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.

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