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.

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GATE Electronics and Communication-Topic wise Previous Year Questions

GATE 2022 Electronics and Communications Syllabus

Revised syllabus of GATE 2022 Electronics and Communications by IIT.

Practice GATE Electronics and Communications previous year questions

Year wise | Subject wise | Topic wise

Section 1: Engineering Mathematics

Linear Algebra: Vector space, basis, linear dependence and independence, matrix algebra, eigenvalues and eigenvectors, rank, solution of linear equations- existence and uniqueness.
Calculus: Mean value theorems, theorems of integral calculus, evaluation of definite and improper integrals, partial derivatives, maxima and minima, multiple integrals, line, surface and volume integrals, Taylor series.
Differential Equations: First order equations (linear and nonlinear), higher order linear differential equations, Cauchy’s and Euler’s equations, methods of solution using variation of parameters, complementary function and particular integral, partial differential equations, variable separable method, initial and boundary value problems.
Vector Analysis: Vectors in plane and space, vector operations, gradient, divergence and curl, Gauss’s, Green’s and Stokes’ theorems.
Complex Analysis: Analytic functions, Cauchy’s integral theorem, Cauchy’s integral formula, sequences, series, convergence tests, Taylor and Laurent series, residue theorem.
Probability and Statistics: Mean, median, mode, standard deviation, combinatorial probability, probability distributions, binomial distribution, Poisson distribution, exponential distribution, normal distribution, joint and conditional probability.
Section 2: Networks, Signals and Systems

Circuit analysis: Node and mesh analysis, superposition, Thevenin’s theorem, Norton’s theorem, reciprocity. Sinusoidal steady state analysis: phasors, complex power, maximum power transfer.
Time and frequency domain analysis of linear circuits: RL, RC and RLC circuits, solution of network equations using Laplace transform.
Linear 2-port network parameters, wye-delta transformation. Continuous-time signals: Fourier series and Fourier transform, sampling theorem and applications.
Discrete-time signals: DTFT, DFT, z-transform, discrete-time processing of continuous-time signals. LTI systems: definition and properties, causality, stability, impulse response, convolution, poles and zeroes, frequency response, group delay, phase delay.
Section 3: Electronic Devices

Energy bands in intrinsic and extrinsic semiconductors, equilibrium carrier concentration, direct and indirect band-gap semiconductors.
Carrier transport: diffusion current, drift current, mobility and resistivity, generation and recombination of carriers, Poisson and continuity equations.
P-N junction, Zener diode, BJT, MOS capacitor, MOSFET, LED, photo diode and solar cell.
Section 4: Analog Circuits

Diode circuits: clipping, clamping and rectifiers.
BJT and MOSFET amplifiers: biasing, ac coupling, small signal analysis, frequency response. Current mirrors and differential amplifiers.
Op-amp circuits: Amplifiers, summers, differentiators, integrators, active filters, Schmitt triggers and oscillators.
Section 5: Digital Circuits

Number representations: binary, integer and floating-point- numbers.
Combinatorial circuits: Boolean algebra, minimization of functions using Boolean identities and Karnaugh map,
logic gates and their static CMOS implementations, arithmetic circuits, code converters, multiplexers, decoders.
Sequential circuits: latches and flip-flops, counters, shift-registers, finite state machines, propagation delay, setup and hold time, critical path delay.
Data converters: sample and hold circuits, ADCs and DACs.
Semiconductor memories: ROM, SRAM, DRAM.
Computer organization: Machine instructions and addressing modes, ALU, data-path and control unit, instruction pipelining.
Section 6: Control Systems

Basic control system components; Feedback principle; Transfer function; Block diagram representation; Signal flow graph; Transient and steady-state analysis of LTI systems; Frequency response; Routh-Hurwitz and Nyquist stability criteria; Bode and root-locus plots; Lag, lead and lag-lead compensation; State variable model and solution of state equation of LTI systems.
Section 7: Communications

