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The Hilbert transform is given by the following: $$ \mathscr H[a(t)] =\frac 1π \times a(t) *\frac 1t $$ Now, I'm trying to do envelope detection on an audio signal and I'm hoping you math folks can help me out on the conceptual level. It has been ages since I took any sort of signal processing class, which was the only place I saw convolution discussed. I'm aware that convolution in the time domain is equivalent to multiplication in the frequency domain.

So if I have a Fourier transform of $a(t) = A(t)$ and I know that the transform of $\frac 1t = j \sqrt \frac π2 \times \mathrm {sgn}(f)$, then I'm pretty close to knowing the Hilbert transform. The problem is that I don't remember, and cannot get Google to reveal to me the secret of how a constant times a signal in the time domain affects the Fourier transform in the frequency domain. In other words, if $\Bbb F[a(t)] = A(t)$, what effect does multiplying by a constant, in this case $\frac 1π$, have on the Fourier transform? Google told me $\Bbb F[c\times a(t)]$ where c is some constant but that doesn't really help me in my haven't-taken-a-math-course-in-four-years mind.

TL;DR: How does multiplying by a constant in the time domain affect the Fourier transform (in the frequency domain)?

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The fourier transform is linear. This implies that if you multiply by a constant in the time domain, you multiply by the same constant in the frequency domain.

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