# Network Theory

 Question 1
A sample and hold circuit is implemented using a resistive switch and a capacitor with a time constant of $1 \mu \mathrm{s}$. The time for sampling switch to stay closed to charge a capacitor adequately to a full scale voltage of $1 \mathrm{~V}$ with 12-bit accuracy is ____ $\mu \mathrm{s}$. (rounded off to two decimal places)
 A 2.32 B 6.65 C 8.32 D 12.36
GATE EC 2023      Transient Analysis
Question 1 Explanation:
Given: Time constant $(\tau)$ of $1 \mu \mathrm{sec}$.
Full scale voltage $=1 \mathrm{~V}$ The voltage across capacitor is given as
$\quad \quad V_{c}(t)=V_{\text {in }}\left(1-e^{-t / \tau}\right)$
$\therefore \quad V_{c}(t)=\left(1-e^{-t / 1 \mu \mathrm{sec}}\right) \quad ...(i)$
To calculate the voltage to stay closed to charge capacitor adequately to a full scale voltage with 12-bit accuracy is given by
\begin{aligned} V_{c}(t) & =V_{\text {ref }}\left[1-\frac{1}{2^{n}}\right] \\ n & =\text { Number of bit }=12 \\ V_{c}(t) & =\left[1-\frac{1}{4096}\right] \quad ...(ii) \end{aligned}
Comparing equations (i) and (ii), we get,
\begin{aligned} e^{-t / 1 \mu \mathrm{sec}} & =\frac{1}{4096} \\ -\frac{t}{1 \mu \mathrm{sec}} & =\ln \left\{\frac{1}{4096}\right\} \\ t & =8.3177 \mu \mathrm{sec} \end{aligned}
 Question 2
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 A $\left[\begin{array}{cc}S_{11} e^{-j 2 \theta_{1}} & S_{12} e^{-j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{-j 2 \theta_{2}}\end{array}\right]$ B $\left[\begin{array}{cc}S_{11} e^{j 2 \theta_{1}} & S_{12} e^{-j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{j 2 \theta_{2}}\end{array}\right]$ C $\left[\begin{array}{cc}S_{1} e^{j 2 \theta_{1}} & S_{12} e^{e j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{j 2 \theta_{2}}\end{array}\right]$ D $\left[\begin{array}{cc}S_{11} e^{-j 2 \theta_{1}} & S_{12} e^{e j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{-j 2 \theta_{2}}\end{array}\right]$
GATE EC 2023      Two Port Networks
Question 2 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,
$V_{2}=\left(V_{2}^{-}\right)^{\prime}=V_{2}^{-} e^{-j \beta l_{2}}$
$\Rightarrow \quad V_{2}^{-}=\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}}$

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}

 Question 3
The h-parameters of a two port network are shown below. The condition for the maximum small signal voltage gain $\frac{V_{\text {out }}}{V_{s}}$ is A $h_{11}=0, h_{12}=0, h_{21}=$ very high and $h_{22}=0$ B $h_{11}=$ very high, $h_{12}=0, h_{21}=$ very high and $h_{22}=0$ C $h_{11}=0, h_{12}=$ very high, $h_{21}=$ very high and $h_{22}=0$ D $h_{11}=0, h_{12}=0, h_{21}=$ very high and $h_{22}=$ very high
GATE EC 2023      Two Port Networks
Question 3 Explanation:
Dependent current source should have $h_{21} I_{1}$ instead of $h_{21} V_{1}$ according to $h$-parameter.
$A_{V}=\frac{V_{\text {out }}}{V_{s}}=\frac{-h_{21} I_{1} \times\left(\frac{1}{h_{22}} \| R_{L}\right)}{h_{11} I_{1}+h_{12} V_{2}}$

To achieve maximum $\frac{V_{\text {out }}}{V_{s}}$
\begin{aligned} & h_{11}=0, \quad h_{12}=0 \\ & h_{21}=\text { Very high, } \quad h_{22}=0 \end{aligned}

Hence, answer should be (A) according to $h_{21} I_{1}$.
 Question 4
The switch $S_{1}$ was closed and $S_{2}$ was open for a long time. At $t=0$, switch $S_{1}$ is opened and $S_{2}$ is closed, simultaneously. The value of $i_{c}\left(0^{+}\right)$, in amperes, is A 1 B -1 C 0.2 D 0.8
GATE EC 2023      Transient Analysis
Question 4 Explanation:
At $t=0^{-} ; S_{1} \rightarrow$ closed, $S_{2} \rightarrow$ opened \begin{aligned} i_{L}\left(0^{-}\right)&=\frac{1 \times 25}{100+25}=0.2 \mathrm{~A} \\ v_{C}\left(0^{-}\right)&=\frac{1}{5} \times 100=20 \mathrm{~V} \end{aligned}

At $t=0^{+} ; S_{1} \rightarrow$ opened, $S_{2} \rightarrow$ closed \begin{aligned} i_{x} & =\frac{20}{25}=\frac{4}{5} \mathrm{~A}=0.8 \mathrm{~A} \\ \text{By KCL: }-i_{c} & =i_{x}+0.2=0.8+0.2 \\ \Rightarrow \quad i_{c} & =-1 \mathrm{~A} \end{aligned}
 Question 5
In the circuit shown below, switch $S$ was closed for a long time. If the switch is opened at $t=0$, the maximum magnitude of the voltage $V_{R}$, in volts, is ____ (rounded off to the nearest integer). A 2 B 4 C 6 D 8
GATE EC 2023      Transient Analysis
Question 5 Explanation:
At $t=0^{-}$ $i_{L}\left(0^{-}\right)=\frac{2}{1}=2 \mathrm{~A}$

At $t=0^{+}$ $V_{R}=-2 \times 2=-4$,
Magnitude of voltage $V_{R}$,
$\left|V_{R}\right|=4$

There are 5 questions to complete.