HW2_solutions
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Homework 2
EECS 230 - Fall 2023
Solutions
For each part, half-credit is given when the right approach is used, and the other
half-credit is given for the correct answer.
1. (8 points: 4 points each) Given a transmission line modeled with:
R'=0
G'=0
L'=6mH/m
C'=700nF/m
𝜔
=2.4X10
9
rad/sec
Find: (a) Z
0
(b)
?
Solution:
(a) Z
0
=
=
=
=
,
so
Z
0
=
92.6Ω (1a)
𝑅'+𝑗ω𝐿'
𝐺'+𝑗ω?'
0+𝑗ω𝐿'
0+𝑗ω?'
𝐿'
?'
6𝑋10
−3
700𝑋10
−9
(b)
?
=
=
=j
=j(2.4X10
9
)
(𝑅' + 𝑗𝜔𝐿')(𝐺' + 𝑗𝜔?')
𝑗ω𝐿'𝑗ω?'
ω 𝐿'?'
(6𝑋10
−3
)(700𝑋10
−9
)
?
=j155.5Krad/m (1b) (units of 1/m is ok too)
2. (27pts, 3pts each) Given a coaxial transmission line with the following parameters:
inner conductor diameter: 0.15cm
outer conductor diameter: 0.4cm
insulator:
𝜀
r
=5.5,
𝜎
=2X10
-6
S/m
frequency: 1 GHz
the conductors are made of copper.
Find: (a) R'
(b) L'
(c) G'
(d) C'
(e)
?
(f)
?
(g) u
p
(h) Z
0
(i) Using module 2.2 on the website, verify your results.
Include a screen-capture of the module showing the results.
Solution:
From Table B-2: conductivity of copper = 5.8X10
7
S/m
radii: a=0.00075m, b=0.002m
𝜔
=2
𝜋
f=(2
𝜋
)(1X10
9
)=6.28X10
9
rad/sec
R
s
=
=
=8.25 mΩ (milli-ohms)
π?µ
?
/σ
?
π(1𝑋10
9
)(4π𝑋10
−7
)/(5. 8𝑋10
7
)
(a) R' =
(
) =
𝑅
?
2π
1
?
+
1
?
8.25?Ω
2π
(
1
0.00075?
+
1
0.002?
)
=1.3 mΩ(1833.3m
-1
)
so
R' =2.38 Ω/m (2a)
(b) L' =
ln(b/a) =
ln(2.667)=(2X10
-7
)0.98
µ
2π
4π𝑋10
−7
2π
so
L' = 196 nH/m (2b)
(c) G' =
=
2πσ
??(?/?)
2π(2𝑋10
−6
)
0.98
so
G' =12.8 µS/m (2c)
(d) C' =
=
,
2πε
??(?/?)
2π(5.5)(8.85𝑋10
−12
)
0.98
so
C'=312 pF/m (2d)
(e,f)
?
=
(𝑅' + 𝑗𝜔𝐿')(𝐺' + 𝑗𝜔?')
=
[0. 238 + 𝑗(6. 28𝑋10
9
)(196𝑋10
−9
)] [(12. 8𝑋10
−6
+ 𝑗(6. 28𝑋10
9
)(312𝑋10
−12
)]
=
[0. 238
+
𝑗1231.] [(12. 8𝑋10
−6
+ 𝑗1. 96]
=
3. 05𝑋10
−6
+ 𝑗0. 47 + 𝑗0. 016 + 𝑗
2
2413
=
− 2413 + 𝑗0. 486
=
(2413)
2
+ (0. 486)
2
?
????(0.486/−2413)
=
2413 ?
+180
?
= 49.1
= 49.1 cos(90
o
) + j 49.1 sin(90
o
)
?
+90
?
= 0.0 + j49.1 m
-1
so
?
= 0.0 Np/m (2e)
?
= 49.1 rad/m (2f)
(g) u
p
=
𝜔
/
?
