Thermal noise phasor amplitudes are Rayleigh distributed. How are voltages at the antenna distributed?

Question

Background. Consider an ideal antenna. For ideal polarized thermal noise treated as a random phasor sum, bivariate Gaussian statistics apply to the resultant phasor (call it $\vec{v}$) that is observed at any given moment in the antenna. That is, the phasor sum's real and imaginary parts are indepent Gaussian normal variables. The amplitude of said phasor sum thus follows (in some books by definition) the Rayleigh distribution:

$$P_A = {a\over \sigma^2}e^{-a^2/2\sigma^2}$$

...where $a = \sqrt{\mathrm{Re}({\vec{v}}) + \mathrm{Im}({\vec{v}})}$, and the scale parameter $\sigma^2 = \overline{\mathrm{Re( \vec{v}})^2} + \overline{\mathrm{Im( \vec{v}})^2}$, which works out to be the square of the $\mathrm{RMS}$ voltage at the antenna.

---

I'd like to model the actual voltage distribution observed at an ideal antenna due to thermal noise. While there are a multitude of sources describing the physics of the Rayleigh-distributed thermal noise amplitude, most approximate the voltage distribution as a simple Gaussian, which is imperfect (a folded Gaussian differs significantly from a Rayleigh distribution).

Question. How are the voltages observed at the antenna distributed? The amplitudes are Rayleigh-distributed... what does that mean for the actual voltage values? I've found one source that states the following:

In the absence of a signal, the electric field due to [...] thermal noise can be represented as a sum of [complex numbers] whose projections on the real and imaginary axes are independent Gaussian random variables. Summing these phasors and taking the real part to get a realized voltage results in a Gaussian two-sided voltage distribution, or equivalently a Rayleigh [amplitude] distribution.

It isn't clear to me what is meant by a "two-sided Gaussian", nor is it clear to me what parameters that Gaussian would take (e.g. how to find $\sigma$?).

Answer 1

Let $\mathbf z = \mathbf x + \mathfrak j \mathbf y$ where the random variates $\mathbf {x,y}$ are independent and normal distributed with zero mean, that is: $\mathbf {x,y} := \mathcal N (0, \sigma)$. If your signal, say voltage, at the antenna is $s(t)e^{\mathfrak j \omega t}$ and noise bandwidth $W$ at noise temperature $T$ then set $\sigma^2 = k_B T W$ and your sampled signal in the additive noise will be $r_n = s(n/2W) + \mathbf z_n$

---

If you only have noise and measure the noise samples with an I/Q detector then on both arms the pdf of the instantaneous voltage will be $n(x) = \frac{1}{\sqrt{2\pi}} exp\big(\frac {-x^2}{2\sigma^2}\big)$