The document discusses two-port networks and their parameters. It provides examples and solutions to calculate various two-port network parameters like z-parameters, h-parameters, y-parameters, and ABCD parameters. It also discusses conditions for networks to be reciprocal and provides examples to determine if a given network is reciprocal or not based on its parameter matrix. The examples cover calculation of impedance, admittance and other parameters from the given two-port network circuit diagrams.

1. Two Port Networks
1. Two Two-port networks are connected in cascade. The combination is to the represented
as a single two – port network, by multiplying the individual
(a) z-parameter matrices
(b) h-parameter matrices
(c) y-parameter matrices
(d) ABCD parameter
[GATE 1991: 2 Marks]
Soln. ABCD parameters relate the voltage and current at one port to voltage and current
at the other port
Option (d)
2. For a 2-port network to be reciprocal,
(a) z11 = z22
(b) y21 = y12
(c) h21 = -h12
(d) AD – BC = 0
[GATE 1992: 2 Marks]
Soln. 𝒚 𝟐𝟏 = 𝒚 𝟏𝟐
𝒉 𝟐 = −𝒉 𝟏𝟐
Option (b) & (c)
3. The condition, that a 2-port network is reciprocal, can be expressed in terms of its ABCD
parameters as…….
[GATE 1994: 1 Mark]
Soln. AD – BC = 1

2. 4. In the circuit of figure, the equivalent impedance seen across terminals A, B is
4
42
2
Zeq
A
B
j 2
-j 2
a b
(a) (16/3) Ω
(b) (8/3) Ω
(c) (8/3 + 12j) Ω
(d) None of the above
[GATE 1997: 2 Marks]
Soln. The product of the opposite arms are equal, so the bridge is balanced.
The point a and b are at the same potential
𝒁 𝒆𝒒 = (𝟐‖ 𝟒) + (𝟐‖ 𝟒)
=
𝟒
𝟑
+
𝟒
𝟑
=
𝟖
𝟑
Option (b)
5. The short-circuit admittance matrix of a two-port network is
[
0 −1
2⁄
1
2⁄ 0
]
The two – port network is
(a) non – reciprocal and passive
(b) non – reciprocal and active
(c) reciprocal and passive
(d) reciprocal and active
[GATE 1998: 1 Marks]

3. Soln. 𝒚 𝟏𝟐 ≠ 𝒚 𝟐𝟏 so the network is nonreciprocal. And active networks are
nonreciprocal
Option (b)
6. A 2-port network is shown in the figure. The parameter h21 for this network can be given
by
R
RR
I1 I2
V2V1
++
--
(a) -1/2
(b) +1/2
(c) -3/2
(d) +3/2
[GATE 1999: 1 Mark]
Soln. 𝑽 𝟏 = 𝒉 𝟏𝟏 𝑰 𝟏 + 𝒉 𝟏𝟐 𝑽 𝟐
𝑰 𝟐 = 𝒉 𝟐𝟏 𝑰 𝟏 + 𝒉 𝟐𝟐 𝑽 𝟐
𝒉 𝟐𝟏 =
𝑰 𝟏
𝑰 𝟐
| 𝑽 𝟐=𝟎
𝑽 𝟐 = 𝑹𝑰 𝟐 + 𝑹( 𝑰 𝟏 + 𝑰 𝟐)
Or 𝟐𝑹𝑰 𝟐 + 𝑹𝑰 𝟏 = 𝑽 𝟐
When 𝑽 𝟐 = 𝟎, 𝟐𝑹𝑰 𝟐 + 𝑹𝑰 𝟏 = 𝟎
So,
𝑰 𝟐
𝑰 𝟏
=
−𝑹
𝟐𝑹
= −
𝟏
𝟐
Option (a)

4. 7. The admittance parameter Y12 in the 2-port network in figure is
E1 E2
I1 I2
5 5
20
(a) -0.2 nho
(b) 0.1 mho
(c) -0.05 mho
(d) 0.05 mho
[GATE 2001: 1 Mark]
Soln. 𝒚 𝟏𝟐 = −
𝟏
𝟐𝟎
= −𝟎. 𝟎𝟓 𝒎𝒉𝒐𝒔
Option (c)
8. The Z parameters Z11 and Z21 for the 2-port network in the figure are
I1 I2
E2E1
2
4
10E1
(a) 𝑍11 =
−6
11
Ω, 𝑍21 =
16
11
Ω
(b) 𝑍11 =
6
11
Ω, 𝑍21 =
4
11
Ω
(c) 𝑍11 =
6
11
Ω, 𝑍21 =
−16
11
Ω
(d) 𝑍11 =
4
11
Ω, 𝑍21 =
4
11
Ω
[GATE 2001: 2 Marks]
Soln. For z – parameters
𝑬 𝟏 = 𝒁 𝟏𝟏 𝑰 𝟏 + 𝒁 𝟏𝟐 𝑰 𝟐
𝑬 𝟐 = 𝒁 𝟐𝟏 𝑰 𝟏 + 𝒁 𝟐𝟐 𝑰 𝟐
Writing KVL in LHS loop