Random processes: autocorrelation and power spectral density, properties of white noise, filtering of random signals through LTI systems.
Analog communications: amplitude modulation and demodulation, angle modulation and demodulation, spectra of AM and FM, superheterodyne receivers.
Information theory: entropy, mutual information and channel capacity theorem.
Digital communications: PCM, DPCM, digital modulation schemes (ASK, PSK, FSK, QAM), bandwidth, inter-symbol interference, MAP, ML detection, matched filter receiver, SNR and BER.
Fundamentals of error correction, Hamming codes, CRC.
Section 8: Electromagnetics

Maxwell’s equations: differential and integral forms and their interpretation, boundary conditions,wave equation, Poynting vector.
Plane waves and properties: reflection and refraction, polarization, phase and group velocity, propagation through various media, skin depth.
Transmission lines: equations, characteristic impedance, impedance matching, impedance transformation, Sparameters, Smith chart.
Rectangular and circular waveguides, light propagation in optical fibers, dipole and monopole antennas, linear antenna arrays.

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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 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}
For Matched load,
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 EC 2011

Question 1
Consider the following statements regarding the complex Poynting vector \vec{P} for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(\vec{P}) denotes the real part of \vec{P}, S denotes a spherical surface whose centre is at the point source, and \hat{n} denotes the unit surface normal on S. Which of the following statements is TRUE?
A
Re(\vec{P}) remains constant at any radial distance from the source
B
Re(\vec{P}) increases with increasing radial distance from the source
C
\oint \oint_{S}Re(\vec{P}).\hat{n} dS remains constant at any radial distance from the source
D
\oint \oint_{S}Re(\vec{P}).\hat{n} dS decreases with increasing radial distance form the source
Electromagnetics   Antennas
Question 1 Explanation: 
Power density of any point source decreases with distance i.e. the density decreases and area of cross-over increases with the product being constant.
Question 2
A transmission line of characteristic impedance 50 W is terminated by a 50 \Omega load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to \frac{\pi }{4} radians. The phase velocity of the wave along the line is
A
0.8 \times 10^{8} m/s
B
1.2 \times 10^{8} m/s
C
1.6 \times 10^{8} m/s
D
3 \times 10^{8} m/s
Electromagnetics   Transmission Lines
Question 2 Explanation: 
\begin{aligned} \text { Phase difference }&=\frac{2 \pi}{\lambda} \text { (path difference) } \\ \Rightarrow \frac{\pi}{4}&=\frac{2 \pi}{\lambda}\left(2 \times 10^{-3}\right) \\ \therefore \quad &=8 \times 2 \times 10^{-3} \\ &=16 \times 10^{-3} \mathrm{m} \\ f&= 10 \mathrm{GHz}=10 \times 10^{9} \mathrm{Hz} \end{aligned}
Hence the phase velocity of wave along the line is,
\begin{aligned} & v=f \lambda=10 \times 10^{9} \times 16 \times 10^{-3} \mathrm{m} / \mathrm{sec} \\ \therefore \quad v &=1.6 \times 10^{8} \mathrm{m} / \mathrm{s} \end{aligned}
Question 3
An analog signal is band-limited to 4 kHz, sampled at the Nyquist rate and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is ________ bits / second.
A
1
B
2
C
3
D
4
Communication Systems   Digital Communications
Question 3 Explanation: 
Quantized levels are equiprobable; hence
\begin{aligned} H&=\log _{2} 4=2 \text { bits/sample }\\ r&=2 \text{samples/sec} \\ \end{aligned}
Hence information rate R=r \cdot H=2 \text{ samples/sec}
\times 2\text{ bits/sample}
\Rightarrow R=4 \mathrm{bits} / \mathrm{sec}
Question 4
The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by
A
G(S)H(S)=k\frac{s(s+1)}{(s+2)(s+3)}
B
G(S)H(S)=k\frac{(s+1)}{(s+2)(s+3)^{2}}
C
G(S)H(S)=k\frac{1}{s(s-1)(s+2)(s+3)}
D
G(S)H(S)=k\frac{(s+1)}{(s+2)(s+3)}
Control Systems   Root Locus
Question 4 Explanation: 
From plot we can observe that one pole terminates at one zero at position -1 and three poles terminates to \infty. It means there are four poles and 1 zero. Pole at -3 goes on both sides. It means there are two poles at -3.
Question 5
A system is defined by its impulse response h(n) =2^{n} u(n - 2). The system is
A
stable and causal
B
causal but not stable
C
stable but not causal
D
unstable and non-causal
Signals and Systems   LTI Systems Continuous and Discrete
Question 5 Explanation: 
h(n)=2^{n} u(n-2)
For causal system h(n)=0 for n \lt 0
Hence given system is causal.
For stability:
\sum_{n=2}^{\infty} 2^{n}=\infty, so given system is not stable.
Question 6
If the unit step response of a network is (1-e^{-at}), then its unit impulse response is
A
\alpha e^{-at}
B
\alpha^{-1} e^{-at}
C
(1-\alpha ^{-1})e^{at}
D
(1-\alpha)e^{-at}
Signals and Systems   Laplace Transform
Question 6 Explanation: 
Unit step response s(t)=\left(1-e^{-\alpha t}\right)
So, unit impulse response is
\begin{aligned} h(t) &=\frac{d s(t)}{d t}=\frac{d}{d t}\left(1-e^{-\alpha t}\right) \\ &=\alpha e^{-\alpha t} \end{aligned}
Question 7
The output Y in the circuit below is always '1' when
A
two or more of the inputs P, Q, R are '0'
B
two or more of the inputs P, Q, R are '1'
C
any odd number of the inputs P, Q, R is '0'
D
any odd number of the inputs P, Q, R is '1'
Digital Circuits   Logic Gates
Question 7 Explanation: 


