Transmission Lines 8.6

Transmission Lines
8.6
Statement for Q.7–8:
Statement for Q.1–3:
A telephone line has the following parameters:
The propagation constant of a lossy transmission
R = 60 W m, G = 600 mS m, L = 0.3 mH m, C = 0.75 nF m
line g = 1 + j2 m -1 and characteristic impedance is 20 W
at frequency 1 Mrad s.
1. If the line operate at 10 MHz, the characteristic
7. The values of G and C are respectively
impedance Z o is
(A) 210
. - j17.6 W
(B) 29.6 - j21.4 W
(A) 0.2 S m, 0.1 mF m
(B) 0.05 S m, 1 mF m
(C) 10.8 - j7.9 W
(D) 14.0 - j 42 W
(C) 0.05 S m, 0.1 mF m
(D) 0.2 S m, 1 mF m
8. The values of R and L are respectively
2. The velocity v will be
(A) 18 ´ 106 m s
(B) 46 ´ 106 m s
(C) 23 ´ 106 m s
(D) 36 ´ 106 m s
(A) 20 W m, 40 mH m
(B) 40 W m, 50 mH m
(C) 40 W m, 40 mH m
(D) 20 W m, 50 mH m
9. A line comprised of two copper wires of diameter 1.2
3. The voltage will drop by 30 dB in the line after
mm that have 3.2 mm center to center spacing. If the
(A) 1.46 m
(B) 3.36 m
wires are separated by a dielectric material with
(C) 4.39 m
(D) 6.43 m
. , the value of characteristic impedance Z o is
e r = 35
Statement for Q.4–6:
A distortion less line operating at 120 MHz has
the following parameters: R = 18 W m, L = 0.9 mH m,
C = 21 pF m .
(A) 96 W
(B) 150 W
(C) 74 W
(D) 105 W
Statement for Q.10–11:
Consider
the
following
parameters
air-filled planer line:
Width w = 40 cm
4. The voltage will drop 20% in the line after
(A) 93.2 m
(B) 18.5 m
Distance between plane d = 1.6 cm
(C) 2.6 m
(D) 0.67 m
Thickness of conducting plane t = 4 mm
Conductivity of plane sc = 7 ´ 10 7 S m
5. The propagation constant is
(A) (7.56 + j14.25)10 -9 m -1
(B) 0.056 + j 4.37 m -1
(C) 8.08 ´ 10 -3 - j0.304 m -1
(D) 0.087 + j 3.28 m -1
Operating frequency f = 500 MHz.
10. The values of R and L are respectively
(A) 26.55 mW m, 50.3 nH m
6. To suffer a 45° phase shift it will travel
(B) 13.28 mW m, 50.3 nH m
(A) 2.46 m
(B) 2.01 m
(C) 26.55 mW m, 100.3 nH m
(C) 1.48 m
(D) 0.24 m
(D) 13.28 mW m, 100.3 nH m
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for
an
Chap 8.6
Transmission Lines
2
11. The values of C and G are respectively
18. A lossless transmission line operating at 4.5 GHz
(A) 167 mF m, 175
. ´ 10 S m
(B) 167 mF m, 0
has L = 2.6 mH m and Z o = 80 W. The phase constant b
(C) 221 pF m, 175
. ´ 10 9 S m
(D) 221 pF m, 0
and the phase velocity v is
9
(A) 148 rad m, 274 ´ 10 5 m s
12. A 81 W lossless planer line was designed but did
(B) 714 rad m, 30.8 ´ 10 4 m s
not meet a requirement. To get the characteristic
(C) 919 rad m, 30.8 ´ 106 m s
impedance of 75 W the fraction of the width of the
strip should be
(D) None of the above
(A) added by 4%
(B) removed by 4%
19. A 60 W coaxial cable feeds a 75 + j25 W dipole
(C) added by 8%
(D) removed by 8%
antenna. The voltage reflection coefficient G and
standing wave ratio s are respectively
13.
