodo H. Rectal lymphogranuloma venereum in Jamaica. Ann R Coll Surg Engl. 1961 Sep;29:141-59.Full text
Abstract
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10273 | https://math.stackexchange.com/questions/4423872/eulers-infinite-descent-proof-of-sums-of-two-squares | Skip to main content
Euler's infinite descent proof of sums of two squares
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I started reading a book on history of algebra, after wondering how we ever got to groups, and I have been stuck on this proof ever since.
I specifically don't really understand where the contradiction lies in this proof.
Its supposed to be that you cannot have an infinite sequence of primes smaller than some q that do not divide x2+y2, by the well ordering principle of natural numbers.
However it seems to me that you couldn't have an infinite sequence of primes smaller than some q, even if they divided x2+y2.
I'm not really understanding how this can be a contradiction, because you can never have an infinite descending sequence of primes, by the Well Ordering Principle.
Can anyone help me out?
The proof is on Wikipedia (similar to the book I'm reading), and I'm going to copy the applicable bit here:
"we shall be able to find a strictly decreasing infinite sequence q,q1,q2,… of positive integers which are not themselves the sums of two squares but which divide into a sum of two relatively prime squares. Since such an infinite descent is impossible, we conclude that q must be expressible as a sum of two squares, as claimed."
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The contradiction lies in that if no qi is produced that is the sum of two squares, then the process used to produce each qi always works. That is, there must always be a next smaller qi+1.
It is exactly because there can not be an infinite downward sequence that this is a contradiction (prime or not, sequence of sums of squares or not, there are no infinite downward sequences of natural numbers at all). And what it contradicts is the assumption that each new qi is not a sum of two squares. Obviously one of the qi has to be a sum of two squares, and at that point the procedure for producing the next qi+1 breaks down.
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AR.5: Number and algebraic methods
AR.6: Number and algebraic methods
AR.7: Modeling from data
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Represent domain and range of a function using interval notation, inequalities, and set (builder) notation.
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10276 | https://www.math.utoronto.ca/barbeau/vig25.pdf | Two periodic sequences.
A mathematical vignette Edward Barbeau 0. Introduction.
This is an investigation that would be suitable for middle school students. It in-volves some facility with arithmetic, particularly in dealing with fractions. Students get some experience in recognizing patterns and expressing them mathematically.
Finally, there is sharpening the ability to perform accurate algebraic manipulations.
The project involves three related sequences A, B, and C.
1. The sequence A.
Suppose that you start with any two numbers a and b. These are the first two entries of a sequence a, b, c, d, e, . . . for which each successive term is the previous term plus 1 divided by the term before than. Thus c = b + 1 a , d = c + 1 b , e = d + 1 c and so on. Start with any two numbers and see what the generated sequence looks like.
More formally, the (bilateral) sequence can be defined by xn+1 = xn + 1 xn−1 , where we allow n to range over all integers.
If a and b are positive, then all subsequent terms will be positive and so the sequence continues indefinitely. However, it may happen that 0 turns up, in which case you can proceed no further.
Here are some questions: A.1. To get a sense of the generic behaviour of these sequences, begin with a pair of starting values and see what happens.
A.2. How does the sequences whose first two terms are a, b compare with that obtained by taking the terms in the oppositve order b, a?
A.3. What are forbidden terms who presence may lead at some point to the value 0 in the sequence?
A.4. What are the sequences all of whose terms are integers?
1 2 A.5. Which sequences are symmetrical in the sense that they are the same when read backwards?
2. The sequence B.
This time, let us start with three numbers a, b and c, and make them the first three terms of a sequence a, b, c, d, e, f each of whose terms in the sum of the two previous terms plus 1 divided by the third previous term: d = b + c + 1 a , e = c + d + 1 b , f = d + e + 1 c and so on. More formally, we can define the sequence bilaterally by xn+1 = xn + xn−1 + 1 xn−2 , where n ranges over all the integers. As before, if we arrive at the value 0, the sequence stops, so we need to prevent this from occurring.
B.1. Try various triples of starting values and see what happens.
B.2. What happens if you take the starting values a, b, c in the order c, b, a?
B.3. What restrictions should be placed on the starting values a, b, c to prevent the term 0 from appearing?
B.4. What are the sequences all of whose terms are integers?
B.5. What happens if the first three terms are consecutive integers in either increasing or decreasing order?
B.6. Which sequences are symmetrical in the sense that they are the same when read backwards?
3. The sequence C.
In the same vein, we can begin with four numbers a, b, c, d and continue the sequence by adding three consecutive terms plus 1 and dividing by the previous term: e = b + c + d + 1 a and so on. More formally, we have xn+1 = xn + xn−1 + xn−2 + 1 xn−3 , where n ranges over all the integers. Once again, if we arrive at a 0 term, then we must stop.
C.1. Experiment with various quadruples of starting values.
C.2. What happens if you take the starting values in opposite order?
3 C.3. What restrictions should be placed on the starting values in order to avoid arriving at a term equal to 0?
C.4. Are there any sequences all of whose terms are integers?
C.5. Classify sequences as being symmetrical (the same when read backwards) and non-symmetrical.
At this point, you may wish to stop reading, explore the situation more, and perhaps formulate some questions of your own.
4. The sequence A.
In every case, unless the sequence gets 0 at some point, it has period 5, i.e., it repeats itself over and over after the first five terms. This can be checked alge-braically; we have the period a, b, b + 1 a , a + b + 1 ab , a + 1 b after which the terms continue a, b, . . . . If we start with b, a, then we can just read this sequence in the opposite direction.
In particular, in examining possible sequences, we need only consider only those for which |a| ≤|b|. (For the other case, we only have to read the sequences from right to left.) If we wish to have integer entries, then a and b must be integers and the de-nominator of each fraction in the foregoing list must divide the numerator. If we assume that |a| ≤|b|, then, since |b| is a divisor or |a + 1|, |a| ≤|b| ≤|a + 1|.
This forces a to be positive and b to take one of the four values a, a + 1, −a, −(a + 1). Look at these possibilities in turn: If b = a, then the sequence begins a, a, (a+1)/a. The only way that a can divide a + 1 is for a = 1, and we obtain the sequence 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, . . . .
If b = a + 1, then the sequence begins a, a + 1, (a + 2)/a so a must be either 1 or 2. In this case we get either the period (1, 2, 3, 2, 1) or the period (2, 3, 2, 1, 1), which gives the same sequence with a different starting point.
If b = −a, then the sequence begins a, −a, −(a −1)/a, so a = 1. But this leads to 0 and is impossible.
4 Finally, if b = −a −1, then the sequence begins a, −a −1, −1, 0, so this case is impossible.
4. The sequence B.
In this situation, we find that unless 0 appears, the sequence has period 8. If we start with a, b, c, we obtain the period a, b, c, b + c + 1 a .
a + b + c + 1 + ac ab , (a + b + 1)(b + c + 1) abc , a + b + c + 1 + ac bc , a + b + 1 c .
Note that we can read this sequence in the opposite direction and it will still satisfy the rule of formation.
B.7. There is the possibility that the sequence B might have a shorter period.
The possibilities for the length of the period are 1, 2 and 4. Determine all the possible periodic sequences.
Let us investigate when all the entries are integers; this requires that only the numbers in each period are integers. To begin with, we suppose that all the entries are positive and that a is a maximum entry. Then 0 < b+c+1 ≤2a+1. It follows from this that the fourth term of the period, (b + c + 1)/a, being an integer, does not exceed 2. Hence, either b + c = 2a −1 or b + c = a −1.
In the first instance, the only possibilities are (b, c) = (a, a −1) or (b, c) = (a −1, a). The fifth term is respectively 1 + (2/a) or 1 + (4/(a −1)) so that a is one of 2, 3, 5. Checking these out leads to sequences with one of these periods: (5, 4, 5, 2, 2, 1, 2, 2) and (3, 2, 3, 2, 3, 2, 3, 2) Otherwise b + c = a −1, and the period is (a, b, a −b −1, 1, a + 1 b −1, a + b + 1 b(a −b −1) = 1 b 1 + 2(b + 1) a −(b + 1) , . . . ).
Since the sixth term is an integer, we must have ab −b2 −b ≤a + b + 1, or a ≤(b + 1)2 b −1 = b + 3 + 4 b −1.
When b = 2, a must be odd and a −3 must divide 6. Hence a = 5 or a = 9 and we obtain the periods: (5, 2, 2, 1, 2, 2, 5, 4) and (9, 2, 6, 1, 4, 1, 6, 2).
When b = 3, a −2 must be a multiple of 3 and a −4 must divide 8. Hence a = 5, 8 and we obtain the periods: (5, 3, 1, 1, 1, 3, 5, 9) and (8, 3, 4, 1, 2, 1, 4, 3).
When b = 4, a must be either 7 or 15, but both these fail on the sixth term.
When b = 5, then a = 9 and we obtain (9, 5, 3, 1, 1, 1, 3, 5).
5 If b ≥6, then 4/(b −1) < 1 and so a ≤b + 3. Hence a −3 ≤b ≤a. Since b + c = a −1 and c ≥1, we must have b = a −2 or b = a −1. We are led to the periodic sequences: (a, a −2, 1, 1, 3 a −2, . . . ) and (a, a −3, 2, 1, 4 a −3).
Since a −2 ≥6, a ≥8 and the fifth term in not an integer. Thus, this case is not possible.
Consider periods that have at least one negative number. If the sequence has three negative terms in a row, the next term must be positive. Thus, the period must have at least one positive entry. Therefore, wolog, we assume that a ≥1 > −1 ≥b.
First suppose that c > 0; we can also suppose that c ≤a, since reversing the sequence gives a valid sequence. Let b + c + 1 > 0. Since b + c + 1 is divisible by a, then a ≤b + c + 1 = c + (b + 1) ≤c ≤a, from which a = c and b = −1. This leads to the period (a, −1, a, 1, −(a + 2), −1, −(a + 2), 1) for a ̸= 0, −2.
The other possibility is that a ≥c ≥1 > −1 ≥b + c + 1 > b.
Since a divides b+c+1, then b+c+1 = −ka where k is a positive integer. Therefore b = −ka −c −1 and we are led to the period (a, −ka −c −1, c, −k, −c + 1 −k c + 1 + ka. . . . ).
The denominator of the fraction in the fifth entry is positive and exceeds the nu-merator. so if the fifth entry is an integer, k −c −1 must be positive and at least as large as c + 1 + ka.
But this would imply that k −c −1 ≥c + 1 + ka or 0 ≥2(c + 1) + k(a −1), an impossibility.
If a > 0 > −1 ≥b, c, then b + c + 1 < 0, and the fourth term is negative. But then the fifth term in the period must be positive. Noting that we cannot have two negative terms immediately preceded and followd by positive terms, we have these possibilities for the signs of the terms in the period: (+, −, −, −, +, +, −, +) (+, −, −, −, +, −, +, ?) (+, −, −, −, +, −, −, −) Since the first and second of these involve the subsequence +, −, +, which can be placed at the front, these cases have been considered. Only the third possibility remains to be considered.
Suppose that the period is (p, −q, −r, −s, t, −u, −v, −w) with all of p, q, r, s, t, u, v, w positive. Then p −q + 1 = rw and −q −r + 1 = −ps, whence p + r = rw + ps or p(s−1)+r(w−1) = 0. Hence w = s = 1. Also p−w+1 = qv and −v−w+1 = −pu, 6 whence p(u −1) + v(q −1) = 0. Hence u = q = 1. It follows that p = r = v and we are led to (p, −1, −p, −1, p, −1, −p, −1).
In summary, we have the following possible periods all of whose entries are integers: (9, 5, 3, 1, 1, 1, 3, 5) (9, 2, 6, 1, 4, 1, 6, 2) (8, 3, 4, 1, 2, 1, 4, 3) (5, 4, 5, 2, 2, 1, 2, 2) (3, 2, 3, 2, 3, 2, 3, 2) (a, −1, a, 1, −(a + 2), −1, −(a + 2), 1) (a ̸= 0, −2) (p, −1, −p, −1, p, −1, −p, −1) (p ̸= 0).
Note that all of the sequences obtained with these periods are symmetrical, i.e., they are the same when they are read backwards.
If the first three terms are consecutive integers b −1, b, b + 1, then the terms are b −1, b, b + 1, 2(b + 1) b −1 , b + 3 b −1.
4 b −1, b + 3 b + 1, 2(b −1) b + 1 , b −1, . . . .
Note that the consecutive pairs of the fourth, fifth and sixth terms differ by 1.
5. The sequence C.
For the sequence C, there seems to be no periodcity for all sequences.
It is possible to have quite long chains of integers; here are some maximal subsequences of integers: {. . . , −1, −2, −1, 5, −3, −1, −2, −1, 1, 1, −1, −2, −1, −3, 5, −1, −2, −1, . . . } (18 terms) {. . . , −4, −3, −1, −5, 2, 1, 1, −1, 1, 2, 3, −7, −1, −2, −3, . . . } (15 terms; non −symmetric) {. . . , 16, 17, 9, 5, 2, 1, 1, 1, 2, 5, 9, 17, 16, . . . } (13 terms; all positive) {. . . , 5, 3, 4, 2, 2, 3, 2, 4, 5, 4, 7, . . . } {. . . , 7, 8, 3, 2, 2, 1, 2, 2, 3, 8, 7, . . . } {. . . , 4, 5, 3, 3, 3, 2, 3, 3, 3, 5, 4, . . . } C.6. If we want to find a sequence all of whose entries are integers, we might consider a periodic sequence. At this point, all positive integers are available as possible lengths of a period. Consider the possibility of periods of length 1, 2, 3, 4, etc..
6. Invariant functions.
7 For each sequence, we can define an associated function. For the sequence A, In the case of the period 5 sequence, we can construct a transformation in the plane which takes each pair of consecutive terms to the next pair of consecutive terms: TA(x, y) = (y, (y + 1)/x).
A.6. Determine a function h(x, y) which is invariant under this transformation, i.e. h(T(x, y)) = h(y, (y + 1)/x) = h(x, y).
We can do the same for the sequence B: TB(x, y, z) = (y, z, (y + z + 1)/x).
B.8. Determine a function h(x, y, z) which is invariant under this transformation.
Likewise, for the sequence C, we can define: TC(x, y, z, w) = (y, z, w, (y +z +w+ 1)/x).
C.7. Determine a function h(x, y, z, w) which is invariant under this transforma-tion.
7. Periodic sequences.
The only constant sequences A has each term equal to 1 2(1 ± √ 5), one of the roots of the quadratic function t2 −t −1.
The only constant sequences B has each term equal to 1 ± √ 2, one of the roots of the quadratic function t2 −2t −1.
For a sequence B with a period (a, b) of length 2, we require that ab = a + b + 1, so that (a −1)(b −1) = 2. We obtain the possibilities (a, b) = (1 + u, 1 + 2/u) for the period, where u ̸= 1, 2. An integer example is (2, 3).
For a sequence B with a period (a, b, c, d) of length 4, we require that b+c+1 = da, c + d + 1 = ab, d + a + 1 = bc and a + b + 1 = cd. Subtracting adjacent pairs of these equations leads to (b −d)(a + 1) = (c −a)(b + 1) = (d −b)(c + 1) = (c −a)(d + 1) = 0.
Also, we have that ad + bc = a + b + c + d + 2 = ab + cd, whence (a −c)(b −d) = 0.
The case a = c and b = d takes us back to the period of length 2, so we may suppose that a ̸= c. Then we have b = d = −1, from which c = −a and we are led to the period (a, −1, −a, −1) for any nonzero real a.
For the sequence C, the constant sequence has terms equal to one of 1 2(3± √ 13), the roots of t2 −3t −1.
For period 2, we require that 2a + b + 1 = b2 and a + 2b + 1 = a2, which leads to (a −b)(a + b + 1) = 0. If the sequence is nonconstant, then a = b2 and b = a2, so that (a, b) = (ω, ω2), (ω2, ω) where ω is an imaginary cube root of unity, i.e. a root of t2 + t + 1.
8 If (c, d, a, b, c, d, a, b) is the period of a period 4 sequence, then b + c + 1 = ad, a + b + 1 = cd, c + d + 1 = ab and a + d + 1 = bc. These lead to (a −c)(1 + d) = 0 and (a −c)(1 + b) = 0.
If a = c, then we must have a + b + 1 = ad and a + d + 1 = ab, whence (b −d)(1 + a) = 0. The case b = d leads to a period 2 sequence. If a = −1, then d = −b. This leads to the period 4 (−1, b, −1, −b). The other possibility is that b = d = −1, which leads to c = −a and the period 4 (a, −1, −a, −1), which is essentially the same as before.
For period 5 sequences with the period (a, b, c, d, e), we require that a+b+c+1 = de, b+c+d+1 = ea, c+d+e+1 = ab, d+e+a+1 = bc, e+a+b+1 = cd, from which (a −d)(e + 1) = (b −e)(a + 1) = (c −a)(b + 1) = (d −b)(c + 1) = (e −c)(d + 1) = 0.
Wolog, we may suppose that a ̸= d. Then e = −1, so that c + d = ab, a + d = bc and a + b = cd. Also a + b + c + d + 1 = a + ea = a −a = 0. Therefore (c + 1)(d + 1) = cd + c + d + 1 = a + b + c + d + 1 = 0, and either c = e = −1 or d = e = −1.
In the first case, (a + 1)(b + 1) = ab + a + b + 1 = a + b + c + d + 1 = 0, so that a = −1 or b = −1. If a = c = e = −1, then b + d = 1 and we obtain the period (−1, b, −1, 1 −b, −1) which contains only integers when b is an integer not equal to 0 or 1. b = c = e = −1 leads essentially to the same conclusion.
In the second case, d = e = −1, whereupon a + b + c = de −1 = 0, c −1 = ab and a −1 = bc. Thus (a −c)(b + 1) = 0. However, neither of these work.
8. Invariant functions.
To find an invariant function h(x, y) for the sequence A, we can simply add together the five terms of the period to get h(x, y) = x+y+(y+1)/x+(x+y+1)/xy+(x+1)/y = x2y + xy2 + x2 + y2 + 2(x + y) + 1 xy which remains invariant under the transformation TA. Alternatively, we can mul-tiply the five terms of the period to obtain (x + 1)(y + 1)(x + y + 1) xy = h(x, y) + 3.
In the case of the sequence B, we have the related transformation T(x, y, z) = (y, z, (y + z + 1)/x in 3-space with invariant function h(x, y, z) = (xyz)−1[xyz(x + y + z) + (x2y + y2z + z2x + xy2 + yz2 + zx2) + (x2 + y2 + z2) + 3(xy + yz + zx) + 2(x + y + z) + 1]. |
10277 | https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Intermediate_Physical_Organic_(Morsch)/01%3A_Models_of_Chemical_Bonding/1.02%3A_Intermolecular_Forces_-_Introduction_and_London_Dispersion | Skip to main content
1.2: Intermolecular Forces - Introduction and London Dispersion
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: Dec 31, 2022
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1.1.5: Molecular Polarity
1.2.1: Specific Interactions
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: 401721
Layne Morsch
University of Illinois Springfield
( \newcommand{\kernel}{\mathrm{null}\,})
Intermolecular forces are forces between molecules. Depending on its strength, intermolecular forces cause the forming of three physical states: solid, liquid and gas. The physical properties of melting point, boiling point, vapor pressure, evaporation, viscosity, surface tension, and solubility are related to the strength of attractive forces between molecules. These attractive forces are called Intermolecular Forces or Van der Waals forces.
Introduction
There are four types of intermolecular forces. The weakest of these are induced dipole forces (London Dispersion Forces). Most of the intermolecular forces are similar to bonding between atoms in a single molecule. Intermolecular forces just extend the thinking to forces between molecules and follows the patterns already set by the bonding within molecules.
Induced Dipole - London Dispersion Forces
London Dispersion Force is the weakest intermolecular force. It is the only attractive interaction between two nonpolar molecules.
The chance that an electron of an atom is in a certain area in the electron cloud at a specific time is called the "electron charge density." Since there is no way of knowing exactly where the electron is located and since they do not all stay in the same area 100 percent of the time, if the electrons all go to the same area at once, a dipole is formed momentarily. Even if a molecule is nonpolar, this displacement of electrons causes a nonpolar molecule to become polar for a moment, this is called an instantaneous dipole.
Since the molecule is now polar, this means that all the electrons are concentrated at one end and the molecule is partially negatively charged on that end. This negative end makes the surrounding molecules have an instantaneous dipole also, attracting the surrounding molecules' positive ends. This process is known as the London Dispersion Force of attraction.
Figure 1: Induced dipoles between iodine molecules.
The ability of a molecule to become polar and displace its electrons is known as the molecule's "polarizability." The more electrons a molecule contains, the higher its ability to become polar. Polarizability increases in the periodic table from the top of a group to the bottom and from right to left within periods. This is because the higher the molecular mass, the more electrons an atom has. With more electrons, the outer electrons are easily displaced because the inner electrons shield the nucleus' positive charge from the outer electrons..
When the molecules become polar, the melting and boiling points are raised because it takes more heat and energy to break these intermolecular forces. Therefore, the greater the mass, the more electrons present, and the more electrons present, the higher the melting and boiling points of these substances.
London dispersion forces are stronger in those molecules that are not compact, but long chains of elements. This is because these molecules have greater surface area and therefore have more points of contact to interact with other molecules.
We can calculate the potential energy between the two identical nonpolar molecules using the following formula:
α is the polarizability of nonpolar molecule.
r is the distance between the two molecule.
I = the first ionization energy of the molecule.
Negative sign indicates the attractive interaction.
Dipole Forces
Molecular dipoles occur due to the unequal sharing of electrons between atoms in a molecule. Those atoms that are more electronegative pull the bonded electrons closer to themselves. The buildup of electron density around an atom or discreet region of a molecule can result in a molecular dipole in which one side of the molecule possesses a partially negative charge and the other side a partially positive charge. Molecules with dipoles that are not canceled by their molecular geometry are said to be polar.
In the figure below, hydrochloric acid is a polar molecule with the partial positive charge on the hydrogen and the partial negative charge on the chlorine. A network of partial + and - charges attract molecules to each other.
Dipole-Dipole Interactions
When a polar molecule encounters another polar molecule, the positive end of one molecule is attracted to the negative end of the other polar molecule. Many molecules have dipoles, and their interaction occur by dipole-dipole interaction. For example: SO2 ↔SO2. (approximate energy: 15 kJ/mol). Polar molecules have permanent dipole moments, so in this case, we consider the electrostatic interaction between the two dipoles:
µ is the permanent dipole moment of the molecule 1 and 2.
Ion-Dipole Interactions
Ion-Dipole interaction is the interaction between an ion and polar molecules. For example, the sodium ion/water cluster interaction is approximately 50 KJ/mol.
Because the interaction involves in the charge of the ion and the dipole moment of the polar molecules, we can calculate the potential energy of interaction between them using the following formula:
r is the distance of separation.
q is the charge of the ion ( only the magnitude of the charge is shown here.)
is the permanent dipole moment of the polar molecule.
Hydrogen Bonding
The hydrogen bond is really a special case of dipole forces. A hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine. In other words - The hydrogen on one molecule attached to O or N that is attracted to an O or N of a different molecule.
In the graphic below, the hydrogen is partially positive and attracted to the partially negative charge on the oxygen or nitrogen. Because oxygen has two lone pairs, two different hydrogen bonds can be made to each oxygen. This is a very specific bond as indicated. Some combinations that are not hydrogen bonds include: hydrogen to another hydrogen or hydrogen to a carbon.
Coulombic Forces
The forces holding ions together in ionic solids are electrostatic forces. Opposite charges attract each other. These are the strongest intermolecular forces. Ionic forces hold many ions in a crystal lattice structure. According to Coulomb’s law:
is the charges.
is the distance of separation.
Based on Coulomb’s law, we can find the potential energy between different types of molecules.
References
Kotz, Treichel, Weaver. Chemistry and Chemical reactivity, sixth ed. Thompson, 2006.
Donald Allan McQuarrie, John Douglas Simon. Physical Chemistry: a molecular approach. University Science Books, 1997.
Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications. Upper Saddle River, NJ: Prentice Hall, 2007
Contributors and Attributions
Kathryn Rashe, Lisa Peterson, Seila Buth, Irene Ly
1.1.5: Molecular Polarity
1.2.1: Specific Interactions |
10278 | https://www.sciencedirect.com/science/article/abs/pii/S0012369216616574 | Adverse Effect of Propranolol on Airway Function in Nonasthmatic Chronic Obstructive Lung Disease - ScienceDirect
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Volume 79, Issue 5, May 1981, Pages 540-544
Clinical Investigations
Adverse Effect of Propranolol on Airway Function in Nonasthmatic Chronic Obstructive Lung Disease
Author links open overlay panel Edward H.Chester M.D., F.C.C.P., Howard J.Schwartz M.D., F.C.C.P., Gerald M.Fleming M.D., F.C.C.P.
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Bronchomotor tone is, in part, under β-adrenergic control, and β-adrenergic agonists are commonly used in the therapy for chronic obstructive pulmonary disease (COPD). β-adrenergic blockade with propranolol is contraindicated in asthmatic patients, yet little is known of its effect in patients with COPD. We studied 13 patients with COPD in a random-entry, double-blind crossover comparison of oral propranolol, 40 mg, and oral placebo on separate days. Pulmonary function worsened after administration of propranolol. Significant differences were present between the drugs' effect on heart rate, airway resistance, specific resistance, and flow rates at one hour, and persisting through four hours (p < 0.01). Propranolol may have a deleterious effect on pulmonary function in nonasthmatic COPD. We conclude that when propranolol is to be used in patients with COPD, the short- and long-term effects on airways should be measured sequentially.
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Materials and Methods
We selected 13 adult male patients who had nonasthmatic obstructive lung disease (Table 1). A diagnosis of chronic bronchitis or emphysema or both was made in accordance with the diagnostic standards of the American Thoracic Society.13 Informed consent was provided by each subject, and the study was approved by the committee on human research at our institution.
Baseline pulmonary function tests included forced vital capacity (FVC), forced expired volume in one second (FEV 1), and forced
Results
Since there was considerable variation in baseline airway dynamics among patients, we chose to normalize all data by reducing the measurements taken following treatment by the baseline value, so that the subsequent analysis of variance could be performed on all values, taking the difference between each observed measurement and the baseline value. A two-way multivariate analysis of variance was performed, with time as one factor (with four levels) and treatment as the second factor (with two
Discussion
In this study, we have documented that patients with COPD demonstrate a significant and sustained increase in airway resistance and diminution of expiratory flow rates following a single, oral dose of 40 mg of propranolol compared with oral placebo.
Propranolol had previously been reported by one laboratory to result in an increased R aw in normal subjects.6, 7 In these studies, MacDonald et al measured R aw in 18 normal subjects at three- to five-minute intervals for 30 minutes following
ACKNOWLEDGMENT
The authors are grateful to Ronald Lendvay for invaluable technical assistance.
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Effect of a β-adrenergic-blocking agent, propranolol, on asthmatics
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β-adrenergic blockade of the lung: dose-dependent cardio-selectivity of tolamolol in asthma
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Effect of propranolol on ventilatory function
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The effect of propranolol on airway resistance
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Effect of propranolol on respiratory function and exercise tolerance in patients with chronic obstructive lung disease
Chest (1975)
RC Kory et al.
The Veterans Administration cooperative study of pulmonary function: I. Clinical spirometry in normal men
Am J Med (1961)
HG Boren et al.
The Veterans Administration cooperative study of pulmonary function: II. The lung volume and its subdivisions in normal men
Am J Med (1966)
CB Payne et al.
Airway responsiveness in chronic obstructive pulmonary disease
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G Zaid et al.
Bronchial response to β-adrenergic blockade
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PS Richardson et al.
Effects of β-adrenergic receptor blockade on airway conductance and lung volumes in normal and asthmatic subjects
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J Meier et al.
The action of β-receptor blockers on ventilatory functions in obstructive lung diseases
German Med Monthly (1966)
View more references
Cited by (31)
Drug-induced pulmonary disease
1994, Disease A Month Show abstract Drug-induced disease of any system or organ can be associated with high morbidity and mortality, and it is tremendously costly to the health care of our country. More than 100 medications are known to affect the lungs adversely, including the airways in the form of cough and asthma, the interstitium with interstitial pneumonitis and noncardiac pulmonary edema, and the pleura with pleural effusions. Patients commonly do not even know what medications they are taking, do not bring them to the physician's office for identification, and usually do not relate over-the-counter medications with any problems they have. They assume that all nonprescription drugs are safe. Patients also believe that if they are taking prescription medications at their discretion, meaning on an as-needed basis, then these medications are also not important. This situation stresses just how imperative it is for the physician to take an accurate drug history in all patients seen with unexplained medical situations. Cardiovascular drugs that most commonly produce a pulmonary abnormality are amiodarone, the angiotensin-converting enzyme inhibitors, and β-blockers. Pulmonary complications will develop in 6% of patients taking amiodarone and 15% taking angiotensin converting enzyme inhibitors, with the former associated with interstitial pneumonitis that can be fatal and the latter associated with an irritating cough that is not associated with any pathologic or physiologic sequelae of consequence. The β-blockers can aggrevate obstructive lung disease in any patient taking them. Of the antiinflammatory agents, acetylsalicylic acid can produce several different airway and parenchymal complications, including aggravation of asthma in up to 5% of patients with asthma, a noncardiac pulmonary edema when levels exceed 40 mg/dl, and a pseudosepsis syndrome. More than 200 products contain aspirin. Low-dose methotrexate is proving to be a problem because granulomatous interstitial pneumonitis develops in 5% of those patients receiving it. This condition occurs most often in patients receiving the drug for rheumatoid arthritis, but it has been reported in a few patients receiving it for refractory asthma. Chemotherapeutic drug-induced lung disease is almost always associated with fever, thus mimicking opportunistic infection, which is the most common cause of pulmonary complications in the immunocompromised host. However, in 10% to 15% of patients, the pulmonary infiltrate is due to an adverse effect from a chemotherapeutic agent. This complication is frequently fatal even when recognized early. Some medications produce a cytotoxic effect, meaning an atypia of the type I and II pneumocytes, whereas others produce a noncardiac pulmonary edema, microangiopathic hemolytic anemia with pulmonary edema, an eosinophilic pneumonitis, or a granulomatous reaction. The clinician responsible for these patients must relate to the pathologist what medications the patient may be taking. Illicit drugs, especially heroin and cocaine, are tremendous problems to the physician in the emergency department. The use of these drugs must be kept in mind when the patient has unexplained acute pulmonary symptoms. Nitrofurantoin is by far the most common antibiotic-induced lung disease, with the acute reaction having been reported in more than 1000 cases around the world. It is associated with acute onset of dyspnea, cough, and fever but rapidly resolves with discontinuation. The long-term side effect is a separate reaction that mimics idiopathic interstitial pneumonitis and fibrosis except that it usually responds slowly to discontinuation and sometimes to the addition of corticosteroids. Numerous other medications are associated with adverse drug reactions affecting the lungs, airways, and pleura, and they must be kept in mind when confronted with the patient who has an unexplained pulmonary problem.
### Cardioselective beta-blockers for chronic obstructive pulmonary disease
2005, Cochrane Database of Systematic Reviews
### Metoprolol, a β-1 selective blocker, can be used safely in coronary artery disease patients with chronic obstructive pulmonary disease
2003, Heart and Vessels
### Effects of single oral doses of bisoprolol and atenolol on airway function in nonasthmatic chronic obstructive lung disease and angina pectoris
1986, European Journal of Clinical Pharmacology
### The additional properties of beta adrenoceptor blocking drugs
1986, Journal of Cardiovascular Pharmacology
### Influence of cardioselectivity and respiratory disease on pulmonary responsiveness to beta-blockade
1984, European Journal of Clinical Pharmacology
View all citing articles on Scopus
This study was supported in part by the Medical Research Service of the Veterans Administration and by the Northeast Ohio Lung Association. This study was approved by the Subcommittee on Human Studies at the Cleveland Veterans Administration Medical Center. Informed consent was obtained from the patients.
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Copyright © 1981 The American College of Chest Physicians. Published by Elsevier Inc. All rights reserved.
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10279 | https://civilcalc.com/beam-calculator | Free Beam Calculator Online
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10280 | https://www.quora.com/Why-is-there-no-2n-1-of-odd-numbers-form | Why is there no 2n-1 of odd numbers form? - Quora
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Mathematics
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Prime Number Theory
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Why is there no 2n-1 of odd numbers form?
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Alex Moon
BS in Pure Mathematics, Michigan State University · Author has 3.4K answers and 2.1M answer views
·Updated 1y
There is,
if you are dealing with positive odd numbers then the form O(n)=2 n−1 O(n)=2 n−1 makes tons of sense as it defines 1 1 as the first odd number as O(1)=2(1)−1=1 O(1)=2(1)−1=1
where as defining your rule as O(n)=2 n+1 O(n)=2 n+1 instead yields 1 1 as your zeroth odd number which can seem weird, this O(1)=2(1)+1=3 O(1)=2(1)+1=3 which just feels unnatural.
At the end of the day people use whatever is more circumstantially convenient for whatever they are doing in the moment, some people find the comfort of only using one form invariably as convenient while others can see ahead of time what makes algebra easier in a given
Continue Reading
There is,
if you are dealing with positive odd numbers then the form O(n)=2 n−1 O(n)=2 n−1 makes tons of sense as it defines 1 1 as the first odd number as O(1)=2(1)−1=1 O(1)=2(1)−1=1
where as defining your rule as O(n)=2 n+1 O(n)=2 n+1 instead yields 1 1 as your zeroth odd number which can seem weird, this O(1)=2(1)+1=3 O(1)=2(1)+1=3 which just feels unnatural.
At the end of the day people use whatever is more circumstantially convenient for whatever they are doing in the moment, some people find the comfort of only using one form invariably as convenient while others can see ahead of time what makes algebra easier in a given case and take the easier route.
If you were to make it a sequence of odd numbers I’d use the 2 n−1 2 n−1 form because it takes the positive integers as a domain and sends them to the odd numbers nicely, 1 1 maps to 1 1 and the rest follows suit. I do not like sending the natural numbers to the odd numbers by sending 0 0 to 1 1 and 1 1 to 3 3 as it just feels gross to me.
However, if you were to ask me to prove the product of two odd numbers is odd you’re far more apt to see me do this:
∃n,m∈Z ST x=2 n+1,y=2 m+1∃n,m∈Z ST x=2 n+1,y=2 m+1
x y x y
by definition of “odd”==by definition of “odd”
(2 n+1)(2 m+1)(2 n+1)(2 m+1)
==
4 m n+2 m+2 n+1 4 m n+2 m+2 n+1
==
2(2 m n+m+n)+1 2(2 m n+m+n)+1
by closure of Z under×&+==by closure of Z under×&+
2 k+1 2 k+1
for some integer k k.
This is an odd number by defintion.
Either definition is valid since the odd numbers are all congruent to 1 mod 2 1 mod 2 and 1 mod 2=−1 1=mod 2−1
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What is the odd number 1 to 100?
Why are odd numbers always of the form 2n+1 and not 2n-1?
Both (2n+1) and (2n-1) represent the odd number series. What is the difference between both and why do we prefer (2n-1)?
Is it true or false that for every positive integer n, (2n-1) (2n+1) is an odd number? Why?
Is 1/2 an even or an odd number?
Kishan Panaganti Badrinath
PhD student at TAMU · Author has 566 answers and 2.5M answer views
·8y
Related
Why are odd numbers always of the form 2n+1 and not 2n-1?
