# 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 $\frac{-j4\omega }{(1+t ^2)^2} \leftrightarrow 2 \pi(-\omega )e^{-|-\omega |}$
$\Rightarrow \frac{t}{(1+t^2)^2} \underrightarrow{FT} \frac{-2\pi}{-j4}\omega e^{-|\omega |}$
$\Rightarrow \;\;\;\rightarrow\frac{\pi}{j2} \omega e^{-|\omega |}$

There are 5 questions to complete.