Fourier transform of power

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:

$$P(t)=U(t)I(t).$$

This is then Fourier transformed to the following:

$$P(\omega)=U(\omega)I^*(\omega),$$

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

• 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 – hyportnex Nov 15 '18 at 15:42
• You just need to move the star (*) between the voltage and the current, and it is correct ... :) – David Nov 15 '18 at 15:44

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.