Both {2 n−1,|∀n∈N}{2 n−1,|∀n∈N} and {1}∪{2 n+1,|∀n∈N}{1}∪{2 n+1,|∀n∈N} represent the set of positive odd numbers. Use any notation as per convenience.
For example:
Question 1: Sum of first n n positive odd numbers is n 2 n 2.
Proof using the 2 n−1 2 n−1 notation but not limited to this notation:
Sum of first n n positive odd numbers =1+3+5+⋯+(2 n−1)=1+3+5+⋯+(2 n−1). Since n n is finite, add this sequence with itself in the reverse order, i.e., (1+(2 n−1))+(2+(2 n−2))+⋯+((2 n−1)+1)(1+(2 n−1))+(2+(2 n−2))+⋯+((2 n−1)+1) which is equal to 2 n 2.■2 n 2.◼
Question 2: If n n is odd, then n 2 n 2 is odd.
Proof using the 2 n+1 2 n+1 notation
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Both {2 n−1,|∀n∈N}{2 n−1,|∀n∈N} and {1}∪{2 n+1,|∀n∈N}{1}∪{2 n+1,|∀n∈N} represent the set of positive odd numbers. Use any notation as per convenience.
For example:
Question 1: Sum of first n n positive odd numbers is n 2 n 2.
Proof using the 2 n−1 2 n−1 notation but not limited to this notation:
Sum of first n n positive odd numbers =1+3+5+⋯+(2 n−1)=1+3+5+⋯+(2 n−1). Since n n is finite, add this sequence with itself in the reverse order, i.e., (1+(2 n−1))+(2+(2 n−2))+⋯+((2 n−1)+1)(1+(2 n−1))+(2+(2 n−2))+⋯+((2 n−1)+1) which is equal to 2 n 2.■2 n 2.◼
Question 2: If n n is odd, then n 2 n 2 is odd.
Proof using the 2 n+1 2 n+1 notation but not limited to this notation:
Let n n be equal to 2 m+1 2 m+1, there exists such an m m since n n is odd. Now n 2=(2 m+1)2=4 m(m+1)+1.■n 2=(2 m+1)2=4 m(m+1)+1.◼
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·
Sep 6
Interpretation: the question asks why an arithmetic progression of the form 2n−1 (with integer n) does not produce every odd number or perhaps why there is "no 2n−1 of odd numbers form." Clarifying the intended meaning, the standard formula for odd integers is 2n+1 or 2n−1 (with integer n). Explain why this represents all odd integers and how indexing conventions work.
Definition: an integer m is odd iff m = 2k+1 for some integer k. Equivalently m = 2k−1 for some integer k — both forms generate the same set because shifting k by 1 maps one form to the other: 2(k−1)+1 = 2k−1.
Why these linear for
Continue Reading
Interpretation: the question asks why an arithmetic progression of the form 2n−1 (with integer n) does not produce every odd number or perhaps why there is "no 2n−1 of odd numbers form." Clarifying the intended meaning, the standard formula for odd integers is 2n+1 or 2n−1 (with integer n). Explain why this represents all odd integers and how indexing conventions work.
Definition: an integer m is odd iff m = 2k+1 for some integer k. Equivalently m = 2k−1 for some integer k — both forms generate the same set because shifting k by 1 maps one form to the other: 2(k−1)+1 = 2k−1.
Why these linear forms capture all odd numbers:
Every integer is either even or odd. Even integers are divisible by 2, so they equal 2t. Odd integers leave remainder 1 when divided by 2, so they equal 2t+1 for some integer t (the division algorithm).
Writing 2n−1 is just a reindexing: let n = t+1, then 2n−1 = 2(t+1)−1 = 2t+1. Thus both forms enumerate the same set {..., −3, −1, 1, 3, 5, ...}.
Common confusions:
If you restrict n to positive integers, 2n−1 lists the positive odd numbers starting at 1 (n=1 →1, n=2 →3, ...). If you allow n to be all integers, 2n−1 gives all odd integers including negatives and zero is not included because it is even.
Some people prefer 2n+1 with n∈Z or n∈Z≥0; others prefer 2n−1 with n∈Z≥1. Both are correct conventions.
Conclusion: There is indeed a valid "2n−1" form for odd numbers; it is equivalent to 2n+1 up to an index shift. The reason one sees 2n+1 more often is merely convention about how n is indexed, not a mathematical obstruction.
Upvote ·
Simha Srivatsa
Studied at Sri Bhagawan Mahaveer Jain College
·8y
Related
Why are odd numbers always of the form 2n+1 and not 2n-1?
It's the same thing.
If you use 2n-1 then n varies from 1 to infinity ie natural numbers
If you use 2n+1 then n varies from 0 to infinity- whole numbers.
We know that addition of two odd numbers is even so
(2n-1)+(2n+1) = 4n which is even.
Multiplication of two odd numbers gives an odd number.
(2n-1)(2n+1)= 4n²-1, when u divide this by 2 u get remainder as 1 so it's odd.
So both represent an odd number, depending on the limits for n in an equation ( where many other things depend on n)you can write 2n-1 or 2n+1 to represent an odd number.
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More answers below
If 2n+1 is the formula for all odd numbers, what is the formula for all even numbers?
Is the number 100 1/2 even or odd?
Can you add 5 odd numbers to get 30?
Why is 1 considered as odd number?
What is the closed form of ∞∑n=0(−1)n(4 x)2 n(2 n)!(4 n+1)!∑n=0∞(−1)n(4 x)2 n(2 n)!(4 n+1)!?
John Steenbruggen
8y
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Why is the sequence of odd numbers 1, 3, 5, 7 not (n+2) but (2n-1)?
When writing a sequence, the first element is at position n=1 n=1. The second element is at position n=2 n=2, the third at n=3 n=3, and so forth.
If this sequence were defined as n+2 n+2, the first element (n=1 n=1) would be 1+2+3 1+2+3. The second element (n=2 n=2) would be 2+2=4 2+2=4. The third element (n=3 n=3) would be 3+2=5 3+2=5, and so on. Obviously, 2,3,4,5,...2,3,4,5,... is not the same sequence, so n+2 n+2 cannot represent this sequence.
If this sequence was defined as 2 n−1 2 n−1, the first element (n=1 n=1) would be 2(1)−1=1 2(1)−1=1. The second element (n−2 n−2) would be 2(2)−1=3 2(2)−1=3. The third element (n=3 n=3) would be 2(3)−1=5 2(3)−1=5, and so on. This sequence matches 1,3,5,7,.1,3,5,7,.
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When writing a sequence, the first element is at position n=1 n=1. The second element is at position n=2 n=2, the third at n=3 n=3, and so forth.
If this sequence were defined as n+2 n+2, the first element (n=1 n=1) would be 1+2+3 1+2+3. The second element (n=2 n=2) would be 2+2=4 2+2=4. The third element (n=3 n=3) would be 3+2=5 3+2=5, and so on. Obviously, 2,3,4,5,...2,3,4,5,... is not the same sequence, so n+2 n+2 cannot represent this sequence.
If this sequence was defined as 2 n−1 2 n−1, the first element (n=1 n=1) would be 2(1)−1=1 2(1)−1=1. The second element (n−2 n−2) would be 2(2)−1=3 2(2)−1=3. The third element (n=3 n=3) would be 2(3)−1=5 2(3)−1=5, and so on. This sequence matches 1,3,5,7,...1,3,5,7,..., and therefore 2 n−1 2 n−1 does model this sequence.
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Patrick Enfield
Studied Physics&Philosophy (Graduated 1979) · Author has 136 answers and 9.6K answer views
·Apr 4
Related
Why are odd numbers always of the form 2n+1 and not 2n-1?
That is a false dichotomy . Even if you prefer to represent an old number as 2n +1 there will be a value N different from lower case n for which the odd number equals h2N-1.
In the immediately following paragraph I explain how you can prove that any odd number can be expressed as 2n+1. After that I set out the algebra connecting the adding 1to an even number with the alternative way of subtracting 1 from a different even number
Apply the quotient remainder theorem given that the odd number can't be of the form M=2n=2n+0 assuming a number is defined as odd just in the case iit is not even
Given th
Continue Reading
That is a false dichotomy . Even if you prefer to represent an old number as 2n +1 there will be a value N different from lower case n for which the odd number equals h2N-1.
In the immediately following paragraph I explain how you can prove that any odd number can be expressed as 2n+1. After that I set out the algebra connecting the adding 1to an even number with the alternative way of subtracting 1 from a different even number
Apply the quotient remainder theorem given that the odd number can't be of the form M=2n=2n+0 assuming a number is defined as odd just in the case iit is not even
Given that he quotient (divisor) is 2 the remainder must be less than 2 but greater than or equal to 0. I mean the theorem guarantees the existence (and uniqueness) of such a remainder given the choice of 2 for the quotient. However from this it follows that there is a value of N for which m=2N - 1 namely N =n+ 1
To avoid having to introduce Greek letters I am distinguishing between lower and upper case n.
If odd number m =2 n +1 for some positive integer n
Then p=2(n+1)-1=2n+2–1=2n+ 1=m
also p=2N-1
Btw I have got nothing against Greek letters other than the difficulty accessing them on mobiles
One of my best friends is Greek so I am definitely not hellenophobic.
You are just subtracting the 1 from a different even number than for the other expression when you add 1 to an even number so you assume a false dichotomy in the question.i wrote this before it's restatement above
Representing odd numbers as 2n+1 follows directly from quotient remainder . 2n-1 is not a direct application but follows fairly easily from the even number plus 1 formulation
Upvote ·
Lance Berg
Author has 28K answers and 54.7M answer views
·Updated 9mo
Related
If all prime numbers are odd, why isn't 2 an odd number?
All prime numbers are not odd.
Two is prime, and not odd.
All prime numbers other than two are odd.
Oddness means not-divisible by two
Primeness means not divisible by numbers other than itself and 1
Two is not divisible by numbers other than itself and 1.
Two is prime.
There is nothing unique about this property set.
3 is prime. But all numbers divisible by 3 other than 3 itself are not prime, since they are divisible by 3 as well as themselves and 1.
5 is prime. But all numbers divisible by 5 other than 5 itself are not prime, for the same reason.
7 is prime. But all numbers divisible by 7 other than
Continue Reading
All prime numbers are not odd.
Two is prime, and not odd.
All prime numbers other than two are odd.
Oddness means not-divisible by two
Primeness means not divisible by numbers other than itself and 1
Two is not divisible by numbers other than itself and 1.
Two is prime.
There is nothing unique about this property set.
3 is prime. But all numbers divisible by 3 other than 3 itself are not prime, since they are divisible by 3 as well as themselves and 1.
5 is prime. But all numbers divisible by 5 other than 5 itself are not prime, for the same reason.
7 is prime. But all numbers divisible by 7 other than 7 itself are not prime, for the same reason.
In fact for any given prime number n, n is prime, but all numbers divisible by n other than n itself are not prime, for the same reason.
The thing that is confusing you is that we have a special name for “not divisible by two” and not for those other not-divisible properties.
But they are all the same thing.
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Sandipan Dutta
Lives in Roorkee, Uttarakhand, India (2006–present)
·7y
Related
Why are odd numbers always of the form 2n+1 and not 2n-1?
Odd numbers are expressed by the following formula: 2n+1 where n belongs to whole numbers
That means for every whole number you enter, you will get a corresponding odd number.
For eg.
For n=0, the odd number is: 2(0)+1=1
For n=1, 2(1)+1= 3
This way, the series continues.
Hope it helps. :D
Upvote ·
9 4
Anonymous
9y
Related
Why is the sequence of odd numbers 1, 3, 5, 7 not (n+2) but (2n-1)?
Your Point is Right if n belongs to a set of odd natural numbers.Here n belongs to a set of natural numbers which can be odd or even.
For example if you take n=5 then n+2 = 7 is odd which is the right answer.
if n=2 then n+2 = 4 is even which is wrong according to our assumption that n+2 is a sequence of odd numbers.
But 2n-1 would always give an odd number under the assumption that n belongs to natural numbers.
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Aditya Tripathi
5y
Related
Both (2n+1) and (2n-1) represent the odd number series. What is the difference between both and why do we prefer (2n-1)?
Hey,
First of all I am considering as a simple thought process must go rather than going via facts… what I mean is that.
(2n-1) is preferred because it goes with what we simply think that is if we need first positive odd number we will simply think of putting n=1 and we will get the answer which is nothing but “1” , n=2 for second(3) and so on…
If we consider (2n+1) though it also frame required odd numbers for desired n but here what happens is that it gives us first positive odd number ”1″ at n=0 which is fine but doesn't go with the normal flow of our thought.
So may be to keep our thinking str
Continue Reading
Hey,
First of all I am considering as a simple thought process must go rather than going via facts… what I mean is that.
(2n-1) is preferred because it goes with what we simply think that is if we need first positive odd number we will simply think of putting n=1 and we will get the answer which is nothing but “1” , n=2 for second(3) and so on…
If we consider (2n+1) though it also frame required odd numbers for desired n but here what happens is that it gives us first positive odd number ”1″ at n=0 which is fine but doesn't go with the normal flow of our thought.
So may be to keep our thinking straight and smooth we consider (2n-1) over (2n+1).
It's a math fact that both gives us odd number for respective n but it's just a simple thought process of our learning and executions gives us a preference to go with(2n-1).
Hope so it's useful. It just came to my mind while searching the same question 😅😆
Overall conclusion is that
It is just for convenience .🙄
Upvote ·
99 11
Nirmit Shah
Btech from Dhirubhai Ambani Institute of Information and Communication Technology (DA-IICT) (Graduated 2021) · Author has 170 answers and 471.7K answer views
·7y
Related
Why are odd numbers always of the form 2n+1 and not 2n-1?
It's the same thing.
If you use 2n-1 then n varies from 1 to infinity ie natural numbers
If you use 2n+1 then n varies from 0 to infinity- whole numbers.
Thanks for reading!!
Nirmit
Upvote ·
9 4
Hari Narayanan
8y
Related
Why are odd numbers always of the form 2n+1 and not 2n-1?
U can represent but n does not belong to natural numbers if u represent odd numbers as 2n+1. n starts form 0. But if u represent odd numbers as 2n-1. n belongs to natural numbers. People dont like anything to start from 0. So people represent odd numbers as 2n-1.
Upvote ·
9 3
Mark Linton
5y
Related
Why are odd numbers always of the form 2n+1 and not 2n-1?
I found a counter example. In Google Books I found a book called “An Elementary investigation of the Theory of Numbers…” by Peter Barlow (1811). on Page 5 it says: Every Odd Number is of the form 2n±l.
Upvote ·
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Both (2n+1) and (2n-1) represent the odd number series. What is the difference between both and why do we prefer (2n-1)?
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10281 | https://en.wikipedia.org/wiki/Neural_plate | Jump to content
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Contents
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1 Involvement in primary neurulation
2 Development
2.1 Cell signaling and essential proteins
3 Other animals
4 Research
4.1 Cell labelling
4.2 Cell grafting
5 References
6 External links
Neural plate
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From Wikipedia, the free encyclopedia
Structure in an embryo which will become the nervous system
| Neural plate |
| Neural crest |
| Details |
| Carnegie stage | 9 |
| Days | 19 |
| Precursor | Ectoderm |
| Gives rise to | Neural folds |
| System | Central nervous system |
| Identifiers |
| Latin | lamina neuralis |
| MeSH | D054258 |
| TE | plate_by_E5.13.1.0.1.0.1 E5.13.1.0.1.0.1 |
| Anatomical terminology [edit on Wikidata] |
In embryology, the neural plate is a key developmental structure that serves as the basis for the nervous system. Cranial to the primitive node of the embryonic primitive streak, ectodermal tissue thickens and flattens to become the neural plate. The region anterior to the primitive node can be generally referred to as the neural plate. Cells take on a columnar appearance in the process as they continue to lengthen and narrow. The ends of the neural plate, known as the neural folds, push the ends of the plate up and together, folding into the neural tube, a structure critical to brain and spinal cord development. This process as a whole is termed primary neurulation.
Signaling proteins are also important in neural plate development, and aid in differentiating the tissue destined to become the neural plate. Examples of such proteins include bone morphogenetic proteins and cadherins. Expression of these proteins is essential to neural plate folding and subsequent neural tube formation.
Involvement in primary neurulation
[edit]
Generally divided into four, the process of primary neurulation involves the neural plate in the first three steps. The formation and folding of the neural plate is the first step in primary neurulation. This is followed by the refinement and growth of neural plate cells. The third step of primary neurulation does not involve the neural plate per se, but rather the edges of the neural plate, which come together, turning the plate into the start of the neural tube. With the neural plate having folded into a tube, the neural folds come together to complete the fusion of the neural tube. This process is illustrated in the figure to the right, where the neural plate is shown in purple. The lime green marks the edges of the neural plate, which become the neural folds, involved in the folding of the plate to create the neural tube. The figure demonstrates the development of the neural plate into the neural tube, which is where the neural crest cells are derived from as well.
In primary neurulation, the layer of ectoderm divides into three sets of cells: the neural tube (future brain and spinal cord), epidermis (skin), and neural crest cells (connects epidermis and neural tube and will migrate to make neurons, glia, and skin cell pigmentation).
Development
[edit]
During the stage of neural plate formation the embryo consists of three cell layers: the ectoderm that eventually forms the skin and neural tissues, the mesoderm that forms muscle and bone, and the endoderm that will form the cells lining the digestive and respiratory tracts. The progenitor cells that make up the precursors to neural tissues in the neural plate are called neuroepithelial cells.[citation needed]
Stretched over the notochord, the ectodermal cells on the dorsal portion of the embryo are ultimately the ones that form the neural plate. Approximately half of those cells will be induced to remain ectoderm, while the other half will form the neural plate.
There are four stages of neural plate and neural tube formation: formation, bending, convergence, and closure. The formation of the neural plate starts when dorsal mesoderm signals ectodermal cells above it to lengthen into columnar neural plate cells. This different shape distinguishes the cells of the presumptive neural plate from other pre-epidermal cells. If the neural plate is separated by itself, it will still develop to make a thinner plate but will not form a neural tube. If the region containing presumptive epidermis and neural plate tissue is isolated, small neural folds will form. Elongation that occurs throughout the formation of the neural plate and closure of the neural tube is vital; the closing areas of the neural tube are seen to have very increased elongation activity in the midline compared to already closed areas when the plate was beginning to shape itself into a tube.
The bending of the neural plate involves the formation of hinges, where the neural plate is connected to surrounding tissues. The midline of the neural plate is referred to the median hinge point (MHP). Cells in this area, known as medial hinge point cells because of their involvement with this structure, are stabilized and connected to the notochord. They are derived from the area of the neural plate anterior to primitive knot. The notochord will begin the shape changes in MHP cells. These cells will decrease in height and become wedge-shaped. Another type of hinge point occurs dorsal-laterally, referred to as dorsal-lateral hinge point (DLHP). These regions furrow and change shape in the same way as MHP cells do before connecting together to form the neural tube. It was seen in an experiment that without the notochord, the MHP characteristics did not develop correctly, so the neural plate and neural tube formation did not happen properly. The communication between the neural plate and the notochord is important for the future induction and formation of the neural tube.
Closure of the neural tube is completed when the neural folds are brought together, adhering to each other. While the cells that remain as the neural tube form the brain and spinal cord, the other cells that were part of the neural plate migrate away from the tube as neural crest cells. After an epithelial–mesenchymal transition, these cells form the autonomic nervous system and certain cells of the peripheral nervous system.
Cell signaling and essential proteins
[edit]
Critical to the proper folding and function of the neural plate is N-cadherin, a type of cadherin protein associated with the nervous system. N-cadherin is critical to holding neural plate cells together. Additionally, cells destined to become neural plate cells express nerve cell adhesion molecule (NCAM) to further neural plate cohesion. Another cadherin, E-cadherin, is expressed by ectodermal cells in the process of neural plate development.
Bone morphogenetic protein 4, or BMP4, is a transforming growth factor that causes the cells of the ectoderm to differentiate into skin cells. Without BMP4 the ectoderm cells would develop into neural cells. Axial mesoderm cells under the ectoderm secrete inhibitory signals called chordin, noggin and follistatin. These inhibitory signals prevent the action of BMP4, which would normally make the cells ectoderm; as a result, the overlying cells take their normal course and develop into neural cells. The cells in the ectoderm that circumscribe these neural cells do not receive the BMP4 inhibitor signals and as a result BMP4 induces these cells to develop into skin cells.
Neural plate border specifiers are induced as a set of transcription factors. Distalless-5, PAX3 and PAX7 prevent the border region from becoming either neural plate or epidermis. These induce a second set of transcription factors called neural crest specifiers, which cause cells to become neural crest cells.
In a newly formed neural plate, PAX3 mRNA, MSX1 mRNA, and MSX1/MSX2 proteins are expressed mediolaterally. When the neural plate begins to fold, rostral areas of the neural plate do not express Pax3 and MSX proteins. Areas caudal to neural tube closure have PAX3 and MSX expression restricted to lateral regions of the neural folds. These fluctuations in mRNA and protein expression allude to how they play a role in differentiation of neural plate cells.
Low pSMAD 1, 5, 8 levels allow a greater mobility at the median hinge point than in lateral neural plate cells. This flexibility allows for the pivoting and hinging that allows the buckling and lifting of the neural plate when formatting the neural tube. The neural plate has to be rigid enough for morphogenic movements to occur while being flexible enough to undergo shape and position changes for the transformation to the neural tube.
Other animals
[edit]
The neural tube closes differently in various species, the distinctions between humans and chickens being some of the most studied. In humans, the neural tube fuses together from a central region of the embryo and moves outwards. In chickens, neural tube closure begins at the future midbrain region and it closes in both directions. In birds and mammals, the closure does not occur at the same time.
In newt and general amphibian embryos, cell division is not a driving role in morphogenesis. Newt embryo cells are much larger and exhibit egg pigmentation to distinguish cells from each other. The newt neural plate doubles in length, decreases in apical width, and increases in thickness. The plate edges rise dorsally and fold toward the midline to form the neural tube. The apical surface area decreases.
In chicken embryos, while the neural plate increases in length and decreases in apical width, the thickness of the plate does not change drastically. As the neural plate progresses through the Hamburger-Hamilton stages, the plate thickens until about HH6-7, when the neural plate begins to fold into tube form. The apical surface area increases during neurulation, unlike amphibian embryos. In mouse embryos, there is a large convex-shaped curve to each side of the middle of the plate. This curve has to be reversed as the plate rolls together to form the neural tube.
Research
[edit]
Research on the neural plate began in earnest by looking into the determination of the ectoderm and its commitment to the neuronal path. With the development of research and laboratory techniques there have been major advances in the study of neurulation and the development and role of the neural plate in a growing embryo. The use of such techniques vary with the stage of development and overall research goals, but include such methods as cell labeling and grafting.
Cell labelling
[edit]
The process of in situ hybridization (ISH) follows the labeling of a DNA or RNA sequence to serve as an antisense mRNA probe, complementary to a sequence of mRNA within the embryo. Labeling with a fluorescent dye or radioactive tag allows for the visualization of the probe and their location within the embryo. This technique is useful as it reveals specific areas of gene expression in a tissue as well as throughout an entire embryo through whole-mount in situ hybridization. This technique is often used in determination of gene expression necessary for the proper development of the embryo. Marking certain genes in a developing embryo allows for the determination of the exact time and place in which the gene is activated, offering information in the role of the particular gene in development.
Similar to the process of in situ hybridization, immunofluorescence (IF) also allows for the determination of particular cell element's roles in development. In contrast to in situ hybridization however, immunofluorescence uses a fluorophore attached to an antibody with biomolecule target, such as proteins, rather than DNA and RNA sequences. The allows for the visualization of biomolecule elements of the cell. In the study of embryogenesis immunofluorescence may be used for purposes similar to hybridization, for the tracking of proteins that are involved in the development of the embryo and their specific time and place of production and use. Current research has expanded on the immunofluorescence technique to combined it with the methods of in situ hybridization, either fluorescent or radioactive. This combination is believed to increase specificity and take away for the limitations of each individual technique. For example, this method with enhance counterstaining in a tissue and multiple protein labeling.
Cell grafting
[edit]
Cell grafting in the early stages of embryo development has provided crucial information on cell fates and the processes of determination. Grafting at specific stages of neurulation has advanced research on the signaling necessary for the proper development of the neural plate and other structures. The grafting of the ectoderm and neural structures is very specialized and delicate procedure, requiring the removal and marking of a desired group of cells, followed by their transplantation, for example, into a new area of the embryo.
Grafting experiments done in Xenopus and chicken embryos show the neural plate's capability to induce other regions of cells, including the pre-placodal region, a group of ectodermal cells essential to the function of sensory organs.
References
[edit]
This article incorporates text in the public domain from the 20th edition of Gray's Anatomy (1918)
^ a b c d e f Gilbert, Scott F. (2010). Developmental biology (9th. ed.). Sunderland, Mass.: Sinauer Associates. pp. 333–338. ISBN 978-0878933846.
^ Browder, Leon (1980). Developmental Biology. Philadelphia: Saunders College. p. 457. ISBN 0-03-056748-3.
^ Human Embryology, Module 7, Section 7.2, Archived 2013-01-16 at the Wayback Machine.
^ Keller, Ray; Shih, John; Sater, Amy K (1 March 1992). "Planar induction of convergence and extension of the neural plate by the organizer Xenopus". Developmental Dynamics. 193 (3): 218–234. doi:10.1002/aja.1001930303. PMID 1600241. S2CID 39722561.
^ a b c d Jacobson, Antone G. (1991). "Experimental Analyses of the Shaping of the Neural Plate and Tube". American Zoologist. 31 (4): 628–643. doi:10.1093/icb/31.4.628. JSTOR 3883562.
^ Smith, Jodi L.; Schoenwolf, Gary C. (1 April 1989). "Notochordal induction of cell wedging in the chicken neural plate and its role in neural tube formation". Journal of Experimental Zoology. 250 (1): 49–62. doi:10.1002/jez.1402500107. PMID 2723610.
^ Wolpert, Lewis (1998). Principles of Development. London: Current Biology. p. 345. ISBN 0-19-850263-X.
^ Wilson, PA; Lagna, G; Suzuki, A; Hemmati-Brivanlou, A (Aug 1997). "Concentration-dependent patterning of the Xenopus ectoderm by BMP4 and its signal transducer Smad1". Development. 124 (16): 3177–84. doi:10.1242/dev.124.16.3177. PMID 9272958.
^ a b Liem, Karel F; Tremml, Gabi; Roelink, Henk; Jessell, Thomas M (1 September 1995). "Dorsal differentiation of neural plate cells induced by BMP-mediated signals from epidermal ectoderm". Cell. 82 (6): 969–979. doi:10.1016/0092-8674(95)90276-7. PMID 7553857. S2CID 17106597.
^ Eom, Dae S; Amarnath, Smita; Agarwala, Seema (20 December 2012). "Apicobasal Polarity and neural tube closure". Development, Growth & Differentiation. 55 (1): 164–172. doi:10.1111/dgd.12030. PMC 3540145. PMID 23277919.
^ de Vellis J, Carpenter E. General Development of the Nervous System. In: Siegel GJ, Agranoff BW, Albers RW, et al., editors. Basic Neurochemistry: Molecular, Cellular and Medical Aspects. 6th edition. Philadelphia: Lippincott-Raven; 1999. Available from:
^ a b Pineau, Isabelle (2006). "A Novel Method for Multiple Labeling Combining In Situ Hybridization With Immunofluorescence". Journal of Histochemistry & Cytochemistry. 54 (11): 1303–1313. doi:10.1369/jhc.6a7022.2006. PMID 16899759.
^ Sadler, T.W. (1986). "A potential role for spectrin during neurulation". J. Embryol. 94 (1): 73–82. Retrieved 27 April 2013.
^ Tan, SS (1986). "Analysis of cranial neural crest cell migration and early fates in postimplantation rat chimaeras". J. Embryol. 98 (1): 21–58. PMID 3655649. Retrieved 27 April 2013.
^ Bailey, Andrew P.; Andrea Streit (2006). "Sensory Organs: Making and Breaking the Pre-Placodal Region". Current Topics in Developmental Biology. 72: 177. doi:10.1016/s0070-2153(05)72003-2. ISBN 9780121531720. PMID 16564335.
External links
[edit]
Swiss embryology (from UL, UB, and UF) hdisqueembry/triderm10
Embryology at Temple EMBIII97/sld010
Overview and diagram at umich.edu Archived 2018-12-28 at the Wayback Machine
| v t e Development of the nervous system |
| Neurogenesis | | | | --- | | General | Neural development Neurulation Neurula Notochord Neuroectoderm Neural plate + Neural fold + Neural groove Neuropoiesis Adult neurogenesis | | Neural crest | Cranial neural crest + Cardiac neural crest complex Truncal neural crest | | Neural tube | Rostral neuropore Neuromere / Rhombomere Cephalic flexure Cervical flexure Pontine flexure Alar plate Basal plate Glioblast Neuroblast Germinal matrix | |
| Eye | Neural tube Optic vesicle Optic stalk Optic cup Surface ectoderm + Lens placode |
| Ear | Otic placode + Otic pit + Otic vesicle |
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Even ions that don't chemically react in the solution interact physically - they are charged, so they either attract or repel each other and their motions are not completely independent. In the effect presence of other ions always interferes and changes the ion behavior. The higher the concentration of ions, the higher the discrepancy between the real and ideal behavior of the ions. As in typical equilibrium calculations we assume ideal behavior of the ions, the higher the concentration, the higher the difference between observed reality and results of our calculations. Sadly, these differences are observable even in solutions in the range 0.001 M, and in the case of 0.1 M solutions they can't be ignored.
The most often used theory that treats discrepancies between ideal and real solutions is the Debye-Hückel theory, proposed in 1923. Lets assume we have a solution containing several ions, with concentrations c and charge z. In all equilibrium formulas we replace concentrations of ions with their activities, calculated by multiplying concentrations by activity coefficients:
6.1
where a ion is the ion activity and f|z| is the activity coefficient of an ion with a charge z (regardless of whether the charge is positive or negative).
For ideal solution the activity coefficient equals 1, for non-ideal solutions it is calculated as a function of ionic strength, defined as
6.2
Ionic strength is a collective parameter, describing presence of all ions in the solution (hence we sum over all ions present). Knowing ionic strength we calculate logarithm of activity coefficients from the formula
6.3
(activity coefficients for all ions of the same charge are assumed to be identical; in more precise versions of the formula they are also a function of the ionic radius).
Debye-Hückel theory works reasonably well for solutions with ionic strength below 0.1. Despite over 100 years of research we still don't have a good and easy to use theory that would allow correct calculations of ion activities for every solution. There are several extensions to Debye-Hückel theory that use experimentally determined equations and/or parameters, some of them work even for ionic strengths up to 5 or 6. However, they are difficult to use, as they require extensive tables of interaction coefficients for all ion pairs present in the solution, and number of these interaction coefficients measured and published in literature is relatively small. Solution of ammonium hydrogenphosphate, containing 8 different ions/molecules, requires 28 coefficients - some of them can be ignored, as concentrations of some ions are very low, still, (NH 4)2 HPO 4 solution is quite simple compared with blood or sea water.
Despite the fact we don't have good theory allowing calculation of activity coefficients from the first principles, Debye-Hückel theory suggests a convenient way of dealing with corrections to ideal solution behavior. We can write acid dissociation constant as
6.4
where a x is activity of ion/molecule x, and f x is its activity coefficient. We left a H+ untouched, as in fact we are interested in the H+ activity, not in H+ concentration (despite the fact pH is often listed as -log([H+]) it is in fact minus logarithm and of the H+ activity, and it is the latter that is measured by pH electrodes or induces color changes of the indicators). If we move now activity coefficient to the LHS of the equation, we will get
6.5
K a' is known as a mixed, or practical dissociation constant. Despite the fact it looks a little bit strange, it is very useful, as it can be determined experimentally for a given ionic strength, and then used in calculations mixing pH (activity) and concentrations of the acid and conjugate base. This way we can solve any buffer problem using exactly the same approach we used earlier, and all corrections related to the ionic strength of the solution will be taken care of automatically.
Sometimes also concentrations constant are used, in which all activity coefficients are moved into the dissociation constant value (so we can easily use these constants to calculate concentration of H+, but not its activity). These are quite useful in finding the equilibrium, as mass and charge balance require real concentrations, not activities, and using concentration constants makes it easier to systematically approach the calculations. Unfortunately, it is not always clear what kind of constants is listed in the published tables.
However, to make sure our calculations are correct, we need to make sure ionic strength of the solution we are working with is the one for which we have the value of the practical dissociation constant. Sometimes it requires additional effort when preparing buffer solution. See next page for details on how to keep ionic strength under control.
Buffers
Table of contents
Introduction to buffers
Composition calculation
pH of buffer with ICE table
Change of the buffer pH
Buffer capacity
Thermodynamic corrections
Constant ionic strength
Three component buffers
pH questions
Table of contents
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Concepts of Biology
Chapter Summary
Concepts of BiologyChapter Summary
Contents Contents
Highlights
Table of contents
Preface
The Cellular Foundation of Life
Cell Division and Genetics
Molecular Biology and Biotechnology
Evolution and the Diversity of Life
Animal Structure and Function
16 The Body’s Systems
17 The Immune System and Disease
18 Animal Reproduction and Development
Introduction
18.1 How Animals Reproduce
18.2 Development and Organogenesis
18.3 Human Reproduction
Key Terms
Chapter Summary
Visual Connection Questions
Review Questions
Critical Thinking Questions
Ecology
A | The Periodic Table of Elements
B | Geological Time
C | Measurements and the Metric System
Index
Search for key terms or text.
Close
18.1 How Animals Reproduce --------------------------
Reproduction may be asexual when one individual produces genetically identical offspring, or sexual when the genetic material from two individuals is combined to produce genetically diverse offspring. Asexual reproduction in animals occurs through fission, budding, fragmentation, and parthenogenesis. Sexual reproduction may involve fertilization inside the body or in the external environment. A species may have separate sexes or combined sexes; when the sexes are combined they may be expressed at different times in the life cycle. The sex of an individual may be determined by various chromosomal systems or environmental factors such as temperature.
Sexual reproduction starts with the combination of a sperm and an egg in a process called fertilization. This can occur either outside the bodies or inside the female. The method of fertilization varies among animals. Some species release the egg and sperm into the environment, some species retain the egg and receive the sperm into the female body and then expel the developing embryo covered with shell, while still other species retain the developing offspring throughout the gestation period.
18.2 Development and Organogenesis ----------------------------------
The early stages of embryonic development begin with fertilization. The process of fertilization is tightly controlled to ensure that only one sperm fuses with one egg. After fertilization, the zygote undergoes cleavage to form the blastula. The blastula, which in some species is a hollow ball of cells, undergoes a process called gastrulation, during which the three germ layers form. The ectoderm gives rise to the nervous system and the epidermal skin cells, the mesoderm gives rise to the muscle cells and connective tissue in the body, and the endoderm gives rise to the digestive system and other internal organs. Organogenesis is the formation of organs from the germ layers. Each germ layer gives rise to specific tissue types.
18.3 Human Reproduction -----------------------
The reproductive structures that evolved in land animals allow them to mate, fertilize internally, and support the growth and development of offspring. Gametogenesis, the production of sperm (spermatogenesis) and eggs (oogenesis), takes place through the process of meiosis.
