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)
The maximum frequency present in the filtered signal in \mathrm{Hz} is ____ (Round off to the nearest integer)
95 | |
80 | |
110 | |
115 |
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}.
\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).
\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).
0.038 | |
0.025 | |
0.068 | |
0.012 |
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}
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)
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)
16 | |
32 | |
48 | |
64 |
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}
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).
2 | |
3 | |
4 | |
5 |
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}
\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?
If an LTI system is causal, it is stable | |
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 | |
If a discrete time LTI system has an impulse response h[n] of finite duration the system is stable | |
If the impulse response 0 \lt |h[n]| \lt 1 for all n, then the L T I system is stable. |
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
h(n)=0 ; \quad n \lt 0
Therefore, for step input also
h(n)=0 ; n \lt 0
There are 5 questions to complete.