I'm currently struggling to convert the electric field to the intensity in the frequency domain.

In principle it seems like I need to do the following: $$ I(\omega)=\mathcal{F}[I(t)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dt|\mathcal{R}(E(t))|^2e^{-i\omega t} $$ where $$ E(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}d\omega E(\omega)e^{i\omega t} $$ with the given electric field $E(\omega)$ in the frequency domain.

This is in the first place not really doable analytically and also not very easy numerically. That's why I wanted to ask if there is a better way to do this transformation.


2 Answers 2


It's actually remarkably easy. No need to go into the time domain. Use Ohm's law and the impedance of free space, $Z_0$ (about 377Ω).

$$I(\omega)= E(\omega)^2 /Z_0 $$

  • $\begingroup$ Okay, but what is then my electric field spectrum consists of deltas? Delta squared is always a bit weird. $\endgroup$
    – Benji
    Commented Sep 9, 2022 at 11:43

Taking square of EM signal in time domain is a pointwise product leading to a convolution in the Fourier domain: F(x×x)=F(x)∗F(x)

As EM signals are mostly sinosuidal, trigonometric identities defines the change in frequency e.g:

$$cos^2x=1/2(1+cos2x)$$ $$sin^2x=1/2(1−cos2x)$$ or

enter image description here

From the above formulae, one can see that:

Trignometric conversions explains the peaks shifts in freq domain. Scalar components yield frequency peaks at reference point


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