A
lossless
line
has
a
voltage
wave
V ( z, t) = 10 sin ( wt - bz). The line has parameter L = 0.2
mH m,
C = 0.5 nF m. The corresponding current wave
(A) 0.212 Ð48.55 °, 1.538
(B) 0.486 Ð68.4 °, 2.628
(C) 0.486 Ð41.45 °, 2.682
(D) 0.212 Ð68.4 °, 1.538
20. For a short-circuited coaxial transmission line:
is
(A) 20 sin ( wt - bz)
(B) 0.5 sin ( wt - bz)
Characteristic impedance Z o = 35 + j 49 W ,
(C) 200 sin ( wt - bz)
(D) sin ( wt - bz)
Propagation constant g = 1.4 + j5 m -1
Length of line l = 0.4 m.
Statement for Q.14–15:
On a distrortion less line, the voltage wave is
V(z, t) = 180 e2 .5 ´10
-3
z
cos (108 t + 2z) + 90 e-2 .5 ´10
-3
z
cos (108 t - 2z)
The input impedance of short-circuited line is
(A) 82 + j 39 W
(B) 41 + j78 W
(C) 68 + j 46 W
(D) 34 + j23 W
where z is the distance from the load. The load
impedance is Z L = 300 W .
Statement for Q.21–22:
Consider the lossless transmission line shown in
14. The phase velocity is
(A) 12.6 ´ 10 9 m s
(B) 314
. ´ 108 m s
(C) 20 ´ 10 7 m s
(D) 5 ´ 10 7 m s
fig. P8.6.21–22.
l/6
Zg
15. The characteristic impedance Z o is
(A) 400 W
(B) 200 W
(C) 100 W
(D) 70 W
Zo = 60 W
Vg
Z L 125 W
~
Fig. P8.6.21–22
Statement for Q.16–17:
21. The SWR s is
A 8.4m long coaxial line has following distributed
parameters: R = 13 W m ,
L = 6.8 mH m,
G = 4.2
mS m, C = 10.75 pF m. The line operates at 2 MHz.
(A) 2.08
(B) 1.63
(C) 2.44
(D) 1.93
16. The characteristic impedance is
22. The input impedance Z in at the generator is
(A) 23 + j186 W
(B) 46 + j93 W
(A) 36 + j 32.4 W
(B) 18 + j16.2 W
(C) 109.9 + j915
. W
(D) 55 + j 46 W
(C) 35 - j24.8 W
(D) 54 - j26.4 W
17. The end-to-end propagation time delay is
23. The quarter-wave lossless 100 W line is terminated
(A) 260 ns
(B) 130 ns
by load Z L = 210 W. If the voltage at the receiving end
(C) 180 ns
(D) 90 ns
is 60 V, the voltage at the sending end is
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3
Transmission Lines
Chap 8.6
(A) 126 V
(B) 28.6 V
29. At 3 m from the source is
(C) 21.3 V
(D) 169 V
(A) j18.2 W, 5.75 Ð59 ° V
(B) - j18.2 W, 5.75 Ð59 ° V
(C) - j18.2 W, 5.75 Ð - 59 ° V
(D) j18.2 W , 5.75 Ð59 ° V
24. A 60 W lossless line has VL = 15 e j 25 V
and
Z L = 75 e j 30 °. The current at l 8 from the load is
(A) 0.2 e
j 40
(C) 0.2 e
j 50
A
A
30. A lossless transmission line with a characteristic
(B) 0.46 e
j 40
A
impedance of 80 W is terminated by a load of 125 W.
(D) 0.46 e
j 50
A
The length of line is 1.25l. The input impedance is
25. Consider a 300 W quarter-wave long transmission
line operating at 1 GHz. It is connected to 10 V, 50 W
source at one end and is left open circuited at the other
(A) 80 W
(B) 51.2 W
(C) 125 W
(D) 45 W
Statement for Q.31–32:
end. The magnitude of the voltage at the open
circuited end of line is
A 50 W, 8.4 m long lossless line operates at 150
MHz . The input impedance at the middle of the line
(A) 10
(B) 5
(C) 0
(D) 7.707
is 80 - j 60 W. The phase velocity is 0.8 c.
