Impulse Response of real coefficient, LTI System

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?

• 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. Feb 27 at 0:30
• @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 Feb 27 at 2:13
• Not sure if I am correct, but check my answer. drive.google.com/file/d/1BvT-LfSuy_lgYgoyBs0NhOxJj7pmp7AO/…
– Rima
Feb 27 at 4:02
• 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.
– Rima
Feb 27 at 4:02
• @Rima Thank you very much I understood the answer given the link! 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:

$$r[n]=r[-n]\tag{1}$$

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

$$h[n]=r[n-25]\tag{2}$$

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

• 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 Feb 27 at 14:54
• @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 $h[n]$. Feb 27 at 16:22