Noise equivalent bandwidth; matched filter is time reversed

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 $\lvert H(f_0)\rvert$.

Example: RC low-pass filter.

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: ideal filter. For an ideal LPF of bandwidth $B$, $B_N=B$ exactly; for practical filters with steep skirts (e.g. high-order Butterworth), $B_N$ approaches the 3 dB bandwidth.

Impulse response of matched filter is time-reversed, delayed input

Set-up. The received signal is $x(t)=s(t)+w(t)$, $0\le t\le T$, where $s(t)$ is a known pulse and $w(t)$ is white noise of two-sided PSD $N_0/2$. It passes through an LTI filter $h(t)\leftrightarrow H(f)$ and is sampled at $t=T$:

 s(t)+w(t)    +--------+   y(t)    sample     y(T)
 ----------->|  h(t)  |---------->--/ ---->  decision
              +--------+           at t=T

Output $y(t)=s_o(t)+n_o(t)$ with

$$s_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df$$

We want $H(f)$ that maximises the output peak signal-to-noise ratio

$$\eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}$$

Schwarz inequality: $\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df$, with equality only when $\phi_1(f)=k\,\phi_2^*(f)$. Put $\phi_1=H(f)$ and $\phi_2=S(f)e^{j2\pi fT}$:

$$\eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0}$$

(using Rayleigh's theorem, $E$ = pulse energy). The maximum is reached when

$$H_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT}$$

Impulse response. Take the inverse transform. For real $s(t)$, $S^*(f)=S(-f)$, which is the transform of $s(-t)$; the factor $e^{-j2\pi fT}$ is a delay of $T$:

$$\begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned}$$

So the optimum (matched) filter impulse response is the input pulse reversed in time and delayed by $T$. Choosing the sampling instant $T$ equal to the pulse duration makes $h(t)$ causal.

   s(t)                      h(t) = s(T - t)
    |\                          /|
    | \                        / |
    |  \                      /  |
 ---+---+---> t          ---+---+---> t
    0   T                   0   T

Results: maximum SNR $=2E/N_0$ depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it gives the most weight to the frequencies where the signal is strongest.