In some publication I have found a formula that for some reason squares signal in time domain. I am analyzing it now.

What should be the Fourier transform of a signal that was squared in time domain? I do not understand why $\mathcal{F}[sin(x)^2]$ results in peak in zero and why side peaks are moved.

I fully understand the fact that multiplication in time domain is convolution in frequency domain and squaring is just $x\times x$.

  • 4
    $\begingroup$ $\cos^2(\theta) = 0.5 + 0.5\cos(2\theta)$. $\endgroup$
    – MBaz
    May 8, 2018 at 0:09

1 Answer 1


If somebody squares a real signal in the time domain, he probably deals with the signal's energy. Since this is a pointwise product, this results in a convolution in the Fourier domain:

$$ \mathcal{F}(x\times x) = \mathcal{F}(x)\ast \mathcal{F}(x) $$

In some cases, as stated by @MBaz, sometimes trigonometric identities can explain the frequency shift, as:

  • $\cos^2 x=\frac{1}{2}(1+\cos2x)$
  • $\sin^2 x=\frac{1}{2}(1-\cos2x)$

From the above formulae, one can see that:

  • they are positive, and the non-oscillatory term is $1/2 \cos(0x)$, explaining a peak at frequency zero,
  • the scalar before $x$ has doubled, explaining the peak moves.

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