## 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.

## GATE Notes – Electronics and Communications (EC)

GATE Electronics and Communications notes for all subjects as per syllabus of GATE 2022 Electronics and Communications.

We always encourage GATE aspirants to read standard books to score good ranks in GATE. Notes can be used as one of the reference while preparing for GATE exam.

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• GATE Notes – Analog Circuits
• GATE Notes – Control Systems
• GATE Notes – Digital Circuits
• GATE Notes – Electromagnetics
• GATE Notes – Electronic Devices and Circuits
• GATE Notes – Engineering Mathematics
• GATE Notes – Network Theory
• GATE Notes – Signals and Systems

NOTE : These GATE Electronics and Communications notes are not design by us and we encourage to read standard books for GATE Electronics and Communications preparation. All notes are available on internet. We have just organize it properly to helps students who are interested to refer the GATE Electronics and Communications notes. All GATE Electronics and Communications aspirants are suggested to use these notes as just reference for quick revision but not entirely depends on it.

## GATE 2022 Electronics and Communications Syllabus

Revised syllabus of GATE 2022 Electronics and Communications by IIT.

Practice GATE Electronics and Communications previous year questions

Download the GATE 2022 Electronics and Communications Syllabus pdf from the official site of IIT Bombay. Analyze the GATE 2022 revised syllabus for Electronics and Communications.

## GATE EC 2018

 Question 1
Two identical nMOS transistors $M_{1}$ and $M_{2}$ are connected as shown below. The circuit is used as an amplifier with the input connected between G and S terminals and the output taken between D and S terminals. $V_{bias}$ and $V_{D}$ are so adjusted that both transistors are in saturation. The transconductance of this combination is defined as $g_{m}=\frac{\partial i_{D}}{\partial V_{GS}}$ while the output resistance is $r_{0}=\frac{\partial V_{GS}}{\partial i_{D}}$ , where $i_{D}$ is the current flowing into the drain of $M_{2}$. Let $g_{m1}$ , $g_{m2}$ be the transconductances and $r_{01}$ , $r_{02}$ be the output resistances of transistors $M_{1}$ and $M_{2}$ , respectively.

Which of the following statements about estimates for $g_{m}$ and $r_{0}$ is correct?
 A $g_{m}\approx g_{m1}\cdot g_{m2}\cdot r_{02} \;and \; r_0 \approx r_{01}+r_{02}.$ B $g_{m}\approx g_{m1}\ + g_{m2} \; and \; r_{0} \approx r_{01}+r_{02}.$ C $g_{m}\approx g_{m1} \; and \; r_{0}\approx r_{01} \cdot g_{m2}\cdot r_{02}.$ D $g_{m}\approx g_{m1} \; and \; r_{0}\approx r_{02}$.
Analog Circuits   FET and MOSFET Analysis
Question 1 Explanation:

$g_{m}=\frac{\Delta I_{D}}{\Delta V_{\text {in }}}=\frac{i_{D}}{v_{g s}}=\frac{i_{D 1}}{v_{g s}}=g_{m 1}$
To calculate $r_{o}$ :

\begin{aligned} v_{\pi 2} &=-I_{x} r_{01} \\ I_{x} &=g_{m 2} v_{\pi 2}+\frac{\left(V_{x}-I_{x} r_{01}\right)}{r_{02}} \\ I_{x} &=-g_{m 2} r_{01} I_{x}+\frac{V_{x}}{r_{02}}-I_{x} \frac{r_{01}}{r_{02}} \\ V_{x} &=r_{02}\left[1+r_{01} g_{m 2}+\frac{r_{01}}{r_{02}}\right] I_{x} \\ r_{0} &=\frac{V_{x}}{I_{x}}=r_{01}+r_{02}+r_{01} r_{02} g_{m 2} \\ & \approx r_{01} r_{02} g_{m 2} \end{aligned}
 Question 2
In the circuit shown below, the op-amp is ideal and Zener voltage of the diode is 2.5 volts. At the input, unit step voltage is applied, i.e. $v_{IN}(t)= u(t)$ volts. Also, at t= 0, the voltage across each of the capacitors is zero.
The time t, in milliseconds, at which the output voltage $v_{OUT}$ crosses -10 V is
 A 2.5 B 5 C 7.5 D 10
Analog Circuits   Operational Amplifiers
Question 2 Explanation:
$\text{For} \quad t \gt 0,$

