0
$\begingroup$

What qualities of $h[n]$ are necessary for:

$$ H(e^{j\omega}) = DTFT\{h_{even}[n]\} + j\ DTFT\{h_{odd}[n]\} $$

Do all real / causal h[n] have the property that:

$$ H(e^{j\omega}) = DTFT\{h_{even}[n]\} + j\ DTFT\{h_{odd}[n]\} $$

where:

$$ h_{even}[n] = \frac{1}{2}(h[n] + h[-n]) $$

$$ h_{odd}[n] = \frac{1}{2}(h[n] - h[-n]) $$

$\endgroup$
4
$\begingroup$

The DTFT relationships

$$x_{even}[n]=\frac12\left(x[n]+x^*[-n]\right)\Longleftrightarrow\textrm{Re}\left\{X(e^{j\omega})\right\}$$

and

$$x_{odd}[n]=\frac12\left(x[n]-x^*[-n]\right)\Longleftrightarrow j\,\textrm{Im}\left\{X(e^{j\omega})\right\}$$

hold for any sequence $x[n]$ for which the DTFT exists. There is no assumption about $x[n]$ being real-valued or causal (note the complex conjugation $^*$ in the definition of even and odd signals). If $x[n]$ is real-valued you can leave out the conjugation.

Note that the DTFT of the odd part $x_{odd}[n]$ equals $j$ times the imaginary part of the DTFT $X(e^{j\omega})$, so you have

$$X(e^{j\omega})=\textrm{DTFT}\{x_{even}[n]\}+\textrm{DTFT}\{x_{odd}[n]\}$$

(without a $j$ on the right-hand side).

$\endgroup$
  • $\begingroup$ thanks, makes sense now. any suggestion for title? $\endgroup$ – MrCasuality Jan 12 at 15:55
  • $\begingroup$ @MrCasuality: If your question has been answered you can accept this answer by clicking on the green check mark to its left, thanks. $\endgroup$ – Matt L. Jan 12 at 17:07

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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