The reproductive cycles are controlled by hormones released from the hypothalamus and anterior pituitary and hormones from reproductive tissues and organs. The hypothalamus monitors the need for FSH and LH production and release from the anterior pituitary. FSH and LH affect reproductive structures to cause the formation of sperm and the preparation of eggs for release and possible fertilization. In the male, FSH and LH stimulate Sertoli cells and interstitial cells of Leydig in the testes to facilitate sperm production. The Leydig cells produce testosterone, which in human males is also responsible for a deepening of the voice, the growth of facial, axillary, and pubic hair, and an increase in muscle bulk. In females, FSH and LH cause estrogen and progesterone to be produced. They regulate the female reproductive cycle, which is divided into the ovarian cycle and the menstrual cycle.
Human pregnancy begins with fertilization of an egg and proceeds through the three trimesters of gestation. The first trimester lays down the basic structures of the body, including the limb buds, heart, eyes, and the liver. The second trimester continues the development of all of the organs and systems. The third trimester exhibits the greatest growth of the fetus and culminates in labor and delivery. The labor process has three stages (contractions, delivery of the fetus, and expulsion of the placenta), each propelled by hormones.
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3: Chemical Formulas and Bonding
CHM 120 - Survey of General Chemistry(Neils)
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3.2 Composition of Compounds
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Contents
1. Home
2. Campus Bookshelves
3. Grand Rapids Community College
4. CHM 120 - Survey of General Chemistry(Neils)
5. 3: Chemical Formulas and Bonding
6. 3.2 Composition of Compounds
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CHM 120 - Survey of General Chemistry(Neils)
Front Matter
1: Matter and Energy
2: Atomic Structure
3: Chemical Formulas and Bonding
4: Intermolecular Forces, Phases, and Solutions
5: The Numbers Game - Solutions and Stoichiometry
6: Reaction Kinetics
7: Equilibrium and Thermodynamics
8: Acids and Bases
9: Electrochemistry
Back Matter
3.2 Composition of Compounds
Last updated Jul 29, 2019
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3.1 An Atomic-Level Perspective of Ionic and Covalent Compounds
3.3 Chemical Bonds
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( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Formulas on the Atomic and the Molar Level
2. Structural Formulas
1. Condensed Structural Formula
2. Expanded Structural Formula
3. Line-Angle Formula
4. Isomers
5. Polymers
Contributors
Skills to Develop
To understand the definition and difference between empirical formulas and molecular formulas
To understand how combustion analysis can be used to identify molecular formulas
Chemistry is the experimental and theoretical study of materials at both the macroscopic and microscopic levels. Understanding the relationship between properties and structures/bonding is a major aspect of these theoretical and experimental studies. A chemical formula is a format used to express the structure of compounds. The formula tells which elements and how many of each element are present in a compound. Formulas are written using the elemental symbol of each atom and a subscript to denote the number of elements.
Molecular formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's molecular formula cannot be reduced any more, then the empirical formula is the same as the molecular formula. Once the molar mass of the compound is known, the molecular formula can be calculated from the empirical formula.
Formulas on the Atomic and the Molar Level
A chemical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well as the atomic level. Thus, H 2 O is composed of two atoms of hydrogen and 1 atom of oxygen. One molecule of water has a mass of 18.02 amus. Likewise, 1.000 mole of H 2 O molecules is composed of 2.000molesof hydrogen and 1.000mole of oxygen. The mass of 1.000 mole of water molecules has a mass of 18.02 grams.
Example 1: Acetone
An acetone molecule has the chemical formula of C 3 H 6 O.
Questions:
a) Is C 3 H 6 O an empirical formula, a molecular formula, or both?
b) How many H atoms are in one acetone molecule?
c) How many total atoms are in one acetone molecule?
d) What is the mass, in amus, of one acetone molecule?
e) What is the mass, in grams, of 1.000 mole of acetone molecules?
f) How many H atoms are in 1.000 mole of acetone molecules?
g) How many total atoms are in 1.000 mole of acetone molecules?
Solutions:
a) The formula is both an empirical formula (because it is the simplest ratio of elements) and also a molecular formula (because the formula tells the actual number of atoms of each element in one acetone molecule).
b) There are 6 H atoms in one acetone molecule.
c) There are 10 total atoms (6 H + 3 C + O) in one acetone molecule.
d) The mass of one acetone molecule: 6Hatoms×1.008 amus Hatom+3 C atoms×12.01 amus C atom+1 O atom×16.00 amus O atom=58.08amus
e) The mass of 1.000 mole of acetone molecules: (1)6 molesHatoms×1.008grams 1.000 molesHatoms+3 moles C atoms×12.01grams 1.000 moles C atoms+1 mole O atoms×16.00grams 1.000 mole O atoms=58.08grams
f) H atoms in 1.00 mole of acetone molecules: (2)1.000 mole acetone molecules×6 molesHatoms 1.000 mole acetone molecules×6.022x10 23Hatoms 1.000 molesHatoms=3.613x10 24Hatoms
g) Total atoms in 1.000 moles of acetone molecules: (3)1.000 mole acetone molecules×10 moles atoms 1.000 mole acetone molecules×6.022x10 23 atoms 1.000 mole atoms=6.022x10 24 atoms
Exercise 1
An certain molecule has the chemical formula of C 8 H 8 O 4
Questions:
a) Is C 8 H 8 O 4 an empirical formula, a molecular formula, or both?
b) How many H atoms are in one C 8 H 8 O 4 molecule?
c) How many total atoms are in one C 8 H 8 O 4 molecule?
d) What is the mass, in amus, of one C 8 H 8 O 4 molecule?
e) What is the mass, in grams, of 1.000 mole of C 8 H 8 O 4 molecules?
f) How many H atoms are in 1.000 mole of C 8 H 8 O 4 molecules?
g) How many total atoms are in 1.000 mole of C 8 H 8 O 4 molecules?
Answer
a) The formula is a molecular formula, because it is not the simplest ratio of elements. The empirical formula is C 2 H 2 O 1
b) There are 8 H atoms in one acetone molecule.
c) There are 20 total atoms (8 H + 8 C + 4 O) in one C 8 H 8 O 4 molecule.
d) The mass of one C 8 H 8 O 4 molecule: 8Hatoms×1.008 amus Hatom+8 C atoms×12.01 amus C atom+4 O atom×16.00 amus O atom=168.1amus
e) The mass of 1.000 mole of C 8 H 8 O 4 molecules: (4)8 molesHatoms×1.008grams 1.000 molesHatoms+8 moles C atoms×12.01grams 1.000 moles C atoms+4 mole O atoms×16.00grams 1.000 mole O atoms=168.1grams
f) H atoms in 1.00 mole of C 8 H 8 O 4 molecules: (5)1.000 mole C 8H 8O 4 molecules×8 molesHatoms 1.000 mole C 8H 8O 4 molecules×6.022x10 23Hatoms 1.000 molesHatoms=4.818x10 24Hatoms
g) Total atoms in 1.000 moles of C 8 H 8 O 4 molecules: (6)1.000 mole C 8H 8O 4 molecules×20 moles atoms 1.000 mole C 8H 8O 4 molecules×6.022x10 23 atoms 1.000 mole atoms=1.204x10 25 atoms
Structural Formulas
There are several ways to write and draw chemical formulas for covalently-bonded compounds, depending on how much information about the structure is needed.
Condensed Structural Formula
Condensed structural formulas show a bit more information about the bonding order of atoms in a molecule than a simple molecular formula. A condensed structural formula is written in a single line to save space and make it more convenient and faster to write out. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together.
Ex. Condensed Structural Formula for ethanol: CH 3 CH 2 OH (molecular Formula: C 2 H 6 O).
Condensed Structural Formula for dimethyl ether: (CH 3)2 O (molecular formula: C 2 H 6 O).
Expanded Structural Formula
A structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. Structural formulas are helpful because they explain the properties and structure of the compound which condensed formulas cannot always represent.
Structural Formula for CH 3 CH 2 OH, ethanol
Line-Angle Formula
Because organic compounds can be complex, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by omitting the symbol of H atoms attached to C atoms, and by replacing the C element symbols with the intersection of lines. Thus, a carbon atom is present wherever one line intersects another line. Hydrogen atoms are assumed to complete each of carbon's four bonds. All other atoms that are connected to carbon atoms are written out. Line angle formulas help show structure and order of the atoms in a compound making the advantages and disadvantages similar to structural formulas.
Line-Angle Formula for Ethanol:
Isomers
Understanding how atoms in a molecules are arranged and how they are bonded together is very important in giving the molecule its identity. Isomers are compounds in which two molecules have the same number of atoms, and thus the same molecular formula, but have completely different physical and chemical properties because of differences in structural formula.
Methylpropane and butane have the same molecular formula of C 4 H 10, but are structurally different (methylpropane on the left, butane on the right).
Polymers
A polymer is formed when small molecules of identical structure, monomers, combine into a large cluster. The monomers are joined together by covalent bonds. When monomers repeat and bind, they form a polymer. While they can be comprised of natural or synthetic molecules, polymers often include plastics and rubber. When a molecule has more than one of these polymers, parenthesis are used to show that all the elements within the polymer are multiplied by the subscript outside of the parenthesis. The subscript (shown as n in the example below) denotes the number of monomers present in the macromolecule (or polymer).
Ethylene becomes the polymer polyethylene.
O
Contributors
Mike Blaber (Florida State University)
Modified by Tom Neils (Grand Rapids Community College)
3.2 Composition of Compounds is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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3.1 An Atomic-Level Perspective of Ionic and Covalent Compounds
3.3 Chemical Bonds
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10286 | https://www.eurominority.eu/index.php/en/nations-states-and-stateless-nations/ | Skip to content
Nations, states and stateless nations
Why talk of “stateless nations”
rather than “national minority”? The term “stateless nations” largely
transcends that of “minority” which is often very restrictive and perceived in
a derogatory manner. The term “minority” defines a state relative to a
majority. However, it does not refer specifically to what it aspires to be,
that is to say a minority people with the characteristics of a nation without
having the attributes of a sovereign state. Accordingly, defining the terms
“nation” and “stateless nation” seems essential.
In the beginning, the tribe?
“Minority” refers to many terms
which often have political overtones. Whenever one speaks of “minority”, one
often thinks of tribes and ethnicities. In fact tribe is a type of minority,
more specifically qualified as indigenous or first people. Also, these
minorities can be native or foreign.
The nation, a dual concept
The nation, for its part, is a
human community having the belief of belonging to the same group. It falls
within defined geographical boundaries. Thus, numerous minorities are defined
as people or nations. The distinctive characteristics of a nation are language,
culture, religion, history… In contrast, certain nations recognise themselves
without being homogenous. In fact, even if language constitutes one of the
corner stones of nations, some are multilingual (and respect this fact) or they
are comprised of distinct religious communities.
The Council of Europe tried to find
a consensus definition of the concept of nation. This exercise is complicated
where, according to the languages and the states, this word is perceived
differently. Certain countries use it to define their citizenship, i.e. the
legal relationship between citizens and the state. Others use it in the sense
of a united ethnolinguistic community.
In Latin, however, the word “natio”
is derived from “nascere” (to be born). It was used in the Middle Ages to
describe membership of a community. In the 18th century, this concept evolved
into being related to a community of individuals enjoying equal rights
independent of ethnic origin. This concept is very close to the definition of
the State-nation or civic nation. In contrast, the notion of the cultural
nation is, for its part, a concept defining the nation as a homogenous
ethno-linguistic entity.
In its conclusions concerning the
definition of the term “nation”, the Council of Europe considers that these two
definitions are “once again valid today”. True, but can we compel a community
to accept these two definitions? In fact, the text of the Council of Europe
quotes from Ernest Renan, a Breton author considered to be one of the
theoreticians of the nation. “‘The Nation’ is a daily plebiscite”. This
sentence sums up the situation. We cannot force people to live in a state
without their consent and, worse, make them deny their own identity in order to
be a citizen of that state, such as the Kurds in Turkey.
The state and state-nation
The state is a form of political
organisation, based by nature on the sovereignty of the nation. The state is
the setting in which the people or the nations can legislate and vote for their
own territorial laws. The state-nation characterises a sovereign state
constituted objectively or artificially by a desire to live together. We can
distinguish between the state-nation and the nation-state. The state-nation can
enforce a national feeling. France, Greece and Turkey correspond to the
archetypeal state-nation. Conversely, a group can be recognised as a nation and
manifest its desire to live together in establishing a state, which we would
call a nation-state.
Stateless nations
A stateless nation is a
non-sovereign nation with no state structures. Its people often live in varying
degrees of attachment to their original nation. Thus there exists in all the
stateless nations feelings of identity that can be qualified as contradictory.
Certain people do not exclusively feel part of their nation of origin.
Others, on the contrary, exclusively feel they are members of the state of which they are citizens.
Finally, a fraction of the population is often divided by two identities, the citizenship acquired by the state and the nationality which attaches it to the nation of origin. Indeed, strictly speaking, citizenship is the link which heightens the authority of the state, contrary to nationality, which is relative to the feeling of belonging to a community in a nation (without necessarily being a constituted state).
From nation to nationalism
Nationalism is the expression which
consists of imposing one’s view of the world. It is expressed either by the
desire to impose one vision of the nation, the so-called negative nationalism,
or it can be characterised by the desire to assert one’s existence as a nation.
This is positive nationalism. These claims may lead to two scenarios: the
introduction of a regional autonomy system or the desire for
self-determination.
Autonomy
Autonomy allows people to have
their own powers and to legislate. Autonomy is characterised by devolution,
i.e. the transfer of powers from central state to a more local level. Thus
Catalonia, the Basque Country, Scotland or, to a lesser degree, Friesland and
Sardinia are today autonomous regions. Unfortunately, many people do not have
any status to express their individuality. This is particularly the case of the
stateless nations present in France, such as Brittany, Alsace and Corsica, or
Kashubia in Poland.
Self-determination
More radical forms of demands can
be expressed, such as the quest for independence, i.e. the total detachment
from the central state. This is separatism or secessionism. Despite its often
sustained character, this is recognised by international law. The right to
self-determination is written in the UN Charter, whose objective is “to develop
friendly relations among nations based on the principle of equal rights of
people and their right to self-determination”. In the English version, the term
“self-determination” is more explicit.
Ethnocentrism, no to dialogue!
Others consider that this right corresponds to
communitarianism or ethocentricism, i.e. the desire to live closed in on
oneself and to refuse the dogma of the indivisible state. Denying the existence
of minorities, identities and regional languages tends to impose a sectarian
and imperial vision of the nation.
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6: Acids and bases
Introduction to General Chemistry (Malik)
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6.3: Strength of acids and bases
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Contents
1. Home
2. Bookshelves
3. Introductory, Conceptual, and GOB Chemistry
4. Introduction to General Chemistry (Malik)
5. 6: Acids and bases
6. 6.3: Strength of acids and bases
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Introduction to General Chemistry (Malik)
Front Matter
1: Matter energy and their measurements
2: Elements
3: Compounds
4: Stoichiometry –the quantification of chemical reactions
5: Solutions
6: Acids and bases
7: Gases
8: Nuclear chemistry
Back Matter
6.3: Strength of acids and bases
Last updated Apr 3, 2023
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6.2: Brønsted–Lowry acids and bases
6.4: Acid-base equilibrium
Page ID 371800
Muhammad Arif Malik
Hampton University, Hampton, VA
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Strong acids
2. Weak acids
3. Strong bases
4. Weak bases
5. The relative strength of acid-conjugate base pair
1. Direction of acid-base equilibrium
The strength of acid HA is the extent to which the acid dissociates into H+ and A- ions, as illustrated in Fig. 6.3.1.
Figure 6.3.1: Image of strong acid mostly dissociating (left) and a weak acid partially dissociating into ions in water (right). Source: Cwszot / CC0
Strong acids
Strong acids, like HCl, almost 100% dissociate into ions when they dissolve in water.
HCl(g)+H 2O(l)→H 3O+(aq)+Cl−(aq)
One arrow is used to indicate that the reaction is nearly 100% complete.
Strong acids include HClO 4, H 2 SO 4, HI, HBr, HCl, and HNO 3
Weak acids
Weak acids dissolve in water but partially dissociate into ions.
For example, acetic acid (CH 3 COOH) is a weak acid, 1 M acetic acid dissolves in water, but only 0.4% of the dissolved molecules dissociate into ions, the remaining 99.6% remain undissociated, as illustrated in Fig. 6.3.2. and equation of the dissociation equilibrium below.
CH 3COOH(aq)+H 2O(I)⟵→H 3O+(aq)+CH 3COO−(aq)
Two arrows pointing in opposite directions are used for the dissociation of weak acids to indicate that the reaction is an equilibrium, i.e., two ways.
Often the arrows are not equal in size -the longer arrow points to acid-base pair that is weaker and present in a larger concentration at equilibrium than their conjugate pair.
Figure 6.3.2: After dissolution in water. HCl –the strong acid is 100% dissociated into H+ and Cl- ions leaving no dissolved HCl molecules in water, but acetic acid (HAc) –the weak acid has a high concentration of HAc molecules and low concentration of H+ and Ac- ions.
Strong bases
Strong bases almost %100 dissociate into ions when dissolved in water. For example, NaOH is a strong base, and it dissociates almost 100% into ions in water.
Strong bases almost %100 dissociate into ions when dissolved in water. For example, NaOH is a strong base, and it dissociates almost 100% into ions in water.
NaOH(s)⟶Water Na+(aq)+OH−(aq)
One arrow is used for the dissolution of strong bases to indicate that the reaction is almost complete.
Strong bases include hydroxides of alkali metals, i.e., LiOH, NaOH, KOH, RbOH, CsOH, and hydroxides of heavy alkaline earth metals, i.e., Ca(OH)2, Sr(OH)2, and Ba(OH)2.
The last three, i.e., the hydroxides of heavy alkaline earth metals, have low solubility in water, but the dissolved fraction exists as ions.
Weak bases
Weak bases partially dissociate into ions when dissolved in water.
For example, ammonia is a weak base –only 0.42% of the dissolved ammonia molecules dissociate into ammonium ions and hydroxide ions in water from a 1 M solution of ammonia.
NH 3(aq)+H 2O(I)⟵→NH 4+(aq)+OH−(aq)
Weak bases in household use include ammonia (NH 3) in window cleaners, NaClO in bleach, Na 2 CO 3 and Na 3 PO 4 in laundry detergent, NaHCO3 in tooth past, Na 2 CO 3 in baking powder, CaCO 3 for use in lawns, Mg(OH)2 and Al(OH)3 in antacids and laxatives.
The weak bases mentioned above are all ionic compounds except ammonia. Ionic compounds are strong electrolytes, i.e., they dissociate into ions almost 100% upon dissolution in water. It appears to contradict the fact that these ionic compounds are weak bases. It does not actually contradict, because the base properties do not refer to these ionic compounds, the base properties refer to the reactions of their polyatomic anions, i.e., ClO-, CO 3 2-, and PO 4 3- with water, as shown in the reactions below:
ClO−+H 2O⟵→HClO+OH−CO 3 2−+2H 2O⟵→H 2CO 3+2OH−, and PO 43−+3H 2O⟵→H 3PO 4+3OH−
The above reactions are equilibrium reactions that are more favored in the revers than the forward direction, producing a small number of OH- ions compared to the anion on the reactant sides. The last two examples, i.e., Mg(OH)2 and Al(OH)2 are classified as weak bases because they are considered insoluble in water. The solubility of Mg(OH)2 is 0.00064 g/100 mL (25 °C), and the solubility of Al(OH)3 is 0.0001 g/100 mL, which are in the range of insoluble ionic compounds.
The solubility and the strength of acids and bases are two different things. A strong base may be less soluble, and a weak base may be more soluble or vice versa, but a dissolved strong base exists as ions only, and a dissolved weak base exists both as molecules and ions.
The relative strength of acid-conjugate base pair
A general rule is that the stronger the acid, the weaker the conjugate base, and vice versa.
The conjugate bases of strong acids have negligible base strength, and the conjugate acids of strong basses have negligible acid strength. Fig. 6.3.3. illustrates the relative strengths of some acids and their conjugated bases.
Figure 6.3.3: Strength of acid-conjugate base pairs relative to water as a reference.
Direction of acid-base equilibrium
In any Brønsted–Lowry acid-base reaction, the general rule is that a stronger acid and a stronger base tend to form a weaker acid and a weaker base.
For example, a dissociation reaction between HCl and H 2 O is almost 100% complete because HCl is a stronger acid than H 3 O+ and H 2 O is a stronger base than Cl-:
HCl+H 2O→H 3O++Cl−
The dissolution of acetic acid (CH 3 COOH) and ammonia (NH 3) are equilibrium reactions because all the acids, bases, and their conjugates are in the weak acids or weak bases category. However, acetic acid and water dominate over their conjugates H 3 O+ and CH 3 COO- by 99.6:0.4 ratio (in 1 M acetic acid solution) because the conjugate acid H 3 O+is a stronger acid than CH 3 COOH, and conjugate base CH 3 COO- is a stronger base than H 2 O.
CH 3COOH(aq)+H 2O(l)⟵→H 3O+(aq)+CH 3COO−(aq)
The longer arrow, in the unbalanced equilibrium arrows, points to the acid-base pair in the reaction that exists in a higher concentration relative to their conjugates.
Similarly, ammonia (NH 3) and water (H 2 O) dominate over their conjugates NH 4+ and OH- by ~99.6:0.4 ratio (1M ammonia solution) because the conjugate acid NH 4+is a stronger acid than H 2 O and conjugate base OH- is a stronger base than NH 3.
NH 3(aq)+H 2O(I))⟵→NH 4+(aq)+OH−(aq)
This page titled 6.3: Strength of acids and bases is shared under a Public Domain license and was authored, remixed, and/or curated by Muhammad Arif Malik.
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6.2: Brønsted–Lowry acids and bases
6.4: Acid-base equilibrium
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10288 | https://resources.healthgrades.com/right-care/vascular-conditions/jugular-vein-distention | Menu
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An Overview of Jugular Vein Distention
What is it?
Causes
Other symptoms
Diagnosis
Treatments
Outlook
Complications
Risk factors
Prevention
FAQ
Summary
Medically Reviewed By Dr. Payal Kohli, M.D., FACC
— Written By Healthgrades Editorial Staff
Updated on May 24, 2022
Key Takeaways
Your jugular veins carry blood from your head to your heart. Jugular vein distention (JVD) is increased pressure in a jugular vein. It is caused by cardiovascular conditions like heart failure or high blood pressure.
Symptoms of JVD include changes in mental status, like confusion or memory loss, coughing, fatigue, and swelling. Get emergency medical care for any possible JVD symptoms.
Treatment for JVD focuses on addressing the underlying cause, which may include medications, lifestyle changes, or, in severe cases, surgery or heart transplant.
JVD can indicate several cardiovascular issues, including heart failure, high blood pressure, and fluid accumulation in the blood vessels. Doctors typically treat JVD by prescribing medications to reduce the pressure inside the heart and recommending lifestyle changes.
Read on to learn more about JVD, including what causes it and how doctors treat it.
What is jugular vein distention?
JVD is a condition that occurs when your central venous pressure (CVP), which is the pressure inside your vena cava, increases. The vena cava is the largest vein in the body, and it carries blood to the heart from other body areas. There are two parts of the vena cava:
The superior vena cava moves blood from the head, neck, chest, and arms.
The inferior vena cava moves blood from the abdomen, legs, and feet.
Your internal and external jugular veins carry blood Trusted Source PubMed Central Highly respected database from the National Institutes of Health Go to source from your head and neck to the superior vena cava, which empties into the heart. The external jugular vein is close to the skin on the side of your neck.
The right internal jugular vein connects to the heart through the superior vena cava. This means that doctors can use the jugular vein to assess the pressure inside the heart.
What causes jugular vein distention?
Many cardiovascular conditions can cause an increase in your CVP and lead to JVD. Some of these conditions include:
Right sided heart failure: This occurs when the right side of your heart cannot pump blood effectively, causing the jugular vein to bulge from fluid accumulation.
Pulmonary hypertension: High blood pressure in the arteries between the lungs and the heart can cause the jugular vein to bulge.
Constrictive pericarditis Trusted Source PubMed Central Highly respected database from the National Institutes of Health Go to source : This condition occurs when the pericardium, which is the sac around the heart, becomes infected or inflamed and scarred. When this happens, the heart’s ability to fill with blood is restricted, leading to an accumulation of blood in the veins.
Superior vena cava obstruction Trusted Source PubMed Central Highly respected database from the National Institutes of Health Go to source : This is a rare condition wherein the superior vena cava becomes partially or fully blocked. It is usually due to a blood clot or tumor.
Tricuspid valve stenosis Trusted Source American Heart Association Highly respected national organization Go to source : This condition occurs when the opening of the tricuspid valve, which regulates blood flow in the heart, becomes narrowed. As a result, blood cannot flow effectively.
Cardiac tamponade Trusted Source PubMed Central Highly respected database from the National Institutes of Health Go to source : This is a serious condition that occurs when fluid accumulates in the pericardium. This buildup compresses the heart and restricts blood flow.
What other symptoms might occur with jugular vein distention?
Depending on the underlying cause, JVD can occur along with other symptoms. These can include:
confusion or memory loss
a cough
fatigue
nausea
nocturia, or frequent urination at night
appetite loss
swelling, especially in the lower extremities or abdomen
weakness
unexpected weight gain
abdominal bloating or fullness
In some cases, JVD may be a symptom of a life threatening condition. Seek immediate medical care or call 911 if you are experiencing serious symptoms, which may include:
anxiety and excessive sweating
bluish coloration of the lips or fingernails
a change in consciousness or alertness
chest problems, such as pain, tightness, or pressure
tachycardia, or a rapid heart rate
respiratory issues, such as shortness of breath or wheezing
dizziness or lightheadedness
How do doctors diagnose jugular vein distention?
Doctors can typically diagnose JVD with a physical examination.
To measure your jugular venous pressure (JVP), your doctor will measure Trusted Source PubMed Central Highly respected database from the National Institutes of Health Go to source the bulge. To do this, they will have you lie on an examination table with your upper body elevated and your head turned to the left. They will then measure the height of the right internal jugular vein in various places along your neck, which will help them determine your JVP.
In some cases, doctors may not be able to perform the standard measurement due to physical characteristics such as a thick neck. Therefore, they may rely on tools such as ultrasound scans to assess your jugular vein and the pressure in your heart.
In addition to measuring your JVP, your doctor may order further tests, including:
Echocardiogram: This imaging test helps doctors assess the heart’s size, structure, and function.
Electrocardiogram: This test measures the heart’s electrical activity. Doctors can use it to diagnose heart conditions such as arrhythmias, which are irregular heartbeats, and coronary artery disease.
Blood tests: Doctors may want to analyze your blood for markers of stretch or signs of kidney or liver disease that may be affecting your heart.
What are the treatments for jugular vein distention?
Treatment for JVD will involve treating the underlying cause. These treatments can include:
Diuretics: These medications reduce blood pressure by increasing the amount of water and salt expelled in your urine.
ACE inhibitors: These medications reduce blood pressure by relaxing your veins and arteries.
Beta-blockers: These medications reduce blood pressure by reducing stress on the blood vessels and the heart.
Lifestyle changes: Your doctor may recommend making changes to your diet and exercise routines to help lower your blood pressure. Such modifications may include changes in your fluid intake.
In severe cases, surgery and even a heart transplant may be necessary, depending on the underlying cause.
You may also need additional medications to treat certain underlying conditions, such as pulmonary hypertension or heart failure.
What is the outlook for someone with jugular vein distention?
The outlook for someone with JVD depends on the underlying cause. Adhering to your doctor’s treatment plan is essential.
JVD can often be one of the first signs of a serious underlying condition. Talking with your doctor and starting treatment promptly will help improve your outlook.
What are some potential complications of jugular vein distention?
The conditions that cause JVD can also cause complications, some of which may be serious. These include:
fatigue
memory or cognitive problems
liver problems
kidney problems
What are the risk factors for jugular vein distention?
Certain factors may increase your risk of developing JVD, including:
high blood pressure
arrhythmias
coronary artery disease
heart attack
certain medications
sleep apnea
excess fluid or sodium intake
Can you prevent jugular vein distention?
Preventing JVD involves taking steps to reduce your risk of developing the underlying conditions that can cause it and monitoring your weight and fluid intake carefully.
You can reduce your risk of developing JVD by talking with your doctor about lifestyle changes you can make and any medications you are taking or might need to take.
FAQ
Here are some frequently asked questions about JVD.
How do you check for jugular vein distention?
To check for JVD, your doctor will have you lie on a table with your head elevated and turn your head to the left. They will measure Trusted Source PubMed Central Highly respected database from the National Institutes of Health Go to source the height of the jugular vein on the right side of your neck. The height will help your doctor determine if increased pressure in your veins is causing JVD.
What does jugular vein distention indicate?
Many underlying conditions can cause JVD, including Trusted Source PubMed Central Highly respected database from the National Institutes of Health Go to source heart failure, narrowed blood vessels, and fluid accumulation around the heart.
What is the difference between jugular venous pressure and jugular vein distention?
JVP reflects the pressure Trusted Source PubMed Central Highly respected database from the National Institutes of Health Go to source in your jugular vein. An elevated JVP leads to JVD, or a bulging of the jugular vein.
Summary
JVD is a condition wherein increased blood pressure causes your jugular vein to bulge. Many cardiovascular conditions — including heart failure, high blood pressure, and fluid accumulation around the heart — can cause JVD.
Doctors typically treat JVD by prescribing medications to reduce the pressure and fluid in your heart and recommending lifestyle changes. In serious cases, surgery or a heart transplant may be necessary.
Talk with your doctor if you think that you are experiencing JVD.
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Gopal, S., et al. (2022). Jugular venous distention.
Problem: Tricuspid valve stenosis. (2021).
Seligson, M. T., et al. (2021). Superior vena cava syndrome.
Stashko, E., et al. (2021). Cardiac tamponade.
Vena cava. (n.d.).
Yadav, N. K., et al. (2022). Constrictive pericarditis.
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Paul's Online Notes Home / Calculus III / 3-Dimensional Space / Quadric Surfaces
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Section 12.4 : Quadric Surfaces
In the previous two sections we’ve looked at lines and planes in three dimensions (or ({\mathbb{R}^3})) and while these are used quite heavily at times in a Calculus class there are many other surfaces that are also used fairly regularly and so we need to take a look at those.
In this section we are going to be looking at quadric surfaces. Quadric surfaces are the graphs of any equation that can be put into the general form
[A{x^2} + B{y^2} + C{z^2} + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0]
where (A), … , (J) are constants.
There is no way that we can possibly list all of them, but there are some standard equations so here is a list of some of the more common quadric surfaces.
Ellipsoid
Here is the general equation of an ellipsoid.
[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1]
Here is a sketch of a typical ellipsoid.
If (a = b = c) then we will have a sphere.
Notice that we only gave the equation for the ellipsoid that has been centered on the origin. Clearly ellipsoids don’t have to be centered on the origin. However, in order to make the discussion in this section a little easier we have chosen to concentrate on surfaces that are “centered” on the origin in one way or another.
Cone
Here is the general equation of a cone.
[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = \frac{{{z^2}}}{{{c^2}}}]
Here is a sketch of a typical cone.
Now, note that while we called this a cone it is more of an hour glass shape rather than what most would call a cone. Of course, the upper and the lower portion of the hour glass really are cones as we would normally think of them.
That brings up the question of what if we really did just want the upper or lower portion (i.e. a cone in the traditional sense)? That is easy enough to answer. All we need to do is solve the given equation for (z) as follows,
[{z^2} = {c^2}\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}} \right) = \frac{{{c^2}}}{{{a^2}}}{x^2} + \frac{{{c^2}}}{{{b^2}}}{y^2} = {A^2}{x^2} + {B^2}{y^2}\,\,\,\,\, \to \,\,\,\,\,\,\,z = \pm \sqrt {{A^2}{x^2} + {B^2}{y^2}} ]
We simplified the coefficients a little to make it the equation(s) easier to deal with. Now, we know that square roots always return positive numbers and so we can then see that (z = \sqrt {{A^2}{x^2} + {B^2}{y^2}} ) will always be positive and so be the equation for just the upper portion of the “cone” above. Likewise, (z = - \sqrt {{A^2}{x^2} + {B^2}{y^2}} ) will always be negative and so be the equation of just the lower portion of the “cone” above.
Also, note that this is the equation of a cone that will open along the (z)-axis. To get the equation of a cone that opens along one of the other axes all we need to do is make a slight modification of the equation. This will be the case for the rest of the surfaces that we’ll be looking at in this section as well.
In the case of a cone the variable that sits by itself on one side of the equal sign will determine the axis that the cone opens up along. For instance, a cone that opens up along the (x)-axis will have the equation,
[\frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = \frac{{{x^2}}}{{{a^2}}}]
For most of the following surfaces we will not give the other possible formulas. We will however acknowledge how each formula needs to be changed to get a change of orientation for the surface.
Cylinder
Here is the general equation of a cylinder.
[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1]
This is a cylinder whose cross section is an ellipse. If (a = b) we have a cylinder whose cross section is a circle. We’ll be dealing with those kinds of cylinders more than the general form so the equation of a cylinder with a circular cross section is,
[{x^2} + {y^2} = {r^2}]
Here is a sketch of typical cylinder with an ellipse cross section.
The cylinder will be centered on the axis corresponding to the variable that does not appear in the equation.
Be careful to not confuse this with a circle. In two dimensions it is a circle, but in three dimensions it is a cylinder.
Hyperboloid of One Sheet
Here is the equation of a hyperboloid of one sheet.
[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1]
Here is a sketch of a typical hyperboloid of one sheet.
The variable with the negative in front of it will give the axis along which the graph is centered.
Hyperboloid of Two Sheets
Here is the equation of a hyperboloid of two sheets.
[ - \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1]
Here is a sketch of a typical hyperboloid of two sheets.
The variable with the positive in front of it will give the axis along which the graph is centered.
Notice that the only difference between the hyperboloid of one sheet and the hyperboloid of two sheets is the signs in front of the variables. They are exactly the opposite signs.
Also note that just as we could do with cones, if we solve the equation for (z) the positive portion will give the equation for the upper part of this while the negative portion will give the equation for the lower part of this.
Elliptic Paraboloid
Here is the equation of an elliptic paraboloid.
[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = \frac{z}{c}]
As with cylinders this has a cross section of an ellipse and if (a = b) it will have a cross section of a circle. When we deal with these we’ll generally be dealing with the kind that have a circle for a cross section.
Here is a sketch of a typical elliptic paraboloid.
In this case the variable that isn’t squared determines the axis upon which the paraboloid opens up. Also, the sign of (c) will determine the direction that the paraboloid opens. If (c) is positive then it opens up and if (c) is negative then it opens down.
Hyperbolic Paraboloid
Here is the equation of a hyperbolic paraboloid.
[\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = \frac{z}{c}]
Here is a sketch of a typical hyperbolic paraboloid.
These graphs are vaguely saddle shaped and as with the elliptic paraboloid the sign of (c) will determine the direction in which the surface “opens up”. The graph above is shown for (c) positive.
With both of the types of paraboloids discussed above note that the surface can be easily moved up or down by adding/subtracting a constant from the left side.
For instance
[z = - {x^2} - {y^2} + 6]
is an elliptic paraboloid that opens downward (be careful, the “-” is on the (x) and (y) instead of the (z)) and starts at (z = 6) instead of (z = 0).
Here are a couple of quick sketches of this surface.
Note that we’ve given two forms of the sketch here. The sketch on the left has the standard set of axes but it is difficult to see the numbers on the axis. The sketch on the right has been “boxed” and this makes it easier to see the numbers to give a sense of perspective to the sketch. In most sketches that actually involve numbers on the axis system we will give both sketches to help get a feel for what the sketch looks like.
