# 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.