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The well-known relationships for zero-mean circularly-symmetric complex Gaussian $z = a + jb = |z| \exp(j\varphi)$ signals are

  • the amplitudes $|z| = \sqrt{a^2 + b^2}$ are Rayleigh-distributed
  • the phases $\varphi = \arg(z)$ are uniformly distributed
  • the pdf of such signals is fully described by their complex covariance matrix $\mathbf{\Sigma}$, which is Wishart-distributed in case it is estimated from a number of independent samples ($\rightarrow \mathbf{\hat{\Sigma}}$)

My question is two-fold:

1) What's the distribution of $z$ in case the amplitudes are constant, e.g. all normalized to 1?

2) What's the distribution of $\mathbf{\hat{\Sigma}}$ in case the same normalization is applied, i.e. the amplitudes of its elements are removed?

I figure that it's then $z = \exp(j\varphi)$, with $\varphi \sim \mathcal{U}$, i.e. I am possibly looking for the distribution the exponential of a uniformly distributed variable follows. For $\mathbf{\Sigma}$, however, I have no idea.

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  • $\begingroup$ I think you answered your own question when it comes to #1. If you have a complex random variable $z$ whose magnitude is always unity and whose phase is uniform, then by definition $z = e^{j\phi}$ with $\phi \sim \mathcal{U}$. $\endgroup$ – Jason R Jan 28 '14 at 15:29
  • $\begingroup$ I second that. But I am still wondering: $z \sim ???$ $\endgroup$ – Michael Jan 28 '14 at 15:55
  • $\begingroup$ The covariance matrix is not a random variable. It is a parameter of the random distribution. Only if you generate the sample covariance matrix, it exhibits a Wishart distribution. $\endgroup$ – jan Jan 28 '14 at 16:26
  • $\begingroup$ @jan: You are right of course. I am sorry for not having mentioned that my problem is based on the sample covariance matrix. I will update the question correspondingly. $\endgroup$ – Michael Jan 28 '14 at 16:49
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    $\begingroup$ @Michael: I'm not sure what your remaining question is. The pdf of $z$ would be trivial; it is a constant value along the unit circle in the complex plane and zero everywhere else. I don't know of a classically-named distribution that has this form, if that's what you're wondering. $\endgroup$ – Jason R Jan 28 '14 at 19:38

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