= (6.28X10
9
rad/sec ) / (49.1 rad/m), so
u
p
= 128 X 10
6
m/sec (2g)
(h)
Z
0
=
=
𝑅'+𝑗ω𝐿'
𝐺'+𝑗ω?'
0.238+𝑗(6.28𝑋10
9
)(196𝑋 10
−9
)
12.8𝑋10
−6
+𝑗(6.28𝑋10
9
)(312𝑋10
−12
)
=
=
0.238 + 𝑗1231
12.8𝑋10
−6
+𝑗1.96
0.238 + 𝑗1231
12.8𝑋10
−6
+𝑗1.96
𝑋
12.8𝑋10
−6
−𝑗1.96
12.8𝑋10
−6
−𝑗1.96
=
3.05𝑋10
−6
−𝑗0.466+𝑗0.016 −𝑗
2
2413.
3.84
=
=
2413. −𝑗0.45
3.84
628. 4 − 𝑗0. 117
=
(628. 4)
2
+ (0. 117)
2
?
????(−0.117/628.4)
=
628. 4?
0
?
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Z
0
= 25.07Ω (2h)
(i) module2,2 screen-capture:
3. (27pts, 3pts each) Given a 2-wire transmission line with the following parameters:
each wire has a diameter: 1mm
the wire centers are separated by 5mm
insulator:
𝜀
r
=2.5,
𝜎
=2X10
-3
S/m
frequency: 3 GHz
the conductors are made of copper
Find: (a) R'
(b) L'
(c) G'
(d) C'
(e)
?
(f)
?
(g) u
p
(h) Z
0
(i) Using module 2.1 on the website, verify your results.
Include a screen-capture of the module showing the results.
Solution:
From Table B-2: conductivity of copper = 5.8X10
7
S/m
d=0.001m, D=0.005m
𝜔
=2
𝜋
f=(2
𝜋
)(3X10
9
)=18.85X10
9
rad/sec
m=ln[(D/d)+
] = ln[5+
]=ln[5+
]=ln[5+4.9]=2.29
(?/?)
2
− 1
5
2
− 1
24
R
s
=
=
=14.3 mΩ (milli-ohms)
π?µ
?
/σ
?
π(3𝑋10
9
)(4π𝑋10
−7
)/(5. 8𝑋10
7
)
(a) R' =
=
2𝑅
?
π?
2(14.3?Ω)
π(0.001?)
=28.6 mΩ(318.3m
-1
)
so
R' =9.1 Ω/m (3a)
(b) L' =
m
=
2.29=(9.16X10
-7
)
µ
π
4π𝑋10
−7
π
so
L' = 916 nH/m (3b)
(c) G' =
=
πσ
?
π(2𝑋10
−3
)
2.29
so
G' =2.74 mS/m (3c)
(d) C' =
=
πε
?
π(2.5)(8.85𝑋10
−12
)
2.29
so
C'=30.35 pF/m (3d)
(e)
?
=
(𝑅' + 𝑗𝜔𝐿')(𝐺' + 𝑗𝜔?')
=
[9. 1 + 𝑗(18. 85𝑋10
9
)(916𝑋10
−9
)] [(2. 74𝑋10
−3
+ 𝑗(18. 85𝑋10
9
)(30. 35𝑋10
−12
)]
=
[9. 1
+
𝑗17266.] [(2. 74𝑋10
−3
+ 𝑗572. 1𝑋10
−3
]
=
0. 025 + 𝑗5. 21 + 𝑗47. 31 + 𝑗
2
9877. 88
=
− 9877. 85 + 𝑗52. 5
=
(9877. 85)
2
+ (52. 5)
2
?
????(52.5/−9877.85)
=
9878 ?
179.7
?
= 99.4
= 99.4 cos(89.85
o
) + j 99.4 sin(89.85
o
)
?
+89.85
?
= 0.26 + j99.4 m
-1
so
?
= 0.26 Np/m (3e)
?
= 99.4 rad/m (3f)
(g) u
p
=
𝜔
/
?
