The signal $x(t)=e^{-t^2}\text{sinc}(t)$ was sampled at interval $T$. It was then found that the discrete time Fourier transform of the sampled signal is: $X(e^{j\omega})=1$. What is the minimum $T$ for which such a result is possible? If this is impossible for any $T$, explain why.

I started it but didn't how to continue , any help? My steps so far:

\begin{align} x[n] &= e^{-n^2}\text{sinc}[n] \\ &= e^{-n^2}\frac{\text{sin}(\pi n)}{\pi n} \\ &= \frac{e^{-n^2}}{2j \pi n}\big(e^{j\pi n} - e^{-j\pi n}\big) \\ &= \frac{1}{2j \pi n}\big( e^{j\pi n - n^2} - e^{-j\pi n - n^2} \big) \\ \end{align}

\begin{align} X(e^{j\omega}) &= 1 \\ &= \sum_{n=0}^{\infty} x[n]e^{-j\omega n} \\ &= \sum_{n=0}^{\infty} \frac{1}{j2\pi n} \big( e^{j\pi n - n^2} - e^{-j\pi n - n^2} \big) \end{align}


The best answer I could give is just a hint: think about what is the inverse discrete time Fourier transform of $X(e^{j\omega})$. Constants in the frequency domain are what in the time domain? Answering that will lead you to the answer of this question.

  • $\begingroup$ I got x[n] = sinc(pi*n) , then exp(-(nT)^2) = 1 and that is not possible for any value of T unless n = 0. Is that correct ? $\endgroup$
    – John_HB
    Oct 25 '19 at 20:33
  • 1
    $\begingroup$ No. A constant in frequency is not a sinc in time domain. $\endgroup$
    – Engineer
    Oct 26 '19 at 1:33
  • $\begingroup$ It is a dirac, but how can i compare it to the first signal ? $\endgroup$
    – John_HB
    Oct 26 '19 at 20:20
  • $\begingroup$ I think it is true for T = pi and n=0 !! $\endgroup$
    – John_HB
    Oct 26 '19 at 20:27
  • $\begingroup$ Yes, it is an impulse. So, how can you sample $x(t)$ so that $x(nT)=1$ for $n=0$ and $x(nT)=0$ for all other values of n? HINT: try plotting it $\endgroup$
    – Engineer
    Oct 27 '19 at 21:25

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