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10290 | https://artofproblemsolving.com/downloads/printable_post_collections/72097?srsltid=AfmBOop81J6iewXwEwlqWJPc-hQOoSKy-AZvDfc_m8oWB_ZF881NB6OM | AoPS Community 100 Geometry Problems
A collection of these problems:
www.artofproblemsolving.com/community/c72097 by djmathman, abishek99, CaptainFlint
1 [MA ????] In the gure shown below, circle B is tangent to circle A at X, circle C is tangent to circle A at Y , and circles B and C are tangent to each other. If AB = 6 , AC = 5 , and BC = 9 ,what is AX ?A BCYX
2 [AHSME ????] In triangle ABC , AC = CD and ∠CAB − ∠ABC = 30 ◦. What is the measure of
∠BAD ?A BCD
3 [AMC 10A 2004] Square ABCD has side length 2. A semicircle with diameter AB is con-structed inside the square, and the tangent to the semicircle from C intersects side AD at
E. What is the length of CE ?
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1AoPS Community 100 Geometry Problems A BCDE
4 [AMC 10B 2011] Rectangle ABCD has AB = 6 and BC = 3 . Point M is chosen on side AB
so that ∠AM D = ∠CM D . What is the degree measure of ∠AM D ?
5 [AIME 2011] On square ABCD , point E lies on side AD and point F lies on side BC , so that
BE = EF = F D = 30 . Find the area of the square.
6 Points A,B, and C are situated in the plane such that ∠ABC = 90 ◦. Let D be an arbitrary point on AB , and let E be the foot of the perpendicular from D to AC . Prove that ∠DBE = ∠DCE .
7 [AMC 10B 2012] Four distinct points are arranged in a plane so that the segments connecting them have lengths a, a, a, a, 2a, and b. What is the ratio of b to a?
8 [Britain 2010] Let ABC be a triangle with ∠CAB a right angle. The point L lies on the side BC
between B and C. The circle BAL meets the line AC again at M and the circle CAL meets the line AB again at N . Prove that L, M , and N lie on a straight line.
9 [OMO 2014] Let ABC be a triangle with incenter I and AB = 1400 , AC = 1800 , BC = 2014 . The circle centered at I passing through A intersects line BC at two points X and Y . Compute the length XY .
10 [India RMO 2014] Let ABC be an isosceles triangle with AB = AC and let Γ denote its circum-circle. A point D is on arc AB of Γ not containing C. A point E is on arc AC of Γ not containing
B. If AD = CE prove that BE is parallel to AD .
11 A closed planar shape is said to be equiable if the numerical values of its perimeter and area are the same. For example, a square with side length 4 is equiable since its perimeter and area are both 16 . Show that any closed shape in the plane can be dilated to become equiable. (A dilation is an ane transformation in which a shape is stretched or shrunk. In other words, if A
is a dilated version of B then A is similar to B.)
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2AoPS Community 100 Geometry Problems
12 [David Altizio] Triangle AEF is a right triangle with AE = 4 and EF = 3 . The triangle is in-scribed inside square ABCD as shown. What is the area of the square? AB CDEF
13 Points A and B are located on circle Γ, and point C is an arbitrary point in the interior of Γ.Extend AC and BC past C so that they hit Γ at M and N respectively. Let X denote the foot of the perpendicular from M to BN , and let Y denote the foot of the perpendicular from N to
AM . Prove that AB ‖ XY .
14 [AIME 2007] Square ABCD has side length 13 , and points E and F are exterior to the square such that BE = DF = 5 and AE = CF = 12 . Find EF 2.
15 Let Γ be the circumcircle of 4ABC , and let D, E, F be the midpoints of arcs AB , BC , CA
respectively. Prove that DF ⊥ AE .
16 [AIME 1984] In tetrahedron ABCD , edge AB has length 3 cm. The area of face ABC is 15 cm 2
and the area of face ABD is 12 cm 2. These two faces meet each other at a 30 ◦ angle. Find the volume of the tetrahedron in cm 3.
17 Let P1P2P3P4 be a quadrilateral inscribed in a quadrilateral with diameter of length D, and let X
be the intersection of its diagonals. If P1P3 ⊥ P2P4 prove that D2 = XP 21 + XP 22 + XP 23 + XP 24 .
18 [iTest 2008] Two perpendicular planes intersect a sphere in two circles. These circles intersect in two points, A and B, such that AB = 42 . If the radii of the two circles are 54 and 66 , nd R2,where R is the radius of the sphere.
19 [AIME 2008] In trapezoid ABCD with BC ‖ AD , let BC = 1000 and AD = 2008 . Let ∠A = 37 ◦,
∠D = 53 ◦, and M and N be the midpoints of BC and AD , respectively. Find the length M N .
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3AoPS Community 100 Geometry Problems
20 [Sharygin 2014] Let ABC be an isosceles triangle with base AB . Line touches its circumcircle at point B. Let CD be a perpendicular from C to, and AE , BF be the altitudes of ABC . Prove that D, E, F are collinear.
21 [Purple Comet 2013] Two concentric circles have radii 1 and 4. Six congruent circles form a ring where each of the six circles is tangent to the two circles adjacent to it as shown. The three lightly shaded circles are internally tangent to the circle with radius 4 while the three darkly shaded circles are externally tangent to the circle with radius 1. The radius of the six congruent circles can be written k+√mn , where k, m, and n are integers with k and n relatively prime. Find k + m + n.
22 Let A, B, C, and D be points in the plane such that ∠BAC = ∠CBD . Prove that the circumcir-cle of 4ABC is tangent to BD .NOTE: The problem should also state that D is on the same side of AB as C. dj
23 [Britain 1995] Triangle ABC has a right angle at C. The internal bisectors of angles BAC and
ABC meet BC and CA at P and Q respectively. The points M and N are the feet of the per-pendiculars from P and Q to AB . Find angle M CN .
24 Let ABCD be a parallelogram with ∠A obtuse, and let M and N be the feet of the perpendic-ulars from A
to sides BC and CD . Prove that 4M AN ∼ 4 ABC .
25 For a given 4ABC , let H denote its orthocenter and O its circumcenter. (a) Prove that ∠HAB = ∠OAC .(b) Prove that ∠HOA = |∠B − ∠C|.
26 Suppose P, A, B, C, and D are points in the plane such that 4P AB ∼ 4 P CD . Prove that
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4AoPS Community 100 Geometry Problems
4P AC ∼ 4 P BD .
27 [AMC 12A 2012] Circle C1 has its center O lying on circle C2. The two circles meet at X and Y .Point Z in the exterior of C1 lies on circle C2 and XZ = 13 , OZ = 11 , and Y Z = 7 . What is the radius of circle C1?
28 Let ABCD be a cyclic quadrilateral with no two sides parallel. Lines AD and BC (extended) meet at K, and AB and CD (extended) meet at M . The angle bisector of ∠DKC intersects
CD and AB at points E and F , respectively; the angle bisector of ∠CM B intersects BC and
AD at points G and H, respectively. Prove that quadrilateral EGF H is a rhombus.
29 [David Altizio] In 4ABC , AB = 13 , AC = 14 , and BC = 15 . Let M denote the midpoint of
AC . Point P is placed on line segment BM such that AP ⊥ P C . Suppose that p, q, and r are positive integers with p and r relatively prime and q squarefree such that the area of 4AP C
can be written in the form p√qr . What is p + q + r.
30 [All-Russian Mo 2013] Acute-angled triangle ABC is inscribed in circle Ω. Lines tangent to
Ω at B and C intersect at P . Points D and E are on AB and AC such that P D and P E are perpendicular to AB and AC respectively. Prove that the orthocenter of triangle ADE is the midpoint of BC .
31 For an acute triangle 4ABC with orthocenter H, let HA be the foot of the altitude from A to
BC , and define HB and HC similarly. Show that H is the incenter of 4HAHB HC .
32 [AMC 10A 2013] In 4ABC , AB = 86 , and AC = 97 . A circle with center A and radius AB
intersects BC at points B and X. Moreover BX and CX ave integer lengths. What is BC ?
33 [APMO 2010] Let ABC be a triangle with ∠BAC 6 = 90 ◦. Let O be the circumcircle of the triangle
ABC ad Γ be the circumcircle of the triangle BOC . Suppose that Γ intersects of line segment
AB at P different from B, and the line segment AS at Q different from C. Let ON be on the diameter of the circle Γ. Prove that the quadrilateral AP N Q is a parallelogram.
34 [AMC 10A 2013] A unit square is rotated 45 ◦ about its center. What is the area of the region swept out by the interior of the square?
35 [Canada 1986] A chord ST of constant length slides around a semicircle with diameter AB .
M is the midpoint of ST and P is the foot of the perpendicular from S to AB . Prove that angle
SP M is constant for all positions of ST .
36 [Sharygin 2012] On side AC of triangle ABC an arbitrary point is selected D. The tangent in D
to the circumcircle of triangle BDC meets AB in point C1; point A1 is defined similarly. Prove that A1C1 ‖ AC .
©2019 AoPS Incorporated
5AoPS Community 100 Geometry Problems
37 In triangle ABC , AB = 13 , BC = 14 , and CA = 15 . Distinct points D, E, and F lie on segments
BC , CA , and DE , respectively, such that AD ⊥ BC , DE ⊥ AC , and AF ⊥ BF . The length of segment DF can be written as mn , where m and n are relatively prime positive integers. What is m + n?
38 [Mandelbrot] In triangle ABC , AB = 5 , AC = 6 , and BC = 7 . If point X is chosen on BC so that the sum of the areas of the circumcircles of triangles AXB and AXC is minimized, then determine BX .
39 [Sharygin 2014] Given a rectangle ABCD . Two perpendicular lines pass through point B. One of them meets segment AD at point K, and the second one meets the extension of side CD
at point L. Let F be the common point of KL and AC . Prove that BF ⊥ KL .
40 [AIME unused] In the figure, In the figure, ABC is a triangle and AB = 30 is a diameter of the circle. If AD = AC/ 3 and BE = BC/ 4, then what is the area of the triangle? A BCDE
41 [MOSP 1995] An interior point P is chosen in rectangle ABCD such that ∠AP D + ∠BP C =180 ◦. Find the sum of the angles ∠DAP and ∠BCP .
42 Let ABC be a triangles and P , Q, R points on the sides of AB , BC , and CA respectively. Prove
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6AoPS Community 100 Geometry Problems
that the circumcircles of 4AP R , 4BQP , and 4CRQ intersect in a common point. This point is named the Miquel point of the configuration.
43 [AIME 2013] Let 4P QR be a triangle with ∠P = 75 o and ∠Q = 60 o. A regular hexagon
ABCDEF with side length 1 is drawn inside 4P QR so that side AB lies on P Q , side CD
lies on QR , and one of the remaining vertices lies on RP . There are positive integers a, b, c,
and d such that the area of 4P QR can be expressed in the form a+b√cd , where a and d are relatively prime, and c is not divisible by the square of any prime. Find a + b + c + d.
44 [”Fact 5”] Let Γ be the circumcircle of an arbitrary triangle 4ABC . Furthermore, denote I its incenter and M the midpoint of minor arc BC
_. Prove that M is the circumcenter of 4BIC .
45 In triangle ABC , angles A and B measure 60 degrees and 45 degrees, respectively. The bisector of angle A intersects BC at T , and AT = 24 . The area of triangle ABC can be written in the form a + b√c, where a, b, and c are positive integers, and c is not divisible by the square of any prime. Find a + b + c.
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10292 | https://www.index-f.com/lascasas/documentos/lc0256.pdf | El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 1 Í n d i c e Páginas 1.- Introducción 1.1. Definición del Método Científico 1.2. Método Analítico. 1.3. Método Sintético. 1.4. Método Inductivo. 1.5. Método Deductivo. 1.6. El Pensamiento Cartesiano. 1.6.1. Las Reglas del Método Cartesiano. 1.6.2. La duda Metódica. 1.6.3. EL Primer Principio Cartesiano. 1.7. Los Procedimientos de la Inducción según J. Mill Stuart. 2.- Etapas del Método Científico 2.1. La Elección del Tema. 2.2. Planteamiento del Problema. 2.2.1. Delimitación del Problema. 2.3. Justificación del problema de Investigación. 2.4. Objetivos de la Investigación. 2.5. Estructuración del Esquema de Investigación. 2.6. Marco Teórico. 2.7. Elaboración de la Hipótesis. 2.8. Metodología. 2.9. Cronograma. 2.10. Anexos o gráficos. 2.11. Glosario de términos. 2.12. Bibliografía. 3. Leyes Científicas. 3.1. Función de la Ley Científica. 3.1.1. Clases de Leyes Científicas. 4.- anexos 5.- bibliografía 3 13 15 17 20 26 26 29 30 31 33 35 36 39 45 46 48 50 53 61 62 69 71 73 75 79 El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 2 El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 3 1.- Introducción Al hablar del método científico es referirse a la ciencia (básica y aplicada) como un conjunto de pensamientos universales y necesarios, y que en función de esto surgen algunas cualidades importantes, como la de que está constituida por leyes universales que conforman un conocimiento sistemático de la realidad. Y es así que el método científico procura una adecuada elaboración de esos pensamientos universales y necesarios. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 4 Sin embargo, mientras que los representantes del “camino más elevado hacia la verdad” se afanan para demostrar que los procedimientos –disciplinados y positivos- de la ciencia limitan su radio de acción hasta el punto de excluir los indubitables aspectos de la realidad. ¿Y en qué fundamentan éstos tal punto de vista? Su argumento estriba, en primer lugar, en la presentación del método científico como interesado únicamente en la física y en la química (ciencias experimentales), es decir, en lo mensurable (lo que se puede medir, pesar y contar), excluyendo aspectos de la realidad como la vida y la mente humana, las cuales quedan reducidas –y a esto lo dan por descontado- exclusivamente a lo material, a lo corpóreo, a lo externo. En segundo lugar, tienen que demostrar que el razonamiento científico constituye un estricto proceso de deducción, proceso del que están excluidos la imaginación y el pensamiento intuitivo. En otras palabras, el método científico tiene su base y postura sobre la teoría mecanicista (todo es considerado como una máquina, y para entender el todo debemos descomponerlo en partes pequeñas que permitan estudiar, analizar y comprender sus nexos, interdependencia y conexiones entre el todo y sus partes), y, por consiguiente también ese mismo carácter. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 5 Si ello fuera realmente así, está claro que quedarían fuera del alcance, del razonamiento científico vastos campos o parcelas de la realidad, de la verdad; siendo entonces necesario hallar un nuevo camino que nos lleve hasta esta misma verdad. Mas la ciencia no está en modo alguno circunscrita a lo mensurable. “El papel desempeñado por la medición y por la cantidad (cualidades cuantitativas) en la ciencia –dice Bertrand Russell- es en realidad muy importante, pero creo que a veces se le supervalora. Las leyes cualitativas pueden ser tan científicas como la leyes cuantitativas.” Tampoco la ciencia está reducida a la física y a la química; mas a los defensores del “elevado camino hacia la verdad” les conviene creer que ello es así. Para ellos es necesario, en efecto, presentar a la ciencia como estando limitada, por su misma naturaleza, a la tarea de preparar el escenario para que la entrada en él una forma más elevada de conocimiento. Pero la esfera de la acción de la ciencia es ya bastante amplia, no ya para incluir a la biología y a la psicología, a la economía y a la antropología, a la sociología y a la historia, sino que también sus métodos son capaces de ir modificándose a si mismos, al objeto de mejor adecuarse a cada uno de los campos estudiados. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 6 Lo que hace que el razonamiento científico es, en primer lugar, el método de observación, el experimento y el análisis, y, después, la construcción de hipótesis y la subsiguiente comprobación de éstas. Este procedimiento no sólo es válido para las ciencias físicas, sino que es perfectamente aplicable a todos los campos del saber.1 A lo largo de la historia, el hombre se ha enfrentado a un sinnúmero de obstáculos y problemas para desentrañar los secretos de la naturaleza, tanto para vivir con ella, como de ella en “perfecta” armonía. Para superar esos problemas ha empleado muy diversas estrategias, las cuales dieron paso a la formalización de procedimientos que, en última instancia, no son sino el propio método científico. El método científico es el procedimiento planteado que se sigue en la investigación para descubrir las formas de existencia de los procesos objetivos, para desentrañar sus conexiones internas y externas, para generalizar y profundizar los conocimientos así adquiridos, para llegar a demostrarlos con rigor racional y para comprobarlos en el experimento y con las técnicas de su aplicación. 1 Pág. 34-35. Lewis John. Ciencia, fe y Escepticismo. Editorial Grijalbo. México 1969. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 7 Al referirse a las formas de existencia de los procesos objetivos, Elí de Gortari lo está haciendo a las diversas maneras en que los procesos de por sí existentes se desarrollan y sólo a ellos; y cuando dice que la finalidad es desempeñar sus conexiones internas y externas, se está refiriendo fenomenológicamente al proceso natural de los acontecimientos de la naturaleza, pero no a todos, solo a aquellos que aún no tienen una explicación acabada que den cuenta precisamente del cómo suceden tales o cuales fenómenos, y de los que una vez desentrañados y explicados sus procesos, se derivan leyes, teorías, modelos, que más tarde serán punto de partida para la búsqueda de nuevos conocimientos. El método científico se emplea con el fin de incrementar el conocimiento y en consecuencia aumentar nuestro bienestar y nuestro poder (objetivamente extrínsecos o utilitarios). En sentido riguroso, el método científico es único, tanto en su generalidad como en su particularidad. Al método científico también se le caracteriza como un rasgo característico de la ciencia, tanto de la pura como de la aplicada; y por su familiaridad puede perfeccionarse mediante la estimación de los resultados a los que lleva mediante el análisis directo. Otra característica es que, no es autosuficiente: no puede operar en un vació de El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 8 conocimiento, si no que requiere de algún conocimiento previo que pueda luego reajustarse y reelaborarse; y que posteriormente pueda complementarse mediante métodos especiales adaptados a las peculiaridades de cada tema, y de cada área, sin embargo en lo general el método científico se apega a las siguientes principales etapas para su aplicación: 1. Enunciar preguntas bien formuladas y verosímilmente fecundas. 2. Arbitrar conjeturas, fundadas y contrastables con la experiencia para contestar a las preguntas. 3. Derivar consecuencias lógicas de las conjeturas. 4. Arbitrar técnicas para someter las conjeturas a contrastación. 5. Someter a su vez a contrastación esas técnicas para comprobar su relevancia y la fe que merecen. 6. Llevar a cabo la contrastación e interpretar sus resultados. 7. Estimar la pretensión de la verdad de las conjeturas y la fidelidad de las técnicas. 8. Determinar los dominios en los cuales valen las conjeturas y las técnicas, y formular los nuevos problemas originados por la investigación. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 9 Descrito desde otro punto de vista, podemos decir que el método científico es el medio por el cual tratamos de dar respuesta a las interrogantes acerca del orden de la naturaleza. Las preguntas que nos hacemos en una investigación generalmente están determinadas por nuestros intereses, y condicionadas por los conocimientos que ya poseemos. De estos dos factores depende también la “clase” de respuesta que habremos de juzgar como “satisfactoria”, una vez encontrada. El método científico es la lógica general2 empleada, tácita o explícitamente para valorar los meritos de una investigación. Es, por tanto, útil pensar acerca del método científico como constituido por un conjunto de normas, las cuales sirven como patrones que deben ser satisfechos si alguna investigación es estimada como investigación responsablemente dirigida cuyas conclusiones merecen confianza racional.3 El método científico sigue una direccionalidad univoca que le es característica, porque el método como tal es en sí un procedimiento encaminado a un objetivo, el intentar lograrlo lleva implícita una dinámica que para el caso del método científico se inicia con la Fase de la Observación, donde el 2 La Lógica estudia las Leyes del Raciocinio (inductivo, deductivo y analógico) o las Leyes de la Razón. 3 Pág. 53-55. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación Editorial Limusa. México 2005. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 10 sujeto conocedor (científico) entra en contacto con el fenómeno, y sabe de él algo, algo que lo induce a continuar buscando; en un segundo gran momento, supone de ése fenómeno cierto nivel de verdad, esto es, en una segunda fase, o Fase del Planteamiento de la hipótesis, que fundamentada en conocimientos previos y en los datos por recoger, podría ser demostrada; por último tenemos la Fase de Comprobación, la cual depende del grado de generalidad y sistematicidad de la hipótesis. Las evidencias que comprueban o desaprueban son igualmente estimables. Es preferible, denominar a la teoría la concepción teórica o teoría general, que es un conjunto de conceptos, categorías y leyes generales sobre los procesos y objetos de la realidad. De esta teoría general se deriva – aunque de hecho se encuentra inserto en ella – el método general de conocimiento concebido éste como la manera de abordar el objeto de estudio y el cual es general para una determinada concepción teórica. Sí se considera a los fenómenos de la naturaleza y de la sociedad en movimiento, en desarrollo constante, es decir en su pasado, presente y futuro; en sus conexiones e interacción; en sus contradicciones internas, y se considera que los cambios cuantitativos se transforman en determinado momento y condiciones, en cambios cualitativos, el método de conocimiento será dialéctico materialista; pero si se El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 11 concibe a los fenómenos y objetos como algo acabado, inmutable, es decir, sin cambio, y cada uno de los aspectos de la realidad se analizan en forma aislada, y no existe interés por conocer las causas esenciales por las cuales los fenómenos surgen, se desarrollan y transforman, entonces en enfoque será metafísico.4 Cualquier teoría general o concepción teórica involucra determinados conceptos y sus interrelaciones que dan cuenta de la forma como se conciben los procesos y objetos. En el caso del materialismo dialéctico, los conceptos, categorías, principios y leyes generales, son: la materia, el movimiento, la contradicción, causa y efecto, esencia y fenómeno, forma y contenido, apariencia y realidad; el principio del historicismo, y de la conexión e interacción de los fenómenos, las leyes de la dialéctica, entre otros. Estas categorías y leyes generales – que forman parte de la filosofía marxista: el materialismo dialéctico – dan cuenta de una determinada concepción de la realidad y, a su vez, son instrumentos metodológicos que orientan la aprehensión de los fenómenos de la realidad concreta. 4 Pág. 58-60. Raúl Rojas Soriano. El Proceso de la Investigación Científica. Editorial Trillas. México 2004. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 12 Asimismo, las teorías, leyes e hipótesis que se elaboran en los distintos campos de la ciencia (por ejemplo, la teoría de la mecánica clásica, la teoría marxista de las clases sociales), permiten explicar las causas de los fenómenos o la relación entre ellos, pero a la vez, tales leyes o teorías se convierten en instrumentos metodológicos que guían el proceso de conocimiento de los fenómenos particulares objeto de estudio. El asunto de la relación entre la teoría y método debe ser abordado, en su primer momento y nivel, como la relación entre la concepción teórica o teoría general de los procesos y objetos, y la forma de abordar el estudio de tales procesos (método general de conocimiento, que para nosotros es el dialéctico que posee un carácter verdaderamente científico en cuanto que permite descubrir la esencia de los objetos y procesos para formular leyes científicas. El materialismo dialéctico supone que todo se halla vinculado y en interacción. 5 En el proceso de la investigación científica se utiliza diversos métodos y técnicas según la ciencia particular de que se trate y de acuerdo a las características concretas del objeto de estudio. Existen, sin embargo, métodos que pueden considerarse generales para todas las ramas de la ciencia en tanto que son procedimientos que se aplican en las distintas 5 Pág. 61. Raúl Rojas Soriano. El Proceso de la Investigación Científica, Editorial Trillas. México 2004. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 13 etapas del proceso de investigación con mayor o menor énfasis, según el momento en que éste se desarrolle. Estos métodos son el análisis y la síntesis, la inducción y la deducción.6 1.2. El Método Analítico El Método analítico es aquel método de investigación que consiste en la desmembración de un todo, descomponiéndolo en sus partes o elementos para observar las causas, la naturaleza y los efectos. El análisis es la observación y examen de un hecho en particular. Es necesario conocer la naturaleza del fenómeno y objeto que se estudia para comprender su esencia. Este método nos permite conocer más del objeto de estudio, con lo cual se puede: explicar, hacer analogías, comprender mejor su comportamiento y establecer nuevas teorías.7 ¿Qué significa Analizar? Analizar significa desintegrar, descomponer un todo en sus partes para estudiar en forma intensiva cada uno de sus elementos, así como las relaciones entre si y con el todo. La importancia del análisis reside en que para comprender la esencia de un todo hay que conocer la naturaleza de sus partes. El todo puede ser de diferente índole: un todo 6 Pág. 78. Ibíd. 7 Pág. 64. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación. Editorial Limusa. México 2005. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 14 material, por ejemplo, determinado organismo, y sus partes constituyentes: los sistemas, aparatos, órganos y tejidos, cada una de las cuales puede separarse para llevar a cabo un análisis mas profundo (esto no significa necesariamente que un aparato u órgano tenga que separarse físicamente del resto del organismo; en otras palabras, aislar un órgano o aparato significa aquí que no se tomen en cuenta las demás partes del todo). Otros ejemplos de un todo material es: la sociedad y sus partes: base económica (fuerzas productivas y relaciones sociales de producción) y la superestructura (política, jurídica, religiosa, moral). La sociedad es un todo material en tanto que existe fuera e independientemente de nuestra conciencia. El todo puede ser también racional, por ejemplo, los productos de la mente: las hipótesis, leyes y teorías. Descomponemos una teoría según las leyes que la integran; una ley o hipótesis, según las variables o fenómenos que vinculan y el tipo de relaciones que establecen, por lo tanto, puede hablarse de análisis empírico y análisis racional. El primer tipo de análisis conduce necesariamente a la utilización del segundo tipo; por ello se le considera como un procedimiento auxiliar del análisis racional. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 15 El análisis va de los concreto a lo abstracto ya que mantiene el recurso de la abstracción puede separarse las partes (aislarse) del todo así como sus relaciones básicas que interesan para su estudio intensivo (una hipótesis no es un producto material, pero expresa relaciones entre fenómenos materiales; luego, es un concreto de pensamiento). 1.3. El Método Sintético El método sintético es un proceso de razonamiento que tiende a reconstruir un todo, a partir de los elementos distinguidos por el análisis; se trata en consecuencia de hacer una explosión metódica y breve, en resumen. En otras palabras debemos decir que la síntesis es un procedimiento mental que tiene como meta la comprensión cabal de la esencia de lo que ya conocemos en todas sus partes y particularidades.8 La síntesis significa reconstruir, volver a integrar las partes del todo; pero esta operación implica una superación respecto de la operación analítica, ya que no representa sólo la reconstrucción mecánica del todo, pues esto no permitirá avanzar en el conocimiento; implica llegar a comprender la esencia del mismo, conocer sus aspectos y relaciones básicas en una perspectiva de totalidad. No hay síntesis sin 8 Pág. 64. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación. Editorial Limusa. México 2005. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 16 análisis sentencia Engels, ya que el análisis proporciona la materia prima para realizar la síntesis. Respecto de las síntesis racionales, por ejemplo, una hipótesis, ellas vinculan dos o más conceptos, pero los organiza de una forma determinada; los conceptos desnutrición y accidentes de trabajo al vincularse pueden dar por resultado una hipótesis: a medida que aumenta la desnutrición de los obreros, se incrementa la tasa de accidentes de trabajo. La hipótesis es una síntesis que puede ser simple o compleja. Asimismo, todos los materiales pueden ser simples (un organismo unicelular) o complejos (un animal mamífero); las sociedades pueden ser relativamente simples (una comunidad primitiva) o complejas (una sociedad industrial). La síntesis, sea material o racional, se comprende en el pensamiento; por ello, es necesario señalar que el pensamiento, si no quiere incurrir en arbitrariedades, no puede reunir en una unidad sino aquellos elementos de la consciencia en los cuales – o en cuyos prototipos reales – existía ya previamente dicha unidad. La síntesis va de lo abstracto a lo concreto, o sea, al reconstruir el todo en sus aspectos y relaciones esenciales permite una mayor comprensión de los elementos constituyentes. Cuando se dice que va de lo abstracto a lo El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 17 concreto significa que los elementos aislados se reúnen y se obtiene un todo concreto real (por ejemplo, el agua) o un todo concreto de pensamiento (una hipótesis o ley). En otros términos, Lo concreto (es decir el movimiento permanente hacia una comprensión teórica cada vez más concreta) es aquí el fin específico del pensamiento teórico, en tanto que es un fin de tal naturaleza, lo concreto define como ley la manera de actuar del teórico (se trata de una acción mental naturalmente) en cada caso particular, por cada generalización tomada aparte. El análisis y la síntesis se contraponen en cierto momento del proceso, pero en otro se complementan, se enriquecen; uno sin el otro no puede existir ya que ambos se encuentran articulados en todo el proceso de conocimiento.9 1.4. Inducción y deducción Debemos de tener en cuenta que, en cualquier área del conocimiento científico el interés radica en poder plantear hipótesis, leyes y teorías para alcanzar una comprensión mas amplia y profunda del origen, desarrollo y transformación de los fenómenos y no quedarse solamente con los hechos empíricos captados a través de la experiencia sensible 9 Pág. 78-82. Raúl Rojas Soriano. El Proceso de la Investigación Científica. Editorial Trillas. México 2004. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 18 (recuérdese que en la ciencia no es cierto aquello de que los hechos hablan por sí solos). Además, a la ciencia le interesa confrontar sus verdades con la realidad concreta ya que el conocimiento, como se ha dicho, no puede considerarse acabado, definitivo, tiene que ajustarse continuamente, en menor o mayor grado según el área de que se trate, a la realidad concreta la cual se encuentra en permanente cambio. En este proceso de ir de lo particular a lo general y de éste regresar a lo particular tenemos la presencia de dos métodos: la inducción y la deducción. La inducción se refiere al movimiento del pensamiento que va de los hechos particulares a afirmaciones de carácter general. Esto implica pasar de los resultados obtenidos de observaciones o experimentos (que se refieren siempre a un numero limitado de casos) al planteamiento de hipótesis, leyes y teorías que abarcan no solamente los casos de los que se partió, sino a otros de la misma clase; es decir generaliza los resultados (pero esta generalización no es mecánica, se apoya en las formulaciones teóricas existentes en la ciencia respectiva) y al hacer esto hay una superación, un salto en el conocimiento al no quedarnos en los hechos particulares sino que buscamos su comprensión más profunda en síntesis racionales (hipótesis, leyes, teorías). Esta generalización no se logra sólo a partir de los hechos empíricos, pues de conocimientos ya alcanzados se El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 19 pueden obtener (generalizar) nuevos conocimientos, los cuales serán mas complejos. Insistimos otra vez: el trabajo científico no va del paso mecánico de los hechos empíricos al pensamiento abstracto; existen niveles de intermediación y a medida que se asciende, las generalizaciones van perdiendo contacto con la realidad inmediata ya que se apoyan en otros conocimientos los cuales sí tienen relación directa o indirecta con la realidad. Para poder pensar en la posibilidad de establecer leyes y teorías con base en la inducción, es necesario partir del principio de la regularidad e interconexión de los fenómenos de la naturaleza y la sociedad, lo cual permite pasar de la descripción (que se refiere fundamentalmente a los hechos empíricos) a otros niveles de la ciencia: la explicación y predicción a través de leyes y teorías. Puede decirse que las conclusiones obtenidas a través de la inducción tienen un carácter probable, el cual aumenta a medida que se incrementa el número de hechos particulares que se examinan. Cabe destacar que los procedimientos de la inducción sólo permiten establecer relaciones entre hechos empíricos (leyes empíricas); para formular leyes teóricas que expliquen a aquéllas, es necesario apoyarse en otros planteamientos teóricos existentes en los marcos de la ciencia de que se trate. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 20 La deducción es el método que permite pasar de afirmaciones de carácter general a hechos particulares. Proviene de deductivo que significa descender. Este método fue ampliamente utilizado por Aristóteles en la silogística en donde a partir de ciertas premisas se derivan conclusiones: por ejemplo, todos los hombres son mortales, Sócrates es hombre, luego entonces, Sócrates es mortal. No obstante, el mismo Aristóteles atribuía gran importancia a la inducción en el proceso de conocimiento de los principios iniciales de la ciencia. Por tanto es claro que tenemos que llegar a conocer las primeras premisas mediante la inducción; porque el método por el cual, hasta la percepción sensible implanta lo universal, es inductivo.” El método deductivo está presente también en las teorías axiomáticas, por ejemplo en la Geometría de Euclides en donde los teoremas se deducen de los axiomas que se consideran principios que no necesitan demostración. Existen otro método afín desde el punto de vista lógico: el hipotético- deductivo. La diferencia con respecto al axiomático estriba en que las hipótesis de las que se deducen planteamientos particulares se elaboran con base en el material empírico recolectado a través de diversos procedimientos como la observación y el experimento.10 10 Pág. 83-84. Raúl Rojas Soriano. El Proceso de la Investigación Científica. Editorial Trillas. México 2004. