Question

The signal $\cos (10\pi t+\frac{\pi }{4})$ is ideally sampled at a sampling frequency of 15 Hz. The sampled signal is passed through a filter with impulse response $\left(\frac{\sin \left(\pi t\right)}{\pi t}\right)$ $\cos (40\pi t-\frac{\pi }{2})$. The filter output is

• A. $\frac{15}{2}\cos (10\pi t-\frac{\pi }{4})$
• B. $\frac{15}{2}\left(\frac{sin\left(\pi t\right)}{\pi t}\right)\cos (10\pi t+\frac{\pi }{4})$
• C. $\frac{15}{2}\cos (40\pi t-\frac{\pi }{4})$
• D. $\frac{15}{2}\left(\frac{\sin \left(\pi t\right)}{\pi t}\right)\cos (10\pi t-\frac{\pi }{2})$

Answer 1

The correct option is C $\frac{15}{2}\cos (40\pi t-\frac{\pi }{4})$

$x\left(t\right)=\cos (10\pi t+\frac{\pi }{4})$

$X\left(f\right)=\frac{1}{2}\left[\delta \right(f-5)+\delta (f+5\left)\right]e^{j\pi /4}$

$\text{After sampling x(t) by 15 Hz, we get}$
$X_s\left(f\right)=15\sum _{n=-\infty }^{\infty }X(f-15n)$
Given, $h\left(t\right)=\left[\frac{sin\left(\pi t\right)}{\pi t}\right]cos(40\pi t-\frac{\pi }{2})$

$H\left(f\right)=\frac{1}{2}[rect\frac{(f-20)}{2\pi }+rect\frac{(f+20)}{2\pi }]e^{-j\frac{\pi }{2}}$

$\text{Output}$ $Y\left(f\right)=X_s\left(f\right)\times H\left(f\right)$

$y\left(t\right)=\frac{15}{2}cos(40\pi t-\frac{\pi }{4})$