I'm trying to obtain the impulse response $h[n]$ of a system whose frequency response is $H(e^{j\omega})=R(\omega)e^{-25j\omega}$. I believed that $h[n]=h[n-25]$, would be the correct answer, however I was told it would be something along the lines of $h[n]= h[50-n], 0<=n<=25$. Can somebody explain why?

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    $\begingroup$ What is $R(\omega)$? Or is it a typo? Because if $H(\mathrm{e}^{j\omega})=R(\omega)\mathrm{e}^{-j25\omega}$, the impulse response should be related to $r[n]$ at the very least. $\endgroup$ – cjferes Feb 27 at 0:30
  • $\begingroup$ @cjferes $R(w)$ like the a filters linear phase expression $H(e^{jw})=R(w)e^{-aw+b}$, some books refer to It as $A(w)$ I believe $\endgroup$ – HelpMeBro Feb 27 at 2:13
  • $\begingroup$ Not sure if I am correct, but check my answer. drive.google.com/file/d/1BvT-LfSuy_lgYgoyBs0NhOxJj7pmp7AO/… $\endgroup$ – Rima Feb 27 at 4:02
  • $\begingroup$ I derived the answer from this link,ccrma.stanford.edu/~jos/fp/… I am out of my computer right now, so wrote it in my copy instead. $\endgroup$ – Rima Feb 27 at 4:02
  • $\begingroup$ @Rima Thank you very much I understood the answer given the link! $\endgroup$ – HelpMeBro Feb 27 at 14:57

It must be added to the problem that $R(\omega)$ is a real-valued, possibly bipolar function. In that case, its inverse discrete-time Fourier transform must be even:


From the given relation between $H(e^{j\omega})$ and $R(\omega)$ it is clear that


must hold. I'm sure that you'll manage to combine $(1)$ and $(2)$ to come up with the desired result.

  • $\begingroup$ So my combining I'll get something like $h[n]=r[-(n-25)]$, $h[n]=r[25-n]$, would this be a correct final answer? Or am I missing something? Thank you $\endgroup$ – HelpMeBro Feb 27 at 14:54
  • $\begingroup$ @HelpMeBro: You didn't state the actual question, so I don't know the expected answer, but I guess you should express the symmetry $(1)$ in terms of $h[n]$, i.e., state the symmetry property satisfied by $h[n]$. $\endgroup$ – Matt L. Feb 27 at 16:22

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