# GATE Electronics and Communication 2021

 Question 1
The vector function $F\left ( r \right )=-x\hat{i}+y\hat{j}$ is defined over a circular arc C shown in the figure.

The line integral of $\int _{C} F\left ( r \right ).dr$ is
 A $\frac{1}{2}$ B $\frac{1}{4}$ C $\frac{1}{6}$ D $\frac{1}{3}$

Question 1 Explanation:
\begin{aligned} \bar{F} &=-x i+y j \\ \int \vec{F} \cdot \overrightarrow{d r} &=\int_{c}-x d x+y d y \\ &=\int_{\theta=0}^{45^{\circ}}(-\cos \theta(-\sin \theta)+\sin \theta \cos \theta) d \theta \\ \int_{\theta=0}^{\pi / 4} \sin 2 \theta d \theta &\left.=-\frac{\cos 2 \theta}{2}\right]_{0}^{\pi / 4} \\ &=-\frac{1}{2}[0-1]=\frac{1}{2} \end{aligned}

 Question 2
Consider the differential equation given below.
$\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}$
The integrating factor of the differential equation is
 A $\left ( 1-x^{2} \right )^{-3/4}$ B $\left ( 1-x^{2} \right )^{-1/4}$ C $\left ( 1-x^{2} \right )^{-3/2}$ D $\left ( 1-x^{2} \right )^{-1/2}$

Question 2 Explanation:
\begin{aligned} \frac{d y}{d x}+\frac{x}{1-x^{2}} y&=x \sqrt{y}, \quad \text { IF }=?\\ \text{Divided by }\sqrt{y}\\ \frac{1}{\sqrt{y}} \frac{d y}{d x}+\frac{x}{1-x^{2}} \sqrt{y}&=x \\ 2 \frac{d u}{d x}+\frac{x}{1-x^{2}} u&=x\\ \text{Let }\qquad x \sqrt{y}&=u\\ \frac{1}{2 \sqrt{v}} \frac{d y}{d x}&=\frac{d u}{d x}\\ \Rightarrow \qquad \frac{d u}{d x}+\frac{x}{2\left(1-x^{2}\right)} u&=\frac{x}{2} \rightarrow \text{ lines diff. equ.} \\ \text { I. } F&=e^{\int \frac{x}{2\left(1-x^{2}\right)} d x}=e^{-\frac{1}{4} \log \left(1-x^{2}\right)}&=e^{\log \left(1-x^{2}\right) \frac{-1}{4}} \\ \text { I.F }&=\frac{1}{\left(1-x^{2}\right)^{\frac{1}{4}}} \end{aligned}
 Question 3
Two continuous random variables X and Y are related as
$Y=2X+3$
Let $\sigma ^{2}_{X}$ and $\sigma ^{2}_{Y}$ denote the variances of X and Y, respectively. The variances are related as
 A $\sigma ^{2}_{Y}=2 \sigma ^{2}_{X}$ B $\sigma ^{2}_{Y}=4 \sigma ^{2}_{X}$ C $\sigma ^{2}_{Y}=5 \sigma ^{2}_{X}$ D $\sigma ^{2}_{Y}=25 \sigma ^{2}_{X}$

Question 3 Explanation:
\begin{aligned} Y &=2 X+3 \\ \operatorname{Var}[Y] &=E\left[(Y-\bar{Y})^{2}\right] \\ E[Y] &=\bar{Y}=2 \bar{X}+3 \\ \operatorname{Var}[Y] &=E\left[(2 X+3-2 \bar{X}-3)^{2}\right] \\ &=E\left[4(X-\bar{X})^{2}\right] \\ &=4 \cdot E\left[(X-\bar{X})^{2}\right] \\ \sigma_{Y}^{2} &=4 \cdot \sigma_{X}^{2} \end{aligned}
 Question 4
Consider a real-valued base-band signal x(t), band limited to $\text{10 kHz}$. The Nyquist rate for the signal $y\left ( t \right )=x\left ( t \right )x\left ( 1+\dfrac{t}{2} \right )$ is
 A $\text{15 kHz}$ B $\text{30 kHz}$ C $\text{60 kHz}$ D $\text{20 kHz}$

Question 4 Explanation:

