text stringlengths 1 1k ⌀ | source stringclasses 12
values |
|---|---|
to the resistance R; i.e.,
X ≈ R≈ l
σ(δs2πa) (4.20)
This
is unique to good conductors at AC; that is, we
see no such reactance at DC. Because this reactance is
positive, it is often referred to as an inductance.
However, this is misleading since inductance refers to
the ability of a structure to store energy in a magne... | Electromagnetics_Vol2.pdf |
δs ≪ a. The utility of this description is that it
facilitates the modeling of wire reactance as an
inductance in an equivalent circuit. Summarizing:
A practical wire may be modeled using an equiv-
alent
circuit consisting of an ideal resistor (Equa-
tion 4.17) in series with an ideal inductor (Equa-
tion 4.22). Wherea... | Electromagnetics_Vol2.pdf |
explained in Section 3.11.
Example 4.2. Equi valent inductance of the
inner conductor of RG-59.
Elsewhere in the book we worked out that the
inductance per unit length L′ of RG-59 coaxial
cable was about 370 nH/m. W e calculated this
from magnetostatic considerations, so the
reactance associated with skin effect is not... | Electromagnetics_Vol2.pdf |
54 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS
associated with skin effect is as important as the
magnetostatic
inductance in the kHz regime, and
becomes gradually less important with
increasing frequency .
Recall that the phase velocity in a low-loss
transmission line is approximately 1/
√
L′C′. This
means
that sk... | Electromagnetics_Vol2.pdf |
for
the impedance Zof a good conductor having
width W, length l, and which is infinitely deep:
Z ≈ 1 + j
σδs
· l
W (A C case) (4.25)
where σis conductivity (SI base units of S/m) and δs
is skin depth. Note that δs and σare constitutive
parameters of material, and do not depend on
geometry; whereas land W describe geomet... | Electromagnetics_Vol2.pdf |
Surface impedance ZS (Equation 4.26) is a ma-
terials property having units of Ω/□, and which
characterizes the AC impedance of a material in-
dependently of the length and width of the mate-
rial.
Surface impedance is often used to specify sheet
materials used in the manufacture of electronic and
semiconductor devices... | Electromagnetics_Vol2.pdf |
4.3. SURF ACE IMPEDANCE 55
Image Credits
Fig. 4.1: c⃝ Sevenchw (C. W ang),
https://commons.wikimedia.org/wiki/File:Current flow in cylinder new .svg,
CC
BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).
Fig. 4.2: Biezl, https://commons.wikimedia.org/wiki/File:Skin depth.svg, public domain. Modified from
origin... | Electromagnetics_Vol2.pdf |
Chapter 5
W a
ve Reflection and T ransmission
5.1 Plane W aves at Normal
Incidence on a Planar
Boundary
[m0161]
When a plane wave encounters a discontinuity in
media,
reflection from the discontinuity and
transmission into the second medium is possible. In
this section, we consider the scenario illustrated in
Figure 5.1:... | Electromagnetics_Vol2.pdf |
intensity ˜Ei of this wave is given by
˜Ei(z) = ˆxEi
0e−jβ1z , z≤ 0 (5.1)
where β1 = ω√
µ1ǫ1 is the phase propagation
constant in Region 1 and Ei
0 is a complex-valued
constant. ˜Ei serves as the “stimulus” in this problem.
That is, all other contributions to the total field may
be expressed in terms of ˜Ei. In fact, al... | Electromagnetics_Vol2.pdf |
geometrical symmetry of the problem, which
precludes waves traveling in any other directions. The
symmetry of the problem also precludes a change of
polarization, so the reflected wave should have no ˆy
component. Therefore, we may be confident that the
reflected electric field has the form
˜Er(z) = ˆxBe+jβ1z , z≤ 0 (5.3)
... | Electromagnetics_Vol2.pdf |
5.1. PLANE W A VES A T NORMAL INCIDENCE ON A PLANAR BOUNDAR Y 57
where Bis a complex-valued constant that remains to
be determined. Since the direction of propagation for
the reflected wave is −ˆz, we have from the plane
wave relationships that
˜Hr(z) = −ˆy B
η1
e+jβ1z , z≤ 0 (5.4)
Similarly
, we infer the existence of ... | Electromagnetics_Vol2.pdf |
respectively , in Region 2. The constant C, like B, is a
complex-valued constant that remains to be
determined.
At this point, the only unknowns in this problem are
the complex-valued constants Band C. Once these
values are known, the problem is completely solved.
