GATE Electronics and Communication 2022

Question 1
Consider the two-dimensional vector field \vec{F}(x,y)=x\vec{i}+y\vec{j}, where \vec{i} and \vec{j} denote the unit vectors along the x-axis and the y-axis, respectively. A contour C in the x-y plane, as shown in the figure, is composed of two horizontal lines connected at the two ends by two semicircular arcs of unit radius. The contour is traversed in the counter-clockwise sense. The value of the closed path integral
\oint _c \vec{F}(x,y)\cdot (dx\vec{i}+dy\vec{j})

A
0
B
1
C
8+2 \pi
D
-1
Engineering Mathematics   Calculus
Question 1 Explanation: 
\oint \vec{F} (x,y)\cdot [dx\vec{i}+dy\vec{j}]
Given \vec{F} (x,y)=x\vec{i}+y\vec{j}
\therefore \int_{c}xdx+ydy=0
Because here vector is conservative.
If the integral function is the total derivative over the closed contoure then it will be zero
Question 2
Consider a system of linear equations Ax=b, where
A=\begin{bmatrix} 1 & -\sqrt{2} & 3\\ -1& \sqrt{2}& -3 \end{bmatrix},b=\begin{bmatrix} 1\\ 3 \end{bmatrix}
This system of equations admits ______.
A
a unique solution for x
B
infinitely many solutions for x
C
no solutions for x
D
exactly two solutions for x
Engineering Mathematics   Linear Algebra
Question 2 Explanation: 
Here equation will be
x-\sqrt{2}y+3z=1
-x+\sqrt{2}y-3z=3
therefore inconsistant solution i.e. there will not be any solution.
Question 3
The current I in the circuit shown is ________

A
1.25 \times 10^{-3}A
B
0.75 \times 10^{-3}A
C
-0.5 \times 10^{-3}A
D
1.16 \times 10^{-3}A
Network Theory   Basics of Network Analysis
Question 3 Explanation: 


Applying Nodal equation at Node-A
\begin{aligned} \frac{V_A}{2k}+\frac{V_A-5}{2k}&=10^{-3}\\ \Rightarrow 2V_A-5&=2k \times 10^{-3}\\ V_A&=3.5V\\ Again,&\\ I&=\frac{5-V_A}{2k}\\ &=\frac{5-3.5}{2k}\\ &=0.75 \times 10^{-3}A \end{aligned}
Question 4
Consider the circuit shown in the figure. The current I flowing through the 10\Omega resistor is _________.

A
1A
B
0A
C
0.1A
D
-0.1A
Network Theory   Basics of Network Analysis
Question 4 Explanation: 
Here, there is no any return closed path for Current (I) . Hence I=0
Current always flow in loop.
Question 5
The Fourier transform X(j\omega ) of the signal x(t)=\frac{t}{(1+t^2)^2} is _________.
A
\frac{\pi}{2j}\omega e^{-|\omega|}
B
\frac{\pi}{2}\omega e^{-|\omega|}
C
\frac{\pi}{2j} e^{-|\omega|}
D
\frac{\pi}{2} e^{-|\omega|}
Signals and Systems   DTFS, DTFT and DFT
Question 5 Explanation: 
x(t)=\frac{t}{(1+t^2)^2}
As we know that FT of te^{-|t|} \; \underleftrightarrow{FT} \;\frac{-j4\omega }{(1+\omega ^2)^2}
Duality \frac{-j4\omega }{(1+t ^2)^2} \leftrightarrow 2 \pi(-\omega )e^{-|-\omega |}
\Rightarrow \frac{t}{(1+t^2)^2} \underrightarrow{FT} \frac{-2\pi}{-j4}\omega e^{-|\omega |}
\Rightarrow \;\;\;\rightarrow\frac{\pi}{j2} \omega e^{-|\omega |}
Question 6
Consider a long rectangular bar of direct bandgap p-type semiconductor. The equilibrium hole density is 10^{17}cm^{-3} and the intrinsic carrier concentration is 10^{10}cm^{-3}. Electron and hole diffusion lengths are 2\mu mand 1\mu m, respectively. The left side of the bar (x=0) is uniformly illuminated with a laser having photon energy greater than the bandgap of the semiconductor. Excess electron-hole pairs are generated ONLY at x=0 because of the laser. The steady state electron density at x=0 is 10^{14}cm^{-3} due to laser illumination. Under these conditions and ignoring electric field, the closest approximation (among the given options) of the steady state electron density at x=2 \mu m, is _____
A
0.37 \times 10^{14} cm^{-3}
B
0.63 \times 10^{13} cm^{-3}
C
3.7 \times 10^{14} cm^{-3}
D
0^{3} cm^{-3}
Electronic Devices   Basic Semiconductor Physics
Question 6 Explanation: 


