Question 1 |
Consider the signal f(t)=1+2cos(\pi t)+3sin\left ( \frac{2 \pi}{3}t \right )+4cos\left (\frac{\pi}{2}t+\frac{\pi}{4} \right ), where t is in seconds. Its fundamental time period, in seconds, is ____________
8 | |
12 | |
16 | |
20 |
Question 1 Explanation:
\begin{aligned} f(t)=1+2 \cos (\pi t)&+3 \sin \left(\frac{2 \pi}{3} t\right)+4 \cos \left(\frac{\pi}{2} t+\frac{\pi}{4}\right) \\ \omega_{1}&=\pi \\ \omega_{2}&=\frac{2 \pi}{3} \\ \omega_{3}&=\frac{\pi}{2} \\ \omega_{0}&=G C D\left(\pi, \frac{2 \pi}{3}, \frac{\pi}{2}\right)=\frac{\pi}{6} \end{aligned}
Fundamental period,
N=\frac{2 \pi}{\omega_{0}}=\frac{2 \pi}{(\pi / 6)}=12
Fundamental period,
N=\frac{2 \pi}{\omega_{0}}=\frac{2 \pi}{(\pi / 6)}=12
Question 2 |
Let the input be u and the output be y of a system, and the other parameters are real
constants. Identify which among the following systems is not a linear system:
\frac{d^{3}y}{dt^{3}} + a_{1} \frac{d^{2}y}{dt^{2}} + a_{2}\frac{dy}{dt} + a_{3}y = b_{3}u+b_{2}\frac{du}{dt}+b_{1}\frac{d^{2}u}{dt^{2}} (with initial rest conditions) | |
y(t)=\int_{0}^{t}e^{a(t-r)}\beta u(\tau)d \tau | |
y= au +b, b \neq 0 | |
y=au |
Question 2 Explanation:
y=a u+b, b \neq 0 is a non-linear system.
Question 3 |
The input x(t) and the output y (t) of a continuous-time system are related as y(t)=\int_{t-T}^{t}x(u)du. The system is
Linear and time-variant | |
Linear and time-invariant | |
Non-linear and time-variant | |
Non-linear and time-invariant |
Question 3 Explanation:
Given that, y(t)=\int_{t-T}^{t} x(u) d u
since the given system satisfies both homogeneity and additivity properties, the system is linear.
Check for time invariance:
y\left(t-t_{0}\right)=\int_{t-t_{0}-T}^{t-t_{0}} x(u) d u
When the applied input is x\left(t-t_{0}\right)
\begin{aligned} y_{1}(t) &=\\ \int_{t-T}^{t} x\left(u-t_{0}\right) d u &=\int_{t-t_{0}-T}^{t-t_{0}} x(\tau) d \tau \\ &=y\left(t-t_{0}\right) \end{aligned}
\Rightarrow System is time invariant
since the given system satisfies both homogeneity and additivity properties, the system is linear.
Check for time invariance:
y\left(t-t_{0}\right)=\int_{t-t_{0}-T}^{t-t_{0}} x(u) d u
When the applied input is x\left(t-t_{0}\right)
\begin{aligned} y_{1}(t) &=\\ \int_{t-T}^{t} x\left(u-t_{0}\right) d u &=\int_{t-t_{0}-T}^{t-t_{0}} x(\tau) d \tau \\ &=y\left(t-t_{0}\right) \end{aligned}
\Rightarrow System is time invariant
Question 4 |
Consider a single input single output discrete-time system with x[n] as input and y[n] as output, where the two are related as
y[n]=\left\{\begin{matrix} n|x[n]| & for 0\leq n\leq 10\\ x[n]-x[n-1] & othrwise \end{matrix}\right.
Which one of the following statements is true about the system?
y[n]=\left\{\begin{matrix} n|x[n]| & for 0\leq n\leq 10\\ x[n]-x[n-1] & othrwise \end{matrix}\right.
Which one of the following statements is true about the system?
It is causal and stable | |
It is causal but not stable | |
It is not causal but stable | |
It is neither causal nor stable |
Question 4 Explanation:
Since present output does not depend upon future values of input, the system is causal and also every bounded input produces bounded output, so it is stable.
Question 5 |
A continuous-time function x(t) is periodic with period T. The function is sampled uniformly with a sampling period T_{s}. In which one of the following cases is the sampled signal periodic?
T=\sqrt{2}T_{s} | |
T=1.2T_{s} | |
Always | |
Never |
Question 5 Explanation:
A signal is said to be periodic if \frac{T}{T_{s}} is a rational
number.
Here, T=1.2 T_{s}
\Rightarrow \frac{T}{T_{s}}=\frac{6}{5} \quad Which is a rational number
Here, T=1.2 T_{s}
\Rightarrow \frac{T}{T_{s}}=\frac{6}{5} \quad Which is a rational number
Question 6 |
Two sequences x_{1}[n] \; and \; x_{2}[n] have the same energy. Suppose x_{1}[n] = \alpha 0.5^{n} u[n] , where \alpha is a positive real number and u[n] is the unit step sequence. Assume
x_{2}[n]=\left\{\begin{matrix} \sqrt{1.5} &for n=0,1 \\ 0 & otherwise. \end{matrix}\right.
Then the value of \alpha is _________.
x_{2}[n]=\left\{\begin{matrix} \sqrt{1.5} &for n=0,1 \\ 0 & otherwise. \end{matrix}\right.
Then the value of \alpha is _________.
1 | |
1.5 | |
2 | |
2.5 |
Question 6 Explanation:
\begin{array}{l} \text { Energy of } x_{1}[n]=\sum_{n=-\infty}^{\infty}\left|x_{1}[n]\right|^{2} \\ \begin{aligned} =\sum_{n=0}^{\infty} \alpha^{2} &\left(\frac{1}{4}\right)^{n}=\alpha^{2} \cdot \frac{1}{1-\frac{1}{4}}=\alpha^{2} \cdot \frac{4}{3} \\ \text { Energy of } x_{2}[n] &=1.5+1.5=3 \\ \Rightarrow \quad \alpha^{2} \frac{4}{3} &=3 \\ \alpha^{2}=\frac{9}{4} \\ \quad \alpha=1.5 \end{aligned} \end{array}
Question 7 |
The waveform of a periodic signal x(t) is shown in the figure.

