# 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 $2\mu m$and $1\mu m$, 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.