Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition

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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

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Solution Manual for Engineering Electromagnetics and Waves:
Advanced Topics
Umran S. Inan, Aziz S. Inan, Ryan K. Said

Contents
A Transmission Lines: Advanced Topics 1
C Cylindrical Waveguides 9
D Cavity Resonators 37
E Field-Matter Interactions: Advanced Topics 45
F Electromagnetic Radiation and Elementary Antennas 58

Solutions to Addendum A
Transmission Lines: Advanced Topics
A.1 Nonlinear termination.
We start by plotting (see Figure A.1) the current-voltage characteristics of the source and load ends
of the 50Ω line given by
5 = 10)s + 9s and )L = 0:35(1 e9L=2)
At t = 0, a voltage of amplitude
9+
1 = Z0
Rs + Z0
V0 = 50
10 + 50 (5 V) ≃ 4:17 V
and an accompanying current )+
1 = 9+
1 =Z0 ≃ 0:0833 A are launched at the source end of the line,
as indicated by point A which is the intersection of the line 9+
1 = Z0)+
1 (i.e., line OA) with the
source-end line, defined by 5 = 10)s + 9s. Next, we draw a straight line (line AB) with slope
[Z0]1 = [50]1 passing through point A and find its intersection with the current-voltage
characteristics of the nonlinear load which is denoted as point B, as shown. The coordinates of
point B which correspond to the values of 9L and )L at t = t+
d = 2+ ns are approximately given
by 9L ≃ 1:072 V and )L ≃ 0:145 A respectively. Using the value of 9L, the amplitudes of the
reflected voltage and current can be determined as 9
1 = 9L 9+
1 ≃ 1:072 4:167 = 3:095 V
and )
1 = 9
1 =Z0 ≃ +0:0619 A respectively. We then draw a straight line (line BC) with slope
+[Z0]1 and passing through point B and find its intersection with the source-end line indicated
as point C, as shown in Figure A.1. This process continues on until steady-state is reached at
approximately point G or at t ≃ 8td = 16 ns. Note that at steady-state, the source- and load-end
voltages and currents become 9s = 9L ≃ 2:50 V and )s = )L ≃ 0:25 A which correspond to the
intersection point of the source- and load-end voltage-current characteristics as indicated in Figure
A.1.

2 Addendum A/ Transmission Lines: Advanced Topics
Figure A.1 Figure for Problem A.1. Application of the graphical Bergeron tech-
nique to Problem A.1. The points A,B,C,D,E,F, and G correspond respectively to times
td; 2td; 3td; 4td; 5td; 6td, and 7td.
A.2 Nonlinear source.
(a) We start by plotting the current-voltage characteristics of the source and load ends of the 50Ω
line given by
)s = 8(5 9s) (5 9s)2 and )L = R1
L 9L ≃ 0
Note that the current-voltage characteristics of the load end of the line is approximately the
horizontal voltage axis since RL ≃ 10 MΩ ≃ 1. At t = 0, an incident voltage of amplitude
9+
1 ≃ 0:798 V and its current counterpart of amplitude )+
1 = Z019+
1 ≃ 16:0 mA (which are
the coordinates of point A, the intersection point between line OA and the nonlinear source-end
current-voltage characteristics as shown) are launched at the source end of the 50Ω line. Next, we
draw a straight line (line AB) with slope [Z0]1 = [50]1 passing through point A and find its
intersection with the load-end current-voltage characteristics which is denoted as point B on the
horizontal voltage axis as shown. The coordinates of point B ()B = 0 mA and 9B ≃ 1:60 V)
correspond to the values of )L and 9L at t = 2+ ns from which the amplitudes of the reflected
current and voltage from the load end can be determined as )
1 = )L )+
1 = 0 16 = 16 mA

3
and 9
1 = 9L 9+
1 ≃ 1:6 0:798 ≃ 0:798 V respectively. We then draw another straight line
(line BC) with slope +[Z0]1 passing through point B and find its intersection with the nonlinear
source-end current-voltage characteristics indicated as point C as shown in Figure A.2. Using these
coordinates (9C ≃ 2:31 V; )C ≃ 14:3 mA), the amplitudes of the current and voltage reflected
from the source end can be determined as )+
2 = )C )+
1 )
1 ≃ 14:3 16 + 16 = 14:3 mA
and 9+
2 = 9C 9+
1 9
1 ≃ 2:31 2 0:798 ≃ 0:714 V respectively. This process continues
on. Using the voltage coordinates of points B, D, F, etc., the load-end voltage can be sketched with
respect to time as shown.
(b) It can be approximately observed from the Bergeron diagram in Figure A.2 that the steady state
condition is reached in t ≃ 12td = 24 ns, or at point M.
Figure A.2 Figure for Problem A.2. Application of the graphical Bergeron technique
and 9L versus t.
A.3 Effects of source risetime.
The one-way time delay along the trace is given by td = (80 ps-cm1)(25 cm) = 2 ns. At t = 0,
an incident voltage of amplitude

