# Inverse DTFT of phase shifted complex exponential

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

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

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.

• 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.
– Dom
Aug 22 '20 at 2:24
• @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 $x[n]$. Aug 22 '20 at 8:22

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

• Except for $n=10$ ... Aug 27 '20 at 9:17