If one has some oscillating voltage signal, for example: $$V(t) = V_{max}\cos(2 \pi \nu_{0}t) e^{-\gamma t}$$ and you take the Fourier transform of this in the usual way to get: $$\hat{V}(\nu) = V_{max}\int^{+\infty}_{0} \cos(2 \pi \nu_{0}t) e^{-\gamma t} \ dt$$ Now I often see in literature the absolute value squared taken of the result, so $\left|\hat{V}(\nu) \right|^{2}$ the physical interpretation of which it is the energy per unit frequency.

My question is, if $\left|\hat{V}(\nu) \right|^{2}$ is energy per unit frequency, then what is the interpretation of $\left|\hat{V}(\nu) \right|$?

  • $\begingroup$ Your FT algorithm is not correct? $\endgroup$
    – MathArt
    Dec 4, 2023 at 9:02

1 Answer 1


This would be the so called magnitude spectrum, its unit is the same as the signal it's generated from, in the case of voltage its interpretation is "voltage by time over frequency" as opposed to "power by time over frequency" of the squared magnitude spectrum. In the ususal logarithmic representation, the two are proportional by a factor of 2 and essentially give you the same information.

The $\cos$ function in your example is a so called power signal, not an energy signal, as it is infinitly long it has infinite energy, so to assess its energy in a practical way, one refers to its power (energy per time).

  • 1
    $\begingroup$ if $x(t)$ has units of Volts, then the unit of $X(f)$ is $Volts \cdot s$ (Volts-second). $\endgroup$
    – Fat32
    Apr 11, 2019 at 14:07
  • $\begingroup$ You are right of course, the continuous Fourier transform yields Vs. I was thinking of the DFT, which yields V. $\endgroup$
    – Max
    Apr 12, 2019 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.