# Question on N point DTFT - Fourier transform

I have been trying to use the logic that both X and Y should have same Z transform, but according to the definition, Y is not anti causal.

Hint: Question says $$y[n]$$ has length $$10$$, but the alternate DFT coefficients of $$y[n]$$ i.e. $$Y(e^{j\omega})|_{\omega = 2\pi \frac{k}{5}}$$, matches with $$X(e^{j\omega})$$ evaluated at those $$\omega$$ exactly.

This should draw your attention towards upsampling of 5-point DFT $$X(e^{j\omega})|_{\omega = 2\pi \frac{k}{5}}$$ or equivalently periodization of a length $$5$$ segment of $$x[n]$$.

Like upsampling of time-domain sequence by $$N$$, by inserting $$(N-1)$$ zeros between samples, shrinks the spectrum in frequency domain and brings $$N-1$$ more copies of spectrum inside $$[-\pi, \pi]$$, similarly, upsampling in frequency domain by inserting $$(N-1)$$ zeros between DFT samples will create more copies of time-domain sequence.

• Thanks! I was only thinking in terms of 5 point DFT. Upsampling the same makes much sense! – Ruhi Jul 17 '20 at 11:26

Hint: Solve for the Z-transform of $$x(n)$$ which is $$X(z)$$. Then from this solve from $$X(e^{j\omega})$$ with $$\omega = 2\pi k/5$$ (the missing j is certainly a typo) and the k's given.

If you are unable to solve the Z-transform, try to use the geometric series directly with the equation for the z-transform and this should help:

$$\sum_{n=0}^\infty r^n = \frac{1}{1-r} \space \space r<1$$

This should get you past the clueless point.