How can I determine the DTFT of $x[-n-1]$? I searched for DTFT problems and checked several references but I couldn't find a similar case. My background is a little lacking, so excuse me if it's too trivial.

What I tried is to substitute $t=(n+1)$, then we just have to determine the DTFT of $x[-t]=X(e^{-j \omega})$ which is trivial.

  • But don't we lose information this way?
  • Don't we have to "change the variables back" or something?

You need to use two properties of DTFT:

  • Time reversal $$\mathcal{F}(x[-n])=X(e^{-j\omega})$$

  • Time shifting $$\mathcal{F}(x[n-n_0])=X(e^{j\omega})e^{-j\omega n_0}$$

Do the time shift at first $$\mathcal{F}(x[n-1])=X(e^{j\omega})e^{-j\omega}$$ then time reversal $$\mathcal{F}(x[-n-1])=X(e^{-j\omega})e^{j\omega}$$

Alternatively, you may calculate it directly through definition of DTFT $X(e^{j\omega})=\sum_{n=-\infty}^{n=+\infty}x[n]e^{-j\omega n}$:

$$\mathcal{F}(x[-n-1])=\sum_{n=-\infty}^{n=+\infty}x[-n-1]e^{-j\omega n}$$ let $n'=-n-1\Rightarrow n=-n'-1$

$$\mathcal{F}(x[-n-1])=\sum_{n'=-\infty}^{n'=+\infty}x[n']e^{-j\omega (-n'-1)}=X(e^{-j\omega})e^{j\omega}$$

  • $\begingroup$ But I don't understand how that works. Would it be correct to think of $X(e^{j\omega})$ in the second property as corresponding to the DTFT of $x[-n]$? $\endgroup$ – Val9265 Oct 3 '16 at 9:00
  • $\begingroup$ Any reason why the time shift has to be applied first? How would you determine the order for other operations? $\endgroup$ – Val9265 Oct 3 '16 at 14:09
  • $\begingroup$ @Val9265 With the time shift at first: $\mathcal{F}(x[n-1])=X(e^{j\omega})e^{-j\omega}$. Then time reversal $\mathcal{F}(x[-n-1])=X(e^{-j\omega})e^{j\omega}$. The point is that the time reversal affects the shifts as well but not the opposite. Draw the two cases on a piece of paper. $\endgroup$ – msm Oct 3 '16 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.