Noise Voltage and Power Distributions

Noise Voltage

The voltage $V$ of random noise has a Gaussian probability distribution
$$P(V) = {1 \over (2 \pi)^{1/2} \sigma} \exp \biggl( {-V^2 \over 2 \sigma^2} \biggr) ~,$$
where $P(V)dV$ is the differential probability that the voltage will be within the infinitesimal range $V$ to$V + dV$ and $\sigma$ is the root mean square (rms) voltage.  The probability of measuring some voltage must be unity; that is, any probability distribution must be normalized to unity:
$$\int_{-\infty}^\infty P(V)dV = 1$$ To confirm the normalization of our noise distribution, we evaluate the integral
$$\int_{-\infty}^\infty {1\over (2 \pi)^{1/2} \sigma} \exp \biggl( {-V^2 \over 2 \sigma^2} \biggr) dV $$ $$= 2 \int_0^\infty {1\over (2 \pi)^{1/2} \sigma} \exp \biggl( {-V^2 \over 2 \sigma^2} \biggr) dV $$ $$= {2^{1/2} \over \pi^{1/2} \sigma} \int_0^\infty \exp \biggl( {-V^2 \over 2 \sigma^2} \biggr) dV $$ Using the definite integral $$\int_0^\infty \exp(-a^2 x^2) dx = {\pi^{1/2} \over 2a}$$ with $a^2 = (2 \sigma^2)^{-1}$ we get the desired result $$\int_{-\infty}^\infty P(V)dV = {2^{1/2} \over \pi^{1/2} \sigma} {\pi^{1/2} 2^{1/2}\sigma \over 2} = 1$$

The rms (root mean square) $\Sigma$ of a normalized distribution is defined by $$\Sigma^2 = \langle V^2 \rangle - \langle V \rangle^2$$ For the symmetric Gaussian distribution, $\langle V \rangle = 0$ so $$\Sigma^2 = \langle V^2 \rangle = \int_{-\infty}^\infty V^2 P(V) dV$$ $$\Sigma^2 = 2\int_0^\infty V^2 {1 \over (2 \pi)^{1/2}\sigma} \exp\biggl({-V^2 \over 2\sigma^2}\biggr) dV$$  $$\Sigma^2 = {2^{1/2} \over \pi^{1/2} \sigma} \int_0^\infty V^2 \exp \biggl( {-V^2 \over 2 \sigma^2}\biggr) dV~.$$ Using the definite integral $$\int_0^\infty x^2 \exp(-ax^2) dx = {1 \over 4a} \biggl( {\pi \over a} \biggr)^{1/2}$$ with $a = (2 \sigma^2)^{-1}$ yields $$\Sigma^2 = {2^{1/2} \over \pi^{1/2} \sigma} {2 \sigma^2 \over 4} 2^{1/2} \sigma \pi^{1/2} = \sigma^2~,$$ demonstrating that $\sigma$ is really the rms.

Noise Power

A square-law detector multiplies the input voltage $V$ by itself to yield an output voltage $V_{\rm o} = V^2$ that is proportional to the input power. What is the probability distribution $P_{\rm o}(V_{\rm o})$ of detector output voltage when the input voltage distribution is Gaussian?  For simplicity, we set $\sigma = 1$.  The same value of $V_{\rm o}$ is produced by both positive and negative values of $V$ and the probability distribution of $V$ is symmetric, so $$P_{\rm o} (V_{\rm o}) dV_{\rm o} = 2 P(V) dV$$ for $V \geq 0$.   Since $dV_{\rm o} = 2 V dV$, $dV / dV_{\rm o} = V_{\rm o}^{-1/2}/2$ and $$P_{\rm o} (V_{\rm o}) = {1 \over (2 \pi)^{1/2}} V_{\rm o}^{-1/2} \exp (-V_{\rm o}/ 2)$$ for $0 \leq V_{\rm o} < \infty$.  Notice that the distribution of detector output voltages is sharply peaked near $V_{\rm o} = 0$ and has a long exponentially decaying tail, so it looks quite different from a Gaussian distribution.

