# Sampling period

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.
• 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 Oct 27 '19 at 21:25