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 21 En este proceso deductivo tiene que tomarse en cuenta la forma como se definen los conceptos (los elementos y relaciones que comprenden) y se realiza en varias etapas de intermediación que permite pasar de afirmaciones generales a otras más particulares hasta acercarse a la realidad concreta a través de indicadores o referentes empíricos. Este procedimiento es necesario para poder comprobar las hipótesis con base en el material empírico obtenido a través de la práctica científica.11 La deducción desempeña un papel muy importante en la ciencia. Mediante ella se aplican los principios descubiertos a casos particulares. El papel de la deducción en la investigación científica es doble: a) Primero consiste en encontrar principios desconocidos, a partir de otros conocidos. Una ley o principio puede reducirse a otra más general que la incluya. Si un cuerpo cae, decimos que pesa porque es un caso particular de la gravitación. b) También la deducción sirve científicamente para describir consecuencias desconocidas, de principios conocidos. Si sabemos que la formula de la velocidad es , t d V = podremos calcular con facilidad la velocidad que desarrolla un avión. La matemática es la ciencia 11 Pág. 85, Raúl Rojas Soriano. El Proceso de la Investigación Científica. Editorial Trillas. México 2004. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 22 deductiva por excelencia; parte de axiomas y definiciones. Inferencias inmediatas y medianas. En el razonamiento deductivo se reconocen dos clases de inferencias (tomado como sinónimo de conclusión, aunque algunos autores reservan el nombre de conclusión para las inferencias complejas). La inferencia inmediata de un juicio extrae otro a partir de una sola premisa. En la inferencia mediata la conclusión se obtiene a partir de dos o más premisas.12 Ejemplo de inferencia inmediata: “Los libros son cultura.” “En consecuencia, algunas manifestaciones culturales son libros.” Ejemplo de inferencia mediata: “Los ingleses son puntuales.” “Por tanto, William es puntual.” A partir de Rene Descartes, la Filosofía sigue dos corrientes principales, claramente opuestas: el racionalismo (centrado en la razón) y el empirismo (cuya base es la experiencia). Mientras que los alemanes y franceses cultivan preferentemente el racionalismo, los autores ingleses son los clásicos empiristas, los cuales, ya desde Roger Bacon, en la 12 Pág. 43-44, José L. López Cano. Método e Hipótesis científicos. Editorial Trillas. México 2001. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 23 Edad Media (1210-1292), muestran una decidida inclinación hacia ese tipo de pensamiento. En el Renacimiento, Francis Bacon (1561-1626) es el promotor del empirismo inglés, luego se continúa con John Locke y George Berkeley, hasta a su culminación, con David Hume, en el siglo XVIII. La idea central de Bacon, es la crítica contra el silogismo y la apología de la inducción. Dice que lo primero que hay que criticar y rechazar, si se intenta una sólida certeza en la investigación científica, es la serie de prejuicios que suelen colarse en nuestros conocimientos ordinarios. Bacon acierta, al señalar con toda precisión cuatro tipos de prejuicios, que plásticamente, son llamados ídolos: ídolos de la especie, ídolos de la caverna, ídolos del foro y ídolos del teatro. Bacon detecta el abuso del silogismo aristotélico como la principal causa del estancamiento de las ciencias. Critica claramente a Aristóteles y su obra. En su lugar, proclama el método inductivo (generalización a partir de la observación de casos particulares) como la clave para hacer progresar a las ciencias. El método inductivo en versión moderna fue desarrollado por el inglés Francis Bacon (1561-1626) y se encuentra ligado a las investigaciones empíricas. Bacon rechazo la silogística de Aristóteles en la que se apoyaba la escolástica (doctrina del medievo) y la cual desdeñaba la experiencia sensible. En El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 24 su lugar, Bacon destacó la importancia de la observación y el experimento en la obtención del conocimiento, pero minimizó el papel de las hipótesis por lo cual ha sido ampliamente criticado. Acerca de la ciencia, Bacon tiene una idea completamente utilitarista (john Dewey). Mientras que los empiristas –afirma- son como hormigas, que sólo acumulan hechos sin ningún orden; los racionalistas o teóricos son como arañas, pues sólo construyen bellas teorías, pero sin solidez. El verdadero científico debe ser como la abeja, que digiere lo que capta, y produce miel para la comunidad a la que pertenece. Para construir ciencia se debe proceder a base de experimentación, con el fin de observar las causas de los fenómenos, y poder comprender los procesos de la naturaleza y sociedad. Para interpretarla, primero hay que ser dócil a ella. La observación puede darnos la forma, o la ley de comportamiento del fenómeno estudiado. La forma es como la esencia íntima del fenómeno; pero no es de orden metafísico, sino físico y social, o sea, observable experimentalmente. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 25 Indudablemente, se debe dar un voto a favor del método inductivo. Gracias a él como pueden descubrir las leyes que rigen a la naturaleza y a la sociedad.13 Sin embargo, no había que deslumbrarse tanto por la eficacia de la inducción, como para menospreciar o dejar de lado el raciocinio deductivo. Lo correcto es saber utilizar cada uno de los dos procesos: deducción e inducción, según sea la naturaleza de la ciencia y del asunto tratado. Nótese cómo es el tema metodológico el que incide con frecuencia en el pensamiento filosófico de estos tiempos. Mientras que Descartes se inclina hacia el método deductivo, la corriente empirista se inclinará hacia el método experimental-inductivo. Lo cierto es que cada uno tiene su propia zona de aplicación, sin que sea necesario desvirtuar uno u otro método en cuanto tal.14 13 Págs. 112-113. Gutiérrez S. Raúl. Historia de las Doctrinas Filosóficas. Editorial Esfinge S.A., Mexico 1990. 14 Págs. 113. Gutiérrez S. Raúl. Historia de las Doctrinas Filosóficas. Editorial Esfinge S.A., México 1990. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 26 1.6. El pensamiento Cartesiano en el mundo Contemporáneo Rene Descartes (1596-1650), ha sido el más famoso genio del siglo XVII. Con él se coloca en la Historia una primera piedra divisoria, con respecto al pensamiento antigua y medieval, y por eso se le suele llamar el “Padre de la Filosofía moderna”, no obstante de él brillaron otros pensadores también revolucionarios, como, por ejemplo, Nicolás de Cusa (1401-1464) y Francis Bacon (1561-1626). Su idea central es la creación de un sistema filosófico completamente inexpugnable, libre de las críticas de los pensadores subsecuentes, y perfectamente garantizado en su verdad y en su orden lógico, similarmente a lo que sucedía en las Matemáticas, edificio mental sólidamente estructurado e inmune a las simples opiniones de cualquier profano en la materia. 1.6.1. Las reglas del método Para evitar el error, no basta con la inteligencia, es necesario saber aplicarla adecuadamente, es decir, se requiere un método. Descartes pone especial énfasis en la necesidad de un Método Racional, que por principio libere al hombre de la fácil caída en el error. En el Discurso del Método (Segunda Parte) describe sus famosas cuatro reglas metódicas, como sigue: El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 27 a) Regla de la Evidencia: No aceptar como verdadero sino lo que es evidente. O, en otros términos: tratar de captar intuitivamente el objeto propio de la inteligencia, a saber, las ideas claras y distintas. Cuando se logra percibir las notas características de una idea y cuando se logra distinguir esas notas con respecto de las demás ideas, se posee una idea clara y distinta, y esto ya es una garantía de la verdad del conocimiento poseído. Para eso hay que evitar la prevención y la precipitación. En una palabra, sólo se puede poseer la verdad cuando el espíritu capta las ideas con toda su evidencia, de un modo fácil, inmediatas, serenas y claras. Esta evidencia ya no puede encerrar la duda y el error. b) Regla del Análisis:”Dividir cada una de las dificultades que se van a examinar, en tantas partes como sea posible y necesario para resolverlas mejor.” Es decir, descomponer las ideas complejas en sus partes más simples; pero, además, remontarse a los principios más simples, de los cuales depende el asunto que se está examinando. c) Regla de la Síntesis: “Conducir por orden los pensamientos, empezando por los objetos más sencillos, más fáciles de conocer, para subir gradualmente hasta el conocimiento de los más complejos…” Se trata de la operación contraria a la anterior, y es El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 28 complementación. Una vez dividido en partes un asunto, para su mejor comprensión, es necesario reconstruir el todo, a partir de los principios encontrados. Coincide, tal como se ha estudiado en Lógica, con la Deducción. Lo importante consiste en el procedimiento gradual que avanza lógicamente (con encadenamiento y congruencia natural), desde lo simple de los principios, a lo complejo de las conclusiones, teoremas y demás consecuencias de las primeras verdades. d) Regla de las Enumeraciones y Repeticiones: “Hacer enumeraciones tan completas, y revisiones tan generales, como para estar seguro de no omitir nada.” Con esto se persigue una intuición global del asunto tratado, de tal manera que la inteligencia posea y domine la materia desde el principio hasta el fin, lo cual supone la repetición o repaso del camino andado. De esta manera, sencilla y coherentemente, Descartes propone a la inteligencia las cuatro reglas más importantes que hay que tener en cuenta si se quiere un resultado eficaz en su funcionamiento. Por tanto, debemos permitir que la mente se percate, por sí misma, del asunto tratado, que el esfuerzo se divida en partes suficientes como para simplificar El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 29 el trabajo, que se reconstruya la totalidad del esfuerzo, y que se revise globalmente el resultado.15 1.6.2. La duda Metódica Una vez establecido el método a seguir, Descartes se propone edificar una Filosofía perfectamente estructurada, al modo de las ciencias matemáticas. Para ello será necesario partir de una verdad absolutamente indubitable, y de la cual se pueda derivar todo el edificio filosófico. Para encontrar esa primera verdad, es preciso borrar, con anterioridad, todo conocimiento que no esté debidamente fundamentado. Por lo tanto, hay que hacer caso omiso, o mejor, dudar, de todo lo que percibimos por los sentidos, y de todos los conocimientos científicos. La duda que propone Descartes tiene como finalidad la fundamentación de la nueva filosofía sobre bases indubitables. Por lo tanto, no se trata de una duda escéptica, en donde el fin es dudar por dudar. Es una duda metódica, puesta solamente como un método o medio, para llegar a un principio completamente evidente. 15 Págs. 100-101. Gutiérrez S. Raúl. Historia de las Doctrinas Filosóficas. Editorial Esfinge S.A., México 1990. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 30 En estas condiciones, con una cierta ambigüedad respecto a la seriedad de la duda metódica y universal, Descartes se lanza a la búsqueda de su primer principio. Si dudo (reflexiona así en la Cuarta parte del Discurso del Método), es que pienso, y si pienso, es que existo. De esto modo llega a lo que le parece su primer principio fundamental: “Pienso, luego existo” (Cogito, ergo sum). 1.6.3. El primer principio Cartesiano No es tan original Descartes al anunciar su principio fundamental: “Cogito, ergo sum”. Ya San Agustín había esgrimido un arma semejante, en contra de los escépticos: “Si fallor, sum” (si me equivoco, existo). Sin embargo, la novedad, en Descartes, consiste en que, por primera vez, se pretende erigir sobre esta verdad todo el cuerpo de verdades filosóficas. Su principio funcionará a la manera de los axiomas de las ciencias matemáticas. El “Cogito” (así se suele llamar al primer principio cartesiano, por brevedad) es, pues, una intuición fundamental. Todo el mundo podrá dudar sobre lo que quiera, pero no podrá dudar de su propia existencia. Si duda, es que piensa, y si piensa, es que existe. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 31 Por su parte, Santo Tomás jamás habla de esa intuición del propio yo; lo que se conoce es el efecto, los frutos, y por medio de ellos, pero ya de un modo mediato, podemos retroceder hasta las substancia, la cual es inferida como se infiere la causa a partir de los efectos, y no por intuición intelectual (directa e inmediata visión del objeto).16 1.7. Los procedimientos de la inducción según John Stuart Mill (1806-1873), quien los expuso en forma de reglas: 1. Método de semejanzas: “Si dos o mas casos del fenómeno sometido a investigación tienen de común sólo una circunstancia, entonces esta circunstancia – en la que sólo concuerdan todos estos casos – es la causa (o consecuencia) del fenómeno dado.” La importancia de este procedimiento radica en que permite una aproximación al conocimiento de la verdadera causa ya que ayuda a eliminar diversos factores, porque no guardan relación, aunque es posible incurrir en error en este punto. En segundo lugar, indica que ciertos factores parecen darse conjuntamente. En tercer lugar, nos permite observar que, en la situación concreta, el factor. 2. Método de la diferencia: “Si el caso en el que aparece el fenómeno dado y el caso en que no aparece son 16 Págs. 102-103. Gutiérrez S. Raúl. Historia de las Doctrinas Filosóficas. Editorial Esfinge S.A., México 1990. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 32 semejantes en todas las circunstancias, excepto en una, que se encuentran en el primer caso, esta circunstancia en la cual se diferencian únicamente estos dos casos, es la consecuencia o la causa, o la parte necesaria de la causa del fenómeno.” 3. Método combinado de semejanza y diferencia: “Si dos o mas casos de surgimiento del fenómeno tienen en común una sola circunstancia, y dos o más casos en que no surge ese fenómeno tienen en común sólo la ausencia de esa misma circunstancia, entonces tal circunstancia en la que sólo se diferencian ambos tipos de casos, es la consecuencia o la causa, o la parte necesaria del fenómeno investigado.” 4. Método de variaciones concomitantes: “Todo fenómeno que varia de alguna manera siempre que otro fenómeno varia de una manera particular, o bien es la causa o es el efecto de este fenómeno, o está conectado con él por alguna causa.” 5. Método de residuos: “Separar del fenómeno una parte tal, que se sabe por inducciones anteriores, que es el efecto de ciertos antecedentes y el resto del fenómeno es el efecto de los demás antecedentes.” El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 33 2. LAS ETAPAS DEL METODO CIENTIFICO Caracterización de los problemas Las expresiones del pensamiento constituyen preguntas y problemas por resolver, o bien, respuestas y soluciones a las indagaciones realizadas. En este sentido, el curso del conocimiento científico consiste en una sucesión ininterrumpida de problemas que surgen a partir de los resultados obtenidos en las investigaciones anteriores y se resuelven mediante el razonamiento y la experimentación. Para encontrar la solución de esos problemas, la actividad científica ha establecido procedimientos adecuados y desenvuelve continuamente otros nuevos. Entre ellos se encuentran los experimentos que nos informan, tan exacta y completamente como es posible, acerca de los procesos naturales y sociales, lo mismo que sobre sus conexiones activas y su mutua causalidad. También se encuentran las teorías, que nos permiten reunir los resultados de los experimentos en una explicación común, necesaria y suficiente. Por último, tenemos la aplicación de dichas teorías para intervenir, de manera directa y concreta, en el comportamiento de los procesos de la sociedad y de la naturaleza, haciendo que produzcan la satisfacción de las necesidades humanas y resolviendo prácticamente, de esta El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 34 manera, los problemas que impulsan la propia actividad científica. En términos generales, por problema entendemos cualquier dificultad que no se puede resolver automáticamente, es decir, con la sola acción de nuestros reflejos instintivos y condicionados, o mediante el recuerdo de los que hemos aprendido anteriormente. Por otra parte, además de los problemas que nos imponen directamente las condiciones naturales y sociales en que vivimos, constantemente estamos creando o inventando otros problemas; como con, por ejemplo, la explicación de los procesos recién descubiertos, la demostración de teoremas, la verificación de hipótesis, la decisión entre dos o más teorías de pugna, o bien, la transformación de la naturaleza y la sociedad, etcétera.17 17 Pág. 223. Eli de Gortari. Lógica General. Edit. Grijalbo. S.A., vigésima sexta edición. México 1965. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 35 Diseño de la investigación Ésta consiste en señalar con toda claridad y precisión el rumbo y la meta. Así que precisar el campo al que pertenece el problema sería en principio el primer paso; determinar con todas sus características el problema a resolver; sería el segundo paso; fijar el objetivo que se busca alcanzar, o mejor dicho establecer cuál será el fin que se pretende alcanzar con la investigación; para esto se deberán definir los procedimientos, esto es, la metodología y todo tipo de requerimientos que permitirán obtener la información mediante los procesos si ese fuera el caso. 18 El método científico consta de las siguientes etapas, las cuales se describen a continuación: 2.1. ELECCION DEL TEMA En la elección del tema se concretará, tanto como sea posible el objeto de conocimiento; además habrá de estructurarse el título tentativo del proyecto de investigación, tentativo porque podría hacérsele algunas pequeñas precisiones durante el proceso de la investigación. ¿Qué se va a investigar? ¿Cómo se realizara la investigación? 18 Pág. 98. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación. Editorial Limusa. Mexico 2005. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 36 ¿Por qué es importante la temática a investigar? Las preguntas son cómo, por qué, cuándo y dónde Explican el surgimiento de un tema de investigación, la razón de un trabajo de investigación. ¿Qué elementos forman parte de la Elección del Tema de investigación? Se recomienda, enlistar los principales problemas locales, estatales, nacionales e internacionales que se identifican desde el lugar donde se va a llevar a cabo la investigación, y así se puede deducir el Tema de Investigación. 2.2.- Planteamiento del Problema de Investigación El problema es la fijación de las contradicciones que se dan en la propia realidad, contradicciones que se fijan en la teoría y que concluyen una vez “esclarecidas” con el planteamiento de un nuevo problema, cuya solución podría ser resuelta por otros investigadores. Para un adecuado planteamiento del problema se requiere de, eliminar del problema cualquier adición engañosa, o sea, identificar aquellas dificultades que chocan con la teoría. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 37 El proceso de solución de todo problema, supone como condición necesaria, la formulación adecuada y científica de la interrogante que se encuentra en la base del problema. Si el problema está formulado científicamente, el camino para la solución está más claramente definido. Un correcto planteamiento del problema, además debe poner de manifiesto las premisas que permitan resolverlo, a partir de la realidad como condición para su solución, aunada al supuesto de un examen teórico, fijando determinadas formas lógico-metodológicas. Una de las reglas heurísticas más importantes para la solución de problemas consiste en que éste pueda resolverse utilizando idealizaciones iniciales, claramente comprendidas y estipuladas, que simplifiquen su complejidad sin tergiversar la realidad mostrando la tendencia general del desarrollo del objeto investigado, ya que es en la realidad en la que se encuentra su posible solución. Otra regla heurística, es la exigencia de resolver por partes los problemas, esta condición es la relativa a la diferenciación del aparato conceptual (marco conceptual) que consiste en hacer desde ahora una clara distinción entre los conceptos involucrados en el problema mismo, ya que la ausencia de diferenciaciones conceptúales hace posible el tratamiento científico del problema. Esta diferenciación conceptual por su esencia, representa el proceso previo para El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 38 la elaboración de la hipótesis, que en sí misma da cuenta del problema. Una vez seleccionado el Tema de Investigación, se debe de enunciar la problemática de la investigación, para buscarle las alternativas de solución a través de las diferentes disciplinas del conocimiento (ciencias empíricas o formales, según sea el tipo de investigación) científico o de la que corresponda dependiendo del problema a investigar. Por Problema se entiende “la cuestión que trata de resolver por medio de procedimientos científicos. El inicio de la investigación es el Problema. Ahora bien, el planteamiento del problema en forma general significa, “la presentación clara y directa de la relación entre dos o mas variables contenidas en el problema, que se pueden comprobar empíricamente y que permiten encontrar las vías de solución o respuestas”, Es decir, plantear un problema es minimizar todos sus efectos y relaciones fundamentales o entre mas particular sea el problema a investigar esto, facilita el proceso de la investigación, en cambio si es muy general dicha investigación pierde el rumbo, por que existirá diferentes líneas de investigación. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 39 En el planteamiento del problema existen tres aspectos básicos que se deben de reflexionar, analizar y conceptualizar y son los siguientes: 1. Descripción del problema de investigación. 2. Elementos del problema de investigación. 3. Formulación del problema de investigación. 2.2.1. Delimitación y ubicación del problema Mario Bunge refiere que: “no se conocen recetas falibles para preparar soluciones correctas a problemas de investigación mediante el mero manejo de los ingredientes del problema”. Sin embargo se pueden tomar en cuenta algunas sugerencias que permitan delimitar y ubicar el problema de investigación como las siguientes: 2.2.2. Elementos del problema Los problemas como tal no existen, es el investigador quien los plantea dadas sus inquietudes, capacidad de observación y conocimientos. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 40 Esta afirmación se apoya en el hecho de que ante un fenómeno o situación dada, todos podríamos pasarlos por alto, pero sólo uno se detiene y se plantea las interrogantes que ésta le despierta. Son elementos aquellas características de la situación problemática imprescindibles para el enunciado del problema, es decir, sumados los elementos del problema se tiene como resultado la estructura de la descripción del problema. Para poder abarcar la búsqueda de una solución a un problema, el investigador debe precisar la naturaleza y las dimensiones del mismo. Para ello, se requiere reunir datos empíricos que se puedan relacionar con el problema y posibles explicaciones del mismo. Para que la lista obtenida de los elementos del problema adquiera verdadero significado, el investigador procurará hallar las relaciones que existen entre los hechos empíricos, por una parte, y entre las explicaciones racionales por la otra, y tratara de relacionar aquellos con estas. Luego de incorporar nuevos datos a la lista de elementos, eliminar los que considere carentes de importancia, el investigador realizará un profundo examen de los supuestos en que se basan los hechos, explicaciones y relaciones halladas. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 41 2.2.3. Descripción del Problema de investigación. Este aspecto nos indica describir de manera objetiva la realidad del problema que se esta investigando. En la descripción se señalan todas las características de la problemática, los hechos y los acontecimientos que están en entorno social, al mismo tiempo se debe mencionar los antecedentes del problema. Antecedentes del estudio o problema de investigación. Las técnicas en las que se basó, las categorías de análisis o ejes centrales que permiten guiar el proceso de investigación. Los supuestos básicos en los que se apoya el enunciado del problema. Un enunciado completo del problema incluye todos los hechos, relaciones y explicaciones que sean importantes en la investigación. Hay que encuadrarlos en un enunciado descriptivo o en una pregunta que indique con claridad que información ha de obtener el investigador para resolver el problema de investigación. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 42 Por ejemplo: ¿De que manera influye la preparación académica (V. Independiente) en un sujeto (cambio en la conducta del sujeto V. Dependiente) determinado? ¿Cómo puede utilizar los conocimientos académicos un sujeto que le permitan lograr movilidad social, económica y cultural dentro de una sociedad determinada? ¿Cómo influye la preparación académica en un sujeto para el desarrollo de una conciencia flexible y racional? El investigador deberá de reconocer e identificar, que datos empíricos e intelectuales (teorías, conceptos, axiomas, postulados, principios, etc.) conducen a la solución del problema de investigación. En el desarrollo del planteamiento del problema, es conveniente ubicarlo en un contexto geopolítico, socioeconómico, histórico y geográfico, etc., ya que dicha problemática no se presenta en forma aislada, esto significa que necesariamente tenemos que ubicarlo en el tiempo y en el espacio. Conocimiento de la problemática, manejar los conceptos, definiciones, elaborar preguntas sobre el objeto de investigación. Tener un pensamiento lateral, es decir, creativo, imaginario. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 43 2.2.4. Elementos que integran a la Descripción del Problema. Antecedentes del estudio. Hechos y acontecimientos. Las características y sus elementos (relaciones y explicaciones, y la importancia dentro del lugar, y el beneficio que traerá consigo). Contexto (político, socioeconómico, histórico, geográfico). 2.2.5.- Formulación del Problema de investigación. De acuerdo con lo anterior, es de suma importancia de conocer cómo se define y se formula el Problema de investigación, con su entorno y sus relaciones de la manera más concreta posible, En la formulación del problema, la definición es la fase mas importante y se debe de realizar con elementos de la problemática que se investiga, definir un problema es señalar todos los elementos, aspectos, características en forma entendible y precisa, con el fin de que otras personas (lectores) puedan entender el proceso de la investigación. Cuando se halla definido la problemática es necesario formularlo y redactarlo para contar con todos los elementos del proceso de la investigación. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 44 Es decir, el proceso de división conceptual del problema consiste en fijar la idea principal, los supuestos básicos en que se fundamente la argumentación inicial en relación con el problema planteado, los supuestos accesorios y accidentales que van surgiendo al usar la lógica para precisar el razonamiento. Es recomendable al finalizar el planteamiento del problema redactar algunas preguntas que surgen de la problemática, es decir, una gran pregunta central como eje de la investigación y de ahí derivar las preguntas secundarias. Las preguntas bases son: ¿Qué? ¿Cuando? ¿Para que? ¿Quien? ¿Donde? ¿Con que? ¿Como? ¿Por que? ¿Cuanto? ¿Que relaciones se pueden establecer? ¿Cuáles son los puntos esenciales de la problemática? ¿Cuáles serian las alternativas de solución de la investigación? ¿Cómo establecer las relaciones con las variables (dependiente, independiente)? El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 45 2.3.- Justificación del problema de investigación. En este apartado se explica las razones o los motivos por los cuales se pretende realizar la investigación por lo general es breve y concisa. Por justificación se entiende sustentar, con argumentos convincentes, la realización de un estudio, en otras palabras, es señalar por qué y para qué se va a llevar a cabo dicha investigación. Para elaborar la justificación primero se tiene que conocer bien el problema, posteriormente se requiere de: Explicar por qué es importante realizar la investigación. Que beneficios se obtendrían al resolver la problemática que se plantea. En el desarrollo de la investigación se puede dimensionar en diferentes tipos de interés como son los siguientes: Intereses personales. Intereses institucionales. Intereses políticos. Es decir, explicar el tipo de interés que se tenga sobre el tema que sé esta investigando, con la finalidad de conocer esas razones, que por la cual se ha interesado. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 46 2.4.- Objetivos de la Investigación. Los objetivos es parte fundamental en el proceso de la investigación científica o de cualquier estudio que se realizar, nos permite, predecir, explicar y describir los fenómenos y adquirir conocimientos de esos fenómenos estudiados. Con los objetivos se busca la finalidad de la investigación, es decir, es la referencia, que guía o permite el desarrollo de la propia investigación. Los objetivos deben estar claramente redactados o bien formulados, para lograr transmitir lo que sé esta investigando y evitar confusiones o desviaciones en la investigación. Con objetivos claros, precisos nos va a permitir a) Extender y desarrollar los conocimientos de un tema. b) Profundizar y preguntar acerca de tesis o argumentos científicos. c) Llevar la práctica los conocimientos adquiridos en el diseño de una investigación. d) Con los objetivos sabremos los alcances, las limitaciones de la investigación y nos va a permitir dirigir todos los esfuerzos hacia una misma dirección la investigación. Los objetivos deben de tener congruencia con las demás fases de la investigación, ya que una de las características propia del proyecto de investigación, los objetivos se tienen El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 47 que estar revisando en el proceso de la investigación, para evitar desviaciones o fallas. En la elaboración de los objetivos es valido plantear un objetivo general que debe de ser más amplio que cualquiera de los objetivos particulares y lo más preciso para lograr las metas que se propone el investigador, de este objetivo general se desprenden los objetivos particulares, que son las fases del proceso de la investigación, es decir, de lo que se va a investigar. En la formulación de los objetivos se utilizan verbos, en infinitivo, es decir, con verbos no conjugados, aunque en la obra de Benjamín Bloom (1960) es una clasificación de metas educativas y no como una guía para la redacción de objetivos para la investigación, se puede sugerir la utilización, para facilitar la redacción, estos pueden ser: El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 48 Objetivos generales y específicos. conocimiento comprensión aplicación Análisis síntesis evaluación Definir Repetir Apuntar Inscribir Registrar Marcar Recordar Nombrar Relatar Subrayar Enlistar Enunciar Traducir Reafirmar Discutir Describir Explicar Expresar Identificar Localizar Transcribir Interpretar Aplicar Usar Emplear Demostrar Dramatizar Practicar Ilustrar Operar Inventariar Esbozar Trazar Distinguir Analizar Diferenciar Calcular Experimentar Probar Comparar Criticar Investigar Componer Planear Proponer Diseñar Formular Arreglar Ensamblar Reunir Construir Crear Organizar Dirigir Aprestar Juzgar Evaluar Tasar Seleccionar Escoger Valorar Estimar Medir 2.5.- Estructuración del esquema El esquema es la representación grafica sistematizada, que tiene como función principal estructurar un conjunto de ideas y los datos necesarios e imprescindibles de manera sintetizada con el menor número de palabras, en un orden lógico, que permita captar en un solo golpe de vista la temática desglosada. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 49 Inmediatamente después de haber sido diseñado y aprobado el proyecto de la investigación, se estructura el esquema que también se le conoce como plan de trabajo o bosquejo; la importancia de esta sección reside en que mediante su estructura dividida en capítulos y éstos a su vez en subcapítulos, permiten de manera ordenada desarrollar sus partes con un cierto orden, o tomarlo como base para posibles modificaciones. Generalmente el primer apartado del esquema se destina a una introducción, los inmediatos siguientes capítulos, hacen una revisión de los antecedentes, esto es de investigaciones que preceden a la que se está realizando. Los capítulos intermedios corresponden al desarrollo de la investigación en sí, y los últimos capítulos se destinan a concluir sobre los resultados de la investigación.19 Es indispensable elaborar el esquema de la investigación o el índice de lo que va hacer la investigación, el esquema puede ajustarse en el proceso de la investigación, conforme a los resultados que genere la propia investigación. El esquema es la ordenación temática probable de la problemática a investigar. 19 Pág. 96. Pág. 86. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación Editorial Limusa. México 2005. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 50 2.6. MARCO TEORICO El marco teórico es el conjunto de principios teóricos que guían la investigación estableciendo unidades relevantes para cada problema a investigar, Cabe mencionar que con cierta frecuencia en la literatura se usa indistintamente los términos: Marco Teórico, Marco Conceptual, Marco Teórico Conceptual, y Marco de Referencia. Si bien es cierto que unos están comprendidos en otros o que se relacionan entre sí, vale la pena hacer una precisión al respecto. El Marco Teórico es el apartado que comprende la delimitación teórica relativa y exclusiva que da sustento a un tema de investigación de forma lógica, donde sus elementos conceptuales son inherentes a la teoría(s) en estudio. Tamayo y Tamayo establece que el Marco Teórico cumple las siguientes funciones. • Delimitación el área de la investigación; para ello habrá que seleccionar los hechos que tengan relación entre sí, mediante una teoría que dé respuesta al problema en cuestión. • Sugerir guías de investigación, para encontrar nuevas alternativas de solución del problema. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 51 • Compendiar conocimientos existentes en el área que se esté investigando. • Expresar proposiciones teóricas generales, postulados, leyes que habrán de servir como base para la formulación mas “adecuada” de la hipótesis, su operacionalización, e incluso para la determinación de los indicadores. Los puntos antes referidos se pueden conjuntar para decir que la función principal del Marco Teórico la constituye el propósito de dar consistencia, unidad y coherencia a las teorías con la investigación en proceso. El Marco Teórico, es pues un instrumento conceptual metodológico que se construye sobre la base de la información pertinente al problema de investigación, más precisamente con la o las teorías que dieron sustento a otras investigaciones. A la información seleccionada que nos muestra el avance de lo logrado en investigaciones anteriores y que están relacionadas con el problema de investigación, se le denomina, Estado del Arte, y que será el que sirva de base para la construcción del Marco Teórico. Del Estado del Arte se precisa saber cuál será la teoría que servirá de base para sustentar el trabajo en cuestión. