# Units of Fourier Transform?

## Main Question or Discussion Point

If i have a signal S(t) (the plot would be voltage vs time) and I take its fourier transform, what are the units of the vertical axis? The horizontal axis can either be frequency in hertz or in radians, but what about the other axis? I guess generally I plot the magnitude of the transform since its not always real, but it should still have a unit, no?

Thanks,

David

Dale
Mentor
This is an interesting question. I use Fourier transforms all the time, but never stopped to think about that. If you look at the definition of the Fourier transform:
$$X(\omega )=\int_{-\infty }^{\infty } e^{-i t \omega } x(t) \, dt$$
then you see that $$e^{-i t \omega }$$ is unitless and $$dt$$ has units of time, so it would seem that if $$x(t)$$ has units of volts then $$X(\omega )$$ must have units of volt seconds.

This is consistent with Parseval's theorem $$\int_{-\infty }^{\infty } |x(t)|^2 \, dt=\int_{-\infty }^{\infty } |X(\omega )|^2 \, d\omega$$ where each side would wind up with units of volt^2 seconds.

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Is Volt*seconds somehow equal to energy? Non-technically speaking, the Fourier transform shows you the "strength" (energy or power or something else?) at each frequency, right? So then the units should be related to energy in a very straight forward way. I've just never seen Volt*seconds before.

Thoughts?

David

chroot
Staff Emeritus
Gold Member
If you take the squared modulus of the Fourier transform result, you get power per Hertz. The transform result itself is imaginary, and thus has no physical units.

- Warren

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Dale
Mentor
Power in a resistive circuit is volt^2/ohm so you cannot take the units too literally for energy. But as chroot said a volt^2 second is proportional to a volt^2/Hertz. So the squared magnitude of the transform is proportional to power/Hertz.