Noise equivalent bandwidth of RC low pass filter

Noise equivalent bandwidth

The noise equivalent bandwidth (NEB) $B_N$ of a filter is the bandwidth of an ideal rectangular filter, with the same peak (usually zero-frequency or centre) gain $\lvert H(0)\rvert$, that passes the same total noise power as the actual filter when both are fed with white noise.

White noise $N_0/2$ into $H(f)$ gives output power

$$P_{actual}=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df=N_0\int_0^{\infty}\lvert H(f)\rvert^2df$$

The ideal filter of gain $\lvert H(0)\rvert$ and bandwidth $B_N$ gives $P_{ideal}=N_0B_N\lvert H(0)\rvert^2$. Equating:

$$B_N=\frac{\int_0^{\infty}\lvert H(f)\rvert^2df}{\lvert H(0)\rvert^2}$$

 |H(f)|^2
 |H(0)|^2 +---------+     equal areas
          |  ...    |
          |      .. |
          |        .|..
          |         |  ......
        --+---------+----------> f
          0        B_N

NEB lets us write the output noise power simply as $N=N_0B_N\lvert H(0)\rvert^2$ (or $kTB_N$ per unit gain), which is used in noise figure and SNR calculations. For a band-pass filter $\lvert H(0)\rvert$ is replaced by the centre-frequency gain $\lvert H(f_0)\rvert$.

NEB of first-order RC low-pass filter

  vin o---/\/\/\---+---o vout
           R       |
                  === C
                   |
  gnd o------------+---o

For the RC low-pass filter (series $R$, shunt $C$):

$$H(f)=\frac{1}{1+j2\pi fRC},\qquad \lvert H(f)\rvert^2=\frac{1}{1+(2\pi fRC)^2},\qquad \lvert H(0)\rvert=1$$ $$\begin{aligned} B_N &= \int_0^{\infty}\frac{df}{1+(2\pi fRC)^2}\qquad (\text{let } u=2\pi fRC,\ df=\tfrac{du}{2\pi RC}) \\ &= \frac{1}{2\pi RC}\int_0^{\infty}\frac{du}{1+u^2}=\frac{1}{2\pi RC}\cdot\frac{\pi}{2} \\ &= \frac{1}{4RC} \end{aligned}$$

Since the 3 dB bandwidth is $f_{3dB}=\frac{1}{2\pi RC}$,

$$B_N=\frac{\pi}{2}f_{3dB}\approx1.57\,f_{3dB}$$

The NEB is larger than the 3 dB bandwidth because the RC response rolls off slowly (only $-20$ dB/decade) and passes noise well beyond $f_{3dB}$. Output noise power $=N_0B_N=\frac{N_0}{4RC}$.

Example: $R=1\ \text{k}\Omega$, $C=1\ \mu\text{F}$: $f_{3dB}=159.2$ Hz, $B_N=\frac{1}{4\times10^{-3}}=250$ Hz.