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I want to test for the presence of broadband noise in a snapshot 1000 complex baseband samples recorded by a software defined radio.

As a follow-up to this post, why was the $\chi^2$ test used? How many degrees of freedom should be used?

Also, how would one extend this approach to complex baseband data? I would assume that I and Q are iid Gaussian random variables. The magnitude of the complex data would then be Rayleigh distributed not Gaussian. Is there a generalization of the $\chi^2$ test for Rayleigh random variables? Or, would I just pick I or Q to operate on?

Update: I was able to find a paper: A test for whiteness. The author outlines a similar process.

enter image description here

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  • $\begingroup$ So, as the post you link to answers the question of how, you're looking for a why with the $\chi^2$ test? $\endgroup$ Aug 2, 2018 at 20:22
  • $\begingroup$ Yes, the post doesn't explain why to use the chi-squared test. $\endgroup$ Aug 2, 2018 at 21:09
  • $\begingroup$ I changed the title to reflect that. However, the text from the answer to that question is pretty clear: because $R$ from cited formula $(16.32)$ is $\chi^2$-distributed. So, I'm still not sure what your question is? $\endgroup$ Aug 2, 2018 at 21:12
  • $\begingroup$ I'm looking for a little more context. Why did he pick 10 degrees of freedom? $\endgroup$ Aug 2, 2018 at 22:14

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If I and Q are normal, then their magnitude square is $\chi^2$ distributed.

Hence, if you check the distribution of the instantaneous power against that distribution, you get information about how much the complex baseband signal is circularly complex normal.

The autocorrelation function has power of the signal as its value for zero shift. For whiteness, the sample autocorrelation approaches an accordingly weighted Dirac impulse with growing observation length.

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    $\begingroup$ So, the magnitude is Rayleigh but the magnitude^2 is chi^2? $\endgroup$ Aug 3, 2018 at 16:46
  • $\begingroup$ can you please provide a reference? I thought the magnitude squared would be an exponential distribution not a chi-squared distribution? $\endgroup$ Aug 3, 2018 at 20:44
  • $\begingroup$ It's the definition of the chi² distribution. $\endgroup$ Aug 3, 2018 at 21:51
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This table shows the $\chi^2$ test values for various degrees of freedom.

enter image description here

Your additional question is

I'm looking for a little more context. Why did he pick 10 degrees of freedom?

That other question had 1,000 original data points and formed the autocorrelation estimate of them. I decided that a lag of 10 (and so 10 degrees of freedom) was going to be accurate enough for my purposes.

However, you might want to make it larger: 1,000 original samples will allow a reasonably accurate autocorrelation estimate out to several hundred lags.

How you choose your lags (and therefore degrees of freedom) is up to you.

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