Chapter 19, Solution 1. To get
z
and , consider the circuit in Fig. (a).
11 21
z
1
Ω
4
Ω
I
2
= 0
+
V
2

−
I
o
+
V
1

−
2
Ω
6
Ω

zTo get
z
and , consider the circuit in Fig. (b).
22 12
z
1
Ω
4
Ω
I
1
= 0
I
o
'
+
V
2
−
+
V
1

−
2
Ω
6
Ω== 1
2
1
12
I
V
zHence,
=][z
Ω






667.11
14
Chapter 19, Solution 2. Consider the circuit in Fig. (a) to get and .
11

1 Ω 1 Ω 1 Ω 1 Ω
(a)

])12(||12[||12
1
1
11
+++==
I
V
z733.2
15
11
2
4111
)411)(1(
2
4
3
2||12
11
=+=
+
+=




15
1
15
4
4
1
III =⋅=

1o2
15
1
IIV ==

06667.0
15
1
12
1
2
21
==== z
I
V
z

To get
z

==++== z
I
V
z

Thus,
=][z Ω






733.206667.0
06667.0733.2
Chapter 19, Solution 3. (a)

To find and , consider the circuit in Fig. (a).
11
z
21
z
I
o

V
zBy current division,
11o
j
j1j
j
III =
−+
=

1o2
jIIV ==

j
1
2
21
==
I
V
zTo get
z
and , consider the circuit in Fig. (b).
22 12


21
j
IV
=j
2
1
12
==
I
V
z

Thus,
=][z Ω






+
0j
jj1(b)

1

(c)

5.0j5.1
j1
j-
j1-j)(||11j
1
1
11
+=
−
++=++==
I
V
z12
)5.0j5.1(
IV
−=5.0j5.1
1
2
21
−==

1
= 0
j Ω -j Ω
I
2

(d)

j1.5-1.5(-j)||11-j
2
2
22
=++==
I
V
z21
)5.0j5.1(
IV
−=5.0j5.1
2
1
12
−==
I

Z
25j12
120j
5j10j12
)10j)(12(
1
+
=
−+
=Zj512
j60-
2
+
=Z5j12
50
3
+
=ZThe z parameters are







−−
−+
739.5j775.126.4j775.1-
26.4j775.1-26.4j775.1
Chapter 19, Solution 5. Consider the circuit in Fig. (a).
s
1
I
2
= 0
I
o
1/s1/s
+
V
2

−

1
s1||
s
1
||1
11
+++






+






++






+
=


1ss
1s
s
1s
s
s
1
s1
1s
1
1s
1
s
1
s1
s
1
||1
s
1
||1
IIII
+++
+
+
=
+++
+
+
=

2
21
+++
==
I
V
zConsider the circuit in Fig. (b).
s
1
I
1
= 0
1/s1/s 1
+
V
1

−
+
V
2
−
I
2

(b)


z1s
s
ss1
1s
1
s1
1s
1
s1
s
1
1s
1
s1
s
1
2
22
+
+++
+
++
=
+
+++



][
z










+++
++
+++
++++++
++
1s3s2s
2s2s
1s3s2s
1
1s3s2s
1
1s3s2s
1ss
23
2
23
2323
2

−
10 Ω
V
1

(a)

50
5.0
10
o
2
o1
V
V
VV
+=
−
, where
oo2
5
3
3020
30
VV =
+
V =o
Ω=== 125.13
32.0
2.4
o
o
1
1
11
V
V
I
V
zΩ=== 875.1
32.0
6.0
o
o
1
2
21
V
V
I
V
z


(b)
2
2
22
5333.0
30
5.0 V
V
VI =+=

Ω=== 875.1
5333.0
1
2
2
22
I
V
z2221
-9)5.0)(20( VVVV =−=Ω=== -16.875
5333.0
9-
2

and z
21
, we consider the circuit below.
I
2
=0
I
1
20 100
Ω Ω
+
+ +
v
x
50
Ω
60
Ω

V
1
- V
2

- -

12v

(
121
81
20
VV
I
1
1
11
1x1
1
==→=
−
=37.70
I
V
zI37.70
I
81
121x20
)
121
40
(
8
57
V)

I
1
=0 20 100
Ω Ω
+
+ +
v
x
50
Ω
60
Ω

V
1
- V
2

- -

12v
x
- +



704.3
I
V
zI704.3I
3
11.11
V
3
1
VV
2
1
12222x1
==→====

Thus,
Ω






−
=
11.1137.70
704.388.29
]z[


V
1

10
Ω -
-

4j10
I
V
zI)6j2j10(V
1
1
1111
+==→+−=

)4j10(
I
V
zI4jI10V
1
2
21112
+−==→−−=

To get z
22

10
Ω -
-
8j15
I
V
zI)8j105(V
2
2
2222
+==→++=

)4j10(
I
V
zI)4j10(V
2
1
1221
+−==→+−=
Thus,
Ω





=−=−= 410R
12111
zz
Ω
6

=−=−= 46R
12222
zz
Ω
2

===
21123
R
zz
Ω
4
Chapter 19, Solution 10. (a)

This is a non-reciprocal circuit so that
the two-port looks like the one

I
2
z
11

(a)

(b)

This is a reciprocal network and
the two-port look like the one shown in
Figs. (c) and (d)
.
+
V
1

