I have read in this newsgroup that the distribution of the phase of complex Gaussian noise is uniform. I have no doubt this is true but I would like to find an "authoritative" referernce for this that I can quote, rather than "Some guy on the Internet said..." ;) Can anyone help? Thanks.
Noise distribution of phase of complex signal
Started by ●November 29, 2006
Reply by ●November 30, 20062006-11-30
spasmous wrote:> I have read in this newsgroup that the distribution of the phase of > complex Gaussian noise is uniform. I have no doubt this is true but I > would like to find an "authoritative" referernce for this that I can > quote, rather than "Some guy on the Internet said..." ;)Both the real and imaginary parts of random noise on a complex signal taken separately are Gaussian. Assuming that Z + I + jQ and that I and Q are of equal average power, the instantaneous noise (What a phase detector would put out) has a Rayleigh distribution. The phases of both Rayleigh and Gaussian noise are uniformly distributed. The first place I would look is Mischa Schwartz: "Information Transmission Modulation and Noise". I can't cite pages for you because I have the first edition and I'm sure you have -- or will get -- a later one. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 30, 20062006-11-30
Reply by ●November 30, 20062006-11-30
"spasmous" <spasmous@gmail.com> wrote in message news:1164861897.959774.320440@l12g2000cwl.googlegroups.com...> Thanks Jerry. I will look for that book in the library - they have the > 4th edition! >yes - thanks Jerry I've not read that one either. does it give a compelling reason for the variance on I and Q being the same? thanks again - Mike
Reply by ●November 30, 20062006-11-30
Mike Yarwood wrote:> "spasmous" <spasmous@gmail.com> wrote in message > news:1164861897.959774.320440@l12g2000cwl.googlegroups.com... >> Thanks Jerry. I will look for that book in the library - they have the >> 4th edition! >> > yes - thanks Jerry I've not read that one either. > > does it give a compelling reason for the variance on I and Q being the same?I don't recall. I can give you one. There is only one actual signal at any given instant. We may decompose it into I and Q along orthogonal axes of our choice. Suppose that I had greater variance than Q. Then by choosing other axes, the variances could be made equal or even reversed. If we can talk about *the variance*, rather than *one of the variances*, then I and Q must be the same. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 30, 20062006-11-30
"Mike Yarwood" <mpyarwood@btopenworld.com> writes:> "spasmous" <spasmous@gmail.com> wrote in message > news:1164861897.959774.320440@l12g2000cwl.googlegroups.com... >> Thanks Jerry. I will look for that book in the library - they have the >> 4th edition! >> > yes - thanks Jerry I've not read that one either. > > does it give a compelling reason for the variance on I and Q being the same?That is by definition of "complex Gaussian." -- % Randy Yates % "Bird, on the wing, %% Fuquay-Varina, NC % goes floating by %%% 919-577-9882 % but there's a teardrop in his eye..." %%%% <yates@ieee.org> % 'One Summer Dream', *Face The Music*, ELO http://home.earthlink.net/~yatescr
Reply by ●November 30, 20062006-11-30
spasmous wrote:> I have read in this newsgroup that the distribution of the phase of > complex Gaussian noise is uniform. I have no doubt this is true but I > would like to find an "authoritative" referernce for this that I can > quote, rather than "Some guy on the Internet said..." ;) > > Can anyone help? Thanks.What the person was probably referring to is a property of the Gaussian distribution. In a complex Gaussian distribution, the real and imaginary parts of the variable are jointly Gaussian. If it can be assumed that the real and imaginary parts are independent (i.e. if the complex noise is white), then you can take advantage of the following relationship between the Gaussian distribution and the Cauchy distribution: If X and Y are independent Gaussian distributions, and you let Z = X / Y, then Z will have a Cauchy distribution; therefore, the ratio of the imaginary part Q to the real part I will have a Cauchy distribution. Then, you can take advantage of a property of the Cauchy distribution: if Z is a Cauchy random variable, then arctan(Z) will have a uniform distribution. I can't find any good references that explain this; if you work out the mathematics, you can confirm this fact. Therefore, arctan(Z) = arctan(Q/I) = angle(I + j * Q) has a uniform distribution. Some references on the Gaussian and Cauchy distributions: http://en.wikipedia.org/wiki/Gaussian_distribution http://en.wikipedia.org/wiki/Cauchy_distribution http://mathworld.wolfram.com/CauchyDistribution.html Jason
