text stringlengths 1 1k ⌀ | source stringclasses 12
values |
|---|---|
142 CHAPTER 8. OPTICAL FIBER
c⃝ S. Lally CC BY -SA 4.0
Figure 8.6: A digital signal that might be applied to
the input of a fiber optic cable.
the cable. Figure 8.5(b) represents the continuum of
possibilities between the extreme cases of (a) and (c),
with associated propagation times greater than that of
case (a) but l... | Electromagnetics_Vol2.pdf |
ton <T may be present. Let the minimum
propagation time through the fiber (as in
Figure 8.5(a)) be τmin. Let the maximum propagation
time through the fiber (as in Figure 8.5(c)) be τmax. If
τmax − τmin ≪ ton, we see little degradation in the
signal output from the fiber. Otherwise, we observe a
“smearing” of pulses at the... | Electromagnetics_Vol2.pdf |
8.3. DISPERSION IN OPTICAL FIBER 143
overlap problem in greater detail. T o avoid overlap:
τmax − τmin + ton <T (8.18)
Let us define the quantity τ ≜ τmax − τmin. This is
sometimes referred to as the delay spread. 1 Thus, we
obtain the following requirement for overlap-free
transmission:
τ <T − ton (8.19)
Note that the ... | Electromagnetics_Vol2.pdf |
tmin = l
vp
= l
c/√ǫr
= l
c/nf
= lnf
c (8.20)
The
maximum propagation time tmax is different
because the maximum path length is different.
Specifically , the maximum path length is not l, but
rather l/cos θ2 where θ2 is the angle between the axis
and the direction of travel as light crosses the axis.
So, for example, θ2... | Electromagnetics_Vol2.pdf |
ysis.
Note that τ increases linearly with l. Therefore, the
rate at which pulses can be sent without overlap
decreases linearly with increasing length. In practical
applications, this means that the maximum
supportable data rate decreases as the length of the
cable increases. This is true independently of media
loss wi... | Electromagnetics_Vol2.pdf |
T >2τ; therefore, T greater than about 204 ps
is required. The modulation scheme allows one
bit per period, so the maximum data rate is
1/T ∼= 4.9 × 109 bits per second; i.e., ∼
=
4.9 Gb/s
.
The
finding of 4.9 Gb/s may seem like a pretty high
data rate; however, consider what happens if the
length increases to 1 km. It ... | Electromagnetics_Vol2.pdf |
next repeater. Alternatively , one may employ “single
mode” fiber, which has intrinsically less dispersion
than the multimode fiber presumed in our analysis.
Additional Reading:
• “Optical fiber” on Wikipedia.
[m0209] | Electromagnetics_Vol2.pdf |
144 CHAPTER 8. OPTICAL FIBER
Image Credits
Fig. 8.1: c⃝ S. Lally , https://commons.wikimedia.org/wiki/File:Figure 8.1-01.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modified by author.
Fig. 8.2: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Internal Reflection in Optical Fiber.... | Electromagnetics_Vol2.pdf |
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 8.6: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Digital Signal on Fiber Optic Cable.svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 8.7: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:D... | Electromagnetics_Vol2.pdf |
Chapter 9
Radiation
9.1
Radiation from a Current
Moment
[m0194]
In this section, we begin to address the following
problem:
Given a distribution of impressed current
density J(r), what is the resulting electric field
intensity E(r)? One route to an answer is via
Maxwell’s equations. V iewing Maxwell’s equations
as a sys... | Electromagnetics_Vol2.pdf |
eventually address the more general problem.
Furthermore, the results presented in this section will
turn out to be sufficient to tackle many
commonly-encountered applications.
The simple current distribution considered in this
section is known as a current moment. An example of
a current moment is shown in Figure 9.1 a... | Electromagnetics_Vol2.pdf |
form of volume current density , since – as indicated
by Equation 9.2 – it exists only at the origin and
nowhere else.
Although some current distributions approximate the
current moment, current distributions encountered in
common engineering practice generally do not exist
1 Als o a form of the Dirac delta function; s... | Electromagnetics_Vol2.pdf |
146 CHAPTER 9. RADIA TION
in precisely this form. Nevertheless, the current
moment turns out to be generally useful as a “building
block” from which practical distributions of current
can be constructed, via the principle of superposition.
Radiation from current distributions constructed in
this manner is calculated si... | Electromagnetics_Vol2.pdf |
current density is
∆ ˜J(r) = ˆz ˜I ∆lδ (r) (9.4)
where ˜I ∆l is simply the scalar current moment
expressed as a phasor.
Now we are ready to address the question “What is
∆ ˜E(r) due to ∆ ˜J(r)?” Without doing any math, we
know quite a bit about ∆ ˜E(r). For example:
• Since electric fields are proportional to the
curren... | Electromagnetics_Vol2.pdf |
loss due to the medium is negligible, then we
expect the phase of ∆ ˜E(r) to change
approximately at rate βwhere βis the phase
propagation constant 2π/λ. Since we expect
spherical phasefronts, ∆ ˜E(r) should therefore
contain the factor e−jβr.
