# Sampling Theorem

$x(t) = \frac{\sin(2000\pi t)}{\pi t}\cos(1000\pi t)$ is sampled with sampling period $T_{s} = \frac{1}{8000}$, to obtain the sampled signal $x_{p}(t)=x(t)p(t)$, where $p(t)= \sum_{n = -\infty}^{\infty} \delta(t - nT_{s})$ . The sampling signal can also be represented in the discrete time domain as $x_{d}[n]=x(nT_{s})$.

I find $$X_{p}(\omega) = \frac{1}{T} \sum_{n = -\infty}^{\infty} X(\omega-16000\pi t) \quad\textrm{where} \quad T = \frac {1}{8000}$$ and

$$X(\omega) = \frac{1}{2} \left[X_{1}(\omega-1000\pi)-X_{1}(\omega+1000\pi)\right].$$

Also $$x_{1}(t) = \frac{\sin(2000\pi)}{\pi t}\quad \textrm{and} \quad X_{1}(\omega) = \begin{cases} 1, & |\omega| < 2000\pi,\\ 0, & |\omega|>2000\pi. \end{cases}$$

Now, I need to find the DTFT of $x_{d}[n]$ and got stuck there.

Need some help, thanks!

• Your equations do not look entirely correct. What's preventing you from applying the continuous to discrete time equation you've already given? Jan 7, 2014 at 19:20
• I don't understand what you mean by "applying continuous to discrete time equation"? Jan 7, 2014 at 19:28
• The discrete time domain is itself a sampled version of the continuous time, sampled by the function x_d[n] = x(nT_s), the equation you gave in your question. Jan 7, 2014 at 19:33
• I was expecting it to be different. I got it know, thanks! Jan 7, 2014 at 21:09

Apply the continuous to discrete time equation you've already given. The discrete time domain is itself a sampled version of the continuous time, sampled by the function $x_d[n] = x(nT_s)$
• I don't understand what you are trying to do, but thanks for the afford. Thanks to Samuel I got the answer; substituting $x_{d}$ in to DTFT formula, will give the answer. Jan 7, 2014 at 21:13