Optimum detector; matched filter; autocorrelation

Optimum detector

An optimum detector is a receiver (filter followed by sampler and threshold device) that makes the decision with the minimum probability of error. For a known pulse in additive white Gaussian noise, this is achieved by the filter that maximises the output signal-to-noise ratio at the sampling instant; this optimum filter is the matched filter.

Impulse response of matched filter is reverse 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.

Autocorrelation function

The autocorrelation (AC) function measures how similar a signal is to a time-shifted copy of itself. For a power signal or a wide-sense stationary (WSS) random process $X(t)$:

$$R_X(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}x(t)\,x(t+\tau)\,dt \quad\text{or}\quad R_X(\tau)=E[X(t)X(t+\tau)]$$

For an energy signal, $R_g(\tau)=\int_{-\infty}^{\infty}g(t)g(t+\tau)\,dt$.

Main properties: $R(-\tau)=R(\tau)$ (even); $R(0)=$ average power (energy); $\lvert R(\tau)\rvert\le R(0)$; $R(\tau)\leftrightarrow S(f)$ (Wiener–Khinchin). Example: white noise has $R(\tau)=\frac{N_0}{2}\delta(\tau)$. Note that the matched filter output for the signal is $s_o(t)=kR_s(t-T)$, the pulse autocorrelation shifted to $T$, which peaks at $t=T$ because $R_s$ is maximum at zero shift.