Chapter 19, Solution 1.
To get z 11 and z 21 , consider the circuit in Fig. (a).
1Ω 4Ω I2 = 0
+ Io +
I1 V1 6Ω 2Ω V2
− −
(a)
V1
z 11 = = 1 + 6 || (4 + 2) = 4 Ω
I1
1
Io = I , V2 = 2 I o = I 1
2 1
V2
z 21 = = 1Ω
I1
To get z 22 and z 12 , consider the circuit in Fig. (b).
I1 = 0 1Ω 4Ω
Io '
+ +
V1 6Ω 2Ω V2 I2
− −
(b)
V2
z 22 = = 2 || (4 + 6) = 1.667 Ω
I2
2 1
Io' = I2 = I2 , V1 = 6 I o ' = I 2
2 + 10 6
V1
z 12 = = 1Ω
I2
Hence,
4 1
[z ] = Ω
1 1.667
Chapter 19, Solution 2.
Consider the circuit in Fig. (a) to get z 11 and z 21 .
1Ω Io ' 1Ω 1Ω 1Ω I2 = 0
+ Io +
I1 V1 1Ω 1Ω 1Ω V2
− −
1Ω 1Ω 1Ω 1Ω
(a)
V1
z 11 = = 2 + 1 || [ 2 + 1 || (2 + 1) ]
I1
3 (1)(11 4) 11
z 11 = 2 + 1 || 2 + = 2 + = 2 + = 2.733
4 1 + 11 4 15
1 1
Io = Io' = Io'
1+ 3 4
1 4
Io' = I1 = I1
1 + 11 4 15
1 4 1
Io = ⋅ I1 = I1
4 15 15
1
V2 = I o = I
15 1
V2 1
z 21 = = = z 12 = 0.06667
I 1 15
To get z 22 , consider the circuit in Fig. (b).
I1 = 0 1Ω 1Ω 1Ω 1Ω
+ +
V1 1Ω 1Ω 1Ω V2 I2
− −
1Ω 1Ω 1Ω 1Ω
(b)
V2
z 22 = = 2 + 1 || (2 + 1 || 3) = z 11 = 2.733
I2
Thus,
2.733 0.06667
[z ] = Ω
0.06667 2.733
Chapter 19, Solution 3.
(a) To find z 11 and z 21 , consider the circuit in Fig. (a).
-j Ω I2 = 0
Io
+ +
I1 V1 jΩ 1Ω V2
− −
(a)
V1 j (1 − j)
z 11 = = j || (1 − j) = = 1+ j
I1 j +1− j
By current division,
j
Io = I = j I1
j +1− j 1
V2 = I o = jI 1
V2
z 21 = =j
I1
To get z 22 and z 12 , consider the circuit in Fig. (b).
I1 = 0 -j Ω
+ +
V1 jΩ 1Ω V2 I2
− −
(b)
V2
z 22 = = 1 || ( j − j) = 0
I2
V1 = j I 2
V1
z 12 = =j
I2
Thus,
1+ j j
[z ] = Ω
j 0
(b) To find z 11 and z 21 , consider the circuit in Fig. (c).
jΩ -j Ω I2 = 0
+ +
1Ω
I1 V1 V2
-j Ω 1Ω
− −
(c)
V1 -j
z 11 = = j + 1 + 1 || (-j) = 1 + j + = 1.5 + j0.5
I1 1− j
V2 = (1.5 − j0.5) I 1
V2
z 21 = = 1.5 − j0.5
I1
To get z 22 and z 12 , consider the circuit in Fig. (d).
I1 = 0 jΩ -j Ω
+ +
1Ω
V1 V2 I2
-j Ω 1Ω
− −
(d)
V2
z 22 = = -j + 1 + 1 || (-j) = 1.5 - j1.5
I2
V1 = (1.5 − j0.5) I 2
V1
z 12 = = 1.5 − j0.5
I2
Thus,
1.5 + j0.5 1.5 − j0.5
[z ] = Ω
1.5 − j0.5 1.5 − j1.5
Chapter 19, Solution 4.
Transform the Π network to a T network.
