# ch16

```1
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
2
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
3
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
4
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
In the above circuit at node 1,
π1 − ππ
ππ ππ − 0
= 2πΌ1 + +
8
2
4
π1 −ππ
But πΌ1 =
therefore
8
π1 − ππ 3ππ
0=
+
8
4
0 = π1 − ππ + 6ππ =&gt; π1 = −5ππ
−5ππ − ππ
πβπ’π , πΌ1 =
= −0.75ππ
8
πΌ1
πππ π¦11 = = 0.15π
π1
At node 2,
(ππ − 0)
+ 2πΌ1 + πΌ2 = 0
4
−πΌ2 = 0.25ππ − 1.5ππ = −1.25ππ
πΌ2
π»ππππ π¦21 = = −0.25π
π1
Chapter 16: Two-Port Networks
5
Irwin, Engineering Circuit Analysis, 11e ISV
Similarly we get y21 and y22 using the figure.
We get
π¦12 = −0.05π πππ π¦22 = 0.25π
Chapter 16: Two-Port Networks
6
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
7
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
8
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
9
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
10
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
11
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
12
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
This reciprocal network is shown as below:
Replacing the blocks with equivalent elements
Chapter 16: Two-Port Networks
13
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
14
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
15
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
16
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
17
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
18
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
19
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
20
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
21
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
22
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
This is not a reciprocal network. We may use the equivalent circuit in the figure.
Writhing the equations
π1 = 40πΌ1 + π20πΌ2
π2 = π30πΌ1 + 50πΌ2
Substituting V1 and V2 as
π1 = 100∠0, π2 = −10πΌ1
Thus the equation becomes
100 = 40πΌ1 + π20πΌ2
−10πΌ1 = π30πΌ1 + 50πΌ2 =&gt; πΌ1 = π2πΌ2
Thus solving these two equations
100 = π80πΌ2 + π20πΌ2 =&gt; πΌ2 = −π
Thus πΌ1 = 2∠0π΄, πΌ2 = 1∠ − 90π΄
Chapter 16: Two-Port Networks
23
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
24
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
25
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Determinant of the matrix is β= 37
π΄ 10
β 37
π§11 = =
= 5, π§12 = =
= 18.5
πΆ
2
πΆ
2
1 1
π· 4
π§21 = = = 0.5, π§22 = = = 2
πΆ 2
πΆ 2
5
18.5
[π] = [
]
0.5
2
Chapter 16: Two-Port Networks
26
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
27
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
28
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
29
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
30
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
31
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
32
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
33
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
34
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
35
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
36
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
To find h11 and h21, we short circuit the output port and connect a current source I1 and
to the input port as shown.
V1 = I1 (2 + 3||6 ) = 4I1
π
Hence β11 = 1 = 4β¦
πΌ1
Also by current division
6
2
−πΌ2 =
πΌ1 = πΌ1
6+3
3
πΌ2
2
π»ππππ β21 = = −
πΌ1
3
To obtain h12 and h22, we open circuit the input port and connect voltage source V2 to
the output port as given in figure.
By voltage division
6
2
π1 =
π2 = π2
6+3
3
π1
π»ππππ β12 =
π2
Also π2 = (3 + 6)πΌ2 = 9πΌ2
Chapter 16: Two-Port Networks
37
Irwin, Engineering Circuit Analysis, 11e ISV
β22 =
πΌ2 1
= π
π2 9
Chapter 16: Two-Port Networks
38
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
39
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
40
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
41
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
42
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
43
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
44
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
45
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
46
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
47
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
48
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
49
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
50
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
51
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
52
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
53
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
54
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
55
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
56
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
57
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
58
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
59
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Let us refer upper network as Na and lower as Nb. The two networks are connected in
parallel.
π¦12π = −π4 = π¦21π , π¦11π = 2 + π4, π¦22π = 3 + π4
Or
2 + π4 −π4
[π¦π ] = [
]π
−π4
3 + π4
And
π¦12π = −4 = π¦21π , π¦11π = 4 − π2, π¦22π = 4 − π6
4 − π2
−4
[π¦π ] = [
]π
−4
4 − π6
6 + π2 −4 − π4
[π¦] = [π¦π ] + [π¦π ] = [
]π
−4 − π4 7 − π2
Chapter 16: Two-Port Networks
60
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
61
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
62
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
63
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
64
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
65
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
π§22
0.4
=
2.5, βπ = π¦11 π¦22 − π¦21 π¦12 = 0.5 &times; 0.4 − 0.2 &times; 0.2 = 0.16
βπ 0.16
π¦12
0.2
π§12 = −
=
= 1.25 = π21
βπ 0.16
π¦11
0.5
π22 =
=
= 3.125
βπ 0.16
π§11 =
Chapter 16: Two-Port Networks
66
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
67
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
68
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
69
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
70
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The circuit is the cascade connection of the two T networks.
A T network has the following network parameters.
π1
π1 (π2 + π3 )
1
π3
π΄ = 1 + , π΅ = π3 +
,πΆ =
,π· = 1 +
π2
π2
π2
π2
Applying this to cascaded networks, we get
π΄π = 1 + 4 = 5, π΅π = 8 + 4 &times; 9 = 44, πΆπ = 1, π·π = 1 + 8 = 9
And
5 44
[ππ ] = [
]
1 9
Similarly
1 6
[ππ ] = [
]π
0.5 4
Thus for total network
27 206
π] = [ππ ][ππ ] = [
]
5.5 42
Chapter 16: Two-Port Networks
71
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
To determine A and C, we leave the output port open as shown in figure so that I2 = 0
and place a voltage source V1 at the input port.
We have
π1 = (10 + 20)πΌ1 = 30πΌ1
π2 = 20πΌ1 − 3πΌ1 = 17πΌ1
π1
πΌ1
πβπ’π  π΄ =
= 1.765 πππ πΆ = = 0.0588π
π2
π2
To obtain B and D we short circuit the output port so that V2 = 0 as shown in circuit.
We place a voltage source V1 at the input port.
At node a in the circuit,
π1 − ππ ππ
−
+ πΌ2 = 0
10
20
π −π
But ππ = 3πΌ1 and πΌ1 = 1 π .
10
Combining these gives
Chapter 16: Two-Port Networks
72
Irwin, Engineering Circuit Analysis, 11e ISV
ππ = 3πΌ1 πππ π1 = 13πΌ1
Substituting Va = 3I1 and replacing first term with I1 in the previous equation
3πΌ1
17πΌ1
πΌ1 −
+ πΌ2 = 0 =&gt;
= −πΌ2
20
20
πΌ1 20
Therefore, π· = − =
= 1.176
πΌ2 17
π1
13πΌ1
π΅=− =−
= 15.29β¦
17
πΌ2
− πΌ1
20
Chapter 16: Two-Port Networks
73
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
74
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
75
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
76
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
77
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
78
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
79
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
80
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
81
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The circuit which is obtained by following the definitions of h parameters is following.
Chapter 16: Two-Port Networks
```