HW3_solutions
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Jan 9, 2024
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Homework 3
EECS 230 - Fall 2023
Solutions
1. (15pts: 5pts each) Given: a lossless transmission line, with:
phase velocity on the line: 0.6c,
V
g
= 5 cos(10.X10
9
rad/sec t) Volts,
R
g
= 10 Ω,
Z
0
= 50 Ω,
l = 3.2 m,
Z
L
= 75 - j25 Ω
(hint: use at least 4 significant figures for lambda throughout)
Find: (a)
Ⲅ
,
(b)
𝜆
,
(c) Z
in
Solution:
(a)
Ⲅ
=
=
=
?
𝐿
−?
0
?
𝐿
+?
0
75 − ?25 − 50
75 − ?25 + 50
25 − ? 25
125 − ? 25
125 + ? 25
125 + ? 25
=
=
3125 + ? 625 − ? 3125 − ?
2
625
125
2
+ 25
2
3750 −? 2500
16250
Ⲅ
= 0.231 -j 0.154 or:
mag=0.2774, phase=-0.558rad, -33.7degrees(1a)
(b)
𝝀
= u
p
/f
𝜔
= 2 f, so: f=
𝜔
/2
= (
10.X10
9
rad/sec) /
2
= 1.59
X10
9
Hz
𝜋
𝜋
𝜋
𝝀
=
,
so:
𝝀
= 0.1131m (1b)
(0.6)(3𝑋10
8
?/???)
1.59𝑋10
9
𝐻𝑧
(c) Z
in
= Z
0
𝑧
𝐿
+????(β?)
1 + ?𝑧𝐿
???(β?)
where:
z
L
=Z
L
/Z
0
= (75 - j 25) / 50 = 1.5 -j 0.5
𝛽
l = (2 /
𝝀
)l = (2 /
𝝀
) (3.2m) = (2 /0.1131m)(3.2m)
𝜋
𝜋
𝜋
= 177.77 radians (1.8486 rad, 105.9degrees)
tan(
𝛽
l ) = -3.5067
so:
Z
in
= Z
0
= 50Ω
𝑧
𝐿
+????(β?)
1 + ?𝑧
𝐿
???(β?)
1.5 −? 0.5 − ? 3.5067
1+?(1.5−?0.5)(−3.5067)
=50Ω
1.5 − ? 4.0067
1−? 5.26005 + +?
2
1.75335
= 50Ω
=
1.5 − ? 4.0067
−0.75335−? 5.26005 −0.75335 + ? 5.26005
−0.75335 + ? 5.26005
50Ω
−1.130025 +? 7.89005 +? 3.01845 − ?
2
21.07544
0.75335
2
+ 5.26005
2
= 50Ω
= 50Ω (0.70639 -j 0.38634)
19.9454 −? 10.9085
28.23566
Z
in
= 35.3 + j 19.3
Ω
or: 40.2e
j28.7degr
(1c)
2. (10 pts) Given a lossless transmission line with operating frequency of 3 GHz,
Z
01
=50 Ω
Z
L
=125 Ω
Find: Z
02
of a lossless quarter-wave transformer to match the load.
Solution:
In general, for a quarter-wave transformer,
Z
in
= Z
0
2
/Z
L
For this problem, to match the load,
Z
in
= Z
01
, so:
Z
01
= Z
02
2
/ Z
L
So;
Z
02
=
=
?
01
?
𝐿
(50Ω)(125Ω)
Z
02
= 79Ω (2)
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3. (10pts, 5pts each) Given a lossless transmission line
operating at frequency 13 GHz
phase velocity of 0.35c
Z
0
= 75 Ω
Z
L
= 0 Ω (a short)
Find: (a) lambda
(b) the minimum length of the line, in meters, for a Z
in
of: 0 + j 50 Ω
Solution:
(a)
𝝀
= u
p
/f
𝝀
=
,
so:
𝝀
= 0.0081m (or 8.1mm) (3a)
(0.35)(3𝑋10
8
?/???)