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4 Addendum A/ Transmission Lines: Advanced Topics
9+
1 (t) =
8
><
>:
Z0V0
(Rs + Z0)tr
t =
( 4
tr
)
t t tr
Z0V0
Rs + Z0
= 4 V t tr
is launched from the source end of the trace. This voltage reaches the short-circuited end of the
trace at t = td, when a voltage of amplitude
9
1 (t) =
8
><
>:
L9+
1 (t td) =
( 4
tr
)
(t td) td < t td + tr
4 V t td + tr
(where td = 2 ns) reflects back towards the source since L = 1. This reflected voltage arrives
at the source end of the line at t = 2td and a new voltage of amplitude
9+
2 (t) =
8
><
>:
s9
1 (t td) =
( 2:4
tr
)
(t 2td) 2td < t 2td + tr
2:4 V t 2td + tr
is launched back towards the load since
s = 12:5 50
12:5 + 50 = 0:6
At t = 3td, a new reflected voltage 9
2 (t) of amplitude
9
2 (t) =
8
><
>:
L9+
2 (t td) =
( 2:4
tr
)
(t 3td) 3td < t 3td + tr
2:4 V t 3td + tr
is launched towards the source. This process continues indefinitely, with the voltages and currents
along the trace gradually reaching their steady-state values. The source-end voltage 9s(t) can be
written in terms of the above voltages as
9s(t) =
( 4
tr
)
[tu(t) (t tr)u(t tr)] (A.1)

( 4
tr
)
[(t 2td)u(t 2td) (t 2td tr)u(t 2td tr)] (A.2)
+
( 2:4
tr
)
[(t 2td)u(t 2td) (t 2td tr)u(t 2td tr)] (A.3)

( 2:4
tr
)
[(t 4td)u(t 4td) (t 4td tr)u(t 4td tr)] (A.4)
+
( 1:44
tr
)
[(t 4td)u(t 4td) (t 4td tr)u(t 4td tr)] (A.5)
+ : : : (A.6)

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5
Figure A.3 Figure for Problem A.3. 9s versus t for tr = 1 ns and tr = 0:25 ns.
(a) For tr = 1 ns, the source-end voltage 9s(t) given by the above expression is sketched as a
function of t in Figure A.3.
(b) For tr = 250 ps, the source-end voltage 9s(t) is sketched in Figure A.3. We see that all transi-
tions (e.g., those starting at 0, 4, or 8 seconds) are sharper for tr = 250 ps.
A.4 Effects of source risetime.
The phase velocity along the line can be found as vp = [Z0C]1 = [(50Ω)(1 pF-cm1)]1 =
20 cm-ns1. At t = 0, a voltage of amplitude
9+
1 (t) =
8
><
>:
Z0V0
(Rs + Z0)tr
t =
( 2
tr
)
t 0 < t tr
Z0V0
Rs + Z0
= 2 V t tr
is applied from the source end of the line with source rise time tr = 1 ns. This voltage reaches the
load end of the line t = td, when a voltage of amplitude
9
1 (t) =
8
><
>:
L9+
1 (t td) =
( 2
tr
)
(t td) td < t td + tr
2 V t td + tr
reflects back towards the source since L ≃ +1. This voltage arrives at the source end at t = 2td
when a new voltage of amplitude
9+
2 (t) =
8
><
>:
s9
1 (t td) =
( 2
3tr
)
(t 2td) 2td < t 2td + tr
2
3 V t 2td + tr
is launched back towards the load since the source reflection coefficient is
s = 25 50
25 + 50 = 1
3
At t = 3td, a new voltage of amplitude

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6 Addendum A/ Transmission Lines: Advanced Topics
9
2 (t) =
8
><
>:

( 2
3tr
)
(t 3td) 3td < t 3td + tr
2
3 V t 2td + tr
is launched back towards the source. This process continues on indefinitely. The source- and the
load-end voltage 9s(t) and 9L(t) can be written in terms of the above voltages as
9s(t) =
( 2
tr
)
[tu(t) (t tr)u(t tr)] (A.7)
+
( 2
tr
)
[(t 2td)u(t 2td) (t 2td tr)u(t 2td tr)] (A.8)

( 2
3tr
)
[(t 2td)u(t 2td) (t 2td tr)u(t 2td tr)] (A.9)

