Two Port Networks

Question 1
For a 2-port network consisting of an ideal lossless transformer, the parameter S_{21} (rounded off to two decimal places) for a reference impedance of 10 \Omega, is _____.
A
1
B
0.5
C
1.5
D
0.8
GATE EC 2020   Network Theory
Question 1 Explanation: 
For ideal transformer of n : 1, the scattering matrix is
\begin{bmatrix} S_{11} &S_{12} \\ S_{21} & S_{22} \end{bmatrix}=\begin{bmatrix} \frac{n^{2}-1}{n^{2}+1} &\frac{2n}{n6{}} \\ & \end{bmatrix}
S_{21}=\frac{2n}{n^{2}+1}=\frac{2(2)}{2^2+1}=\frac{4}{5}=0.8
Question 2
In the given circuit, the two-port network has the impedance matrix[Z] =\begin{bmatrix} 40 & 60\\ 60 & 120 \end{bmatrix}. The value of Z_L for which maximum power is transferred to the load is _________ \Omega.
A
48
B
38
C
40
D
54
GATE EC 2020   Network Theory
Question 2 Explanation: 
From maximum power transfer theorem
Z_{L}=Z_{th}
Z_{th}=Z_{22}-\frac{Z_{12}\times Z_{21}}{R_{S}+Z_{11}}
For given data,
Z_{th}=120-\frac{60\times 60}{10+40}=48\Omega
Z_{L}=48\Omega
Question 3
The ABCD matrix for a two-port network is defined by:

\begin{bmatrix} V_{1}\\ I_{1} \end{bmatrix}=\begin{bmatrix} A & B\\ C & D \end{bmatrix}=\begin{bmatrix} V_{2}\\ - l_{2} \end{bmatrix}


The parameter B for the given two-port network (in ohms, correct to two decimal places) is _______.
A
4.8
B
5.5
C
3.4
D
6.2
GATE EC 2018   Network Theory
Question 3 Explanation: 
B=-\left.\frac{V_{1}}{I_{2}}\right|_{V_{2}=0}
When V_{2}=0 (i.e., when port-2 is short circuited)


\begin{aligned} I_{1}&=\frac{V_{1}}{2 \Omega+(5 \Omega \| 2 \Omega)}=\frac{7 V_{1}}{24 \Omega}\\ I_{2}&=-I_{1} \times \frac{5 \Omega}{5 \Omega+2 \Omega}=\frac{-5 V_{1}}{24 \Omega}\\ \text { So, } \quad B&=-\frac{V_{1}}{I_{2}}=\frac{24}{5} \Omega=4.80 \Omega \end{aligned}
Question 4
The z-parameter matrix \begin{bmatrix} z_{11} &z_{12} \\ z_{21}& z_{22} \end{bmatrix} for the two-port network shown is
A
\begin{bmatrix} 2 & -2\\ -2&2 \end{bmatrix}
B
\begin{bmatrix} 2 & 2\\ 2&2 \end{bmatrix}
C
\begin{bmatrix} 9 & -3\\ 6&9 \end{bmatrix}
D
\begin{bmatrix} 9 & 3\\ 6&9 \end{bmatrix}
GATE EC 2016-SET-3   Network Theory
Question 4 Explanation: 
Redrawing the circuit,


\begin{aligned} Z_{11} &=\frac{V_{1}}{I_{1}}=3 \| 6=2 \Omega \\ V_{2} &=-3 \times I_{3 \Omega}=-3 \times I_{1} \times \frac{6}{9}=-2 I_{1} \\ \therefore Z_{21} &=\frac{V_{2}}{I_{1}}=-2 \\ \text { Now, } Z_{22} &=\left.\quad \frac{V_{2}}{I_{2}}\right|_{I_{1}=0}=3 \| 6=2 \Omega \\ V_{1} &=-6 \times I_{6 \Omega} \\ &=-6 \times I_{2} \times \frac{3}{9} \\ &=-2 \\ \therefore[Z] &=\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right] \end{aligned}

Question 5
The z-parameter matrix for the two-port network shown is
\begin{bmatrix} 2j\omega & j\omega \\ j\omega & 3+2j\omega \end{bmatrix}
where the entries are in \Omega. Suppose Z_{b}(j\omega )=R_{b}+j\omega. Then the value of R_{b} (in \Omega) equals ________
A
1
B
2
C
3
D
4
GATE EC 2016-SET-2   Network Theory
Question 5 Explanation: 
For T-network,


\begin{aligned} Z_{11} &=Z_{a}+Z_{c} \\ Z_{22} &=Z_{b}+Z_{c} \\ \text { and, } \quad Z_{12} &=Z_{21}=Z_{c} \\ \text { Given, } &[Z]=\left[\begin{array}{cc} 2 j \omega & j \omega \\ j \omega & 3+2 j \omega \end{array}\right] \\ \text { Therefore } Z_{12} &=j \omega \\ \text { and, } \quad Z_{22} &=3+2 j \omega \\ &=3+j \omega+j \omega \\ &=Z_{b}+Z_{c} \\ &=R_{b}+j \omega+Z_{c} \\ & R_{b}=3 \Omega \end{aligned}
Question 6
Consider a two-port network with the transmission matrix: T=\begin{pmatrix} A & B\\ C& D \end{pmatrix}. If the network is reciprocal, then
A
T^{-1}=T
B
T^{2}=T
C
Determinant (T) = 0
D
Determinant (T) = 1
GATE EC 2016-SET-1   Network Theory
Question 6 Explanation: 
For reciprocal network AD-BC= 1
|T|=1
Question 7
The ABCD parameters of the following 2-port network are
A
\begin{bmatrix} 3.5+j2 &20.5 \\ 20.5 & 3.5-j2 \end{bmatrix}
B
\begin{bmatrix} 3.5+j2 &30.5 \\ 0.5 & 3.5-j2 \end{bmatrix}
C
\begin{bmatrix} 10 &2+j0 \\ 2+j0 & 10 \end{bmatrix}
D
\begin{bmatrix} 7+j4 &0.5 \\ 30.5 & 7-j4 \end{bmatrix}
GATE EC 2015-SET-3   Network Theory
Question 7 Explanation: 


