Question

A continuous time signal x(t) is defined as x(t) = 2 + cos$\left(50\pi t\right)$ is sampled with sampling interval $t_s=0.025$ sec and passed through an ideal low pass filter whose frequency response is as shown in figure below



The spectrum of output will be

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given, $x\left(t\right)=2+cos\left(50\pi t\right)$

Frequency of signal ${\omega }_m=50\pi rad/sec$

$T_s=0.025sec$

$\therefore$ sampling frequency ${\omega }_s=\frac{2\pi }{T_s}=80\pi$ rad/sec

then, $X\left(j\omega \right)=4\pi \delta \left(\omega \right)+\pi [\omega +50\pi +\delta (\omega -50\pi \left)\right]$

Let the sampled signal be represented as $X_s\left(j\omega \right),when{X}_p\left(j\omega \right)$ is given as

$X_s\left(j\omega \right)=\frac{1}{T_s}{\sum }_{n=-\infty }^{\infty }X\left(j\right(\omega -n{\omega }_s\left)\right)$


$X_s\left(j\omega \right)=40{\sum }_{n=-\infty }^{\infty }\left[4\pi \delta \right(\omega -80\pi \left)\right]+\pi \delta (\omega -50\pi -80\pi n)-\pi \delta (\omega +50\pi -80\pi n)$

now, the sampled input $X\left(j\omega \right)$ is passed through a low passed filter having cut-off frequency at $\omega =40\pi$.

Therefore the output $Y\left(j\omega \right)$ will contain only the components which are less than $\omega =40\pi$

Now, by putting $T_s=0.025$ , we will get