# Signals and Systems

 Question 1
A signal $x(t)=2 \cos (180 \pi t) \cos (60 \pi t)$ is sampled at $200 \mathrm{~Hz}$ and then passed through an ideal low pass filter having cut-off frequency of $100 \mathrm{~Hz}$.
The maximum frequency present in the filtered signal in $\mathrm{Hz}$ is ____ (Round off to the nearest integer)
 A 95 B 80 C 110 D 115
GATE EE 2023      Sampling
Question 1 Explanation:
Given :
\begin{aligned} x(\mathrm{t})&=2 \cos (180 \pi \mathrm{t}) \cos (60 \pi \mathrm{t}) \\ &=\cos 240 \pi \mathrm{t}+\cos 120 \pi \mathrm{t} \end{aligned}
$\mathrm{F}_{1}=120 \mathrm{~Hz}$ and $\mathrm{F}_{2}=60 \mathrm{~Hz}$

Now, present frequency component,
\begin{aligned} & =\mathrm{nf}_{\mathrm{s}} \pm \mathrm{f}_{1} \\ & =\mathrm{f}_{1}, \mathrm{f}_{\mathrm{s}} \pm \mathrm{f}_{1}, 2 \mathrm{f}_{\mathrm{s}} \pm \mathrm{f}_{1} \\ & =120 \mathrm{~Hz}, 200 \pm 120, \ldots . \\ & =120 \mathrm{~Hz}, 80 \mathrm{~Hz}, 320 \mathrm{~Hz}, \ldots . \end{aligned}

and other frequency component.
\begin{aligned} & =n f_{s} \pm f_{2} \\ & =f_{2}, f_{s} \pm f_{2}, 2 f_{s} \pm f_{2}, \ldots \\ & =60,200 \pm 60, \ldots \\ & =60 \mathrm{~Hz}, 140,260 \mathrm{~Hz}, \ldots \end{aligned}

Given cut-off frequency of LPF $=100 \mathrm{~Hz}$
$\therefore$ It is pas only $60 \mathrm{~Hz}$ and $80 \mathrm{~Hz}$ frequency component.
$\therefore$ Max. frequency $=80 \mathrm{~Hz}$.
 Question 2
The discrete-time Fourier transform of a signal $x[n]$ is $X(\Omega)=1(1+\cos \Omega) e^{-j \Omega}$. Consider that $x_{p}[n]$ is a periodic signal of period $N=5$ such that

\begin{aligned} x_{p}[n] & =x[n], \text { for } n=0,1,2 \\ & =0, \text { for } n=3,4 \end{aligned}

Note that $x_{p}[n]=\sum_{k=0}^{N-1} a_{k} e^{j \frac{2 \pi}{N} k m}$. The magnitude of the Fourier series coefficient $a_{3}$ is ____ (Round off to 3 decimal places).
 A 0.038 B 0.025 C 0.068 D 0.012
GATE EE 2023      Fourier Transform
Question 2 Explanation:
Given : $x_{p}(n)$ is a period signal of period $N=5$.
$x_{p}(n)=\left\{\begin{array}{cc} x(n), & \text { for } n=0,1,2 \\ 0, & \text { for } n=3,4 \end{array}\right.$

and $\mathrm{x}(\mathrm{n}) \stackrel{\text { DTFT }}{\longrightarrow} \mathrm{X}(\Omega)=(1+\cos \Omega) \mathrm{e}^{-\mathrm{j} \Omega}$

We have,
\begin{aligned} a_{K} & =\frac{X\left(\frac{2 \pi}{N} K\right)}{N} \\ & =\frac{\left[1+\cos \left(\frac{2 \pi}{N} K\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{N} R}}{N} \end{aligned}

For, $\mathrm{N}=5$ and $\mathrm{K}=3$
\begin{aligned} a_{3} & =\frac{\left[1+\cos \left(\frac{2 \pi \times 3}{5}\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{5} \times 3}}{5} \\ \left|a_{3}\right| & =\frac{1}{5}\left[1+\cos \frac{6 \pi}{5}\right] \\ & =0.0382 \end{aligned}

 Question 3
The period of the discrete-time signal $x[n]$ described by the equation below is $N=$ ___ (Round off to the nearest integer).

$x[n]=1+3 \sin \left(\frac{15 \pi}{8} n+\frac{3 \pi}{4}\right)-5 \sin \left(\frac{\pi}{3} n-\frac{\pi}{4}\right)$
 A 16 B 32 C 48 D 64
GATE EE 2023      Introduction of C.T. and D.T. Signals
Question 3 Explanation:
Given :
$x[n]=1+3 \sin \left(\frac{15 \pi}{8} n+\frac{3 \pi}{4}\right)-5 \sin \left(\frac{\pi}{3} n-\frac{\pi}{4}\right)$
\begin{aligned} \omega_{1}&=\frac{15 \pi}{8} \\ \therefore \quad \mathrm{N}_{1}&=\frac{2 \pi}{\omega_{1}} \mathrm{~K}-=\frac{2 \pi}{15 \pi} \times 8 \times \mathrm{K}=16 \\ \text { and } \quad \mathrm{N}_{2}&=\frac{2 \pi}{\omega_{2}} \mathrm{~K}=\frac{2 \pi}{\pi} \times 3 \times \mathrm{K}=6 \\ \therefore \quad \text { Period }&=\operatorname{LCM}\left(\mathrm{N}_{1}, \mathrm{~N}_{2}\right) \\ &=\operatorname{LCM}(16,6) \\ &=48 \end{aligned}
 Question 4
For the signals $x(t)$ and $y(t)$ shown in the figure, $z(t)=x(t) * y(t)$ is maximum at $t=T_{1}$. Then $T_{1}$ in seconds is (Round off to the nearest integer).

 A 2 B 3 C 4 D 5
GATE EE 2023      Introduction of C.T. and D.T. Signals
Question 4 Explanation:
Using convotution property
\begin{aligned} z(t) & =x(t) * y(t) \\ & =\int_{-\infty}^{\infty} \mathrm{y}(\tau) \mathrm{x}(\mathrm{t}-\tau) \mathrm{d} \tau \end{aligned}

For max. overlap

\begin{aligned} \therefore \quad -1+t&=3\\ \Rightarrow \quad t&=4 sec \\ or, \quad \quad T_1&=4 sec \end{aligned}
 Question 5
Which of the following statement(s) is/are true?
 A If an LTI system is causal, it is stable B A discrete time LTI system is causal if and only if its response to a step input $u[n]$ is 0 for $\mathrm{n} \lt 0$ C If a discrete time LTI system has an impulse response $h[n]$ of finite duration the system is stable D If the impulse response $0 \lt |h[n]| \lt 1$ for all $n$, then the $L T I$ system is stable.
GATE EE 2023      Linear Time Invariant Systems
Question 5 Explanation:
For causal system, impuse response
$h(n)=0 ; \quad n \lt 0$

Therefore, for step input also
$h(n)=0 ; n \lt 0$

There are 5 questions to complete.