# 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,
$y_{t}(t)=u(t) \Rightarrow \text { option }(D)$

There are 5 questions to complete.