31. The input impedance at the generator is
Statement for Q.26–29:
The loss line line shown in fig. P8.6.26-29 has
following parameter: V g = 10 Ð0 ° V, Z g = 50 - j 40 W,
b = 0.25 rad m. Determine the input impedance Z in and
(A) 40.3 + j 38.4 W
(B) 21.6 - j20.3 W
(C) 43.2 - j 40.3 W
(D) 80.3 + j76.8 W
32. The voltage reflection coefficient at the load is
voltage at the point given in question and choose
(A) 0.468 Ð - 6.34 °
(B) 0.468 Ð6.34 °
correct option
(C) 0.468 Ð - 38.66 °
(D) 0.468 Ð51.34 °
33. Three lossless lines are connected as shown in fig.
100 m
P8.6.33. The input impedance Z in at A is
Zg
Zo = 60 W
Vg
~
ZL
A
j40
Zin
C
B
Zo = 50 W
Zo = 100 W
Zo = 75 W
5l/8
3l/4
l/2
Z L 60 - j35 W
Fig. P8.6.26–29
26. At sending end is
(A) j29.4 W, 2. 68 Ð102 ° V
(B) j40.3 W, 5.75 Ð102 ° V
(C) j29.4 W, 5.75 Ð102 ° V
(D) j40.3 W, 2. 68 Ð102 ° V
Fig. P8.6.33
(A) 46 - j 69 W
(B) 39 - j57 W
(C) 67 + j 48 W
(D) 61 + j52 W
27. At receiving end is
(A) j40 W, 5.75 Ð109.6 ° V
(B) j40 W, 5.75 Ð - 109.6 ° V
(C) 60 - j 43 W, 5.75 Ð109.6 ° V
(D) 60 - j 43 W, 5.75 Ð - 109.6 ° V
34. Two l 4 transformer in tandem are to connect a 50
W line to a 75 W load as shown in fig. P8.6.34. If
Z o 2 = 30 W and there is no reflected wave to the left of
A, then the characteristic impedance Z o1 is
28. At 4 m from the load is
(A) - j1436 W, 6.75 Ð167 ° V
Zo = 50 W
Zo1
Zo2
(B) j1436 W, 6.75 Ð167 ° V
(C) - j3471 W, 5.75 Ð167 ° V
l/4
Fig. P8.6.34
(D) j3471 W, 5.75 Ð167 ° V
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l/4
75 W
Chap 8.6
Transmission Lines
(A) 28 W
(C) 49 W
4
(B) 56 W
38. The Z in looking into line 3 is
(D) 24.5 W
(A) 100 W
(B) 50 W
(C) 25 W
(D) ¥
35. Two identical antennas, each of input impedance
74 W are fed with three identical 50 W quarter-wave
lossless transmission lines as shown in fig. P8.6.35.
The input impedance at the source end is
Statement for Q.39–40:
For
the
transmission
line
shown
in
fig.
P8.6.39–40 the Z o = 100 W.
l/4
l/4
74 W
Z L 35 - j47.5 W
Zin
d
S.C.
l
74 W
Fig. P8.6.39–40
l/4
39. If Z L = 0 the Z in is
Fig. P8.6.35
(A) 148 W
(B) 106 W
(C) 74 W
(D) 53 W
Consider the three 50 W lossless line shown in fig.
l/4
100
e1
(C) 48.23 - j 68.2 W
(D) 48.23 + j 68.2 W
(A) 39 + j183 W
(B) 39 - j183 W
(C) 64 + j148 W
(D) 64 - j148 W
41. The 300 W lossless line shown in fig. P8.6.41 is
l/4
e2
(B) 94.11 + j76.45 W
40. If Z L = ¥, then Z in is
Statement for Q.36–38:
Line 3 Zo
(A) 94.11 - j76.45 W
W
matched to the left of the stub. The value of Z L is
Zo
Lin
n
Zo
Li
ZL
l/4
S.C.
S.C.
0
l/1
l/4
Fig. P8.6.41
Fig. P8.6.36–38
36. The input impedance Z in looking into line 1 is
(A) o
(B) 25 W
(C) ¥
(D) 50 W
(A) 1 - j1.37
(B) 1 + j1.37
(C) 300 + j 413
(D) 300 - j 413
42. A short-circuited stub is connected to a 50 W
37. The Z in looking into line 2 is
transmission line as shown in fig. P8.6.42. The
(A) 25 W
(B) 0
admittance seen at the junction of the stub and the
(C) ¥
(D) 100 W
transmission line is
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5
Transmission Lines
Chap 8.6
Solutions
Z L 100 W
1. (B) R + jwL = 60 + j2 p (10 ´ 106 )( 3 ´ 10 -7)
= 62.7 Ð17.44 °
l/2
S.C.
