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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
In the above circuit at node 1,
𝑉1 − 𝑉𝑜
𝑉𝑜 𝑉𝑜 − 0
= 2𝐼1 + +
8
2
4
𝑉1 −𝑉𝑜
But 𝐼1 =
therefore
8
𝑉1 − 𝑉𝑜 3𝑉𝑜
0=
+
8
4
0 = 𝑉1 − 𝑉𝑜 + 6𝑉𝑜 => 𝑉1 = −5𝑉𝑜
−5𝑉𝑜 − 𝑉𝑜
𝑇ℎ𝑢𝑠, 𝐼1 =
= −0.75𝑉𝑜
8
𝐼1
𝑎𝑛𝑑 𝑦11 = = 0.15𝑆
𝑉1
At node 2,
(𝑉𝑜 − 0)
+ 2𝐼1 + 𝐼2 = 0
4
−𝐼2 = 0.25𝑉𝑜 − 1.5𝑉𝑜 = −1.25𝑉𝑜
𝐼2
𝐻𝑒𝑛𝑐𝑒 𝑦21 = = −0.25𝑆
𝑉1
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Similarly we get y21 and y22 using the figure.
We get
𝑦12 = −0.05𝑆 𝑎𝑛𝑑 𝑦22 = 0.25𝑆
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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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
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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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
𝑉1 = 40𝐼1 + 𝑗20𝐼2
𝑉2 = 𝑗30𝐼1 + 50𝐼2
Substituting V1 and V2 as
𝑉1 = 100∠0, 𝑉2 = −10𝐼1
Thus the equation becomes
100 = 40𝐼1 + 𝑗20𝐼2
−10𝐼1 = 𝑗30𝐼1 + 50𝐼2 => 𝐼1 = 𝑗2𝐼2
Thus solving these two equations
100 = 𝑗80𝐼2 + 𝑗20𝐼2 => 𝐼2 = −𝑗
Thus 𝐼1 = 2∠0𝐴, 𝐼2 = 1∠ − 90𝐴
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Determinant of the matrix is ∆= 37
𝐴 10
∆ 37
𝑧11 = =
= 5, 𝑧12 = =
= 18.5
𝐶
2
𝐶
2
1 1
𝐷 4
𝑧21 = = = 0.5, 𝑧22 = = = 2
𝐶 2
𝐶 2
5
18.5
[𝑍] = [
]
0.5
2
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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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
𝑉
Hence ℎ11 = 1 = 4Ω
𝐼1
Also by current division
6
2
−𝐼2 =
𝐼1 = 𝐼1
6+3
3
𝐼2
2
𝐻𝑒𝑛𝑐𝑒 ℎ21 = = −
𝐼1
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
𝑉1 =
𝑉2 = 𝑉2
6+3
3
𝑉1
𝐻𝑒𝑛𝑐𝑒 ℎ12 =
𝑉2
Also 𝑉2 = (3 + 6)𝐼2 = 9𝐼2
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
ℎ22 =
𝐼2 1
= 𝑆
𝑉2 9
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
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Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Chapter 16: Two-Port Networks
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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.
𝑦12𝑎 = −𝑗4 = 𝑦21𝑎 , 𝑦11𝑎 = 2 + 𝑗4, 𝑦22𝑎 = 3 + 𝑗4
Or
2 + 𝑗4 −𝑗4
[𝑦𝑎 ] = [
]𝑆
−𝑗4
3 + 𝑗4
And
𝑦12𝑏 = −4 = 𝑦21𝑎 , 𝑦11𝑎 = 4 − 𝑗2, 𝑦22𝑎 = 4 − 𝑗6
4 − 𝑗2
−4
[𝑦𝑏 ] = [
]𝑆
−4
4 − 𝑗6
6 + 𝑗2 −4 − 𝑗4
[𝑦] = [𝑦𝑎 ] + [𝑦𝑏 ] = [
]𝑆
−4 − 𝑗4 7 − 𝑗2
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Chapter 16: Two-Port Networks
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SOLUTION:
Chapter 16: Two-Port Networks
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Chapter 16: Two-Port Networks
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Chapter 16: Two-Port Networks
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SOLUTION:
𝑧22
0.4
=
2.5, ∆𝑌 = 𝑦11 𝑦22 − 𝑦21 𝑦12 = 0.5 × 0.4 − 0.2 × 0.2 = 0.16
∆𝑌 0.16
𝑦12
0.2
𝑧12 = −
=
= 1.25 = 𝑍21
∆𝑌 0.16
𝑦11
0.5
𝑍22 =
=
= 3.125
∆𝑌 0.16
𝑧11 =
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Chapter 16: Two-Port Networks
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SOLUTION:
The circuit is the cascade connection of the two T networks.
A T network has the following network parameters.
𝑅1
𝑅1 (𝑅2 + 𝑅3 )
1
𝑅3
𝐴 = 1 + , 𝐵 = 𝑅3 +
,𝐶 =
,𝐷 = 1 +
𝑅2
𝑅2
𝑅2
𝑅2
Applying this to cascaded networks, we get
𝐴𝑎 = 1 + 4 = 5, 𝐵𝑎 = 8 + 4 × 9 = 44, 𝐶𝑎 = 1, 𝐷𝑎 = 1 + 8 = 9
And
5 44
[𝑇𝑎 ] = [
]
1 9
Similarly
1 6
[𝑇𝑏 ] = [
]𝑆
0.5 4
Thus for total network
27 206
𝑇] = [𝑇𝑎 ][𝑇𝑏 ] = [
]
5.5 42
Chapter 16: Two-Port Networks
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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
𝑉1 = (10 + 20)𝐼1 = 30𝐼1
𝑉2 = 20𝐼1 − 3𝐼1 = 17𝐼1
𝑉1
𝐼1
𝑇ℎ𝑢𝑠 𝐴 =
= 1.765 𝑎𝑛𝑑 𝐶 = = 0.0588𝑆
𝑉2
𝑉2
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,
𝑉1 − 𝑉𝑎 𝑉𝑎
−
+ 𝐼2 = 0
10
20
𝑉 −𝑉
But 𝑉𝑎 = 3𝐼1 and 𝐼1 = 1 𝑎 .
10
Combining these gives
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
𝑉𝑎 = 3𝐼1 𝑎𝑛𝑑 𝑉1 = 13𝐼1
Substituting Va = 3I1 and replacing first term with I1 in the previous equation
3𝐼1
17𝐼1
𝐼1 −
+ 𝐼2 = 0 =>
= −𝐼2
20
20
𝐼1 20
Therefore, 𝐷 = − =
= 1.176
𝐼2 17
𝑉1
13𝐼1
𝐵=− =−
= 15.29Ω
17
𝐼2
− 𝐼1
20
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 16: Two-Port Networks
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 16: Two-Port Networks
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SOLUTION:
Chapter 16: Two-Port Networks
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Chapter 16: Two-Port Networks
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SOLUTION:
Chapter 16: Two-Port Networks
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SOLUTION:
Chapter 16: Two-Port Networks
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SOLUTION:
Chapter 16: Two-Port Networks
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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