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I have been working on this problem for a few days now and I think this is the closest I have gotten. I am getting an Answer of zero and I would like to know if that is correct and if someone could check my work please? Thanks for your time, here is what I have so far (also Im quite sure evaluating at n = -10 isnt correct, could someone please give me the correct process?)

Not looking for an answer, just how to start/continue to process. Thanks

I put all my steps in the attatched picture. (sorry, Im new to Stack Exchange and couldnt figure out how to format correctly)

Thanks for your time :) -Dom

enter image description here

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  • $\begingroup$ The expression at the bottom of the pic is the one I am integrating and getting and answer of zero on. $\endgroup$
    – Dom
    Aug 21 '20 at 2:41
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HINTS:

  1. $e^{j\pi/2}=j$, so $X(e^{j\omega})=j\omega e^{-j10\omega}$
  2. The term $e^{-j10\omega}$ corresponds to a delay of $10$ samples in the time domain, so you can just ignore it for now, and then add a delay of $10$ samples. So you compute the IDTFT of $j\omega$, and then replace the variable $n$ of the result by $n-10$.
  3. The IDTFT of $j\omega$ is $$\frac{1}{2\pi}\int_{-\pi}^{\pi}j\omega e^{jn\omega}d\omega$$ That integral is most easily computed using integration by parts.

Note that $X(e^{j\omega})$ is the frequency response of an ideal discrete-time differentiator with a delay of $10$ samples.

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  • $\begingroup$ Thanks very much Matt. Very helpful hints. I got the answer zero again, however with much more confidence this time. Is that what you got? I also checked with a calculator. $\endgroup$
    – Dom
    Aug 22 '20 at 2:24
  • $\begingroup$ @DominicMeads: Zero for all values of $n$ can't be the right answer, because then also the DTFT would be zero. Of course, for $n=10$ you get the value zero, but you're supposed to solve for any value of $n$, because you want a general expression for the sequence $x[n]$. $\endgroup$
    – Matt L.
    Aug 22 '20 at 8:22
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After thoroughly checking analytically, using MATLAB, and a calculator, and accounting for the time domain shift, the final answer is:

cos(pi*[n-10])/[n-10]

Thanks @MattL

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  • $\begingroup$ Except for $n=10$ ... $\endgroup$
    – Matt L.
    Aug 27 '20 at 9:17

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