How to recover $f(t)$ from Fourier Transform of its absolute value $mathcal{F}|f(t)|$?

Let the Fourier Transform of a real signal, $f(t)$, be $\mathcal{F}(\omega)$. And the FT of the absolute value of the same signal, $|f(t)|$, be $\mathcal{F}(u)$.

Can $\mathcal{F}(w)$ be recovered from $\mathcal{F}(u)$?

For instance, the FT of $a \cdot \cos(ft)$ returns a spectrum in which the frequency $f$ has amplitude $a$.

Can $f$ and $a$ be recovered from the FT of $a \cdot \cos(ft)$?

Answer

I recently was pointed to a very nice trick by Robert Bristow Johnson which possibly applies here too to demonstrate this "inability" of recovery. I thought I'd share it here, in addition to the accepted answer.

The trick is to see $|x(n)|$ as $sgn(x(n)) \cdot x(n)$ where $sgn$ is a function that returns 1 for positive sign and -1 for negative sign. In the case of a sinusoid, this equals modulation of the sinusoid at some frequency $f_{sin}$ with a square waveform at the same frequency.

Multiplication in the time domain is equivalent to convolution in the frequency domain. The spectrum of a sinusoid is a spike at $\pm f_{sin}$. The spectrum of a square wave is a series of spikes starting at $f_{sin}$ and repeating at odd harmonics. Therefore, the spectrum of $|x(n)|$ when $x(n)$ is a sinusoid, is a shifted version of the square wave spectrum by $f_{sin}$. This gives us a component at double the $f_{sin}$ with a bit of DC. In other words, it does full rectification to the sinusoid and it now sounds at double the frequency. We will come back to this.

To ask whether we could recover the $x(n)$ from the $\mathcal{F}(|x(n)|) = \mathcal{F}(x(n) \cdot sgn(x(n)))$ is to ask if there is a monotonic function that realises the mapping:

$$\mathcal{F}(x(n)) = g\left(\mathcal{F}(x(n) \cdot sgn(x(n)))\right)$$

Which if we take one step further, it becomes:

$$\mathcal{F}(x(n)) = g\left(\mathcal{F}(x(n)) * \mathcal{F}(sgn(x(n)))\right)$$

And now we are in trouble, because:

$$\mathcal{F}(x(n)) * \mathcal{F}(sgn(x(n))) = \mathcal{F}(sgn(x(n))) * \mathcal{F}(x(n))$$

Therefore, our $g$ would produce the same output for two different values, which is not the definition of a monotonic function. In other words, you can synthesize the same spectrum in more than one ways.

Now, if you fix $x(n)$ to be a sinusoid, then you could say, I will deconvolve $x(n)$ and a square wave and I will recover the original signal provided that i could also fix the phase. It doesn't necessarily have to start from 0. But it doesn't matter, say you employ some iterative method and after a lot of prespiration you recover $x(n)$. You see here you only have two "actors" and you know both of them very well, so you can tell them appart easily.

BUT, in the general case of some $|x(n)|=sgn(x(n)) \cdot x(n)$, where you only have $|x(n)|$, you can't really tell what its $sgn(x(n))$ was before it was lost!

It's like looking at a photograph where the camera is shooting a scene through a mirror. Can you tell, just by looking at the photograph, if the camera was looking at the real scene OR the real scene through a mirror? Can you recover the "truth"? The mirror here is effected by the modulation function.

So, it is impossible to perform this recovery because it is the product of two components, one of which you have lost forever.

Hope this helps.