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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 1.
Obtain the z parameters for the network in Fig. 19.65.
Figure 19.65
For Prob. 19.1 and 19.28.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 1.To get 11z and 21z , consider the circuit in Fig. (a).
=++== 4)24(||611
111 I
Vz
1o 2
1
II = , 1o2 2 IIV == == 1
1
221 I
Vz
To get 22z and 12z , consider the circuit in Fig. (b).
=+== 667.1)64(||22
222 I
Vz
22'
o 61
1022
III =+= , 2o1 '6 IIV ==
== 121
12 I
V
z
Hence, =][z
667.11
14
4
+
V1
+
V2
I1 = 0 1
(b)
I o'
I2 = 04
+
V1
1
I1
(a)
Io +
V2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 2.
* Find the impedance parameter equivalent of the network in Fig. 19.66.
Figure 19.66
For Prob. 19.2.
* An asterisk indicates a challenging problem.
Chapter 19, Solution 2.
Consider the circuit in Fig. (a) to get 11z and 21z .
I2 = 01
+
V1
1
I1
(a)
Io +
V2
1 I o'
1 1 1 1
1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
])12(||12[||121
111 +++== I
Vz
733.21511
24111)411)(1(
243
2||1211 =+=++=
++=z
'o
'oo 4
131
1III =+=
11'
o 154
41111
III =+=
11o 151
154
41
III ==
1o2 151
IIV ==
06667.0151
121
221 ==== zI
Vz
To get 22z , consider the circuit in Fig. (b).
733.2)3||12(||12 112
222 ==++== zI
Vz
Thus,
=][z
733.206667.006667.0733.2
1 1
(b)
1
1 1 1 1
1
+
V1
+
V2
I1 = 0
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 3.
Find the z parameters of the circuit in Fig. 19.67.
Figure 19.67
For Prob. 19.3.
Chapter 19, Solution 3.
12 216 z j z= = 11 12 11 124 4 4 6 z z z z j = = + = + 22 12 22 1210 10 4 z z j z z j j = = =
4 6 6[ ]
6 4 j j
z j j
+ = =
+
4 j6 j6 j6 j4
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 4.
Calculate the z parameters for the circuit in Fig. 19.68.
Figure 19.68
For Prob. 19.4.
Chapter 19, Solution 4.Transform the network to a T network.
5 j12120 j
5 j10 j12)10 j)(12(
1 +=+=Z
j512 j60-
2 +=Z
5 j12
503 +=Z
The z parameters are
j4.26--1.77525144
j5)--j60)(12(22112 =+=== Zzz
26.4 j775.1169
)5 j12)(120 j(1212111 +=+
=+= zzZz
739.5 j7758.1169
)5 j12)(50(2121322
=+=+= zzZz Thus,
=][z
739.5 j775.126.4 j775.1-26.4 j775.1-26.4 j775.1
Z 3 Z 1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Consider the circuit in Fig. (b).
+++=
++==
1s1
s1||s1
s1
||1s1||s1
2
222 I
Vz
1ssss1
1s1
s1
1s1s1
s1
1s1
s1s1
222
+++++++
=++++
+++
=z
1s3s2s2s2s
23
2
22 +++++=z
2112 zz =
Hence,
=][z
1s3s2s2s2s
1s3s2s1
1s3s2s1
1s3s2s1ss
23
2
23
2323
2
+
V2
+
V1
I1 = 0 1
(b)
s
1/s
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 6.
Compute the z parameters of the circuit in Fig. 19.70.
Figure 19.70
For Prob. 19.6 and 19.73.
Chapter 19, Solution 6.
To find z 11 and z 21 , consider the circuit below.
I1 5 10 4I1 I2=0Vo
++
V1 20 V2
+ _
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1 111
1 1
(20 5)25
V I z
I I
+= = =
1 120
2025o
V V I = =
2 2 2 1 1 1 14 0 4 20 4 24o oV I V V V I I I I + = = + = + =
221
1
24V
z I
= =
To find z 12 and z 22, consider the circuit below.
I1=0 5 10 4I1 I2
++
V1 20 V2
2 2 2(10 20) 30V I I = + =
222
1
30V
z I
= =
1 220V I = 1
122
20V
z I
= =
Thus,25 20
[ ]24 30
z =
+ _
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
1xxxxx1 V
12140
V160
V12V50V
20VV = ++=
88.29I
Vz)
20
V(
121
81
20
VVI
1
1
11
1x1
1 == ==
37.70IV
zI37.70
I81
121x20)
12140
(8
57V)
12140
(8
57V
857
V12)160
V13(60V
1
2211
11xxx
2
== =
====
To get z 12 and z 22, we consider the circuit below.I2
I1=0 20 100
++ +
vx 50 60 V1
- V2 - -
12v x -
+
2x22
222x V09.060V12V
150V
I,V31
V50100
50V =++==
+=
11.1109.0/1IV
z2
222 ===
704.3I
VzI704.3I
3
11.11V
3
1VV
2
112222x1 == ====
Thus,
=
11.1137.70704.388.29
]z[
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 8.
Find the z parameters of the two-port in Fig. 19.72.
Figure 19.72
For Prob. 19.8.
Chapter 19, Solution 8.
To get z 11 and z 21, consider the circuit below.
j4 I1 -j2 5 I2 =0
+ j6 j8
+V2
V1
10 --
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4 j10IV
zI)6 j2 j10(V1
11111 +== +=
)4 j10(IV
zI4 jI10V
1
221112 +== =
To get z 22 and z 12, consider the circuit below.
j4 I1=0 -j2 5 I2
+
j6 j8 +
V2 V1
10 --
8 j15IV
zI)8 j105(V2
22222 +== ++=
)4 j10(IV
zI)4 j10(V2
11221 +== +=
Thus,
++++
= )8 j15()4 j10()4 j10()4 j10(
]z[
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Chapter 19, Problem 9.