Z1 Z3
Z2
(12)( j10) j120
Z1 = =
12 + j10 − j5 12 + j5
- j60
Z2 =
12 + j5
50
Z3 =
12 + j5
The z parameters are
(-j60)(12 - j5)
z 12 = z 21 = Z 2 = = -1.775 - j4.26
144 + 25
( j120)(12 − j5)
z 11 = Z1 + z 12 = + z 12 = 1.775 + j4.26
169
(50)(12 − j5)
z 22 = Z 3 + z 21 = + z 21 = 1.7758 − j5.739
169
Thus,
1.775 + j4.26 - 1.775 − j4.26
[z ] = Ω
- 1.775 − j4.26 1.775 − j5.739
Chapter 19, Solution 5.
Consider the circuit in Fig. (a).
1 s I2 = 0
+ Io +
I1 V1 1 1/s 1/s V2
− −
(a)
1 1 1
1 + s +
1 1 s 1 s + 1 s
z 11 = 1 || || 1 + s + = || 1 + s + =
s s 1 s 1 1
1+ +1+ s +
s s + 1 s
s2 + s +1
z 11 =
s 3 + 2s 2 + 3s + 1
1 1 s
1 ||
s s +1 s +1
Io = I1 = I1 = I1
1 1 1 1 s
1 || + 1 + s + +1+ s + + s2 + s +1
s s s +1 s s +1
s
Io = I1
s + 2s + 3s + 1
3 2
1 I1
V2 = I o = 3
s s + 2s 2 + 3s + 1
V2 1
z 21 = = 3
I 1 s + 2s + 3s + 1
2
Consider the circuit in Fig. (b).
I1 = 0 1 s
+ +
V1 1 1/s 1/s V2 I2
− −
(b)
V2 1 1 1 1
z 22 = = || 1 + s + 1 || = || 1 + s +
I2 s s s s + 1
1 1 1
1 + s + 1+ s +
s s + 1 s +1
z 22 = =
1 1 s
+1+ s + 1+ s + s2 +
s s +1 s +1
s 2 + 2s + 2
z 22 =
s 3 + 2s 2 + 3s + 1
z 12 = z 21
Hence,
s2 + s + 1 1
s 3 + 2s 2 + 3s + 1 s 3 + 2s 2 + 3s + 1
[z ] =
1 s 2 + 2s + 2
s 3 + 2s 2 + 3s + 1 s 3 + 2s 2 + 3s + 1
Chapter 19, Solution 6.
To find z 11 and z 21 , connect a voltage source V1 to the input and leave the output
open as in Fig. (a).
I1 10 Ω Vo 20 Ω
+
+
V1 0.5 V2 30 Ω V2
−
−
(a)
V1 − Vo Vo 30 3
= 0.5 V2 + , where V2 = Vo = Vo
10 50 20 + 30 5
3 V
V1 = Vo + 5 Vo + o = 4.2 Vo
5 5
V1 − Vo 3.2
I1 = = V = 0.32 Vo
10 10 o
V1 4.2 Vo
z 11 = = = 13.125 Ω
I 1 0.32 Vo
V2 0.6 Vo
z 21 = = = 1.875 Ω
I 1 0.32 Vo
To obtain z 22 and z 12 , use the circuit in Fig. (b).
10 Ω 20 Ω I2
+
+
V1 0.5 V2 30 Ω V2
−
−
(b)
V2
I 2 = 0.5 V2 + = 0.5333 V2
30
V2 1
z 22 = = = 1.875 Ω
I 2 0.5333
V1 = V2 − (20)(0.5 V2 ) = -9 V2
V1 - 9 V2
z 12 = = = -16.875 Ω
I 2 0.5333 V2
Thus,
13.125 - 16.875
[z ] = Ω
1.875 1.875
Chapter 19, Solution 7.
To get z11 and z21, we consider the circuit below.
I2=0
I1 20 Ω 100 Ω
+
+ +
vx 50 Ω 60 Ω
V1
- V2
- -
12vx -
+
V1 − Vx Vx Vx + 12Vx 40
= +
→ Vx = V1
20 50 160 121
V − Vx 81 V1 V1
I1 = 1 = ( ) → z11 = = 29.88
20 121 20 I1
13Vx 57 57 40 57 40 20x121
V2 = 60( ) − 12Vx = − Vx = − ( )V1 = − ( ) I1
160 8 8 121 8 121 81
V
= −70.37 I1 → z 21 = 2 = −70.37
I1
To get z12 and z22, we consider the circuit below.