• Ampere’s law indicates that a ˆz-directed current
at the origin should giv... | Electromagnetics_Vol2.pdf |
at θ= 0 and return to zero at θ= π. The
symmetry of the problem suggests
⏐
⏐⏐∆ ˜E(ˆr)
⏐⏐⏐is
maximum at θ= π/2. This magnitude must
vary in the simplest possible way , leading us to
conclude that ∆ ˜E(ˆr) is proportional to sin θ.
Furthermore, the radial symmetry of the problem
means that ∆ ˜E(ˆr) should not depend at a... | Electromagnetics_Vol2.pdf |
analysis reaches a dead end, so we shall simply state
the result from the rigorous solution: C = jηβ/4π.
The units are correct, and we finally obtain:
∆ ˜E(r) ≈ ˆθjηβ
4π
(
˜I ∆l
)
(s in θ) e−jβr
r (9.6)
Additional
evidence that this solution is correct
comes from the fact that it satisfies the wave equation
∇2∆ ˜E(r) + β... | Electromagnetics_Vol2.pdf |
9.2. MAGNETIC VECTOR POTENTIAL 147
Note that the expression we have obtained for the
radiated electric field is approximate (hence the “≈ ”).
This is due in part to our presumption of a simple
spherical wave, which may only be valid at distances
far from the source. But how far? An educated guess
would be distances much... | Electromagnetics_Vol2.pdf |
and the Hertzian dipole as effectively the same in
practical engineering applications.
Additional Reading:
• “Dirac delta function” on Wikipedia.
• “Dipole antenna” (section entitled “Hertzian
Dipole”) on Wikipedia.
9.2 Magnetic V ector Potential
[m0195]
A common problem in electromagnetics is to
determine
the fields ra... | Electromagnetics_Vol2.pdf |
engineering problems, one is concerned with
propagation through media which are well-modeled
as homogeneous media with neutral charge, such as
free space. 5 Therefore, in this section, we shall limit
our scope to problems in which ˜ρv = 0. Thus,
Equation 9.7 simplifies to:
∇ · ˜E = 0 (9.11)
T o solve the linear system o... | Electromagnetics_Vol2.pdf |
with the inhomogeneous part representing the source
current.
4 Recall that there is no loss of generality in doing so, since any
other time-domain variation in the current distribution can be repre-
sented using sums of time-harmonic solutions via the Fourier trans-
form.
5 A counter-example would be propagation throug... | Electromagnetics_Vol2.pdf |
148 CHAPTER 9. RADIA TION
The magnetic vector potential ˜A is defined by the
following relationship:
˜B ≜ ∇ × ˜A (9.12)
where ˜B = µ˜H is
the magnetic flux density . The
magnetic field appears in three of Maxwell’s
equations. For Equation 9.12 to be a reasonable
definition, ∇ × ˜A must yield reasonable results when
substit... | Electromagnetics_Vol2.pdf |
∇ ×
(
˜E + jω˜A
)
= 0 (9.15)
Now , for reasons that will become apparent in just a
moment, we define a new scalar field ˜V and require it
to satisfy the following relationship:
− ∇ ˜V ≜ ˜E + jω˜A (9.16)
Using this definition, Equation 9.15 becomes:
∇ ×
(
−∇ ˜V
)
= 0 (9.17)
which is simply
∇ × ∇ ˜V = 0 (9.18)
Once again we... | Electromagnetics_Vol2.pdf |
is an enhanced version of that relationship that
accounts for the coupling with H (here, represented
by A) in the time-varying (decidedly non-static) case.
That assessment is correct, but let’s not get too far
ahead of ourselves: As demonstrated in the previous
paragraph, we are not yet compelled to make any
particular... | Electromagnetics_Vol2.pdf |
∇ × ∇ × ˜A ≡ ∇
(
∇ · ˜A
)
− ∇2 ˜A (9.23)
Equation 9.22 becomes:
∇
(
∇ · ˜A
)
− ∇2 ˜A = ω2µǫ˜A − jωµǫ∇˜V + µ˜J
(9.24)
Now multiplying both sides by −1 and rearranging
terms:
∇2 ˜A + ω2µǫ˜A = ∇
(
∇ · ˜A
)
+ jωµǫ∇˜V − µ˜J
(9.25)
Combining terms on the right side:
∇2 ˜A+ω2µǫ˜A = ∇
(
∇ · ˜A + jωµǫ˜V
)
−µ˜J (9.26)
Now consid... | Electromagnetics_Vol2.pdf |
9.2. MAGNETIC VECTOR POTENTIAL 149
equation. W e established earlier that ˜V can be
essentially any scalar field – from a mathematical
perspective, we are free to choose. Invoking this
freedom, we now require ˜V to satisfy the following
expression:
∇ · ˜A + jωµǫ˜V = 0 (9.27)
Clearly this is advantageous in the sense tha... | Electromagnetics_Vol2.pdf |
electromagnetic fields radiated by a current
distribution. The procedure is simply as follows:
1. Solve the partial differential Equation 9.28 for ˜A
along with the appropriate electromagnetic
boundary conditions.