The y parameters of a network are:
[ ]
=
4.02.02.05.0y
Determine the z parameters for the network.
Chapter 19, Solution 9.
2211
0.42.50.16
y z
y = = = , 11 22 21 12 05 0.4 0.2 0.2 0.16y y y y y x x = = =
1212 21
0.21.25
0.16y
z z y
= = = =
1122
0.53.125
0.16y
z y
= = =
Thus,2.5 1.25
[ ]1.25 3.125
z =
125.325.1
25.15.2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 10.
Construct a two-port that realizes each of the following z parameters.
(a) [ ]
=105
2025z
(b) [ ]
+
+=
ss
s
ss1
21
131
z
Chapter 19, Solution 10.
(a) This is a non-reciprocal circuit so that the two-port looks like the oneshown in Figs. (a) and (b) .
z11 z22
z12 I 2
(a)
+
V2 +
I 1 I 2
+
V1 +
25 10
20 I 2
(b)
+
V2 +
I 1 I 2
+
V1 +
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
(b) This is a reciprocal network and the two-port look like the one shown inFigs. (c) and (d) .
s5.01
1s2
11211 +=+=zz
s21222 =zz
s1
12 =z
z11 z 12 z22 z 12
(c)
+V2
I1 I2
+V1
(d)
+
V2
I1 I 2
+
V1
2 H0.5 F1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 11.
Determine a two-port network that is represented by the following z parameters:
[ ]
+= j j
j j
825
2536z
Chapter 19, Solution 11.
This is a reciprocal network, as shown below.1+j5 3+j 1 j5 3
5-j2 5
-j2
j1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 12.
For the circuit shown in Fig. 19.73, let
[z]
=124
610
Find 121 ,, V I I , and 2V .
Figure 19.73
For Prob. 19.12.
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Chapter 19, Solution 12.
1 1 210 6V I I = (1)2 2 24 12V I I = + (2)
2 210V I = (3)
If we convert the current source to a voltage source, that portion of the circuit becomeswhat is shown below.
4 2 I1
+
V112 V
1 1 1 112 6 0 12 6 I V V I + + = = (4)
Substituting (3) and (4) into (1) and (2), we get
1 1 2 1 212 6 10 6 12 16 6 I I I I I = = (5)2 1 2 1 2 1 210 4 12 0 4 22 5.5 I I I I I I I = + = + = (6)
From (5) and (6),
2 2 2 212 88 6 82 0.1463 A I I I I = = = 1 25.5 0.8049 A I I = =
2 210 1.463 VV I = = 1 112 6 7.1706 VV I = =
+ _
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 13.
Determine the average power delivered to 45 j Z L += in the network of Fig. 19.74. Note: The voltage is rms.
Figure 19.74
For Prob. 19.13.
Chapter 19, Solution 13.
Consider the circuit as shown below.
10 I1 I2
++
500 V V 1V2 ZL
_
1 1 240 60V I I = + (1)2 1 280 100V I I = + (2)2 2 2 (5 4)LV I Z I j = = + (3)
1 1 1 150 10 50 10V I V I = + = (4)Substituting (4) in (1)
1 1 2 1 250 10 40 60 5 5 6I I I I I = + = + (5)
Substituting (3) into (2),2 1 2 1 2(5 4) 80 100 0 80 (105 4)I j I I I j I + = + = + + (6)
Solving (5) and (6) givesI2 = 7.423 + j3.299 A
[z]+ _
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We can check the answer using MATLAB.
First we need to rewrite equations 1-4 as follows,
U
5000
0
X*A
IIV
V
010014 j5010
1008010
604001
2
1
2
1
=
==
+
>> A=[1,0,-40,-60;0,1,-80,-100;0,1,0,(5+4i);1,0,10,0]A =
1.0e+002 *0.0100 0 -0.4000 -0.6000
0 0.0100 -0.8000 -1.00000 0.0100 0 0.0500 + 0.0400i
0.0100 0 0.1000 0>> U=[0;0;0;50]U =
000
50>> X=inv(A)*UX =-49.0722 +39.5876i50.3093 +13.1959i
9.9072 - 3.9588i-7.4227 + 3.2990i
P = |I 2|25 = 329.9 W .
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 14.
For the two-port network shown in Fig. 19.75, show that at the output terminals,
sZz
zzzZ
+=
11
211222Th
and
ss
VZz
zV
+=
11
21Th
Figure 19.75
For Prob. 19.14 and 19.41.
Chapter 19, Solution 14.
To find ThZ , consider the circuit in Fig. (a).
2121111 IzIzV += (1)2221212 IzIzV += (2)
+
(a)
+
V1
I1 I2
Z S
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
But12 =V , 1s1 - IZV =
Hence, 2s1112
12121s11
-)(0 IZz
z
IIzIZz += ++=
222s11
1221-1 IzZz
zz
++=
===22
2Th
1II
VZ
s11
122122 Zz
zzz
To find ThV , consider the circuit in Fig. (b).
02 =I , s1s1 ZIVV =
Substituting these into (1) and (2),
s11
s1111s1s Zz
VIIzZIV += =
s11
s211212 Zz
VzIzV +==
== 2Th VVs11
s21
Zz
Vz
(b)
+
V1
I1
VS
+
V2 = V Th
Z S I2 = 0
+
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 15.
For the two-port circuit in Fig. 19.76,
[z]
=12080
6040
(a) Find LZ for maximum power transfer to the load.(b) Calculate the maximum power delivered to the load.
Figure 19.76
For Prob. 19.15.