$I=\frac{1 V}{1 \mathrm{k} \Omega}=1 \mathrm{mA}$
Till $t=2.5 \mathrm{msec}$, both $V_{1}$ and $V_{2}$ will increase and after $t=2.5 \mathrm{msec}$, $V_{2}=2.5 \mathrm{V}$ and $V_{1}$ increases with time.
\begin{aligned} \text { when } v_{\text {out }}(t) &=-10 \mathrm{V} \\ & V_{1}=7.5 \mathrm{V}\\ \text{So,}\\ \frac{1}{1 \mu F} \int_{0}^{t}(1 \mathrm{m} \mathrm{A}) d t &=7.5 \mathrm{V} \\ 10^{3} t &=7.5 \\ t &=7.5 \mathrm{msec} \end{aligned}
 Question 3
A good transimpedance amplifier has
 A low input impedance and high output impedance. B high input impedance and high output impedance. C high input impedance and low output impedance. D low input impedance and low output impedance.
Analog Circuits   Feedback Amplifiers
Question 3 Explanation:
A good transimpedance amplifier should have low input impedance and low output impedance
 Question 4
Let the input be u and the output be y of a system, and the other parameters are real constants. Identify which among the following systems is not a linear system:
 A $\frac{d^{3}y}{dt^{3}} + a_{1} \frac{d^{2}y}{dt^{2}} + a_{2}\frac{dy}{dt} + a_{3}y =$ $b_{3}u+b_{2}\frac{du}{dt}+b_{1}\frac{d^{2}u}{dt^{2}}$ (with initial rest conditions) B $y(t)=\int_{0}^{t}e^{a(t-r)}\beta u(\tau)d \tau$ C $y= au +b$, $b \neq 0$ D $y=au$
Signals and Systems   Basics of Signals and Systems
Question 4 Explanation:
$y=a u+b, b \neq 0$ is a non-linear system.
 Question 5
The Nyquist stability criterion and the Routh criterion both are powerful analysis tools for determining the stability of feedback controllers. Identify which of the following statements is FALSE:
 A Both the criteria provide information relative to the stable gain range of the system. B The general shape of the Nyquist plot is readily obtained from the Bode magnitude plot for all minimum-phase systems. C The Routh criterion is not applicable in the condition of transport lag, which can be readily handled by the Nyquist criterion. D The closed-loop frequency response for a unity feedback system cannot be obtained from the Nyquist plot.
Control Systems   Frequency Response Analysis
 Question 6
Consider p(s) = $s^{3}+a_{2}s^{2}+a_{1}s+a_{0}$ with all real coefficients. It is known that its derivative ${p}'(s)$ has no real roots. The number of real roots of ${p}(s)$ is
 A 0 B 1 C 2 D 3
Engineering Mathematics   Numerical Methods
Question 6 Explanation:
If p(s) has "r" real roots, then $p^{\prime}(s)$ will have atleast $"r-1^{\prime \prime}$ real roots.
 Question 7
In a p-n junction diode at equilibrium, which one of the following statements is NOT TRUE?
 A The hole and electron diffusion current components are in the same direction. B The hole and electron drift current components are in the same direction. C On an average, holes and electrons drift in opposite direction. D On an average, electrons drift and diffuse in the same direction.
Electronic Devices   PN-Junction Diodes and Special Diodes
Question 7 Explanation:

$\longrightarrow$ Hole diffusion
$\longleftarrow$ Electron diffusion
$\longleftarrow$ Hole drift
$\longrightarrow$ Electron drift
$\longrightarrow$ Hole diffusion current
$\longrightarrow$ Electron diffusion current
$\longleftarrow$ Hole drift current
$\longleftarrow$ Electron drift current
 Question 8
The logic function f(X,Y) realized by the given circuit is
 A NOR B AND C NAND D XOR
Digital Circuits   Logic Families
Question 8 Explanation:
From pull-down network,
\begin{aligned} \overline{f(X, Y)}&=\bar{X} \bar{Y}+X Y=X \odot Y \\ f(X, Y)&=\overline{X \odot Y}=X \oplus Y \end{aligned}
 Question 9
A function F(A,B,C) defined by three Boolean variables A, B and C when expressed as sum of products is given by

$F=\bar{A}\cdot \bar{B} \cdot \bar{C} + \bar{A}\cdot B \cdot \bar{C} + A\cdot \bar{B} \cdot \bar{C}$

where,$\bar{A},\bar{B} \;and \; \bar{C}$ are complements of the respective variable. The product of sums (POS) form of the function F is
 A $F=(A+B+C)\cdot (A+\tilde{B}+C)\cdot (\bar{A}+B+C)$ B $F=(\bar{A}+\bar{B}+\bar{C})\cdot (\bar{A}+B+\bar{C})\cdot (A+\bar{B}+\bar{C})$ C $F=(A + B + \bar{C}) \cdot (A + \bar{B} + \bar{C} ) \cdot (\bar{A} + B + \bar{C}) \cdot$ $(\bar{A}+\bar{B}+C) \cdot (\bar{A}+\bar{B}+\bar{C})$ D $F=(\bar{A} + \bar{B} + C) \cdot (\bar{A} + B + C) \cdot$ $(A + B + \bar{C}) \cdot (A+B+C)$
Digital Circuits   Boolean Algebra
Question 9 Explanation:
\begin{aligned} F(A, B, C, D) &=\bar{A} \bar{B} \bar{C}+\bar{A} B \bar{C}+A \bar{B} \bar{C} \\ &=\Sigma m(0,2,4)=\Pi M(1,3,5,6,7) \\ =&(A+B+\bar{C})(A+\bar{B}+\bar{C})(\bar{A}+B+\bar{C}) \\ &(\bar{A}+\bar{B}+C)(\bar{A}+\bar{B}+\bar{C}) & \end{aligned}
 Question 10
The points P, Q, and R shown on the Smith chart (normalized impedance chart) in the following figure represent:
 A P: Open Circuit, Q: Short Circuit, R: Matched Load B P: Open Circuit, Q: Matched Load, R: Short Circuit C P: Short Circuit, Q: Matched Load, R: Open Circuit D P: Short Circuit, Q: Open Circuit, R: Matched Load
Electromagnetics   Transmission Lines
Question 10 Explanation:
For Short circuit,
$r=x=0 \quad \Rightarrow \text { Point } " P^{\prime \prime}$
For Open circuit,
$r=x=\infty \quad \Rightarrow \text { Point }^{\prime \prime} R^{\prime \prime}$
$r=1 \text { and } x=0 \Rightarrow \text { Point " } Q^{\prime \prime}$
P: Short Circuit, Q: Matched Load R: Open circuit
There are 10 questions to complete.