2. ˜H = (1/µ)∇ × ˜A
3. ˜E may now be determined from ˜H using
Equation 9.10.
Summarizing:
The magnetic vect... | Electromagnetics_Vol2.pdf |
Equation 9.27. This equation is known as the Lorenz
gauge condition. This constraint is not quite as
arbitrary as the preceding derivation implies; rather,
there is some deep physics at work here. Specifically ,
the Lorenz gauge leads to the classical interpretation
of ˜V as the familiar scalar electric potential, as no... | Electromagnetics_Vol2.pdf |
four fundamental forces recognized in modern
physics; the others being gravity , the strong nuclear
force, and the weak nuclear force. For more
information on that concept, an excellent starting
point is the video “Quantum Invariance & The Origin
of The Standard Model” referenced at the end of this
section.
Additional ... | Electromagnetics_Vol2.pdf |
150 CHAPTER 9. RADIA TION
9.3 Solution of the W ave
Equation for Magnetic V ector
Potential
[m0196]
The magnetic vector potential ˜A due to a current
density ˜J is given by the following wave equation:
∇2 ˜A − γ2 ˜A = −µ˜J (9.29)
where γis the propagation constant, defined in the
usual manner 7
γ2 ≜ −ω2µǫ (9.30)
Equatio... | Electromagnetics_Vol2.pdf |
˜J(r) = ˆl ˜I ∆lδ (r) (9.31)
where ˜I has units of current (SI base units of A), ∆l
has units of length (SI base units of m), ˆl is the
7 Alternati vely , γ2 ≜ −ω2µǫc accounting for the possibility of
lossy media.
c⃝ C. W ang CC BY -SA 4.0
Figure 9.2: An example of a current moment located
at the origin. In this case, ... | Electromagnetics_Vol2.pdf |
obtain:
∇2 ˜A − γ2 ˜A = −µˆl ˜I ∆lδ (r) (9.34)
The general solution to this equation presuming
homogeneous and time-invariant media (i.e., µand ǫ
constant with respect to space and time) is:
˜A(r) = ˆl µ ˜I ∆l e±γr
4πr (9.35)
T
o confirm that this is a solution to Equation 9.34,
substitute this expression into Equation ... | Electromagnetics_Vol2.pdf |
follows:
˜A(r) = ˆl µ ˜I ∆l e±αre±jβr
4πr (9.37)
8 Als o a form of the Dirac delta function; see “ Additional Read-
ing” at the end of this section.
9 Remember, the density of a volume current is with respect to
the area through which it flows, therefore the units are A/m 2. | Electromagnetics_Vol2.pdf |
9.3. SOLUTION OF THE W A VE EQUA TION FOR MAGNETIC VECTOR POTENTIAL 151
Now consider the factor e±αr/r, which determines
the dependence of magnitude on distance r. If we
choose the negative sign in the exponent, this factor
decays exponentially with increasing distance from
the origin, ultimately reaching zero at r→ ∞.... | Electromagnetics_Vol2.pdf |
applies at r→ ∞. Invoking the radiation condition,
Equation 9.35 becomes:
˜A(r) = ˆl µ ˜I ∆l e−γr
4πr (9.38)
In
the loss-free (α = 0) case, we cannot rely on the
radiation condition to constrain the sign of γ.
However, in practical engineering work there is
always some media loss; i.e., αmight be negligible
but is not ... | Electromagnetics_Vol2.pdf |
version of the problem in which the current moment
is no longer located at the origin, but rather is located
at r′. This is illustrated in Figure 9.3. This current
distribution can be expressed as follows:
˜J(r) = ˆl ˜I ∆lδ (r − r′) (9.40)
The solution in this case amounts to a straightforward
modification of the existi... | Electromagnetics_Vol2.pdf |
following a path C through space. This is illustrated in
Figure 9.4. Such a filament may be viewed as a
collection of a large number N of discrete current
moments distributed along the path. The contribution
of the nth current moment, located at rn, to the total
magnetic vector potential is:
∆ ˜A(r; rn) = ˆl(rn) µ ˜I(rn... | Electromagnetics_Vol2.pdf |
152 CHAPTER 9. RADIA TION
c⃝ C. W ang CC BY -SA 4.0
Figure 9.4: A filament of current lying along the path
C. This current distribution may be interpreted as a
collection of current moments lying along C.
magnetic vector potential:
˜A(r) ≈
N∑
n=1
∆ ˜A(r; rn) (9.43)
≈ µ
4π
N∑
n=1
ˆl( rn) ˜I(rn) e−γ|r−rn|
|r − rn| ∆l (9.4... | Electromagnetics_Vol2.pdf |
Additional Reading:
• “Magnetic potential” on Wikipedia.