( 2
3tr
)
[(t 4td)u(t 4td) (t 4td tr)u(t 4td tr)] (A.10)
+
( 2
9tr
)
[(t 4td)u(t 4td) (t 4td tr)u(t 4td tr)] (A.11)
+ : : : (A.12)
and
9L(t) =
( 2
tr
)
[(t td)u(t td) (t td tr)u(t td tr)] (A.13)
+
( 2
tr
)
[(t td)u(t td) (t td tr)u(t td tr)] (A.14)

( 2
3tr
)
[(t 3td)u(t 3td) (t 3td tr)u(t 3td tr)] (A.15)

( 2
3tr
)
[(t 3td)u(t 3td) (t 3td tr)u(t 3td tr)] (A.16)
+
( 2
9tr
)
[(t 5td)u(t 5td) (t 5td tr)u(t 5td tr)] (A.17)
+ : : : (A.18)
(a) For l = 5 cm, the one-way time delay is td = l=vp = (5 cm)=(20 cm-ns1) = 0:25 ns.
Substituting td = 0:25 ns and tr = 1 ns in the above expressions yield the source- and the load-end
voltages 9s(t) and 9L(t) which are both sketched as a function of t in Figure A.4.
(b) For l = 50 cm, the one-way time delay is td = 2:5 ns. Again substituting td = 2:5 ns and
tr = 1 ns in the above expressions yield the source-end and load-end voltages 9s(t) and 9L(t)
which are both sketched as a function of t in Figure A.4. We note that, as expected, the steady-state
is achieved much quicker for td = 0:25, and with much less ringing.

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7
Figure A.4 Figure for Problem A.4. 9s(t) and 9rmL(t) versus t for td = 0:25 ns and
td = 2:5 ns.
A.5 Half-wave coaxial line resonator.
(a) Using Table 2.2 of the text with a = 1 mm, b = 4 mm, and f = 3 GHz, the line parameters of
the air-filled coaxial resonator can be evaluated as
L = 0:2ln4 ≃ 0:277 H-m1
C = 55:6
ln4 ≃ 40:1 pF-m1
R = 4:15 108 5 103p3 109
1 4 106 ≃ 2:84 Ω-m1
and G = 0. Since !L ≃ 5226 ≫ 2:84 ≃ R, low-loss approximation is valid and the real and
imaginary parts of the propagation constant can be calculated using (3.73) as
≃ R
2
√ C
L ≃ 0:0171 np-m1
≃ !pLC ≃ 62:9 rad-m1
Using these values, the quality factor follows as
Qair ≃ 2 ≃ 1839

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8 Addendum A/ Transmission Lines: Advanced Topics
(b) For teflon-filled coaxial line, the line parameters C and G have new values given by
Cteflon) c2
v2
p
Cair ≃
( 30
20:7
)2
(40:1) ≃ 84:2 pF-m1
G ≃ 3:3 104
ln b
a
≃ 2:38 104 S-m1
Again low-loss approximation is valid since !L ≫ R and !C ≫ G. Using (3.73), we find ≃
0:0632 and ≃ 91:1 from which the Q factor follows as Qteflon ≃ 1442.
(c) Using the line parameters in part (a) along with (3.72), the characteristic impedance of the air-
filled coaxial resonator can be calculated as Z0 ≃ 83:14 j0:0226 ≃ 83:1Ω. From Section A.2.2,
the element values of the equivalent series RLC circuit can be computed as
Req = Z0

2 ≃ 0:0710 Ω
Leq = Z0
2! ≃ 6:93 nH
Ceq = 2
Z0! ≃ 0:406 pF
respectively.

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Solutions to Addendum C
Cylindrical Waveguides
C.1 Frequency range of the dominant mode.
(a) The cutoff frequency of the dominant TE10 mode in an air-filled hollow WR-10 rectangular
waveguide with a = 0:254 cm and b = 0:127 cm is given by
fc10 = c
2

a ≃ 3 1010
2(0:254) ≃ 59:1 GHz
(b) Since a = 2b, the next higher-order modes are TE11, TE20, and TM11. The cutoff frequency of
these three modes is
fc11 = fc20 = 2fc10 ≃ 118 GHz
(c) Therefore, the frequency band over which only the dominant TE10 mode can propagate is fc20
fc10 ≃ 59:1 GHz.
C.2 AM waves through a tunnel?
The minimum tunnel dimension amin needed for the dominant TE10 mode to propagate at 1 MHz
can be calculated as
fc10 ≃ 3 108
2amin
= 106 ! amin = 150 m!
As long as a > amin = 150 m, TE10 will propagate regardless of the value of the other dimension
b. In practice, since there is no tunnel which has such a large dimension, AM transmission through
a tunnel is not feasible.