## GATE Electronics and Communication 2019

 Question 1
Which one of the following functions is analytic over the entire complex plane?
 A $ln(z)$ B $e^{1/z}$ C $\frac{1}{1-z}$ D $cos(z)$
Engineering Mathematics   Complex Analysis
Question 1 Explanation:
$f(z) = \cos z$ is analytic every where.
 Question 2
The families of curves represented by the solution of the equation

$\frac{dy}{dx}=-\left (\frac{x}{y} \right )^n$

for $n = -1$ and $n = +1$, respectively, are
 A Parabolas and Circles B Circles and Hyperbolas C Hyperbolas and Circles D Hyperbolas and Parabolas
Engineering Mathematics   Differential Equations
Question 2 Explanation:
\begin{aligned} \frac{d y}{d x} &=-\left(\frac{x}{y}\right)^{n} \\ n=-1\quad\quad \frac{d y}{d x} &=-\frac{y^{\prime}}{x} \\ \frac{d y}{y} &=-\frac{d x}{x} \\ \int \frac{1}{y} d y &=-\int \frac{1}{x} d x \\ \ln y &=-\ln x+\ln c \\ \ln (y x) &=\ln c \end{aligned}
$x y=c \quad$ (Represents rectangular hyporbola)
\begin{aligned} n=1, \quad \frac{d y}{d x}&=-\frac{x}{y} \\ y d y &=-x d x \\ y d y &=-\int x d x \\ \frac{y^{2}}{2} &=-\frac{x^{2}}{2}+c \end{aligned}
$x^{2}+y^{2}=2 c \quad$ (Represents family of circles)
 Question 3
Let H(z) be the z-transform of a real-valued discrete-time signal h[n]. If $P(z)=H(z)H\left (\frac{1}{z} \right )$ has a zero at $z=\frac{1}{2}+\frac{1}{2}j$, and P(z) has a total of four zeros, which one of the following plots represents all the zeros correctly?
 A A B B C C D D
Signals and Systems   Z-Transform
Question 3 Explanation:
$P(Z)=H(Z)H\left ( \frac{1}{Z} \right )$
(i) $h(n)$ is real. Som $p(n)$ will be also real
(ii) $P(z)=P(z^{-1})$
From (i) : if $z_1$ is a zero of $P(z)$, then $z_1^*$ will be also a zero of $P(z)$.
From (ii): If $z_1$ is a zero of $P(z)$, then $\frac{1}{z_1}$ will be also a zero of $P(z)$.
So, the 4 zeros are,
\begin{aligned} z_1&= \frac{1}{2}+\frac{1}{2}j\\ z_2&= z_1^*=\frac{1}{2}-\frac{1}{2}j\\ z_3&=\frac{1}{z_1}=\frac{1}{\frac{1}{2}-\frac{1}{2}j}=1-j \\ z_4&=\left ( \frac{1}{z_1} \right )^*=z_3^*=1+j \end{aligned}
 Question 4
Consider the two-port resistive network shown in the figure. When an excitation of 5 V is applied across Port 1, and Port 2 is shorted, the current through the short circuit at Port 2 is measured to be 1 A (see (a) in the figure).
Now, if an excitation of 5 V is applied across Port 2, and Port 1 is shorted (see(b) in the figure), what is the current through the short circuit at Port 1?
 A 0.5 A B 1.0 A C 2.0 A D 2.5 A
Network Theory   Network Theorems
Question 4 Explanation:
According to reciprocity theorem,
In a linear bilateral single source network the ratio of response to excitation remains the same even after their positions get interchanged.
$\therefore \quad \frac{I}{5}=\frac{1}{5} \Rightarrow I=1 \mathrm{A}$
 Question 5
Let Y(s) be the unit-step response of a causal system having a transfer function
$G(s)=\frac{3-s}{(s+1)(s+3)}$

that is, $Y(s)=\frac{G(s)}{s}$. The forced response of the system is
 A $u(t)-2e^{-t}u(t)+e^{-3t}u(t)$ B $2u(t)-2e^{-t}u(t)+e^{-3t}u(t)$ C $2u(t)$ D $u(t)$
Signals and Systems   Laplace Transform
Question 5 Explanation:
Given, $\quad G(s)=\frac{3-s}{(s+1)(s+3)}$
$\therefore \quad Y(s)=\frac{G(s)}{s}=\frac{3-s}{s(s+1)(s+3)}$
Using partial fractions, we get,
\begin{aligned} Y(s)&=\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+3)} \\ A\left(s^{2}+4 s+3\right)&+B\left(s^{2}+3 s\right)+C\left(s^{2}+s\right)=3-s \\ A+B+C&=0\\ 4 A+3 B+C&=-1 \\ \text{and }3 A&=3 \\ \text{Therefore, }&\text{we get,}\\ A=1, B&=-2 \text { and } C=1\\ \text{So, }\quad Y(s)&=\frac{1}{s}-\frac{2}{(s+1)}+\frac{1}{(s+3)} \\ \text{and}\quad \mathrm{y}(t)&=u(t)-2 e^{-t} u(t)+e^{-3 t} u(t) \\ \end{aligned}
Forced response,
$y_{t}(t)=u(t) \Rightarrow \text { option }(D)$
 Question 6
For an LTI system, the Bode plot for its gain is as illustrated in the figure shown. The number of system poles $N_p$ and the number of system zeros $N_z$ in the frequency range $1Hz\leq f\leq 10^7Hz$ is
 A $N_p=5,N_z=2$ B $N_p=6,N_z=3$ C $N_p=7,N_z=4$ D $N_p=4,N_z=2$
Control Systems   Frequency Response Analysis
Question 6 Explanation:

Number of poles $(N_{P})$= 6
Number of zeros $(N_{Z})$ = 3
 Question 7
A linear Hamming code is used to map 4-bit messages to 7-bit codewords. The encoder mapping is linear. If the message 0001 is mapped to the codeword 0000111, and the message 0011 is mapped to the codeword 1100110, then the message 0010 is mapped to
 A 10011 B 1100001 C 1111000 D 1111111
Communication Systems   Information Theory and Coding
Question 7 Explanation:

 Question 8
Which one of the following options describes correctly the equilibrium band diagram at T=300 K of a Silicon $pnp^+p^{++}$ configuration shown in the figure?
 A A B B C C D D
Electronic Devices   PN-Junction Diodes and Special Diodes
 Question 9
The correct circuit representation of the structure shown in the figure is
 A A B B C C D D
Electronic Devices   IC Fabrication
Question 9 Explanation:

 Question 10
The figure shows the high-frequency C-V curve of a MOS capacitor (at T = 300 K) with $\Phi _{ms}=0V$ and no oxide charges. The flat-band, inversion, and accumulation conditions are represented, respectively, by the points
 A P, Q, R B Q, R, P C R, P, Q D Q, P, R
Electronic Devices   BJT and FET Basics
Question 10 Explanation:
Since $\phi_{ms}= 0$, the MOS-capacitor is ideal.
Point P Represents accumulation
Point Q Represents flat band
Point R Represents Inversion
There are 10 questions to complete.