These values can be determined by the application of
bo... | Electromagnetics_Vol2.pdf |
tangential component of the electric field across the
boundary requires ˜E1(0) = ˜E2(0), and therefore
˜Ei(0) + ˜Er(0) = ˜Et(0) (5.9)
Now employing Equations 5.1, 5.3, and 5.5, we
obtain:
Ei
0 + B = C (5.10)
Clearly a second equation is required to determine
both Band C. This equation can be obtained by
enforcing the bo... | Electromagnetics_Vol2.pdf |
steps are the same. W e define ˜H1 and ˜H2 to be the
total magnetic field intensities in Regions 1 and 2,
respectively . The total field in Region 1 is
˜H1(z) = ˜Hi(z) + ˜Hr(z) (5.11)
The field in Region 2 is simply
˜H2(z) = ˜Ht(z) (5.12)
The boundary condition requires ˜H1(0) = ˜H2(0),
and therefore
˜Hi(0) + ˜Hr(0) = ˜Ht(... | Electromagnetics_Vol2.pdf |
58 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
incidence from Region 1 toward Region 2. W e may
now solve for Cby substituting Equation 5.15 into
Equation 5.10. W e find:
C = (1 + Γ 12) Ei
0 (5.17)
Now summarizing the solution:
˜Er(z) = ˆxΓ 12Ei
0e+jβ1z , z≤ 0 (5.18)
˜Et(z) = ˆx (1
+ Γ 12) Ei
0e−jβ2z , z≥ 0 (5.19)
Equ... | Electromagnetics_Vol2.pdf |
A second case of practical interest is when Region 2
is a perfect conductor. First, note that this may seem
at first glance to be a violation of the “lossless”
assumption made at the beginning of this section.
While it is true that we did not explicitly account for
the possibility of a perfect conductor in Region 2,
let... | Electromagnetics_Vol2.pdf |
Thus, we obtained the correct answer because we
were able to independently determine that η2 = 0 in a
perfect conductor.
When Region 2 is a perfect conductor, the reflec-
tion
coefficient Γ 12 = −1 and the solution de-
scribed in Equations 5.18 and 5.19 applies.
It may be helpful to note the very strong analogy
between r... | Electromagnetics_Vol2.pdf |
load, and the special case η2 = 0, also considered
previously , is analogous to a short-circuit load.
In the case of transmission lines, we were concerned
about what fraction of the power was delivered into a
load and what fraction of the power was reflected
from a load. A similar question applies in the case of
plane w... | Electromagnetics_Vol2.pdf |
5.1. PLANE W A VES A T NORMAL INCIDENCE ON A PLANAR BOUNDAR Y 59
equal to the incident power density minus the
reflected power density . Thus:
St
ave = Si
ave − Sr
ave
=
(
1 − |Γ 12|2)
Si
ave (5.23)
In other words, the ratio of power density transmitted
into Region 2 to power density incident from
Region 1 is
St
ave
Sia... | Electromagnetics_Vol2.pdf |
of permittivity for these media. Assuming µ= µ0,
we find:
Γ 12 = η2 − η1
η2 + η1
=
√
µ0/ǫ2 −
√
µ0/ǫ1√
µ0/ǫ2 +
√
µ0/ǫ1
=
√ǫ1 − √ǫ2
√ǫ1 + √ǫ2
(5.25)
Moreo
ver, recall that permittivity can be expressed in
terms of relative permittivity; that is, ǫ= ǫrǫ0.
Making the substitution above and eliminating all
extraneous factors... | Electromagnetics_Vol2.pdf |
Moon’s surface as a planar boundary between
free space and a semi-infinite region of lossy
dielectric material. The reflection coefficient is
given approximately by Equation 5.26 with
ǫr1 ≈ 1 and ǫr2 ∼ 3. Thus, Γ 12 ∼ − 0.27.
Subsequently , the fraction of power reflected
from the Moon relative to power incident is
|Γ 12|2... | Electromagnetics_Vol2.pdf |
60 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.2 Plane W aves at Normal
Incidence on a Material Slab
[m0162]
In Section 5.1, we considered what happens when a
uniform
plane wave is normally incident on the planar
boundary between two semi-infinite media. In this
section, we consider the problem shown in Figure 5.2:
... | Electromagnetics_Vol2.pdf |
For consistency of terminology , let us refer to the
problem considered in Section 5.1 as the
“single-boundary” problem and the present (slab)
problem as the “double-boundary” problem. Whereas
there are only two regions (“Region 1” and “Region
2”) in the single-boundary problem, in the
double-boundary problem there is ... | Electromagnetics_Vol2.pdf |
In both problems, we presume an incident wave in
Region 1 incident on the boundary with Region 2
having electric field intensity
˜Ei(z) = ˆxEi
0e−jβ1z (Region 1) (5.28)
where β1 = ω√
µ1ǫ1 is the phase propagation
constant in Region 1. ˜Ei serves as the “stimulus” in
this problem. That is, all other contributions to the
... | Electromagnetics_Vol2.pdf |
be determined; and subsequently the reflected
magnetic field intensity is:
˜Hr(z) = −ˆy B
η1
e+jβ1z (Re gion 1) (5.31)
Similarly , we infer the existence of a transmitted
plane wave propagating in the +ˆz direction in
Region 2. The electric and magnetic field intensities
of this wave are given by:
˜Et2(z) = ˆxCe−jβ2z (Reg... | Electromagnetics_Vol2.pdf |
5.2. PLANE W A VES A T NORMAL INCIDENCE ON A MA TERIAL SLAB 61
complex-valued constant that remains to be
determined.