From continuity equation of electrons
\frac{dn}{dt}=n\mu _n\frac{dE}{dx}+\mu _nE\frac{dn}{dx}+G_n-R_n+x_n\frac{d^2x}{dx^2} \;\;\;...(i)
[Because \vec{E} is not mentioned hence
\frac{dE}{dx}=0
For x \gt 0, G_n is also zero
n=\frac{n_i^2}{N_A}=\frac{10^{20}}{10^{17}}=10^3
n=n_0+\delta n=10^3+10^{14}=10^{14}
at steady state, \frac{db}{dt}=0
Hence equation (i) becomes:
O=D_n\frac{d^2\delta n}{dx^2}-\frac{\delta n}{\tau _n}
\frac{d^2\delta n}{dx^2}=\frac{\delta n}{L_n^2} \;\;\;...(ii)
From solving equation (ii)
\delta _n(x)=\delta _n(0)e^{-x/L_n}
at x=2\mu m
\delta _n(2\mu m)=10^{14}e^{-2/2}=10^{14}e^{-1}=0.37 \times 10^{14}
Question 7
In a non-degenerate bulk semiconductor with electron density n=10^{16}cm^{-3}, the value of E_C-E_{Fn}=200meV, where E_C and E_{Fn} denote the bottom of the conduction band energy and electron Fermi level energy, respectively. Assume thermal voltage as 26 meV and the intrinsic carrier concentration is 10^{10}cm^{-3}. For n=0.5 \times 10^{16}cm^{-3}, the closest approximation of the value of (E_C-E_{Fn}), among the given options, is ______.
A
226 meV
B
174 meV
C
218 meV
D
182 meV
Electronic Devices   Basic Semiconductor Physics
Question 7 Explanation: 
Here we have to find the value of E_c-E_{fn}
As we know,
E_C-E_F=kT \ln\left ( \frac{N_c}{n} \right ) \;\;\;...(i)
E_C-E_{F1}=kT \ln\left ( \frac{N_c}{n_1} \right ) \;\;\;...(ii)
E_C-E_{F2}=kT \ln\left ( \frac{N_c}{n_2} \right ) \;\;\;...(iii)
Equation (ii) - Equation (iii)
(E_C-E_{F1})-(E_C-E_{F2})=kT \ln \left ( \frac{\frac{N_c}{n_1}}{\frac{N_c}{n_2}} \right )=kT \ln \frac{n_2}{n_1}
\Rightarrow 200meV-(E_C-E_{F2})=26meV \times \ln \left ( \frac{0.5 \times 10^{16}}{1 \times 10^{16}} \right )
200meV-(E_C-E_{F2})=+26meV \ln (0.5)=-18
(E_C-E_{F2})=200+8=218meV
Question 8
Consider the CMOS circuit shown in the figure (substrates are connected to their respective sources). The gate width (W) to gate length (L) ratios \frac{W}{L} of the transistors are as shown. Both the transistors have the same gate oxide capacitance per unit area. For the pMOSFET, the threshold voltage is -1 V and the mobility of holes is 40\frac{cm^2}{V.s}. For the nMOSFET, the threshold voltage is 1 V and the mobility of electrons is 300\frac{cm^2}{V.s}. The steady state output voltage V_o is ________.

A
equal to 0 V
B
more than 2 V
C
less than 2 V
D
equal to 2 V
Analog Circuits   FET and MOSFET Analysis
Question 8 Explanation: 


\begin{aligned} \mu _PCO_x\left ( \frac{\omega }{L} \right )_1[4-V_0-1]^2&=\mu _nCO_x\left ( \frac{\omega }{L} \right )_2[V_0-0-1]^2\\ \Rightarrow \frac{300}{40}\times \frac{1}{5}(V_0-1)^2&=(3-V_0)^2\\ \Rightarrow \sqrt{1.5} (V_0-1)&=3-V_0\\ \Rightarrow V_0&=\frac{3+\sqrt{1.5}}{\sqrt{1.5}+1} \lt 2V \end{aligned}
Question 9
Consider the 2-bit multiplexer (MUX) shown in the figure. For OUTPUT to be the XOR of C and D, the values for A_0,A_1,A_2 \text{ and }A_3 are _______