## GATE Electronics and Communication 2020

 Question 1
If $v_1, v_2,..., v_6$ are six vectors in $\mathbb{R}^4$, which one of the following statements is False?
 A It is not necessary that these vectors span $\mathbb{R}^4$. B These vectors are not linearly independent. C Any four of these vectors form a basis for $\mathbb{R}^4$. D If {$v_1, v_3,v_5, v_6$} spans $\mathbb{R}^4$, then it forms a basis for $\mathbb{R}^4$.
Engineering Mathematics   Calculus
Question 1 Explanation:
$v_1, v_2,..., v_6$ are six vectors in $\mathbb{R}^4$.
For a 4-dimensional vector space,
(i) any four linearly independent vectors form a basis (or)
(ii) Any set of four vectors in $\mathbb{R}^4$ spans $\mathbb{R}^4$, then it forms a basis.
Therefore, clearly options (A), (B), (D) are true.
Option (C) is FALSE
 Question 2
For a vector field $\vec{A}$, which one of the following is False?
 A $\vec{A}$ is solenoidal if $\bigtriangledown \cdot \vec{A}=0$ B $\bigtriangledown \times \vec{A}$ is another vector field. C $\vec{A}$ is irrotational if $\bigtriangledown ^2 \vec{A}=0$. D $\bigtriangledown \times(\bigtriangledown \times \vec{A})=\bigtriangledown (\bigtriangledown \cdot \vec{A})-\bigtriangledown ^2 \vec{A}$
Engineering Mathematics   Calculus
Question 2 Explanation:
Divergence and curl operator is performed on a vector field $\vec{A}$
Curl operation provides a vector orthogonal to the given vector field $\vec{A}$
$\bigtriangledown \times(\bigtriangledown \times \vec{A})=\bigtriangledown (\bigtriangledown \cdot \vec{A})-\bigtriangledown ^2 \vec{A}$
If a vector field is irrortational then $\bigtriangledown \times \vec{A}=0$
If a vector field is solenoidal then $\bigtriangledown \cdot \vec{A}=0$
If a field is scalar A, then $\bigtriangledown ^2 \vec{A}=0$, is a laplacian equation.
Hence option (C) is incorrect
 Question 3
The partial derivative of the function

$f(x,y,z)=e^{1-x \cos y}+xze^{-1/(1+y^2)}$

with respect to x at the point (1,0,e) is
 A -1 B 0 C 1 D $\frac{1}{e}$
Engineering Mathematics   Calculus
Question 3 Explanation:
\begin{aligned} \text{Given, } f(x,y,z)&=e^{1-x\cos y}+xze^{-1/(1+y^{2})} \\ \frac{\partial f}{\partial x}&=e^{1-x\cos y}(0-\cos y)+ze^{-1/1+y^{2}} \\ \left ( \frac{\partial f }{\partial x} \right )_{(1,0,e)}&=e^{0}(0-1)+e\cdot e^{-1/(1+0)} \\ &=-1+1=0 \end{aligned}
 Question 4
The general solution of $\frac{d^2y}{dx^2}-6\frac{dy}{dx}+9y=0$ is
 A $y=C_1e^{3x}+C_2e^{-3x}$ B $y=(C_1+C_2x)e^{-3x}$ C $y=(C_1+C_2x)e^{3x}$ D $y=C_1e^{3x}$
Engineering Mathematics   Differential Equations
Question 4 Explanation:
Taking $\frac{\mathrm{d} }{\mathrm{d} x}=D$
Given, $D^{2}-6D+9=0$
$(D-3)^2=0$
$D=3,3$
So, Solution of the given Differential equation
$y=(C_{1}+C_{2}x)e^{3x}$
 Question 5
The output y[n] of a discrete-time system for an input x[n] is

$y[n]=\begin{matrix} max\\ -\infty \leq k\leq n \end{matrix}\; |x[k]|$.

The unit impulse response of the system is
 A 0 for all n B 1 for all n C unit step signal u[n]. D unit impulse signal $\delta$[n].
Signals and Systems   LTI Systems Continuous and Discrete
 Question 6
A single crystal intrinsic semiconductor is at a temperature of 300 K with effective density of states for holes twice that of electrons. The thermal voltage is 26 mV. The intrinsic Fermi level is shifted from mid-bandgap energy level by
 A 18.02 meV B 9.01 meV C 13.45 meV D 26.90 meV
Electronic Devices   Basic Semiconductor Physics
Question 6 Explanation:
$\frac{E_{c}+E_{v}}{2}-E_{F_{i}}=\frac{KT}{2}\ln \left ( \frac{N_{C}}{N_{V}}\right )\, \, \, \, \, \, \left ( \because N_{C}=\frac{N_{V}}{2} \right )$
$=\frac{0.026}{2}\ln 0.5=-9.01\, meV$
 Question 7
Consider the recombination process via bulk traps in a forward biased $pn$ homojunction diode. The maximum recombination rate is $U_{max}$. If the electron and the hole capture cross-section are equal, which one of the following is False?
 A With all other parameters unchanged, $U_{max}$ decreases if the intrinsic carrier density is reduced. B $U_{max}$ occurs at the edges of the depletion region in the device. C $U_{max}$ depends exponentially on the applied bias. D With all other parameters unchanged,$U_{max}$ increases if the thermal velocity of the carriers increases.
Electronic Devices   PN-Junction Diodes and Special Diodes
 Question 8
The components in the circuit shown below are ideal. If the op-amp is in positive feedback and the input voltage $V_i$ is a sine wave of amplitude 1 V, the output voltage $V_o$ is
 A a non-inverted sine wave of 2 V amplitude B an inverted sine wave of 1 V amplitude C a square wave of 5 V amplitude D a constant of either +5 or -5V
Analog Circuits   Operational Amplifiers
Question 8 Explanation:

Given circuit is a Schmitt trigger of non-inverting type.
$V_{o}=\pm 5\, V$
$V^{+}=\frac{V_{o}\times 1+V_{i}\times 1}{1+1}=\frac{V_{o}+V}{2}$
let,$V_{o}=-5\, V,\, \, \, \,V^{+}=\frac{-5+V_{i}}{2}$
$V_{o}$ can change from -5 V to +5 V if $V^{+} \gt 0$
i.e. $\frac{-5+V_{i}}{2} \gt 0\Rightarrow V_{i} \gt 5\, V$
similarly, $V_{o}$ can change from -5 V to +5 V if $V_{i} \lt -5\, V$
But given input has peak value 1 V. Hence output cannot change from +5 V to -5 V or -5 V to +5 V.
Output remain constant at +5 V or -5 V.
 Question 9
In the circuit shown below, the Thevenin voltage $V_{TH}$ is
 A 2.4 V B 2.8 V C 3.6 V D 4.5 V
Network Theory   Network Theorems
Question 9 Explanation:
By applying the Source Transformation
 Question 10
The figure below shows a multiplexer where $S_1 \; and \; S_0$ are the select lines, $I_0 \; to \; I_3$ are the input data lines, EN is the enable line, and $F(P, Q, R)$ is the output, F is
 A $PQ+\bar{Q}R$ B $P+Q\bar{R}$ C $P\bar{Q}R+\bar{P}Q$ D $\bar{Q}+PR$
Digital Circuits   Combinational Circuits
Question 10 Explanation:
Output,$F=\bar{P}\bar{Q}R+P\bar{Q}R+PQ\, \, \, \,$
$F=\bar{Q}R+PQ$
There are 10 questions to complete.

## GATE EC 2014 SET-2

 Question 1
The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is ______.
 A 200 B 100 C 50 D 45
Engineering Mathematics   Linear Algebra
Question 1 Explanation:
Determinant of A=5
Determinant of B=40
Determinant of AB=|A||B|
$\begin{array}{l} \quad=5 \times 40 \\ \quad=200 \end{array}$
 Question 2
Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E [X], is _____.
 A 100 B 50 C 25 D 10
Engineering Mathematics   Probability and Statistics
Question 2 Explanation:
$E[X]=\frac{1+2+3+\cdots 99}{50}=\frac{2500}{50}=50$
 Question 3
For $0\leq t \leq \infty$ , the maximum value of the function $f(t) =e^{-t}-2e^{-2t}$ occurs at
 A $t=log_{e}4$ B $t=log_{e}2$ C t=0 D $t=log_{e}8$
Engineering Mathematics   Calculus
Question 3 Explanation:
\begin{aligned} f(t)&=e^{-t}-2 e^{-2 t} \\ f^{\prime}(t)&=-e^{-t}+4 e^{-2 t} \\ \text{For maximum value} &P(t)=0 \\ f^{\prime}(t)&=0=-e^{-t}+4 e^{-2 t} \\ \Rightarrow \quad 4 e^{-2 t}&=e^{t} \\ 4 e^{t}&=1 \\ \therefore \quad t&=\log _{e} 4 \end{aligned}
 Question 4
The value of
$\lim_{x\rightarrow \infty }(1+\frac{1}{x})^{x}$
is
 A ln 2 B 1 C e D $\infty$
Engineering Mathematics   Calculus
Question 4 Explanation:
$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e^{\lim _{x \rightarrow \infty} \frac{1}{x} \cdot x}=e^{1}=e$
 Question 5
If the characteristic equation of the differential equation
$\frac{d^{2}y}{dx^{2}}+2\alpha \frac{dy}{dx}+y=0$
has two equal roots, then the values of a are
 A $\pm 1$ B 0,0 C $\pm j$ D $\pm 1/2$
Engineering Mathematics   Differential Equations
Question 5 Explanation:
$\frac{d^{2} y}{d x^{2}}+2 \alpha\frac{d y}{d x}+y=0$
The characteristic equation is given as
\begin{aligned} \left(m^{2}+2(x)+1\right) &=0 \\ m_{1}, m_{2} &=\frac{-2 x_{1} \pm \sqrt{4 x^{2}-4}}{2} \end{aligned}\\ \text{since both roots are equal i.e.} \\ \begin{aligned} m_{1}=& m_{2} \\ \frac{-2 \alpha+\sqrt{4 \alpha^{2}-4}}{2} &=\frac{-2\alpha \cdot-\sqrt{4 a^{2}-4}}{2} \\ \sqrt{4\left(1^{2}-4\right.} &=-\sqrt{4 c^{2}-4} \\ 2 \sqrt{4 c^{2}-4} &=0 \\ 4 \alpha^{2}-4 &=0 \\ \alpha^{2} &=1 \\ \alpha &=\pm 1 \end{aligned}
 Question 6
Norton's theorem states that a complex network connected to a load can be replaced with an equivalent impedance
 A in series with a current source B in parallel with a voltage source C in series with a voltage source D in parallel with a voltage source
Network Theory   Network Theorems
Question 6 Explanation:
Norton's theorem states that a complex network connected to a load can be replaced with an equivalent impedance in parallel with a current source

 Question 7
In the figure shown, the ideal switch has been open for a long time. If it is closed at t = 0, then the magnitude of the current (in mA) through the 4 k$\Omega$ resistance at t = $0^{+}$ is _____.