Now let us consider the boundary between Regions 2
and 3. Note that ˜Et2 is incident on this boundary in
precisely the same manner as ˜Ei is incident on the
boundary between Regions 1 and 2. Therefore, ... | Electromagnetics_Vol2.pdf |
propagation constant and wave impedance,
respectively , in Region 3. The constant F, like D, is a
complex-valued constant that remains to be
determined. W e infer no wave traveling in the (− ˆz) in
Region 3, just as we inferred no such wave in
Region 2 of the single-boundary problem.
For convenience, T able 5.1 shows a... | Electromagnetics_Vol2.pdf |
by enforcing boundary conditions on the electric and
magnetic fields at each of the two boundaries. As in
the single-boundary case, the relevant boundary
condition is that the total field should be continuous
across each boundary . Applying this condition to the
boundary at z= 0, we obtain:
˜Et2(0) + ˜Er2(0) = ˜Et(0) (5.... | Electromagnetics_Vol2.pdf |
Equations
5.42–5.45 are recognizable as a system of 4
simultaneous linear equations with the number of
unknowns equal to the number of equations. W e
could simply leave it at that, however, some very
useful insights are gained by solving this system of
equations in a particular manner. First, note the
resemblance betwe... | Electromagnetics_Vol2.pdf |
62 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
Electric Field Intensity Magnetic Field Intensity Region of V alidity
Region 1 ˜Ei(z) = ˆxEi
0e−j β1z ˜Hi(z) = + ˆy
(
Ei
0/η1
)
e−jβ1z z≤ − d
˜Er(z) = ˆxBe+jβ1z ˜Hr(z) = −ˆy (B/η1) e+jβ1z
Region 2 ˜Et2(z) = ˆxC e−jβ2z ˜Ht2(z) = + ˆy (C/η2) e−jβ2z −d≤ z≤ 0
˜Er2(z) = ˆxDe+... | Electromagnetics_Vol2.pdf |
two unknowns – a dramatic simplification!
However, even this is more work than is necessary ,
and a little cleverness at this point pays big dividends
later. The key idea is that we usually have no interest
in the fields internal to the slab; in most problems, we
are interested merely in reflection into Region 1 from
the ... | Electromagnetics_Vol2.pdf |
in Regions 2 or 3 of the original problem, so we
define a new wave impedance ηeq to represent this
new condition. If we can find an expression for ηeq,
then we can develop a solution to the original
(two-boundary) problem that looks like a solution to
the equivalent (simpler, single-boundary) problem.
c⃝ C. W ang CC BY -... | Electromagnetics_Vol2.pdf |
Bringing the factor of η2 to the front and substituting
D= Γ 23C:
ηeq = η2
Ce+jβ2d + Γ 23Ce−jβ2d
Ce+jβ2d − Γ 23C e−jβ2d (5.54)
Finally , we divide out the common factor of Cand
multiply numerator and denominator by e−jβ2d, | Electromagnetics_Vol2.pdf |
5.2. PLANE W A VES A T NORMAL INCIDENCE ON A MA TERIAL SLAB 63
yielding:
ηeq = η2
1 + Γ 23e−j2β2d
1 − Γ 23e−j2 β2d (5.55)
Equation 5.55 is the wave impedance in the re-
gion
to the right of the boundary in the equiva-
lent scenario shown in Figure 5.3. “Equivalent”
in this case means that the incident and reflected
field... | Electromagnetics_Vol2.pdf |
the complex value of the wave impedance is not
indicating loss. Instead, the non-zero phase of ηeq
represents the ability of the standing wave inside the
slab to impart a phase shift between the electric and
magnetic fields. This is precisely the same effect that
one observes at the input of a transmission line: The
inp... | Electromagnetics_Vol2.pdf |
Figure 5.3:
Γ 1,eq ≜ ηeq − η1
ηeq + η1
(5.56)
The
quantity Γ 1,eq may now be used precisely in the
same way as Γ 12 was used in the single-boundary
problem to find the reflected fields and reflected
power density in Region 1.
2 See the section “Input Impedance of a T erminated Lossless
Transmission Line” for a reminder. Th... | Electromagnetics_Vol2.pdf |
3 as approximately free space, and Region 2 as
the pane. Thus
η1 = η3 = η0 ∼= 376.7 Ω (5.57)
and
η2 = η0
√ǫr
∼
= 188
.4 Ω (5.58)
The reflection coefficient from Region 2 to
Region 3 is
Γ 23 = η3 − η2
η3 + η2
= η0 − η2
η0 + η2
∼= 0.3333 (5.59)
Gi
ven f = 2.45 GHz, the phase propagation
constant in the glass is
β2 = 2π
λ ... | Electromagnetics_Vol2.pdf |
64 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
The ratio of reflected power density to incident
po
wer density is simply the squared magnitude
of this reflection coefficient, i.e.:
Sr
ave
Siav e
= |Γ 1,eq |2 = 0.292 ∼= 29.2% (5.63)
where Sr
av
e and Si
ave are the reflected and
incident power densities, respectively .
Si... | Electromagnetics_Vol2.pdf |
from the z= −dinterface must be transmitted into
Region 3. In other words:
St
ave
Siav e
= 1 − |Γ 1,eq |2 (5.65)
where St
av
e is the transmitted power density .
