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. $\endgroup$ Feb 20 '15 at 17:26
  • $\begingroup$ @robertbristow-johnson: "there might be an answer"? This would just be $X(f)\star X^*(-f)$. $\endgroup$
    – Matt L.
    Feb 20 '15 at 18:50
  • $\begingroup$ so @MattL., i wonder if there is something that does the Fourier transform of $\sqrt{x[n]}$? $\endgroup$ Feb 21 '15 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.
    Feb 21 '15 at 9:30
  • $\begingroup$ i agree, @MattL. sometimes i ask rhetorical questions. also, sometimes, i understate an answer. $\endgroup$ Feb 21 '15 at 18:47

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. $\endgroup$
    – kkm
    Oct 23 '14 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
    Feb 20 '15 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? $\endgroup$
    – kkm
    Feb 23 '15 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 $|f(x)|$. $\endgroup$
    – jojek
    Oct 23 '14 at 6:56
  • $\begingroup$ @Mikhail Could you link a reference or add a proof/derivation to your answer? $\endgroup$
    – Jim Clay
    Oct 23 '14 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. $\endgroup$
    – jojek
    Oct 23 '14 at 21:52
  • 1
    $\begingroup$ Indeed, but I feel its still useful! $\endgroup$
    – Mikhail
    Oct 26 '14 at 0:02

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