Y= PO+ PR+ RO
Question 8
In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. v_{i} is a small signal input. The gain magnitude |\frac{v_{0}}{v_{i}}| at 10 M rad/s is
A
maximum
B
minimum
C
unity
D
zero
Analog Circuits   BJT Analysis
Question 8 Explanation: 
Whenever we use a bypass capacitor in parallel R_E then it always increases voltage gain. As compare to "CE with R_E without capacitor". Hence only answer (A) is possible. and
\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10 \times 10^{-6} \times 10^{-9}}}
=10 M rad/sec and gain is maximum ar resonance frequency.
Question 9
Drift current in the semiconductors depends upon
A
only the electric field
B
only the carrier concentration gradient
C
both the electric field and the carrier concentration
D
both the electric field and the carrier concentration gradient
Electronic Devices   Basic Semiconductor Physics
Question 9 Explanation: 
\begin{aligned} j &= n\;e\;v_{d}\\ \text{Put,}\quad v_{d} &= \mu E\\ \therefore J&=n e \mu E\\ \text{Hence}\quad l&=n e \mu EA\\ \end{aligned}
So, 1 depends upon carrier concentration and electric field.
Question 10
A Zener diode, when used in voltage stabilization circuits, is biased in
A
reverse bias region below the breakdown voltage
B
reverse breakdown region
C
forward bias region
D
forward bias constant current mode
Electronic Devices   PN-Junction Diodes and Special Diodes
Question 10 Explanation: 


In breakdown region only, zener diode is useful.
In reverse bias region below the breakdown voltage it will behave like an open circuit.
There are 10 questions to complete.