Chapter 19, Solution 15.
(a) From Prob. 18.12,
241040
60x80120
Zz
zzzZ
s11
211222Th =
+=
+=
== 24ZZ ThL
(b) 192)120(1040
80V
Zzz
V ss11
21Th =+
=+
=
==
= 24x4R
R 2V
P 2Th
2
Th
Thmax 384W
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 16.
For the circuit in Fig. 19.77, at 2= rad/s, z =1011 , z 12 = z = 621 j , z = 422 . Obtainthe Thevenin equivalent circuit at terminals a -b and calculate ov .
Figure 19.77
For Prob. 19.16.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 16.
As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).
At terminals a-b,)6 j105(||6 j)6 j4(Th ++=Z
6 j4.26 j415
)6 j15(6 j6 j4Th ++=+=Z
=ThZ 4.6
==++= 6 j)015(6 j1056 j6 j
ThV V906
The Thevenin equivalent circuit is shown in Fig. (b).
From this,
=+= 14818.3)6 j(4 j4.64 j
oV
=)t(v o V)148t2cos(18.3
4 j6 5
150 V
(a)
a
b
+
10 j6
6.4
690 V
(b)
+
+
Vo
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 17.
* Determine the z and y parameters for the circuit in Fig. 19.78.
Figure 19.78
For Prob. 19.17.
* An asterisk indicates a challenging problem.
Chapter 19, Solution 17.
To obtain 11z and 21z , consider the circuit in Fig. (a).
In this case, the 4- and 8- resistors are in series, since the same current, oI , passesthrough them. Similarly, the 2- and 6- resistors are in series, since the same current,
'oI , passes through them.
===++== 8.420
)8)(12(8||12)62(||)84(
1
111 I
Vz
11o 52
1288
III =+= 1'
o 53
II =
I2 = 0
8
4
+
V1 I1
(a)
+
V2
I o'
Io
6
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But 024- 'oo2 =+ IIV
111'
oo2 52-
56
58-
2-4 IIIIIV =+=+=
=== -0.45
2-
1
221
I
Vz
To get 22z and 12z , consider the circuit in Fig. (b).
===++== 2.420
)14)(6(14||6)68(||)24(222
2I
Vz
== -0.42112 zz
Thus,
=][z
4.20.4-
0.4-8.4
We may take advantage of Table 18.1 to get [ y] from [ z].20)4.0()2.4)(8.4( 2z ==
21.020
2.4
z
2211 ===
zy 02.0
204.0-
z
1212 ===
zy
02.020
4.0-
z
2121 ===
zy 24.0
208.4
z
1122 ===
zy
Thus,
=][y S24.002.0
02.021.0
I1 = 0
8
4
+
V1
(b)
+
V2
6
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 18.
Calculate the y parameters for the two-port in Fig. 19.79.
Figure 19.79
For Prob. 19.18 and 19.37.
Chapter 19, Solution 18.
To get 11y and 21y , consider the circuit in Fig.(a).
111 8)3||66( IIV =+=
81
1
111 == V
Iy
12-
832-
366- 11
12
VVII ==+=
121-
1
221 == V
Iy
3 6
(a)
+
V2 = 0V1 +
I 1 I 2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 19.
Find the y parameters of the two-port in Fig. 19.80 in terms of s.
Figure 19.80
For Prob. 19.19.
Chapter 19, Solution 19.
Consider the circuit in Fig.(a) for calculating 11y and 21y .
1111 1s22
)s1(2s2
2||s1
IIIV +=+=
=
5.0s2
1s2
1
111 +=
+==V
Iy
2-
1s2-
2)s1()s1-( 11
12
VIII =+=+=
-0.51
221 == VIy
1
(a)
+
V2 = 0V1 +
I1 I2
1
1/s
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To get 22y and 12y , refer to the circuit in Fig.(b).
222 2ss2
)2||s( IIV +==
s1
5.0s22s
2
222 +=
+==V
Iy
2-
s22s
2ss-
2ss- 2
221
VVII =
++=+=
-0.52
112 == V
Iy
Thus,
=][y Ss10.50.5-
0.5-5.0s
(b)
+
V1 = 0 +
I 1 I21
1
1/s
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 20.
Find the y parameters for the circuit in Fig. 19.81.
Figure 19.81
For Prob. 19.20.
Chapter 19, Solution 20.
To get y 11 and y 21, consider the circuit below.
3ix
I1 2 I2
+ ix +
I1 V1 4 6 V2 =0- -
Since 6-ohm resistor is short-circuited, i x = 0
75.0
V
IyI
6
8)2//4(IV
1
111111 == ==
5.0VI
yV21
)V86
(32
I24
4I
1
2211112 == ==+
=
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
To get y 22 and y 12, consider the circuit below.
3ix
I1 2
+ ix +
V1=0 4 6 V2 I2 -
-
1667.061
VI
y6
V2
Vi3iI,
6V
i2
222
22xx2
2x === =+==
0VI
y02
Vi3I
2
112
2x1 == ==
Thus,
S1667.05.0075.0
]y[
=
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 21.
Obtain the admittance parameter equivalent circuit of the two-port in Fig. 19.82.
Figure 19.82
For Prob. 19.21.
Chapter 19, Solution 21.
To get 11y and 21y , refer to Fig. (a).
At node 1,
4.04.02.05 1
11111
11 == =+= V
IyVV
VI
-0.2-0.21
22112 == = V
IyVI
0.2 V 1
(a)
+
V2 = 0V1 +
I1 I 2
10
V1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
To get 22y and 12y , refer to the circuit in Fig. (b).
Since 01 =V , the dependent current source can be replaced with an open circuit.
