GATE ECE · Network Theory

Two Port Networks Practice Questions

Generate GATE-level questions on Two Port Networks. Focus on: 1. Z, Y, h, and ABCD parameters. 2. Interconnection of two-port networks. 3. Reciprocity and Symmetry conditions.

30 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025MCQ

The Z-parameter matrix of a two-port network relates the port voltages and port currents as follows:

\left[\begin{array}{l} \mathrm{V}_{1} \\ \mathrm{~V}_{2} \end{array}\right]=\mathrm{Z}\left[\begin{array}{l} \mathrm{I}_{1} \\ \mathrm{I}_{2} \end{array}\right]

The Z-parameter matrix (with each entry in ohms) of the network shown below is __________.

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📖 Explanation

$\Delta \rightarrow \mathrm{Y}$ conversion ( $\mathrm{Z} \rightarrow \mathrm{Z} / 3$ ) For T-network, Z-parameters [latex]

\begin{aligned} & {\left[\begin{array}{ll} \mathrm{Z}_{11} & \mathrm{Z}_{12} \mathrm{Z}_{21} & \mathrm{Z}_{22} \end{array}\right]=\left[\begin{array}{cc} \frac{8}{3}+\frac{2}{3} & \frac{2}{3} \frac{2}{3} & \frac{8}{3}+\frac{2}{3} \end{array}\right]} & {[\mathrm{Z}]=\left[\begin{array}{cc} \frac{10}{3} & \frac{2}{3} \frac{2}{3} & \frac{10}{3} \end{array}\right]} \end{aligned}

[/latex]

Q5GATE 2023MCQ

The S-parameters of a two port network is given as

[S]=\left[\begin{array}{ll} S_{11} & S_{12} \\ S_{21} & S_{22} \end{array}\right]

with reference to $Z_{0}$ . Two lossless transmission line sections of electrical lengths $\theta_{1}=\beta l_{1}$ and $\theta_{2}=\beta l 2$ are added to the input and output ports for measurement purposes, respectively. The S-parameters $\left[S^{\prime}\right]$ of the resultant two port network is

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📖 Explanation

Let us evaluate $S_{11}$ and $S_{21}$ first at $V_{2}^{+}=0$ (A) $S11:$

\begin{aligned} V_{1}&=V_{1}^{+}+V_{1}^{-}=\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}+\left(V_{1}^{-}\right) e^{+j \beta l_{1}} \\ \therefore \quad V_{1}^{+}&=\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}} \\ V_{1}^{-}&=\left(V_{1}^{-}\right) e^{+j \beta l_{1}} \\ S_{11}&=\frac{V_{1}^{-}}{V_{1}^{+}}=\frac{\left(V_{1}^{-}\right) e^{+j \beta l_{1}}}{\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}}=S_{11}^{\prime} e^{+j 2 \beta l_{1}} \\ \Rightarrow \quad S_{11}^{\prime}&=S_{11} e^{-j 2 \beta l_{1}}=S_{11} e^{-j 2 \theta_{1}} \\ \therefore \quad S_{11}^{\prime}&=S_{11} e^{-j 2 \theta_{1}} \end{aligned}

(B) $S_{21}$ : $V_{2}=V_{2}^{+} e^{+j \beta l_{2}}+V_{2}^{-} e^{-j \beta l_{2}}=\left(V_{2}^{+}\right)^{\prime}+\left(V_{2}^{-}\right)^{\prime}$ Here, $V_{2}^{+}=0$ Hence, $V_{2}=\left(V_{2}^{-}\right)^{\prime}=V_{2}^{-} e^{-j \beta l_{2}}$ [latex]\Rightarrow \quadV_{2}^{-}=\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}} [/latex] From previous discussion in $S_{11}$ ,

\begin{aligned} V_{1}^{+}&=\left(V_{1}^{+}\right) e^{-j l_{1}} \\ \therefore \quad S_{21}&=\frac{V_{2}^{-}}{V_{1}^{+}}=\frac{\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}}}{\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}}=S_{21}^{\prime} e^{+j\left(\beta l_{2}+\beta l_{1}\right)} \\ \Rightarrow S_{21}^{\prime}&=S_{21} e^{-j\left(\beta l_{2}+\beta l_{1}\right)} \\ \Rightarrow \quad S_{21}^{\prime}&=S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} \end{aligned}

Q18GATE 2014MCQ

Consider the building block called 'Network N ' shown in the figure. Let C = 100 $\mu$ F 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

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📖 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}

Q19GATE 2012MCQ

With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed: (i) 1 $\Omega$ connected at port B draws a current of 3 A (ii) 2.5 $\Omega$ connected at port B draws a current of 2 A For the same network, with 6 V dc connected at port A, 1 $\Omega$ connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is

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📖 Explanation

Case (i):

\begin{aligned} V_{\mathrm{DC}}&=10 \mathrm{V}, R_{L}=1 \Omega, I=3 \mathrm{A} \\ V_{\mathrm{Th}}&=I\left(R_{\mathrm{Th}}+R_{L}\right)=3\left(R_{\mathrm{Th}}+1\right)\\ \Rightarrow \quad V_{\mathrm{Th}}&=3 R_{\mathrm{Th}}+3 &\ldots(i) \end{aligned}

Case (ii):

\begin{aligned} R_{L}&=2.5 \Omega, I=2 \mathrm{A}, V_{\mathrm{DC}}=10 \mathrm{V} \\ V_{\mathrm{Th}}&=2\left(R_{\mathrm{Th}}+2.5\right) \\ V_{\mathrm{Th}}&=2 R_{\mathrm{Th}}+5 &\ldots(ii) \end{aligned}

From equations (i) and (ii) $V_{\mathrm{Th}}=9 \mathrm{V} and R_{\mathrm{TH}}=2 \Omega \quad\ldots(iii)$ Now, $V_{Th}$ depends on independent voltage source and varies with applied voltage. $R_{Th}$ does not depend on voltage source and is same for any applied voltage source, since voltage_ source is short circuited while calculating $R_{Th}$

\begin{aligned} \text{For }V_{\mathrm{DC}}&=6 \mathrm{V}, R_{\mathrm{L}}=1 \Omega, I=7 / 3 \\ V_{T h}&=I\left(R_{T h}+R_{L}\right) \\ V_{T h}&=\frac{7}{3}(2+1)=7 V &\ldots(iv) \end{aligned}

$\because$ The network is linear and non reciprocal, it may contain dependent voltage source

\begin{aligned} \therefore \quad V_{T h}&=a V+b \\ \Rightarrow \quad 9&=a 10+b \quad \text{from equation (iii)} \\ \text{and }\quad 7&=a 6+b \quad\text{from equation (iv)} \\ \end{aligned}

Solving, we get a=1/2 and b=4 $\therefore \quad V_{\mathrm{Th}}=\frac{V}{2}+4=\frac{8}{2}+4=8 \mathrm{V}$