How do you calculate the power and energy of a signal given only the frequency domain form of the signal function? For the purposes of this question, please do not assume that it is possible to find a closed form representation of the inverse fourier transform time domain function. Or, in other words, assuming you cannot look at the time domain at all, how can you derive power and energy from the frequency domain representation of a function?

To add some clarification:

Rayleigh's Property:

$$\int_{-\infty}^{\infty}|x(t)|^2dt = \int_{-\infty}^{\infty}|X(f)|^2df$$

Definitions of Power and Energy:

$$E_x = \lim_{T\to\infty}\int_{-T}^{T}|x(t)|^2dt$$

$$P_x = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2dt$$

Since the limit approximates Rayleigh's property, it seems it should be possible to find Energy and maybe power even if you cannot access the time domain function.

  • \$\begingroup\$ I think you need to show where you are stuck in this. It looks like homework so, how far did you get? \$\endgroup\$
    – Andy aka
    Oct 27, 2020 at 14:05
  • \$\begingroup\$ It isn't homework. I've been trying to understand how to use frequency domain to find these values. Take, for example, Rayleigh's Property. It suggests the energy of a time domain signal equals the energy of a frequency domain signal when the absolute value squared of either is integrated over the domain (for which it is non-zero, at any rate). But it isn't immediately obvious if that's true, nor am I sure how to incorporate that into the definition for power, since power incorporates a factor of 1/2T into the definition. \$\endgroup\$ Oct 27, 2020 at 14:09
  • \$\begingroup\$ The form of Parseval's I'm familiar with requires access to the time domain of the function, which I have explicitly prohibited in my question. Since I can't access the time domain, I can't (as far as I know) use the integral to find the fourier coefficients. I don't know another method to find those coefficients. Edit: the comment I was replying to seems to have disappeared. \$\endgroup\$ Oct 27, 2020 at 14:14
  • 1
    \$\begingroup\$ Not in the time domain does not rule out time. If your spectrum changes over time, you have to add a third dimension — time — to your frequency domain calculations. Otherwise multiply the static power taken from the integration of the spectrum by the duration to get the energy. \$\endgroup\$
    – Janka
    Oct 27, 2020 at 14:20
  • \$\begingroup\$ @Janka, so, let's say I have a function with an area of A in the frequency domain. The power would be simply the area of that function? And then the energy would be taking that power and multiplying it by the duration of the signal in the time domain (if that duration were known)? \$\endgroup\$ Oct 27, 2020 at 14:24

1 Answer 1


How do you calculate the power and energy of a signal given only the frequency domain form of the signal function?

For power you do it in exactly the same way that you would in the time domain;

Time domain: \$ P(t)=V(t)I(t) \$

S-Domain: \$ \tilde{P}(\omega)=\tilde{V}(\omega)\tilde{I}(\omega) \$

If you don't know \$ \tilde{V}(\omega) \$ and \$ \tilde{I}(\omega) \$ Then the situation is the same as if you were to try and calculate the power in the time domain without knowing \$ V(t) \$ and \$ I(t) \$, it is not possible.

For energy you pick any number you like..

The s-domain does not care about time. If we assume that you have a signal at 1kHz with a power of 1W. if you observe this signal for 1s in the time domain then the energy is 1J, is you observe it for 1000s in the time domain then the energy is 1kJ.


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