# system function $H(\omega)$ relationship to odd and even components of h[n]

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:

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

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

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).

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