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What are the statistics of the cepstrum of gaussian white noise?

\begin{align}\newcommand{\Nfft}{ {N_{\mathrm{FFT}} }}\DeclareMathOperator{\FFT}{FFT}\DeclareMathOperator{\IFFT}{IFFT} x_i &\sim \mathcal{N}(0, \sigma^2)& i&\in\{0,\ldots,\Nfft-1\}\\[1.2em] % X &= \FFT(x)\\ X_i &\sim \mathcal{N}\left(0, \frac{\sigma^2}{\Nfft}\right)\\[1.2em] % \left\lvert X_i\right\rvert^2 &\sim \chi^2\\[1.2em] % C &= \IFFT\left(\log{\left\lvert X\right\rvert^2}\right)\\ C_i &\sim ~\ ??? \end{align}

What if I apply a window function?

\begin{align} X &= \FFT(w \cdot x)\\ &= \FFT(w)*\FFT(x) \\[1.2em] % C &= \IFFT\left(\left\lvert X\right\rvert^2\right)\\ C_i &\sim ??? \end{align}

Even an approximation would be nice. I'm working on a detector and understanding the noise structure is important for my approach.

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  • $\begingroup$ how large is your IFFT? $\endgroup$ – Marcus Müller May 22 at 11:30
  • $\begingroup$ The same as the FFT. In my case 1024 samples. $\endgroup$ – Alister Trabattoni May 22 at 12:32
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The logarithm of a $\chi^2$-distributed random variable (RV) is a Log Chi-Squared RV.

Bartlett (of the frequency estimator fame) has a famous result that the sum of independent identically distributed $\ln\chi^2$ variables (and every logarithm is just a scaled version of any other logarithm) converges to a normal distribution much faster than a sum of $\chi^2$-distributed RVs.

Thus, I'm pretty confident that your sum of independent random variables will end up being normal in both real and imaginary part. But I'm too lazy to do a proof: I don't think this will help you that much!

I'm working on a detector and understanding the noise structure is important for my approach.

The noise is additive to $x$, and since the Fourier transform is linear, it's also additive in $X$.

White noise being uncorrelated to the signal, the signal and noise powers are additive, too.

However, latest with the logarithmic transform, you lose that additivity, and knowing what the cepstrum of white noise in isolation looks like alone will not help you, because the cepstrum of (noise + signal) is not the (cepstrum of noise) + (cepstrum of signal).

That means your noise leads to a strange non-linear rotation thing of your complex cepstrum. It's hard to deal with additive noise in a nonlinear transform, which is why practically all approaches to that from low-SNR communications theory very much try to first invert the nonlinearity, before then dealing with the noise/signal separation. (see: digital predistortion, non-linear DFT for fiber optics, you name it)

Since the logarithm is a invertible operation, I'd strongly propose the first step of your cepstrum-based detector to be that of reverting the logarithm. That's kind of funny, because it just dumps you back to using the power spectrum instead of the cepstrum – but sometimes tools make it harder to work with statistically easy things.

Shot in the blue: As for a detector: Maybe do a noise-variance estimator on the power spectrum (which, again, is easy to calculate from the cepstrum), and do a signal power estimator on the cepstrum, and combine these results cleverly. Since a sensible combination for signal detection of these will be relatively smooth, but inherently pretty nonlinear operation, a reasonable approach to finding an optimum detector is probably stochastic gradient descent; "Neural networks" come to mind.

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  • $\begingroup$ Thank you for your detailed answer. I took me some time to diggest all this information. I wasn't aware about this Bartlett result. I ran some simulations, and I do observe that the noise end up to look very close to a normal distribution. Now I'm wondering: can analyticaly know the expected variance ? The distribution seems to be with zero mean. But as you say the log nonlinearity makes things complicated :(. I will think at how I can use the power spectrum instead. It seems more complicated but if it's the cost to keep linearity... I still need to think about your last idea. $\endgroup$ – Alister Trabattoni May 25 at 8:29

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