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Given the absolute value of the Fourier transform of a signal $g(t)$:

$|G(f)|$

If I compute the inverse Fourier transform of $|G(f)|$,

$$\int_{-\infty}^{+\infty} |G(f)|\, e^{j2\pi ft}df$$

do I obtain $|g(t)|$?

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    $\begingroup$ No. The absolute value operator is non-linear. For one, you lose all phase information in the FT. $\endgroup$ – MBaz Apr 16 at 14:04
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    $\begingroup$ so is the inverse Fourier Transform of $-G(f)$ also $g(t)$? $\endgroup$ – robert bristow-johnson Apr 16 at 14:19
  • $\begingroup$ good catch, @MattL. but you missed one. $\endgroup$ – robert bristow-johnson Apr 16 at 15:16
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    $\begingroup$ @robertbristow-johnson: I left one for you ... :) $\endgroup$ – Matt L. Apr 16 at 16:01
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One first important matter is to understand how much an absolute value may change a signal, and his amplitude spectrum. Below are two examples: $s_1$ with alternating $-1/1$ (blue), and $|s_1|$ (red) which is constant flat; and a smooth sine $s_2$, with its bumpy magnitude counterpart. In plain text: an oscillating time signal can become either very calm or much wilder after taking an absolute value. Of course, you can reverse the argument: a magnitude on a spectrum can drastically change the underlying signal.

signal, magnitude, spectra

The above entails that a non-linear operator can change a great deal of things inside and outside an integral formula (which is linear).

To show that even with the simplest example one cannot recover the absolute signal amplitudes from the magnitude spectrum, let us consider discrete Fourier transformations. Let $[x_0,x_1]$ denote a two-sample signal. Its discrete Fourier transform (up to a factor) is given by $[x_0+x_1,x_0-x_1]$ (sum and difference). From their absolute values $[|x_0+x_1|,|x_0-x_1|]$, we see that the mirrored two-sample signal $[x_1,x_0]$ would have the same the magnitude spectrum. Hence we cannot recover the absolute value of the original value.

This only gets worse when the signals are longer, the combinatorics allow many more signals with the same amplitude spectrum, but with a very distinct modulus in time.

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The answer is clearly 'no', except for trivial cases. Note that

$$G(f)=|G(f)|e^{j\phi(f)}\tag{1}$$

where $\phi(f)$ is the phase of $G(f)$. Rewriting $(1)$ as

$$|G(f)|=G(f)e^{-j\phi(f)}\tag{2}$$

and taking the inverse Fourier transform gives

$$\tilde{g}(t)=(g\star a)(t)\tag{3}$$

where $\tilde{g}(t)$ is the inverse Fourier transform of $|G(f)|$, and $a(t)$ is the impulse response of a (generally non-causal) all-pass with frequency response $A(f)=e^{-j\phi(f)}$.

If we are given $|G(f)|$, we do not know $\phi(f)$ and, consequently, we don't know $a(t)$, so we can't compute $\tilde{g}(t)$ from $(3)$. Anyway, $(3)$ shows the relation between the inverse Fourier transform of $|G(f)|$ and $g(t)$.

Note that since $|G(f)|$ is real-valued, $\tilde{g}(t)$ must be even

$$\tilde{g}(t)=\tilde{g}(-t)\tag{4}$$

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Short answer, as a general rule NO, but depending on $g(t)$, in special cases YES,

e.g. when $g(t)=0$ or $g(t)=Const$ !

Maybe if you find the Fourier of $Sign(G(f))$, e.g. $G(f)$ turns out to be all positive or all negative, you might get a representation for $g(t)$, using properties such as convolution property.

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  • $\begingroup$ you need to fix your notation. i fixed it for the OP. $g(f)$ is exactly the same function as $g(t)$. it's just a different dummy variable. $\endgroup$ – robert bristow-johnson Apr 16 at 14:21
  • $\begingroup$ corrected thanks $\endgroup$ – MimSaad Apr 16 at 18:53
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Consider the transform of a simple sin function. you have two delta's with opposite sign's, when you take the inverse of its absolute value you get back a cosine function. Their absolute values differ with phase.

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