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
(c) y-parameter matrices
(b) h-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
Soln.
(c) h21 = -h12
(d) AD – BC = 0
[GATE 1992: 2 Marks]
𝒚𝟐𝟏 = 𝒚𝟏𝟐
𝒉𝟐 = −𝒉𝟏𝟐
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
4. In the circuit of figure, the equivalent impedance seen across terminals A, B is
A
2
4
j2
Zeq
a
b
-j 2
2
4
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
𝒁𝒆𝒒 = (𝟐‖𝟒) + (𝟐‖𝟒)
=
𝟒 𝟒 𝟖
+ =
𝟑 𝟑 𝟑
Option (b)
5. The short-circuit admittance matrix of a two-port network is
0 −1⁄2
[
]
1⁄
0
2
The two – port network is
(a) non – reciprocal and passive
(c) reciprocal and passive
(b) non – reciprocal and active
(d) reciprocal and active
[GATE 1998: 1 Marks]
Soln.
𝒚𝟏𝟐 ≠ 𝒚𝟐𝟏
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
I1
I2
+
+
R
V1
R
R
V2
-
(a) -1/2
(b) +1/2
(c) -3/2
(d) +3/2
[GATE 1999: 1 Mark]
Soln.
𝑽𝟏 = 𝒉𝟏𝟏 𝑰𝟏 + 𝒉𝟏𝟐 𝑽𝟐
𝑰𝟐 = 𝒉𝟐𝟏 𝑰𝟏 + 𝒉𝟐𝟐 𝑽𝟐
𝒉𝟐𝟏 =
𝑰𝟏
|
𝑰𝟐 𝑽𝟐=𝟎
𝑽𝟐 = 𝑹𝑰𝟐 + 𝑹(𝑰𝟏 + 𝑰𝟐 )
Or 𝟐𝑹𝑰𝟐 + 𝑹𝑰𝟏 = 𝑽𝟐
When 𝑽𝟐 = 𝟎, 𝟐𝑹𝑰𝟐 + 𝑹𝑰𝟏 = 𝟎
So,
𝑰𝟐
𝑰𝟏
=
−𝑹
𝟐𝑹
Option (a)
=−
𝟏
𝟐
7. The admittance parameter Y12 in the 2-port network in figure is
I1
E1
20
I2
5
(a) -0.2 nho
(b) 0.1 mho
Soln.
𝒚𝟏𝟐 = −
E2
5
(c) -0.05 mho
(d) 0.05 mho
[GATE 2001: 1 Mark]
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
2
4
E2
E1
10E1
(a) 𝑍11 =
−6
16
Ω, 𝑍21 = 11 Ω
11
6
4
(b) 𝑍11 = 11 Ω, 𝑍21 = 11 Ω
6
−16
4
11
4
(c) 𝑍11 = 11 Ω, 𝑍21 =
Ω
(d) 𝑍11 = 11 Ω, 𝑍21 = 11 Ω
[GATE 2001: 2 Marks]
Soln. For z – parameters
