In the notes I am reading they use the following. Let $U$ be the voltage (depends on time) and $I$ the current in a circuit with some resistor with resistance $R$. Then the power is given by:


This is then Fourier transformed to the following:


where $*$ denotes complex conjugated. Could somebody give a proof of this?

  • 2
    $\begingroup$ it will be difficult to prove this because it is untrue; instead read up on convolution in the frequency domain en.wikipedia.org/wiki/Fourier_transform , eqs. 108 and 109 $\endgroup$
    – hyportnex
    Nov 15, 2018 at 15:42
  • $\begingroup$ You just need to move the star (*) between the voltage and the current, and it is correct ... :) $\endgroup$
    – EigenDavid
    Nov 15, 2018 at 15:44

1 Answer 1


I think you are confusing complex power with instantaneous power.

The instantaneous power is simply $p(t)=v(t)i(t)$ as you've mentioned. The complex power however is not the Fourier transform of $P$. For two harmonic signals with frequency $\omega$, The complex power is instead defined as: $$S:=\frac 12 VI^*$$ where $V$ and $I$ are the phasors of the voltage and current, respectively. Written in rms values $V_\mathrm{rms}=V/\sqrt2$ and $I_{rms}=I/\sqrt2$, this is simply: $$S = V_\mathrm{rms}I^{*}_{\mathrm{rms}}$$ which is what you're referring to as $P=UI^*$.

But if these quantities are not the same, why are they both called power? This is simply because if I take the real part of the complex power, I get the time -averaged instantaneous power. To see why, let $v(t) = |V| \cos(\omega t + \phi_v)$ and $i(t) = |I| \cos(\omega t + \phi_i)$ be our two harmonic voltage and current signals. the time-averaged power is: $$P_\mathrm{avg} = \frac{1}{T}\int_{0}^T v(t)i(t)dt \\ = \frac{1}{T}\int_{0}^T |V| \cos(\omega t + \phi_v)\times |I| \cos(\omega t + \phi_i)dt \\ = \frac 12 |V||I| \cos(\phi_v-\phi_i) \\ =\frac 12 \mathrm{Re}(|V| e^{-i\phi_v}|I|e^{-i\phi_i}) \\ = \frac 12 \mathrm{Re}(VI^*)$$ which finally gives: $$P_\mathrm{avg} = \mathrm{Re}(S)$$ Note that the imaginary part of $S$ is usually called the reactive power, which shows a constant back-and-forth transmission of power between the source and the load.

In conclusion, the complex power $S$ is not the Fourier transform of the instantaneous power. As hyportnex and David point out in the comments, Taking the Fourier transform of $v(t)i(t)$ gives $V(\omega)*I(\omega)$, where $*$ is understood as convolution.

You may be curious about what happens if you use a Fourier transform instead of phasors in the above definition of $S$. Even though these two concepts are related, they are not the same. Even dimensionally, the phasor $V$ of a voltage has units of Volts, while its Fourier transform has units of Volts/Hz.

You can easily see that the quantity $\varepsilon(\omega) = V(\omega)I^*(\omega)$ is nothing but the energy spectral density. where $V(\omega)$ and $I(\omega)$ are the Fourier transforms of $v$ and $i$, respectively (as opposed to phasors).

This is an immediate consequence of Parseval's theorem, which says: $$\int_{-\infty}^{\infty}\varepsilon(\omega) d\omega = \int_{-\infty}^{\infty} V(\omega)I^*(\omega) d\omega \\ = \int_{-\infty}^{\infty} v(t)i(t) dt = \int_{-\infty}^{\infty} p(t) dt= E$$ which is the total energy.


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