A
A_0=0,A_1=0,A_2=1,A_3=1
B
A_0=1,A_1=0,A_2=1,A_3=0
C
A_0=0,A_1=1,A_2=1,A_3=0
D
A_0=1,A_1=1,A_2=0,A_3=0
Digital Circuits   Combinational Circuits
Question 9 Explanation: 


f=\bar{C}\bar{D}I_0+\bar{C}DI_1+C\bar{D}I_2+CDI_3
For this
A_0=A_3=0
A_1=A_2=1
Question 10
The ideal long channel nMOSFET and pMOSFET devices shown in the circuits have threshold voltages of 1 V and -1 V, respectively. The MOSFET substrates are connected to their respective sources. Ignore leakage currents and assume that the capacitors are initially discharged. For the applied voltages as shown, the steady state voltages are ______

A
V_1=5 V, V_2=5 V
B
V_1=5 V, V_2=4 V
C
V_1=4 V, V_2=5 V
D
V_1=4V, V_2=-5 V
Analog Circuits   FET and MOSFET Analysis
Question 10 Explanation: 


There are 10 questions to complete.

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}
Engineering Mathematics   Calculus
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}
Engineering Mathematics   Differential Equations
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}
Communication Systems   Random Signals and Noise
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}
Signals and Systems   Sampling
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
Signals and Systems   DTFS, DTFT and DFT
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}
Electronic Devices   Basic Semiconductor Physics
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}
Electronic Devices   Basic Semiconductor Physics
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 )
Electronic Devices   BJT and FET Basics
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
Analog Circuits   Operational Amplifiers
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}}
Analog Circuits   Operational Amplifiers
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

Prepare for GATE 2023 with practice of GATE Electronics previous year questions and solution

Prepare for GATE 2023 with practice of GATE Electronics previous year questions and solution

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 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 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-1

Question 1
For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT ALWAYS hold ?
A
(M^{T})^{T}=M
B
(cM)^{T}=c(M)^{T}
C
(M+N)^{T}=M^{T}+N^{T}
D
MN=NM
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
Matrix multiplication is not commutative.
Question 2
In a housing society, half of the families have a single child per family, while the remaining half have two children per family. The probability that a child picked at random, has a sibling is _____
A
1.5
B
0.67
C
2
D
3
Engineering Mathematics   Probability and Statistics
Question 2 Explanation: 
Required probability
=\frac{\frac{1}{2} \times \frac{2}{3}}{\frac{1}{2} \times \frac{1}{3}+\frac{1}{2} \times \frac{2}{3}}=\frac{2}{3}=0.666
Question 3
C is a closed path in the z -plane by |z| = 3. The value of the integral \oint_{c}(\frac{z^{2}-z+4j}{z+2j})dz is
A
-4\pi (1+ j2 )
B
4\pi (3- j2 )
C
-4\pi (3+ j2 )
D
4\pi (1- j2)
Engineering Mathematics   Complex Analysis
Question 3 Explanation: 


\oint_{c} \frac{z^{2}-z+4 j}{z+2 j}
Pole =z=-2 j
which is inside of |z|=3
From Cauchy integral formula
\begin{aligned} \oint \frac{z^{2}-z+4 j}{z+2 j} &=2 \pi i\left[\lim _{z \rightarrow-2 j} z^{2}-z+4 j\right] \\ &=2 \pi j[-4+2 j+4 j] \\ &=2 \pi j[-4+6 j] \\ &=-4 \pi[2 j+3] \end{aligned}
Question 4
A real (4x4) matrix A satisfies the equation A_{2} = I, where I is the (4x4) identity matrix. The positive eigen value of A is _____.
A
1
B
2
C
3
D
4
Engineering Mathematics   Linear Algebra
Question 4 Explanation: 
\begin{aligned} \text{since, }A^{2}=l, \text{eig}\left(A^{2}\right)&=\text{eig}(I)=1 \\ \Rightarrow \quad \text{eig}(A)^{2}&=1\\ \Rightarrow \quad \text{eig}(A)&=\pm 1 \end{aligned}
Therefore, the positive eigen value of A is +1.
Question 5
Let X_{1} , X_{2}, \; and \; X_{3} be independent and identically distributed random variables with the uniform distribution on [0,1]. The probability P{X_{1} is the largest} is
A
0.5
B
0.33
C
0.25
D
0.75
Communication Systems   Random Processes
Question 5 Explanation: 
If multiple independent random variables are uniformly distributed in the same interval then each random variable will have equal chances to be largest and to be lowest.
P\left(X_{1} \text { is the largest) }=\frac{1}{3}\right.
Question 6
For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance Z_{1} of the first section to the input impedance Z_{2} of the second section is
A
Z_{2}=Z_{1}
B
Z_{2}=-Z_{1}
C
Z_{2}=Z_{1}^{*}
D
Z_{2}=-Z_{1}^{*}
Network Theory   Transient Analysis
Question 7
Consider the configuration shown in the figure which is a portion of a larger electrical network

