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Ask a simple question:

we know $F[h(t)] = H(f)$, where $h(t)$ is the impulse response.
How to show $F[h(t)] = H^*(f)$?

My answer is just $H(-f)$.

sleeve chen
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  • You may like to read this http://www.math.ubc.ca/~feldman/m267/ft.pdf from page 11. If you still need help, just ask and we can go through it. – Matthew Cassell Dec 06 '14 at 09:57

1 Answers1

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You are absolutely right: the Fourier transform of $h(-t)$ is $H(-f)$ (if $H(f)$ is the Fourier transform of $h(t)$). $H^*(f)$ is the Fourier transform of $h^*(-t)$. Of course, if $h(t)$ is real-valued we have $h(-t)=h^*(-t)$ and, consequently, $H(-f)=H^*(f)$. But in general you have the pairs

$$h(-t)\Longleftrightarrow H(-f)\\ h^*(-t)\Longleftrightarrow H^*(f)$$

Matt L.
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