# DTFT of even and odd samples

Here to find DTFT of $$h(2n)$$ they have scaled omega, while in RHS to find DTFT $$x(2n+1)$$ they didn't, why is that?

• Where does that text come from? – Matt L. Sep 16 '19 at 8:24

$$\textrm{DTFT}\big\{x[2n+1]\big\}=\sum_{n=-\infty}^{\infty}x[2n+1]e^{-jn\omega}=\sum_{n\textrm{ odd}}x[n]e^{-j(n-1)\omega/2}\tag{1}$$
Using the trick they suggested, $$(1)$$ can be written as
$$\textrm{DTFT}\big\{x[2n+1]\big\}=\frac{e^{j\omega /2}}{2}\left[\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega/2}-\sum_{n=-\infty}^{\infty}(-1)^nx[n]e^{-jn\omega/2}\right]\tag{2}$$