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 52 Para la elaboración del Marco Teórico, analícese la teoría o las teorías más afines, mismas que le permitirán formalizar el trabajo mediante la reducción de los fenómenos a proposiciones lógicas, y de esta manera poder relacionar lo más preciso posible el cuerpo teórico con la realidad para orientar la búsqueda. Resumiendo, para la elaboración del Marco Teórico se habrá de considerar básicamente lo siguiente: a) El problema de investigación. b) La referencia a los estudios afines de investigaciones fundamentales y recientes, relacionadas con el problema de investigación. c) Ubicación de la teoriza o teorías base para dar sustento a la investigación en proceso. d) Definición conceptual. e) Las implicaciones teóricas y metodológicas que podrían permitir determinar las limitaciones teóricas, metódicas y metodológicas. f) De trabajos anteriores, establecer el sistema de hipótesis que les dio sustento, y el papel que desempeñaron en El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 53 ellos, y considerarlos al momento de estructurar la propia hipótesis de trabajo. g) Esbozar las variables y de ser viable, los indicadores.20 2.7. ELABORACION DE HIPOTESIS En toda investigación se debe establecer la hipótesis de investigación. La hipótesis debe concordar con la definición del problema, así como con los demás elementos del diseño. Su función principal es la de operar como un eje guía de la investigación, porque en torno a ella deberán girar todas las operaciones que se realicen, esto significa, que durante el proceso no se deberá perder de vista su funcionalidad.21 2.7.1.- Formulación de Hipótesis. La siguiente fase son las hipótesis, y son los elementos importantes de toda investigación que sirven como guías precisas y orientan al investigador, a comprobar la problemática que sé esta investigando, las hipótesis son de gran importancia, se construyen tanto en la vida cotidiana como en el proceso de la investigación científica, las hipótesis surgen en la elaboración del planteamiento del problema. 20 Pág. 85-86. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación. Editorial Limusa, México 2005. 21 Pág. 86. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación. Editorial Limusa. México 2005. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 54 Todo el tiempo se plantean soluciones tentativas a los problemas que se presentan en nuestro entorno social. Sin embargo, las conjeturas que se establecen cuando se actúa científicamente, son creaciones mentales (intelectuales) construidas conscientemente. Es decir, no surgen de la espontaneidad sino se formulan de acuerdo con criterios que se les permitan ser útiles en el proceso de la investigación científica. Las hipótesis constituyen una herramienta que ayuda a ordenar, estructurar y sistematizar el reconocimiento mediante una proposición. La hipótesis implica una serie de conceptos y juicios tomados de la realidad estudiada, que llevan la esencia del conocimiento. Una hipótesis clara, concreta, viable puede ser la guía de la investigación, por que establece los límites, ayuda a organizar las ideas, y da un enfoque al procedimiento de la problemática estudiada. Ahora bien la palabra hipótesis se deriva de hipo: bajo y thesis: suposición, podemos conceptualizar de la siguiente manera: Hipótesis Es una suposición que establece relaciones entre los hechos o fenómenos, mediante dos o más variables (v. independiente y v. dependiente), y a la que todavía falta una comprobación. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 55 Para enunciar hipótesis científicas, así como para comprobarlas, se deben seguir una serie de reglas y procedimientos, que constituyen, en parte, la investigación científica. Arias Galicia nos señala las siguientes reglas. Dar la esencia. La definición debe dar la esencia de lo que intenta definir, es decir, su naturaleza, sus límites. Evitar tautologías. No debe directa o indirectamente contener el objetivo. o Ejemplo: La psicología es la ciencia que estudia a los fenómenos psicológicos. (tautología) o La psicología es la ciencia que estudia la conducta y los procesos cognoscitivos (lenguaje, pensamiento, ideas, conocimiento, inteligencia, etc.) del sujeto. (forma correcta) Debe ser afirmativa. Toda definición debe expresarse siempre en términos afirmativos, nunca en términos negativos. Empleo del lenguaje claro. Debe de expresarse en palabras claras y asequibles, no debe contener metáforas o figuras literarias. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 56 Variables Las hipótesis están compuestas por variables, y las variables son atributos que se miden en las hipótesis o también pueden ser conceptos operacionales que adquieren diferentes valores y se refieren a las cualidades o características, como por ejemplo: Masa (m), velocidad (v), aceleración (a), inteligencia, sexo, edad, estrato social, tasa de interés, escolaridad, peso, longitud, etc. La investigación gira en torno de las variables, debido a que la finalidad del trabajo científico es descubrir la existencia de ellas y su magnitud, así, como probar las relaciones que las unen entre sí. Esto quiere decir que después de haber establecido una descripción clara y científica del objeto de estudio de la investigación, el investigador procede a explicar dicho objeto. Dicha explicación costa de dos elementos como son: Variables independiente (X), se identifica como la Causa o antecedente. Variable dependiente (Y), se considera el Efecto o resultado. ¿Qué es variable independiente? Son todos los elementos o factores que explican un fenómeno científico. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 57 Esta variable puede ser manipulada por el investigador o científico. ¿Qué es variable dependiente? Son los efectos o resultados del fenómeno que se intenta investigar. ¿Cómo se determinan las variables? No es el propio investigador, quien va a determinar las variables, sino el objeto de estudio va hacer quien lo determine. 2.7.2. Definición de hipótesis científica La palabra “hipótesis” deriva del hipo: bajo, y thesis: posición o situación. Ateniéndose a sus raíces etimológicas, hipótesis significa una explicación supuesta que ésta bajo ciertos hechos, a los que sirve de soporte. La hipótesis es aquella explicación anticipada que le permite al científico asomarse a la realidad. Otra definición de hipótesis que amplía la anterior, nos dice: Una hipótesis es una suposición que permite establecer relaciones entre hechos. El valor de una hipótesis reside en su capacidad para establecer esas relaciones entre El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 58 los hechos, y de esa manera explicarnos por qué se produce. La hipótesis es una suposición de la existencia de una entidad, la cual permite la explicación de los fenómenos o del fenómeno estudiado. Las hipótesis son las proposiciones tentativas que relacionaran los datos empíricos con el conjunto de teorías adoptadas y provisionalmente analizadas en el Marco Teórico. En sí al elaborar la hipótesis, el investigador no tiene la total certeza de poderla comprobar. “Las hipótesis deberán ser proposiciones elaboradas correctamente desde el punto de vista formal (no tautológicas, coherentes y contradictorias, etc.) y deben, a partir de la corrección formal, proporcionar algún significado, es decir, deben decir algo en relación con los hechos a que se hace referencia. En segundo lugar, deben estar basadas en el conocimiento científico preexistente o, en última instancia, no estar en abierta contradicción con lo que la ciencia ya sabe acerca de la estructura y comportamiento de la naturaleza y de la sociedad. En tercer lugar al plantear una hipótesis, deberá tenerse en cuenta que pueda ser verificada apelando a los procedimientos metodológicos y técnicos de que la ciencia dispone.22 22 Pág. 86. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación. Editorial Limusa. México 2005. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 59 En efecto, las hipótesis fraguadas por los científicos pueden estar encaminadas a explicar un conjunto de fenómenos, como en el caso del éter, o bien a explicar un solo hecho, como la hipótesis que permitió descubrir la existencia de Neptuno y Plutón. La finalidad de estas hipótesis no es otra que la de explicar, de dar razón de los acontecimientos por medio de la interpolación de hechos que podrían haber sido observados, en condiciones adecuadas. ¿Qué es una explicación? Podemos definirla como un conjunto de enunciados de los cuales deducimos el hecho o los hechos que se desea explicar. La explicación nos permite eliminar el carácter problemático de las cosas. La función de una hipótesis descriptiva consiste en simbolizar la conexión ordenada de los hechos. Un ejemplo de este tipo de hipótesis lo encontraremos en Ptolomeo, en la medida en que este astrónomo proporcionó una representación geométrica de los cuerpos celestes, y, por otro lado, la hipótesis del éter, concebido como un fluido sin fricción y como sólido completamente elástico, es en realidad una hipótesis descriptiva. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 60 La hipótesis analógicas son aquellas mediante las cuales formulamos una hipótesis basándonos en que lo que es verdadero en un conjunto de fenómenos, puede ser también verdadero acerca de otro conjunto, debido a que ambos tienen en común ciertas propiedades formales. La hipótesis es una verdad provisional y nunca definitiva. En realidad, la ciencia toda puede considerarse, en última instancia, como una continua hipótesis susceptible de verificarse y de ser corregida (un sentido amplio del termino hipótesis). Sin embargo, en el proceso de la ciencia, es preciso distinguir entre hipótesis, ley y teoría. La hipótesis tiene carácter provisional; pero puede irse depurando y ajustando hasta convertirse en una ley y después en una teoría científica, la cual viene siendo una explicación más completa de un conjunto de fenómenos, y a su vez, puede abarcar varias leyes. Cuando una hipótesis es comprobada, adquiere el carácter de ley que puede definirse como aquella “relación constante y necesaria entre ciertos hechos” como acontece, por ejemplo, con las leyes del movimiento de Newton. Es claro que antes de llegar a ser comprobadas estas leyes, Newton formuló hipótesis en las cuales presumía lo que debía acontecer, y lo cual quedó confirmado al hacer los experimentos.23 23 Págs. 76-77. José L. López Cano. Método e Hipótesis científicos. Editorial Trillas. Mexico 2001. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 61 La investigación científica, no se queda con los aspectos externos de los procesos o problemas, sino que trata de descubrir los elementos esenciales que expliquen estas hipótesis empíricas, lo cual sólo puede realizarse planteando hipótesis teóricas que, por lo mismo, son más generales y en las cuales se destacan aquellas relaciones fundamentales entre los fenómenos. Como se ha visto, el problema descriptivo se refiere fundamentalmente a las manifestaciones o aspectos externos de los procesos y estructuras y la hipótesis que trate de responder a este tipo de problemas puede vincular dos o mas variables, pero, esto no es suficiente para determinar sus causas.24 2.8.- Metodología La metodología es un procedimiento general para obtener de una manera más precisa el objetivo de la investigación, dependiendo de la problemática que se vaya a estudiar se determina el tipo de investigación, es decir: a) Bibliográfica. b) De Campo. c) Experimental. 24 Pág. 109. Raúl Rojas Soriano. El Proceso de la Investigación Científica. Editorial Trillas. México 2004. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 62 2.9.- CRONOGRAMA Es el apartado del diseño de la investigación elaborado por quien habrá de realizar la investigación, y en el que se señala las diferentes etapas de realización del proyecto en relación con los tiempos estimados. Al cronograma se le conoce también como: Grafico de Gantt o Calendario de Actividades; sea cual fuere el nombre, lo más importante es que en él queden registradas todas las actividades de la investigación y el tiempo estimado para realizar cada una de ellas, debiendo estar colocadas en un orden lógico, de acuerdo al proceso y a los requerimientos de la propia investigación.25 25 Pág. 98. Ortiz Frida, García Maria del Pilar. Metodología de la Investigación Editorial Limusa. México 2005. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 63 CRONOGRAMA 2.10. ANEXOS 2.11. GLOSARIO 2.12. BIBLIOGRAFIA El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 64 ELEMENTOS DE UN PROYECTO DE INVESTIGACION INTRODUCCION RESUMEN O ABSTRACT CAPITULO I PLANTEAMIENTO DEL PROBLEMA 1.1. ANTECEDENTES 1.2. DELIMITACION DEL PROBLEMA 1.3. JUSTICACION 1.4. OBJETIVO GENERAL Y OBJETIVOS ESPECIFICOS. 1.5. DISEÑO Y CONTRUCCION DE HIPOTESIS (PRELIMINAR. TRABAJO, DESCRIPTIVA, ANALOGICA, NULA, ETC.) CAPITULO II MARCO TEORICO CAPITULO III ENFOQUE EPISTEMOLOGICO O ESCUELA DE ALGUNA CORRIENTE FILOSOFICA (METODO CUALITATIVO O CUANTITATIVO). METODOLOGIA TECNICAS DE INVESTIGACION (PARA LA RECOLECCION DE LA INFORMACION EMPIRICA; CUESTIONARIOS, ENTREVISTAS, ETC.) CRONOGRAMA BIBLIOGRAFIA GLOSARIO DE TERMINOS ANEXOS El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 65 ¿Cómo seleccionar un Tema o tópico de Investigación? ¿Cuál es el Problema que desea Resolver o Solucionar? ¿Cuáles son los elementos que forman parte del Problema? Los objetivos se formulan, para establecer y definir, lo que se pretende alcanzar o conseguir. Dentro de ellos se contaran un Objetivo General y objetivos Específicos. ¿El investigador deberá de definir y precisar, cual es la meta o el fin del proyecto de investigación? Describir con precisión la naturaleza y características del fenómeno o hecho. Identificar y determinar el periodo de retorno o frecuencia con que ocurre o se presenta el fenómeno estudiado. Comprobar la Hipótesis en función de los Datos Empíricos y teóricos, asimismo identificar la relación causal entre las variables. Presentar los Juicios o proposiciones que sirven como argumentos y justificación del problema. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 66 Por ejemplo: A continuación se presenta el siguiente objeto de estudio sobre la neurosis en las mujeres de 20 a 30 años quienes se encuentran laborando como profesoras de educación primaria en la Ciudad X, y en el grado de segundo año grupo A y B respectivamente. Las preguntas a las que tendría que dar respuesta la presente investigación serian las siguientes: 1. ¿Cómo se desarrolla la Neurosis en las profesoras? 2. ¿Qué tipos o clases de Neurosis se identifican en las profesoras? 3. ¿Cuál es la Causa o factor (s) que producen la Neurosis? 4. ¿Quiénes están propensos o pueden ser afectados por este Fenómeno o problema? 5. ¿Cómo se puede evitar y ayudar a las personas que se encuentran afectados con este problema? 6. ¿Qué niveles de neurosis existen en las profesoras? 7. ¿Qué métodos psicológicos existen para el tratamiento de la neurosis? 8. ¿Qué condiciones se requiere en las profesoras para que se genere la Neurosis? El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 67 Objetivo principal ¿Deberá describir y definir el objetivo principal y meta final (la causa material y formal) de la investigación? Es importante desarrollar la capacidad para ordenar las ideas y la información recabada; y de esta manera conectar y relacionar unos datos con otros, dándole sentido coherente y forma, el cual se traduzca en información significativa para la investigación. ¿Deberá describir y definir los objetivos específicos de la investigación? Debemos lograr que la investigación tenga o al menos cuente con la información siguiente: • Una estructura interna o cuerpo de la investigación. • Su ordenamiento de la información de forma coherente y sistemática. • Y su conexión y relación de los elementos que la integran, pueden contar con relaciones entre si. ¿Debemos saber que tipo de investigación será? ¿Cuál será el método que sustentara dicha investigación? ¿Definir adecuadamente la metodología de la investigación? ¿Definir el cuerpo teórico con alguna Teoría(s) que se identifique o varias teorías que permitan identificar y definir el objeto de estudio? El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 68 Con base en los distintos “tipos de ideas” a los que se ha hecho referencia, se hacen las siguientes sugerencias para su aplicación. Idea General: Títulos o apartados generales… Idea Principal: Títulos de preguntas o ideas importantes… Idea secundaria: Partes de un párrafo, clasificaciones… Detalles: Subdivisiones de la partes… El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 69 3.- Leyes objetivas y leyes científicas Los cambios y las transformaciones a que se encuentran sujetos los procesos existentes están regulados por ciertas relaciones constantes a las cuales denominamos leyes. Las leyes objetivas constituyen así las formas generales de las relaciones de cambio y representan las conexiones internas y necesarias en que se produce la variación de los procesos y de sus propiedades. Por lo tanto, en las leyes se pone de manifiesto lo único que es invariable dentro del flujo continuo de cambios y transformaciones, que es la relación de su variación. De esta manera tenemos que el comportamiento de los procesos está regulado según leyes y, por eso mismo, las leyes exhiben la regularidad del universo. Desde luego, las leyes objetivas, rigen independientemente de nuestra voluntad o nuestra conciencia, porque son inherentes a la naturaleza y la sociedad. Ahora bien, cuando el hombre logra descubrir una ley objetiva, la expresa en la forma de una ley científica. En consecuencia, la ley científica es una reconstrucción racional que refleja a la ley objetiva. Dicha reconstrucción se mejora con el avance del conocimiento, aproximadamente cada vez más a la ley objetiva correspondiente, pero sin que pueda El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 70 llegar nunca a coincidir por completo con ella. Una vez establecida, la ley científica expresa una relación necesaria que se cumple en ciertas condiciones y cuyos efectos se manifiestan en acciones determinadas que se producen en los procesos. Debemos comprender que, el comportamiento de los procesos no está determinado por las leyes, sino simplemente regulado por ellas. Así, el hombre transforma los efectos de una ley cambiando las condiciones de los procesos afectados. Por su parte, las leyes científicas no determinan a los procesos, sino que constituyen las pautas de su determinación. Esto es, que la ley científica no expresa lo que ocurrirá en un cierto proceso, sino lo que sucederá cuando se cumplan tales y cuales condiciones. En este sentido, las leyes científicas desempeñan la función de predecir lo desconocido, con base en lo conocido. Igualmente, las leyes científicas sirven como instrumentos de las investigaciones ulteriores y, en tanto cumplen esta función, se constituyen en partes integrantes del método científico. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 71 En todo caso, las leyes científicas permiten explicar el comportamiento de los procesos, cuando se conocen las condiciones de su cumplimiento. En otras palabras, las leyes científicas nos sirven para contestar los principales interrogantes de la ciencia, o sea, el qué, el dónde, el cuándo, el cómo y el por qué de los procesos existentes.26 3.1. Función de la Ley científica Puesto que las leyes se formulan una vez que se ha hecho la comprobación y expresan relaciones constantes entre los fenómenos, su principal función es explicar un hecho con base en la relación que éste guarda con otro. Un hecho singular se explica mediante una ley, en el sentido de que tal hecho es un caso particular de ella; se deduce de ella. En otras palabras, un hecho singular es una interpretación de un esquema de ley o formula legaliforme y, por tanto, toda formula legaliforme puede recibir una multitud de interpretaciones, ya que especifica una clase de hechos posibles. 26 Pág. 46, Eli de Gortari. Lógica General. Edit. Grijalbo. S.A., vigésima sexta edición. Mexico 1965. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 72 Las leyes se descubren (no se inventan) y nos muestran una relación que se da en la realidad, esto es, son esquemas objetivos. Las formulas en cambio, se construyen pero no arbitrariamente sino expresando esos esquemas objetivos. Referida a los hechos, una formula legaliforme tiene un dominio de validez limitado, más allá del cual resulta falsa. Ejemplos: Un movimiento imposible para un avión que vuele a velocidad uniforme. Un movimiento posible para ese mismo objeto. Esto significa que, aunque es lógicamente posible la trayectoria A, físicamente es imposible; lo cual limita el dominio de validez de la fórmula. Las leyes condensan nuestro conocimiento de lo actual (lo que es) y lo que (lo que puede ser), y gracias a esto nos permiten predecir lo que sucederá con un fenómeno determinado que tenga las características necesarias para ser un elemento de la relación expresada por la fórmula. Resumiendo lo anterior, se puede decir que, las funciones de la ley son las propias del conocimiento científico: explicar y predecir el curso de los fenómenos o hechos que ocurren en la naturaleza y en la sociedad. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 73 3.1.1. Clases de Ley Científicas Puesto que la formula es el reflejo de la realidad objetiva, mientras más cercana se encuentre a esa realidad, y mejor la exprese, en la medida en que fielmente la refleje, se considerará como una ley más profunda o, para decirlo con lenguaje técnico, se considerará como una Ley de nivel alto (axioma o postulado). Puesto que la ciencia tiene como meta la objetividad, debe aspirar a leyes de nivel alto, a formulas legaliformes que no dependan de las circunstancias. En cambio las leyes de nivel bajo (teoremas) se limitan al marco de referencia; es decir, se formulan en función de las circunstancias en que se da el fenómeno que es el elemento de la relación. A pesar de que son leyes de bajo nivel y su alcance es limitado, encajan en un sistema científico y se derivan de leyes de alto nivel, en las cuales se fundamentan. Como conclusión, se puede decir que, el concepto de ley puede significar lo siguiente: Esquema objetivo. Formula (función proposicional) que intenta reproducir un esquema objetivo. Fórmula que refiere (o relaciona) a un esquema objetivo con experiencia. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 74 Metaenunciado (enunciado de otro enunciado) que se refiere a un enunciado legaliforme. Regla basada en un enunciado legaliforme. Por último, puesto que ya se ha dicho que, todo hecho cumple con un conjunto de leyes o, si se prefiere, que todo hecho podría explicarse mediante un conjunto de fórmulas legaliformes y, por supuesto, a través de un conjunto de datos empíricos, entonces, más que una ley suelta, se necesita un sistema (encadenamiento, cohesión) de leyes para explicar un hecho. Además se puede agregar que, un sistema de leyes constituye lo que se llama “teoría”.27 27 Pág. 25-28. Yuren Camarena M. Teresa. Leyes, teorías y modelos (área: metodología de la ciencia). Quinta reimpresión. Editorial Trillas. México 1984. El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 75 4. ANEXOS El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 76 El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 77 El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 78 El Método Científico y sus Etapas, Ramón Ruiz, México 2007. 79 5. BIBLIOGRAFIA 1. Gutiérrez S. Raúl. Introducción al Método científico. Decimoctava edición, editorial Esfinge, México, 2006. 2. Ruiz L. Ramón.: (2006). Historia y Evolución del Pensamiento Científico. en línea a partir de 28 Marzo 2007, primera versión en español e inglés. |
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5.7: Binomial Distribution
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5.6: Birthday Demo
5.8: Binomial Demonstration
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David Lane
Rice University
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Learning Objectives
Define binomial outcomes
Compute the probability of getting X successes in N trials
Compute cumulative binomial probabilities
Find the mean and standard deviation of a binomial distribution
When you flip a coin, there are two possible outcomes: heads and tails. Each outcome has a fixed probability, the same from trial to trial. In the case of coins, heads and tails each have the same probability of 1/2. More generally, there are situations in which the coin is biased, so that heads and tails have different probabilities. In the present section, we consider probability distributions for which there are just two possible outcomes with fixed probabilities summing to one. These distributions are called binomial distributions.
A Simple Example
The four possible outcomes that could occur if you flipped a coin twice are listed below in Table 5.7.1. Note that the four outcomes are equally likely: each has probability 1/4. To see this, note that the tosses of the coin are independent (neither affects the other). Hence, the probability of a head on Flip 1 and a head on Flip 2 is the product of P(H) and P(H), which is 1/2×1/2=1/4. The same calculation applies to the probability of a head on Flip 1 and a tail on Flip 2. Each is 1/2×1/2=1/4.
Table 5.7.1: Four Possible Outcomes
| Outcome | First Flip | Second Flip |
| 1 | Heads | Heads |
| 2 | Heads | Tails |
| 3 | Tails | Heads |
| 4 | Tails | Tails |
The four possible outcomes can be classified in terms of the number of heads that come up. The number could be two (Outcome 1, one (Outcomes 2 and 3) or 0 (Outcome 4). The probabilities of these possibilities are shown in Table 5.7.2 and in Figure 5.7.1.
Since two of the outcomes represent the case in which just one head appears in the two tosses, the probability of this event is equal to 1/4+1/4=1/2. Table 5.7.2 summarizes the situation.
Table 5.7.2: Probabilities of Getting 0, 1, or 2 Heads
| Number of Heads | Probability |
| 0 | 1/4 |
| 1 | 1/2 |
| 2 | 1/4 |
Figure 5.7.1 is a discrete probability distribution: It shows the probability for each of the values on the X-axis. Defining a head as a "success," Figure 5.7.1 shows the probability of 0, 1, and 2 successes for two trials (flips) for an event that has a probability of 0.5 of being a success on each trial. This makes Figure 5.7.1 an example of a binomial distribution.
The Formula for Binomial Probabilities
The binomial distribution consists of the probabilities of each of the possible numbers of successes on N trials for independent events that each have a probability of π (the Greek letter pi) of occurring. For the coin flip example, N=2 and π=0.5. The formula for the binomial distribution is shown below:
P(x)=N!x!(N−x)!πx(1−π)N−x(5.7.1)
where P(x) is the probability of x successes out of N trials, N is the number of trials, and π is the probability of success on a given trial. Applying this to the coin flip example,
P(0)=2!0!(2−0)!(0.5)0(1−0.5)2−0=22(1)(0.25)=0.25(5.7.2)(5.7.3)(5.7.4)
P(0)=2!1!(2−1)!(0.5)1(1−0.5)2−1=21(0.5)(0.5)=0.50(5.7.5)(5.7.6)(5.7.7)
P(0)=2!2!(2−2)!(0.5)2(1−0.5)2−2=22(0.25)(1)=0.25(5.7.8)(5.7.9)(5.7.10)
If you flip a coin twice, what is the probability of getting one or more heads? Since the probability of getting exactly one head is 0.50 and the probability of getting exactly two heads is 0.25, the probability of getting one or more heads is 0.50+0.25=0.75.
Now suppose that the coin is biased. The probability of heads is only 0.4. What is the probability of getting heads at least once in two tosses? Substituting into the general formula above, you should obtain the answer 0.64.
Cumulative Probabilities
We toss a coin 12 times. What is the probability that we get from 0 to 3 heads? The answer is found by computing the probability of exactly 0 heads, exactly 1 head, exactly 2 heads, and exactly 3 heads. The probability of getting from 0 to 3 heads is then the sum of these probabilities. The probabilities are: 0.0002, 0.0029, 0.0161, and 0.0537. The sum of the probabilities is 0.073. The calculation of cumulative binomial probabilities can be quite tedious. Therefore we have provided a binomial calculator to make it easy to calculate these probabilities.
Mean and Standard Deviation of Binomial Distributions
Consider a coin-tossing experiment in which you tossed a coin 12 times and recorded the number of heads. If you performed this experiment over and over again, what would the mean number of heads be? On average, you would expect half the coin tosses to come up heads. Therefore the mean number of heads would be 6. In general, the mean of a binomial distribution with parameters N (the number of trials) and π (the probability of success on each trial) is:
μ=Nπ(5.7.11)
where μ is the mean of the binomial distribution. The variance of the binomial distribution is:
σ2=Nπ(1−π)(5.7.12)
where σ2 is the variance of the binomial distribution.
Let's return to the coin-tossing experiment. The coin was tossed 12 times, so N=12. A coin has a probability of 0.5 of coming up heads. Therefore, π=0.5. The mean and variance can therefore be computed as follows:
μ=Nπ=(12)(0.5)=6(5.7.13)
σ2=Nπ(1−π)=(12)(0.5)(1.0−0.5)=3.0(5.7.14)
Naturally, the standard deviation (σ) is the square root of the variance (σ2).
σ=Nπ(1−π)−−−−−−−−√(5.7.15)
5.6: Birthday Demo
5.8: Binomial Demonstration |
10294 | https://www.zhihu.com/question/337877928 | 如何直观简洁地理解动态规划问题？ - 知乎
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动态规划是由二叉搜索树一步一步优化得来的。
先让我们来看一下，二叉搜索树是什么：
正常的二叉树
但是，我们更常遇到的是二进制的二叉搜索树:
二进制的二叉搜索树
这就是我们人类解决问题时的思考过程，也就是数据结构。
遇到问题，我们人类一般是用中序遍历的方式来解决。这样我们就来到了第一个算法：
DFS(深度优先搜索算法 Depth-First Search algorithm）
搜索过程是这样的
中序遍历
所以，你看到所有的算法教程，都会讲到DFS算法，这本来是图论上的内容。就是说，你在学数据结构之前，最好先学一下图论。而图论又是离散数学的一部分，那学完图论，顺便把离散数学也学了吧。
这其实也是枚举法，他们也叫暴力搜索，就是说，你把所有的可能性都列举出来了，你从中图中也能看得出来：
你把思维树中的所有节点都访问了一遍。
ABCDEFGH, 都被你访问到了。
这样做的缺点也比较明显。
这种暴力搜索，虽然算无遗策，但是有个问题，它的时间复杂度是 2^n , 这句话什么意思呢？就是随着数量n 的增加，你搜索整棵树的时间，会呈现出一种火箭式的窜起场景，整个数据的增加量是一条直飞冲天的陡峭曲线轨迹，就是说，呈指数级增涨的方式，增加你的程序运行的时间。
你自己算一下就明白了，
2^5= 32
2^10=1024
2^100=?
2^1000=?
...
..
.
子问题的数量，只是增加到十个，子树就增加到了1024个，而我们实际生活中，n 的数量一般是上千个，工厂的订单数量一般就几百到上千个，然后，每个订单的数量和产品种类也就几种，几个数字，那么，所以实际生活中，n 的数量一般就上万，那就带来一个问题。
DFS, 暴力搜索，枚举法，暴力枚举，或者回溯法，这些只是叫法不同，其实都是同一种算法。虽然这几种算法的原理是一样，但差别还是有的：
1）回溯法更接近于还原中序遍历的过程。
2）DFS为了提高效率，已经抽象过了。就是说，它在回溯法的基础上，对回溯法进行了改进，用到栈这个抽象出来的数据结构。
所以，我们一开始就学DFS，是有点理解不了的。当然，你要是不看中序遍历的这个图，你第一次看回溯法，你也根本搞不清楚这个算法到底在干什么，为什么要回到那个点。这个疑惑，你看一下中序遍历的这个图，就明白了。
例如，B 这个节点，它访问了两次：
1）由A到B再到D, 这是第一次访问节点B
2) 因为D已经到头了，它只能回头，于是，由D回到B再到E,这是第二次访问B.
回溯法有一段代码，就是写怎么样保存B节点的状态，等再次回到B时，用之前保存的数据，恢复B点的状态。这就是回溯算法的代码中，那个特别让人疑惑的地方，它为什么要么操作？你要是不看中序遍历的图，是搞不清楚的。
理解了这点，你就明白了为什么DFS对回溯法进行改进的点在那里了，它用栈这个数据结构来存储原来状态的数据，比你原来的，回溯法所采用的 反复 访问同一个节点的手动操作的方式，改进的地方在于，CPU的硬件制造公司英特尔，会为栈这种数据结构，特地在最靠近CPU的地方，也就是CPU内部，计算单元的旁边，制造出缓存这么个硬件。缓存跟内存相比，离CPU的距离要近得多，已经做CPU内部去了。相当于东西就在你手边，都不用伸手就能够到。所以，缓存的存入和读取出来的速度，要比内存的速度要快得多。你写的回溯法，没有特别注明的话，或者是编译器没有优化的话，是写到内存去的。内存离CPU还是有10厘米左右的物理距离的，它不比缓存快太正常了。
用了栈之后，程序的编译器在看到这类代码时，能很明确的知道这类代码是想干什么的，会在背后，悄悄地把数据的存取位置转移到缓存上，悄悄地为你提速。谁把这么多钱放你们家冰箱的。
但是，
这些算法的效率仍然很低，数据量一大时，这点特别明显：低到你买英伟达的H800计算卡，也无法在一年内算出来。等你算出来，都十年以后了，这种算法对你的实际工作来说，毫无意义。你也不可能用超级计算机来算，开机就是几万块钱的电费，还什么都没干，几万块钱电费就没了，谁吃得消？
于是，你决定，我要优化。
让我们先来到，第2种算法：
递归
好，先不说第2种算法，我们先说，第3种算法：
剪枝
剪枝，也就是排除法，把树的一些不必要的分枝给剪掉，这在运筹学中，叫约束条件。符合约束条件的分枝留下来，继续进行运算；不符合约束条件的分枝，直接去掉，这种约束条件，有时候是隐式的，就是说不是那么明显，要你自己去找出来，由你来决定，那些分枝可以剪掉。
还是举个例子吧，例如，背包算法
text
有三样东西装进一个书包，问，怎么样装才能让包里东西最值钱？
这种背包问题，一般都会设置几个条件，不设置你也能想到
1）什么都不装，肯定不用考虑（000）
2）只装一件东西，肯定不用考虑（010，100，001）
3）三件东西都装，这显然不可能，因为没人这么出题。（111）
如果我们用二进制来求这个背包问题，数字1 表示拿走这件东西，装进背包；数字 0 表示不装进背包，个位数，十位数，百位数的位置，各对应一件东西，那么我们可以对这个问题进行下面的剪枝操作。
看明白了吗？我们通过眼睛看的方式，把这颗答案树，直接剪掉了一大半，画红叉叉的节点，全被我人为的剪掉了，只剩下几个答案要计算了，（101，110，011）
我只要计算这三种组合的装入背包的价值，再比较一下大小，我就知道了应该装入那两件东西。
而且，剪枝的代码非常简单，就 if 条件判断语句就好，就把枝给剪掉了
```text
假设这是你的二进制数据
binary_data = '111'
判断二进制数据是否等于 '111' 或 '000'
if binary_data == '111' or binary_data == '000':
print("二进制数据等于 (111) 或 (000)，这条分枝剪掉。")
# 可以在这里添加更多的结束操作，例如返回特定值
# 如果是在函数中，可以使用 return 语句
else:
print("二进制数据不等于 (111) 或 (000)，继续搜索二叉树操作。")
```
有多个分枝要剪，那多设几个if判断语句就完了。反正这个剪枝的条件也是你用眼睛反复在那里看，看出来的，看出来一个写一条判断句就行了。
但是，
实际上这种算法有两个缺点：
1） 剪枝剪不了这么多，剪枝效果有限。
原因是，我这个例子只有3件东西，所以效果特别明显，实际没有这么强的效果。要是东西多了，7到8件东西，你还用眼睛去剪枝，剪到你头秃都剪不完，刚才也说了，东西的数量有可能是一万件，这么大的数量，你就是找人来帮你剪，也要剪上十天半个月。
2）剪枝只针对特定问题有效，对别的问题无效。
像这个背包问题，可以抽象成二进制问题，取是1，不取是0. 实际上，在现实生活中，有很多的问题很现实，没办法抽象成1或0，取或者不取。
例如上面，中序遍历的那个图，那是搜索ABCDEFGH的，跟二进制的关系不大，这个问题都不是二进制，你怎么剪？
你只能老老实实的：
2.递归
递归就是中序遍历的一种特殊形式，从根节点出发，向下探索，一直到最底层，到达最底层后，由最底层，再回头往上走，一直走回根（root) 节点.
它跟剪枝，有点不太一样，我画图你就明白了。它有的地方不去。
让我们来看一下，一个递归求阶乘的程序是怎么回事
```text
def factorial(n):
if n==1:
return 1; # 出口
return factorial(n-1)n
factorial(3)
```
就是说，右边一大块节点，它都不访问，因为不存在为0的节点，它只访问最左这条边。
递归
就是说这个程序，在走到出口时，f(1)==1, 发现前面没路了，于是，它就往回走了，一直走。
最终，走回到出发点 f(3)
递归程序有什么问题吗？
眼睛视力很好的你，一看就发现了，左边这条路是不是走了两回？是的，被你发现了，于是我们来到：
递推
就是说，我直接从出口f(1)==1 走到 根节点 f(3), 不就好了，画出图来，就是
你看，现在，这条路你只走了一次，直接从出口走到根节点。而且，我把递推的程序写出来，能把你笑开了花
text
factorial=1
for n in (1,2,3):
factorial=factorialn
print(factorial)
啊，这就是递推？这不就是最简单 for 循环，是的，最简单的for循环，就是递推。
整个for循环的流程图，是这样的
递推
看到这里，你觉得你又行了，你觉得这么简单的阶乘，你完全可以凭手来算，不用计算机，于是，就到了下一个算法
记忆化数组
记忆化数组就是你小学时学的99乘法表，3的阶乘，你手写出来就是
1 x 2 = 2
2 x 3 = 6
所以，factorial(3) =6
要是画成一维数组，就是下面这个
记忆化数组
这就相当于一个草稿纸，或者是99乘法表，或者是小抄，把一些公式，或者数据提前算好，这样要结果的时候，直接过来抄结果就行了，
----- 不用算，
算多慢啊，还要费脑子，计算机都是直接抄答案。这个一维的数组，就是记忆化数组。
这的确可以提高速度，你别管我做了什么，怎么做到的，你就说快不快吧，因为，闭卷考试可能要1个多小时，才能把答案写完，直接抄试卷答案，可能只要半小时就抄完了。那肯定快。
问题在于，99乘法表，只能个位数，要是123154546 x 5564548 =? 这个怎么算啊，简单。欧洲人几百年前就已经发明了对数表，直接查对数表，进行加减就行了。这时空格太小，我就不写怎么样通过查对数表去计算大数，多位数了。写不下了。
学到这里，你已经脱离树，来到一维的表格了，越来越抽象了。但是，还差一步，差一步，就到：
动态规划
动态规划就是把一维表格升级为二维表格，就是下面这样
动态规划的二维数组
就是说，刚才的背包问题，我完全可能从树中抽象出来，抽象成二维的数组，只要计算这个二维数组的最大值max，就可以求出到底要装那两件东西，包里的东西最值钱。
走到这一步的时候，实际上我们仍然可以回到树上面，看一下动态规划到底在干什么
背包问题的动态规划
相当于，你不光眼睛的视力很好，而且眼光如闪电，你一眼就看出了我提出的那个背包问题的实质，就是直接看两件东西的各种组合就行了，直接比较一下各个组合的价值大小就能得出答案了。
至于，动态规划的二维数组是怎么对应到树上节点的，这里地方太小，我不想写了，你自己拿一道习题来做一遍就能看到了：动态规划有些操作仍然是重复的，但是在那几个关键的地方时，例如，两两进行数值上的比较，都能在树上找到相应节点。
自己做一遍吧。
手工写一遍二维数组里面的数据，再对照着二叉树来看一下，这步操作是在二叉树上的那个节点上。
动态规划这样的抽象操作有什么好处，动态规划能极大的改善文章一开始提出的那个问题：
DFS 随着数量n 的增加，计算机的运算时间会指数级增涨。
动态规划让这个如火箭般飞起的计算量降了下来，计算量的规模变得可控，不再是指数级。由此，动态规划的计算时间，跟DFS相比，是能明显减少的。
但是，
这种算法只针对特定的问题有效，面对其他问题，无效，同样是解决不了。实际上，动态规划也只是扩大了解决问题的种类的范围，动态规划并不万能。DP(dynamic programming) 同样有其解决不了的问题，也就是，这类问题并不存在动态规划需要的动态转移方程。对于这类问题，你还是要回头，用DFS算法。例如，8皇后问题。
8皇后问题，
它的时间复杂度是 n的阶乘， n!=8!=8x7x6x5x4x3x2x1, 而且这个问题，跟我们想象的还不太一样，以至于你都不能简单地用排列组合公式，来计算总共有多少摆法。原因是，这个问题除了分类，分步以外，还有第三种情况，走不到。这一点，你只有自己试着摆一下，才会有所体会。
1） 把Queen摆在第一行的最左边
2）把Queen摆在第二行的最右边
3）把Quenn摆在第三行的左边
4）把Queen摆在第四行的右边
5）把Quenn摆在第五行的左边
6）把Queen摆在第六行的右边
然后摆不下去了，没地方了。退回去再找另一个空格摆吧，退到第5步还不行，就继续退，退到第4步。
不好，摆成9皇后了
这个问题，除了剪技以外，目前人类没有找到别的优化方法。
好了，你看到了，我们怎么样从一棵完整的树，一步一步优化，抽象出二维数组的动态规划的过程，但实际上，我发现这些人并不会一样一样的来，而是,成年人，全部都要。
他们会采用所有的优化方法：
前处理（预处理）
例如，这个背包问题，东西一多，这帮人会把东西的价值，体积，或者是价值与体积之比直接进行排序，这样前处理的好处，显而易见。不重要的东西，没必要拿的东西，我在前处理的阶段，就已经把它们给排除掉了，这样做，n 的数量不就减少了吗？n的值减小，计算机的计算量就大大减少了。
就像你出门旅行的时候，不会把冰箱，彩电，洗衣机一起带出去，因为背包根本就装不下。同理，你去动物园，大象那里根本就不设铁丝网，动物园知道你的背包装不进大象，这件东西你根本就背不走。带点能带的吧！
而且这样做，又没什么难度，就简单的排一下序就完成了。
99乘法表，对数表也算是一种预处理。对于电脑来说，算一个问题可能要半天，抄答案1秒都用不了。别人算圆周率等于3.1415926，算了半个世纪，你拿过来直接用就行了。
剪枝
排完序，就是剪枝，不必要的分枝直接不进行计算。
子问题去重。
去重，就是去掉重复步骤的意思。递推就是对递归的去重优化，从上面的图中，你可以明显看到，递归的那个图中，一个节点访问了两次，那就可以根据找出的这个规律，每个节点只访问一次就够了，如递推图展示的那样。
记忆化数组也是一样的优化原理，在求fibonacci数组的过程中，你只要把fibonacci(3)的计算结果，放入到记忆化数组中，那后续再遇到这个子问题时，就不用反复，重复地去计算这个fibonacci(3)的值了，直接从数组中秒抄过来就行了。这一点，只有你自己亲自画出fibonacci的展开树时，才会有深切的体会，我反正是不会帮你画出来的。
这就是他们常说的，用空间换时间。这一点，具体对应到计算机来说，就是用访问CPU缓存空间的纳秒级的时间来换取CPU烧自己大脑计算出来的，以秒为单位的时间。
原因就是英特尔公司，在设计x86时，会主动针对此类重复的问题进行cpu物理结构上的专门设计。设计缓存的结构和算法，让你更快地得到结果。
啊，赞美一下，伟大的英特尔，虽然它现在有点过不下去了。
数学
举个例子，在做求模的运算时，计算机要是不用公式，硬算，在那里算来算去，还真的不如你直接套用费马小定理得出答案的结果来得快。这样会形成一个局部的奇观，你手算比计算机死板的算，速度更快。
数学公式毕竟是数学家对数学规律的总结，不是接近本质，而是它就是本质。数学公式的快，不言而喻。
用这帮人的话来说，数学的时间复杂度是O(1),是常数级的。
英特尔不会真的倒闭吧？应该是很有可能。
看出去，是没有复活的风险了。
本来应该写代码的，但是，没别人写得好，还是看一下人家是怎么写的吧
动态规划入门系列课程全集（数学建模清风主讲）_哔哩哔哩_bilibili
在此，致敬一下，想出动态规划算法的黑人科学家
美国科学院首位黑人院士、加州大学伯克利分校首位黑人终身教授、杰出统计学家戴维·布莱克维尔（David Blackwell）因病于2010年7月8日逝世，享年91岁。
他独立发明的“动态规划”（dynamic programming）广泛应用于金融以及多个科学领域；他的“更新定理”（renewal theorem）如今也依然运用在工程学多个领域；他创立的“Rao-Blackwell”定理是现代统计学的基础性概念。
来说一下，这位黑人科学家的伟大之处吧，因为恰好跟主题有关。
简单来说，Blackwell院士把问题降维了。他发现如果某类问题存在状态转移方程，可以把时间复杂度由o（n! ) 降低到 o( nxm), 差不多类似于 n 的 n 次方， 降低到 n x m 的程度，也就是差不多 n 的二次方的程度。乘法次数由n次，直接降到 2次。
他这一发现，太历害了。我们还是举实例说明吧，还是以一开始的背包问题来举例。我们要是还按暴力枚举的思路来解决背包问题，那么，时间复杂度降不下来，也就是 o ( n! ),那么
三个东西装一个包里，要算 n!=3!=3x2x1=6次
4个东西装一个包里，要算 n!=4!=4x3x2x1=24次
...