= (18.85X10
9
rad/sec ) / (99.4 rad/m), so
u
p
= 181. X 10
6
m/sec (3g)
(h)
Z
0
=
=
𝑅'+𝑗ω𝐿'
𝐺'+𝑗ω?'
9.1+𝑗(18.85𝑋10
9
)(916𝑋 10
−9
)
2.74𝑋10
−3
+𝑗(18.85𝑋10
9
)(30.35𝑋10
−12
)
=
9.1 + 𝑗17266.
2.74𝑋10
−3
+𝑗572.1𝑋10
−3
=
9.1 + 𝑗17266
2.74𝑋10
−3
+𝑗572.1𝑋10
−3
𝑋
2.74𝑋10
−3
−𝑗572.1𝑋10
−3
2.74𝑋10
−3
−𝑗572.1𝑋10
−3
=
0.025 −𝑗5.21+𝑗47.3 −𝑗
2
9877.88
0.33
=
=
9877.9 −𝑗42.09
0.33
29933. − 𝑗127. 54
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=
(29933)
2
+ (127. 54)
2
?
????(−127.54/29933.)
=
29933. ?
−0.24
?
= 173.
=173 cos(-0.12
o
) + 173. sin(-0.12
o
)
?
−0.12
?
Z
0
= 173 - j0.36 Ω (3h)
(i) module2,1 screen-capture:
4. (15pts, 3pts each) Given a lossless transmission line with Z
0
=100 Ω, terminated with an
impedance of 55 +j 125 Ω, operating at a frequency such that the wavelength is 10 cm.
Find:
(a) reflection coefficient,
Ⲅ
(b) voltage standing wave ratio, S
(c) the position of the voltage maximum nearest the load
(d) the position of the current maximum nearest the load
(e) Using Module 2.4 on the website, verify your results.
Include a screen-capture of the module showing the results.
Solution:
(a)
Ⲅ
=
=
=
𝑧
𝐿
−1
𝑧
𝐿
+1
0.55+𝑗1.25−1
0.55+𝑗1.25+1
−0.45+𝑗1.25
1.55+𝑗1.25
1.55−𝑗1.25
1.55−𝑗1.25
=
=
−0.6975+𝑗0.5625+𝑗1.9375−𝑗
2
1.5625
1.55
2
+1.25
2
0.865+𝑗2.5
3.965
Ⲅ
=
0.218 + j0.631
or:
0.668 e
j70.9degrees
(4a)
(b) VSWR = S =
1+|Γ|
1−|Γ|
|
Ⲅ
|=
=
0.668
0. 218
2
+ 0. 631
2
0. 446
=
S= 1.668/0.332,
so
S= 5.02 (4b)
(c) for a voltage maximum:
𝜽
r
= atan(0.631/0.218) = 70.94
o
, or 1.24 rad.
so:
d
max
= n
𝜆
/2 + (1.24rad)
𝜆
/(4
𝜋
)
= n
𝜆
/2 + .0985
𝜆
nearest voltage dmax = 0.0985
𝜆
= 0.0985(10cm) =
0.985 cm (4c)
(d) current: d
max
occurs
𝜆
/4 from voltage d
max
:
nearest current dmax = 0.985cm + 10cm/4 =
3.485cm (4d)
(e) module 2.4 does not have
𝜆
as an input, instead it has f and
𝜀
r
.
choose
𝜀
r
=1, so need f=c/
𝜆
=3.X10
8
m/sec / 0.1m = 3.GHz
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5. (15pts, 3pts each) Given a lossless transmission line with Z
0
=100 Ω, terminated with a
resistor and inductor in parallel. The resistor has a value of 10 KΩ, and the inductor has a value
of 10 µH. The transmission line is operating at a frequency of 60 GHz, and
𝜀
r
=2.
Find:
(a) reflection coefficient, gamma
(b) voltage standing wave ratio, S
(c) the position of the voltage maximum nearest the load
(d) the position of the current maximum nearest the load
(e) Using Module 2.4 on the website, verify your results.