1.0101
102
22222 === = V
IyIV
02
112 == V
Iy
Thus,
=][y S1.02.0-
04.0
Consequently, the y parameter equivalent circuit is shown in Fig. (c) .
+
V1 = 0+
I 1 I20.2 V 1
(b)
10
V1
+
V2
+
V1
I 1 I 2
(c)
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 22.
Obtain the y parameters of the two-port network in Fig. 19.83.
Figure 19.83
For Prob. 19.22.
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Chapter 19, Solution 22.
To obtain y 11 and y 21, consider the circuit below.
5 I2
+ +
I1 5 0.5V 2 2 V2=0V1
The 2- resistor is short-circuited.
= = = =1 11 111
25 0.4
2 5I I
V y V
12
2 1 211 1
11 2 0.22 2.5
I I I I y
V I = = = =
To obtain y 12 and y 22, consider the circuit below.
I1 5
+ +
5 0.5V 2 2 V2 I2V1=0
At the top node, KCL gives2 2 2
2 2 2 222
0.5 1.2 1.22 5
V V I I V V y
V = + + = = =
2 11 2 12
2
0.2 0.25
V I I V y
V = = = =
Hence, =
0.4 0.2
[ ] S0.2 1.2
y
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Chapter 19, Problem 23.
(a) Find the y parameters of the two-port in Fig. 19.84.(b) Determine V ( )s2 for ( )t uvs 2= V.
Figure 19.84
For Prob. 19.23.
Chapter 19, Solution 23.
(a)
)1s(ys1s1
//1/1y 1212 += +=
=
2s1s1y1y1yy 12111211 +=++== =+
s1ss
1ss1
ys1
ysyys2
12221222++=++== =+
+++
++=
s1ss
)1s(
)1s(2s]y[ 2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
(b) Consider the network below.
I1 I2 1
+ + +
Vs V1 V2 2- - -
11s VIV += or V s V 1 = I1 (1)
22 I2V = (2)
2121111 VyVyI += (3)
2221212 VyVyI += (4)
From (1) and (3)
212111s2121111s VyV)y1(VVyVyVV ++= += (5)
From (2) and (4),
22221
12221212 V)y5.0(y1VVyVyV5.0 += += (6)
Substituting (6) into (5),
++= =
+++=
)y5.0)(y1(y1
y
s/2V
s2
VyVy
)y5.0)(y1(V
221121
12
2
212221
2211s
( ) )2.1s8.1s()1s(8.0
s1ss
21
2s11s
1)1s(
s/2V
222 +++=
++++++
++=
[y]
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Chapter 19, Problem 24.
Find the resistive circuit that represents these y parameters:
[y]
=
83
41
41
21
Chapter 19, Solution 24.
Since this is a reciprocal network, a network is appropriate, as shown below .
S41
41
21
12111 ==+= yyY , =1Z 4
S41
- 122 == yY , =2Z 4
S8
1
4
1
8
321223 ==+= yyY , =3Z 8
Y1
Y2
(a)
1/4 S
1/4 S
(b)
4
4
(c)
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 25.
Draw the two-port network that has the following y parameters:
[y]
=5.15.0
5.01S
Chapter 19, Solution 25.
This is a reciprocal network and is shown below.
0.5 S
0.5S 1S
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 26.
Calculate [ y] for the two-port in Fig. 19.85.
Figure 19.85
For Prob. 19.26.
Chapter 19, Solution 26.
To get 11y and 21y , consider the circuit in Fig. (a).
At node 1,
x1xx
xx1 -2
412
2VV
VVV
VV = +=+ (1)
But 5.15.122
2 11
11111x1
1 == =+
=
=V
IyV
VVVVI
Also,1x2x
x
2-3.575.12
4VVIV
VI == =+
-3.51
221 == V
Iy
2
(a)
+
V2 = 0V1 +
I21
2 V x
4
2
+
Vx
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
To get 22y and 12y , consider the circuit in Fig.(b).
At node 2,
42 x2x2
VVVI
+= (2)
At node 1,
x2xxxx2
x -23
1242 VVV
VVVVV = =+=
+ (3)
Substituting (3) into (2) gives
2xxx2 -1.55.121
2 VVVVI ===
-1.52
222 == V
Iy
5.022
-
2
112
2x1 == == V
Iy
VVI
Thus,
=][y S5.1-5.3-
5.05.1
2
(b)
I21
2 V x
4
2
+
Vx +
I1
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Chapter 19, Problem 27.
Find the y parameters for the circuit in Fig. 19.86.
Figure 19.86
For Prob. 19.27.
Chapter 19, Solution 27.
Consider the circuit in Fig. (a).
25.04
41
1
1
11111 === = I
IVI
yIV
55201
221112 == == V
IyVII
0.1 V 2+ V1
(a)
I 1 I 2
+
V2 = 0
4
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 28.
In the circuit of Fig. 19.65, the input port is connected to a 1-A dc current source.Calculate the power dissipated by the 2 - resistor by using the y parameters. Confirmyour result by direct circuit analysis.
Chapter 19, Solution 28.
We obtain 11y and 21y by considering the circuit in Fig.(a).
4.34||61in =+=Z
2941.01
in1
111 === ZV
Iy
11
12 346-
4.3106-
106-
VV
II =
==
-0.176534
6-
1
221 === V
Iy
To get 22y and 12y , consider the circuit in Fig. (b).
4
+
V1
1
I1
(a)
+
V2 = 0
I2
4
+
V1 = 0
+
V2
1
(b)
I oI 1
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2
2
22 IV
2434
)734(2)734)(2(
76
4||2)1||64(||21 ==
+=
+=+=
y
7059.03424
22 ==y
o1 76-
II = 222o 347
4814
)734(22
VIII ==+=
-0.176534
6-34
6-
2
11221 === = V
IyVI
Thus,
=][y S0.70590.1765-0.1765-2941.0
The equivalent circuit is shown in Fig. (c). After transforming the current source to avoltage source, we have the circuit in Fig. (d).