• “Dirac delta function” on Wikipedia.
9.4 Radiation from a Hertzian
Dipole
[m0197]
Section 9.1 presented an informal derivation of the
electromagnetic
field radiated by a Hertzian dipole
represented by a zero-length current moment. In this
section, we provide a r... | Electromagnetics_Vol2.pdf |
method is to model these relatively complex
distributions of current as the sum of Hertzian dipoles,
which reduces the problem to that of summing the
contributions of the individual Hertzian dipoles, with
each Hertzian dipole having the appropriate (i.e.,
different) position, magnitude, and phase.
T o facilitate use of... | Electromagnetics_Vol2.pdf |
current moment, ˆl is the direction of current flow , and
δ(r) is the volumetric sampling function defined as
follows:
δ(r) ≜ 0 for r ̸= 0; and (9.47)∫
V
δ(r) dv≜ 1 (9.48)
where V is any volume which includes the origin
(r = 0). In this description, the Hertzian dipole is
located at the origin. | Electromagnetics_Vol2.pdf |
9.4. RADIA TION FROM A HER TZIAN DIPOLE 153
The solution for the magnetic vector potential due to a
ˆz-directed Hertzian dipole located at the origin was
presented in Section 9.3. In the present scenario, it is:
˜A(r) = ˆz µ ˜I ∆l e−γr
4πr (9.49)
where
the propagation constant γ = α+ jβ as usual.
Assuming lossless medi... | Electromagnetics_Vol2.pdf |
∇ × ˆze−jβr
r = ∇ ×
[
ˆr (cos θ) e−j
βr
r
−ˆθ(s
in θ) e−jβr
r
]
(9.55)
c⃝ C. W ang CC BY -SA 4.0
Figure 9.5: A Hertzian dipole located at the origin,
represented as a current moment. In this case, ˆl = ˆz.
At this point, it is convenient to make the following
definitions:
Cr ≜ (cos θ) e−jβr
r (9.56)
Cθ ≜ − (
sin θ) e−jβ... | Electromagnetics_Vol2.pdf |
antenna. Specifically , let us assume r≫ λ. Now we
use the relationship β = 2π/λ and determine the
following:
jβ + 1
r = j2π
λ + 1
r (9.62)
≈ j2
π
λ = jβ (9.63)
Equation
9.61 becomes:
˜H ≈ ˆφj
˜I· β∆l
4π (s in θ) e−jβr
r (9.64)
where
the approximation holds for low-loss media
and r≫ λ. This expression is known as a far ... | Electromagnetics_Vol2.pdf |
154 CHAPTER 9. RADIA TION
• Notice the factor β∆l has units of radians; that
is, it is electrical length. This tells us that the
magnitude of the radiated field depends on the
electrical length of the current moment.
• The factor e−jβr/rindicates that this is a
spherical wave; that is, surfaces of constant
phase corresp... | Electromagnetics_Vol2.pdf |
Ampere’s law . That is,
˜E = 1
jωǫ∇ × ˜H (9.65)
where ˜H is
given by Equation 9.64. At field points far
from the dipole, the radius of curvature of the
spherical phasefronts is very large and so appear to be
locally planar. That is, from the perspective of an
observer far from the dipole, the arriving wave
appears to be... | Electromagnetics_Vol2.pdf |
9.5. RADIA TION FROM AN ELECTRICALL Y -SHOR T DIPOLE 155
9.5 Radiation from an
Electrically-Short Dipole
[m0198]
The simplest distribution of radiating current that is
encountered
in common practice is the
electrically-short dipole (ESD). This current
distribution is shown in Figure 9.6. The two
characteristics that de... | Electromagnetics_Vol2.pdf |
apparent, recall the behavior of transmission lines:
11 A potential source of confusion is that the Hertzian dipole is
also a “dipole” which is “electrically-short. ” The distinction is that
the current comprising a Hertzian dipole is constant over its length.
This condition is rarely and only approximately seen in pra... | Electromagnetics_Vol2.pdf |
Hertzian dipoles.
The current standing wave on a transmission line
exhibits a period of λ/2, regardless the source or
termination. For the ESD, L≪ λ/2 and so we expect
an even simpler variation. Also, we know that the
current at the ends of the dipole must be zero, simply
because the dipole ends there. These considerat... | Electromagnetics_Vol2.pdf |
calculate ˜E from ˜H using Ampere’s law . W e shall
employ a simpler approach, shown in Figure 9.7.
Imagine the ESD as a collection of many shorter
segments of current that radiate independently . The
total field is then the sum of these short segments.
Because these segments are very short relative to the
12 A more rig... | Electromagnetics_Vol2.pdf |
156 CHAPTER 9. RADIA TION
length of the dipole as well as being short relative to a
wavelength, we may approximate the current over
each segment as approximately constant. In other
words, we may interpret each of these segments as
being, to a good approximation, a Hertzian dipole.