GATE EC 2012

Question 1
The current i_{b} through the base of a silicon npn transistor is 1 + 0.1 cos(10000 \pi t) mA. At 300 K, the r_{\pi} in the small signal model of the transistor is
A
250 \Omega
B
27.5 \Omega
C
25 \Omega
D
22.5 \Omega
Analog Circuits   BJT Analysis
Question 1 Explanation: 
We know that
\begin{array}{l} r_{\pi}=(\beta+1) r_{e} \\ r_{\pi}=(\beta+1) \frac{V_{T}}{I_{e}} \\ r_{\pi}=(\beta+1) \frac{V_{T}}{(\beta+1) I_{b}} \\ r_{\pi}=\frac{V_{T}}{I_{b}} \end{array}
Where I_{b} is d.c. current through base so
I_{b}=1 \mathrm{mA}
V_{T}=25 \mathrm{mV} at room temperature
So, r_{\pi}=\frac{25 \times 10^{-3}}{1 \times 10^{-3}}=25 \Omega
Question 2
The power spectral density of a real process X(t) for positive frequencies is shown below. The values of E[X^{2}(t)] and |E[X(t)]|, respectively, are
A
6000/\pi,0
B
6400/\pi,0
C
6400/\pi,20(\pi /\sqrt{2})
D
6000/\pi,20(\pi /\sqrt{2})
Communication Systems   Random Processes
Question 2 Explanation: 


We know that E\left[X^{2}(t)\right] represent the total power in random signal. So
\begin{aligned} P=E\left[X^{2}(t)\right]&=2 \times \frac{1}{2 \pi} \int_{9 \times 10^{3}}^{11 \times 10^{3}} S_{x}(\omega) d \omega \\ E\left[X^{2}(t)\right]&=\frac{1}{\pi}\left[400+\frac{1}{2} \times 6 \times 2 \times 10^{3}\right] \\ E\left[X^{2}(t)\right]&=\frac{6400}{\pi} \end{aligned}
At \omega=0, there is no any frequency component present, hence dc value of the process is zero.
Question 3
In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is
A
1750
B
2625
C
4000
D
5250
Communication Systems   Digital Communications
Question 3 Explanation: 
\begin{aligned} \mathrm{BW} &=\frac{R_{b}}{2}(1+\alpha) \\ 3500 &=\frac{R_{b}}{2}(1+0.75) \\ R_{b} &=4000 \mathrm{bits} / \mathrm{sec} \end{aligned}
In base band transmission \rightarrow
Symbol rate = Bit rate =4000 symbols/sec
Question 4
A plane wave propagating in air with \vec{E}=(8\hat{a}_{x}+6\hat{a}_{y}+5\hat{a}_{z})e^{j(\omega t+3x-4y)} V/m is incident on a perfectly conducting slab positioned at 0 \leq x. The E field of the reflected wave is
A
(-8\hat{a}_{x}-6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t+3x+4y)} V/m
B
(-8\hat{a}_{x}+6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t+3x+4y)} V/m
C
(-8\hat{a}_{x}-6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t-3x-4y)} V/m
D
(-8\hat{a}_{x}+6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t-3x-4y)} V/m
Electromagnetics   Plane waves and Properties
Question 4 Explanation: 


Incident wave propagation \rightarrow-3 \hat{a}_{x}+4 \hat{a}_{y}
Reflected wave propagation \rightarrow 3 \hat{a}_{x}+4 \hat{a}_{y}
E_{i} incident direction \rightarrow 8 \hat{a}_{x}+6 \hat{a}_{y}+5 \hat{a}_{z}
\left.\begin{array}{l}E_{i} \text { tangential } \rightarrow 6 \hat{a}_{y}+5 \hat{a}_{z} \\ E_{r} \text { tangential } \rightarrow-6 \hat{a}_{y}-5 \hat{a}_{z}\end{array}\right\} E_{\text {tang }}=0
\left.\begin{array}{rl}E_{i} \text { normal } & =8 \hat{a}_{x} \\ E_{r} \text { normal } & =8 \hat{a}_{x}\end{array}\right\} E_{\text {normal }}=E_{\max }
E_{r} reflected direction =8 \hat{a}_{x}-6 \hat{a}_{y}-5 \hat{a}_{z}
Question 5
The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction, is given by \vec{E}=10(\hat{a}_{y}+j\hat{a}_{z})e^{-j25x}. The frequency and polarization of the wave, respectively, are
A
1.2 GHz and left circular
B
4 Hz and left circular
C
1.2 GHz and right circular
D
4 Hz and right circular
Electromagnetics   Plane waves and Properties
Question 5 Explanation: 
Out of phase by 90^{\circ}and equal amplitude the wave is circularly polarized.
Check the trace of E with the time which gives left circularly polarized.
\begin{aligned} a_{y}&=\sin \omega t \\ a_{z}&=\cos \omega t \end{aligned}
Question 6
Consider the given circuit.