5454.0167.149714.0
)5.8)(9714.0(667.55.8889.1||2)5.8)(889.1||2(V =+=++=
===2
)5454.0(R
VP
22
W1487.0
6/34 S
1 A
(c)
5.667 8.5
(d)
+ 8.5 V
+
V
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Chapter 19, Problem 29.
In the bridge circuit of Fig. 19.87, 101 = I A and 42 = I A(a) Find 1V and 2V using y parameters.(b) -Confirm the results in part (a) by direct circuit analysis.
Figure 19.87
For Prob. 19.29.
Chapter 19, Solution 29.
(a) Transforming the subnetwork to Y gives the circuit in Fig. (a).
It is easy to get the z parameters22112 == zz , 32111 =+=z , 322 =z
54921122211z === zzzz
22z
2211 5
3y
zy === , 5
2--
z
122112 ===
zyy
1 1
10 A
Vo
+
V1
+
V2
(a)
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Thus, the equivalent circuit is as shown in Fig. (b).
21211 235052
53
10 VVVVI = == (1)
21212 3-220-53
52-
-4 VVVVI += +==
2121 5.1105.110 VVVV += = (2)Substituting (2) into (1),
= += 222 25.43050 VVV V8
=+= 21 5.110 VV V22
(b) For direct circuit analysis, consider the circuit in Fig. (a).
For the main non-reference node,
122
410 oo = = VV
=+=
= o1o1 10
110 VV
VVV22
==
= 41
4- o2o2 VV
VVV8
2/5 S
10 A
+
V1
+
V2
(b)
1/5 S
I1 I2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 30.
Find the h parameters for the networks in Fig. 19.88.
Figure 19.88
For Prob. 19.30.
Chapter 19, Solution 30.
(a) Convert to z parameters; then, convert to h parameters using Table 18.1.=== 60211211 zzz , =10022z
24003600600021122211z === zzzz
241002400
22
z11 ==
=
zh , 6.0
10060
22
1212 === z
zh
-0.6-
22
2121 == z
zh , 01.0
1
2222 == zh
Thus,
=][h
S0.010.6-
6.024
(b) Similarly,= 3011z === 20222112 zzz
200400600z ==
1020
20011 ==h 120
2012 ==h
-121 =h 05.020
122 ==h
Thus,
=][h
S0.051-
110
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Adding (1) and (2),
1441 6.3518 IVVI = = 1143 8.283 IIVV ==
1131 8.3 IIVV =+=
== 8.31
111 I
Vh
-3.6-3.61
-
1
2211
42 == == I
IhI
VI
To get 22h and 12h , refer to the circuit in Fig. (b). The dependent current source can bereplaced by an open circuit since 04 1 =I .
4.052
1222
2
112221 == =++= V
VhVVV
S2.051
5122 22
2222
2 === =++= VI
hVV
I
Thus,
=][h
S0.23.6-
4.038
1 1
(b)
+
V1
2 I2
+
I 1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 32.
Find the h and g parameters of the two-port network in Fig. 19.90 as functions of s.
Figure 19.90
For Prob. 19.32.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 32.(a) We obtain 11h and 21h by referring to the circuit in Fig. (a).
1211 1ss
s1s1
||ss1 IIV
+++=
++=
1ss
1s 21
111 +++== I
Vh
By current division,
1s1-
1s-
s1ss1-
21
221
112 +== +=+= I
Ih
III
To get 22h and 12h , refer to Fig. (b).
1s1
1ss1ss1
22
1122
221 +== +=+= V
Vh
VVV
1ss
s1s1
s1
s 22
22222 +=+==
+=V
IhIV
Thus,
=][h
1ss
1s1-
1s1
1ss
1s
22
22
s1
I 1
(a)
I 2
+
V2 = 0
+
V1
s
s1
(b)
I 2
+
V1
sI1 = 0
+
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
(b) To get 11g and 21g , refer to Fig. (c).
1sss
s1s11
s1
s1 21
11111 ++=++==
++=V
IgIV
1ss1
1sss1s1s1
21
2212
112 ++== ++=++= V
Vg
VVV
To get 22g and 12g , refer to Fig. (d).
222 s1s1s)1s(
s)1s(||s1
s IIV
+++
+=
++=
1ss1s
s 22
222 ++
++==I
Vg
1ss1-
1ss-
s1s1s1-
22
1122
221 ++== ++=++= I
Ig
III
Thus,
=][g
1ss1s
s1ss
11ss
1-1ss
s
22
22
s1
V1
(c)
+
V2
s I2 = 0
+
I1
s1
(d)
s I2
+
V1 = 0
+
V2
I 1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 33.
Obtain the h parameters for the two-port of Fig. 19.91.
Figure 19.91
For Prob. 19.33.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 33.
To get h 11 and h 21, consider the circuit below.4 j6 I2
+-j3 +
I1 V1 5 V2=0- -
2821.1 j0769.3IV
h6 j9
I)6 j4(5I)6 j4//(5V
1
111
111 +==+
+=+=
Also, 2564.0 j3846.0IIhI
6 j95I
1
22112 +== +
=
To get h 22 and h 12, consider the circuit below.
4 j6 I2 I1
+ +-j3 +
V1 5 V2 - -
2564.0 j3846.06 j9
5VV
hV6 j9
5V
2
11221 =+
== +
=
2821.0 j0769.0
)6 j9(3 j3 j9
)6 j9//(3 j1
VI
hI)6 j9//(3 jV 2
222
22+=
++=
+==
+=
Thus,
+++=
2821.0 j0769.02564.0 j3846.02564.0 j3846.02821.1 j077.3
]h[
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 34.
