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Suppose $w(t)$ is a white noise with uniform spectral density $N_0/2$. It is bandpass filtered to get a narrowband white noise $n(t)$ for $f_c - B_T/2 \leqslant |f| \leqslant f_c + B_T/2$. We write this in polar form by splitting into inphase and quad-phase wrt $2\pi f_ct$. $n(t) = n_I \cos(2\pi f_ct) - n_Q \sin(2\pi f_ct)$ such that we get a phase $\phi_n = \tan^{-1} (n_Q/n_I) $. Is this phase uniformly distributed on $[ 0, 2\pi ]$? If yes, why?

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...And now for a differing opinion....

The OP's representation of bandpass white noise as $$n(t) = n_I \cos(2\pi f_ct) - n_Q \sin(2\pi f_ct)\tag{1}$$ is inadequate; because each sample path of this noise process is a pure sinusoid of fixed frequency $f_c$ Hz which is not noise-like at all. Why so? Well, a sample path is what one gets when all the random variables are assumed to have taken on appropriate values, and so with $n_I = \frac{\sqrt{3}}{2}$ and $n_Q = \frac 12$ (say), we get the sample path $$\frac{\sqrt{3}}{2}\cos(2\pi f_ct) - \frac 12 \sin(2\pi f_ct) = \cos\left(2\pi f_ct+\frac{\pi}{6}\right)$$ which is a pure sinusoid as claimed. A better representation of $n(t)$ is $$n(t) = n_I(t) \cos(2\pi f_ct) - n_Q(t) \sin(2\pi f_ct)\tag{2}$$ where $\{n_I(t)\}$ and $\{n_Q(t)\}$ are low-pass uncorrelated white noise processes with identical power spectral densities that have constant value for $|f| \leq \frac{B_T}{2}$. Note that a typical sample path from $(2)$ is "sort of" sinusoidal-looking at frequency approximately $f_c$ Hz, but the instantaneous amplitude and instantaneous phase (as well as instantaneous frequency (derivative of instantaneous phase) are "slowly varying" functions of time. Here, "slowly varying" means in comparison to the approximate time period $f_c^{-1}$ of each cycle of the sinusoid.


All this is fine and dandy but has nothing to do with the question being considered: the distribution of the instantaneous phase $\operatorname{atan2}(n_Q(t),n_I(t))$. The WSS theory tells us nothing about the distributions of the random variables $n_Q(t)$ and $n_I(t)$, except, of course, when $n(t)$ is a Gaussian process in which case $\{n_I(t)\}$ and $\{n_Q(t)\}$ are independent Gaussian processes and we know for sure that $\operatorname{atan2}(n_Q(t),n_I(t))$ is indeed uniformly distributed on $[0,2\pi)$. But, I contend that since we are discussing white noise processes, any distribution other than $\mathcal U(0,2\pi)$ for the phase would mean that the noise exhibits a preference for some phases and a dislike for some other phases, and this idea sticks in my craw as being totally unrepresentative of white noise. Indeed, even the answer by @StanleyPawlukiewicz that $\arctan(n_Q(t)/n_I(t))$ is uniformly distributed on $[-\pi/2,\pi/2]$ is disturbing because it suggests that it might be the case that $n_I(t)$ is always positive while $n_Q(t)$ has a more symmetric distribution. We really need $\operatorname{atan2}(n_Q(t),n_I(t))$ for defining the phase; $\arctan$ is inadequate for the job.

So, there you have it, folks, The phase is uniformly distributed on $[0,2\pi)$ because it offends my sensibilities to have any other distribution for the phase, and that's all there is to it; ymmv.

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  • $\begingroup$ Well presented sensibilities! :-) $\endgroup$ – Peter K. Jul 23 at 23:38
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$$ -\frac{\pi}{2}\le \mathrm{arctan}(x)\le \frac{\pi}{2} \quad x\in (-\infty , \infty) $$

so as asked, no.

Perhaps $$ Y(f)=|H(f)|^2 X(f) $$ where $H(f)$ is your narrow band filter, would imply that the random phase of the input $x(t)$ is the same nature at the output.

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