I2
I1=0 20 Ω 100 Ω
+
+ +
vx 50 Ω 60 Ω
V1
- V2
- -
12vx -
+
50 1 V2 V2 + 12Vx
Vx = V2 = V2 , I2 = + = 0.09V2
100 + 50 3 150 60
V2
z 22 = = 1 / 0.09 = 11.11
I2
1 11.11 V
V1 = Vx = V2 = I 2 = 3.704I 2
→ z12 = 1 = 3.704
3 3 I2
Thus,
29.88 3.704
[z] = Ω
− 70.37 11.11
Chapter 19, Solution 8.
To get z11 and z21, consider the circuit below.
j4 Ω
I1 -j2 Ω 5Ω I2 =0
• •
+
j6 Ω j8 Ω
+
V2
V1
10 Ω -
-
V
V1 = (10 − j2 + j6)I1
→ z11 = 1 = 10 + j4
I1
V2
V2 = −10I1 − j4I1
→ z 21 = = −(10 + j4)
I1
To get z22 and z12, consider the circuit below.
j4 Ω
I1=0 -j2 Ω 5Ω I2
• •
+
j6 Ω j8 Ω
+
V2
V1
10 Ω -
-
V2
V2 = (5 + 10 + j8)I 2
→ z 22 = = 15 + j8
I2
V
V1 = −(10 + j4)I 2
→ z12 = 1 = −(10 + j4)
I2
Thus,
(10 + j4) − (10 + j4)
[z] = Ω
− (10 + j4) (15 + j8)
Chapter 19, Solution 9.
It is evident from Fig. 19.5 that a T network is appropriate for realizing the z
parameters.
R1 R2 6Ω 2Ω
R3 4Ω
R 1 = z 11 − z 12 = 10 − 4 = 6 Ω
R 2 = z 22 − z 12 = 6 − 4 = 2 Ω
R 3 = z 12 = z 21 = 4 Ω
Chapter 19, Solution 10.
(a) This is a non-reciprocal circuit so that the two-port looks like the one
shown in Figs. (a) and (b).
I1 z11 z22 I2
+ +
+ +
V1 z12 I2 − − z21 I1 V2
− −
(a)
(b) This is a reciprocal network and the two-port look like the one shown in
Figs. (c) and (d).
I1 z11 – z12 z22 – z12 I2
+ +
V1 z12 V2
− −
(c)
I1 25 Ω 10 Ω I2
+ +
+ +
V1 20 I2 − −
5 I1 V2
− −
(b)
2 1
z 11 − z 12 = 1 + = 1+
s 0.5 s
z 22 − z 12 = 2s
1
z 12 =
s
I1 1Ω 0.5 F 2H I2
+ +
V1 1F V2
− −
(d)
Chapter 19, Solution 11.
This is a reciprocal network, as shown below.
1+j5 3+j 1Ω j5 Ω 3Ω j1 Ω
5-j2 5Ω
-j2 Ω
Chapter 19, Solution 12.
This is a reciprocal two-port so that it can be represented by the circuit in Figs. (a)
and (b).
I1 z11 – z12 z22 – z12 I2
+ +
V1 z12 V2 2Ω
− −
(a)
I1 8Ω 2Ω I2
Io
+ +
V1 4Ω V2 2Ω
− −
(b)
From Fig. (b),
V1 = (8 + 4 || 4) I 1 = 10 I 1
By current division,
1
Io = I , V2 = 2 I o = I 1
2 1
V2 I1
= = 0 .1
V1 10 I 1
Chapter 19, Solution 13.
This is a reciprocal two-port so that the circuit can be represented by the circuit below.
40 Ω 50 Ω 20 Ω
+ I1 + + I2
120∠0° V 10 I2 − −
30 I1 100 Ω
−
rms
We apply mesh analysis.
For mesh 1,
- 120 + 90 I 1 + 10 I 2 = 0
→ 12 = 9 I 1 + I 2 (1)
For mesh 2,
30 I 1 + 120 I 2 = 0 → I 1 = -4 I 2 (2)
Substituting (2) into (1),
- 12
12 = -36 I 2 + I 2 = -35 I 2 → I 2 =
35
2
1 2 1 12
P = I 2 R = (100) = 5.877 W
2 2 35
Chapter 19, Solution 14.
To find Z Th , consider the circuit in Fig. (a).
I1 I2
+
+
ZS V1 Vo = 1
−
−
(a)
V1 = z 11 I 1 + z 12 I 2 (1)
V2 = z 21 I 1 + z 22 I 2 (2)
But
V2 = 1 , V1 = - Z s I 1
- z 12
Hence, 0 = (z 11 + Z s ) I 1 + z 12 I 2
→ I 1 = I
z 11 + Z s 2
- z 21 z 12
1= + z 22 I 2
z 11 + Z s
V2 1 z z
Z Th = = = z 22 − 21 12
I2 I2 z 11 + Z s
To find VTh , consider the circuit in Fig. (b).