Obtain the h and g parameters of the two-port in Fig. 19.92.
Figure 19.92
For Prob. 19.34.
Chapter 19, Solution 34.
Refer to Fig. (a) to get 11h and 21h .
At node 1,
x1xx
1 43003000
100VI
VVI =
+= (1)
11x 754300
IIV ==
But == =+= 8585101
1111x11 I
VhIVIV
10
I 1
1
+
Vx
+
V1
50 2
+
V2 = 0
300
I2
(a)
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
At node 2,
111xxxx
2 75.1430075
575
300530050100
IIIVVVV
I ===+
=
75.141
221 == I
Ih
To get 22h and 12h , refer to Fig. (b).
At node 2,
x22x22
2 8094005010
400VVI
VVVI +=
++=
But 4400100 2
2x
VVV ==
Hence, 2222 29209400 VVVI =+=
S0725.040029
2
222 === V
Ih
25.041
4 21
122
x1 === == VV
hV
VV
=][h
S0725.075.1425.085
10 1
+
Vx
+
V1
50 2
300
(b)
I1 = 0
+
I 2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
To get 11g and 21g , refer to Fig. (c).
At node 1,
x1xxx
1 5.1435035010
100 VIVVV
I = +
+= (2)
But x11x1
1 1010VVI
VVI =
=
or 11x 10 IVV = (3)
Substituting (3) into (2) gives
11111 5.144951455.14350 VIIVI = =
S02929.0495
5.14
1
111 ===
V
Ig
At node 2,
xxx2 -8.42861035011
)50( VVVV =
=
1111 4955.14
)286.84(-8.4286286.84-8.4286 VVIV
+=+=
-5.96-5.961
22112 == = V
VgVV
10 1+
Vx
+
V2
50 2
300
(c)
+
I2 = 0I 1
V1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
To get 22g and 12g , refer to Fig. (d).
091.9100||10 =
091.93005010 2x2
2 +++
=VVV
I
x22 818.611818.7091.309 VVI += (4)
But 22x 02941.0091.309091.9
VVV == (5)
Substituting (5) into (4) gives
22 9091.309 VI =
== 34.342
222 I
Vg
091.30934.34
091.30922
o
IVI ==
)091.309)(1.1(34.34-
110100- 2
o1
III ==
-0.1012
112 ==
I
Ig
Thus,
=][g
34.345.96-0.101-S02929.0
10
+
Vx
50
300
(d)
+
V2
Io I o
+
V1 = 0
I 1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 35.To get 11h and 21h consider the circuit in Fig. (a).
144
n4
Z 2R ===
== =+= 22)11(1
111111
I
VhIIV
-0.521-
2- N N-
1
221
1
2
2
1 === ==II
hII
To get 22h and 12h , refer to Fig. (b).
Since 01 =I , 02 =I .Hence, 022 =h .
At the terminals of the transformer, we have 1V and 2V which are related as
5.021
2n N N
2
112
1
2
1
2
=== === VV
hV
V
Thus,
=][h
05.0-
5.02
4 1
I 1
(a)
I 2
+
V2 = 0
+
V1
1 : 2
4 1
(b)
I 2
+
V1
1 : 2I 1 = 0
+
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 36.
For the two-port in Fig. 19.94,
[h ]
S01.02
316
Find:
(a) 12 /V V (b) 12 / I I (c) 11 /V I (d) 12 / I V
Figure 19.94
For Prob. 19.36.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 36.
We replace the two-port by its equivalent circuit as shown below.
= 2025||100
112 40)2)(20( IIV == (1)
032010- 21 =++ VI 111 140)40)(3(2010 III =+=
141
1 =I , 1440
2 =V
14136
316 211 =+= VIV
708-
)2(125100
12 =
= II
(a) ==13640
1
2
VV
2941.0
(b) =1
2
I
I1.6-
(c) ==136
1
1
1
V
IS10353.7 -3
(d) ==1
40
1
2
IV 40
2 I 1
++
16
10 V 3 V 2
4 I 1 I 2
+
V1
+
V2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 37.
The input port of the circuit in Fig. 19.79 is connected to a 10-V dc voltage source whilethe output port is terminated by a 5 - resistor. Find the voltage across the 5- resistor
by using h parameters of the circuit. Confirm your result by using direct circuit analysis.
Chapter 19, Solution 37.
(a) We first obtain the h parameters. To get 11h and 21h refer to Fig. (a).
26||3 =
== =+= 88)26(1
111111
I
VhIIV
32-
32-
636-
1
221112 == =+= I
IhIII
3
+
V1
6
I1
(a)
I 2
+
V2 = 0
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
To get 22h and 12h , refer to the circuit in Fig. (b).
49
9||3 =
94
49
2
22222 == = V
IhIV
32
32
366
2
112221 == =+= V
VhVVV
=][h
S94
32-
32
8
The equivalent circuit of the given circuit is shown in Fig. (c).
1032
8 21 =+ VI (1)
1112 2930
2945
32
49
||532
IIIV =
=
=
21 3029
VI = (2)
3
+
V1
I1 = 0 6
(b)
+
I2
(c)
++
8
10 V 2/3 V 2
I 1 I 2
+
V2
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Substituting (2) into (1),
1032
3029
)8( 22 =+
VV
== 252300
2V V19.1
(b) By direct analysis, refer to Fig.(d).
Transform the 10-V voltage source to a6
10-A current source. Since
= 36||6 , we combine the two 6- resistors in parallel and transform
the current source back to V536
10 = voltage source shown in Fig. (e).
