# Implications of $X( -j\omega ) = X^*(j\omega)$

What are the implications of:

If $x(t)$ is real and $x(-t) = x^*(t)$, then $X(-j\omega) = X^*(j\omega)$ and $X(j\omega)$ is real.

I am trying to understand it and I would like to research it further (what's the theorem that guarantees this?).

I'd like to understand why it's important aside from the fact that it guarantees a symmetric Fourier transform or frequency response if we're talking about that of the impulse response.

Thanks,

The fact is that if $x(t)$ is real and then $X(-j\omega) = X^*(j\omega)$ (it's easy to prove) and, because of duality, if $X(j\omega)$ is real, so also then $x(-t) = x^*(t)$. in your case, you have both $x(t)$ and $X(j\omega)$ being real. keep in mind that anything that is purely real is also equal to its complex conjugate.
"if $x(t)$ is real and even (that is $x(-t) = x(t)$) so also is $X(j\omega)$ real and even ($X(-j\omega) = X(j\omega)$)."
• for me, when i'm considering a windowed segment of audio, i like having my window $w(t)$ be real and even symmetry, so $w(-t) = w(t)$ and so also is $W(-j\omega)=W(j\omega)$. also, when considering linear-phase FIR filter responses, it is convenient to consider the impulse response as real and even: $$h[-n]=h[n]$$ so that $$H(e^{-j\omega})=H(e^{j\omega})$$ and $H(e^{j\omega})$ is real, so i can sketch it more easily. then to realize the FIR in a real-time system, i have to delay $h[n]$ it sufficiently so that $h[n]$ is causal. but that's a last step. – robert bristow-johnson Dec 11 '14 at 21:45