𝑬𝟏 = 𝒁𝟏𝟏 𝑰𝟏 + 𝒁𝟏𝟐 𝑰𝟐
𝑬𝟐 = 𝒁𝟐𝟏 𝑰𝟏 + 𝒁𝟐𝟐 𝑰𝟐
Writing KVL in LHS loop
𝑬𝟏 = 𝟐𝑰𝟏 + 𝟒𝑰𝟏 + 𝟒𝑰𝟐 − 𝟏𝟎𝑬𝟏
Or
𝟏𝟏𝑬𝟏 = 𝟔𝑰𝟏 + 𝟒𝑰𝟐 − − − −(𝑰)
𝒁𝟏𝟏 =
𝒁𝟏𝟐 =
𝑬𝟏
𝑰𝟏
𝑬𝟏
𝑰𝟐
|𝑰𝟐=𝟎 =
|𝑰𝟏=𝟎 =
𝟔
𝟏𝟏
𝟒
𝟏𝟏
Ω
Ω
Writing KVL in RHS Loop
𝑬𝟐 = 𝟒(𝑰𝟏 + 𝑰𝟐 ) − 𝟏𝟎𝑬𝟏 − − − − − (𝑰𝑰)
𝟔𝑰𝟏 +𝟒𝑰𝟐
Substituting 𝑬𝟏 =
𝟏𝟏
𝑬𝟐 = 𝟒(𝑰𝟏 + 𝑰𝟐 ) −
𝒁𝟐𝟏 =
𝑬𝟐
𝑰𝟏
|𝑰𝟐=𝟎 =
𝒊𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 − −(𝑰𝑰𝑰)
𝟏𝟎(𝟔𝑰𝟏 +𝟒𝑰𝟐 )
−𝟏𝟔
𝟏𝟏
𝟏𝟏
Ω
Option (c)
9. The impedance parameters Z11 and Z12 of the two-port network in the figure are
2
2
2
1
2
1
1
(a) 𝑍11 = 2.75Ω 𝑎𝑛𝑑 𝑍12 = 0.25 Ω
(b) 𝑍11 = 3 Ω 𝑎𝑛𝑑 𝑍12 = 0.5Ω
1
2
(c) 𝑍11 = 3 Ω 𝑎𝑛𝑑 𝑍12 = 0.25Ω
(d) 𝑍11 = 2.25Ω 𝑎𝑛𝑑 𝑍12 = 0.5 Ω
[GATE 2003: 2 Marks]
Soln. Using ∆ − 𝒀 conversion, the circuit reduces to
2
0.5
0.5
Z1
2
Z2
Z3
0.25
𝒁𝟏𝟏 = 𝒁𝟏 + 𝒁𝟑
= 𝟐. 𝟓 + 𝟎. 𝟐𝟓
= 𝟐. 𝟕𝟓Ω
𝒁𝟏𝟐 = 𝒁𝟑 = 𝟎. 𝟐𝟓Ω
Option (a)
10. The h parameter of the circuit shown in the figure are
I1
I2
10
+
V1
0.1 0.1
]
−0.1 0.3
10 −1
(b) [
]
1 0.05
(a) [
+
20
V2
30 20
]
20 30
10
1
(d) [
]
−1 0.05
[GATE 2005: 2 Marks]
(c) [
Soln.
𝑽𝟏 = 𝒉𝟏𝟏 𝑰𝟏 + 𝒉𝟏𝟐 𝑽𝟐
𝑰𝟐 = 𝒉𝟐𝟏 𝑰𝟏 + 𝒉𝟐𝟐 𝑽𝟐
Writing KVL in LHS and RHS Loop
𝑽𝟏 = 𝟏𝟎𝑰𝟏 + 𝟐𝟎(𝑰𝟏 + 𝑰𝟐 ) − − − − − −(𝒊)
𝑽𝟐 = 𝟐𝟎(𝑰𝟐 + 𝑰𝟏 ) − − − − − −(𝒊𝒊)
Or 𝑽𝟏 = 𝟏𝟎𝑰𝟏 + 𝑽𝟐
𝒉𝟏𝟏 =
𝒉𝟐𝟏 =
𝒉𝟏𝟐 =
𝒉𝟐𝟐 =
𝑽𝟏
|
𝑰𝟏 𝑽𝟐 =𝟎
𝑰𝟐
|
𝑰𝟏 𝑽𝟐 =𝟎
𝑽𝟏
|
𝑽𝟐 𝑰𝟏 =𝟎
𝑰𝟐
𝑽𝟐
= 𝟏𝟎
= −𝟏
=𝟏
|𝑰𝟏=𝟎 =
𝟏
𝟐𝟎
= 𝟎. 𝟎𝟓Ω
Option (d)
11. In the two network shown in the figure below Z12 and Z21 are, respectively
I1
I2
βI1
re
(a) 𝑟𝑒 𝑎𝑛𝑑 𝛽𝑟0
(b) 0 𝑎𝑛𝑑 − 𝛽𝑟0
r0
(c) 0 𝑎𝑛𝑑 𝛽𝑟0
(d) 𝑟𝑒 𝑎𝑛𝑑 −𝛽𝑟0
[GATE 2006: 1 Mark]
Soln.