5. 𝑬 𝟏 = 𝟐𝑰 𝟏 + 𝟒𝑰 𝟏 + 𝟒𝑰 𝟐 − 𝟏𝟎𝑬 𝟏
Or 𝟏𝟏𝑬 𝟏 = 𝟔𝑰 𝟏 + 𝟒𝑰 𝟐 − − − −(𝑰)
𝒁 𝟏𝟏 =
𝑬 𝟏
𝑰 𝟏
| 𝑰 𝟐=𝟎 =
𝟔
𝟏𝟏
Ω
𝒁 𝟏𝟐 =
𝑬 𝟏
𝑰 𝟐
| 𝑰 𝟏=𝟎 =
𝟒
𝟏𝟏
Ω
Writing KVL in RHS Loop
𝑬 𝟐 = 𝟒( 𝑰 𝟏 + 𝑰 𝟐) − 𝟏𝟎𝑬 𝟏 − − − − − (𝑰𝑰)
Substituting 𝑬 𝟏 =
𝟔𝑰 𝟏+𝟒𝑰 𝟐
𝟏𝟏
𝒊𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 − −(𝑰𝑰𝑰)
𝑬 𝟐 = 𝟒( 𝑰 𝟏 + 𝑰 𝟐) −
𝟏𝟎(𝟔𝑰 𝟏+𝟒𝑰 𝟐)
𝟏𝟏
𝒁 𝟐𝟏 =
𝑬 𝟐
𝑰 𝟏
| 𝑰 𝟐=𝟎 =
−𝟏𝟔
𝟏𝟏
Ω
Option (c)
9. The impedance parameters Z11 and Z12 of the two-port network in the figure are
1
1 1
2
2 2 2
1 2
(a) 𝑍11 = 2.75Ω 𝑎𝑛𝑑 𝑍12 = 0.25 Ω
(b) 𝑍11 = 3 Ω 𝑎𝑛𝑑 𝑍12 = 0.5Ω
(c) 𝑍11 = 3 Ω 𝑎𝑛𝑑 𝑍12 = 0.25Ω
(d) 𝑍11 = 2.25Ω 𝑎𝑛𝑑 𝑍12 = 0.5 Ω
[GATE 2003: 2 Marks]

6. Soln. Using ∆ − 𝒀 conversion, the circuit reduces to
Z1
2 0.5
0.25
0.5 2
Z2
Z3
𝒁 𝟏𝟏 = 𝒁 𝟏 + 𝒁 𝟑
= 𝟐. 𝟓 + 𝟎. 𝟐𝟓
= 𝟐. 𝟕𝟓Ω
𝒁 𝟏𝟐 = 𝒁 𝟑 = 𝟎. 𝟐𝟓Ω
Option (a)
10. The h parameter of the circuit shown in the figure are
+
20
10
+
- -
I1 I2
V1 V2
(a) [
0.1 0.1
−0.1 0.3
]
(b) [
10 −1
1 0.05
]
(c) [
30 20
20 30
]
(d) [
10 1
−1 0.05
]
[GATE 2005: 2 Marks]

7. Soln. 𝑽 𝟏 = 𝒉 𝟏𝟏 𝑰 𝟏 + 𝒉 𝟏𝟐 𝑽 𝟐
𝑰 𝟐 = 𝒉 𝟐𝟏 𝑰 𝟏 + 𝒉 𝟐𝟐 𝑽 𝟐
Writing KVL in LHS and RHS Loop
𝑽 𝟏 = 𝟏𝟎𝑰 𝟏 + 𝟐𝟎( 𝑰 𝟏 + 𝑰 𝟐) − − − − − −(𝒊)
𝑽 𝟐 = 𝟐𝟎( 𝑰 𝟐 + 𝑰 𝟏) − − − − − −(𝒊𝒊)
Or 𝑽 𝟏 = 𝟏𝟎𝑰 𝟏 + 𝑽 𝟐
𝒉 𝟏𝟏 =
𝑽 𝟏
𝑰 𝟏
| 𝑽 𝟐=𝟎 = 𝟏𝟎
𝒉 𝟐𝟏 =
𝑰 𝟐
𝑰 𝟏
| 𝑽 𝟐=𝟎 = −𝟏
𝒉 𝟏𝟐 =
𝑽 𝟏
𝑽 𝟐
| 𝑰 𝟏=𝟎 = 𝟏
𝒉 𝟐𝟐 =
𝑰 𝟐
𝑽 𝟐
| 𝑰 𝟏=𝟎 =
𝟏
𝟐𝟎
= 𝟎. 𝟎𝟓Ω
Option (d)
11. In the two network shown in the figure below Z12 and Z21 are, respectively