In this circuit, the race around
A
does not occur
B
occurs when CLK = 0
C
occurs when CLK = 1 and A = B = 1
D
occurs when CLK = 1 and A = B = 0
Digital Circuits   Sequential Circuits
Question 6 Explanation: 
Given flipflop is S-R flipflop with A= Sand B = R.
In S-R flipflop race around condition does not occur.
Question 7
The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B. The number of combinations for which the output is logic 1, is
A
4
B
6
C
8
D
10
Digital Circuits   Combinational Circuits
Question 7 Explanation: 
Output will be 1 if A>B.
If B=00 then there will be three combinations for which output will be 1 i.e. when A=01,10 , or 11.
If B=01 there will be two conditions i.e. A=10 and 11.
If B=10 there will be one condition i.e. A=11.
So total 6 combinations are there for which output will be 1 .
Question 8
The i-v characteristics of the diode in the circuit given below are i=\left\{\begin{matrix} \frac{v-0.7}{500}A &v\geq 0.7V \\ 0 A & v \lt 0.7 V \end{matrix}\right.

The current in the circuit is
A
10 mA
B
9.3 mA
C
6.67 mA
D
6.2 mA
Electronic Devices   PN-Junction Diodes and Special Diodes
Question 8 Explanation: 
Applying KVL in given circuit we have
10-1000 i-v=0 \qquad\ldots(1)
since 10 V is greater than 0.7 V so current will flow through diode and is
i=\frac{v-0.7}{500}\qquad\ldots(2)
from (1) and (2) we have,
\begin{aligned} 10-2(v-0.7)-v &=0 \\ 10+1.4-3 v &=0 \\ 3 v &=11.4 \\ v &=\frac{11.4}{3}=3.8 v\\ \text{so}\quad i &=\frac{v-0.7}{500}=\frac{3.8-0.7}{500} \\ &=6.2 \mathrm{mA} \end{aligned}
Question 9
In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i(t) for all t is
A
zero
B
a step function
C
an exponentially decaying function
D
an impulse function
Network Theory   Transient Analysis
Question 9 Explanation: 
Since there is no resistance so time constant will be zero. That means as soon as the switch will be closed voltages at C_{1} and C_{2} will become equal and capacitor allows sudden change of voltage only if impulse of current will pass through it.
Question 10
The average power delivered to an impedance (4 - j3) \Omegaby a current 5cos(100\pi t +100) A is
A
44.2 W
B
50W
C
62.5W
D
125W
Network Theory   Basics of Network Analysis
Question 10 Explanation: 
Average power is same as RMS power.
\begin{array}{l} P=I_{\mathrm{rms}}^{2} R=\left(\frac{5}{\sqrt{2}}\right)^{2} \times 4 \\ =\frac{25}{2} \times 4=50 \mathrm{W} \end{array}
Note: Power is consumed only by resistance i.e. by real part of impedance.
There are 10 questions to complete.

