EE135:
Homework
Set
#1.
Solutions.winter
2012.
Problem 1.28 A series RLC circuit is connected to a generator with a voltage
!s (t) = V0 cos(" t + # /3) (V).
(a) Write the voltage loop equation in terms of the current i(t), R, L, C, and vs (t).
(b) Obtain the corresponding phasor-domain equation.
!
(c) Solve the equation to obtain an expression for the phasor current I.
R
L
Vs(t)
i
C
Figure P1.28: RLC circuit.
Solution:
"
di 1
(a) !s (t) = Ri + L +
i dt.
dt C
I˜
(b) In phasor domain: V!s = RI˜ + j" LI˜ +
.
j" C
#
/3
j
!
Vs
V0 e
" CV0 e j# /3
.
(c) I˜ =
=
=
R + j (" L − 1/" C) R + j (" L − 1/" C) " RC + j(" 2 LC − 1)
Problem 2.1 A transmission line of length l connects a load to a sinusoidal voltage
source with an oscillation frequency f . Assuming the velocity of wave propagation
on the line is c, for which of the following situations is it reasonable to ignore the
presence of the transmission line in the solution of the circuit:
(a) l = 20 cm, f = 20 kHz,
(b) l = 50 km, f = 60 Hz,
(c) l = 20 cm, f = 600 MHz,
(d) l = 1 mm, f = 100 GHz.
Solution: A transmission line is negligible when l/! ≤ 0.01.
lf
l
(20 × 10−2 m) × (20 × 103 Hz)
(a)
=
=
= 1.33 × 10−5 (negligible).
!
up
3 × 108 m/s
3
0
l
(50 × 10 m) × (60 × 10 Hz)
lf
=
= 0.01 (borderline).
(b)
=
!
up
3 × 108 m/s
−2
6
lf
l
(20 × 10 m) × (600 × 10 Hz)
(c)
=
=
= 0.40 (nonnegligible).
!
up
3 × 108 m/s
−3
9
l
(1 × 10 m) × (100 × 10 Hz)
lf
=
= 0.33 (nonnegligible).
(d)
=
!
up
3 × 108 m/s
Problem 2.6 A coaxial line with inner and outer conductor diameters of 0.5 cm
and 1 cm, respectively, is filled with an insulating material with !r = 4.5 and
" = 10−3 S/m. The conductors are made of copper.
(a) Calculate the line parameters at 1 GHz.
(b) Compare your results with those based on CD Module 2.2. Include a printout
of the screen display.
Solution: (a) Given
a = (0.5/2) cm = 0.25 × 10−2 m,
b = (1.0/2) cm = 0.50 × 10−2 m,
combining Eqs. (2.5) and (2.6) gives
!
#
"
1
# f µc 1 1
R# =
+
2#
"c
a b
$
"
#
# (109 Hz)(4# × 10−7 H/m)
1
1
1
+
=
2#
5.8 × 107 S/m
0.25 × 10−2 m 0.50 × 10−2 m
= 0.788 $/m.
From Eq. (2.7),
L# =
" #
µ
4# × 10−7 H/m
b
=
ln
ln 2 = 139 nH/m.
2#
a
2#
From Eq. (2.8),
G# =
2# × 10−3 S/m
2#"
=
= 9.1 mS/m.
ln (b/a)
ln 2
From Eq. (2.9),
C# =
2#!
2#!r !0
2# × 4.5 × (8.854 × 10−12 F/m)
=
=
= 362 pF/m.
ln (b/a) ln (b/a)
ln 2
(b) Solution via Module 2.2:
Problem 2.7 Find ! , " , up , and Z0 for the two-wire line of Problem 2.2. Compare
results with those based on CD Module 2.1. Include a printout of the screen display.
Solution: From Problem 2.2:
R! = 3.71 #/m,
L! = 1.36 × 10−6 H/m,
G! = 1.85 × 10−6 S/m,
C! = 2.13 × 10−11 F/m.
At 2 GHz:
!
$ = (R! + j% L! )(G! + j% C! )
= 0.0076 + j67.54.
Hence
! = 0.0076 Np/m,
" = 67.54 rad/m.
2& × 2 × 109
%
=
= 1.86 × 108 m/s,
"
67.54
"
R! + j % L !
Z0 =
= 253 #.
G! + j% C!
up =
Problem 2.12 Generate a plot of Z0 as a function of strip width w, over the range
from 0.05 mm to 5 mm, for a microstrip line fabricated on a 0.7-mm–thick substrate
with !r = 9.8.
Solution:
Problem 2.13 In addition to not dissipating power, a lossless line has two important
features: (1) it is dispertionless (µp is independent of frequency) and (2) its
characteristic impedance Z0 is purely real. Sometimes, it is not possible to design
a transmission line such that R! " ! L! and G! " ! C ! , but it is possible to choose the
dimensions of the line and its material properties so as to satisfy the condition
R!C ! = L! G!
(distortionless line).
Such a line is called a distortionless line because despite the fact that it is not lossless,
it does nonetheless possess the previously mentioned features of the loss line. Show
that for a distortionless line,
!
!
√
C! √ ! !
L!
!C ! ,
" = R!
R
G
,
#
=
!
L
Z
=
=
.
0
L!
C!
Solution: Using the distortionless condition in Eq. (2.22) gives
"
$ = " + j# = (R! + j! L! )(G! + j! C ! )
#$
%$ !
%
√
R!
G
+ j!
+ j!
= L!C !
!
!
L
C
#$
%$ !
%
!
√
R
R
!
!
+
j
+
j
= L!C !
L!
L!
!
$ !
%
√
√
R
C!
!
= L!C !
!
+
j
+ j! L!C ! .
=
R
L!
L!
Hence,
" = Re($ ) = R!
!
C!
,
L!
√
# = Im($ ) = ! L!C ! ,
up =
1
!
.
=√
#
L!C !
Similarly, using the distortionless condition in Eq. (2.29) gives
#
!
! #
R! + j ! L !
L!
R! /L! + j!
L!
Z0 =
=
=
.
G! + j! C !
C ! G! /C ! + j!
C!
Problem 2.19 A 50-! lossless transmission line is terminated in a load with
impedance ZL = (30 − j50) !. The wavelength is 8 cm. Find:
(a) the reflection coefficient at the load,
(b) the standing-wave ratio on the line,
(c) the position of the voltage maximum nearest the load,
(d) the position of the current maximum nearest the load.
(e) Verify quantities in parts (a)–(d) using CD Module 2.4. Include a printout of
the screen display.
Solution:
(a) From Eq. (2.59),
"=
◦
ZL − Z0 (30 − j50) − 50
=
= 0.57e− j79.8 .
ZL + Z0 (30 − j50) + 50
(b) From Eq. (2.73),
S=
1 + |"| 1 + 0.57
=
= 3.65.
1 − |"| 1 − 0.57
(c) From Eq. (2.70)
dmax =
#r $ n$
−79.8◦ × 8 cm % rad n × 8 cm
+
=
+
4%
2
4%
180◦
2
= −0.89 cm + 4.0 cm = 3.11 cm.
(d) A current maximum occurs at a voltage minimum, and from Eq. (2.72),
dmin = dmax − $ /4 = 3.11 cm − 8 cm/4 = 1.11 cm.
(e) The problem statement does not specify the frequency, so in Module 2.4 we
need to select the combination of f and &r such that $ = 5 cm. With &r chosen as 1,
f=
3 × 108
c
=
= 3.75 GHz.
$
8 × 10−2
The generator parameters are irrelevant to the problem.
The results listed in the output screens are very close to those given in parts (a)
through (d).