Question 1 |
In the table shown below, match the signal type with its spectral characteristics.
\begin{array}{ll} \textbf{Signal type} &\textbf{Spectral Characteristics}\\ \text{(i) Continuous, aperiodic}& \text{(i) Continuous, aperiodic}\\ \text{(ii) Continuous, periodic}& \text{(ii) Continuous, periodic}\\ \text{(iii) Discrete, aperiodic}& \text{(iii) Discrete, aperiodic}}\\ \text{(iv) Discrete, periodic} & \text{(iv) Discrete, periodic}} \end{array}
\begin{array}{ll} \textbf{Signal type} &\textbf{Spectral Characteristics}\\ \text{(i) Continuous, aperiodic}& \text{(i) Continuous, aperiodic}\\ \text{(ii) Continuous, periodic}& \text{(ii) Continuous, periodic}\\ \text{(iii) Discrete, aperiodic}& \text{(iii) Discrete, aperiodic}}\\ \text{(iv) Discrete, periodic} & \text{(iv) Discrete, periodic}} \end{array}
(i) \rightarrow (a), (ii) \rightarrow (b), (iii) \rightarrow (c), (iv) \rightarrow (d) | |
(i) \rightarrow (a), (ii) \rightarrow (c), (iii) \rightarrow , (iv) \rightarrow (d) | |
(i) \rightarrow (d), (ii) \rightarrow (b), (iii) \rightarrow (c), (iv) \rightarrow (a) | |
(i) \rightarrow (a), (ii) \rightarrow (c), (iii) \rightarrow (d), (iv) \rightarrow (b) |
Question 2 |
Let m(t) be a strictly band-limited signal with bandwidth B and energy E. Assuming \omega_{0}=10 B, the energy in the signal m(t) \cos \omega_{0} t is
\frac{E}{4} | |
\frac{E}{2} | |
E | |
2 E |
Question 2 Explanation:
\begin{aligned} \text { Energy }(E)&=\frac{1}{2 \pi} \int_{-B}^{B}(1)^{2} \cdot d \omega \\ E&=\frac{B}{\pi}\\ \text {Now, let } y(t)&=m(t) \cos \omega_{0} t \\ Y(\omega)&=\frac{1}{2}\left[M\left(\omega-\omega_{0}\right)+M\left(\omega+\omega_{0}\right)\right] \\ \text{Here; }\omega_{0}&=10 B \end{aligned}
Now,
\begin{aligned} Energy \left(E^{\prime}\right)&=\frac{1}{2 \pi} \int_{-\infty}^{\infty}|Y(\omega)|^{2} \cdot d \omega\\ &=\frac{1}{2 \pi}\left[\int_{-11 B}^{-9 B}\left(\frac{1}{2}\right)^{2} \cdot d \omega+\int_{9 B}^{11 B}\left(\frac{1}{2}\right)^{2} \cdot d \omega\right]\\ & =\frac{1}{2 \pi}\left[\frac{1}{4} \times 2 B+\frac{1}{4} \times 2 B\right] \\ E^{\prime} & =\frac{B}{2 \pi}=\frac{1}{2}\left(\frac{B}{\pi}\right)=\frac{E}{2} \end{aligned}
Question 3 |
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 3 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 4 |
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 4 Explanation:
y=a u+b, b \neq 0 is a non-linear system.
Question 5 |
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 5 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
There are 5 questions to complete.