−
I
2
I
1
+
V
2
−
z
12

z

10
Ω
25
Ω
(b)s5.0
1
1
s
2
1
1211
+=+=−
zzs2
1222
=− zzs
1
12
=
z

1 F

3
Ω

5-j2 5
Ω
-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).
Ω
2 Ω
2 Ω8 Ω
I
1
+
V

12
z
11
– z
12
j1

I
o
2 Ω
(b)

From Fig. (b),
111
10)4||48( IIV =+=By current division,
1o
2
1
II = ,
1o2
2 IIV ==

==
1
1
1
2

−
100 Ω

We apply mesh analysis.
For mesh 1,
2121
91201090120- IIII +=→=++ (1)
For mesh 2,
2121
-4012030 IIII =→=+ (2)
Substituting (2) into (1),
35
12-
-35-3612
2222
=→=+= IIII

=






== )100(
35
12
2
1
R

S
V
o
= 1
(a)

2121111
IzIzV
+=
(1)
2221212
IzIzV +=
(2)
But
1
2
=V
,
1s1
- IZV =

Hence,
2
s11
12
12121s11
-
)(0 I
Zz
z

Z
s11
1221
22
Zz
zz
z
+
−To find
, consider the circuit in Fig. (b).
Th
V
+
−
I
2
= 0
Z
S

+
V
2
= V
Th

−

s21
1212
Zz
Vz
IzV
+
====
2Th
VV
s11
s21
Zz
Vz
+
Chapter 19, Solution 15. (a) From Prob. 18.12,

24
1040
60x80
120
Zz

=
V W192
24x8
192
R8
V
P
2
Th
Th
2
max
===
Chapter 19, Solution 16. As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).
+
−
j6
Ω
b
a
(a)

Z=
Th
Z
Ω4.6==°∠
−++
= 6j)015(
6j1056j
6j
Th
V
V906 °∠The Thevenin equivalent circuit is shown in Fig. (b).
6.4 Ω
+
V
o
−
+
−
6∠90° V
j4 Ω
(b)

o
'
+
V
2
−
+
V
1

−
2 Ω
8 Ω
I
1

(a)
6 Ω
In this case, the 4-
Ω
and 8-
Ω
resistors are in series, since the same current, , passes
through them. Similarly, the 2-
Ω
and 6-
Ω
resistors are in series, since the same current,
, passes through them.
o

o
5
3
II
=But
024-
'
oo2
=+− IIV

111
'
oo2
5
2-
5
6
5
8-
2-4 IIIIIV =+=+=

Ω===
-0.4
5
2-
1
2

6 Ω

Ω===++==
2.4
20
)14)(6(
14||6)68(||)24(
2
22
2
I
V
zΩ==
-0.4
2112
zz

Thus,
=][z
Ω






4.20.4-

z
12
12
==
∆
=
z
y02.0
20
4.0
-
z
21
21
==
∆
=
z
y

24.0
20
8.4
z
11
22
==

1
+
−
+
V
2
= 0
−
(a)
6 Ω
3 Ω
3 Ω
6 Ω
V
1111
8)3||66( IIV
=+=8
1
1
1
11
==
V
I

and , consider the circuit in Fig.(b).
22 12
y
6 Ω 3 Ω
I
o
I
1
I
2
+
−
+
V
1
= 0
−
3 Ω6 Ω
V
2

(b)

2
1
6||3
1
)6||63(||3
1
2

1
6
1-
6
-
2
2
2
1
V
V
I
I
=












==21



Chapter 19, Solution 19. Consider the circuit in Fig.(a) for calculating and
y
.
11
y
21
1/s
I
2
I
1
+
−
+
V
2
= 0
−
s
1
1
V

1
11
+=
+
==
V
I
y2
-
1s2
-
2)s1(
)s1-(
11
12
VI
II =
+
=
+
=-0.5
1
2
21

222
2s
s2
)2||s(
IIV
+
==s
1
5.0
s2
2s
2
2
22
+=
+
==
V
I
y2
-
s2
2s
2s






+
+

Chapter 19, Solution 20.
To get y
11
and y
21
, consider the circuit below. 3i
x

Since 6-ohm resistor is short-circuited, i
x
= 0

75.0
V
I
yI
6
8
)2//4(IV
1
1
11111
==→==5.0
V
I
yV
2
1
)V
8
6
(


+ i
x
+

V
1
=0 4 6Ω Ω V
2
I
2

-
- 1667.0
6
1
V
I
y
6
V
2
V
i3iI,







−
= Chapter 19, Solution 21. To get and , refer to Fig. (a).
11
y
21
y
At node 1,
10
Ω
I
2
I
1
+
−
+
V
2

2112
==→=
V
I
yV

and
12
y
, refer to the circuit in Fig. (b).
I

o get

ince
, the dependent current source can be replaced with an open circuit.

22
y
0.2 V
1
T
+
V
1
0
+
I
1
I

I
yIV0
2
1
12
==
V
I

y
Thus,

onsequently,
the y parameter equivalent circuit is shown in Fig. (c)
=][y
S
1.02.0-
04.0







C
.

y
Ω Ω
2
1
Ω
(a)
+
V
2
0
V
1

=
−
+
I
1
I
2
V
1
+
V
x

−
3
V
x


ubstituting (2) into (1) gives,
I =
1
V −
S