The advantage of this approach is that... | Electromagnetics_Vol2.pdf |
β = 2π/λ. This expression also assumes field points
far from the dipole; specifically , distances rthat are
much greater than λ. Repurposing this expression for
the present problem, the segment at the origin radiates
the electric field:
˜E(r; z′ = 0) ≈ ˆθjηI0 · β∆l
4π (s in θ) e−jβr
r (9.70)
where
the notation z′ = 0 indi... | Electromagnetics_Vol2.pdf |
Similarly , ˆθis replaced by ˆθ′, since it also varies with
z′. The electric field radiated by the ESD is obtained
by integration over these contributions:
˜E(r) ≈
∫ +L/ 2
−L/ 2
d˜E(ˆr; z′) (9.73)
c⃝ C. W ang CC BY -SA 4.0
Figure 9.8: Parallel ray approximation for an ESD.
yielding:
˜E(r) ≈ jηβ
4π
∫ +L/ 2
−L/ 2
ˆθ′ ˜I( ... | Electromagnetics_Vol2.pdf |
that r≫ L(i.e., the distance to field points is much
greater than the length of the dipole), the vector r is
approximately parallel to the vector r − ˆzz′.
Subsequently , it must be true that
|r − ˆzz′| ≈ r− ˆr · ˆzz′ (9.76)
Note that the magnitude of r− ˆr · ˆzz′ must be
approximately equal to r, since r≫ L. So, insofa... | Electromagnetics_Vol2.pdf |
9.5. RADIA TION FROM AN ELECTRICALL Y -SHOR T DIPOLE 157
Equation 9.75 that exhibits varying phase is
e−jβ|r−ˆzz′|. Using Equation 9.76, we find
e−jβ|r−ˆzz′| ≈ e−jβre+jβˆr·ˆzz′
(9.78)
The worst case in terms of phase variation within the
integral is for field points along the zaxis. For these
points, ˆr · ˆz = ±1 and sub... | Electromagnetics_Vol2.pdf |
r
∫ +L/ 2
−L/ 2
˜I( z′)dz′
(9.80)
The integral in this equation is very easy to evaluate;
in fact, from inspection (Figure 9.6), we determine it
is equal to I0L/2. Finally , we obtain:
˜E(r) ≈ ˆθjηI0 · βL
8π (s in θ) e−jβr
r (9.81)
Summarizing:
The electric field intensity radiated by an ESD lo-
cated
at the origin and ... | Electromagnetics_Vol2.pdf |
one-half the integral over the uniform current
distribution that defines the Hertzian dipole. This
similarly sometimes causes confusion between
Hertzian dipoles and ESDs. Remember that ESDs are
physically realizable, whereas Hertzian dipoles are
not.
It is common to eliminate the factor of βin the
magnitude using the re... | Electromagnetics_Vol2.pdf |
indicate a Poynting vector ˜E × ˜H that is always
directed radially outward from the location of the
dipole. This confirms that power flow is always
directed radially outward from the dipole. Due to the
symmetry of the problem, Figures 9.9–9.12 provide a
complete characterization of the relative magnitudes
and orientatio... | Electromagnetics_Vol2.pdf |
158 CHAPTER 9. RADIA TION
c⃝ S. Lally CC BY -SA 4.0
Figure 9.9: Magnitude of the radiated field in any
plane of constant φ.
c⃝ S. Lally CC BY -SA 4.0
Figure 9.10: Orientation of the electric and magnetic
fields in any plane of constant φ.
c⃝ S. Lally CC BY -SA 4.0
Figure 9.11: Magnitude of the radiated field in any
plane ... | Electromagnetics_Vol2.pdf |
9.6. F AR-FIELD RADIA TION FROM A THIN STRAIGHT FILAMENT OF CURRENT 159
9.6 Far-Field Radiation from a
Thin Straight Filament of
Current
[m0199]
A simple distribution of radiating current that is
encountered
in common practice is the thin straight
current filament, shown in Figure 9.13. The defining
characteristic of thi... | Electromagnetics_Vol2.pdf |
Dipole”), so students may want to review that section
first. This section presents the more general case.
Let the magnitude and phase of current along the
filament be given by the phasor quantity ˜I(z) (SI base
units of A). In principle, the only constraint on ˜I(z)
c⃝ C. W ang CC BY -SA 4.0
Figure 9.13: A thin straight ... | Electromagnetics_Vol2.pdf |
as we shall do here.