A signal g(t) is defined by g(t)=x(\frac{t-1}{2}). The average power of g(t) is _______.

A signal g(t) is defined by g(t)=x(\frac{t-1}{2}). The average power of g(t) is _______.
1 | |
2 | |
3 | |
4 |
Question 7 Explanation:
\begin{array}{l} x(t)=-3 t, \quad-1\lt t \lt 1 \\ x\left(\frac{t-1}{2}\right)=-\frac{3}{2}(t-1) \qquad-1 \lt t \lt 3 \\ \text { and } \quad T=6 \\ \text { Average power }=\frac{1}{6} \int_{-1}^{3}\left(-\frac{3}{2}(t-1)\right)^{2} d t=2 \end{array}
Question 8 |
A discrete time signal x[n]=sin(\pi ^{2 }n), n being an integer, is
periodic with period \pi | |
periodic with period \pi ^{2} | |
periodic with period \pi ^/2 | |
not periodic |
Question 8 Explanation:
\begin{aligned} x[n] &=\sin \left(\pi^{2} n\right) \\ \omega_{0} &=\pi^{2} \\ \therefore \quad N &=\frac{2 \pi}{\omega_{0}} \cdot m \end{aligned}
where m is the smallest integer that converts \frac{2 \pi}{\omega_{0}} into a integer value.
\therefore \quad N=\frac{2 \pi}{\pi^{2}} \cdot m=\frac{2}{\pi} \cdot m
So, there exists no such integer value of m which could make the N integer, so the system is not periodic.
where m is the smallest integer that converts \frac{2 \pi}{\omega_{0}} into a integer value.
\therefore \quad N=\frac{2 \pi}{\pi^{2}} \cdot m=\frac{2}{\pi} \cdot m
So, there exists no such integer value of m which could make the N integer, so the system is not periodic.
Question 9 |
The impulse response of a continuous time system is given by h(t)= \delta (t-1)+\delta (t-3). The value of the step response at t = 2 is
0
| |
1 | |
2 | |
3 |
Question 9 Explanation:
Step response = Integration of impulse response
\int_{-\infty}^{t} \delta(t-1) d t=u(t-1)
\int_{-\infty}^{t} \delta(t-3) d t=u(t-3)

\begin{aligned} \text{at}\quad t&=2\\ y(t)&=1 \end{aligned}
\int_{-\infty}^{t} \delta(t-1) d t=u(t-1)
\int_{-\infty}^{t} \delta(t-3) d t=u(t-3)

\begin{aligned} \text{at}\quad t&=2\\ y(t)&=1 \end{aligned}
Question 10 |
For a periodic signal v\left ( t \right )=30\sin 100t+10\cos 300t+6\sin \left ( 500t+\frac{\pi }{4} \right ), the fundamental frequency in rad/s is
100 | |
300 | |
500 | |
1500 |
Question 10 Explanation:
\begin{array}{l} \omega_{1}=100 \\ \omega_{2}=300 \\ \omega_{3}=500 \end{array}
H.C.F. of \left(\omega_{1}, \omega_{2} \text { and } \omega_{3}\right)= H.C.F. (100,300,500)
\omega=100 \mathrm{rad} / \mathrm{sec}
H.C.F. of \left(\omega_{1}, \omega_{2} \text { and } \omega_{3}\right)= H.C.F. (100,300,500)
\omega=100 \mathrm{rad} / \mathrm{sec}
There are 10 questions to complete.