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Impulse response of matched filter

Short answer

Maximizing the output SNR at sampling instant T via the Schwarz inequality yields H_opt(f)=k G*(f) e^{-j2πfT}. The impulse response is h_opt(t)=k g(T-t), the time-reversed and delayed version of the input pulse, called the matched filter.

Derive the expression for the impulse response of a matched filter for a AWGN channel.

A matched filter is the linear filter that maximises the peak output signal-to-noise ratio at the sampling instant when a known pulse is received in additive white Gaussian noise (AWGN). It is used as the optimum detector in digital receivers.

Derivation

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).

Example

For a rectangular pulse g(t)=Ag(t)=A, 0≤t≤T0\le t\le T, the matched filter is h(t)=kAh(t)=kA for 0≤t≤T0\le t\le T (an integrate-and-dump filter). The output is a triangle peaking at t=Tt=T with value kA2T=kEkA^2T=kE, and ηmax=2A2T/N0\eta_{max}=2A^2T/N_0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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