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10 Addendum C/ Cylindrical Waveguides
C.3 Propagation experiment in a road tunnel.
(a) For the rectangular railway tunnel under consideration (a = 17 m and b = 4:9 m), the cutoff
frequencies of the TE10 and TE01 modes are given by
fc10 = c
2a ≃ 3 108
2(17) ≃ 8:82 MHz
fc01 = c
2b ≃ 3 108
2(4:9) ≃ 30:6 MHz
(b) Using (C.17), the lowest-order modes that do not propagate at the 100 MHz FM frequency are
TE92 and TM92 since the cutoff frequency of these two modes is given by
fc92 ≃ 3 108
2
√( 2
4:9
)2
+
( 9
17
)2
≃ 100:3 MHz
C.4 Radio-wave propagation in railway tunnels.
For tunnel 1 (a = 8 m and b = 4 m), the total number of propagating modes at 100 MHz is 26.
These modes are TE10, TE01, (TE and TM)02, (TE and TM)11, (TE and TM)12, (TE and TM)20,
(TE and TM)21, (TE and TM)22, (TE and TM)30, (TE and TM)31, (TE and TM)32, (TE and TM)40,
(TE and TM)41, and (TE and TM)50. For tunnel 2 (a = 6 m and b = 4 m), the total number of
propagating modes at 100 MHz is 18. Unlike tunnel 1, (TE and TM)32, (TE and TM)40, (TE and
TM)41, and (TE and TM)50 modes do not propagate.
C.5 Rectangular waveguide modes.
(a) For the WR-137 rectangular air waveguide (a = 3:484 cm, b = 1:58 cm), the cutoff frequencies
of the four lowest-order modes TE10, TE20 (and TM20), TE01, and TE11 (and TM11) are respectively
calculated from (C.17) as fc10 ≃ 4:31 GHz, fc20 ≃ 8:61 GHz fc01 ≃ 9:49 GHz, and fc11 ≃ 10:4
GHz.
(b) For the dominant TE10 mode, the phase velocity vp and the guide wavelength at the two end
frequencies (i.e., 5.85 GHz and 8.20 GHz) of the specified frequency range can be calculated using
(C.20) and (C.21) as
vp10 jf=5:85 GHz ≃ 3 108
√
1 ( 4:31
5:85
)2 ≃ 4:43 108 m-s1
vp10 jf=8:20 GHz ≃ 3 108
√
1 ( 4:31
8:20
)2 ≃ 3:52 108 m-s1
10 f=5:85 GHz ≃ 3 108=(5:85 109)
√
1 ( 4:31
5:85
)2 ≃ 7:57 cm
10 f=8:20 GHz ≃ 3 108=(8:20 109)
√
1 ( 4:31
8:20
)2 ≃ 4:30 cm

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11
C.6 Dominant mode in a rectangular waveguide.
(a) The cutoff frequency of the dominant TE10 mode in the WR-137 rectangular air waveguide
(a = 3:484 cm, b = 1:58 cm) can be calculated from (C.17) as
fc10 = c
2a ≃ 3 1010
2(3:484) ≃ 4:31 GHz
Using this value, the propagation constant 10 and wave impedance ZTE10 at 3.5 GHz can be found
using (C.16) and (C.28) as
10 = 10 =
√( fc10
f
)2
1 ≃ 2(3:5 109)
3 108
√( 4:31
3:5
)2
1 ≃ 52:5 np-m1
ZTE10 = air
√
1 (fc10 =f)2
≃ 377
√
1 (4:31=3:5)2
≃ j526:3Ω
Note that the dominant TE10 mode is not propagating at 3.5 GHz as expected since fc10 ≃ 4:31 GHz >
f = 3:5 GHz.
(b) Repeating the same calculations as in part (a) at 7 GHz, we have
10 = j
√
1
( fc10
f
)2
≃ j 2(7 109)
3 108
√
1
( 4:31
7
)2
≃ j115:6 rad-m1
ZTE10 ≃ 377
√
1 (4:31=7)2
≃ 478Ω
Note that the dominant TE10 mode is a propagating mode at 7 GHz since 4:31 GHz < 7 GHz.
C.7 Waveguide wavelength.
(a) Using (C.21), the guide wavelength of the dominant TE10 mode propagating in a rectangular
air waveguide with a = 2b = 7:214 cm can be written as
10 = c=f
√
1 (fc10 =f)2
≃ 30
√
[f(GHz)]2 (2:08)2
cm
where c ≃ 3 1010 cm-s1 and fc10 = c=(2a) ≃ 2:08 GHz are used. Note that the expression
found for 10 above is only valid for f > 2:08 GHz (i.e., propagating case). This expression (i.e.,
10) is plotted as a function of frequency f as shown in Figure C.1.
(b) If the same waveguide is filled with water (assume ϵr ≃ 81), then, fc10 ≃ 2:08=p81 ≃ 0:231
GHz. Therefore, the guide wavelength for this case is given by
10 ≃ 30=p81
√
[f(GHz)]2 (0:231)2
cm