Example 5.3. Transmission of WiFi through a
glass pane.
Continuing Example 5.2: What fraction of
incident power passes completely through the
glass pane?
Solu... | Electromagnetics_Vol2.pdf |
loss, so that the media in each region is characterized
entirely by permittivity and permeability .
W e now focus on a particular class of applications
involving this structure. In this class of applications,
we seek total transmission through the slab. By “total
transmission” we mean 100% of power incident on
the slab... | Electromagnetics_Vol2.pdf |
then shall provide some examples of these
applications.
W e begin with characterization of the media.
Region 1 is characterized by its permittivity ǫ1 and
permeability µ1, such that the wave impedance in
Region 1 is η1 =
√
µ1/ǫ1. Similarly , Region 2 is
characterized by its permittivity ǫ2 and permeability
µ2, such tha... | Electromagnetics_Vol2.pdf |
5.3. TOT AL TRANSMISSION THROUGH A SLAB 65
the reflection coefficient
Γ 1,eq = ηeq − η1
ηeq + η1
(5.67)
where ηeq is
given by
ηeq = η2
1 + Γ 23e−j2β2d
1 − Γ 23e−j2 β2d (5.68)
and where Γ 23 is given by
Γ 23 = η3 − η2
η3 + η2
(5.69)
T
otal transmission requires that Γ 1,eq = 0. From
Equation 5.67 we see that Γ 1,eq is zer... | Electromagnetics_Vol2.pdf |
the numerator and denominator, we
obtain:
η1 = η2
(1 + P)η3 + (1 − P)η2
(1 − P)η3 + (1 + P)η2
(5.74)
The parameters η1, η2, η3, β2, and ddefining any
single-slab structure that exhibits total transmis-
sion must satisfy Equation 5.74.
Our challenge now is to identify combinations of
parameters that satisfy this conditio... | Electromagnetics_Vol2.pdf |
always 1, so the first value of P you might think to try
is P = +1. In fact, this value satisfies Equation 5.75.
Therefore, e−j2β2d = +1. This new condition is
satisfied when 2β2d= 2πm, where m= 1,2,3,...
(W e do not consider m≤ 0 to be valid solutions since
these would represent zero or negative values of d.)
Thus, we fin... | Electromagnetics_Vol2.pdf |
the media in Regions 1 and 3 is that they have equal
wave impedance.
Example 5.4. Radome design by half-wave
matching.
The antenna for a 60 GHz radar is to be
protected from weather by a radome panel
positioned directly in front of the radar. The
panel is to be constructed from a low-loss
material having µr ≈ 1 and ǫr ... | Electromagnetics_Vol2.pdf |
66 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
used?
Solution
. This is a good application for
half-wave matching because the material on
either side of the slab is the same (presumably
free space) whereas the material used for the slab
is unspecified. The phase velocity in the slab is
vp = c
√ǫr
∼= 1.5 × 108 m/s (5.7... | Electromagnetics_Vol2.pdf |
or decreases from the design frequency , there will be
increasing reflection and decreasing transmission.
Quarter-wave matching requires that the wave
impedances in each region are different and related in
a particular way . The quarter-wave solution is
obtained by requiring P = −1, so that
η1 = η2
(1 + P)η3 + (1 − P)η2... | Electromagnetics_Vol2.pdf |
ing.
Example 5.5. Radome design by quarter-wave
matching.
The antenna for a 60 GHz radar is to be
protected from weather by a radome panel
positioned directly in front of the radar. In this
case, however, the antenna is embedded in a
lossless material having µr ≈ 1 and ǫr = 2, and
the radome panel is to be placed betwe... | Electromagnetics_Vol2.pdf |
given by
η2 = η0
√ǫr
⇒ ǫr =
(η0
η2
)2
∼= 1.41 (5.83)
The
phase velocity in the slab will be
vp = c√ǫr
∼= 3 × 108 m/s√
1.41
∼
= 2
.53 × 108 m/s
(5.84)
so the wavelength in the slab is
λ2 = vp
f
∼
= 2.53 × 108 m/s
60 × 109 Hz
∼
= 4
.20 mm (5.85) | Electromagnetics_Vol2.pdf |
5.4. PROP AGA TION OF A UNIFORM PLANE W A VE IN AN ARBITRAR Y DIRECTION 67
Thus, the minimum possible thickness of the
radome
panel is d= λ2/4 ∼= 1.05 mm, and the
relati
ve permittivity of the radome panel must be
ǫr ∼= 1.41.
Additional
Reading:
• “Radome” on Wikipedia.
5.4 Propagation of a Uniform
Plane W ave in an Ar... | Electromagnetics_Vol2.pdf |
in the +ˆz direction through simple lossless media.