..
当 东西的数量是 18个时，要算 18×17×16×...×2×1=259896000000000000000 次。
这意味着，到底要装那些东西，你手算的话，算一年都算不完，最后就是乱装一通，18个东西，看一下那件东西像付钱的样子，往包里装就行了。
用Blackwell院士的方法，列一个二维表，nxm= 18 x 5=90次（我们的背包最大容量也不过装个5~6斤的东西，所以我们这里m取5）, 根据我们从小练就的99乘法表，我们算一下，半小时不就算出来了。
当然，
如果我们仍然觉得Blackwell的这个二维表太抽象时，我们可以把二维表还原为树的结构：
动态规划实际仍是树状结构
那这样，我们更能看清楚的Blackwell先生的伟大之处，他的发现，把一张巨大的n! 的树，成功的降维成了一个二阶树，nxm.太神奇了。原来那么多的节点被直接跳过，直接被压缩成为一个二阶结点，当然，你也可以旋转一下这个图，使得它看上去，更像上面的树：
现在看明白了，人类的思维实际上是树
出发点(根节点）在最上方，走到最下方的终点，你也可以把这个图看成递归的树，由原来的折叠（起点和终点对折到一起）展开回到二维空间。
可惜，他的这一成果，仍然是有局限性，那就是说，要有状态转移方程。没有的话，还是用不了动态规划。
就是说，大部分问题，仍然是n!的问题，甚至是 n^n 的问题，例如围棋，计算量仍然不会少，目前 人类也没办法，或者说大量的问题仍然没有找到一条类似于动态规划一样的法则，来显著降低计算量，实现降维的效果。
或许是我们的目前的计算量仍不够大，以致于，没能发现更高维的公式，定理，实现类似于微积分一样的效果，把人类带到一个更高的层次。
或许真的有一种更通用的状态转移方程，这东西是存在的，只是我们的实践不够，计算量不够，所以才没有发现罢了。
这种发现，可能类似于数论中的哥德巴赫猜想，类似于费曼大定理，需要更多的研究，更多的成果，而那些低垂的果实，类似于费曼小定理这种，孙子定理这种，早就被人们收割了。那些高耸的果实，这类真正的问题，人类目前仍然一点思路都没有，不得其门而入。
这一点，你在观看《算法导论》时，会反复发现。那就是，目前的大部分的数学研究成果，都被计算机科学家们应用到计算机上了。但仍然有一大堆的问题没有什么好的算法，于是，人类能做的，只有进一步从硬件方面想办法，利用摩尔定律，采用更新制程，来获得更快的运算速度。
但英特尔的倒闭，实际上是在告诉我们，这条路也到头了。人类现在所做的，就是把计算机除CPU外的其他部分，也按摩尔定律进行迭代。集合到一起，做成一种类似于目前的SOC一样的东西。
这没办法，就像祖冲之一样，1000多年前，用算筹，最多也就是把PI算到 3.1415926， 后面没了，算不动了。
本来这篇文章已经写完了，但是，你我都知道它有一个不完美的地方，有一个小小的疏漏，现在补上去。
你也看到了，我上面写的时候，有一个思维上的小跳跃，就是从自然数（1234567890，或31415926）跳到到了二进制数01， 然后用二叉树来说明 人类的思维，0或1，阴或阳，正或负，胜或败，投入或产出。那实际上，你也会问一个问题，你心里也有一个坎：
------ 为什么这样操作，凭什么能这么做？
那我一定要1，3，5，7，9，这样来操作不行吗？我不用你的二进制，不用你的二叉树来思维，来表示我思考的过程不行吗？
是的，我们要把这事给说清楚，否则就又变成牛顿与莱布尼茨之争，变成著名的
text
“注意到”，“我们注意到”，或者是垃圾数学中常用句式，“我们由此可知”
变成费曼大定理中的那个著名桥段，
text
老子想到了一个绝妙的证明方法，但这里的空白太小，写不下，我留点事给别人做做
首先，二叉树为什么已经能容纳下所有的可能性？而不必用三叉树，四叉树，红黑树，因为
二进制数可以表示所有的自然数
如3可以表示为0011，5可以表示为0101........你所知道的自然数，都可以表示用二进制来表示，如什么31414926，2714都可以用二进制数来表示，当然这样的话，你画出来的，或者你脑海中的这棵二叉树，展开来的话，就会比较大，这回真变成这里空白太小，画不下这么大一棵参天大树了，这棵二叉树太大了，把天空都给遮住了。
2.数学上的四平方数定理
上面的回答就是终级答案，实际上，前人的智慧远不止步与此，前人如果只是思考到了上面一点，数学也就不是一种思维的游戏了。数学家的作用，就是把这棵大树的分枝，每一片叶子都刮干净，取下来。这就是著名的四平方数定理。
text
四平方数定理（也称为拉格朗日四平方和定理）是数论中的一个重要结论，
由拉格朗日在1770年证明。该定理的内容如下：
任意一个正整数都可以表示为不超过四个整数的平方和。
a,b,c,d（其中某些可以为零），使得： n = a ^2 + b^ 2 + c^2 + d^ 2
1)一个整数： 例如 1=1^2, 4=2^2, 9=3^2,16=4^2.....
2)二个整数：5=1^2+2^2，13=2^2+3^2.....
3)三个整数：3=1^2+1^2+1^2 (1,1,1).....
4)四个整数：7=1^2+2^2+1^2+1^2(1,1,1,2)....
这都能被数学家用定理证明出来，那么上面的二进制能表示所有的自然数，也就更易理解了。
当然，你也可以联想到费马二平方定理：形如 4k+1 的素数可表示为两个平方和（如 5 = 2 ^2 + 1 ^2 ）,这类问题的证明过程，毫无乐趣可言，你要继续深究，还是请出大神梁文锋的作品deepseek，这个更快，这些大部分都是从取模的公式中推导出来的引理。
好了，这个小跳跃，补齐了。
那就是说，动态规划这种降维的方式，是客观存在的，是本来就存在的，你从底层的二进制转换就能看得出来，只不过它的动态转移方程，是因事而异的，这些转移方程实际上是有限状态机的图画出来后，人们总结归纳出来的，每一种状态机的形状不一样，当然他们的状态转移方程自然也就不一样。
这东西，离我们的生活很近，就跟家里的遥控器一样，动一下，电风扇就会挂一档，空调就会打到25度，电视就会转到中央5套，这些特性是这些电器本来就有的。但是，你要深究背后是怎么实现的，就会看到最底层的，最基础的结构，然后是前人在基础结构上的一次次的思维上的小跳跃，从而直到目的。
难怪，梁文锋说他们完全明白nvidia各种GPU的工作原理，原来是，他们穿过软件层，又回到了硬件层，由抽象又回到了具体，直接面对0101的二进制数据层。等于是已经跳上来，到达到抽象的软件层的最高层，又蹦迪回去了，硬件层是怎么实现01010数据的转移，进行加减乘除运算的，就像我上面画的背包问题的那个图一样，直接在101，011，110这三个答案的数之间，跳来跳去。至于怎么在这三个二进制数之间跳来跳去，那就要用到本身nvidia公司自己对外提供的硬件命令了。
原因是，nvidia公司把这些操作的命令，直接做进了电路设计中。不需要你再绕来绕去，通过软件层（编译器如GCC），把软件代码再解释成硬件操作的命令，直接给你命令，编译器不需要了，省去了不必要的编译时间。反正编译器也是要翻译成机器操作的代码，这类代码不一定是汇编代码，就像是Intel的simd命令。原因也很搞笑，你写的汇编代码不一定有原厂的工程师写的高效。如硬件制造商intel,nvidia的工程师自己写的加法功能函数，很多情况下，比你自己现样用汇编功能写的加法函数要快，而且是要快得多，因为电路就是他们设计的。
看有些数据，自己写的加法要比原厂给的加法，要慢25倍左右，原因是原厂工程帅给的代码，只有一条指令，且只要在一个时钟周期就算出来了，而自己写的，经过编译后，反而编译成了15指令，要3~5个时钟周期，才能算出来。15x 3=45 < 1x1=1,这是真正意义上的，写还不如不写，因为你写也是瞎写，你 根本就不知道，这CPU,GPU,是怎么设计出来，为了那项功能专门设计了那部分电路。
这点就像以前vivo，oppo(也就是步步高）让王者荣耀团体，为自己手机的特定机型重新编程，转码一样，这部分特定的代码，真成了vivo手机部分机型所独有的了。跟林妹妹一样，哀家独有，别家没有的。跟格力的遥控器一样，可以设计成自己家独有的专用版本。
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北京大学 电子与通信工程硕士
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动态规划（Dynamic Programming，DP）是运筹学的一个分支，是求解决策过程最优化的过程。20世纪50年代初，美国数学家贝尔曼（R.Bellman）等人在研究多阶段决策过程的优化问题时，提出了著名的最优化原理，从而创立了动态规划。
动态规划算法通常用于求解具有某种最优性质的问题。在这类问题中，可能会有许多可行解。每一个解都对应于一个值，我们希望找到具有最优值的解。动态规划算法与分治法类似，其基本思想也是将待求解问题分解成若干个子问题，先求解子问题，然后从这些子问题的解得到原问题的解。与分治法不同的是，适合于用动态规划求解的问题，经分解得到子问题往往不是互相独立的。若用分治法来解这类问题，则分解得到的子问题数目太多，有些子问题被重复计算了很多次。如果能够保存已解决的子问题的答案，而在需要时再找出已求得的答案，这样就可以避免大量的重复计算，节省时间。我们可以用一个表来记录所有已解的子问题的答案。不管该子问题以后是否被用到，只要它被计算过，就将其结果填入表中。这就是动态规划法的基本思路。具体的动态规划算法多种多样，但它们具有相同的填表格式
分享一个良心老师：
良心老师：
给定一个数组，求不相邻的数字和最大：
给定一个数组，给一个目标值，问你存不存在一个子集满足和等于给定的目标值，输出true或者false（前提，都为正整数）
代码在下一篇文章中分享~
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问题： n n 阶台阶，从 0 0 开始走，每次可以向上走 1 1 级或 2 2 级。问方案总数?
这个问题，我们最暴力的方法就是搜索，枚举记录到底怎么走。但这样太慢了，我们要用更快地方法。
这里就用到动态规划了。假设，我们下面要到第 x x 个台阶，所以我们就一定是从 x−1 x-1 或 x−2 x-2 个台阶上过来的。那么，我们走到第 x 个台阶上的方案数量就为到第 x-1 个台阶上的方案加上到第x-2个台阶上的方案。为什么呢？我们用加法原理就可以解释。
我们现在把到第x个台阶上的方案记作 dp(x) ，那么 dp(x)=dp(x-1)+dp(x-2)
代码怎么写呢？只需要开一个 dp 数组存方案就可以了。我们知道，上前两级台阶的方案数量是 1,1 .就把数组前两项开始就附好值，再按后面的式子循环算术即可。
总结：总方案数量（大问题）->每次的方案数（小问题）
```text
include
include
int t;
__int128 f = {0, 1, 1};
using namespace std;
int main(){
for (int i = 3; i <= 5000; ++i) f[i]=f[i-1]+f[i-2];
scanf ("%d", &t);
printf ("%lld\n",f[t+1]);
}
```
进阶问题：假设我们要吃 n 个面包，我们现在有 1 个面包，我们可以让我们手中的多一个或多 5 个，我们最少几次才能凑够 n 个？
假设每次的代价是一，我们开一个数组记录代价。我们模仿上面的问题，则凑出 x 个面包的次数我们可以分类来求解。我们可以往前推一步来找到前一项。
如果要选一个面包，那么代价就为 1+dp[x−1] ,如果要选五个面包，那么代价就为 1+dp[x−5] ,我么要求最小的步数，那么直接取最小值就好了。最终式子： dp[x]=1+min(dp[x − 1],dp[x − 5])
总结，我们要把大问题拆成有着同样形式的小问题（状态），最后用式子分别表示小问题的几种情况，在进行综合即可。
进阶问题：最长上升子序列。就是给定 n 个数，求出他最多有多少个从小到大的数字。
我们开一个数组，每一项 dp[i] 就记录着原数组 当 a[x] 结尾时的上最长升子序列的长度。
举个例子：一串数 2 4 6 3 4 5 ，枚举到他的第六项 dp 时，他有这么几种选择：
可以直接与 dp 结合，长度 dp=dp+1=2
可以直接与 dp 结合，长度 dp=dp+1=3
可以直接与 dp 结合，长度 dp=dp+1=3
可以直接与 dp 结合，长度 dp=dp+1=4
那么我们就可以发现，枚举在 x 前面的， a[i] 又比 a[x] 小的那些 i , 形成一个更长的上升子序列。
总的式子： f[x] = max (f[i] + 1) 注： i<x,a[i]<a[x]
方法总结：对于一个没有求出解的状态，利用能走到它的状态，来得出它的解。也可以拿一个已经求好了解的状态更新它能走到的状态。
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10295 | https://flexbooks.ck12.org/cbook/ck-12-middle-school-math-concepts-grade-6/section/9.18/related/lesson/similar-figures-and-scale-factor-x-maths/ | Skip to content
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Back To Unknown Measures in Similar FiguresBack
9.18
Similar Figures and Scale Factor
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
After going through this section students will understand the concept of scale factor and can relate it to real life applications.
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Guidance
Similar polygons are two polygons with the same shape, but not necessarily the same size. Similar polygons have corresponding angles that are congruent, and corresponding sides that are proportional.
These polygons are not similar:
Think about similar polygons as enlarging or shrinking the same shape. The symbol @$\begin{align}\sim\end{align}@$ is used to represent similarity. Specific types of triangles, quadrilaterals, and polygons will always be similar. For example, all equilateral triangles are similar and all squares are similar. If two polygons are similar, we know the lengths of corresponding sides are proportional. In similar polygons, the ratio of one side of a polygon to the corresponding side of the other is called the scale factor. The ratio of all parts of a polygon (including the perimeters, diagonals, medians, midsegments, altitudes) is the same as the ratio of the sides.
Example A
Suppose . Based on the similarity statement, which angles are congruent and which sides are proportional?
Just like in a congruence statement, the congruent angles line up within the similarity statement. So, and @$\begin{align}\angle C \cong \angle L\end{align}@$. Write the sides in a proportion: @$\begin{align}\frac{AB}{JK} = \frac{BC}{KL} = \frac{AC}{JL}\end{align}@$. Note that the proportion could be written in different ways. For example, @$\begin{align}\frac{AB}{BC} = \frac{JK}{KL}\end{align}@$ is also true.
Example B
@$\begin{align}MNPQ \sim RSTU\end{align}@$. What are the values of @$\begin{align}x, y\end{align}@$ and @$\begin{align}z\end{align}@$?
In the similarity statement, @$\begin{align}\angle M \cong \angle R\end{align}@$, so @$\begin{align}z = 115^{\circ}\end{align}@$. For @$\begin{align}x\end{align}@$ and @$\begin{align}y\end{align}@$, set up proportions.
@$$\begin{align}\frac{18}{30} &= \frac{x}{25} && \ \frac{18}{30} = \frac{15}{y}\
450 &= 30x && 18y = 450\
x &= 15 && \quad y = 25\end{align}@$$
Example C
@$\begin{align}ABCD \sim AMNP\end{align}@$. Find the scale factor and the length of @$\begin{align}BC\end{align}@$.
Line up the corresponding sides, @$\begin{align}AB\end{align}@$ and @$\begin{align}AM = CD\end{align}@$, so the scale factor is @$\begin{align}\frac{30}{45} = \frac{2}{3}\end{align}@$ or @$\begin{align}\frac{3}{2}\end{align}@$. Because @$\begin{align}BC\end{align}@$ is in the bigger rectangle, we will multiply 40 by @$\begin{align}\frac{3}{2}\end{align}@$ because @$\begin{align}\frac{3}{2}\end{align}@$ is greater than 1. @$\begin{align}BC = \frac{3}{2} (40)=60\end{align}@$.
Watch this video for help with the Examples above.
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Practice
Determine if the following statements are true or false.
All equilateral triangles are similar.
All isosceles triangles are similar.
All rectangles are similar.
All rhombuses are similar.
All squares are similar.
All congruent polygons are similar.
All similar polygons are congruent.
All regular pentagons are similar.
If @$\begin{align}\triangle BIG \sim \triangle HAT\end{align}@$.
a) List the congruent angles and proportions for the sides.
b) If @$\begin{align}BI = 9\end{align}@$ and @$\begin{align}HA = 15\end{align}@$, find the scale factor.
c) If @$\begin{align}BG = 21\end{align}@$, find @$\begin{align}HT\end{align}@$.
d) If @$\begin{align}AT = 45\end{align}@$, find @$\begin{align}IG\end{align}@$.
e) Find the perimeter of @$\begin{align}\triangle BIG\end{align}@$ and @$\begin{align}\triangle HAT\end{align}@$. What is the ratio of the perimeters?
Use the picture to the right to answer questions.
a) Find @$\begin{align}m \angle E\end{align}@$ and @$\begin{align}m \angle Q\end{align}@$.
b) @$\begin{align}ABCDE \sim QLMNP\end{align}@$, find the scale factor.
c) Find @$\begin{align}BC\end{align}@$.
d) Find @$\begin{align}CD\end{align}@$.
e) Find @$\begin{align}NP\end{align}@$.
Determine if the following triangles and quadrilaterals are similar. If they are, write the similarity statement.
a)
b)
c)
d)
@$\begin{align}\triangle ABC \sim \triangle DEF\end{align}@$ Solve for @$\begin{align}x\end{align}@$ and @$\begin{align}y\end{align}@$.
@$\begin{align}\triangle CAT \sim \triangle DOG\end{align}@$ Solve for @$\begin{align}x\end{align}@$ and @$\begin{align}y\end{align}@$.
@$\begin{align}\triangle MNO \sim \triangle XNY\end{align}@$ Solve for @$\begin{align}a\end{align}@$ and @$\begin{align}b\end{align}@$.
Two right triangles are similar. The legs of one of the triangles are 5 and 12. The second right triangle has a hypotenuse of length 39. What is the scale factor between the two triangles?
Similar Polygons and Scale Factors Principles - Basic
This video provides the student with a walk through on similar polygons and scale factors.
Watch the video at:
Real World Example 1 - Super Models
When an engineer designs a new car, how can she tell whether her design is efficient? Computer models can provide some information on how air flows around the car. However, airflow is complicated, and our computer models aren't perfect. And it's too expensive to build a full-sized prototype of the car to test. The solution? In the early stages of the design process, engineers use scale models to test airflow and wind resistance.
As Unpredictable as Weather
Air and water patterns are hard to model with computers. The U.S. Military instead creates physical models of ships and airplanes that are tested by scientists using wind and water tunnels. The tunnels allow the scientists to see how air and water flow over the models. These scale models are geometrically similar to the full-sized military vehicles.
Engineers can build models of natural features too. In Louisiana, engineers built a scale model of parts of the Mississippi River. The model shows how sediments flow down the river and how changes to the river affect the areas near its banks. They have used it to test ideas for restoring parts of the Mississippi Basin. Even though computers continue to get more powerful, engineers will depend on scale models for a long time.
See for yourself:
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Not everything that engineers model becomes a reality. For instance, students from South Dakota recently built a very impractical airplane.
How did the model help the students test their ideas?
Real World Example 2 - Crazy Quilt
Similar triangles aren’t just used for measuring distances and comparing shapes. Artists and craftspeople can make geometric patterns from similar triangles. Quilters, for instance, use similar triangles to create beautiful blankets from scraps of material.
Why It Matters
You could make a quilt out of random pieces of material but you’d likely end up with something that is not so pleasant to look at. Most quilters, however, are artists who take pains taking pride in every scrap they use and every stitch they make to produce a finished product that is not only useful but also a work of art.
To achieve their desired patterns in their quilts, quilters must measure and cut their shapes precisely. Two of the shapes they commonly use are half-square triangles and quarter-square triangles—and these two types of triangles just happen to be similar triangles.
See for yourself:
Explore More
The “Tumbling Triangles” quilt pattern is completely made up of similar triangles. Learn how to piece together one of these quilts at the link below.
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10296 | https://jillianstarrteaching.com/number-paths/ | Skip to main content
Teaching with Jillian Starr
Making Sense of Math for Elementary Teachers
How to Use Number Paths to Transition to Number Lines
Early childhood educators, I’m going to let you in on a little secret. It’s the missing piece of your math teaching puzzle: Number Paths.
Before we jump into the question: What are number paths? Let’s take a step back.
Like many first grade teachers, I used math manipulatives to support one-to-one correspondence and to teach counting on. Then, I transitioned my students to number lines as they seemed ready to move on to visual representation.
As I explained in my post on Comparing Numbers, number lines are an excellent way for students to see the relationship between numbers. But, we often switch our students over the number lines too soon. That’s where number paths enter the scene.
Today, I’ll dive into the difference between number paths and number lines, the ins and outs of number paths, and when and how to transition from one to the other.
Number Paths vs. Number Lines
Number Lines
My guess is that if you landed on this page, you already know about number lines. But if not, have no fear! I’ve got you covered.
A horizontal number line is a line that uses tick marks to represent sequential numbers from left to right. A vertical number line marks sequential numbers from bottom to top.
Quick Facts About Number Lines:
Number lines are “measurement models” that count intervals.
A tick mark represents each milestone number.
Number lines measure lengths in sequence rather than just sequential numbers.
Number lines are often used to visualize numbers (visual representation) in sequence in a way that can translate to operations (additional, subtraction, division, and multiplication) and fractions in later years.
Number Paths
Number paths might be a new concept for you. That’s okay! They were for me too, and then I never stopped using them.
Number paths are a sequential, visual representation of numbers. Comprised of rectangles, they read from left to right, and each rectangle tends to represent one unit or number.
Quick facts about Number Paths:
Number Paths typically start with one.
Number paths follow a “Counting model.”
Rectangles represent each number.
Each rectangle can be counted.
Start with Number Paths
So, this is your big takeaway from today’s blog post. Use number paths– especially in first grade.
Many math experts and professionals believe that it is the best tool for early mathematicians.
Why? Number paths will help your students build an incremental understanding of sequence AND act as a tool for teaching addition and subtraction.
If you start with Number Lines, you might see…
Here is a big reason why I tried number paths after years of using number lines. I noticed three habits that indicated serious misunderstandings over and over again. I noticed:
Students start counting with Zero.
Students count the tick marks.
Students count the spaces in between each number.
What misunderstandings did those habits indicate?
Students needed support with one-to-one correspondence.
They did not understand the value of the tick marks and the spaces between them.
Students need to work on the difference between counting spaces and intervals.
Benefits of Starting with Number Path
In first grade, students start building mathematical understanding beyond the concrete and into the world of visual representation.
Before, students learned to count with objects. As they moved understanding into “sequential” numbers, they would arrange those objects (also known as manipulatives) into lines. They would use one-to-one correspondence to count each object in order (sequence) and learn to count from left to right. The first object equals one. The second object equals two. And so on.
Number Paths transition those same steps into the visual representation plane. Instead of counting objects, students count rectangles.
Two of the significant benefits of using a number path include:
Seamless transition of one-to-one correspondence
Continued practice counting on.
NOTE: You can scaffold this even further by using rectangle manipulatives.
Transitioning from Number Paths to Number Lines
Eventually, we do want students to transition to using number lines.
The WHY is simple: They are incredibly supportive, versatile tools. But their magic is only unlocked once our students understand the concept behind them.
The best way to transition students from using number paths to using number lines involves three steps:
Open number paths
Open number lines
Grouping with benchmark numbers
Open Number Paths
This is the first step. Also referred to as “empty” instead of open, these are very similar to traditional number paths. However, they are called “open” because they don’t include numbers. Instead, students can play games and interact with blank rectangles.
Grouping with Benchmark Numbers
Now, it’s a perfect time to discuss grouping using benchmark figures. At this point, I want students to start seeing numbers in chunks, with distance between numbers.
Ann Carlyle, the author of the “It Makes Sense!” series, begins this process with different colored paper squares (two colors). Students alternate the color of paper rectangles using equal groups, such as fives. With this technique, students begin to see the value of a number based on where the section ends and how many sections came before it.
Open Number Lines
Now, it is time to move to open number lines.
Open number lines start at zero, but unlike their traditional counterparts, they are also a counting model. The distance between each tick mark isn’t necessarily equal or even proportional.
This final step, using an open number line, allows students to express their own sense of numbers while getting a feel for the structure of a traditional number line. The open number line is handy for addition and subtraction.
Have I convinced you yet? Will you incorporate number paths into your first grade math lessons? I hope so! This simple yet effective tool will work wonders for your kiddos who struggle with counting and those who don’t. Let me know how it goes!
Exciting News!
First Grade Teachers, something big is coming! If you’re someone who loves digging into math concepts, or wants to become a better math teacher for your students, I’d love for you to join me. If you’d like to be the first to find out more, enter your info below!
Get on the Waitlist!
1st Grade Teachers! Something big is coming this June and I will be inviting a small group of teachers to get a first glimpse before it becomes available for everyone. If you’d like to be the first to find out more, just enter your info below!
GET ME ON THE WAITLIST
Get on the Waitlist!
1st Grade Teachers! Something big is coming this June and I will be inviting a small group of teachers to get a first glimpse before it becomes available for everyone. If you’d like to be the first to find out more, just enter your info below!
Addition & Subtraction Resources
Addition & Subtraction Centers Pack
Make a Ten to Add Centers
2-Digit Addition & Subtraction Centers
Single Digit Addition Logic Puzzles
You May Also Enjoy These Posts:
10 Domino Addition Games and Activities for Your Math CentersHow to Teach Odd and Even Numbers without WorksheetsTop Addition Strategies: The Transformation Strategy
hello I'm Jillian
I’m so happy you’re here. I want every child to feel confident in their math abilities, and that happens when every teacher feels confident in their ability to teach math.
In my fifteen years of teaching, I sought every opportunity to learn more about teaching math. I wanted to know HOW students develop math concepts, just like I had been taught how students learn to read. I want every teacher to experience the same math transformation I did, and have the confidence to teach any student that steps foot in their classroom. I’m excited to be alongside you in your math journey! |
10297 | https://pressbooks-dev.oer.hawaii.edu/atmo/back-matter/glossary/ | Glossary – Atmospheric Processes and Phenomena
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Contents
Atmospheric Science: ATMO 200 Companion Text
Disclaimer
Preface
About the Contributors
Authors
Contributors
Acknowledgements
Main Body
Chapter 1: Atmospheric Basics2
Alison Nugent
1. Introduction
2. Overview of Earth’s Atmosphere
3. Equation of State— Ideal Gas Law
4. Hydrostatic Balance
5. Layers of the Atmosphere
6. Weather and Climate: What’s the Difference?
7. Chapter 1 Reference Guide: Coordinate Systems, Units, Terminology
2. Chapter 2: Solar and Infrared Radiation2
Alison Nugent
1. Introduction
2. Radiation
3. Application to the Earth-Sun System
3. Chapter 3: Thermodynamics3
Alison Nugent
1. Introduction
2. Energy Transfer
3. Frameworks for Understanding the Atmosphere
4. Defining Changes in the Atmosphere
5. Introduction to Thermodynamic Diagrams
6. Heat Budget at Earth’s Surface
7. Additional Information
4. Chapter 4: Water Vapor1
Alison Nugent and Shintaro Russell
1. Introduction
2. Saturation
3. Humidity Variables
4. Why Do We Care So Much About Moisture?
5. Chapter 5: Atmospheric Stability4
Alison Nugent and David DeCou
1. Introduction
2. Atmospheric Stability & Lapse Rates
3. Skew-T Log-P Diagram
6. Chapter 6: Clouds2
Alison Nugent and Shintaro Russell
1. What is a cloud and how does it form?
2. Cloud Naming Conventions
3. Cloud Identification Examples
7. Chapter 7: Precipitation Processes3
Alison Nugent and David DeCou
1. Introduction
2. Cloud Droplets
3. Raindrops
4. Collision-Coalescence Process
5. Ice Phase Process
6. Precipitation Types
8. Chapter 9: Weather Reports and Map Analysis
Alison Nugent and David DeCou
1. Introduction
2. Meteorological Reports and Observations
3. Weather Observation Locations
4. Sea-Level Pressure Adjustment
5. Synoptic Weather Maps
6. Map Analysis
9. Chapter 10: Atmospheric Forces and Wind3
Alison Nugent and Shintaro Russell
1. Introduction
2. Main forces
3. Force Balances
10. Chapter 11: General Circulation3
Alison Nugent and David DeCou
1. Introduction
2. Differential Heating
3. Single-Cell Model
4. Three-Cell Model
11. Chapter 12: Fronts and Airmasses4
Alison Nugent and Shintaro Russell
1. Introduction
2. Air Masses
3. Surface Fronts
12. Chapter 13: Extratropical Cyclones3
Alison Nugent and David DeCou
1. Introduction
2. Mid-latitude Frontal Cyclones
13. Chapter 14: Thunderstorm Fundamentals4
Alison Nugent and Shintaro Russell
1. Introduction
2. Thunderstorms
3. Atmospheric Instability and Thunderstorms
14. Chapter 15: Thunderstorm Hazards5
Alison Nugent and David DeCou
1. Introduction
2. Keeping Yourself Safe
15. Chapter 22: Atmospheric Optics4
1. Introduction
2. Visibility
Glossary
Atmospheric Processes and Phenomena
Glossary
Absolute humidity
The ratio of the mass of water vapor to the volume of air.
Absolutely stable
The environmental lapse rate is less than the moist adiabatic lapse rate.
Absolutely unstable
The environmental lapse rate is greater than the dry adiabatic lapse rate.
Accretion
In cloud physics, usually the growth of an ice hydrometeor by collision with super-cooled cloud drops that freeze wholly or partially upon contact.
American Meteorological Society, cited 2020:Accretion. Glossary of Meteorology. [Available online at
Adiabatic processes
A process in which a system does not interact with its surroundings by virtue of a temperature difference between them.
American Meteorological Society, cited 2019: Adiabatic processes. Glossary of Meteorology. [Available online at
Advection
The process of transport of an atmospheric property solely by the mass motion (velocity field) of the atmosphere; also, the rate of change of the value of the advected property at a given point.
American Meteorological Society, cited 2020: Advection. Glossary of Meteorology. [Available online at
Aggregation
The process of clumping together of snow crystals following collision as they fall to form snowflakes.
American Meteorological Society, cited 2020: Aggregation. Glossary of Meteorology. [Available online at
Air mass modification
The change of characteristics of an air mass as it moves away from its region of origin.
American Meteorological Society, cited 2020: Air Mass Modification. Glossary of Meteorology. [Available online at
Air parcel
An imaginary volume of air to which may be assigned any or all of the basic dynamic and thermodynamic properties of atmospheric air.
American Meteorological Society, cited 2019: Air parcel. Glossary of Meteorology. [Available online at
Albedo
The ratio of reflected flux density to incident flux density, referenced to some surface.
American Meteorological Society, cited 2019: Albedo. Glossary of Meteorology. [Available online at
Anvil Cloud
The anvil-shaped cloud that comprises the upper portion of mature cumulonimbus clouds; the popular name given to a cumulonimbus capillatus cloud, particularly if it embodies the supplementary feature incus.
American Meteorological Society, cited 2020: Anvil Cloud. Glossary of Meteorology. [Available online at
Atmospheric boundary layer
The bottom layer of the troposphere that is in contact with the surface of the earth.
American Meteorological Society, cited 2019: Atmospheric boundary layer. Glossary of Meteorology. [Available online at
Atmospheric windows
A range of wavelengths over which there is relatively little absorption of radiation by atmospheric gases. The major windows are the visible window, from ∼0.3 to ∼0.9 μm; the infrared window, from ∼8 to ∼13 μm; and the microwave window, at wavelengths longer than ∼1 mm.
American Meteorological Society, cited 2019: Atmospheric Window. Glossary of Meteorology. [Available online at
Bergeron–Findeisen process
A theoretical explanation of the process by which precipitation particles may form within a mixed cloud (composed of both ice crystals and liquid water drops).
American Meteorological Society, cited 2020: Bergeron–Findeisen process. Glossary of Meteorology. [Available online at
Blackbody
A hypothetical body that cannot be excited to radiate by an external source of electromagnetic radiation of any frequency, direction, or state of polarization except in a negligibly small set of directions around that of the source radiation.
American Meteorological Society, cited 2019: Blackbody. Glossary of Meteorology. [Available online at
Centrifugal force
The apparent force in a rotating system, deflecting masses radially outward from the axis of rotation, with magnitude per unit mass (ω^2)(R), where ω is the angular speed of rotation and R is the radius of curvature of the path.
American Meteorological Society, cited 2020: Centrifugal Force. Glossary of Meteorology. [Available online at
Clausius-Clapeyron equation
The differential equation relating pressure of a substance to temperature in a system in which two phases of the substance are in equilibrium.
Climate
The slowly varying aspects of the atmosphere–hydrosphere–land surface system.
American Meteorological Society, cited 2019: Climate. Glossary of Meteorology. [Available online at
Cloud Condensation Nuclei
Hygroscopic aerosol particles that can serve as nuclei of atmospheric cloud droplets, that is, particles on which water condenses (activates) at supersaturations typical of atmospheric cloud formation (fraction of one to a few percent, depending on cloud type).
American Meteorological Society, cited 2019: Cloud Condensation Nuclei. Glossary of Meteorology. [Available online at
Collision–coalescence process
In cloud physics, the process produces precipitation by collision and coalescence between liquid particles (cloud droplets, drizzle drops, and raindrops).
American Meteorological Society, cited 2020: Collision-coalescence process. Glossary of Meteorology. [Available online at
Conditionally unstable
The environmental lapse rate is between the moist and dry adiabatic lapse rates.
Conduction
Transport of energy (charge) solely as a consequence of random motions of individual molecules (ions, electrons) not moving together in coherent groups.
American Meteorological Society, cited 2019: Conduction. Glossary of Meteorology. [Available online at
Contact freezing
When many freezing nuclei cause super-cooled liquid droplets.
Contour
Generally, an outline or configuration of a body or surface. Often, the term is used for one of a set of lines (contour lines) drawn to represent the configuration of a surface.
American Meteorological Society, cited 2020: Contour. Glossary of Meteorology. [Available online at
Convection
The movement within a fluid due to the tendency of lower density fluid to rise over higher density fluid, which sinks due to the force of gravity resulting in heat transfer within the fluid.