GATE EC 2013

Question 1
A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles
A
an AND gate
B
an OR gate
C
an XOR gate
D
a NAND gate
Digital Circuits   Logic Gates
Question 1 Explanation: 
Truth table of XOR gate
\begin{array}{cc|c} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}
So, from the XOR gate truth table it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch.
Question 2
Consider a vector field {\overrightarrow{A}}( \overrightarrow{r}). The closed loop line integral \int \overrightarrow{A}\cdot \overrightarrow{di} can be expressed as
A
\int \int ( \overline{V}\times \overrightarrow{A} )\cdot \overrightarrow{ds} over the closed surface bounded by the loop
B
\int \int \int ( \overline{V}\cdot \overrightarrow{A} )dv over the closed volume bounded by the loop
C
\int \int \int ( \overline{V}\cdot \overrightarrow{A} )dv over the open volume bounded by the loop
D
\int \int ( \overline{V}\times \overrightarrow{A} )\cdot \overrightarrow{ds} over the open surface bounded by the loop
Engineering Mathematics   Calculus
Question 2 Explanation: 
According to Stoke's theorem
\oint_{c} \vec{A} \cdot \vec{d} i=\iint_{s}(\nabla \times \vec{A}) \cdot \overrightarrow{d s}
Question 3
Two systems with impulse responses h_{1}\left ( t \right ) and h_{2}\left ( t \right ) are connected in cascade. Then the overall impulse response of the cascaded system is given by
A
product of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
B
sum of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
C
convolution of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
D
subtraction of h_{2}\left ( t \right ) from h_{1}\left ( t \right )
Signals and Systems   LTI Systems Continuous and Discrete
Question 3 Explanation: 
The overall impulse response h(t) of the cascade system is given by:
h(t)=h_{1}(t) * h_{2}(t)
Question 4
In a forward biased pn junction diode, the sequence of events that best describes the mechanism of current flow is
A
injection, and subsequent diffusion and recombination of minority carriers
B
injection, and subsequent drift and generation of minority carriers
C
extraction, and subsequent diffusion and generation of minority carriers
D
extraction, and subsequent drift and recombination of minority carriers
Electronic Devices   PN-Junction Diodes and Special Diodes
Question 4 Explanation: 
In a forward biased pn-junction diode, the current flow is due to diffusion of majority carriers and recombination of minority carriers.
Question 5
In IC technology, dry oxidation (using dry oxygen) as compared to wet oxidation (using steam or water vapor) produces
A
superior quality oxide with a higher growth rate
B
inferior quality oxide with a higher growth rate
C
inferior quality oxide with a lower growth rate
D
superior quality oxide with a lower growth rate
Electronic Devices   IC Fabrication
Question 5 Explanation: 
Dry oxidation has better quality over wet oxidation.
Dry oxidation is slower over wet oxidation.
Question 6
The maximum value of \theta until which the approximation \sin \theta \approx \theta holds to within 10% error is
A
10^{\circ}
B
18^{\circ}
C
50^{\circ}
D
90^{\circ}
Engineering Mathematics   Numerical Methods
Question 6 Explanation: 
10^{\circ}=\frac{10 \pi}{180}=0.1745
\sin 10^{\circ}=0.1736
So, for 10^{\circ} \rightarrow \sin \cong \theta holds within 10 % error
\begin{aligned} 18^{\circ} &=\frac{18 \times \pi}{180}=0.3142 \\ \sin 18^{\circ} &=0.3090 \end{aligned}
So, for 18^{\circ} \rightarrow \sin \theta \equiv \theta holds within 10% error
50^{\circ}=\frac{50 \times \pi}{180}=0.8727
\sin 50^{\circ}=0.766
So, for 50^{\circ} \rightarrow \sin \theta \cong \theta does not hold within 10 % error