Note that electric field intensity vector is linearly
polarized in a direction parallel to ˆx. Depending on
the position in space (and, for the physical
time-domain waveform, time), ˜E points either in the
+ˆx direction or the −ˆx direction. Let us be a bit more
specifi... | Electromagnetics_Vol2.pdf |
68 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
c⃝ C. W ang CC BY -SA 4.0
Figure 5.5: The same plane wave described in a ro-
tated coordinate system, yielding Equation 5.87.
c⃝ C. W ang CC BY -SA 4.0
Figure 5.6: The same plane wave described in yet an-
other rotation of the coordinate system, yielding Equa-
tion 5.88.... | Electromagnetics_Vol2.pdf |
changed.
Let us now rotate the +zaxis of the coordinate
system around the xaxis into the position originally
occupied by the −zaxis. Now the very same wave is
expressed as
˜E = + ˆyE0e+jβz (5.88)
This is illustrated in Figure 5.6. At first glance, it
appears that the direction of propagation has reversed;
but, again, it... | Electromagnetics_Vol2.pdf |
Cartesian coordinate system. One situation in which
we face this complication is when the wave is
obliquely incident on a surface. In this case, it is
impossible to select a single orientation of the
coordinate system in which the directions of
propagation, reference polarization, and surface
normal can all be describe... | Electromagnetics_Vol2.pdf |
characteristics of the wave, as opposed to being
determined arbitrarily and separately from the
characteristics of the wave. The ray-fixed
representation of a uniform plane wave is:
˜E(r) = ˆeE0e−jk·r
(5.89)
This
is illustrated in Figure 5.7. In this representation,
r is the position at which ˜E is evaluated, ˆe is the
... | Electromagnetics_Vol2.pdf |
5.4. PROP AGA TION OF A UNIFORM PLANE W A VE IN AN ARBITRAR Y DIRECTION 69
Thus, k · r = βz, as expected.
In ray-fixed coordinates, a wave can be represented
by one – and only one – expression, which is the same
expression regardless of the orientation of the
“global” coordinate system. Moreover, only two basis
directio... | Electromagnetics_Vol2.pdf |
ky ≜ βˆk · ˆy (5.94)
kz ≜ βˆk · ˆz (5.95)
Then,
k · r = kxx+ kyy+ kzz (5.96)
With this expression in hand, Equation 5.89 may be
rewritten as:
˜E = ˆeE0e−jkxxe−jky ye−jkz z (5.97)
If desired, one can similarly decompose ˆe into its
Cartesian components as follows:
ˆe = ( ˆe · ˆx) ˆx + (ˆe · ˆy) ˆy + (ˆe · ˆz) ˆz (5.98)
... | Electromagnetics_Vol2.pdf |
Propagating away from the z-axis means
ˆk = ˆx cos φ+ ˆy sin φ (5.99)
where φindicates the specific direction of
propagation. For example, φ= 0 yields
ˆk = + ˆx, φ= π/2 yields ˆk = + ˆy, and so on.
Therefore,
k ≜ βˆk = β(ˆx cos φ+ ˆy sin φ) (5.100)
For completeness, note that the following factor
appears in the phase-de... | Electromagnetics_Vol2.pdf |
70 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.5 Decomposition of a W ave into
TE and TM Components
[m0166]
A broad range of problems in electromagnetics
in
volve scattering of a plane wave by a planar
boundary between dissimilar media. Section 5.1
(“Plane W aves at Normal Incidence on a Planar
Boundary Between Los... | Electromagnetics_Vol2.pdf |
added complexity is easily handled if we take the
effort to represent the incident wave as the sum of two
waves having particular polarizations. These
polarizations are referred to as transverse electric
(TE) and transverse magnetic (TM). This section
describes these polarizations and the method for
decomposition of a ... | Electromagnetics_Vol2.pdf |
propagation ( ˆki) lie.
The TE-TM decomposition consists of finding the
components of the electric and magnetic fields which
are perpendicular (“transverse”) to the plane of
incidence. Of the two possible directions that are
perpendicular to the plane of incidence, we choose
ˆe⊥, defined as shown in Figure 5.8. From the fi... | Electromagnetics_Vol2.pdf |
in the new , ray-fixed coordinate system. W e begin
with the following phasor representation of the
electric field intensity:
˜Ei = ˆeiEi
0e−jki·r (5.105)
where ˆei is a unit vector indicating the reference
polarization, Ei
0 is a complex-valued scalar, and r is a
vector indicating the position at which ˜Ei is
evaluated.... | Electromagnetics_Vol2.pdf |
5.5. DECOMPOSITION OF A W A VE INTO TE AND TM COMPONENTS 71
The electric field vector is always perpendicular to the
direction of propagation, so ˆei · ˆki = 0. This leaves:
ˆei =
(ˆei · ˆe⊥
)ˆe⊥ +
(
ˆei · ˆei
∥
)
ˆei
∥ (5.107)
Substituting this expression into Equation 5.105, we
obtain:
˜Ei = ˆe⊥Ei
TEe−jki·r + ˆei
∥ Ei... | Electromagnetics_Vol2.pdf |
apparent since the magnetic field vector is
perpendicular to both the direction of propagation and
the electric field vector.
Summarizing:
The TE component is the component for which
˜Ei is
perpendicular to the plane of incidence.
