Units of Fourier Transform?

[daviddoria]

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]

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.

 

[daviddoria]

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]

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

 

[Dale]

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.