 A 0.75 B 1.25 C 0.5 D 1.5
Network Theory   Transient Analysis
Question 7 Explanation:
At steady state $t = 0_{-}$

\begin{aligned} \therefore V_{c}(0) &=V_{c}\left(0^{+}\right)=5 \mathrm{V} \\ I_{L}\left(0^{-}\right) &=I_{L}\left(0^{+}\right)=1 \mathrm{mA} \end{aligned}
At, t=0, switch get closed

Thus, the current through $4\Omega$ resistance is
$I=\frac{5}{4 \times 10^{3}}=1.25 \mathrm{mA}$
 Question 8
A silicon bar is doped with donor impurities $N_{D}=2.25\times10^{15} atom/cm^{3}$. Given the intrinsic carrier concentration of silicon at T = 300 K is $n_{i}=1.5\times10^{10}cm^{-3}$. Assuming complete impurity ionization, the equilibrium election and hole concentrations are
 A $n_{0}=1.5 \times 10^{16}cm^{-3},$ $p_{0}=1.5 \times 10^{5}cm^{-3}$ B $n_{0}=1.5 \times 10^{10}cm^{-3},$ $p_{0}=1.5 \times 10^{15}cm^{-3}$ C $n_{0}=2.25 \times 10^{15}cm^{-3},$ $p_{0}=1.5 \times 10^{10}cm^{-3}$ D $n_{0}=2.25 \times 10^{15}cm^{-3},$ $p_{0}=1 \times 10^{5}cm^{-3}$
Electronic Devices   Basic Semiconductor Physics
Question 8 Explanation:
since $N_{D} \gt \gt n_{i}$
therefore equilibrium electron concentration is
$n \simeq N_{D}=2.25 \times 10^{15} \mathrm{cm}^{-3}$
And equilibrium hole concentration is given by mass action law
$p=\frac{n_{i}^{2}}{N_{D}}=\frac{\left(1.5 \times 10^{10}\right)^{2}}{2.25 \times 10^{15}}=1 \times 10^{5} \mathrm{cm}^{-3}$
 Question 9
An increase in the base recombination of a BJT will increase
 A the common emitter dc current gain $\beta$ B the breakdown voltage $BV_{CEO}$ C the unity-gain cut-off frequency $f_{T}$ D the transconductance $g_{m}$
Analog Circuits   BJT Analysis
 Question 10
In CMOS technology, shallow P-well or N -well regions can be formed using
 A low pressure chemical vapour deposition B low energy sputtering C low temperature dry oxidation D low energy ion-implantation
Electronic Devices   IC Fabrication
Question 10 Explanation:
Ion implanation/diffusion is used for well implantation.
There are 10 questions to complete.

## GATE EC 2014 SET-3

 Question 1
The maximum value of the function $f(x)=ln(1+x)-x(\; where \; x>-1)$ occurs at x = ____.
 A 0 B 1 C 0.5 D -1
Engineering Mathematics   Calculus
Question 1 Explanation:
\begin{aligned} f^{\prime}(x) &=\frac{1}{1+x}-1=0 \\ \frac{1-1-x}{1+x} &=0 \\ \frac{x}{1+x} &=0 \\ x &=0\\ f^{\prime \prime}(x)&=\frac{-1}{(1+x)^{2}} \\ f^{\prime}(0)&=-1 \lt 0 \end{aligned}
f(x) have maximum value at x=0
\begin{aligned} f(0)&=\ln (1+0)-0=0 \\ t_{\max }&=0 \end{aligned}
 Question 2
Which ONE of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively ?
 A $\frac{dy}{dx}+xy=e^{-x}$ B $\frac{dy}{dx}+xy=0$ C $\frac{dy}{dx}+xy=e^{-y}$ D $\frac{dy}{dx}+e^{-y}=0$
Engineering Mathematics   Differential Equations
Question 2 Explanation:
General form of linear differential equation
$\frac{d y}{d x}+p y=\theta$ when P and $\theta$ can be function of x
Only option (A) is in this form.
 Question 3
Match the application to appropriate numerical method.
 A P1-M3, P2-M2, P3-M4, P4-M1 B P1-M3, P2-M1, P3-M4, P4-M2 C P1-M4, P2-M1, P3-M3, P4-M2 D P1-M2, P2-M1, P3-M3, P4-M4
Engineering Mathematics   Numerical Methods
 Question 4
An unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss is
 A 0.067 B 0.073 C 0.082 D 0.091
Engineering Mathematics   Probability and Statistics
Question 4 Explanation:
It means 3-head appears in $1^{\text {st }}$ 9 trials.
Probability of getting exactly 3 head in 1^{\text {st }} 9 trials
\begin{aligned} &=\text{ Coefficient } p^{3}\text{ in }(4+p) 9 \\ \text{When, }&[\overline{p+q=1}] \\ &={ }^{9} C_{3} q^{6} p^{3}\\ \text{when, }\quad \rho&= \text{ probability of occure of head}\\ q &=\text { probability of occure of tail } \\ &={ }^{9} C_{3} \times\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{3} \\ &={ }^{9} C_{3} \times\left(\frac{1}{2}\right)^{9} \end{aligned}
and in $10^{\text {th }}$ trial head must appears.
So required probability
\begin{aligned} &={ }^{9} C_{3}\left(\frac{1}{2}\right)^{9} \times \frac{1}{2} \\ &=\frac{9 \times 8 \times 7}{3 !} \times\left(\frac{1}{2}\right)^{10}=\frac{84}{1024}=0.082 \end{aligned}
 Question 5
If z = xy ln(xy), then
 A $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$ B $y\frac{\partial z}{\partial x}=x\frac{\partial z}{\partial y}$ C $x\frac{\partial z}{\partial x}=y\frac{\partial z}{\partial y}$ D $y\frac{\partial z}{\partial x}+x\frac{\partial z}{\partial y}=0$
Engineering Mathematics   Calculus
Question 5 Explanation:
\begin{aligned} \frac{\partial z}{\partial x}&=yln(x y)+\frac{x y}{x y} y\\ \frac{\partial z}{\partial x}&=y[\ln (x y)+1] &\ldots(i)\\ \frac{\partial z}{\partial y}&=x \ln (x y)+\frac{x y}{x y} \times x\\ \frac{\partial z}{\partial y}&=x[\ln (x y)+1]\\ \text{Here}\quad x\frac{\partial z}{\partial x}&=y \frac{\partial z}{\partial y} \end{aligned}
 Question 6
A series RC circuit is connected to a DC voltage source at time t = 0. The relation between the source voltage $V_{S}$, the resistance R, the capacitance C, and the current $i(t)$ is given below :
$V_{S}=Ri(t)+\frac{1}{C}\int_{0}^{t}i(u)du$
Which one of the following represents the current $i(t)$ ?
 A A B B C C D D
Network Theory   Sinusoidal Steady State Analysis
Question 6 Explanation:
Given that:
$V_{s}=R i(t)+\frac{1}{C} \int_{0}^{t} i(t) d t\qquad\ldots(i)$
Using Laplace transform,
\begin{aligned} \text { WS) }&=R I(s)+\frac{1}{C s} I(s) &\ldots(ii)\\ \text { or } I(s)&=\frac{V(s)}{\left(R+\frac{1}{C S}\right)} &\ldots(iii)\\ \text { For, } V(s)&=\frac{1}{s}&\ldots(iv) \end{aligned}
From equation (iii) and (iv),
$I(s)=\frac{C}{(R C s+1)}=\frac{1}{R\left(s+\frac{1}{R C}\right)}\qquad\ldots(v)$
Using inverse Laplace transform in equation (v), we get,
$i(t)=\frac{1}{R} e^{-t / R C}$
Thus, option (A) is correct.
 Question 7
In the figure shown, the value of the current I (in Amperes) is_____.
 A 0 B 0.25 C 0.5 D 1
Network Theory   Network Theorems
Question 7 Explanation:

Using super position theorem, when 5 V source acting alone, we get

$I_{1}=\frac{V}{R_{\mathrm{eq}}}=\frac{5}{10+5+5}=\frac{1}{4} \mathrm{A}\quad \ldots(i)$
When 1 A source acting alone, we get

$I_{2}=\frac{1 \times 5}{5+10+5}=\frac{5}{20}=\frac{1}{4} \mathrm{A}\quad \ldots(ii)$
Therefore,
$I=I_{1}+I_{2}=\frac{1}{2} A=0.5 \mathrm{A}$
 Question 8
In MOSFET fabrication, the channel length is defined during the process of
 A isolation oxide growth B channel stop implantation C poly-silicon gate patterning D lithography step leading to the contact pads
Electronic Devices   IC Fabrication
Question 8 Explanation:
Channel length is defined during the poly-silicon gate pattering.
 Question 9
A thin P-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to
 A the minority carrier mobility B the minority carrier recombination lifetime C the majority carrier concentration D the excess minority carrier concentration
Electronic Devices   Basic Semiconductor Physics
 Question 10
At T = 300 K, the hole mobility of a semiconductor $\mu _{P}=500 cm^{2}/V-s$ and $\frac{KT}{q}=26mV$. The hole diffusion constant $D_{P} \; in \; cm^{2}/s$ is _____.
 A 11 B 12 C 13 D 14
Electronic Devices   Basic Semiconductor Physics
Question 10 Explanation:
$\begin{array}{l} \frac{D_{p}}{\mu_{p}}=v_{T} \\ D_{p}=\mu_{p} V_{T}=500 \times 26 \times 10^{-3} \\ D_{p}=13 \mathrm{cm}^{2} / \mathrm{s} \end{array}$
There are 10 questions to complete.