The TM component is the component for which
˜Hi is
perpendicular to the plane of incidence;... | Electromagnetics_Vol2.pdf |
whereas direct analysis of the scattering of
arbitrarily-polarized waves is relatively difficult.
Note that the nomenclature “TE” and “TM” is
commonly but not universally used. Sometimes “TE”
is referred to as “perpendicular” polarization,
indicated using the subscript “⊥ ” or “s” (short for
senkrecht, German for “perpe... | Electromagnetics_Vol2.pdf |
of the field, and similarly the wave inside a waveguide
may consist of multiple unique TM modes which
collectively comprise the TM component of the field.
Finally , consider what happens when a plane wave is
normally-incident upon the boundary; i.e., when
ˆki = −ˆn. In this case, Equation 5.103 indicates that
ˆe⊥ = 0, so... | Electromagnetics_Vol2.pdf |
72 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.6 Plane W aves at Oblique
Incidence on a Planar
Boundary: TE Case
[m0167]
In this section, we consider the problem of reflection
and
transmission from a planar boundary between
semi-infinite media for a transverse electric (TE)
uniform plane wave. Before attempting this ... | Electromagnetics_Vol2.pdf |
Region 1. The electric field intensity ˜Ei
TE of this
wave is given by
˜Ei
TE(r) = ˆyEi
TEe−jki·r (5.111)
In this expression, r is the position at which ˜Ei
TE is
evaluated, and
ki = ˆkiβ1 (5.112)
c⃝ C. W ang CC BY -SA 4.0
Figure 5.9: A TE uniform plane wave obliquely inci-
dent on the planar boundary between two semi-i... | Electromagnetics_Vol2.pdf |
reflected and transmitted components of the electric
field will have the same polarization as that of the
incident electric field. This is because there is nothing
present in the problem that could account for a change
in polarization. Thus, the reflected and transmitted
fields will also be TE. Therefore, we postulate the
f... | Electromagnetics_Vol2.pdf |
propagation constant in Region 2.
At this point, the unknowns in this problem are the
constants Band C, as well as the directions ˆkr and
ˆkt. W e may establish a relationship between Ei
TE, B,
and Cby application of boundary conditions at
z= 0. First, recall that the tangential component of
the total electric field int... | Electromagnetics_Vol2.pdf |
5.6. PLANE W A VES A T OBLIQUE INCIDENCE ON A PLANAR BOUNDAR Y : TE CASE 73
total field in Region 1 is the sum of incident and
reflected fields, so
˜E1(r) = ˜Ei
TE(r) + ˜Er(r) (5.117)
The total field in Region 2 is simply
˜E2(r) = ˜Et(r) (5.118)
Next, note that all electric field components are
already tangent to the bounda... | Electromagnetics_Vol2.pdf |
boundary , it must be true that
ki · r0 = kr · r0 = kt · r0 (5.122)
Essentially , we are requiring the phases of each field
in Regions 1 and 2 to be matched at every point along
the boundary . Any other choice will result in a
violation of boundary conditions at some point along
the boundary . This phase matching criter... | Electromagnetics_Vol2.pdf |
netic field components associated with the TE electric
field components shown in Figure 5.9.
obtained by applying the appropriate boundary
conditions to the magnetic field. The magnetic field
associated with each of the electric field components
is identified in Figure 5.10. Note the orientations of
the magnetic field vectors... | Electromagnetics_Vol2.pdf |
Thus:
˜Hi(r) =
(ˆz sin ψi − ˆx cos ψi)Ei
TE
η1
e−jki·r (5.126)
Similarly
, we determine that the reflected magnetic
field has the form:
˜Hr(r) = 1
η1
ˆkr × ˜Er (5.127) | Electromagnetics_Vol2.pdf |
74 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
In the global coordinate system:
ˆkr = ˆx sin ψr − ˆz cos ψr (5.128)
Thus:
˜Hr(r) = ( ˆz sin ψr + ˆx cos ψr) B
η1
e−jkr ·r (5.129)
The
transmitted magnetic field has the form:
˜Ht(r) = 1
η2
ˆkt × ˜Et (5.130)
In
the global coordinate system:
ˆkt = ˆx sin ψt + ˆz cos ψt (5.... | Electromagnetics_Vol2.pdf |
where “ ˆx·” selects the component of the magnetic
field that is tangent to the boundary . Evaluating this
expression, we obtain:
−
(
cos ψi)Ei
TE
η1
e−jki·r0
+ (
cos ψr) B
η1
e−jkr ·r0
= −
(
cos ψt)C
η2
e−jkt·r0 (5.136)
No
w employing the phase matching condition
expressed in Equation 5.122, we find:
−
(
cos ψi)Ei
TE
η1... | Electromagnetics_Vol2.pdf |
follows:
B = Γ TEEi
TE (5.140)
where
Γ TE ≜ η2 cos ψi − η1 cos ψt
η2 cos ψr + η1 cos ψt (5.141)
It
is worth noting that Equation 5.141 becomes the
reflection coefficient for normal (TEM) incidence
when ψi = ψr = ψt = 0, as expected.
