277
CHAPTER 8
Section 8-6: Input Impedance
Problem 8.17 At an operating frequency of 300 MHz, a loss less 50-0. air-spaced
transmission line 2.501 in length is terminated with an impedance
Find the input impedance.
4'.
= (60+
j20) o..
f
Solution: Given a lossless transmission line, 2D= 50 0.,
= 300 MHz, I = 2.5 01,
and ZL = (60 + j20) o.. Since the line is air filled, lip = c and therefore, from Eq.
(8.38),
f3
=
ro
=
Up
21t X 300 X 106
3 x IOn
= 21t rad/m.
Since the line is lossless, Eq. (8.69) is valid:
Zm -
0
. -2
----
-
Zo + jZL tan J31
(ZL+jZotanf3l)
50
------------
50 + j( 60 + j20) tan (21t rad/m x 2.5 m)
(60+j20)+j50tan(21trad/mx2.5m))
_
50+
= 50 (60+
j(60+
0
j20) +j20)
j50 x 0)
= (60+
}'20) Q.
line of electrical length I = 0.35A
terminated in a load impedance as shown in Fig. 8-38 (P8.18). Find
S, and Zn.
Problem 8.18 A loss less transmission
r,
Figure P8.18: Loaded transmission line.
From Eq. (8A9a),
r=
ZL -Zo
ZL +Zo
=
(60+ j30) -100
(60+ j30) + 100
= 0.307ei132.5°
Eq. (8.59),
S=
I+
1-
WI
WI
= 1+0.307 = 1.89.
1-0.307
IS
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CHAPTER 8
From Eq. (8.63)
Zo++ jZL
tan f31
Zin =Zo (ZL
jZotanf3l)
100+ j30)
j(60++ jj30)
~ !0O(60+
!00tan (l1t{adO.35A.)
(¥0.35A.»)
= (64.8- j38.3) Q.
Problem 8.19 Show that the input impedance of a quarter-wavelength
line terminated in a short circuit appears as an open circuit.
long loss less
Solution:
Zo + JZL
tanf31 .
~Zotan(31)
Zin = Zo (ZL
For
I
=~,
f31
= 2;:. ~ = ¥. With ZL = 0, we have
Zin = Zo ---Zo
(jZotanrr./2)
=)00
.
(open circuit).
Problem 8.20 Show that at the position where the magnitude of the voltage on the
line is a maximum the input impedance is purely real.
Solution: From Eq. (8.56),
representation for
r,
lmax
= (Sr+2111t)/2f3, so ITom Eq. (8.61), using polar
which is real, provided 20 is real.
Problem 8.21 A voltage generator with vg(t) = 5cos(2rr. x 109t) V and internal
impedance Zg = 50 Q is connected to a 50-Q lossless air-spaced transmission
line. The line length is 5 cm and it is terminated in a load with impedance
ZL = (100- jIOO) Q. Find
(a)
at the load.
(b) Zin at the input to the transmission line.
(c) the input voltage Vi and input current~.
r
CHAPTER
283
8
Section 8-7: Special Cases
Problem 8.24
At an operating frequency of200 MHz, it is desired to use a section
of a lossless 50-Q transmission line terminated in a short circuit to construct an
equivalent load with reactance X = 25 Q. If the phase velocity of the line is 0.75c,
what is the shortest possible line length that would exhibit the desired reactance at its
input?
Solution:
13= rojup
=
(21t rad/cycle) x (200 x 106cycle/s)
0.75 x (3 x 108 m/s)
= 5.59
rad/m.
On a lossless short-circuited transmission line, the input impedance is always purely
imaginary; Le., Zf;
I
=
13
~tan-l
= jXfnc,
Zo
(XinC)
Solving Eq. (8.68) for the line length,
=
5.59 rad/m
50 Q
I
tan-1 (25
Q)
=
5.59 rad/m rad. '
(0.464+n1t)
for which the smallest positive solution is 8.3 em (with n = 0).
Problem 8.25
A lossless transmission line is terminated in a short circuit. How
long (in wavelengths) should the line be in order for it to appear as an open circuit at
its input terminals?
Solution:
Hence,
From Eq. (8.68), Zj~ = jZotanl31. If 131= (1t/2+ n1t), then Zf~= joo (Q).
I
= 21t
~ (~2 + n1t) = ~
4 + n'A.
2 .
This is evident from Figure 8.15(d).
Problem 8.26
The input impedance of a 31-cm-long loss less transmission line of
unknown characteristic impedance was measured at 1 MHz. With the line terminated
in a short circuit, the measurement yielded an input impedance equivalent to an
inductor with inductance of 0.128 pH, and when the line was open circuited, the
measurement yielded an input impedance equivalent to a capacitor with capacitance
of 20 pF. Find Zo of the line, the phase velocity, and the relative permittivity of the
insulating material.