## GATE EC 2014 SET-4

 Question 1
The series $\sum_{^{n=0}}^{\infty }\frac{1}{n!}$ converges to
 A 2 ln 2 B $\sqrt{2}$ C 2 D e
Engineering Mathematics   Calculus
Question 1 Explanation:
Given,
\begin{aligned} \text { Let, } x(n) &=\sum_{n=0}^{\infty} \frac{1}{n !} \\ &=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots \\ &=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots \end{aligned}
Also we know that expression of $e^{x}$
\begin{aligned} e^{x}&=1+x+\frac{1}{2} x^{2}+\frac{1}{6} x^{3}+\frac{1}{24} x^{4}+\cdots\\ \text{Put }x&=1\text{ in above expression}\\ e^{1}&=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots \\ e&=\sum_{n=0}^{\infty} \frac{1}{n !} \end{aligned}
 Question 2
The magnitude of the gradient for the function $f(x,y,z)=x^{2}+3y^{2}+z^{3}$ at the point (1,1,1) is _____.
 A 5 B 6 C 7 D 8
Engineering Mathematics   Calculus
Question 2 Explanation:
\begin{aligned} f(v, y, z) &=x^{3}+3 y^{2}+z^{3} \\ \nabla f &=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{i}+\frac{\partial f}{\partial z} \hat{k} \\ &=2 x \hat{i}+6 y \hat{j}+3 z^{2} \hat{k} \\ (\Delta t)_{(1,1,1)} &=2 \hat{i}+6 \hat{j}+3 \hat{k} \\ |\Delta+|_{(1,1,1)} &=\sqrt{4+36+9}=\sqrt{49}=7 \end{aligned}
 Question 3
Let X be a zero mean unit variance Gaussian random variable. E[|X|] is equal to ______.
 A 0.8 B 1.8 C 0.2 D 1.2
Communication Systems   Random Processes
Question 3 Explanation:
For a Gaussian random variable,
\begin{aligned} f_{x}(x)&=\frac{1}{\sigma_{x} \sqrt{2 \pi}} e^{\left(\frac{\mu-x}{2 \sigma^{2}}\right)^{2}} \\ \text{For mean} (\mu)&=0 \text{ and Variance }\left(\sigma_{x}^{2}\right)=1 \\ f_{X}(x) &=\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} \\ E[|x|] &=\int_{-\infty}^{\infty}|x| f_{X}(x) d x \\ &=\int_{-\infty}^{\infty}|x| \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x \\ =& \int_{-\infty}^{0}-x \cdot \frac{1}{\sqrt{2 \pi}} e^{-x^{2}} d x+\int_{0}^{\infty} x \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x\\ &=2 \int_{0}^{\infty} x \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} x \cdot e^{\frac{-x^{2}}{2}} d x \\ &\text{Let }\quad -\frac{x^{2}}{2}=t \\ &\therefore \quad-\frac{2 x d x}{2}=d t \\ &\text{Hence }\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty}-e^{-t} d t=0.8 \\ &\therefore \quad E[|x|]=0.8 \\ \end{aligned}
 Question 4
If a and b are constants, the most general solution of the differential equation $\frac{d^{2}x}{dt^{2}}+2\frac{dx}{dt}+x=0$ is
 A $ae^{-t}$ B $ae^{-t}+bte^{-t}$ C $ae^{t}+bte^{-t}$ D $ae^{-2t}$
Engineering Mathematics   Differential Equations
Question 4 Explanation:
The differential equation is given as
\begin{aligned} \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+x&=0 \\ y&=C \cdot F+P \cdot I\\ \end{aligned}
since, Q=0, i.e. RHS term is zero, so there will
be no particular integral.
\begin{aligned} \therefore \quad y&=C \cdot F\\ \text{Let, }\quad \frac{\partial}{\partial x}&=D \\ \text{So. }\quad \left(D^{2}+2 D+1\right) x&=0 \\ \therefore \quad(D+1)^{2}&=0 \\ \therefore \quad y&=a e^{-t}+b t e^{t} \end{aligned}
 Question 5
The directional derivative of $f(x,y)=\frac{xy}{\sqrt{2}}(x+y)$ at (1,1) in the direction of the unit vector at an angle of $\frac{\pi }{4}$ with y-axis, is given by____.
 A 1 B 2 C 3 D 4
Engineering Mathematics   Calculus
Question 5 Explanation:
\begin{aligned} f(x, y) &=\frac{x y}{\sqrt{2}}(x+y)=\frac{x^{2} y+x y^{2}}{\sqrt{2}} \\ \nabla f &=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j} \\ &=\frac{2 x y+y^{2}}{\sqrt{2}}\hat{i}+\frac{(x^{2}+2xy)}{\sqrt{2}}\hat{j} \end{aligned}
The direction is $\hat{n}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$
Directional derivative in direction of $\hat{n}$ is
$\begin{array}{l} =\left(\frac{2 x y+y^{2}}{2}+\frac{x^{2}+2 x y}{2}\right)_{(1,1)} \\ =\frac{3}{2}+\frac{3}{2}=3 \end{array}$
 Question 6
The circuit shown in the figure represents a
 A voltage controlled voltage source B voltage controlled current source C current controlled current source D current controlled voltage source
Network Theory   Basics of Network Analysis
 Question 7
The magnitude of current (in mA) through the resistor $R_{2}$ in the figure shown is______.
 A 1.2 B 2.1 C 2.8 D 3.6
Network Theory   Basics of Network Analysis
Question 7 Explanation:
Using source transformation, we get,

Applying KVL in above circuit, we get,
$20-2 I-I-4 I+8-3 I=0$
or $28=10 I$
or $\quad I=2.8 \mathrm{mA}$
 Question 8
At T = 300 K, the band gap and the intrinsic carrier concentration of GaAs are 1.42 eV and $10^{6}cm^{-3}$, respectively. In order to generate electron hole pairs in GaAs, which one of the wavelength $(\lambda _{C})$ ranges of incident radiation, is most suitable ? (Given that : Plank's constant is $6.62 \times 10^{-34}$J-s, velocity of light is $3 \times 10^{10} cm/s$ and charge of electron is $1.6*10^{-19}$C)
 A $0.42 \mu m \lt \lambda _{c} \lt 0.87\mu m$ B $0.87 \mu m \lt \lambda _{c} \lt 1.42\mu m$ C $1.42 \mu m \lt \lambda _{c} \lt 1.62\mu m$ D $1.62 \mu m \lt \lambda _{c} \lt 6.62\mu m$
Electronic Devices   Basic Semiconductor Physics
Question 8 Explanation:
$\begin{array}{l} \lambda_{c} \leq \frac{1.24}{E_{g}(e V)} \mu \mathrm{m} \\ \lambda_{c} \leq \frac{1.24}{1.42} \\ \lambda_{c} \leq 0.87 \mathrm{m} \\ \end{array}$
 Question 9
In the figure, $ln(\rho _{i}$) is plotted as a function of 1/T , where $\rho _{i}$ is the intrinsic resistivity of silicon, T is the temperature, and the plot is almost linear. The slope of the line can be used to estimate
 A band gap energy of silicon $(E_{g})$ B sum of electron and hole mobility in silicon $(\mu _{n}+\mu _{p})$ C reciprocal of the sum of electron and hole mobility in silicon $(\mu _{n}+\mu _{p})^{-1}$ D intrinsic carrier concentration of silicon $(n_{i})$
Electronic Devices   Basic Semiconductor Physics
 Question 10
The cut-off wavelength (in $\mu$ m) of light that can be used for intrinsic excitation of a semiconductor material of bandgap $E_{g}$ = 1.1 eV is _____.
 A 0.12 B 1.12 C 2.12 D 3
Electronic Devices   Basic Semiconductor Physics
Question 10 Explanation:
\begin{aligned} \quad \lambda_{c}&=\frac{1.24}{E_{g}(\mathrm{eV})} \mu \mathrm{m}=\frac{1.24}{1.1} \mu \mathrm{m} \\ \lambda_{c}&=1.12 \mu \mathrm{m} \end{aligned}
There are 10 questions to complete.