90^{\circ}=\frac{90 \times \pi}{180}=1.571
\sin 90^{\circ}=1
So, for 90^{\circ} \rightarrow \sin \theta \cong \theta does not hold within 10 % error.
So, the maximum value of \theta for the approximation
\sin \theta \cong \theta \text{ holds to within } 10 \% \text{ error is } 18^{\circ}
Question 7
The divergence of the vector field \overrightarrow{A}=x\hat{a_{x}}+y\hat{a_{y}}+z\hat{a_{x}} is
A
0
B
1/3
C
1
D
3
Engineering Mathematics   Calculus
Question 7 Explanation: 
\begin{array}{l} \nabla \cdot \vec{A}=\frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z} \\ \nabla \cdot \vec{A}=\frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y)+\frac{\partial}{\partial z}(z)=1+1+1 \\ \nabla \cdot \vec{A}=3 \end{array}
Question 8
The impulse response of a system is h\left ( t \right )= tu\left ( t \right ). For an input u(t-1), the output is
A
\frac{t^{2}}{2}u\left ( t \right )
B
\frac{t(t-1)}{2}u(t-1)
C
\frac{\left ( t-1 \right )^{2}}{2}u\left ( t-1 \right )
D
\frac{t^{2}-1}{2}u\left ( t-1 \right )
Signals and Systems   Laplace Transform
Question 8 Explanation: 
\begin{aligned} h(t)&=t u(t) \\ \text{Taking }&\text{Laplace transform}\\ H(s)&=\frac{1}{s^{2}} \\ x(t) &=u(t-1) \\ \text { Taking }&\text{Laplace transform } \\ X(s)&=\frac{e^{-s}}{s} \\ \frac{Y(s)}{X(s)}&=H(s) \\ Y(s)&=H(s) X(s) \\ Y(s)&=\frac{1}{s^{2}} \cdot \frac{e^{-s}}{s}=\frac{e^{-s}}{s^{3}} \\ \text{Taking } &\text{the inverse Laplace transform } \\ y(t)&=\frac{(t-1)^{2}}{2} u(t-1) \\ \end{aligned}
Question 9
The Bode plot of a transfer function G(s) is shown in the figure below.
The gain \left ( 20\log\left | G\left ( s \right ) \right | \right ) is 32 dB and -8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all \omega. Then G(s) is
A
\frac{39.8}{s}
B
\frac{39.8}{s^{2}}
C
\frac{32}{s}
D
\frac{32}{s^{2}}
Control Systems   Frequency Response Analysis
Question 9 Explanation: 
10 \mathrm{rad} / \mathrm{s} to 1 \mathrm{rad} / \mathrm{s} is 1 decade
32-(-8)=40 d B
So, the slope is 40dB/decade it means there are two poles at origin, it means either option (B) or option (D) is correct put \omega=1rad/sec in both the options.
20 \log \left[\frac{39.8}{(1)^{2}}\right]=32 \mathrm{dB}
20 \log \left[\frac{32}{(1)^{2}}\right]=30.1 \mathrm{dB}
So, option (B) is correct option =\frac{39.8}{s^{2}}
Question 10
In the circuit shown below what is the output voltage \left ( V_{out} \right ) if a silicon transistor Q and an ideal op-amp are used?
A
-15V
B
-0.7V
C
+0.7V
D
+15V
Analog Circuits   Operational Amplifiers
Question 10 Explanation: 
Due to virtual short
V_{c} = V_{B} = 0V
(collector voltage of transistor 0)
and given that base voltage of transistor 0
V_{B} = O_{V}
So, V_{C} = O_{V}
It means collector to base of transistor Oare short circuited.
If any junction of transistor is short circuited then the junction acts as reverse bias. So, the C-B junction is reverse bias. The given op-amp is an inverting configuration which have positive voltage as input. So the output voltage of op-amp will be negative voltage.
V_{\text{out}} = -ve voltage
Emitter voltage of transistor
V_{E} = -ve voltage
given transistor is n-p-ntransistor and
V_{B} = OV
V_{E} = -ve voltage
So, the E-8 junction will be forward bias. Thus, the transistor is in active region and will behave as closed switch.
So,
\begin{aligned} V_{B E} &=0.7 \mathrm{V} \quad \text { (for silicon transistor) } \\ \mathrm{V}_{\mathrm{B}}-\mathrm{V}_{E} &=0.7 \mathrm{V} \\ \mathrm{V}_{\mathrm{E}} &=\mathrm{V}_{\mathrm{B}}-0.7 \\ \mathrm{V}_{\mathrm{E}} &=0-0.7 \\ \mathrm{V}_{\mathrm{E}} &=-0.7 \mathrm{V} \end{aligned}
There are 10 questions to complete.