Given that $H(\omega)$ is the Fourier transform of $h(t)$, what is $H(-\omega)$ the Fourier transform of?
Any help will be much appreciated. Thank you.
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Not sure I understood your question, but I will guess:
If $H(\omega) = \mathcal{F}[h(t)](\omega)$ is the Fourier transform of $h(t)$ then $H(-\omega)$ is the Fourier transform of $h(-t)$:
$$ \begin{aligned} H(-\omega) &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty} dt\; g(t) e^{-i(-\omega) t}\\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty} dt\; g(t) e^{-i\omega (-t)}\\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty} dt'\; g(-t') e^{-i\omega t'}\\ &= \mathcal{F}[h(-t')](\omega) \end{aligned} $$
From second to third row I substituted $t' = -t$ in the integral.
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Thank you! I think you are right. – monkinsane Dec 07 '13 at 05:35