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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?
Gilles
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Val9265
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1 Answers1

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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 $n'=-n-1\Rightarrow n=-n'-1$

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

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msm
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  • 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 $x[-n]$? – Val9265 Oct 03 '16 at 09:00
  • Any reason why the time shift has to be applied first? How would you determine the order for other operations? – Val9265 Oct 03 '16 at 14:09
  • @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. – msm Oct 03 '16 at 15:07