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There is a signal $y[n]$ with a differentiable DTFT $Y(e^{i\omega})$. How do I find the inverse DTFT of $i\frac{dY(e^{i\omega})}{dw}$ in terms of $y[n]$ (where of course $i = \sqrt{-1}$)?

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    $\begingroup$ Begin with the fundamental $$ Y(e^{j \omega }) = \sum_{n=-\infty}^{\infty} y[n] e^{-j \omega n} $$ Then apply differentiation wrt $\omega$ to both sides and try to keep a relation between resulting DTFT and a corresponding new signal $y_2[n]$. Then if you still cannot approach the result, put your progress work here and people will probably help. $\endgroup$ – Fat32 Mar 2 '17 at 13:36
  • $\begingroup$ @Fat32 thank you, I knew about that equation but wasn't sure how to start so your answer is helpful. I will try it and get back to this thread. $\endgroup$ – dierensourcan Mar 2 '17 at 14:00
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You can develop it relatively easily, anyways, here's a screenshot from wikipedia enter image description here

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  • $\begingroup$ Well that's straightforward! Thanks much. $\endgroup$ – dierensourcan Mar 2 '17 at 14:07
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This belongs to the classical results for instance in Engineering Tables/DTFT Transform Properties.

You should find something like $ny[n]$ (up to a potential factor). Basically, integration in one domain (time or frequency) results in a division by the index, and differentiation by the index, with some multiplicative factor involving $i^n$.

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  • $\begingroup$ Yes, I thought it was -2πy[n] because the IDTFT for Y(e^(iw)) is 2πy[n], times -1 from multiplying the i in i**(d/dw)Y(e^(iw)) times the **i from (d/dw)Y(e^(iw)). But that is wrong according to my online course. $\endgroup$ – dierensourcan Mar 2 '17 at 14:04

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