ZS I1 I2 = 0
+ +
+
VS V1 V2 = VTh
−
− −
(b)
I2 = 0 , V1 = Vs − I 1 Z s
Substituting these into (1) and (2),
Vs
Vs − I 1 Z s = z 11 I 1
→ I 1 =
z 11 + Z s
z 21 Vs
V2 = z 21 I 1 =
z 11 + Z s
z 21 Vs
VTh = V2 =
z 11 + Z s
Chapter 19, Solution 15.
(a) From Prob. 18.12,
z12z 21 80x 60
ZTh = z 22 − = 120 − = 24
z11 + Zs 40 + 10
ZL = ZTh = 24Ω
z 21 80
(b) VTh = Vs = (120) = 192
z11 + Zs 40 + 10
V 2Th 192 2
Pmax = = = 192 W
8R Th 8x 24
Chapter 19, Solution 16.
As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).
5Ω 10 – j6 Ω 4 – j6 Ω
a
+
15∠0° V j6 Ω j4 Ω
−
(a) b
At terminals a-b,
Z Th = (4 − j6) + j6 || (5 + 10 − j6)
j6 (15 − j6)
Z Th = 4 − j6 + = 4 − j6 + 2.4 + j6
15
Z Th = 6.4 Ω
j6
VTh = (15∠0°) = j6 = 6∠90° V
j6 + 5 + 10 − j6
The Thevenin equivalent circuit is shown in Fig. (b).
6.4 Ω
+
+
6∠90° V Vo j4 Ω
−
−
(b)
From this,
j4
Vo = ( j6) = 3.18∠148°
6.4 + j4
v o ( t ) = 3.18 cos( 2t + 148°) V
Chapter 19, Solution 17.
To obtain z 11 and z 21 , consider the circuit in Fig. (a).
4Ω Io I2 = 0
Io '
+ 2Ω +
I1 V1 V2
8Ω
− −
6Ω
(a)
In this case, the 4-Ω and 8-Ω resistors are in series, since the same current, I o , passes
through them. Similarly, the 2-Ω and 6-Ω resistors are in series, since the same current,
I o ' , passes through them.
V1 (12)(8)
z 11 = = (4 + 8) || (2 + 6) = 12 || 8 = = 4 .8 Ω
I1 20
8 2 3
Io = I1 = I1 Io' = I
8 + 12 5 5 1
But - V2 − 4 I o + 2 I o ' = 0
-8 6 -2
V2 = -4 I o + 2 I o ' = I1 + I1 = I
5 5 5 1
V2 - 2
z 21 = = = -0.4 Ω
I1 5
To get z 22 and z 12 , consider the circuit in Fig. (b).
I1 = 0 4Ω
+ 2Ω +
V1 V2 I2
8Ω
− −
6Ω
(b)
V2 (6)(14)
z 22 = = (4 + 2) || (8 + 6) = 6 || 14 = = 4 .2 Ω
I2 20
z12 = z 21 = -0.4 Ω
Thus,
4.8 - 0.4
[z ] = Ω
- 0.4 4.2
We may take advantage of Table 18.1 to get [y] from [z].
∆ z = (4.8)(4.2) − (0.4) 2 = 20
z 22 4.2 - z 12 0.4
y 11 = = = 0.21 y 12 = = = 0.02
∆z 20 ∆z 20
- z 21 0.4 z 11 4.8
y 21 = = = 0.02 y 22 = = = 0.24
∆z 20 ∆z 20
Thus,
0.21 0.02
[y ] = S
0.02 0.24
Chapter 19, Solution 18.
To get y 11 and y 21 , consider the circuit in Fig.(a).
I1 6Ω 3Ω I2
+
V1 +
6Ω 3Ω V2 = 0
−
−
(a)
V1 = (6 + 6 || 3) I 1 = 8 I 1
I1 1
y 11 = =
V1 8
-6 - 2 V1 - V1
I2 = I1 = =
6+3 3 8 12
I 2 -1
y 21 = =
V1 12
To get y 22 and y 12 , consider the circuit in Fig.(b).
I1 6Ω Io 3Ω I2
+
+
V1 = 0 6Ω 3Ω V2
−
−
(b)
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