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12 Addendum C/ Cylindrical Waveguides2 4 6 8
5
10
15
20
25
f (GHz)
λ (cm)
TE10
TE20
0.2 0.4 0.6 0.8
5
10
15
20
25
λ (cm)
f (GHz)
TE10
Air-filled
waveguide
Water-filled
waveguide
Figure C.1 Figure for Problem C.7. Right panel: 10 and 20 versus f for air-filled waveg-
uide. Left panel: 10 versus f for water-filled waveguide (i.e., ϵr ≃ 81).
which is valid for f > 0:231 GHz. This expression is plotted in Figure C.1 as a function of the
frequency f.
(c) For the propagating TE20 mode in the air waveguide of part (a), the 20 expression is the same as
the 10 expression except 2.08 should be replaced by 4.16 (i.e., fc20 = 2fc10 ). This expression is valid
for f > 4:16 GHz and is plotted as a function of f in Figure C.1. For the propagating TE20 mode
in the water-filled waveguide of part (b), 20 expression is the same as the 10 expression found
in part (b) except that 0.231 should be replaced by 0.462. This expression is valid for f > 0:462
GHz and is plotted in Figure C.1.
C.8 Rectangular waveguide modes.
(a) For the WR-42 rectangular air waveguide (a = 1:067 cm, b = 0:432 cm), the cutoff frequencies
of the TE10, TE01, TE11, TE20, and TE02 modes are given by
fc10 ≃ 3 1010
2(1:067) ≃ 14:06 GHz < 40 GHz
fc01 ≃ 3 1010
2(0:432) ≃ 34:72 GHz < 40 GHz
fc11 ≃ 3 1010
2
√
(1:067)2 + (0:432)2 ≃ 37:46 GHz < 40 GHz
fc20 ≃ 3 1010
(1:607) ≃ 28:12 GHz < 40 GHz
fc02 ≃ 3 1010
(0:432) ≃ 69:44 GHz > 40 GHz
Note that all these modes except the TE02 mode are propagating at 40 GHz. The propagation
constants and the wave impedances of these modes can be calculated using (C.18) and (C.19) as
10 ≃ j 2(4 1010)
3 108
√
1
( 14:06
40
)2
≃ j784:3 rad-m1

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13
ZTE10 ≃ 377
√
1 ( 14:06
40
)2 ≃ 402:7Ω
01 ≃ j 2(4 1010)
3 108
√
1
( 34:72
40
)2
≃ j415:9 rad-m1
ZTE01 ≃ 377
√
1 ( 34:72
40
)2 ≃ 759:4Ω
11 ≃ j 2(4 1010)
3 108
√
1
( 37:46
40
)2
≃ j293:8 rad-m1
ZTE11 ≃ 377
√
1 ( 37:46
40
)2 ≃ 1075Ω
20 ≃ j 2(4 1010)
3 108
√
1
( 28:1
40
)2
≃ j595:9 rad-m1
ZTE20 ≃ 377
√
1 ( 28:1
40
)2 ≃ 530Ω
02 ≃ 2(4 1010)
3 108
√( 69:44
40
)2
1 ≃ 1189 np-m1
ZTE02 ≃ 377
√
1 ( 69:44
40
)2 ≃ j265:65Ω
(b) For the TE10 mode, we have from (C.20) and (C.21)
vp10 ≃ 3 108
√
1 ( 14:06
40
)2 ≃ 3:20 108 m-s1
10 ≃
31010
41010
√
1 ( 14:06
40
)2 ≃ 0:801 cm
Similarly, for the TE01 mode, we have
vp01 ≃ 3 108
√
1 ( 34:72
40
)2 ≃ 6:04 108 m-s1
01 ≃
31010
41010
√
1 ( 34:72
40
)2 ≃ 1:51 cm
In a similar fashion, for the TE11 mode, we obtain vp11 ≃ 8:56 108 m-s1, 11 ≃ 2:14 cm, and
for the TE20 mode, we obtain vp20 ≃ 4:22 108 m-s1, and 11 ≃ 1:054 cm respectively.

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