For R = 1\Omega and currents i_{1} = 2 A , i_{4} =- 1A , i_{5}=- 4 A , which one of the following is TRUE ?
A
i_{6}=5A
B
i_{3}=-4A
C
Data is sufficient to conclude that the supposed currents are impossible
D
Data is insufficient to identify the currents i_{2},i_{3} \; and \; i_{6}
Network Theory   Basics of Network Analysis
Question 7 Explanation: 
Given data:
\begin{array}{l} i_{1}=2 \mathrm{A}, i_{4}=-1 \mathrm{A}, i_{5}=-4 \mathrm{A} \\ R=1 \Omega \end{array}
To calculate:
i_{6}=?


Using KVL at all the three nodes we get,
At node A
i_{5}-i_{3}+i_{2}=0\qquad \ldots(i)
At node B
i_{4}+i_{1}-i_{2}=0\qquad \ldots(ii)
At node C
i_{6}+i_{3}-i_{1}=0 \qquad \ldots(iii)
By putting the value of i_{3} and i_{2} from equation (i)
and (ii) in equation (iii) we get,
\begin{aligned} i_{6}+\left(i_{2}+i_{5}\right)-i_{1}&=0 \\ i_{6}+\left(i_{1}+i_{4}+i_{5}\right)-i_{1}&=0 \\ \therefore \quad i_{6}+(2-1-4)-2&=0 \\ i_{6}&=5 \mathrm{A} \end{aligned}
Question 8
When the optical power incident on a photodiode is 10\muW and the responsivity is 0.8 A/W, the photocurrent generated (in \mu A) is _____.
A
2
B
4
C
8
D
10
Electronic Devices   PN-Junction Diodes and Special Diodes
Question 8 Explanation: 
\begin{array}{l} \text { Responsivity }(R)=\frac{I_{p}}{P_{o}} \\ \text { where } I_{p}=\text { Photo current } \\ \qquad P_{0}=\text { Incident power } \\ \therefore \quad I_{p}=R \times P_{0}=8 \mu \mathrm{A} \end{array}
Question 9
In the figure, assume that the forward voltage drops of the PN diode D_{1} and Schottky diode D_{2} are 0.7 V and 0.3 V, respectively. If ON denotes conducting state of the diode and OFF denotes non-conducting state of the diode, then in the circuit,
A
both D_{1} \; and \; D_{2} are ON
B
D_{1} is ON and D_{2} are OFF
C
both D_{1} \; and \; D_{2} are OFF
D
D_{1} is OFF and D_{2} are ON
Analog Circuits   Diodes Applications
Question 9 Explanation: 
Consider D_{1} \rightarrow \text{OFF } and D_{2} \rightarrow \text{ON} then


Apply KVL
\begin{aligned} 10 &=1000 I+20 I+0.3 \\ I &=\frac{9.7}{1020} \\ I &=9.5 \mathrm{mA} \end{aligned}
Now, we calculate V_{D_{1}} of
\begin{aligned} & 10=9.5+V_{D_{1}} \\ \therefore V_{D_{1}} &=0.5 \mathrm{V} \end{aligned}
since v_{D_{1}} \lt 0.7 \mathrm{V}, \quad \mathrm{D}_{1} is in OFF state i.e. our assumption is correct and hence (D) is the correct option.
Question 10
If fixed positive charges are present in the gate oxide of an n-channel enhancement type MOSFET, it will lead to
A
a decrease in the threshold voltage
B
channel length modulation
C
an increase in substrate leakage current
D
an increase in accumulation capacitance
Electronic Devices   Ic Fabrication
Question 10 Explanation: 
Fixed charges reduces threshold voltage.
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.