To confirm that $P_{\rm o}$ is properly normalized we evaluate $$\int_0^\infty P_{\rm o} (V_{\rm o}) d V_{\rm o} = {1 \over (2 \pi)^{1/2}} \int_0^\infty V_{\rm o}^{-1/2} \exp (-V_{\rm o} / 2) d V_{\rm o}$$ using the definite integral $$\int_0^\infty x^n \exp(-ax) dx = {\Gamma(n+1) \over a^{n+1}}~,$$ where $\Gamma$ is the Gamma function, $\Gamma(1/2) = \pi^{1/2}$, and $\Gamma(n+1) = n\Gamma(n)$.  Subsituting $n = -1/2$ and $a = 1/2$ yields the correct result $$\int_0^\infty P(V_{\rm o}) dV_{\rm o} = {1 \over (2 \pi)^{1/2}} {\Gamma(1/2) \over (1/2)^{1/2}} = {1 \over (2 \pi)^{1/2}} \pi^{1/2} 2^{1/2} = 1 ~.$$

The mean detector output voltage is $$\langle V_{\rm o} \rangle = \int_0^\infty V_{\rm o} P_{\rm o} (V_{\rm o}) d V_{\rm o}$$ $$ = \int_0^\infty {V_{\rm o} \over (2 \pi)^{1/2}}V_{\rm o}^{-1/2} \exp(-V_{\rm o} / 2) d V_{\rm o}$$ $$={1 \over (2 \pi)^{1/2}} \int_0^\infty V_{\rm o}^{1/2} \exp(-V_{\rm o} / 2) d V_{\rm o}~.$$
Using the definite integral above with $n = 1/2$ and $a = 1/2$ yields $$\langle V_{\rm o}\rangle =  {1 \over (2 \pi)^{1/2}} {\Gamma(3/2) \over (1/2)^{3/2}}~.$$ $\Gamma(3/2) = (1/2)\Gamma(1/2) = \pi^{1/2}/2$ so
$$\langle V_{\rm o} \rangle = {1 \over (2 \pi)^{1/2}} {\pi^{1/2} \over 2} 2^{3/2} = 1~.$$ The average detector output voltage is nonzero; it equals the average input power.  Had we allowed $\sigma \neq 1$ we would have gotten $\langle V_{\rm o} \rangle = \sigma^2$.

What is the rms $\Sigma_0$ of the detector output voltage?
$$\Sigma_0^2 = \langle V_{\rm o}^2 \rangle - \langle V_{\rm o} \rangle^2$$ so we must evaluate $$\langle V_{\rm o}^2 \rangle = \int_0^\infty V_{\rm o}^2 P_{\rm o}(V_{\rm o}) d V_{\rm o}$$ $$={1 \over (2 \pi)^{1/2}} \int_0^\infty V_{\rm o}^2 V_{\rm o}^{-1/2} \exp(-V_{\rm o} / 2) dV_{\rm o}$$ $$= {1 \over (2 \pi)^{1/2}} \int_0^\infty V_{\rm o}^{3/2} \exp(-V_{\rm o} / 2) dV_{\rm o}~.$$ Using the definite integral above with $n = 3/2$ and $a = 1/2$ yields $$\langle V_{\rm o}^2 \rangle = {1 \over (2 \pi)^{1/2}} {\Gamma(5/2) \over (1/2)^{5/2}}~.$$ $\Gamma(5/2) = (3/2)\Gamma(3/2) =  3 \pi^{1/2}/4$ so $$\langle V_{\rm o}^2 \rangle = {1 \over (2\pi)^{1/2}} 2^{5/2} {3 \pi^{1/2} \over 4} = 3$$ and $$\Sigma_{\rm o}^2 = \langle V_{\rm o}^2 \rangle - \langle V_{\rm o} \rangle^2 = 3 - 1 = 2~.$$

Thus the rms $\Sigma_{\rm o} = 2^{1/2}$ of the detector output voltage is $2^{1/2}$ times the mean output voltage.  [If we had kept track of $\sigma \neq 1$, we would have gotten $\Sigma_{\rm o} = 2^{1/2} \sigma^2$.] The rms uncertainty in each independent sample of the measured noise power is $2^{1/2}$ times the mean noise power.  If $N \gg 1$ independent samples are averaged, the fractional rms uncertainty of the averaged power is $(2/N)^{1/2}$.  This result is the heart of the radiometer equation.  According to the central limit theorem, the distribution of these averages approaches a Gaussian as $N$ becomes large.