13.07𝑋10
9
𝐻𝑧
(b)
For a shorted line:
Z
in
= j Z
0
tan (
𝛽
l)
𝛽
= 2 /
𝝀
= 2 /0.0081m = 777.92 rad/m
𝜋
𝜋
so:
j 50 Ω = j 75 Ω tan(777.92 rad/m l)
tan(777.92 rad/m l) = 0. 667
777.92 rad/m l = atan(0.667) = 33.7 degrees = 0.588 radians
l = 0.588rad/777.92rad/m
l= 0.000756m or 0.756mm or 0.093
𝝀
(3b)
4. (20 pts) Given a lossless transmission line,
the phase velocity is 0.75c,
V
g
= 3 cos(3.5X10
6
rad/sec t) Volts
R
g
= 5 Ω,
Z
0
= 50 Ω
Z
L
= 30 -j90 Ω
l = 2.4
𝝀
Find: (a) (3pts)
Ⲅ
(b) (3pts)
𝝀
(c) (4pts) Z
in
(d) (4pts) V
0
+
(e) (3pts) time-avg P
inc
(f) (3pts) time-avg P
refl
Solution:
(a)
Ⲅ
=
=
=
?
𝐿
−?
0
?
𝐿
+?
0
30 −? 90 − 50
30 −? 90 + 50
−20 −? 90
80 − ? 90
80 + ? 90
80 + ? 90
=
=
−1600 −? 1800 −? 7200 − ?
2
8100
80
2
+90
2
6500 −? 9000
14500
Ⲅ
= 0.448 -j 0.621
or: magnitude= 0.7656, angle: -0.9453 rad, -54.2degrees
(4a)
(b)
𝝀
= u
p
/f
𝜔
= 2 f, so: f=
𝜔
/2
= (
3.5X10
6
rad/sec) /
2
= 0.557
X10
6
Hz
𝜋
𝜋
𝜋
𝝀
=
,
so:
𝝀
= 403.95m (4b)
(0.75)(3𝑋10
8
?/???)
0.557𝑋10
6
𝐻𝑧
(c)
Z
in
= Z
0
𝑧
𝐿
+????(β?)
1 + ?𝑧
𝐿
???(β?)
Z
0
= 50 Ω
𝜷
l = (2
𝜋
/
𝝀
)l = (2
𝜋
)(I/lambda) = (2
𝜋
)(2.4) = 15.08 rad
tan(
𝜷
l) = tan(15.08rad) = -0.726
z
L
= Z
L
/Z
0
= (30 - j 90)/50 = 0.6 -j 1.8
So:
Z
in
=
50 Ω
= 50 Ω
= 50 Ω
0.6 − ? 1.8 − ? 0.726
1 + ?(0.6 − ?1.8)(−0.726)
0.6 − ? 2.526
1 + ?(−0.436 + ?1.31)
0.6 − ? 2.526
−0.31 − ? 0.436
−0.31+ ? 0.436
−0.31 + ? 0.436
= 50 Ω
= 50 Ω
= 50 Ω (3.2 + j 3.65)
−0.186+? 0.262 +? 0.783− ?
2
1.1
0.31
2
+ 0.436
2
0.914+? 1.045
0.286
Z
in
= 160 + j 182.5 Ω or: magnitude=243.32, angle = 0.85 rad, 48.5degrees (4c)
(d)
V
g
= 3V
Z
g
=5 Ω
Z
in
= 160 + j 182.5 Ω
𝜷
l = 15.08 rad
Ⲅ
= 0.448 -j 0.621
V
0
+
=
3(160+?182.5)
5+160+?182.5
(
)
1
?
?15.08???
+ (0.448 −? 0.621)?
−?15.08???
(
)
=
480+?547.5
165+?182.5
(
)
1
???(15.08???)+????(15.08???) + [0.448−?0.621][???(15.08???)−????(15.08???)]
(
)
=
480+?547.5
165+?182.5
165−?182.5
165−?182.5
(
)
1
−0.809+?0.587 + [0.448−?0.621][−0.809−?0.587]
(
)
=
79200−?87600+?90338−?
2
99919.
165
2
+182.5
2
(
)
1
−0.809+?0.587 + [−0.362−?0.263+?0.502+?