8. re
I1
r0
I2
βI1
(a) 𝑟𝑒 𝑎𝑛𝑑 𝛽𝑟0
(b) 0 𝑎𝑛𝑑 − 𝛽𝑟0
(c) 0 𝑎𝑛𝑑 𝛽𝑟0
(d) 𝑟𝑒 𝑎𝑛𝑑 −𝛽𝑟0
[GATE 2006: 1 Mark]
Soln.
re
I1
r0
I2
βI1V1 V2
𝑽 𝟏 = 𝒁 𝟏𝟏 𝑰 𝟏 + 𝒁 𝟏𝟐 𝑰 𝟐
𝑽 𝟐 = 𝒁 𝟐𝟏 𝑰 𝟏 + 𝒁 𝟐𝟐 𝑰 𝟐
𝒁 𝟏𝟐 =
𝑽 𝟏
𝑰 𝟐
| 𝑰 𝟏=𝟎
When 𝑰 𝟏 = 𝟎, 𝑽 𝟏 = 𝟎, 𝒔𝒐 𝒁 𝟏𝟐 = 𝟎
𝒁 𝟐𝟏 =
𝑽 𝟐
𝑰 𝟏
| 𝑽 𝟐=𝟎
When 𝑰 𝟐 = 𝟎, 𝑽 𝟐 = −𝜷𝑰 𝟏 𝒓 𝟎
𝒁 𝟐𝟏 = −𝜷𝒓 𝟎
Option (b)

9. 12. For the two-port network shown, the short-circuit admittance parameter matrix is
1
0.5 0.5
2
0.5
1
ya
yb
yc
(a) [
4 −2
−2 4
] 𝑆
(b) [
1 −0.5
−0.5 4
] 𝑆
(c) [
1 −0.5
0.5 1
] 𝑆
(d) [
4 2
2 4
] 𝑆
[GATE 2010: 1 Mark]
Soln. The short circuit admittance parameters of a two port π network:
𝒚 𝟏𝟏 = 𝒚 𝒂 + 𝒚 𝒃 =
𝟏
𝟎.𝟓
+
𝟏
𝟎.𝟓
= 𝟒Ω
𝒚 𝟏𝟐 = 𝒚 𝟐𝟏 = −𝒚 𝒃 = −
𝟏
𝟎.𝟓
= −𝟐Ω
𝒚 𝟐𝟐 = 𝒚 𝒃 + 𝒚 𝒄 =
𝟏
𝟎.𝟓
+
𝟏
𝟎.𝟓
= 𝟒Ω
Option (a)
13. If the scattering matrix [S] of a two port network is
[𝑆] = [0.2∠00
0.9∠00
0.9∠00
0.1∠00]
Then the network is
(a) Lossless and reciprocal
(b) Lossless but not reciprocal
(c) Not lossless but reciprocal
(d) Neither lossless not reciprocal
[GATE 2010: 1 Mark]
Soln. For the reciprocal network | 𝑺 𝟏𝟐| = | 𝑺 𝟐𝟏|
For a loss less network | 𝑺 𝟏𝟏| 𝟐
+ | 𝑺 𝟏𝟐| 𝟐
= 𝟏
( 𝟎. 𝟐) 𝟐
+ ( 𝟎. 𝟗) 𝟐
≠ 𝟏

10. The network is lossy and reciprocal.
Option (c)
14. In the circuit shown below, the network N is described by the following Y matrix:
𝑌 = [
0.1𝑆 −0.01𝑆
0.01𝑆 0.1𝑆
] The voltage gain
𝑉2
𝑉1
is
(a) 1/90
(b) -1/90
(c) -1/99
(d) -1/11
[GATE 2011: 2 Marks]
Soln.
I1 I225
100
-
V1
-
V2
++
N
100V
𝑰 𝟐 = 𝒚 𝟐𝟏 𝑽 𝟏 + 𝒚 𝟐𝟐 𝑽 𝟐
= 𝟎. 𝟎𝟏 𝑽 𝟏 + 𝟎. 𝟏 𝑽 𝟐 − − − − − − − (𝒊)
𝑽 𝟐 = −𝑰 𝟐 𝑹 𝑳 = −𝟏𝟎𝟎𝑰 𝟐
𝑰 𝟐 =
−𝑽 𝟐
𝟏𝟎𝟎
Substituting the value of I2 in equation (i)
−𝑽 𝟐
𝟏𝟎𝟎
= 𝟎. 𝟎𝟏𝑽 𝟏 + 𝟎. 𝟏𝑽 𝟐
Or
𝑽 𝟐
𝑽 𝟏
=
−𝟏
𝟏𝟏
Option (d)