There are two approaches that we might consider in
order to find the electric field radiated by this
distribution. The first approach is to calculate the
magnetic vector potential ˜A by integration over the
current distribution (Section 9.3), calculate
˜H = (1/µ)∇ × ˜A, and finally calculate ˜E from ˜H... | Electromagnetics_Vol2.pdf |
160 CHAPTER 9. RADIA TION
shown that a ˆz-directed Hertzian dipole at the origin
radiates the electric field
˜E(r) ≈ ˆθjη
˜I(β∆l )
4π (s in θ) e−jβr
r (9.84)
where ˜I and ∆
lmay be interpreted as the current and
length of the filament, respectively . In this expression,
ηis the wave impedance of medium in which the
filame... | Electromagnetics_Vol2.pdf |
of this segment shrink to differential length dz′, we
may describe the contribution of this segment to the
field radiated by the ESD as follows:
d˜E(r; z′ = 0) ≈ ˆθjη
˜I(0) (βdz′)
4π (s in θ) e−jβr
r(9.86)
Using
this approach, the electric field radiated by any
segment can be written:
d˜E(r; z′) ≈ ˆθ′jηβ
˜I(z′)
4π (s in ... | Electromagnetics_Vol2.pdf |
Figure 9.15: Parallel ray approximation.
Given some of the assumptions we have already
made, this expression can be further simplified. For
example, note that θ′ ≈ θsince L≪ r. For the same
reason, ˆθ′ ≈ ˆθ. Since these variables are
approximately constant over the length of the
filament, we may move them outside the int... | Electromagnetics_Vol2.pdf |
|r − ˆzz′| determines the magnitude of ˜E(r), we may
use the approximation:
|r − ˆzz′| ≈ r (magnitude) (9.93)
Insofar as |r − ˆzz′| determines phase, we have to be
more careful. The part of the integrand of
Equation 9.90 that exhibits varying phase is
e−jβ|r−ˆzz′|. Using Equation 9.91, we find
e−jβ|r−ˆzz′| ≈ e−jβre+jβz′... | Electromagnetics_Vol2.pdf |
9.7. F AR-FIELD RADIA TION FROM A HALF-W A VE DIPOLE 161
These simplifications are known collectively as a far
field approximation, since they are valid only for
distances “far” from the source.
Applying these simplifications for magnitude and
phase to Equation 9.90, we obtain:
˜E(r) ≈ ˆθjηβ
4π
e−jβ
r
r (s in θ)
·
∫ +L/ 2... | Electromagnetics_Vol2.pdf |
r≫ λ, the wave appears to be locally planar.
Therefore, we are justified using the plane wave
relationship ˜H = (1/η)ˆr × ˜E to calculate ˜H.
As a check, one may readily verify that Equation 9.96
yields the expected result for the electrically-short
dipole (Section 9.5).
9.7 Far-Field Radiation from a
Half-W ave Dipole
... | Electromagnetics_Vol2.pdf |
Note also that this “cosine pulse” distribution is very
similar to the triangular distribution of the ESD, and
is reminiscent of the sinusoidal variation of current in
a standing wave.
Since L= λ/2 for the HWD, Equation 9.97 may
equivalently be written:
˜I(z) ≈ I0 cos
(
2πz
λ
)
(9.98)
The
electromagnetic field radiated ... | Electromagnetics_Vol2.pdf |
162 CHAPTER 9. RADIA TION
of current may be calculated using the method
described in Section 9.6, in particular:
˜E(r) ≈ ˆθjη
2
e−jβ
r
r (s in θ)
·
[
1
λ
∫ +L/ 2
−L/ 2
˜I( z′)e+jβz′ cos θdz′
]
(9.99)
which is valid for field points r far from the dipole;
i.e., for r≫ Land r≫ λ. For the HWD, the quantity
in square bracke... | Electromagnetics_Vol2.pdf |
ηˆr × ˜E (9.103)
This
relationship indicates that the magnetic field will
be + ˆφ-directed.
The magnitude and polarization of the radiated field is
similar to that of the electrically-short dipole (ESD;
Section 9.5). A comparison of the magnitudes in any
radial plane containing the z-axis is shown in
Figure 9.17. For eit... | Electromagnetics_Vol2.pdf |
dipole also oriented along the z-axis. This result is for
any radial plane that includes the z-axis.
9.8 Radiation from Surface and
V olume Distributions of
Current
[m0221]
In Section 9.3, a solution was developed for radiation
from
current located at a single point r′. This current
distribution was expressed mathemati... | Electromagnetics_Vol2.pdf |
9.8. RADIA TION FROM SURF ACE AND VOLUME DISTRIBUTIONS OF CURRENT 163
any distribution of current that is constrained to flow
along a single path through space, as along an
infinitesimally-thin wire.
