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\$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!

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  • \$\begingroup\$ Your equations do not look entirely correct. What's preventing you from applying the continuous to discrete time equation you've already given? \$\endgroup\$ – Samuel Jan 7 '14 at 19:20
  • \$\begingroup\$ I don't understand what you mean by "applying continuous to discrete time equation"? \$\endgroup\$ – cecemelly Jan 7 '14 at 19:28
  • \$\begingroup\$ 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. \$\endgroup\$ – Samuel Jan 7 '14 at 19:33
  • \$\begingroup\$ I was expecting it to be different. I got it know, thanks! \$\endgroup\$ – cecemelly Jan 7 '14 at 21:09
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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)\$

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enter image description here

Please follow these steps. It might take a while, at least you have a way to do it if you are stuck in the time domain.

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  • \$\begingroup\$ 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. \$\endgroup\$ – cecemelly Jan 7 '14 at 21:13
  • \$\begingroup\$ No worries at all! My approach is a bit harder, but it works every time. Good luck! \$\endgroup\$ – Adel Bibi Jan 7 '14 at 21:18

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