Returning to Equation 5.123, we now find
C = (1 + Γ TE) Ei
TE (5.142)
Let us now summarize... | Electromagnetics_Vol2.pdf |
5.6. PLANE W A VES A T OBLIQUE INCIDENCE ON A PLANAR BOUNDAR Y : TE CASE 75
simply state the result, and in Section 5.8 we shall
perform this part of the derivation in detail and with
greater attention to the implications. One finds:
ψr = ψi (5.145)
and
ψt = arcsin
(β1
β2
sin ψi
)
(5.146)
Equation
5.145 is the unsurpris... | Electromagnetics_Vol2.pdf |
finding of this section:
The electric field reflection coefficient for oblique
TE
incidence, Γ TE, is given by Equation 5.147.
The following example demonstrates the utility of this
result.
Example 5.7. Po wer transmission at an
air-to-glass interface (TE case).
Figure 5.11 illustrates a TE plane wave incident
from air ont... | Electromagnetics_Vol2.pdf |
air to glass.
η2 ≈ η0√
2.1
∼= 260 .0 Ω (glass) (5.149)
ψi = 30 ◦ (5.150)
Note
β1
β2
≈ ω√µ0ǫ0
ω√µ0 · 2.1ǫ0
∼
= 0.690 (5.151)
so
ψt =
arcsin
(β1
β2
sin ψi
)
∼
= 20
.2◦ (5.152)
Now substituting these values into
Equation 5.147, we obtain
Γ TE ∼= −0.2220 (5.153)
Subsequently , the fraction of power reflected
relative to po... | Electromagnetics_Vol2.pdf |
76 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
5.7 Plane W aves at Oblique
Incidence on a Planar
Boundary: TM Case
[m0164]
In this section, we consider the problem of reflection
and
transmission from a planar boundary between
semi-infinite media for a transverse magnetic (TM)
uniform plane wave. Before attempting this ... | Electromagnetics_Vol2.pdf |
z= 0 plane. The wave is incident from Region 1. The
magnetic field intensity ˜Hi
TM of this wave is given by
˜Hi
TM(r) = ˆyHi
TMe−jki·r (5.154)
In this expression, r is the position at which ˜Hi
TM is
c⃝ C. W ang CC BY -SA 4.0
Figure 5.12: A TM uniform plane wave obliquely
incident on the planar boundary between two sem... | Electromagnetics_Vol2.pdf |
reflected and transmitted components of the magnetic
field will have the same polarization as that of the
incident electric field. This is because there is nothing
present in the problem that could account for a
change in polarization. Thus, the reflected and
transmitted fields will also be TM. So we postulate
the following... | Electromagnetics_Vol2.pdf |
specific convention, and will be incorrect with respect
to the opposite convention. W e choose −ˆy because it
has a particular advantage which we shall point out at
the end of this section.
Continuing, we postulate the following expression for
the transmitted wave:
˜Ht(r) = ˆyCe−jkt·r (5.158)
where Cis an unknown, possi... | Electromagnetics_Vol2.pdf |
5.7. PLANE W A VES A T OBLIQUE INCIDENCE ON A PLANAR BOUNDAR Y : TM CASE 77
where ˆkt is the unit vector indicating the direction of
propagation and β2 = ω√µ2ǫ2 is the phase
propagation constant in Region 2.
At this point, the unknowns in this problem are the
constants Band C, as well as the unknown directions
ˆkr and ... | Electromagnetics_Vol2.pdf |
˜H2(r) = ˜Ht(r) (5.161)
Also, we note that all magnetic field components are
already tangent to the boundary . Thus, continuity of
the tangential component of the magnetic field across
the boundary requires ˜H1(r0) = ˜H2(r0), where
r0 ≜ ˆxx+ ˆyysince z= 0 on the boundary .
Therefore,
˜Hi
TM(r0) + ˜Hr(r0) = ˜Ht(r0) (5.162... | Electromagnetics_Vol2.pdf |
the boundary . This expression allows us to solve for
the directions of propagation of the reflected and
transmitted fields, which we shall do later. Our
priority for now shall be to solve for the coefficients
Band C.
Enforcing Equation 5.165, we observe that
Equation 5.164 reduces to:
Hi
TM − B = C (5.166)
A second equat... | Electromagnetics_Vol2.pdf |
propagation. Expressions for each of the electric field
components is determined formally below .
