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.

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