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Consider a signal, f(t), with impulse samples taken N times, i.e f[0],f[1],f[2],...f[N-1] Let us perform FFT on it. Now, we have the amplitude on the y-axis and the frequency on the x-axis. I want to know if the unit of the quantity on the y-axis remains the same. If yes, why? If no, what happens to it?

Example: If we consider a voltage signal. What will be the unit of the quantity on y-axis after FFT of f(t)?

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The notation $f(t)$ is usually used for continuous signals. The answer to your question depends on the fact that you're talking about a discrete Fourier transform, so you might want to make that clearer in the notation for the sample values. –  joriki Jul 25 '12 at 12:05
    
Thank you Sir. Is that particular edit okay? –  SJR Jul 25 '12 at 12:22

3 Answers 3

up vote 4 down vote accepted

It's still a voltage. If you do a continuous Fourier transform, you go from signal to signal integrated over time, which is signal per frequency, but in a discrete Fourier transform you're just summing discrete voltages with coefficients, and the result is still a voltage. Of course if you want you can multiply it by the time interval between sample points to get a voltage per frequency unit.

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Depending on the implementation, your FFT routine might output yout = yin/N, or yout = yin/sqrt(N), or if you applied a window, you have to replace N with a coefficient that depends on the chosen window.

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The unit stays the same. More concretely, a Fourier transform changes a function $f: X \to Y$ to a function $\hat f: Z \to Y$ (in physics, $X$ is usually the set of time points and $Z$ the set of frequencies), so the codomain of the function and hence its unit stays the same.

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More precisely, a discrete Fourier transform. A continuous Fourier transform does change the dimensions of the function values. –  joriki Jul 25 '12 at 12:04

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