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).
0.25 | |
0.55 | |
0.75 | |
1.15 |
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).
12 | |
14 | |
15 | |
18 |
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
]\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
20 \pi \times 10^{3} | |
40 \pi \times 10^{3} | |
8 \pi \times 10^{3} | |
32 \pi \times 10^{3} |
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
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
\left(\frac{15 W}{4}\right) \cos (10.5 W t) | |
\left(\frac{15 W}{2}\right) \cos (10.5 W t) | |
\left(\frac{15 W}{8}\right) \cos (10.5 W t) | |
(15 W) \cos (10.5 W t) |
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}
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
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
\delta[n]+\delta[n-1]-\frac{1}{2} \delta[n-2]-\frac{5}{2} \delta[n-3]+\delta[n-5] | |
\delta[n]-\delta[n-1]-\frac{1}{2} \delta[n-2]-\frac{5}{2} \delta[n-3]+\delta[n-5] | |
\delta[n]-\delta[n-1]-\frac{1}{2} \delta[n-2]+\frac{5}{2} \delta[n-3]-\delta[n-5] | |
\delta[n]+\delta[n-1]+\frac{1}{2} \delta[n-2]+\frac{5}{2} \delta[n-3]+\delta[n-5] |
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]
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.