If the DTFT of discrete sequence $x[n]$ is $X(e^{j\omega})$, what is the DTFT of $g[n] = x[2n]$?

I see the textbook answer is

\begin{align*} G(e^{j\omega}) &= \frac{1}{2} \left( X(e^{j\omega/2}) + X(e^{j(\omega-2\pi)/2}) \right) \end{align*}

My start of the problem:

\begin{align*} X(e^{j\omega}) &= \sum\limits_{n=-\infty}^\infty x[n] e^{-j\omega n} \\ G(e^{j\omega}) &= \sum\limits_{n=-\infty}^\infty g[n] e^{-j\omega n} \\ G(e^{j\omega}) &= \sum\limits_{n=-\infty}^\infty x[2n] e^{-j\omega n} \\ \end{align*}

How do I derive the given textbook answer?


I try the obvious $m=2n$ substitution, however I don't see how you can change the variable on the summation so that it counts by 2?

\begin{align*} G(e^{j\omega}) &= \sum\limits_{n=-\infty}^\infty x[m] e^{-j\omega m/2} \\ \end{align*}



If you have a sequence

$$\hat{x}[n]=\begin{cases}x[n],&n\text{ even}\\0,&n\text{ odd}\end{cases}$$

then the DTFT of $x[2n]$ can be written as


You can obtain $\hat{x}[n]$ as


Taking the DTFT of $(2)$ to express $\hat{X}(e^{j\omega})$ in terms of $X(e^{j\omega})$, and using $(1)$ will give the desired result.

  • $\begingroup$ Matt L +1. I have been thinking about this problem. But the way you expressed x^[n] didn't flash. Good one $\endgroup$ Oct 22 '15 at 11:41
  • $\begingroup$ how does $(-1)^{n} \rightarrow e^{-2\pi}$ $\endgroup$
    – panthyon
    Oct 22 '15 at 15:26
  • 3
    $\begingroup$ @panthyon: $(-1)^n=e^{jn\pi}$, then use modulation property of the DTFT: $x[n]e^{jn\omega_0}\Longleftrightarrow X(e^{j(\omega-\omega_0)})$. $\endgroup$
    – Matt L.
    Oct 22 '15 at 16:36

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