815
8)5)(3(
5||3 ==
==+= 6375)5(8156 8152V 1.1905 V
3 6
+ 10 V
(d)
+
V2
3 3
+ 5 V
(e)
+
V2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 38.
The h parameters of the two-port of Fig. 19.95 are:
[h ]
=mS230
04.0600
Given the 2=s Z k and = 400 L Z , find in Z and out Z .
Figure 19.95For Prob. 19.38.
Chapter 19, Solution 38.
From eq. (19.75),
12 2111 3
22
0.04 30 400600 333.33
1 1 1 2 10 400re fe L L
in ieoe L L
h h R h h R x x Z h h
h R h R x x= = = =
+ + +
From eq. (19.79),
113
0 11 22 21 12
2, 000 600650
( ) ( ) 2600 2 10 30 0.04s ie s
out s ie e re fe s
R h R h Z
R h h h h R h h h h x x x+ + += = = =
+ +
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 39.
Obtain the g parameters for the wye circuit of Fig. 19.96.
Figure 19.96
For Prob. 19.39.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 39.
We obtain g 11 and g 21 using the circuit below.I1 R 1 R 3 I2=0
+
R 2V1 V2
1 11 11
1 2 1 1 2
1V I I g
R R V R R= = =
+ +
By voltage division,2 2 2
2 1 211 2 1 1 2
R V RV V g R R V R R
= = =+ +
We obtain g 12 and g 22 using the circuit below.
I1 R 1 R 3
+ +
R 2 I2V1=0 V2
By current division,2 1 2
1 2 121 2 2 1 2
R I R I I g
R R I R R= = =
+ +
Also,
1 22 2 3 1 2 2 3
1 2
( // ) R R
V I R R R I R R R
= + = + + 2 1 222 3
2 1 2
V R Rg R
I R R= = +
+
21
21322
21
221
21
212
2111
R R R R
R g,R R
R g
R R R g,
R R 1g
++=
+=
+=
+=
+ _
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 40.
Find the g parameters for the circuit in Fig. 19.97.
Figure 19.97
For Prob. 19.40.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 41.
For the two-port in Fig. 19.75, show that
g L +
=Zg
g
I
I
11
21
1
2
( )( ) s Ls L
s ZggZgZgZg
VV
12212211
212
1 ++=
where g is the determinant of [ g] matrix.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 42.
The h parameters of a two-port device are given by
h = 60011 , h 312 10= , h 12021 = , h 622 102 = S
Draw a circuit model of the device including the value of each element.
Chapter 19, Solution 42.
With the help of Fig. 19.20, we obtain the circuit model below.
I1 600 I2
+ +
V1 10 -3 V2 + 120I 1 500 k V2
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 43.
Find the transmission parameters for the single-element two-port networks in Fig. 19.98.
Figure 19.98
For Prob. 19.43.
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 43.
(a) To find A and C , consider the network in Fig. (a).
12
121 == = V
VAVV
002
11 == = VICI
To get B and D , consider the circuit in Fig. (b).
11 IZV = , 12 - II =
ZI
IZ
I
VB ===
1
1
2
1
---
1-
2
1 ==I
ID
Hence,
=][T
10
Z1
Z
+
I1 I2
+
V2
V1
(a)
+
V2 = 0
Z
+
I1 I2
V1
(b)
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
(b) To find A and C , consider the circuit in Fig. (c).
12
121 == = V
VAVV
YZV
ICVIZV === == 12
1211
To get B and D , refer to the circuit in Fig.(d).
021 == VV 12 - II =
0-
2
1 ==IV
B , 1-
2
1 ==II
D
Thus,
=][T
1Y
01
+
I1 I2
+V2
V1
(c)
+
V2 = 0
I2
I 1
(d)
+
V1
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 44.
Determine the transmission parameters of the circuit in Fig. 19.99.
Figure 19.99
For Prob. 19.44.
Chapter 19, Solution 44.
To determine A and C , consider the circuit in Fig.(a).
11 ])20 j15 j(||)10 j-(20[ IV +=
111 310
j20 j15-
-j10)(-j5)(20 IIV
=
+=
1'
o II =
11o 32
5 j10 j-10 j-
III
=
=
'20-j20)( oo2 IIV += = 11 I20I340
j + = 1I340
j20
15
- 20
+
(a)
I o'
V1
I2 = 0-j10
+
V2
I1
Io
Io
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
=
==1
1
2
1
I3
40 j20
)310 j20( IVV
A 0.7692 + j0.3461
===3
40 j20
1
2
1VI
C 0.03461 + j0.023
To find B and D , consider the circuit in Fig. (b).
We may transform the subnetwork to a T as shown in Fig. (c).
10 j20 j10 j15 j
-j10))(15 j(1 =
=Z
340-j
j15-)-j10)(-j20(2 ==Z
20 j j15-
-j20))(15 j(3 ==Z
15
- 20
+
(b)
V1
-j10 I1 I 2
+
V2 = 0
j10
+
(c)
I1 I 2
+
V2 = 0V1
20
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
112 j32 j3
20 j340 j20340 j20
- III +=+
=
6923.0 j5385.02 j3
j3-
2
1 +=+==
I
ID
11 20 j340 j20)340 j20)(20 j(
10 j IV
+
+=
)18 j24( j])7 j9(210 j[ 111 =++= IIV
j55)-15(136
j3 j2)-(3-
)18 j24( j--
1
1
2
1 +=
+
==
I
I
I
VB
+= j25.385-6.923B
=][T
j0.69230.5385S j0.0230.03461
j25.3856.923-3461.0 j7692.0
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Solution 45.
To determine A and C, consider the circuit below.
