# 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,

## 1 Answer

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.

so you can say:

"if $x(t)$ is real and even (that is $x(-t) = x(t)$) so also is $X(j\omega)$ real and even ($X(-j\omega) = X(j\omega)$)."

• Thanks, Robert! What made it "click" was the last part of your answer. When do you think that it is necessary or just convenient to have signals/systems with these properties, however? They're definitely helpful when determining the outputs of some systems, but can you think of anything that wouldn't be possible unless these characteristics were met? Thanks again! – aralar Dec 11 '14 at 1:48
• 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
• ye welcome. ... – robert bristow-johnson Dec 11 '14 at 22:13