Prove that time-reverse yields inverse filter for deconvolution?

Question

In reading literature on the construction of impulse responses from sine sweeps (e.g. papers by Farina), I see it stated over & over that the way the impulse response is constructed from a (exponential) sine sweep is to take the original sine sweep signal $S(t)$, time reverse it ( $S(T-t)$ ), scale it somewhat in magnitude, and convolve this with the recorded signal to obtain the impulse response of the (time-invariant, linear) system such as a room's reverberation.. i.e., the claim is that $S^{-1}(t)$ is proportional to $S(T-t)$.

Can anyone explain how to prove this, or point me to a reference that actually shows a 'proof' of this? The closest I've seen is "it can be shown that...", and a few hand-waving comments about how time-reversal inverts the phase -- but it's not obvious that this 'does the job' for producing a proper inverse from a function with nonlinear time dependence.

I've been trying to show this explicitly (in the continuous-time case) for about 7 hours now, and have a few pages of integrals written out, even with a little help from WolframAlpha :-(. The exponential term in the frequency makes it hard to compute the Fourier Transform explicitly (you get Gamma functions with two arguments), let alone try to take the reciprocal of the FT and then take the Inverse FT of that.

Thanks!

This is not a homework problem, I just want to understand why it works. (I know it works because I've coded it up, but it still seems kind of like magic.)

PS- Thought I'd try a linear sine sweep as it might be easier, but you also get a frequency-squared term which again makes it...not at all obvious...that taking $t$ -> $-t$ is going to invert all the phase terms.

P.P.S.- Here are my notes so far; the calculation starts on page 4. Dropbox link to PDF

Answer 1

This is not a general result. It specifically holds, iff the Fourier transform of the signal is of unit magnitude. In that case you can write the signal in frequency domain as

$$S(\omega)=\exp(i \phi(\omega))$$

for a real function $\phi$. The time reversed signal (I'll ignore the constant time offset, it doesn't change anything relevant) is then the conjugate

$$\bar{S}(\omega)=\exp(-i \phi(\omega))$$

Convolving the two by multiplying the frequency domain representation gives

$$\exp(i\phi(\omega))\times \exp(-i\phi(\omega)) = 1$$

Which is the response of the identity system and demonstrates that the signal and its time reversed are the responses of inverses.

This result is only exact in this special case. Sweeps can belong to this class of signal, for example if they have been generated as an allpass filter response. You can also do that by constructing the sweep entirely in frequency domain.

A sweep that is synthesised in time domain using a modulated sinusoid is typically not of that form, but approximates it well. Especially near the endpoints of the frequency range you will get significant deviation from the unit response.