\begin{aligned} \left[\begin{array}{l} V_{1} \\ I_{1} \end{array}\right] &=\left[\begin{array}{ll} A & B \\ C & D \end{array}\right]\left[\begin{array}{l} V_{2} \\ I_{2} \end{array}\right] \\ \text { When } I_{2} &=0 \\ V_{1} &=(5+j 4+2) I_{1} \\ V_{2} &=2 I_{1} \\ A &=\left.\frac{V_{1}}{V_{2}}\right|_{I_{2}=0}=\frac{7+j 4}{2}=3.5+j 2 \\ C &=\left.\frac{I_{1}}{V_{2}}\right|_{I_{2}=0}=\frac{1}{2}=0.5 \end{aligned}
Option (B) matches. From here only
Question 8
The 2-port admittance matrix of the circuit shown is given by

A
\begin{bmatrix} 0.3 & 0.2\\ 0.2& 0.3 \end{bmatrix}
B
\begin{bmatrix} 15 & 5\\ 5& 15 \end{bmatrix}
C
\begin{bmatrix} 3.33 & 5\\ 5& 3.33 \end{bmatrix}
D
\begin{bmatrix} 0.3 & 0.4\\ 0.4& 0.3 \end{bmatrix}
GATE EC 2015-SET-2   Network Theory
Question 8 Explanation: 


\begin{aligned} V_{1} &=6 I_{1}+4 I_{2} \\ V_{2} &=4 I_{1}+6 I_{2} \\ [Z] &=\left[\begin{array}{cc} 6 & 4 \\ 4 & 6 \end{array}\right] \\ Y&=\frac{1}{20}\left[\begin{array}{rr} 6 & -4 \\ -4 & 6 \end{array}\right]=\left[\begin{array}{rr} 0.3 & -0.2 \\ -0.2 & 0.3 \end{array}\right] \\ &\text { Ignoring negative sign: } \\ [Y]&=\left[\begin{array}{cc} 0.3 & 0.2 \\ 0.2 & 0.3 \end{array}\right] \end{aligned}
Question 9
For the two-port network shown in the figure, the impedance (Z) matrix (in \Omega) is
A
\begin{bmatrix} 6 & 24\\ 42&9 \end{bmatrix}
B
\begin{bmatrix} 9 & 8\\ 8&24 \end{bmatrix}
C
\begin{bmatrix} 9 & 6\\ 6&24 \end{bmatrix}
D
\begin{bmatrix} 42 & 6\\ 6&60 \end{bmatrix}
GATE EC 2014-SET-4   Network Theory
Question 9 Explanation: 
converting II-network to Y-network, we get


\therefore \quad Z -parameter
\begin{aligned} Z_{11}&=3 \Omega+6 \Omega=9 \Omega \\ Z_{12}&=Z_{21}=6 \Omega \\ Z_{22}&=18 \Omega+6 \Omega=24 \Omega \\ \left[Z\right]&=\left[\begin{array}{cc}9 & 6 \\ 6 & 24\end{array}\right] \end{aligned}
Question 10
Consider the building block called 'Network N ' shown in the figure. Let C = 100 \muF and R = 10 k\Omega.

Two such blocks are connected in cascade as shown in the figure.

The transfer function \frac{V_{3}(s)}{V_{1}(s)} of the cascaded network is
A
\frac{s}{1+s}
B
\frac{s^{2}}{1+3s+s^{2}}
C
(\frac{s}{1+s})^{2}
D
\frac{s}{2+s}
GATE EC 2014-SET-3   Network Theory
Question 10 Explanation: 
The cascaded network is,


Applying mesh analysis to determine the current
I_{2}(\mathrm{s})
We get,,
\begin{array}{l} \left(R+\frac{1}{C s}\right) I_{1}(s)=R I_{2}(s)=V_{1}(s) \qquad \ldots(i)\\ \left(R+\frac{1}{C s}\right) I_{2}(s)-R I_{2}(s)=0 \qquad \ldots(ii) \end{array}
Also,
V_{3}(s)=I_{2}(s) \times R \qquad \ldots(iii)
From equation (i) and (ii), we get
\begin{aligned} &=\left(R+\frac{1}{C s}\right) \times\left(1+\frac{1}{R C s}\right) I_{2}(s)-R I_{2}(s)=V_{1}(s) \\ &=\left(\frac{R C s+1}{C S}\right)\left(\frac{R C s+1}{R C s}\right) I_{2}(s)-R I_{2}(s)=V_{1}(s) \\ \text{or } I_{2}(s) &=\frac{R C^{2} s^{2} V_{1}(s)}{(1+R C s)^{2}-R^{2} C^{2} s^{2}} \;\ldots(iv) \end{aligned}
Using equation (iii) and (iv), we get
\begin{aligned} \frac{V_{3}(s)}{V_{1}(s)}&=\frac{s^{2} R^{2} C^{2}}{1+3 R C s+s^{2} R^{2} C^{2}} \\ \because \qquad R&=10 \mathrm{k} \Omega, C=100 \mu \mathrm{F} \text{ and } R C=1 \\ \frac{V_{3}(s)}{V_{1}(s)}&=\frac{s^{2}}{1+3 s+s^{2}} \end{aligned}
There are 10 questions to complete.
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