2
0.365]
(
)
=
179119+?2738
60531
(
)
1
−0.809+?0.587 −0.727+?0.239
(
)
=
2.
96
+
? 0.
045 (
)
1
−1.536+?0.826
(
)
=
=
2.96 + ? 0.045
−1.536+?0.826
(
)
−1.536 − ? 0.826
−1.536 − ? 0.826
−4.55 −? 2.44 −?0.069−?
2
0.0372
1.536
2
+ 0.826
2
=
−4.513 −? 2.51
3.042
V
0
+
= -1.48 -j 0.825 Volts
or
magnitude: 1.696 Volts, phase: -2.63 rad, -150,85 degrees(4d)
(e) avg P
inc
=
=
= 2.87/100, so:
avg P
inc
= 0.0287 W (4e)
𝑉
0
+
|
|
|
|
|
|
2
2 ?
0
|
|
1.48
2
+0.825
2
2 (50)
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(f) avg P
refl
= -
-(0.448
2
+ 0.621
2
)(0.0287 W)
Γ
| |
2
(?𝑣𝑔 𝑃
???
) =
avg P
refl
= -(0.586)(0.0287 W)
avg P
refl
= -0.0168 W (4f)
5. (10pts, 2.5pts each) Using a Smith Chart, given the following values for the normalized load
impedance, z
l
,
Find: the voltage reflection coefficient at the load:
Ⲅ
(a) z
l
= 0.3 + j 0.5
(b) z
l
= 0.4 + j 2.5
(c) z
l
= 1.3 - j 0.75
(d) z
l
= 1.5 -j 0.25
You must turn in a separate Smith Chart, annotated with your work, for each value of z
l
.
Solution:
from module 2,6:
(a) gamma = 0.618 angle: 123.4 degrees
(b)
= 0.897
42.74 degr
(c)
= 0.333
-50.1 degr
(d)
= 0.222
-20.85 degr
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6. (10pts, 2.5pts each) Using a Smith Chart: given the following values for the voltage reflection
coefficient at the load, gamma,
Find: normalized load impedance, z
l
.
(a)
Ⲅ
= 0.3 + j 0.3
(b)
Ⲅ
= -0.2 + j 0.7
(c)
Ⲅ
= 0.7 - j 0.1
(d)
Ⲅ
= 0.9 - j 0.4
You must turn in a separate Smith Chart, annotated with your work, for each value of
Ⲅ
.
Solution:
use module 2.6
(a) gamma = 0.424 angle: 45 degr
zL = 1.41 +j 1.03
(b) gamma = 0.728 angle: 106 degr
zL = 0.24 + j 0.725
(c) gamma = 0.707 angle: -8.1 degr
zL = 5 - j2
(d) gamma = 0.985 angle: -24 degr
zL = 0.175 -j4.7
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7. (10pts) Using a Smith Chart,
Given: Z
0
= 25 Ω,
Z
L
= 24 + j 89 Ω
l = 2.7
𝝀
Find: Z
in
You must turn in a Smith Chart, annotated with your work.
Describe your process step-by-step.
Solution:
zL = 0.96 +j 3.56
rotate 2.7lambda - 2.5lambda = 0.2lambda toward load
from module 2-6: zin = 0.093 -j 0.64
Zin = (25Ω)(zin) = 2.325 -j 16 Ω
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8. (15pts) Using a Smith Chart,
Given: Z
0
= 200 Ω
Z
L
= 89 + j 60 Ω
Find: (a) (4pts)
Ⲅ
(b) (4pts) S
(c) (4pts) Z
in
at a distance 0.35
𝝀
from the load
(d) (3pts) d, in wavelengths, at the nearest voltage maximum to the load
You must turn in a Smith Chart, annotated with your work.
Describe your process step-by-step.
Solution:
using module 2.6:
(a) gamma = 0.428 angle 139.9 degr
(b) S = 2.49
(c) zin at .35 lambda = 0.54 -j 0.53, so Zin=(200Ω)(0.54 -j 0.53) = 108 -j 106 Ω
(d) d = 0.194 lambda
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