In this section, we derive an expression for the
radiation from current that is constrained to flow
along a surface and fro... | Electromagnetics_Vol2.pdf |
current can be described as a distribution of current
moments. By the principle of superposition, the
radiation from this distribution of current moments
can be calculated as the sum of the radiation from the
individual current moments. This is expressed
mathematically as follows:
˜A(r) =
∫
d˜A(r; r′) (9.108)
where the... | Electromagnetics_Vol2.pdf |
alternatively as follows:
d˜J(r) = ˆl ˜Js dsδ(r − r′) (9.109)
where ˜Js has units of surface current density (SI base
units of A/m) and dshas units of area (SI base units
of m 2). T o emphasize that this is precisely the same
current moment, note that ˜Js ds, like ˜I dl, has units
of A·m. Similarly ,
d˜J(r) = ˆl ˜J dvδ... | Electromagnetics_Vol2.pdf |
˜A(r) = µ
4π
∫
S
˜Js(r′) e−γ|r−r′|
|r − r′| ds (9.112)
where ˜Js(
r′) ≜ ˆl(r′) ˜Js(r′) and where S is the
surface over which the current flows. Similarly for a
volume current distribution, we obtain:
˜A(r) = µ
4π
∫
V
˜J(r′) e−γ|r−r′|
|r − r′| dv (9.113)
where ˜J
(r′) ≜ ˆl(r′) ˜J(r′) and where V is the volume
in which th... | Electromagnetics_Vol2.pdf |
164 CHAPTER 9. RADIA TION
Image Credits
Fig. 9.1: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:A z-directed current moment located at the origin.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.2: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:A z-directed... | Electromagnetics_Vol2.pdf |
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.6: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Current distribution of the esd.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.7: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Esd a... | Electromagnetics_Vol2.pdf |
Fig. 9.10: c⃝ Offaperry (S. Lally),
https://commons.wikimedia.org/wiki/File:Electric and Magnetic Fields in Plane of Constant Theta.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.11: c⃝ Offaperry (S. Lally), https://commons.wikimedia.org/wiki/File:FHplaneMag.svg,
CC BY -SA 4.0 (https://cre... | Electromagnetics_Vol2.pdf |
9.8. RADIA TION FROM SURF ACE AND VOLUME DISTRIBUTIONS OF CURRENT 165
Fig. 9.15: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Parallel ray approximation for an esd.svg,
CC
BY -SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 9.16: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wi... | Electromagnetics_Vol2.pdf |
Chapter 10
Antennas
10.1
How Antennas Radiate
[m0201]
An antenna is a tr ansducer; that is, a device which
converts signals in one form into another form. In the
case of an antenna, these two forms are (1)
conductor-bound voltage and current signals and (2)
electromagnetic waves. Traditional passive antennas
are capabl... | Electromagnetics_Vol2.pdf |
c⃝ S. Lally CC BY -SA 4.0
Figure 10.1: T win lead transmission line terminated
into an open circuit, giving rise to a standing wave.
applied to the input of an ideal twin lead transmission
line. The spacing between conductors is much less
than a wavelength, and the output of the transmission
line is terminated into an ... | Electromagnetics_Vol2.pdf |
situation over an interval of one-half of the period of
the sinusoidal source. For the other half-period, the
direction of current will be in the direction opposite
that depicted in Figure 10.1. In other words: The
source is varying periodically , so the sign of the
current is changing every half-period. Generally ,
ti... | Electromagnetics_Vol2.pdf |
Figure 10.1, note that each Hertzian dipole
representing current on one conductor has an
associated Hertzian dipole representing the current on
the other conductor, and is only a tiny fraction of a
wavelength distant. Furthermore, these pairs of
Electromagnetics V ol. 2. c⃝ 2020 S.W . Ellingson CC BY SA 4.0. https://do... | Electromagnetics_Vol2.pdf |
10.1. HOW ANTENNAS RADIA TE 167
Hertzian dipoles are identical in magnitude but
opposite in sign. Therefore, the radiated field from
any such pair of Hertzian dipoles is approximately
zero at distances sufficiently far from the transmission
line. Continuing to sum all such pairs of Hertzian
dipoles, the radiated field rem... | Electromagnetics_Vol2.pdf |
points close to the transmission line. Subsequently
the cancellation of fields from Hertzian dipole pairs is
less precise. The resulting sum fields are not
negligible and depend on the separation between the
conductors. For the present discussion, it suffices to
restrict our attention to the simpler “far field” case.
A sim... | Electromagnetics_Vol2.pdf |
fashioned into an electrically-short dipole (ESD).
they are bent. Since the modified section is much
shorter than one-half wavelength, the current
distribution on this section must be very simple. In
fact, the current distribution must be that of the
electrically-short dipole (ESD), exhibiting magnitude
which is maximum... | Electromagnetics_Vol2.pdf |
also that we need not interpret the ESD portion of the
transmission line as a modification of the
transmission line: Instead, we may view this system
as an unmodified transmission line attached to an
antenna, which in this case in an ESD.
The general case. Although we developed this
insight for the ESD specifically , the ... | Electromagnetics_Vol2.pdf |
168 CHAPTER 10. ANTENNAS
10.2 Power Radiated by an
Electrically-Short Dipole
[m0207]
In this section, we determine the total power radiated
by
an electrically-short dipole (ESD) antenna in
response to a sinusoidally-varying current applied to
the antenna terminals. This result is both useful on its
own and necessary as... | Electromagnetics_Vol2.pdf |
2π/λwhere λis wavelength. Note that L≪ λsince
this is an ESD. Also note that Equation 10.1 is valid
only for the far-field conditions r≫ Land r≫ λ,
and presumes propagation in simple (linear,
homogeneous, time-invariant, isotropic) media with
negligible loss.
