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Optimum detector; matched filter impulse response

Short answer

An optimum detector minimizes error probability by maximizing output SNR with a matched filter. For a known pulse g(t), the filter impulse response is h(t)=k g(T-t), i.e., the time-reversed and delayed input signal, derived from the Schwarz inequality.

What do you mean by optimum detector? Show that the impulse response of the matched filter is reverse delayed version of the input signal.

Optimum detector

An optimum detector is a receiver that makes decisions with the minimum probability of error for given signals and noise. In AWGN this is achieved by first passing the received signal through a filter that maximises the output SNR at the sampling instant (the matched filter or an equivalent correlator), then comparing the sample with a threshold (maximum-likelihood decision).

Matched filter impulse response

Consider a known pulse g(t)g(t), 0≤t≤T0\le t\le T, plus white Gaussian noise w(t)w(t) of power spectral density N0/2N_0/2, applied to a linear time-invariant filter with impulse response h(t)h(t) and transfer function H(f)H(f).

 x(t) = g(t) + w(t) --> [ h(t) ] --> y(t) = go(t) + n(t)
                                         |
                                     sample at t = T

Signal output at the sampling instant:

go(T)=∫−∞∞H(f)G(f)ej2πfTdfg_o(T)=\int_{-\infty}^{\infty}H(f)G(f)e^{j2\pi fT}df

Noise output power (white noise through H(f)H(f)):

E[n2(t)]=N02∫−∞∞∣H(f)∣2dfE[n^2(t)]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df

Peak pulse SNR to be maximised:

η=∣go(T)∣2E[n2(t)]=∣∫H(f)G(f)ej2πfTdf∣2N02∫∣H(f)∣2df\eta=\frac{\lvert g_o(T)\rvert^2}{E[n^2(t)]}=\frac{\left|\int H(f)G(f)e^{j2\pi fT}df\right|^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df}

Schwarz inequality: for any ϕ1(f),ϕ2(f)\phi_1(f),\phi_2(f),

∣∫ϕ1ϕ2 df∣2≤∫∣ϕ1∣2df∫∣ϕ2∣2df\left|\int\phi_1\phi_2\,df\right|^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df

with equality only when ϕ1(f)=k ϕ2∗(f)\phi_1(f)=k\,\phi_2^*(f). Put ϕ1=H(f)\phi_1=H(f) and ϕ2=G(f)ej2πfT\phi_2=G(f)e^{j2\pi fT}:

∣∫H(f)G(f)ej2πfTdf∣2≤∫∣H(f)∣2df∫∣G(f)∣2df\left|\int H(f)G(f)e^{j2\pi fT}df\right|^2\le\int\lvert H(f)\rvert^2df\int\lvert G(f)\rvert^2df

Substituting into η\eta:

η≤2N0∫−∞∞∣G(f)∣2df=2EN0\eta\le\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert G(f)\rvert^2df=\frac{2E}{N_0}

using Rayleigh's energy theorem E=∫∣G(f)∣2df=∫g2(t)dtE=\int\lvert G(f)\rvert^2df=\int g^2(t)dt.

Optimum filter: the maximum ηmax=2E/N0\eta_{max}=2E/N_0 is reached when equality holds:

Hopt(f)=k G∗(f) e−j2πfTH_{opt}(f)=k\,G^*(f)\,e^{-j2\pi fT}

Impulse response: take the inverse Fourier transform:

hopt(t)=k∫−∞∞G∗(f)e−j2πf(T−t)df=k[∫−∞∞G(f)ej2πf(T−t)df]∗(real g)=k g(T−t)\begin{aligned} h_{opt}(t)&=k\int_{-\infty}^{\infty}G^*(f)e^{-j2\pi f(T-t)}df\\ &=k\left[\int_{-\infty}^{\infty}G(f)e^{j2\pi f(T-t)}df\right]^*\quad(\text{real }g)\\ &=k\,g(T-t) \end{aligned}

So the impulse response of the optimum filter is the time-reversed and delayed (by TT) version of the input signal, scaled by kk. Because it is "matched" to the signal, it is called the matched filter.

  g(t)                     h(t) = g(T - t)
   |\                          /|
   | \                        / |
   |  \                      /  |
   +---+-----> t        +---+---+----> t
   0   T                0       T
  (flip about t = 0, then shift right by T)

Properties

  • Maximum output SNR =2E/N0=2E/N_0 depends only on the signal energy, not on its shape.
  • ∣Hopt(f)∣=k∣G(f)∣\lvert H_{opt}(f)\rvert=k\lvert G(f)\rvert: the filter passes strongly where the signal spectrum is strong.
  • The output go(t)g_o(t) is proportional to the autocorrelation of g(t)g(t) shifted by TT, peaking at t=Tt=T; hence a matched filter is equivalent to a correlator (multiply by g(t)g(t) and integrate over 00 to TT).
  • The delay TT makes h(t)h(t) causal (zero for t<0t<0).

✓ Answer checked by Claude Opus 5.5 (high effort)

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