$\mathrm{NR}=2 \times f_{\mathrm{max}}=2 \times 15=30 \mathrm{kHz}$
 Question 5
Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k]=y[k] is/are
 A k=0,1,2,,15 B k=0 C k=15 D k=0 and k=15

Question 5 Explanation:
If two' N' point signals x(n) and h(n) are convolving with each other linearly and circularly
then
$y(k)=z(k)$ at $k=N-1$
where, y(n)= Linear convolution of x(n) and h(n)
z(n)= Circular convolution of x(n) and h(n)
Since, $N=16$ (Given)
Therefore, $\quad y(k)=z(k)$ at $k=N-1=15$
 Question 6
A bar of silicon is doped with boron concentration of $10^{16} \text{cm}^{-3}$ and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of $10^{20} \text{cm}^{-3} s^{-1}$. If the recombination lifetime is $100 \;\mu s$, intrinsic carrier concentration of silicon is $10^{10} \text{cm}^{-3}$ and assuming $100\%$ ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is
 A $10^{20} \text{cm}^{-6}$ B $2 \times 10^{20} \text{cm}^{-6}$ C $10^{32} \text{cm}^{-6}$ D $2 \times 10^{32} \text{cm}^{-6}$

Question 6 Explanation:
Boron $\rightarrow$ Acceptor type doping

\begin{aligned} N_{A} &=10^{16} \mathrm{~cm}^{-3} \\ g_{0 p} &=1020 \mathrm{~cm}^{-3} \mathrm{~s}^{-1} \\ \tau &=100 \mu \mathrm{s} \\ n_{i} &=10^{10} \mathrm{~cm}^{-3} \end{aligned}
Product of steady state electron-hole concentration =?
At thermal equilibrium (before shining light)
$\begin{array}{ll} \text { Hole concentration, } & p_{o} \simeq N_{A}=10^{16} \mathrm{~cm}^{-3} \\ \text { Electron concentration, } & n_{0}=\frac{n_{i}^{2}}{p_{0}}=\frac{10^{20}}{10^{16}}=10^{4} \mathrm{~cm}^{-3} \end{array}$
After, illumination of light,
Hole concentration, $p=p_{o}+\delta p$
Electron concentration, $\quad n=n_{o}+\delta n$
Due to shining light, excess carrier concentration,
\begin{aligned} \delta p &=\delta n=g_{o p} \cdot \tau=10^{20} \times 100 \times 10^{-6}=10^{16} \mathrm{~cm}^{-3} \\ \therefore \qquad p &=10^{16}+10^{16}=2 \times 10^{16} \mathrm{~cm}^{-3}\\ n&=10^{4}+10^{16} \simeq 10^{16} \mathrm{~cm}^{-3} \end{aligned}
So, product of steady state electron-hole concentration
\begin{aligned} &=n p=10^{16} \times 2 \times 10^{16} \\ &=2 \times 10^{32} \mathrm{~cm}^{-6} \end{aligned}
 Question 7
The energy band diagram of a p-type semiconductor bar of length L under equilibrium condition (i.e.. the Fermi energy level $E_{F}$ is constant) is shown in the figure. The valance band $E_{V}$ is sloped since doping is non-uniform along the bar. The difference between the energy levels of the valence band at the two edges of the bar is $\Delta$.

If the charge of an electron is q, then the magnitude of the electric field developed inside this semiconductor bar is
 A $\frac{\Delta }{qL}$ B $\frac{2\Delta }{qL}$ C $\frac{\Delta }{2qL}$ D $\frac{3\Delta }{2qL}$

Question 7 Explanation:
The built-in electric field is due to non-uniform doping (the semiconductor is under equilibrium)

\begin{aligned} E &=\frac{1}{q}\frac{ d E_{v}}{d x} \\ &=\frac{1}{q} \frac{\Delta}{L} \\ &=\frac{\Delta}{q L} \end{aligned}
 Question 8
In the circuit shown in the figure, the transistors $M_{1}$ and $M_{2}$ are operating in saturation. The channel length modulation coefficients of both the transistors are non-zero. The transconductance of the $\text{MOSFETs} M_{1}$ and $M_{2}$ are $g_{m1}$ and $g_{m2}$ , respectively, and the internal resistance of the $\text{MOSFETs} M_{1}$ and $M_{2}$ are $r_{01}$ and $r_{02}$ , respectively.

