# Is there a physical reason why phase should be uniformly distributed?

I was thinking a little about how signal phase might be distributed. Usually in such discussions I see people make the assumption that phase is uniformly distributed between $$0$$ and $$2 \pi$$. This only makes sense to me if its a statement about our ignorance of the system in question. That is, if we don't know anything about the phase, then it makes sense to assume all phase values are equally as likely -- uniformly distributed.

Is this the only reason though? Is there a physical reason, or process, as to why this can be the case?

• Assuming a pure sine wave of frequency "f" and assuming the sine wave is sampled at sampling frequency "fs" and fs/f is not an interger. With a sufficient number of samples, the phase of each sample will be uniformly distributed.
– Ben
Apr 27, 2021 at 17:20
• Could you please cite one or two specific examples of where the claims are being made? Apr 27, 2021 at 17:47
• @Ben: assuming the sine wave is sampled at sampling frequency "fs" and fs/f is not a rational number. With fs/f rational, the distribution is still non-uniform. Apr 28, 2021 at 9:26
• yep you're right.. I should have written rational number instead of integer.
– Ben
Apr 28, 2021 at 11:13

A zero mean complex white noise process has a uniform phase distribution. The uniform phase distribution would occur for any complex noise process whose I and Q (real and imaginary) components are zero-mean, independent and identically distributed.

The thermal noise floor common to radio receivers for receiving signals without other specific interference is typically well approximated by a zero-mean complex white noise Gaussian distribution; meaning a white noise Gaussian distribution on I (the real component) and Q (the imaginary component) and these two are independent from each other. This makes sense if you consider that the real and imaginary values for any given sample have no relation to each other and are also therefore independent. Within this restriction of having an independent and identically distributed random process for I and Q, the actual phase of any sample is equally likely as any other phase and is therefore uniformly distributed (regardless if the system is white noise or band-limited).

Complex quantization noise (using two real A/D converters in practice; one for the I channel and the other for the Q channel) will also be well approximated by a white noise process where in this case the phase is uniformly distributed as above but unlike the thermal noise example given above where the signed amplitude was approximated as Gaussian distributed, the amplitude for quantization noise is also uniformly distributed over a quantization level.

There are many effects that take part in forming the signal distribution. The multitude of references are popped up when searching with keywords "signal phase uniform distribution", and many are about the communication channel effects -- for the most part, about the Rayleigh fading in the channel. I think the channel impulse response is the subject of a few discussions that you mention in your question. Because the phase distribution of the Gaussian process per se is analyzed in multiple textbooks and courses of instruction, I am going to consider the contribution of the channel impulse response -- in particular, the scenarios where it can result in a uniform phase distribution.

The impulse response of a wireless channel is formed by the transmission medium. The three key components of the channel response are path loss, shadowing, and multipath. Multipath propagation and shadowing lead to fading. Similar to how stochastic processes in general can be of the Wide Sense Stationary kind, the impulse response of channels can often be described with the Wide Sense Stationary Uncorrelated Scattering model.

In WSSUS, if there is no dominant contribution of a unique propagation path (typically, a line-of-sight path) over the other paths (this is the Rayleigh fading scenario), the channel impulse response (CIR) reduces to a time-variant complex channel coefficient which is a sum over all constituent paths $$c(t) = \sum_i{a_i \exp(j\phi _i (t))}$$

For the quadrature components of the channel coefficient that are i.i.d. Gaussian distributed variables, the probability density functions can be computed in a closed form $$p(r) = {\frac {r} {\sigma^2}} \exp\left(-{\frac {r^2} {2\sigma^2}}\right) \\ p(\phi) = {\frac {1} {2\pi}}$$

If a contribution of the line-of-sight propagation path is strong as compared to the complex Gaussian component from the other propagation paths (the Rician fading scenario), this contribution persists in CIR $$c(t) = a_0 + \sum_i{a_i \exp(j\phi _i (t))}$$

After some tedious computation, for the Rician magnitude pdf one can arrive at $$p(r) = {\frac {r} {\sigma^2}} \exp{\left(-{\frac {r^2 + a_0^2} {2\sigma^2}}\right)} I_0\left({\frac {r·a_0} {\sigma^2}}\right)$$ where $$I_0(·)$$ is the modified Bessel function of the first kind. I found this formula in multiple sources and also myself traced the derivation to some degree. The formula for the Rician phase pdf $$p(\phi)$$ depends on $$\phi$$ (non-uniform distribution), is very complicated, and I am not sure that it exists in the closed form. With MATLAB' Communications Toolbox you can implement a Rician fading channel object for modeling real-world phenomena in wireless communications, though.

The CIR phase distribution is uniform for the Rayleigh fading scenario and non-uniform for the Rician fading scenario. I emphasize that these formulas per se do not describe the phase distribution of a signal received from a multipath communication channel: the received signal is a convolution of CIR and a source signal.

You learned that the phase distribution for a AWGN signal is uniform, and the derivation for the WSS signal is trivial. Now, the CIR analysis for WSSUS fading channels is a hot topic and there is a field for rewarding research. So you are on the right track in your studies.