Moments of random variable; NEB of RC-LPF; output noise power

Moments and central moments

For a continuous random variable $X$ with PDF $f_X(x)$:

• $n$th moment (about the origin):

$$E[X^n] = \int_{-\infty}^{\infty} x^n f_X(x)\,dx$$

The first moment is the mean $\mu = E[X]$; the second is the mean-square value $E[X^2]$.

• $n$th central moment (about the mean):

$$E[(X-\mu)^n] = \int_{-\infty}^{\infty} (x-\mu)^n f_X(x)\,dx$$

The second central moment is the variance $\sigma^2 = E[X^2]-\mu^2$.

First central moment is always zero:

$$\begin{aligned} E[X-\mu] &= \int_{-\infty}^{\infty}(x-\mu)f_X(x)\,dx \\ &= \int_{-\infty}^{\infty}x f_X(x)\,dx - \mu\int_{-\infty}^{\infty}f_X(x)\,dx \\ &= \mu - \mu(1) = 0 \end{aligned}$$

Noise equivalent bandwidth of RC low-pass filter

The NEB $B_N$ is the bandwidth of an ideal LPF, with the same zero-frequency gain $H(0)$, that passes the same white-noise power:

$$B_N = \frac{1}{|H(0)|^2}\int_0^{\infty}|H(f)|^2\,df$$

For the RC-LPF, $H(f) = \dfrac{1}{1+j2\pi fRC}$, so $H(0) = 1$ and

$$\begin{aligned} B_N &= \int_0^{\infty}\frac{df}{1+(2\pi fRC)^2} \\ &= \frac{1}{2\pi RC}\Big[\tan^{-1}(2\pi fRC)\Big]_0^{\infty} \\ &= \frac{1}{2\pi RC}\cdot\frac{\pi}{2} = \frac{1}{4RC} \end{aligned}$$

Since the 3 dB bandwidth is $f_{3dB} = 1/(2\pi RC)$, $B_N = \frac{\pi}{2}f_{3dB} \approx 1.57 f_{3dB}$.

Ideal LPF with $|H(f)| = 1$ for $|f| \le B$ and 0 elsewhere:

$$B_N = \frac{1}{1}\int_0^{B}1\,df = B$$

So the NEB of an ideal LPF equals its actual bandwidth.

Output noise power of the RC-LPF

Input white noise has two-sided PSD $S_X(f) = N_0/2$. Output PSD is $S_Y(f) = |H(f)|^2 N_0/2$, so

$$\begin{aligned} P_{out} &= \int_{-\infty}^{\infty}\frac{N_0}{2}\cdot\frac{df}{1+(2\pi fRC)^2} \\ &= \frac{N_0}{2}\cdot 2B_N = \frac{N_0}{2}\cdot\frac{1}{2RC} = \frac{N_0}{4RC} \end{aligned}$$

Answer: $B_{N,RC} = \dfrac{1}{4RC}$, $B_{N,ideal} = B$, output noise power $= \dfrac{N_0}{4RC}$ (watts, on a 1 $\Omega$ basis).