Include a screen-capture of the module showing the results.
Solution:
First find Z
L
:
𝜔
=2
𝜋
f = 2
𝜋
(60X10
9
Hz)= 377X10
9
rad/sec
j
𝜔
L=j(377X10
9
)(10.X10
-6
)=j3.77X10
6
Ω
Impedances in parallel: Z
L
= Z
1
Z
2
/(Z
1
+Z
2
) = (10X10
3
)(j3.77X10
6
)/((j3.77+0.01)X10
6
)
Z
L
= 37.7X10
3
/3.7 = 10 KΩ
(a)
Ⲅ
=
=
= 99/101, so
Ⲅ
= 0.98 (5a)
𝑧
𝐿
−1
𝑧
𝐿
+1
100−1
100+1
(b) VSWR = S =
1+|Γ|
1−|Γ|
|
Ⲅ
|=0.98
S= 1.98/0.02,
so
S= 99 (5b)
(c) for a voltage maximum:
𝜽
r
= 0, so nearest to load, is the load itself:
d
max
=0 (5c)
or if don't count the load, then:
d
max
is at
𝜆
/2
𝜆
=(c/f)(1/
)= (3.X10
8
m/sec)/(60X10
9
Hz)/(
0.0035m = 3.5mm
ε
?
2
) =
d
max
=1.75mm (5c)
(d) current: d
max
occurs
𝜆
/4 from voltage d
max
:
𝜆
=(c/f)(1/
= (3.X10
8
m/sec)/(60X10
9
Hz)/(
0.0035m = 3.5mm
ε
?
2
) =
nearest current dmax = 0.cm + 3.5mm/4 =
0.875mm (5d)
(e) module 2.4:
6. (8pts) Given a lossless transmission line with Z
0
=50 Ω, 100 meters in length, operating at a
frequency of 900 MHz, terminated with an impedance of 25 - j55 Ω, and
𝜀
r
=4.
Find:
(a) (3pts) reflection coeff,
Ⲅ
(b) (2pts)
?
(c) (3pts) input impedance
Solution:
(a)
Ⲅ
=
=
=
=
𝑧
𝐿
−1
𝑧
𝐿
+1
0.5−𝑗1.1 −1
0.5−𝑗1.1+1
−0.5−𝑗1.1
1.5−𝑗1.1
1.5 +𝑗1.1
1.5+𝑗1.1
−0.75−𝑗0.55−𝑗1.65−𝑗
2
1.21
1.5
2
+1.1
2
=
, so
Ⲅ
= 0.128 -j 0.64
or
0.653 e
j-78.7degrees
(6a)
0.46−𝑗2.2
3.46
(b)
?
=2
𝜋
/
𝜆
𝜆
=
(c/f)(1/
= (3.X10
8
m/sec)/(900X10
6
Hz)/(
0.16667m
ε
?
4
) =
so:
?
=2
𝜋
/
𝜆
= 2
𝜋
/0.167m, so
?
=
37.7 rad/m (6b)
(c) z
in
=
𝑧
𝐿
+𝑗???(β?)
1+𝑗𝑧
𝐿
???(β?)
tan(
?
l) = tan(37.699111rad/m X 100m) = tan(3769.9111rad)= 0.0
z
in
=
=
0.5−𝑗1.1+𝑗0.0
1+𝑗(0.5−𝑗1.1)(0.0)
0.5−𝑗1.1
1
=
0. 5 − 𝑗1. 1
Z
in
=
Z
0
z
in
= 50Ω(0.5-j1.1), so
Z
in
=25-j55. Ω
or: 60.4e
-j65.6degrees
(6c)
Some students may have small number for tan(beta l), instead of 0, and so may end up
with a different value for Zin. As long as the correct eqn for zin was used, this is worth full credit.
from module2.4:
gamma = 0.133 -j .635, or: amp: 0.65, -78.2degrees (close enough)
beta...
lambda=0.167m, so consistent.
zin= 25-j55 ohms
or: 60.4,-1.1rad ohms (good)
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