LTI Systems Continuous and Discrete


Question 1
Let X(t) be a white Gaussian noise with power spectral density \frac{1}{2} \mathrm{~W} / \mathrm{Hz}. If X(t) is input to an LTI system with impulse response e^{-t} u(t). The average power of the system output is ___ \mathrm{W}. (Rounded off to two decimal place).
A
0.25
B
0.55
C
0.75
D
1.15
GATE EC 2023   Signals and Systems
Question 1 Explanation: 


h(t)=e^{-t} u(t)
Given: Input PSD
\Rightarrow \quad S_{x}(f)=\frac{1}{2} \mathrm{~W} / \mathrm{Hz}

We know output PSD,
\begin{aligned} S_{Y}(f) & =S_{X}(f)|H(f)|^{2} \\ S_{Y}(f) & =\frac{1}{2}|H(f)|^{2} \\ \text { Power }[y(t)] & =\int_{-\infty}^{\infty} S_{Y}(f) d f=\int_{-\infty}^{\infty} \frac{1}{2}|H(f)|^{2} d f \\ & =\frac{1}{2} \int_{-\infty}^{\infty} h^{2}(t) d t=\frac{1}{2} \int_{0}^{\infty} e^{-2 t} d t \\ & =\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}=0.25 \mathrm{~W} \end{aligned}
Question 2
Let x(t)=10 \cos (10.5 \mathrm{Wt}) be passed through an LTI system having impulse response h(t)=\pi\left(\frac{\sin W t}{\pi t}\right)^{2} \cos 10 W t. The output of the system is
A
\left(\frac{15 W}{4}\right) \cos (10.5 W t)
B
\left(\frac{15 W}{2}\right) \cos (10.5 W t)
C
\left(\frac{15 W}{8}\right) \cos (10.5 W t)
D
(15 W) \cos (10.5 W t)
GATE EC 2023   Signals and Systems
Question 2 Explanation: 
Given h(t) is Real and Even. When sinusoidal input applied to LTI system having even impulse response, then output will also be sinusoidal.

here, y(t)=\left.H(W)\right|_{W=10.5 W} \cdot 10 \cos (10.5 W t)
let, h(t)=f(t) \cos 10 W t
where, f(t)=\pi\left(\frac{\sin W t}{\pi t}\right)^{2}

Now; H(W)=\frac{1}{2}[F(W+10 W)+F(W-10 W)]

\left.\therefore \quad H(W)\right|_{W=10.5 W}=\frac{3}{8} W
Hence, \begin{aligned} y(t) & =\left(\frac{3}{8} W\right)(10 \cos 10.5 W t) \\ & =\frac{15}{4} W \cos 10.5 W t \end{aligned}


Question 3
Consider a system with input x(t) and output y(t)=x\left(e^{t}\right). The system is
A
Causal and time invariant
B
Non-causal and time varying
C
Causal and time varying
D
Non-causal and time invariant
GATE EC 2023   Signals and Systems
Question 3 Explanation: 
We have, \quad y(t)=x\left(e^{t}\right)
At t=0
y(0)=x(1)
i.e. present value of output depends on future value of input, hence it is non-causal.

For Time Variant: Delay the input,
y(t)=x\left(e^{t}-t_{0}\right) \quad . . . (i)

Delay the output,
y\left(t-t_{0}\right)=x\left(e^{t-t_{0}}\right) \quad . . . (ii)

i.e. equations (i) \neq (ii)
Hence, it is time variant system.
Question 4
The outputs of four systems (S_1,S_2,S_3,S_4) corresponding to the input signal \sin (t), for all time t, are shown in the figure.
Based on the given information, which of the four systems is/are definitely NOT LTI (linear and time-invariant)?

A
S_1
B
S_2
C
S_3
D
S_4
GATE EC 2022   Signals and Systems
Question 4 Explanation: 


Since, LTI systems does not change the frequency of sinusoidal input. So S3 and S4 are definitely not LTI as input and output sinusoidal frequencies are different.
Question 5
Let x_1(t)=e^{-t}u(t) and x_2(t)=u(t)-u(t-2) , where u(\cdot) denotes the unit step function. If y(t) denotes the convolution of x_1(t) and x_2(t), then \lim_{t\rightarrow \infty }y(t)= _________ (rounded off to one decimal place).
A
0
B
0.4
C
0.6
D
0.8
GATE EC 2022   Signals and Systems
Question 5 Explanation: 
\begin{aligned} x_1(t)=e^{-1}U(t)&\underleftrightarrow{L.T.}\frac{1}{(s+1)}\\ x_2(t)=U(t)-U(t-2)&\underleftrightarrow{L.T.}\frac{1}{s}-\frac{e^{-2s}}{s}=\frac{1-e^{-2s}}{s}\\ y(t)=x_1(t)\otimes x_2(t)&\underleftrightarrow{L.T.} \frac{(1-e^{-2s})}{s(s+1)} \\ y(t)&\underleftrightarrow{L.T.}\frac{1-e^{-2s}}{s(s+1)}\\ \lim_{t \to \infty }y(t)&=\lim_{s \to 0 }\frac{s(1-e^{-2s})}{s(s+1)}=0 \end{aligned}


There are 5 questions to complete.