Ignoring the body effect, the ac small signal voltage gain $\left ( \partial V_{out}/\partial V_{in} \right )$ of the circuit is
 A $-g_{m2}\left ( r_{01}\left | \right |r_{02}\right )$ B $-g_{m2}\left ( \frac{1}{g_{m1}}\left | \right |r_{02} \right )$ C $-g_{m1}\left ( \frac{1}{g_{m2}}\left | \right |r_{01}\left | \right |r_{02} \right )$ D $-g_{m2}\left ( \frac{1}{g_{m1}}\left | \right |r_{01}\left | \right |r_{02} \right )$

Question 8 Explanation:
MOSFET $M_2$ acts as common source amplifier.

Drain to gate connected MOSFET $M_1$ acts as load.

For given circuit, AC equivalent is as shown.

Replace $M_2$ with small signal model

\begin{aligned} \frac{V_{\text {out }}}{V_{\text {in }}} &=\frac{-g_{m 2} V_{g s}\left(r_{\infty} \| R_{\text {eq }}\right)}{V_{g s}} \\ A_{V} &=-g_{m 2}\left( \frac{1}{g_{m 1}}|| r_{o1} || r_{o 2} \right) \end{aligned}
 Question 9
For the circuit with an ideal OPAMP shown in the figure. $V_{\text{REF}}$ is fixed.

If $V_{\text{OUT}}=1$ volt for $V_{\text{IN}}-0.1$ volt and $V_{\text{OUT}}=6$ volt for $V_{\text{IN}}=1$ volt, where $V_{\text{OUT}}$ is measured across $R_{L}$ connected at the output of this OPAMP, the value of $R_{F}/R_{\text{IN}}$ is
 A 3.28 B 2.86 C 3.82 D 5.55

Question 9 Explanation:
MARKS TO ALL AS PER IIT ANSWER KEY

\begin{aligned} V &=V^{+} \\ \frac{V_{\text {out }} R_{\text {in }}+V_{\text {in }} R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} \\ \frac{1 \times R_{\text {in }}+0.1 \times R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(i)\\ \frac{6 R_{\text {in }}+1 \times R_{F}}{R_{\text {in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(ii) \end{aligned}
Equate equation (i) and (ii),
\begin{aligned} 1 \times R_{\text {in }}+0.1 \times R_{F} &=6 \times R_{\text {in }}+1 \times R_{F} \\ -5 R_{\text {in }} &=0.9 R_{F} \\ \therefore \quad \frac{R_{F}}{R_{\text {in }}} &=\frac{-5}{0.9}=-5.55 \end{aligned}
(According to the given data magnitude is taken)
 Question 10
Consider the circuit with an ideal OPAMP shown in the figure.

Assuming $\left | V_{\text{IN}} \right |\ll \left | V_{\text{CC}} \right |$ and $\left | V_{\text{REF}} \right |\ll \left | V_{\text{CC}} \right |$ , the condition at which $V_{\text{OUT}}$ equals to zero is
 A $V_{\text{IN}}\:=\:V_{\text{REF}}$ B $V_{\text{IN}}\:=\:0.5\:V_{\text{REF}}$ C $V_{\text{IN}}\:=\:2\:V_{\text{REF}}$ D $V_{\text{IN}}\:=\:2\:+\:V_{\text{REF}}$

Question 10 Explanation:
For ideal op-amp, $V^{\prime}=V^{+}=0$
KCL at node $\mathrm{V}^{-}:$
\begin{aligned} \frac{V_{\text {IN}}-0}{R}+\frac{\left(-V_{\text {REF }}-0\right)}{R}+\frac{V_{\text {OUT }}-0}{R_{F}} &=0 \\ \frac{V_{\text {OUT }}}{R_{F}} &=\frac{1}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right) \\ V_{\text {OUT }} &=\frac{R_{F}}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right)\\ \text { We want, }\qquad V_{\text {OUT }}&=0 \\ \Rightarrow\qquad V_{\text {REF }}-V_{\text {IN }}&=0 \\ \Rightarrow\qquad V_{\text {IN }}&=V_{\text {REF }} \end{aligned}
There are 10 questions to complete.