I have a frequency-domain representation $X(e^{i\omega})$ of the complex discrete one-dimensional signal $x[n]$: $X(e^{i\omega})=\mathcal{F}\{x[n]\}$. Is there a frequency-domain transformation of $X(e^{i\omega})$ into $\hat{X}(e^{i\omega})=\mathcal{F}(|x[n]|)$?

Obviously, $\hat{X}(e^{i\omega})=\mathcal{F}\left\{\left|\mathcal{F}^{-1}\{X(e^{i\omega})\}\right|\right\}$ will do the trick, but I am interested if there is a dual to the absolute value function $|x|$ in the frequency domain.

  • $\begingroup$ if it were about the magnitude-squared of a general function $x[n]$, then there might be an answer. as it is, that discontinuity of the magnitude function at the origin makes it not answerable. $\endgroup$ Commented Feb 20, 2015 at 17:26
  • $\begingroup$ @robertbristow-johnson: "there might be an answer"? This would just be $X(f)\star X^*(-f)$. $\endgroup$
    – Matt L.
    Commented Feb 20, 2015 at 18:50
  • $\begingroup$ so @MattL., i wonder if there is something that does the Fourier transform of $\sqrt{x[n]}$? $\endgroup$ Commented Feb 21, 2015 at 1:03
  • $\begingroup$ @robertbristow-johnson: In general, I don't think so. I'm sure you can construct an $x[n]$ for which the FT of $\sqrt{x[n]}$ doesn't even exist. $\endgroup$
    – Matt L.
    Commented Feb 21, 2015 at 9:30
  • $\begingroup$ i agree, @MattL. sometimes i ask rhetorical questions. also, sometimes, i understate an answer. $\endgroup$ Commented Feb 21, 2015 at 18:47

2 Answers 2


Absolute value is not a linear operation, so no, there is not a consistent dual in the frequency plane.

Having said that, though, it is somewhat similar to calculating the signal power, which involves multiplying $x[n]$ with it's complex conjugate. Multiplication in the time domain is equivalent to convolution in the frequency domain. The result in the time domain is real, so the result in the frequency domain is even (i.e. symmetric around 0 Hz).

Absolute value will probably produce a similar effect to calculating the power since it is the square root of the power. It is also real so it will be even in the frequency plane.

  • $\begingroup$ Thanks, and my hunch is also that there is no closed form dual for the absolute value function. Still, the statement that the operator has to be linear to have a dual is not entirely correct. Take a counterexample, multiplication by a signal is non-linear, but its dual exists, namely the convolution. $\endgroup$ Commented Oct 23, 2014 at 22:40
  • $\begingroup$ @kkm Multiplication of two signals is a linear operation. If you double one of the signals, you will double the result. $\endgroup$
    – Jim Clay
    Commented Feb 20, 2015 at 13:56
  • $\begingroup$ Interesting, you got me thinking. Usually, in DSP we take linear as being all three of degree-one homogenous $F(k x[n])=k F(x[n])$, additive $F(u[n]+v[n])=F(u[n])+F(v[n])$ and shift-invariant $y[n]=F(x[n]) \implies y[n+s]=F(x[n+s])$. Multiplication by a signal satisfies the first 2 criteria, but fails the third. I am assuming therefore that the term linear in your original statement means just homogenous and additive. But then I am still trying to understand the statement--why an operator has a Fourier dual only if (iff?) it is linear in the weaker sense? Is it easy to show? $\endgroup$ Commented Feb 23, 2015 at 22:37

Well the Fourier transform of the absolute value function exists:

$-\frac{\sqrt{\frac{2}{\pi }}}{w^2}$

Source: FourierTransform[Abs[x], x, w]

  • 2
    $\begingroup$ How that answers the question? It is not about FT of $|x|$, but general case of composite function $|f(x)|$. $\endgroup$
    – jojeck
    Commented Oct 23, 2014 at 6:56
  • $\begingroup$ @Mikhail Could you link a reference or add a proof/derivation to your answer? $\endgroup$
    – Jim Clay
    Commented Oct 23, 2014 at 13:37
  • $\begingroup$ @Jim Clay: Judging from a syntax it comes straight out of Mathematica - just confirmed, thus he is not relaying most likely on any literature whatsoever. $\endgroup$
    – jojeck
    Commented Oct 23, 2014 at 21:52
  • 1
    $\begingroup$ Indeed, but I feel its still useful! $\endgroup$
    – Mikhail
    Commented Oct 26, 2014 at 0:02

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