I1
I2
re
V1
βI1
r0
𝑽𝟏 = 𝒁𝟏𝟏 𝑰𝟏 + 𝒁𝟏𝟐 𝑰𝟐
𝑽𝟐 = 𝒁𝟐𝟏 𝑰𝟏 + 𝒁𝟐𝟐 𝑰𝟐
𝒁𝟏𝟐 =
𝑽𝟏
|
𝑰𝟐 𝑰𝟏 =𝟎
When 𝑰𝟏 = 𝟎, 𝑽𝟏 = 𝟎,
𝒁𝟐𝟏 =
𝒔𝒐
𝑽𝟐
|
𝑰𝟏 𝑽𝟐 =𝟎
When 𝑰𝟐 = 𝟎, 𝑽𝟐 = −𝜷𝑰𝟏 𝒓𝟎
𝒁𝟐𝟏 = −𝜷𝒓𝟎
Option (b)
𝒁𝟏𝟐 = 𝟎
V2
12. For the two-port network shown, the short-circuit admittance parameter matrix is
0.5
1
2
yb
yc
ya
0.5
0.5
1
4 −2
]𝑆
−2 4
1
−0.5
(b) [
]𝑆
−0.5
4
1 −0.5
]𝑆
0.5
1
4 2
(d) [
]𝑆
2 4
[GATE 2010: 1 Mark]
(a) [
(c) [
Soln. The short circuit admittance parameters of a two port π network:
𝒚𝟏𝟏 = 𝒚𝒂 + 𝒚𝒃 =
𝟏
𝟎.𝟓
+
𝒚𝟏𝟐 = 𝒚𝟐𝟏 = −𝒚𝒃 = −
𝒚𝟐𝟐 = 𝒚𝒃 + 𝒚𝒄 =
𝟏
𝟎.𝟓
+
𝟏
= 𝟒Ω
𝟎.𝟓
𝟏
𝟎.𝟓
𝟏
𝟎.𝟓
= −𝟐Ω
= 𝟒Ω
Option (a)
13. If the scattering matrix [S] of a two port network is
0
0
[𝑆] = [0.2∠00 0.9∠00 ]
0.9∠0 0.1∠0
Then the network is
(a) Lossless and reciprocal
(c) Not lossless but reciprocal
(b) Lossless but not reciprocal
(d) Neither lossless not reciprocal
[GATE 2010: 1 Mark]
Soln. For the reciprocal network |𝑺𝟏𝟐 |
For a loss less network
= |𝑺𝟐𝟏 |
|𝑺𝟏𝟏 |𝟐 + |𝑺𝟏𝟐 |𝟐 = 𝟏
(𝟎. 𝟐)𝟐 + (𝟎. 𝟗)𝟐 ≠ 𝟏
The network is lossy and reciprocal.
Option (c)
14. In the circuit shown below, the network N is described by the following Y matrix:
𝑉
0.1𝑆 −0.01𝑆
𝑌=[
] The voltage gain 𝑉2 is
0.01𝑆
0.1𝑆
1
(a) 1/90
(b) -1/90
(c) -1/99
(d) -1/11
[GATE 2011: 2 Marks]
Soln.
25
I2
I1
+
+
100V
N
V1
-
𝑰𝟐 = 𝒚𝟐𝟏 𝑽𝟏 + 𝒚𝟐𝟐 𝑽𝟐
= 𝟎. 𝟎𝟏 𝑽𝟏 + 𝟎. 𝟏 𝑽𝟐 − − − − − − − (𝒊)
𝑽𝟐 = −𝑰𝟐 𝑹𝑳 = −𝟏𝟎𝟎𝑰𝟐
𝑰𝟐 =
−𝑽𝟐
𝟏𝟎𝟎
Substituting the value of I2 in equation (i)
−𝑽𝟐
𝟏𝟎𝟎
Or
= 𝟎. 𝟎𝟏𝑽𝟏 + 𝟎. 𝟏𝑽𝟐
𝑽𝟐
𝑽𝟏
=
−𝟏
𝟏𝟏
Option (d)
100
V2
-