I1 -j2 I2 =0
+
V1 4 V2
1 1 2 1(4 2) , 4V j I V I = =
= = = 1
2
4 21 0.54
V j A j V
1 1
2 1
0.254
I I C
V I = = =
To determine B and D, consider the circuit below.
I1 -j2 I2
+
V1 4 V2 =0
The 4- resistor is short-circuited. Hence,1
2 12
, 1I I I D I
= = =
= = = = = 1 21 1 22 2
22 2 2V j I V j I j I B j
I I
Hence,1 0.5 2
0.25 1A B j j
C D S
=
=
1S25.02 j5.0 j1
+ _
+ _
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 46.
Find the transmission parameters for the circuit in Fig. 19.101.
Figure 19.101
For Prob. 19.46.
Chapter 19, Solution 46.To get A and C , refer to the circuit in Fig.(a).
At node 1,
2o12oo
1 23212VVI
VVVI =
+= (1)
At node 2,
2ooo
x2o -2
24
41
VVVV
IVV = ===
(2)
From (1) and (2),
S-2.525-
-522
121 === = V
ICVI
But 21o1
1 1VV
VVI +=
=
21212 -3.52.5- VVVVV = +=
-3.52
1 ==V
VA
1 1
+
1 2
(a)
I x
I 1
+
V2 V1
I2 = 0
+
Vo
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
To get B and D , consider the circuit in Fig. (b).
At node 1,
o1oo
1 3212VI
VVI = += (3)
At node 2,04
1 xo
2 =++ IV
I
o2oo2 -302 VIVVI = =+= (4)
Adding (3) and (4),
2121 -0.502 IIII = =+ (5)
5.0-
2
1 ==I
ID
But o11o1
1 1VIV
VVI +=
= (6)
Substituting (5) and (4) into (6),
2221 65-
31-
21-
IIIV =+=
=== 8333.065-
2
1
I
VB
Thus,
=][T
0.5-S2.5-
0.83333.5-
1 1
+
1 2
(b)
I x
I 1
V1
+
Vo
+
V2 = 0
I 2
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PROPRIETARY MATERIAL . 2007 The McGraw-Hill Companies, Inc. All rights reserved. No partof this Manual may be displayed, reproduced or distributed in any form or by any means, without the priorwritten permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 47.
Obtain the ABCD parameters for the network in Fig. 19.102.
Figure 19.102
For Prob. 19.47.
Chapter 19, Solution 47.
To get A and C, consider the circuit below.
6
I1 1 4 I2=0
+ + + +Vx 2 5Vx V2
V1
x1xxxx1 V1.1V
10V5V
2V
1VV = +=
3235.04.3/1.1VV
AV4.3V5)V4.0(4V21xxx2 === =+=
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
02941.04.3/1.0VI
CV1.0VV1.11
VVI
2
1xxx
x11 === ==
=
To get B and D, consider the circuit below.
6
I1 1 4 I2
0V+ + + +
Vx 2 5Vx V2=0V1
- - - -
x1xxx1 V
610
V2
V6
V1
VV = += (1)
xxx
2 V1217
6V
4V5
I == (2)
x11 VIV += (3)From (1) and (3)
4706.0)1712
(64
II
DV64
VVI2
1xx11 === ==
176.1)1712
(6
10IV
B2
1 ===
=4706.002941.0176.13235.0
]T[
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permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 48.
For a two-port, let A = 4, B = 30 , C = 0.1 S, and D = 1.5. Calculate the inputimpedance, 11 / IVZ =in when:
(a) the output terminals are short-circuited,(b) the output port is open-circuited,(c) the output port is terminated by a 10- load.
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PROPRIETARY MATERIAL . 2007 The McGraw-Hill Companies, Inc. All rights reserved. No partof this Manual may be displayed, reproduced or distributed in any form or by any means, without the priorwritten permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 49.
Using impedances in the s domain, obtain the transmission parameters for the circuit inFig. 19.103.
Figure 19.103
For Prob. 19.49.
Chapter 19, Solution 49.To get A and C , refer to the circuit in Fig.(a).
1s1
s11s1
s1
||1 +=+=
12 s1||1s1s1||1
VV +=
s1s2
1s1
1s1
s1
2
1 +=
+
++
==V
VA
++
+=
++
+= )1s(s1s2
||1s
11s
1s1
||1s
1111 IIV
)1s3)(1s(1s2
)1s(s1s2
1s1
)1s(s1s2
1s1
1
1
+++=
++++
++
+=
IV
1/s
+
(a)
I 1
+
V2
V1
I 2 = 0
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PROPRIETARY MATERIAL . 2007 The McGraw-Hill Companies, Inc. All rights reserved. No partof this Manual may be displayed, reproduced or distributed in any form or by any means, without the priorwritten permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Buts
1s221
+= VV
Hence,)1s3)(1s(
1s2s
1s2
1
2
+++
=+
I
V
s)1s3)(1s(
1
2 ++==IV
C
To get B and D , consider the circuit in Fig. (b).
1s2s21
||1s1
||s1
||1 1111 +=
=
= IIIV
1
1
2 12ss-
s
1
1s
11s
1-
II
I +=
++
+=
s1
2s
1s2-
2
1 +=+==I
ID
s1-
s-s-1s2
1s21
2
1221 == =
+
+=
I
VB
IIV
Thus,
=][T
+++
+
s1
2s
)1s3)(1s(s1
s1s2
1/s
+
(b)
I 1
V1
I2
+
V2 = 0
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PROPRIETARY MATERIAL . 2007 The McGraw-Hill Companies, Inc. All rights reserved. No partof this Manual may be displayed, reproduced or distributed in any form or by any means, without the priorwritten permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.
Chapter 19, Problem 50.
Derive the s-domain expression for the t parameters of the circuit in F
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