G + jwC = 600 ´ 10 -6 + j2 p(10 ´ 106 )(0.75 ´ 10 -9)
l/8
= 0.047 Ð89.27 °
(A) 00.01 - j0.02
(B) 0.02 + j0.01
(C) 0.04 + j0.02
(D) 0.04 - j0.02
connected
62.7 Ð17.44 °
= 36.52 Ð - 35.9 °
0.047 Ð89.27 °
=
2. (B) g = ( R + jwL) ( G + jwC)
Consider the fig. P8.6.43–44. A 50 W transmission
is
G + jwC
= 29.6 - j21.4 W
Statement for Q.43–44:
line
R + jwL
Zo =
Fig. P8.6.42
to
a
load
impedance
Z L = 35 - j 47.5 W. A short-circuited stub is connected
to match the line.
7l/8
= ( 62.7 Ð17.44 ° )(0.047 Ð89.2 ° ) = 172
. Ð53.32 °
= 1027
.
+ j1.38 m -1
a = 1027
.
Np m, b = 1.38 m -1 ,
v=
w 2 p ´ 10 7
=
= 46 ´ 106 m s
b
1.38
l/4
3. (B) a = 1027
.
Np m = 1027
.
´ 8.686 = 8.92 dB m
200+j150 W
Zin
al = 30 dB
Þ
l =
30
= 3.36 m
8.92
4. (B) Vo e - al = 0.2 Vo
ZL
e - al = 0.2
l/8
Þ
al = ln 5 = 1.61
1.61
Þ l =
= 18.5 m
0.087
a = 0.087
Fig. P8.6.43–44
5. (D) For distortion less line,
43. The length d is
(A) 0.111l
(B) 0.06l
(C) 0.13l
(D) 0.02l
G=
R
G
=
L
C
R
18 ´ 21 ´ 10 -12
C=
= 4.2 ´ 10 -4 S m
L
9 ´ 10 -7
a = RG = 18 ´ 4.2 ´ 10 -4 = 0.087
44. The length l is
(A) 0.67 l
(B) 0.13l
(C) 0.53l
(D) 0.86 l
************
b = w LC = 2 p ´ 120 ´ 106 0.9 ´ 10 -6 ´ 21 ´ 10 -12 = 3.28
6. (D) bl = 45 ° =
p
4
b = 328
.
p
= 0.24 m
4( 3.28)
Þ
7. (C) Z o =
Þ
l=
R + jwL
G + jwC
= 20
R + jwL = 400( G + jwC)
g = ( R + jwL)( G + jwC) = (1 + j2) 2
2
400( G + jwC) 2 = (1 + j2) 2 Þ G + jwC = 0.05 + j 0.1
0.1
G = 0.05 S m,
C = 6 = 0.1 mF m
10
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Chap 8.6
Transmission Lines
8. (A) R + jwL = 400( G + jwC) = 400(0.05 + j 0.1)
G=
Þ
R + jwL = 20 + j 40
40
R = 20 W m, L = 6 = 40 mH m
10
6
Z L - Z o 300 - Z o 1
=
=
Z L + Z o 300 + Z o 2
Þ
Z o = 100 W
16. (C) R + jwL = 13 + j(2 p)(2 ´ 106 )( 6.8 ´ 10 -6 )
= 13 + j 85.45
9. (D) 2 a = 1.2 mm, d = 3.2 mm
L=
C=
m
cosh -1
p
æç d ö÷
= 4 ´ 10 -7 cosh -1
çè 2 a ÷ø
pe
.
10 ´ 35
=
= 59.4 pF m
. ö÷
æ d ö÷
æ 32
cosh -1 ç
cosh -1 ç
çè 2 a ÷ø
çè 12
. ÷ø
L=
d=
= 2.69 ´ 10
md 4 p ´ 10 -7 ´ 0.016
=
= 50.3 nH m
w
0.4
ew e o 40
=
= 221 pF
1.6
d
sw
, s = 0 for air thus G = 0
G=
d
L
d m
md d
12. (C) Z o =
=
=
C
w ew
w e
d
d
= 75 W
Z o = ho = 81 W, Z o¢ =
w¢ ho
w
w¢ = 108
. w
V
13. (B) I ( z, t) = o sin ( wt - bz)
Zo
L
0.2 ´ 10 -6
Zo =
=
= 20 W
C
0.5 ´ 10 -9
I ( z, t) =
G + jwC
=
V ( z, t)
= 0.5 sin ( wt - bz)
Zo
14. (D) v =
w 10
=
= 5 ´ 10 7 m s
b
2
15. (C) G =
Vo90
1
=
=
Vo+ 180
2
8
13 + j 85.45
= 109.9 + j915
.