Convective Available Potential Energy
Also known as CAPE, is the maximum buoyancy of an undiluted air parcel, related to the potential updraft strength of thunderstorms.
American Meteorological Society, cited 2019: Convective Available Potential Energy. Glossary of Meteorology. [Available online at
Convective inhibition
Also known as CIN, is the energy needed to lift an air parcel upward adiabatically to the lifting condensation level (LCL) and then as a pseudo-adiabatic process from the LCL to its level of free convection (LFC).
American Meteorological Society, cited 2019: Convective Inhibition. Glossary of Meteorology. [Available online at
Coriolis force
An apparent force on moving particles in a non-inertial coordinate system, that is, the Coriolis acceleration as seen in this (relative) system.
American Meteorological Society, cited 2020: Coriolis Force. Glossary of Meteorology. [Available online at
Cumuliform
Like cumulus; generally descriptive of all clouds, the principal characteristic of which is vertical development in the form of rising mounds, domes, or towers.
American Meteorological Society, cited 2019: Cumuliform. Glossary of Meteorology. [Available online at
Cyclogenesis
Any development or strengthening of cyclonic circulation in the atmosphere; the opposite of cyclolysis.
American Meteorological Society, cited 2020: Cyclogenesis. Glossary of Meteorology. [Available online at
Cyclostrophic wind
That horizontal wind velocity for which the centripetal acceleration exactly balances the horizontal pressure force. The cyclostrophic wind can be an approximation to the real wind in the atmosphere only near the equator, where the Coriolis acceleration is small; or in cases of very great wind speed and curvature of the path (such as a tornado or hurricane), so that the centripetal acceleration is the dominant one.
American Meteorological Society, cited 2020: Cyclostrophic wind. Glossary of Meteorology. [Available online at
Dew point temperature
The temperature to which the air must be cooled to reach saturation, without changing the moisture or air pressure.
Diabatic process
A thermodynamic change of state of a system in which the system exchanges energy with its surroundings by virtue of a temperature difference between them.
American Meteorological Society, cited 2019: Diabatic process. Glossary of Meteorology. [Available online at
Doldrums
A nautical term for the equatorial trough, with special reference to the light and variable nature of the winds.
American Meteorological Society, cited 2020: Doldrums. Glossary of Meteorology. [Available online at
Downburst
An area of strong, often damaging, winds produced by one or more convective downdrafts. Downbursts over horizontal spatial scales ≤ 4 km are referred to as micro-bursts, whereas larger events with horizontal spatial scales > 4 km are termed macro-bursts.
American Meteorological Society, cited 2020: Downburst. Glossary of Meteorology. [Available online at
Downdraft
Sinking air.
Dropsondes
When the radiosonde packages are dropped from an aircraft.
Dry adiabatic lapse rate
A process lapse rate of temperature, the rate of decrease of temperature with height of a parcel of dry air lifted by a reversible adiabatic process through an atmosphere in hydrostatic equilibrium. The adiabatic lapse rate of unsaturated air containing water vapor.
American Meteorological Society, cited 2019: Dry-adiabatic lapse rate. Glossary of Meteorology. [Available online at
Eccentricity
The circularity of a planetary orbit.
Electromagnetic radiation
Energy propagated in the form of an advancing electric and magnetic field disturbance.
American Meteorological Society, cited 2019: Electromagnetic radiation. Glossary of Meteorology. [Available online at
Energy
A measurable physical quantity, with dimensions of mass times velocity squared, that is conserved for an isolated system. Energy of motion is kinetic energy; energy of position is potential energy.
American Meteorological Society, cited 2019: Energy. Glossary of Meteorology. [Available online at
Enhanced-Fujita scale
A scale used to classify tornado strength based on the amount of damage that was caused.
Equation of State
Also known as the ideal gas law or the Charles–Gay–Lussac law. An equation relating temperature, pressure, and volume of a system in thermodynamic equilibrium.
American Meteorological Society, cited 2019: Equation of state. Glossary of Meteorology. [Available online at
Equilibrium level
The level at which an air parcel, rising or descending adiabatically, attains the same density as its environment.
American Meteorological Society, cited 2019: Level of neutral buoyancy. Glossary of Meteorology. [Available online at
Eulerian framework
A fixed framework, relative to a single point on the Earth’s surface.
Ferrel cell
A zonally symmetric circulation that appears to be thermally indirect (when viewed using height or pressure as the vertical coordinate) first proposed by William Ferrel in 1856 as the middle of three meridional cells in each hemisphere.
American Meteorological Society, cited 2020: Ferrel cell. Glossary of Meteorology. [Available online at
First law of thermodynamics
The total internal energy U of an isolated system is constant. Energy cannot be created or destroyed.
American Meteorological Society, cited 2019: First law of thermodynamics. Glossary of Meteorology. [Available online at
Flanking line
An organized lifting zone of cumulus and towering cumulus clouds, connected to and extending outward from the mature updraft tower of a supercell or strong multicell convective storm.
American Meteorological Society, cited 2020: Flanking Line. Glossary of Meteorology. [Available online at
Freezing nuclei
Ice nuclei that are effective at causing the freezing of super-cooled liquid droplets.
Front
In meteorology, generally, the interface or transition zone between two air masses of different density.
American Meteorological Society, cited 2020: Front. Glossary of Meteorology. [Available online at
Frontal wave
A horizontal wave-like deformation of a front in the lower levels, commonly associated with a maximum of cyclonic circulation in the adjacent flow.
American Meteorological Society, cited 2020: Frontal Wave. Glossary of Meteorology. [Available online at
Geometric optics
The application of ray tracing to explain scattering and refractive effects by particles that are very large compared with the wavelength of the radiation.
American Meteorological Society, cited 2020: Geometric Optics. Glossary of Meteorology. [Available online at
Geostrophic adjustment
The process by which an unbalanced atmospheric flow field is modified to geostrophic equilibrium, generally by a mutual adjustment of the atmospheric wind and pressure fields depending on the initial horizontal scale of the disturbance.
American Meteorological Society, cited 2020: Geostrophic Adjustment. Glossary of Meteorology. [Available online at
Geostrophic balance
Describes a balance between Coriolis and horizontal pressure-gradient forces.
American Meteorological Society, cited 2020: Geostrophic Balance. Glossary of Meteorology. [Available online at
Gradient winds
Winds flowing along curved isobars.
Graupel
Heavily rimed snow particles, often called snow pellets; often indistinguishable from very small soft hail except for the size convention that hail must have a diameter greater than 5 mm.
American Meteorological Society, cited 2020: Graupel. Glossary of Meteorology. [Available online at
Greenhouse gases
Those gases, such as water vapor, carbon dioxide, ozone, methane, nitrous oxide, and chlorofluorocarbons, that are fairly transparent to the short wavelengths of solar radiation but efficient at absorbing the longer wavelengths of the infrared radiation emitted by the earth and atmosphere. The trapping of heat by these gases controls the earth's surface temperature despite their presence in only trace concentrations in the atmosphere.
American Meteorological Society, cited 2019: Greenhouse gases. Glossary of Meteorology. [Available online at
Gust front
The leading edge of a meso-scale pressure dome separating the outflow air in a convective storm from the environmental air.
American Meteorological Society, cited 2020: Gust Front. Glossary of Meteorology. [Available online at
Hadley cell
A direct thermally driven and zonally symmetric circulation under the strong influence of the earth's rotation, first proposed by George Hadley in 1735 as an explanation for the trade winds.
American Meteorological Society, cited 2020: Hadley Cell. Glossary of Meteorology. [Available online at
Hailstones
A single unit of hail, ranging in size from that of a pea to, on rare occasions, exceeding that of a grapefruit (i.e., from 5 mm to more than 15 cm in diameter).
American Meteorological Society, cited 2020: Hailstones. Glossary of Meteorology. [Available online at
Heat
The transfer of thermal energy due to the temperature difference between two objects.
Heat capacity
The ratio of the energy or enthalpy absorbed (or released) by a system to the corresponding temperature rise (or fall).
American Meteorological Society, cited 2019: Heat capacity. Glossary of Meteorology. [Available online at
Hook echo
A pendant, curve-shaped region of reflectivity caused when precipitation is drawn into the cyclonic spiral of a meso-cyclone.
American Meteorological Society, cited 2020: Hook Echo. Glossary of Meteorology. [Available online at
Hydrostatic balance
Describes a balance between vertical pressure gradient and buoyancy forces.
American Meteorological Society, cited 2019: Hydrostatic balance. Glossary of Meteorology. [Available online at
Hypsometric equation
An equation relating the thickness, h, between two isobaric surfaces to the mean virtual temperature of the layer. The hypsometric equation is derived from the hydrostatic equation and the ideal gas law.
American Meteorological Society, cited 2019: Hypsometric equation. Glossary of Meteorology. [Available online at
Ice nuclei
Any particle that serves as a nucleus leading to the formation of ice crystals without regard to the particular physical processes involved in the nucleation.
American Meteorological Society, cited 2020: Ice nucleus. Glossary of Meteorology. [Available online at
Instability
A property of the steady state of a system such that certain disturbances or perturbations introduced into the steady state will increase in magnitude, the maximum perturbation amplitude always remaining larger than the initial amplitude.
American Meteorological Society, cited 2020: Instability. Glossary of Meteorology. [Available online at
Intertropical Convergence Zone (ITCZ)
The axis, or a portion thereof, of the broad trade-wind current of the Tropics. This axis is the dividing line between the southeast trades and the northeast trades (of the Southern and Northern Hemispheres, respectively).
American Meteorological Society, cited 2020: Intertropical Convergence Zone. Glossary of Meteorology. [Available online at
Isobars
A line of equal or constant pressure; an isopleth of pressure.
American Meteorological Society, cited 2020: Isobars. Glossary of Meteorology. [Available online at
Isopleths
In common meteorological usage, a line of equal or constant value of a given quantity, with respect to either space or time.
American Meteorological Society, cited 2020: Isopleths. Glossary of Meteorology. [Available online at
Isotherms
A line of equal or constant temperature.
American Meteorological Society, cited 2020: Isotherms. Glossary of Meteorology. [Available online at
Jet stream
Relatively strong winds concentrated within a narrow stream in the atmosphere.
American Meteorological Society, cited 2020: Jet Steam. Glossary of Meteorology. [Available online at
Kinetic energy
The energy that a body possesses as a consequence of its motion, defined as one- half the product of its mass and the square of its speed.
American Meteorological Society, cited 2019: Kinetic energy. Glossary of Meteorology. [Available online at
Lagrangian framework
A framework that is constantly moving and travels with the air.
Lapse rate
The decrease of an atmospheric variable with height, the variable being temperature, unless otherwise specified.
American Meteorological Society, cited 2019: Lapse rate. Glossary of Meteorology. [Available online at
Latent heat
The specific enthalpy difference between two phases of a substance at the same temperature.
American Meteorological Society, cited 2019: Latent heat. Glossary of Meteorology. [Available online at
Lee Cyclogenesis
The synoptic-scale development of an atmospheric cyclonic circulation on the downwind side of a mountain range.
American Meteorological Society, cited 2020: Lee Cyclogenesis. Glossary of Meteorology. [Available online at
Level of free convection
The level at which a parcel of air lifted dry-adiabatically until saturated and saturation-adiabatically thereafter would first become warmer than its surroundings in a conditionally unstable atmosphere.
American Meteorological Society, cited 2019: Level of Free Convection. Glossary of Meteorology. [Available online at
Lifting Condensation Level
The level at which a parcel of moist air lifted dry-adiabatically would become saturated. This is where clouds form.
American Meteorological Society, cited 2019: Lifting Condensation Level. Glossary of Meteorology. [Available online at
Lightning
Lightning is a transient, high-current electric discharge with path lengths measured in kilometers.
American Meteorological Society, cited 2020: Lightning. Glossary of Meteorology. [Available online at
Mesoscale convective complex
A subset of mesoscale convective systems (MCS) that exhibit a large, circular (as observed by satellite), long-lived, cold cloud shield.
American Meteorological Society, cited 2020: Mesoscale Convective Complex. Glossary of Meteorology. [Available online at
Microburst
A convective downdraft (downburst) that covers an area less than 4 km along a side with peak winds that last 2–5 minutes.
American Meteorological Society, cited 2020: Microburst. Glossary of Meteorology. [Available online at
Mie scattering
Scattering of electromagnetic waves by homogeneous spheres of arbitrary size, named after Gustav Mie (1868–1957), whose theory of 1908 explains the process.
American Meteorological Society, cited 2020: Mie Scattering. Glossary of Meteorology. [Available online at
Mixing ratio
The ratio of the mass of water vapor to the mass of dry air.
Moist adiabatic lapse rate
The rate of decrease of temperature with height along a moist adiabat.
American Meteorological Society, cited 2019: Moist adiabatic lapse rate. Glossary of Meteorology. [Available online at
Neutral stability
A condition of a system for which a small perturbation of a parcel of the system causes it to neither depart from its new position nor return to its previous one.
American Meteorological Society, cited 2019: Neutral stability. Glossary of Meteorology. [Available online at
Obliquity
The degree of tilt in the axis of rotation.
Occluded front
A front that forms as a cyclone moves deeper into colder air.
American Meteorological Society, cited 2020: Occluded Front. Glossary of Meteorology. [Available online at
Outflow boundary
A surface boundary formed by the horizontal spreading of thunderstorm-cooled air.
American Meteorological Society, cited 2020: Outflow Boundary. Glossary of Meteorology. [Available online at
Overshooting top
A domelike protrusion above a cumulonimbus anvil, representing the intrusion of an updraft through its equilibrium level (level of neutral buoyancy).
American Meteorological Society, cited 2020: Overshooting Top. Glossary of Meteorology. [Available online at
Planck's radiation law
The distribution law of photon energies for radiation in equilibrium with matter at absolute temperature T.
American Meteorological Society, cited 2019: Planck’s radiation law. Glossary of Meteorology. [Available online at
Polar cell
A weak meridional circulation in the high-latitude troposphere characterized by ascending motion in the sub-polar latitudes (50°–70°), descending motion over the pole, poleward motion aloft, and equatorward motion near the surface.
American Meteorological Society, cited 2020: Polar Cell. Glossary of Meteorology. [Available online at
Polar Easterlies
Air typically flowing from the northeast while in the Antarctic, air flowing from the southeast.
Polar front
According to the polar-front theory, the semi-permanent, semi-continuous front separating air masses of tropical and polar origin.
American Meteorological Society, cited 2020: Polar Front. Glossary of Meteorology. [Available online at
Polar Front theory
A theory originated by the Scandinavian school of meteorologists whereby a polar front, separating air masses of polar and tropical origin, gives rise to cyclonic disturbances that intensify and travel along the front, passing through various phases of a characteristic life history.
American Meteorological Society, cited 2020: Polar Front Theory. Glossary of Meteorology. [Available online at
Polar jet stream
Relatively strong winds concentrated within a narrow stream in the atmosphere. The polar-front jet stream is associated with the polar front of middle and upper-middle latitudes.
American Meteorological Society, cited 2020: Polar Jet Stream. Glossary of Meteorology. [Available online at
Potential energy
The energy a system has by virtue of its position; the negative of the work done in taking a system from a reference configuration, where the potential energy is assigned the value zero, to a given configuration, with no change in kinetic energy of the system.
American Meteorological Society, cited 2019: Potential energy. Glossary of Meteorology. [Available online at
Potential temperature
The temperature that an unsaturated parcel of dry air would have if brought adiabatically and reversibly from its initial state to a standard pressure, p₀, typically 100 kPa.
American Meteorological Society, cited 2019: Potential temperature. Glossary of Meteorology. [Available online at
Precession
The wobble in the rotational axis of a planet that slowly traces out a cone.
Precipitation
All liquid or solid phase aqueous particles that originate in the atmosphere and fall to the earth's surface.
American Meteorological Society, cited 2020: Precipitation. Glossary of Meteorology. [Available online at
Pressure Gradient force
The force due to differences of pressure within a fluid mass.
American Meteorological Society, cited 2020: Pressure Gradient Force. Glossary of Meteorology. [Available online at
Radiation
The process by which electromagnetic radiation is propagated through free space.
American Meteorological Society, cited 2019: Radiation. Glossary of Meteorology. [Available online at
Radiosonde
An expendable meteorological instrument package, often borne aloft by a free-flight balloon, that measures, from the surface to the stratosphere, the vertical profiles of atmospheric variables and transmits the data via radio to a ground receiving system. Radiosondes typically measure temperature, humidity, and, in many cases, pressure.
American Meteorological Society, cited 2020: Radiosonde. Glossary of Meteorology. [Available online at
Rawinsondes
Radiosondes that also infer wind data at different heights.
Rayleigh scattering
Approximate theory for electromagnetic scattering by small particles named for Lord Rayleigh (John William Strutt, 1842–1919), who in 1871 showed that the blue color of the clear sky is explained by the scattering of light by molecules in the atmosphere.
American Meteorological Society, cited 2020: Rayleigh Scattering. Glossary of Meteorology. [Available online at
Relative humidity
The ratio of the amount of water vapor present in the air to the maximum amount of water vapor needed for saturation at a certain pressure and temperature.
Saturation
The condition in which vapor pressure is equal to the equilibrium vapor pressure over a plane surface of pure liquid water, or sometimes ice.
American Meteorological Society, cited 2019: Saturation. Glossary of Meteorology. [Available online at
Shelf cloud
A low-level, horizontal, wedge-shaped arcus cloud associated with a convective storm's gust front (or occasionally a cold front).
American Meteorological Society, cited 2020: Shelf Cloud. Glossary of Meteorology. [Available online at
Snowflake
Colloquially an ice crystal, or more commonly an aggregation of many crystals that falls from a cloud.
American Meteorological Society, cited 2020: Snowflake. Glossary of Meteorology. [Available online at
Sounding
A vertical profile of the environmental lapse rate by releasing a radiosonde attached to a weather balloon.
Specific heat
The heat capacity of a system divided by its mass.
American Meteorological Society, cited 2019: Specific heat. Glossary of Meteorology. [Available online at
Specific humidity
The ratio of the mass of water vapor to the total mass of air (dry air and water vapor combined).
Spontaneous freezing
When liquid water droplets freeze without any sort of nucleus.
Squall line
A line of active deep moist convection frequently associated with thunder, either continuous or with breaks, including contiguous precipitation areas.
American Meteorological Society, cited 2020: Squall Line. Glossary of Meteorology. [Available online at
Stability
The characteristic of a system if sufficiently small disturbances have only small effects, either decreasing in amplitude or oscillating periodically; it is asymptotically stable if the effect of small disturbances vanishes for long time periods.
American Meteorological Society, cited 2019: Stability. Glossary of Meteorology. [Available online at
Stationary front
A type of frontal system that are almost stationary with the winds flowing nearly parallel and from the opposite paths in each side separated by the front.
Steady-state
A fluid motion in which the velocities at every point of the field are independent of time; streamlines and trajectories are identical.
American Meteorological Society, cited 2019: Steady state. Glossary of Meteorology. [Available online at
Stefan-Boltzmann Law
One of the radiation laws, which states that the amount of energy radiated per unit time from a unit surface area of an ideal blackbody is proportional to the fourth power of the absolute temperature of the blackbody.
American Meteorological Society, cited 2019: Stefan-Boltzmann Law. Glossary of Meteorology. [Available online at
Stoke’s Drag Law
Equation to find the terminal velocity of a falling cloud droplet.
Stratiform
Descriptive of clouds of extensive horizontal development, as contrasted to the vertically developed cumuliform types.
American Meteorological Society, cited 2019: Stratiform. Glossary of Meteorology. [Available online at
Subpolar Low
A band of low pressure located, in the mean, between 50° and 70° latitude.
American Meteorological Society, cited 2020: Subpolar Low. Glossary of Meteorology. [Available online at
Subtropical Highs
These highs appear as centers of action on mean charts of sea level pressure, generally between 20° and 40° latitude. They lie over the oceans and are best developed in the summer season.
American Meteorological Society, cited 2020: Subtropical Highs. Glossary of Meteorology. [Available online at
Super-cooled water
Liquid water that exists below the freezing point of water (below 0°C).
Supercell
An often dangerous convective storm that consists primarily of a single, quasi-steady rotating updraft, which persists for a period of time much longer than it takes an air parcel to rise from the base of the updraft to its summit (often much longer than 10–20 min).
American Meteorological Society, cited 2020: Supercell. Glossary of Meteorology. [Available online at
Synoptic
In meteorology, this term has become somewhat specialized in referring to the use of meteorological data obtained simultaneously over a wide area for the purpose of presenting a comprehensive and nearly instantaneous picture of the state of the atmosphere.
American Meteorological Society, cited 2020: Synoptic. Glossary of Meteorology. [Available online at
Terminal velocity
The particular falling speed, for any given object moving through a fluid medium of specified physical properties, at which the drag forces and buoyant forces exerted by the fluid on the object just equal the gravitational force acting on the object.
American Meteorological Society, cited 2020: Terminal Fall Velocity. Glossary of Meteorology. [Available online at
Thermal energy
A form of energy transferred between systems, existing only in the process of transfer. Also the same as enthalpy.
American Meteorological Society, cited 2019: Heat. Glossary of Meteorology. [Available online at
Thunder
The sound emitted by rapidly expanding gases along the channel of a lightning discharge.
American Meteorological Society, cited 2020: Thunder. Glossary of Meteorology. [Available online at
Thunderstorm
In general, a local storm, invariably produced by a cumulonimbus cloud and always accompanied by lightning and thunder, usually with strong gusts of wind, heavy rain, and sometimes with hail.
American Meteorological Society, cited 2020: Thunderstorm. Glossary of Meteorology. [Available online at
Trade winds
The wind system, occupying most of the Tropics, that blows from the subtropical highs toward the equatorial trough; a major component of the general circulation of the atmosphere.
American Meteorological Society, cited 2020: Trade Winds. Glossary of Meteorology. [Available online at
Turbulent drag
The relationship between wind speed and force caused by the wind against objects or along surfaces.
American Meteorological Society, cited 2020: Turbulent Drag. Glossary of Meteorology. [Available online at
Updraft
Rising air.
Vapor pressure
The pressure exerted by the molecules of a given vapor.
American Meteorological Society, cited 2019: Vapor pressure. Glossary of Meteorology. [Available online at
Virtual temperature
Also called Density temperature. The temperature that dry dry air would have if its pressure and density were equal to those of a given sample of moist air.
American Meteorological Society, cited 2019: Virtual temperature. Glossary of Meteorology. [Available online at
Wall cloud
A local, often abrupt lowering from a cumulonimbus cloud base into a low-hanging accessory cloud, normally a kilometer or more in diameter.
American Meteorological Society, cited 2020: Wall Cloud. Glossary of Meteorology. [Available online at
Warm front
Any non-occluded front, or portion thereof, that moves in such a way that warmer air replaces cold air.
American Meteorological Society, cited 2020: Warm Front. Glossary of Meteorology. [Available online at
Warm sector
The region of warmer air between the cold and warm fronts.
Weather
The state of the atmosphere, mainly with respect to its effects upon life and human activities. As distinguished from climate, weather consists of the short-term (minutes to days) variations in the atmosphere. Popularly, weather is thought of in terms of temperature, humidity, precipitation, cloudiness, visibility, and wind.
American Meteorological Society, cited 2019: Weather. Glossary of Meteorology. [Available online at
Westerlies
Specifically, the dominant west-to-east motion of the atmosphere, centered over the middle latitudes of both hemispheres.
American Meteorological Society, cited 2020: Westerlies. Glossary of Meteorology. [Available online at
Wet-Bulb temperature
The lowest temperature that can be achieved if water evaporates within the air.
Wien’s Law
A radiation law that is used to relate the wavelength of maximum emission from a blackbody inversely to its absolute temperature.
American Meteorological Society, cited 2019: Wien’s law. Glossary of Meteorology. [Available online at
Wind shear
The local variation of the wind vector or any of its components in a given direction.
American Meteorological Society, cited 2020: Wind Shear. Glossary of Meteorology. [Available online at
Work
A form of energy arising from the motion of a system against a force, existing only in the process of energy conversion.
American Meteorological Society, cited 2019: Work. Glossary of Meteorology. [Available online at
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14: Fluid Mechanics
University Physics I - Mechanics, Sound, Oscillations, and Waves (OpenStax)
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14.4: Measuring Pressure
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Contents
1. Home
2. Bookshelves
3. University Physics
4. University Physics (OpenStax)
5. University Physics I - Mechanics, Sound, Oscillations, and Waves (OpenStax)
6. 14: Fluid Mechanics
7. 14.4: Measuring Pressure
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University Physics I - Mechanics, Sound, Oscillations, and Waves (OpenStax)
Front Matter
1: Units and Measurement
2: Vectors
3: Motion Along a Straight Line
4: Motion in Two and Three Dimensions
5: Newton's Laws of Motion
6: Applications of Newton's Laws
7: Work and Kinetic Energy
8: Potential Energy and Conservation of Energy
9: Linear Momentum and Collisions
10: Fixed-Axis Rotation Introduction
11: Angular Momentum
12: Static Equilibrium and Elasticity
13: Gravitation
14: Fluid Mechanics
15: Oscillations
16: Waves
17: Sound
18: Answer Key to Selected Problems
Back Matter
14.4: Measuring Pressure
Last updated Sep 12, 2022
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14.3: Fluids, Density, and Pressure (Part 2)
14.5: Pascal's Principle and Hydraulics
Page ID 4056
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Gauge Pressure vs. Absolute Pressure
1. Definition Absolute Pressure
Measuring Pressure
Manometers
Barometers
Example 14.4.1: Fluid Heights in an Open U-Tube
Solution
Exercise 14.4.1
Units of pressure
Learning Objectives
Define gauge pressure and absolute pressure
Explain various methods for measuring pressure
Understand the working of open-tube barometers
Describe in detail how manometers and barometers operate
In the preceding section, we derived a formula for calculating the variation in pressure for a fluid in hydrostatic equilibrium. As it turns out, this is a very useful calculation. Measurements of pressure are important in daily life as well as in science and engineering applications. In this section, we discuss different ways that pressure can be reported and measured.
Gauge Pressure vs. Absolute Pressure
Suppose the pressure gauge on a full scuba tank reads 3000 psi, which is approximately 207 atmospheres. When the valve is opened, air begins to escape because the pressure inside the tank is greater than the atmospheric pressure outside the tank. Air continues to escape from the tank until the pressure inside the tank equals the pressure of the atmosphere outside the tank. At this point, the pressure gauge on the tank reads zero, even though the pressure inside the tank is actually 1 atmosphere—the same as the air pressure outside the tank.
Most pressure gauges, like the one on the scuba tank, are calibrated to read zero at atmospheric pressure. Pressure readings from such gauges are called gauge pressure, which is the pressure relative to the atmospheric pressure. When the pressure inside the tank is greater than atmospheric pressure, the gauge reports a positive value. Some gauges are designed to measure negative pressure. For example, many physics experiments must take place in a vacuum chamber, a rigid chamber from which some of the air is pumped out. The pressure inside the vacuum chamber is less than atmospheric pressure, so the pressure gauge on the chamber reads a negative value. Unlike gauge pressure, absolute pressure accounts for atmospheric pressure, which in effect adds to the pressure in any fluid not enclosed in a rigid container.
Definition Absolute Pressure
The absolute pressure, or total pressure, is the sum of gauge pressure and atmospheric pressure:
(14.4.1)p abs=p g+p atm
where p abs is absolute pressure, p g is gauge pressure, and p atm is atmospheric pressure.
For example, if a tire gauge reads 34 psi, then the absolute pressure is 34 psi plus 14.7 psi (p atm in psi), or 48.7 psi (equivalent to 336 kPa).
In most cases, the absolute pressure in fluids cannot be negative. Fluids push rather than pull, so the smallest absolute pressure in a fluid is zero (a negative absolute pressure is a pull). Thus, the smallest possible gauge pressure is p g = −p atm (which makes p abs zero). There is no theoretical limit to how large a gauge pressure can be.
Measuring Pressure
A host of devices are used for measuring pressure, ranging from tire gauges to blood pressure monitors. Many other types of pressure gauges are commonly used to test the pressure of fluids, such as mechanical pressure gauges. We will explore some of these in this section.
Any property that changes with pressure in a known way can be used to construct a pressure gauge. Some of the most common types include strain gauges, which use the change in the shape of a material with pressure; capacitance pressure gauges, which use the change in electric capacitance due to shape change with pressure; piezoelectric pressure gauges, which generate a voltage difference across a piezoelectric material under a pressure difference between the two sides; and ion gauges, which measure pressure by ionizing molecules in highly evacuated chambers. Different pressure gauges are useful in different pressure ranges and under different physical situations. Some examples are shown in Figure 14.4.1.
Figure 14.4.1: (a) Gauges are used to measure and monitor pressure in gas cylinders. Compressed gases are used in many industrial as well as medical applications. (b) Tire pressure gauges come in many different models, but all are meant for the same purpose: to measure the internal pressure of the tire. This enables the driver to keep the tires inflated at optimal pressure for load weight and driving conditions. (c) An ionization gauge is a high-sensitivity device used to monitor the pressure of gases in an enclosed system. Neutral gas molecules are ionized by the release of electrons, and the current is translated into a pressure reading. Ionization gauges are commonly used in industrial applications that rely on vacuum systems.
Manometers
One of the most important classes of pressure gauges applies the property that pressure due to the weight of a fluid of constant density is given by p = h ρ g. The U-shaped tube shown in Figure 14.4.2 is an example of a manometer; in part (a), both sides of the tube are open to the atmosphere, allowing atmospheric pressure to push down on each side equally so that its effects cancel.
A manometer with only one side open to the atmosphere is an ideal device for measuring gauge pressures. The gauge pressure is p g = h ρ g and is found by measuring h. For example, suppose one side of the U-tube is connected to some source of pressure p abs, such as the balloon in part (b) of the figure or the vacuum-packed peanut jar shown in part (c). Pressure is transmitted undiminished to the manometer, and the fluid levels are no longer equal. In part (b), p abs is greater than atmospheric pressure, whereas in part (c), pabs is less than atmospheric pressure. In both cases, p abs differs from atmospheric pressure by an amount h ρ g, where ρ is the density of the fluid in the manometer. In part (b), p abs can support a column of fluid of height h, so it must exert a pressure h ρ g greater than atmospheric pressure (the gauge pressure p g is positive). In part (c), atmospheric pressure can support a column of fluid of height h, so p abs is less than atmospheric pressure by an amount h ρ g (the gauge pressure p g is negative).
Figure 14.4.2: An open-tube manometer has one side open to the atmosphere. (a) Fluid depth must be the same on both sides, or the pressure each side exerts at the bottom will be unequal and liquid will flow from the deeper side. (b) A positive gauge pressure p g = h ρ g transmitted to one side of the manometer can support a column of fluid of height h. (c) Similarly, atmospheric pressure is greater than a negative gauge pressure p g by an amount h ρ g. The jar’s rigidity prevents atmospheric pressure from being transmitted to the peanuts.
Barometers
Manometers typically use a U-shaped tube of a fluid (often mercury) to measure pressure. A barometer(Figure 14.4.3) is a device that typically uses a single column of mercury to measure atmospheric pressure. The barometer, invented by the Italian mathematician and physicist Evangelista Torricelli (1608–1647) in 1643, is constructed from a glass tube closed at one end and filled with mercury. The tube is then inverted and placed in a pool of mercury. This device measures atmospheric pressure, rather than gauge pressure, because there is a nearly pure vacuum above the mercury in the tube. The height of the mercury is such that h ρ g = p atm. When atmospheric pressure varies, the mercury rises or falls.
Weather forecasters closely monitor changes in atmospheric pressure (often reported as barometric pressure), as rising mercury typically signals improving weather and falling mercury indicates deteriorating weather. The barometer can also be used as an altimeter, since average atmospheric pressure varies with altitude. Mercury barometers and manometers are so common that units of mm Hg are often quoted for atmospheric pressure and blood pressures.
Figure 14.4.3: A mercury barometer measures atmospheric pressure. The pressure due to the mercury’s weight, h ρ g, equals atmospheric pressure. The atmosphere is able to force mercury in the tube to a height h because the pressure above the mercury is zero.
Example 14.4.1: Fluid Heights in an Open U-Tube
A U-tube with both ends open is filled with a liquid of density ρ 1 to a height h on both sides (Figure 14.4.1). A liquid of density ρ 2<ρ 1 is poured into one side and Liquid 2 settles on top of Liquid 1. The heights on the two sides are different. The height to the top of Liquid 2 from the interface is h 2 and the height to the top of Liquid 1 from the level of the interface is h 1. Derive a formula for the height difference.
Figure 14.4.4: Two liquids of different densities are shown in a U-tube.
Strategy
The pressure at points at the same height on the two sides of a U-tube must be the same as long as the two points are in the same liquid. Therefore, we consider two points at the same level in the two arms of the tube: One point is the interface on the side of the Liquid 2 and the other is a point in the arm with Liquid 1 that is at the same level as the interface in the other arm. The pressure at each point is due to atmospheric pressure plus the weight of the liquid above it.
Pressure on the side with Liquid 1 = p 0 + ρ 1 gh 1
Pressure on the side with Liquid 2 = p 0 + ρ 2 gh 2
Solution
Since the two points are in Liquid 1 and are at the same height, the pressure at the two points must be the same. Therefore, we have
p 0+ρ 1gh 1=p 0+ρ 2gh 2.
Hence,
ρ 1h 1=ρ 2h 2.
This means that the difference in heights on the two sides of the U-tube is
h 2−h 1=(1−p 1 p 2)h 2.
The result makes sense if we set ρ 2=ρ 1, which gives h 2 = h 1. If the two sides have the same density, they have the same height.
Exercise 14.4.1
Mercury is a hazardous substance. Why do you suppose mercury is typically used in barometers instead of a safer fluid such as water?
Units of pressure
As stated earlier, the SI unit for pressure is the pascal (Pa), where
(14.4.2)1 Pa=1 N/m 2.
In addition to the pascal, many other units for pressure are in common use (Table 14.4.1). In meteorology, atmospheric pressure is often described in the unit of millibars (mb), where
(14.4.3)1000 mb=1×10 5 Pa.
The millibar is a convenient unit for meteorologists because the average atmospheric pressure at sea level on Earth is 1.013 x 10 5 Pa = 1013 mb = 1 atm. Using the equations derived when considering pressure at a depth in a fluid, pressure can also be measured as millimeters or inches of mercury. The pressure at the bottom of a 760-mm column of mercury at 0 °C in a container where the top part is evacuated is equal to the atmospheric pressure. Thus, 760 mm Hg is also used in place of 1 atmosphere of pressure. In vacuum physics labs, scientists often use another unit called the torr, named after Torricelli, who, as we have just seen, invented the mercury manometer for measuring pressure. One torr is equal to a pressure of 1 mm Hg.
Table 14.4.1: Summary of the Units of Pressure| Unit | Definition |
--- |
| SI unit: the Pascal | (14.4.4)1 Pa=1 N/m 2 |
| English unit: pounds per square inch ( lb/in.2 or psi) | (14.4.5)1 psi=7.015×10 3 Pa |
| Other units of pressure | (14.4.6)1 atm=760 mmHg=1.013×10 5 Pa=14.7 psi=29.9 inches ofHg=1013 mb |
| | (14.4.7)1 bar=10 5 Pa |
| | (14.4.8)1 torr=1 mmHg=122.39 Pa |
This page titled 14.4: Measuring Pressure is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
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14.3: Fluids, Density, and Pressure (Part 2)
14.5: Pascal's Principle and Hydraulics
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