Reply by ●November 30, 20062006-11-30
cincydsp@gmail.com wrote:> What the person was probably referring to is a property of the Gaussian > distribution. In a complex Gaussian distribution, the real and > imaginary parts of the variable are jointly Gaussian. If it can be > assumed that the real and imaginary parts are independent (i.e. if the > complex noise is white), then you can take advantage of the following > relationship between the Gaussian distribution and the Cauchy > distribution: > > If X and Y are independent Gaussian distributions, and you let Z = X / > Y, then Z will have a Cauchy distribution; therefore, the ratio of the > imaginary part Q to the real part I will have a Cauchy distribution. > > Then, you can take advantage of a property of the Cauchy distribution: > if Z is a Cauchy random variable, then arctan(Z) will have a uniform > distribution. I can't find any good references that explain this; if > you work out the mathematics, you can confirm this fact. Therefore, > arctan(Z) = arctan(Q/I) = angle(I + j * Q) has a uniform distribution. > > Some references on the Gaussian and Cauchy distributions: > http://en.wikipedia.org/wiki/Gaussian_distribution > http://en.wikipedia.org/wiki/Cauchy_distribution > http://mathworld.wolfram.com/CauchyDistribution.html > > JasonSorry to nitpick on Jason's otherwise excellent post, but don't forget that the case being considered here may have assumed that the complex Gaussian distribution is spherical-symmetric. For example, if the covariance matrix is a scaled identity matrix. This may be what Jason means by "white" but as far as I know the term is usually applied to random processes, not a random variable. But the process can have a color, yet the distribution of the phase may still be uniform in [0, 2\pi). Now, for the original poster, what happens when the covariance matrix is arbitrary? What is a sufficient condition for the phase to still be uniformly distributed? Is it necessary? Hope this helps, Julius
Reply by ●November 30, 20062006-11-30
> Sorry to nitpick on Jason's otherwise excellent post, but don't forget > that the case being considered here may have assumed that the > complex Gaussian distribution is spherical-symmetric. For example, > if the covariance matrix is a scaled identity matrix. This may be what > Jason means by "white" but as far as I know the term is usually > applied to random processes, not a random variable. But the process > can have a color, yet the distribution of the phase may still be > uniform > in [0, 2\pi).Good catch. I did mean that the distribution was spherically symmetric. I was thinking of the random variable as a sample of a white complex Gaussian noise random process. Jason
Reply by ●November 30, 20062006-11-30
"Jerry Avins" <jya@ieee.org> wrote in message news:r-adnX-s-KlOTvPYnZ2dnUVZ_s2dnZ2d@rcn.net...> Mike Yarwood wrote: >> "spasmous" <spasmous@gmail.com> wrote in message >> news:1164861897.959774.320440@l12g2000cwl.googlegroups.com... >>> Thanks Jerry. I will look for that book in the library - they have the >>> 4th edition! >>> >> yes - thanks Jerry I've not read that one either. >> >> does it give a compelling reason for the variance on I and Q being the >> same? > > I don't recall. I can give you one. There is only one actual signal at any > given instant. We may decompose it into I and Q along orthogonal axes of > our choice. Suppose that I had greater variance than Q. Then by choosing > other axes, the variances could be made equal or even reversed. If we can > talk about *the variance*, rather than *one of the variances*, then I and > Q must be the same. >Thanks Jerry - it's a nice example for real-world additive noise and much appreciated.