( 4.2 + j 0.13) 10 -3
( R + jwL)( G + jwC)
b = 0.39 rad m, t =
pfms
p5 ´ 108 ´ 4 p ´ 10 -7 ´ 7 ´ 10 7
Þ
R + jwL
g = (13 + j 85.45)( 4.2 + j0.13) ´ 10 -3 = 0.45 + j0.39
-6
2
2
=
= 26.55 ´ 10 -3 W
wdsc 0.4 ´ 2.69 ´ 10 -6 ´ 7 ´ 10 7
w¢ 81
=
w 75
Zo =
1
1
11. (D) C =
= ( 4.2 + j0.13) ´ 10 -3
17. (A) g =
L
0.66 ´ 10 -6
=
= 105 W
C
59.4 ´ 10 -12
10. (A) Skin depth d =
R=
æç 3.2 ö÷
= 0.66 mH m
çè 1.2 ÷ø
-9
Zo =
Þ
G + jwC = 4.2 ´ 10 -3 + j(2 p)(2 ´ 106 )(10.7 ´ 10 -12 )
18. (C) Z o =
L
,
C
bl
0.39 ´ 8.4
= 260 ns
=
w 2 p ´ 2 ´ 106
g = jb = jw LC,
Z ob = wL,
b=
wL 2 p ´ 4.5 ´ 10 9 ´ 2.6 ´ 10 -6
=
= 919 rad m
Zo
80
v=
w Zo
80
=
=
= 30.8 ´ 106 m s
b
L 2.6 ´ 10 -6
19. (A) G =
s=
Z L - Z o 75 + j25 - 60
=
= 0.212 Ð48.55 °
Z L + Z o 75 + j25 + 60
1 + |G| 1 + 0.212
=
= 1538
.
1 -|G| 1 - 0.212
20. (A) Z in = Z sc = Z o tanh gl = Z o
sinh gl
cosh gl
gl = (1.4 + j5)(0.4) = 0.56 + j2
sinh gl = -0.245 + j1055,
.
cosh gl = -0.483 + j0.536
( 35 + j 49)( -0.245 + j1055
.
)
Z in =
= 82 + j 39 W
( -0.483 + j0.536)
21. (A) G =
s=
Z L - Z o 125 - 60
=
= 0.35
Z L + Z o 125 + 60
1 + |G| 1 + 0.35
=
= 2.08
1 -|G| 1 - 0.35
22. (C) bl =
2p l
= 60 °
l 6
æ Z + jZ o tan bl ö÷
æ 125 + j 60 tan 60 ° ö÷
= 60ç
Z in = Z o ç L
ç Z o + jZ L tan bl ÷
çè 60 + j125 tan 60 ° ÷ø
è
ø
= 35 - j24.8
23. (A) G =
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Z L - Z o 210 - 100
=
= 0.355
Z L + Z o 210 + 100
7
Transmission Lines
SWR s =
1 + |G| 1 + 0.355
=
= 2.1
1 -|G| 1 - 0.355
if
Chap 8.6
tan bl = ¥, Z in =
Since the line is l 4 long, the sending end will be at
31. (B) bl =
Vmax and the receiving end will be at Vmin
Vmax = sVmin = 2.1 ´ 60 = 126 V
24. (A) bl =
IL =
æ Z ¢ + jZ o tan bl ö÷
æ 80 - j 60 + j50 ö÷
Z in = Z o ç L
= 50ç
ç Z o + jZ ¢L tan bl ÷
çè 50 + 60 + j 80 ÷ø
è
ø
VL 15 e j 25°
=
= 0.2 e - j 5°
Z L 75 e j 30 °
= 21.6 - j203 W
lö
æ
I ç l = ÷ = I L e jbl = 0.2 e - j 5°e j 45° = 0.2 e j 40 ° A
çè
8 ÷ø
|G| = |G¢| = 0.468,
2p
l
, z=
b=
l
4
V = V+ e
+ V- e = V- e
2
V ( Z = 0) = Vo =
= j10 V,
|V | = 10
V
33. (B) At C, bl =
æ j 40 + j 60 tan 352.4 ° ö÷
= 60ç
= j29.4
çè 60 - 40 tan 352.4 ° ÷ø
Z in