Solution:
Now ro = 21tf
Zf;
= 6.28
= jroL = j21t
x 106 rad/s, so
x 106 x 0.128 x 10-6
= jO.804
Q
284
CHAPTER
and Z~c = IjjoX: = IjU2rc x 106 x 20 X 10-12) = -j8000 Q.
From Eq. (8.74), Zo = vZi~Z~c = VUO.804 Q)( - j8000 Q)
Eq. (8.75),
ro
= 80
8
Q. Using
rof
= - = ---:---====
U
p
13
tan-)· V/ -Z~c
m jZ':Jc
m
6.28 X 106 x 0.31
- tan-) (±V-
!
1.95 x 106
(±0.01 +l1rc ) m s,
jO.804j( - j8000))
where 11 2: 0 for the plus sign and 11 2: I for the minus sign.
For 11 = 0,
up
1.94 X 108 mls = 0.65cand £Of = (cjupf = IjO.652 = 2.4. For other values
of 11, up is very slow and Er is unreasonably high.
=
Problem 8.27 A 60-Q resistive load is preceded by a Aj 4 section of a 50-Q loss less
line, which itself is preceded by another Aj 4 section of a IOO-Qline. What is the input
impedance?
Solution: The input impedance of the 1../4 section ofline closest to the load is found
from Eq. (8.77):
z.m--_ Z~
ZL -
502
60
= 41. 7 Q.
The input impedance of the line section closest to the load can be considered as the
load impedance of the next section of the line. By reapplying Eq. (8.77), the next
section of 1../4 line is taken into account:
Zin
Z2
=~ _
ZL -
1 2
00
41.7
= 240
Q.
Problem 8.28 A 100-MHz FM broadcast station uses a 300-Q transmission line
between the transmitter and a tower-mounted half-wave dipole antenna. The antenna
impedance is 73 Q. You are asked to design a quarter-wave transformer to match the
antenna to the line.
(a) Determine the electrical length and characteristic
wave section.
impedance of the quarter-
(b) If the quarter-wave section is a two-wire line with d = 2.5 em, and the spacing
between the wires is made of polystyrene with &- = 2.6, determine the physical
length of the quarter-wave section and the radius of the two wire conductors.
291
CHAPTER 8
ZL,=75 Q
(Antenna I)
B
Generator
ZL,= 75 Q
(Antenna 2)
Figure P8.32: Antenna configuration for Problem 8.32.
This is divided equally between the two antennas.
15~.37 = 76.68 (W).
Hence, each antenna receives
Problem 8.33 For the circuit shown in Fig. 8-42 (P8.33), calculate the average
incident power, the average reflected power, and the average power transmitted into
the infinite 100-0. line. The A/2 line is loss less and the infinitely long line is
slightly lossy. (Hint: The input impedance of an infinitely long line is equal to its
characteristic impedance so long as a =I- 0.)
Solution: Considering the semi-infinite transmission line as equivalent to a load
(since all power sent down the line is lost to the rest of the circuit), L[ = ZI = 100.0.
Since the feed line is A./2 in length, Eq. (8.76) gives .4n = ZL = 100 .0 and
(31 = (2n/A)(A./2)
= n, so e±j(31 = -1. From Eq. (8.49a),
r = ZL-Zo
= 100- 50 =.!..
ZL+Zo
100+50
3
CHAPTER 8 .
292
W=1_.
+~~
2V~_
a
20=50n
~
_
Qf1· 21
= 100n
--- __
P~v-!-p~v
I
00
pr_1 !
av
Figure P8.33: Line terminated in an infinite line.
Also, converting the generator to a phasor gives
results into Eq. (8.66),
o =
v.+
( Zg+Zin
V.Z;, )
( ej\3/+re-jj31
I
) ~
Vg
= 2ejOO
C50+100
x 100 )
= 1~180o = -1
(V). Plugging all these
.
(
(-I)+t(-I)
I
)
(V).
From Eqs. (8.82), (8.83), and (8.84),
i
Pay
r
Pay
IVo+12
=
2Z0
2
l1ejl80012
= ~
i
_n
= 10.0 mW,
1
= -In Pay = -"3
X 10 mW = -1.1
mW,
112
P~v = Pay= P~y+P~v = 10.0 mW -1.1 mW = 8.9 mW.
Problem 8.34 An antenna with a load impedance Li = (75 +j25) Q is connected to
a transmitter through a 50-Q lossless transmission line. If under matched conditions
(50-Q load), the transmitter can deliver lOW to the load, how much power does it
deliver to the antenna? Assume Zg Zoo
=