Units of Fourier Transform?

  • Thread starter daviddoria
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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
 

Answers and Replies

28,493
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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:
[tex]X(\omega )=\int_{-\infty }^{\infty } e^{-i t \omega } x(t) \, dt[/tex]
then you see that [tex]e^{-i t \omega }[/tex] is unitless and [tex]dt[/tex] has units of time, so it would seem that if [tex]x(t)[/tex] has units of volts then [tex]X(\omega )[/tex] must have units of volt seconds.

This is consistent with Parseval's theorem [tex]\int_{-\infty }^{\infty } |x(t)|^2 \, dt=\int_{-\infty }^{\infty }
|X(\omega )|^2 \, d\omega[/tex] where each side would wind up with units of volt^2 seconds.
 
Last edited:
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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
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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
 
Last edited:
28,493
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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.
 

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