Given that we have already limited scope to the far
field, it ... | Electromagnetics_Vol2.pdf |
integrated over an area (m 2) gives power (W).
Anticipating that this problem will be addressed in
spherical coordinates, we note that
ds = ˆrr2 sin θdθdφ (10.4)
and subsequently:
Prad =
∫ π
θ=0
∫ 2π
φ=0
ˆr
⏐⏐⏐˜E(r)
⏐⏐⏐
2
2η
·
(ˆrr2 sin θ dθdφ
)
= 1
2η
∫ π
θ=0
∫ 2
π
φ=0
⏐⏐⏐˜E(r)
⏐⏐⏐
2
r2 sin θdθdφ (10.5)
Return... | Electromagnetics_Vol2.pdf |
units. Recall that βhas SI base units of rad/m, so βL
has units of radians. This leaves η|I0|2, which has SI
base units of Ω · A2 = W , as expected.
The power radiated by an ESD in response to the
current I0 applied
at the terminals is given by
Equation 10.9.
Finally , it is useful to consider how various
parameters af... | Electromagnetics_Vol2.pdf |
10.3. POWER DISSIP A TED BY AN ELECTRICALL Y -SHOR T DIPOLE 169
that is, the length of the antenna expressed in radians,
where 2πradians is one wavelength. Thus, we see
that the power radiated by the antenna increases as the
square of electrical length.
Example 10.1. Po wer radiated by an ESD.
A dipole is 10 cm in leng... | Electromagnetics_Vol2.pdf |
radiated power is ≈ 98.6 µW
.
10.3
Power Dissipated by an
Electrically-Short Dipole
[m0208]
The power delivered to an antenna by a source
connected
to the terminals is nominally radiated.
However, it is possible that some fraction of the
power delivered by the source will be dissipated
within the antenna. In this secti... | Electromagnetics_Vol2.pdf |
along the zaxis. In Section 9.5, it is shown that the
current distribution is:
˜I(z) ≈ I0
(
1 − 2
L|z|
)
(10.10)
where I0 (SI
base units of A) is a complex-valued
constant indicating the maximum current magnitude
and phase, and Lis the length of the ESD. This
current distribution may be interpreted as a set of
discrete... | Electromagnetics_Vol2.pdf |
170 CHAPTER 10. ANTENNAS
calculated using Equation 4.17 (Section 4.2,
“Impedance of a Wire”):
Rseg ≈ 1
2
√
µf
πσ · ∆l
a (10.12)
where µis
permeability , f is frequency , σis
conductivity , and ∆l is the length of the segment.
Substitution into Equation 10.11 yields:
Pseg(zn) ≈ 1
4a
√
µf
πσ
⏐⏐
⏐˜I(zn)
⏐
⏐
⏐
2
∆l (10.13)... | Electromagnetics_Vol2.pdf |
any straight wire antenna of length L. For the ESD
specifically , the current distribution is given by
Equation 10.10. Making the substitution:
Ploss ≈ 1
4a
√
µf
πσ
∫ +L/ 2
z′=−L/ 2
⏐⏐
⏐
⏐I0
(
1 − 2
L|z′|
)⏐
⏐
⏐
⏐
2
dz′
≈ 1
4a
√
µf
πσ |I0|2
∫ +L/ 2
z′=−L/ 2
⏐
⏐⏐⏐1 − 2
L|z′|
⏐
⏐⏐⏐
2
dz′
(10.17)
The
integral is straightfo... | Electromagnetics_Vol2.pdf |
Rloss such that
Ploss = 1
2 |I0|2 Rlo ss (10.19)
Comparing Equations 10.18 and 10.19, we find
Rloss ≈ L
6a
√
µf
πσ (10.20)
The power dissipated within an ESD in response
to
a sinusoidal current I0 applied at the terminals
is 1
2 |I0|2 Rlo ss where Rloss (Equation 10.20) is
the resistance perceived by a source applied to... | Electromagnetics_Vol2.pdf |
comprised of aluminum having conductivity
≈ 3.7 × 107 S/m and µ≈ µ0. A sinusoidal
current having frequency 30 MHz and peak
magnitude 100 mA is applied to the antenna
terminals. What is the power dissipated within
this antenna?
Solution. The wavelength λ= c/f ∼= 10 m, so
L= 10 cm ∼= 0.01λ. This certainly qualifies as
ele... | Electromagnetics_Vol2.pdf |
10.4. REACT ANCE OF THE ELECTRICALL Y -SHOR T DIPOLE 171
find that the loss resistance Rlo ss ≈ 9.49 mΩ.
Subsequently , the power dissipated within this
antenna is
Ploss = 1
2 |I0|2 Rlo ss ≈ 47.5 µW (10.21)
W e
conclude this section with one additional caveat:
Whereas this section focuses on the limited
conductivity of ... | Electromagnetics_Vol2.pdf |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.