# Signals and Systems

 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      LTI Systems Continuous and Discrete
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 $x_{1}(t)=u(t+1.5)-u(t-1.5)$ and $x_{2}(t)$ is shown in the figure below. For $y(t)=x_{1}(t) * x_{2}(t)$, the $\int_{-\infty}^{\infty} y(t) d t$ is ___. (rounded off to the nearest integer).

 A 12 B 14 C 15 D 18
GATE EC 2023      Fourier Transforms, Frequency Response and Correlation
Question 2 Explanation:
$x_{1}(t)=u(t+1.5)-u(t-1.5)$

$]\Rightarrow x_1(t)=rect\left ( \frac{t}{3} \right )$
$] x_1(t)=rect\left ( \frac{t}{3} \right ) \overset{FT}{\leftrightarrow} 3Sa(1.5\omega )$
Now, $x_{2}(t)=\delta(t+3)+\operatorname{rect}\left(\frac{t}{2}\right)+2 \delta(t-2)$

Taking Fourier transform
\begin{aligned} X_{2}(\omega) & =e^{3 j \omega}+2 S a(\omega)+2 e^{-2 j \omega} \\ y(t) & =x_{1}(t) * x_{2}(t) \\ Y(\omega) & =X_{1}(\omega) \cdot X_{2}(\omega) \\ Y(\omega) & =\int_{-\infty}^{\infty} y(t) \cdot e^{-j \omega t} \cdot d t \end{aligned}

We know
$\therefore \quad \int_{-\infty}^{\infty} y(t)=Y(0)$
$\therefore \quad Y(0)=X_{1}(0) \cdot X_{2}(0)$ $=3[1+2+2]=15$

 Question 3
Let $x_{1}(t)$ and $x_{2}(t)$ be two band-limited signals having bandwidth $B=4 \pi \times 10^{3} \mathrm{rad} / \mathrm{s} \mathrm{each}$. In the figure below, the Nyquist sampling frequency, in rad/s, required to sample $y(t)$, is

 A $20 \pi \times 10^{3}$ B $40 \pi \times 10^{3}$ C $8 \pi \times 10^{3}$ D $32 \pi \times 10^{3}$
GATE EC 2023      Sampling
Question 3 Explanation:
Given that, $x_{1}(t)$ and $x_{2}(t)$ are two bandlimited signals having bandwidth $B=4 \pi \times 10^{3} \mathrm{rad} / \mathrm{sec}$.

and $y(t)=x_{2}(t) \cos \left(12 \pi \times 10^{3} t\right)+x_{1}(t) \cos \left(4 \pi \times 10^{3} t\right)$

So, $\text{Nyquist Rate }=2\omega _{max}=2[16 \pi \times 10^3]=32 \pi \times 10^3\; rad/sec$
 Question 4
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      LTI Systems Continuous and Discrete
Question 4 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 5
Let an input $x[n]$ having discrete time Fourier transform.
$X\left(e^{j \Omega}\right)=1-e^{-j \Omega}+2 e^{-3 j \Omega}$ be passed through an LTI system. The frequency response of the LTI system is $H\left(e^{j \Omega}\right)=1-\frac{1}{2} e^{-j 2 \Omega}$. The output $y[n]$ of the system is
 A $\delta[n]+\delta[n-1]-\frac{1}{2} \delta[n-2]-\frac{5}{2} \delta[n-3]+\delta[n-5]$ B $\delta[n]-\delta[n-1]-\frac{1}{2} \delta[n-2]-\frac{5}{2} \delta[n-3]+\delta[n-5]$ C $\delta[n]-\delta[n-1]-\frac{1}{2} \delta[n-2]+\frac{5}{2} \delta[n-3]-\delta[n-5]$ D $\delta[n]+\delta[n-1]+\frac{1}{2} \delta[n-2]+\frac{5}{2} \delta[n-3]+\delta[n-5]$
GATE EC 2023      DTFS, DTFT and DFT
Question 5 Explanation:
\begin{aligned} y[n] & =x[n] * h[n] \\ Y\left(e^{j \Omega}\right) & =X\left(e^{j \Omega}\right) H\left(e^{j \Omega}\right) \\ & =\left[1-e^{-j \Omega}+2 e^{-3 j \Omega}\right]\left[1-\frac{1}{2} e^{-2 j \Omega}\right] \\ & =1-e^{-j \Omega}+2.5 e^{-3 j \Omega}-0.5 e^{-j 2 \Omega}-e^{-j 5 \Omega} \end{aligned}

Taking IDTFT;
$y[n]=\delta[n]-\delta[n-1]-0.5 \delta[n-2]+2.5 \delta[n-3]-\delta[n-5]$

There are 5 questions to complete.