Before asking my question, let me introduce the context:

For spectrum sensing based on energy detection, which has been widely studied in presence of AWGN, the optimal detection threshold is computed as:

$$ \lambda = 2~\sigma^2_\omega ~ \sqrt{N}~ (Q^{-1}(P_{fa}) + \sqrt{N})~~~~[1]$$

where $\sigma^2_\omega$ is the noise variance, and $P_{fa}$ is the design false-alarm rate (false-alarm probability).

The detection algorithm works as follows:

We take an interval of $N$ samples of the received signal $x[n] = k~s[n] + \omega[n]$, where $k = 1$ if the signal of interest is present and $k = 0$ otherwise. Here both $s[n]$ and $\omega[n]$ come from zero-mean Gaussian wide-sense stationary processes.

The received energy is then $\Lambda = \sum_{n = 0}^{N-1} |x[n]|^2$, and $\Lambda$ follows a $\chi^2$ distribution.

Now, regarding the presence or not of a signal of interest during the considered interval, it is assumed that $k = 1$ if $\Lambda > \lambda$, and $k = 0$ otherwise.

Of course, equation [1] is not optimal in practical scenarios, since AWGN is just a model not possible in real life. However, let's just accept it as a good approximation for wide-band noise (nearly-uncorrelated noise). After applying an LTI pass-band filter to such a noise, I would obtain Gaussian colored noise (no longer wide-band), and thus uncorrelated samples is no longer a good approximation. As a consequence, not much can be said about the distribution of $\Lambda$, due to the introduced correlation in $x[n]$ after filtering.

My question is: Is it still justified to assume that the detection threshold $\lambda$ is linearly proportional to the noise variance (after the pass-band filter)?

My results in this regard: A practical application on acoustic signals I am working on suggests a non-linear relation between $\lambda$ and the noise variance, once fixed a false-alarm probability.

I mean, I fix a threshold, then I run the application in absence of a signal of interest, and after a long run I measure the false-alarm rate. By increasing the noise power, for example 10 dB, and then doing the same for the threshold, I obtain a false-alarm rate quite different from the previous run. However, I don't see a proper justification for this behavior other than a non-linear relation between $\lambda$ and the noise variance. And in such a case, I am unable to explain why is this so.

I am using as the receiver a mobile (its microphone), and as the transmitter a speaker.

Should I empirically determine what is this non-linear relation, or is there any theoretical result in this regard that I could use?

  • $\begingroup$ Interesting, could you post an online reference for that formula and its derivation? I presume you are also band-limiting the "signal" and the LTI is applied to the raw data; not the calculated energy? $\endgroup$ – rrogers Apr 3 '18 at 21:04
  • $\begingroup$ Eq. [1] doesn't come from the exact distribution of the test statistic, which is $\chi^2$. The exact equation is: $P_{fa} = \Gamma(N, \lambda/(2\sigma^2_\omega))/\Gamma(N)$, which is derived from the $\chi^2$ pdf. Eq. [1] is obtained by approximating the pdf to a Gaussian one and then solving for $\lambda$. However, this approximation does not change the linear relation between $\lambda$ and $\sigma^2_\omega$ for a fixed $P_{fa}$. You may find the details, for example, here: springer.com/cda/content/document/cda_downloaddocument/… $\endgroup$ – Luis M Gato Apr 4 '18 at 14:45
  • $\begingroup$ Regarding the use of an LTI (band-pass) filter, I apply it to the raw signal. Since I am interested in measuring the false-alarm probability, I am not transmitting any signal of interest, i.e. $k = 0$. The (recorded audio) signal is simply the acoustic noise, which is nearly white in the frequency band of interest. I compute the energy after applying such a filter, thus $x[n]$ is no longer white and Eq. [1] is no longer a good approximation. However, still I expected a linear relation between $\lambda$ and $\sigma^2_\omega$. I haven't found yet a justification of why it is not the case. $\endgroup$ – Luis M Gato Apr 4 '18 at 14:56
  • $\begingroup$ Just to be double check: your signal chain (in particular the digitization) does preserve the S/N? 5:1 signal (well in your case say standard deviation) is 5 times better than the digitizer; wherever that is. In addition, your question is about k=0? What amount of discrepancy are we talking about? Since your equation [1] obviously says that the ration should be constant, I would assume it's the signal processing that's in error. Now I am pretty sure that the non-gaussian distribution alters the equation but doesn't matter for your question. $\endgroup$ – rrogers Apr 5 '18 at 21:28
  • $\begingroup$ If I were you I would dump the signal processing numbers along the signal chain and apply the test at each point. Using a software process to do the squaring for early parts; but examining the data before squaring. If you want you can post them somewhere and I will look but presumably you wouldn't have any problem. The purpose is, where do you lose the ratio. For example: per thousand samples with three SD's threshold you should 5 hits; as my, admittedly faulty, memory kicks up. Tedious but you should know where things are falling apart. $\endgroup$ – rrogers Apr 5 '18 at 21:41

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