Vg
Z in + Z g
= 39 - j57
= (5.75 Ð102 ° ) e
- j 352 .4
= 5.75 Ð109.6 ° V
34. (D) bl =
28. (C) bl = 0.25 ´ 4 = 1 rad = 57.3°
æ j 40 + j 60 tan 57.3° ö÷
= 60ç
= - j 3471 W
çè 60 - 40 tan 57.3° ÷ø
V = VL e
jbl
= (5.75 Ð109.6 ° ) e
j 57.3°
= 5.75 Ð167 ° V
29. (B) 3 m from the source is the same as 97 m from
the load,
bl = 0.25 ´ 97 = 24.25 rad = 309.40 °
Z in
æ j 40 + j 60 tan 309.40 ° ö÷
= 60ç
= - j18.2
çè 60 - 40 tan 309.40 ° ÷ø
V = VL e
jbl
= (5.75 Ð109.6 ° )( e
30. (B) bl =
j 309.4 °
) = 5.75 Ð59 °
2p
p
p
(1.25 l) = + 360 °, tan = ¥
l
2
2
æ Z + jZ o tan bl ö÷
Z in = Z o ç L
ç Z o + jZ L tan bl ÷
è
ø
2 p 5l 5 p
= , tan 225 ° = 1, Z L = Z inB
l 8
4
æ Z + jZ o ö÷
æ 124.4 + j72.5 + j50 ö÷
= 50ç
Z in = Z o ç L
ç Z o + jZ L ÷
çè 50 + j124.4 - 72.5 ÷ø
è
ø
VL = Vs ( z = l), Vo = VL e jbl
Z in
tan p = 0
Z o2
100 2
=
= 124.4 + j72.5 W
Z L 60 - j 35
At A, bl =
27. (A) Z in = Z L = j 40 W
VL = Vo e
2p l
= p,
l 2
Z inC = Z L = 60 - j 35 W
2 p 3l 3p
At B, bl =
= , Z L = Z inC = 60 - j 35 W
l 4
4
Z inB =
æ
ö÷
j29.4
=ç
10 Ð0 ° = 5.75 Ð102 °
çè j29.4 + 50 - j 40 ÷ø
- jbl
p
= - 38.66 ° + 90 ° = 51.34 °
4
G = 0.468 Ð51.34 °
26. (C) bl = 0.25 ´ 100 = 25 rad = 1432.4 ° º 352.4 °
At sending end Z in
q G = q G¢ + 2 ´
But
jp
gz
Z ¢L - Z o 80 - j 60 - 50
=
= 0.468 Ð - 38.66 °
Z L¢ + Z o 80 - j 60 + 50
32. (D) G¢ =
g = a + jb, V+ = 0, V- = 10 V
- gz
wl 2 p ´ 150 ´ 106 ´ 4.2 21p
=
=
v
0.8 ´ 3 ´ 108
4
tan bl = 1
2p l
= 45 °
l 8
25. (A) a = 0 ,
Z o2 80 2
=
= 51.2 W
Z L 125
Z2
2p l p
p
= , tan = ¥, Z in = o ,
ZL
l 4 2
2
Z in 2 =
Z o22
, Z L1 = Z in 2
75
Z in 1 =
Z o21
= Zo
Z L1
35. (A) bl =
Þ
Z o21 = Z o Z L1 =
2p l p
= ,
l 4 2
Z ¢in =
50
( 30) 2 = 24.5 W
75
Z o2 50 2
=
= 3378
.
ZL
74
This act as the load to the left line. But there are two
such loads in parallel. Due to the two lines on the right
Z ¢L = 16.9 W,
Z in =
50 2
= 148 W
16.9
æ 2 p l ö÷
36. (B) tan bl = tan ç
=¥,
çè l 4 ÷ø
Z in =
Z o2 50 2
=
= 25W
Z L 100
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Chap 8.6
Transmission Lines
37. (C) Z L = 0
2
o
Z
0
Z in =
æ 2 p l ö÷
tan bl = tan ç
=0
çè l 2 ÷ø
= ¥ (open circuit).