From the plane wave relationships, we determine that
the incident electric field intensity is
˜Ei(r) = −η1 ˆki × ˜Hi
TM (5.167)
where η1 =
√
µ1/ǫ1 is the wave impedance in
Region 1. T o make progress requires that we express... | Electromagnetics_Vol2.pdf |
78 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
The transmitted magnetic field has the form:
˜Et(r) = −η2 ˆkt × ˜Ht (5.173)
In the global coordinate system:
ˆkt = ˆx sin ψt + ˆz cos ψt (5.174)
Thus:
˜Et(r) =
(ˆx cos ψt − ˆz sin ψt)
η2Ce−jkt·r (5.175)
The total electric field in Region 1 is the sum of
incident and reflect... | Electromagnetics_Vol2.pdf |
= +
(
cos ψt)
η2Ce−jkt·r0 (5.179)
Now employing the “phase matching” condition
expressed in Equation 5.165, we find:
+
(
cos ψi)
η1Hi
TM
+ (cos ψr) η1B
= +
(
cos ψt)
η2C (5.180)
Equations 5.166 and 5.180 comprise a linear system
of equations with unknowns Band C. This system of
equations is easily solved for Bas follows... | Electromagnetics_Vol2.pdf |
TM ≜ η1Hi
TM in Equation 5.169,
yielding:
˜Ei(r) =
(ˆx cos ψi − ˆz sin ψi)
Ei
TMe−jki·r (5.185)
The factor η1Bin Equation 5.172 becomes
Γ TMEi
TM, so we obtain:
˜Er(r) = ( ˆx cos ψr + ˆz sin ψr)
· Γ TMEi
TMe−jkr ·r (5.186)
Thus, we see Γ TM is the reflection coefficient for the
electric field intensity .
Returning to Equa... | Electromagnetics_Vol2.pdf |
5.7. PLANE W A VES A T OBLIQUE INCIDENCE ON A PLANAR BOUNDAR Y : TM CASE 79
can be found using Equation 5.165. Here we shall
simply state the result, and in Section 5.8 we shall
perform this part of the derivation in detail and with
greater attention to the implications. One finds:
ψr = ψi (5.189)
i.e., angle of reflecti... | Electromagnetics_Vol2.pdf |
reflection coefficient for normal (TEM) incidence
when ψi = ψt = 0. If we had chosen +ˆy as the
reference polarization for Hr, we would have instead
obtained an expression for Γ TM that has the opposite
sign for TEM incidence. 3 There is nothing wrong
with this answer, but it is awkward to have different
values of the re... | Electromagnetics_Vol2.pdf |
Thus,
we obtain what is perhaps the most important
finding of this section:
The electric field reflection coefficient for oblique
TM
incidence, Γ TM, is given by Equation 5.191.
3 Obtaining this result is an excellent way for the student to con-
firm their understanding of the derivation presented in this section.
The follo... | Electromagnetics_Vol2.pdf |
Equation 5.191. Calculating the quantities that
enter into this expression:
η1 ≈ η0 ∼= 376.7 Ω (air) (5.192)
η2 ≈ η0
√
2.1
∼
= 260
.0 Ω (glass) (5.193)
ψi = 30 ◦ (5.194)
Note
β1
β2
≈ ω√µ0ǫ0
ω√µ0 · 2.1ǫ0
∼
= 0.690 (5.195)
so
ψt =
arcsin
(β1
β2
sin ψi
)
∼
= 20
.2◦ (5.196)
Now substituting these values into
Equation 5.1... | Electromagnetics_Vol2.pdf |
80 CHAPTER 5. W A VE REFLECTION AND TRANSMISSION
c⃝ C. W ang CC BY -SA 4.0
Figure 5.13: A TM uniform plane wave incident from
air to glass.
|Γ TM|2 ∼= 0 .021; i.e., about 2.1% .
1 − |Γ T
M|2 ∼
= 97.9%
of the power is transmitted
into
the glass.
Note that the result obtained in the preceding example
is different from th... | Electromagnetics_Vol2.pdf |
Let a point on the boundary be represented as the
position vector
r0 = ˆxx+ ˆyy (5.198)
In both Sections 5.6 (“Plane W aves at Oblique
Incidence on a Planar Boundary: TE Case”) and 5.7
(“Plane W aves at Oblique Incidence on a Planar
Boundary: TM Case”), it is found that
ki · r0 = kr · r0 = kt · r0 (5.199)
In this expre... | Electromagnetics_Vol2.pdf |
5.8. ANGLES OF REFLECTION AND REFRACTION 81
respectively; and β1 and β2 are the phase propagation
constants in Region 1 (from which the wave is
incident) and Region 2, respectively . Equation 5.199
is essentially a boundary condition that enforces
continuity of the phase of the electric and magnetic
fields across the bo... | Electromagnetics_Vol2.pdf |
Therefore, we may express Equation 5.199 in the
following form
β1
(ˆx sin ψi + ˆz cos ψi)
· (ˆxx+ ˆyy)
= β1 (ˆx sin ψr − ˆz cos ψr) · (ˆxx+ ˆyy)
= β2
(ˆx sin ψt + ˆz cos ψt)
· (ˆxx+ ˆyy) (5.206)
which reduces to
β1 sin ψi = β1 sin ψr = β2 sin ψt (5.207)
Examining the first and second terms of
Equation 5.207, and noting ... | Electromagnetics_Vol2.pdf |
of the relative values of constitutive parameters; i.e.,
µ1 = µr1µ0, ǫ1 = ǫr1ǫ0, µ2 = µr2µ0, and
ǫ2 = ǫr2ǫ0. In terms of the relative parameters:
√
µr1ǫr1 sin ψi = √µr2ǫr2 sin ψt (5.211)
This
is known as Snell’s law or the law of refraction.
Refraction is simply transmission with the result that
the direction of propag... | Electromagnetics_Vol2.pdf |
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