39. (A) For
Z in 1 =
42. (A) At junction input impedance of line
(short circuit),
38. (A) Z L = 25 || ¥ = 25 W, Z in
æ Z + jZ o tan bl ö÷
Z inL = Z o ç L
= Z L = 100 W
ç Z o + jZ L tan bl ÷
è
ø
50 2
=
= 100 W
25
input impedance of stub Z L = 0
æ 2 p l ö÷
tan bl = tan ç
=1
çè l 8 ÷ø
l
line tan bl = ¥
4
æ Z + jZ o tan bl ö÷
= jZ o = j50 W
Z ins = Z o ç L
ç Z o + jZ L tan bl ÷
è
ø
1
1
1
At junction Y = =
+
= 0.01 - j0.02
Z j50 100
Z o2
100 2
=
= 32 - j24
Z L 200 + j150
l
line, tan bl = 1
8
For
æ 0 + jZ o ö÷
= jZ o = j100 W
Z in 2 = Z o ç
ç Zo + 0 ÷
è
ø
At the
ZL =
43. (B) Normalized load z L = rL + jxL
{
( 32 - j24)( j100)
= 47.06 - j1176
.
32 - j24 + j100
æ 47.06 - j1176
. + j100( -1) ö÷
Z in = 100ç
= 94.11 - j76.45
çè 100 + j( 47.06 - j1176
. )( -1) ÷ø
l
line
40. (D) If Z L = ¥, at the input end of
8
æ Z + jZ o ö÷
= - jZ o = - j100
= Zo ç L
ç Z o + jZ L ÷
è
ø
Z L = ( 32 - j 24)||( - j100) = 19.5 - j24.4
æ 19.5 - j24.4 + j100( -1) ö÷
= 64 - j148 W
Z in = 100ç
çè 100 + j(19.5 - j24.4)( -1) ÷ø
41. (C) For the line to be matched, it is required that
the sum of normalized input admittance of the shorted
stub and main line at the point where the stub is
l
shorted stub, z L = 0 and
connected be unity. For
10
z ins
z + j tan bl
= L
= j tan bl ,
1 + z L j tan bl
ì 1
tan -1 t,
t ³0
ï
d ï 2p
=í
l ï 1
-1
ï 2 p (p + tan t), t > 0
î
Z L = 35 - j 47.5, Z o = 50 W, z L =
ZL
= 0.7 - j0.95
Zo
rL = 0.7, xL = -0.95,
1
t=
( -0.95 ± 0.7{0.7(1 - 0.7) 2 + 0.95 2 } ) = 5.93, 0.4
0.7 - 1
d
1
For t = 5.93,
=
tan -1 5.93 = 0.223
l 2p
d
1
For t = 0.4,
=
tan -1 0.4 = 0.06
l 2p
44. (B) bB =
bB =
rL2 t - (1 - xL t)( xL + t)
,
rL + ( xL + t) 2
For line to be matched at junction normalized input
admittance of line must be 1 + j1.3764
æ 2 p l ö÷
tan bl = tan ç
=¥
çè l 4 ÷ø
For t = 5.93,
(0.7) 2 (5.93) - {1 - ( -0.95)(5.93)}(5.93 - 0.95)
= - 118
.
0.7 + (5.93 - 0.95) 2
For t = 0.4, bB = 0.9526
1
ì 1
bB ³ 0
tan -1
,
ï
2
p
b
l ïï
B
=í
l ï 1 æç
-1 1 ö
÷
ï 2 p ç p + tan b ÷, bB > 0
ïî è
B ø
æ 2 p l ö÷
yins = - j cot bl = - j cot ç
= - j1.3764
çè l 10 ÷ø
yin = z L , z L = 1 + j1.3764,
}
ì 1
xL ± rL ( (1 - rL ) 2 + xL2 ) , rL ¹ 1
ï
rL - 1
ï
t = tan bd = í
ï xL
, rL = 1
ïî 2
7l
line Z in 1 ||Z in 2 will be load.
8
æ 2 p 7 l ö÷
tan bl = tan ç
×
= -1
çè l 8 ÷ø
Z in 2
8
l
= 0.388
l
l
For bB = 0.953,
= 0.13
l
For bB = -118
. ,
Z L = Z o z L = 300 + j 413
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