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Matched filter approximation using RC low pass filter

Short answer

A matched filter maximizes peak SNR for a known signal in white noise, with impulse response h(t)=k s(T-t). For a rectangular pulse, an RC low-pass filter approximates the integrate-and-dump; optimum 3 dB bandwidth f_c ≈ 0.2/T gives SNR within 0.89 dB of the matched filter.

Define Matched filter. Explain the approximation of the matched filter for a rectangular pulse using a single pole RC low pass filter with variable bandwidth.

Matched filter

A matched filter is a linear filter whose impulse response is matched to (a time-reversed, delayed copy of) a known input pulse s(t)s(t), h(t)=k s(T−t)h(t)=k\,s(T-t), so that it gives the maximum possible peak signal-to-noise ratio at the sampling instant t=Tt=T when the pulse is received in additive white noise. It is the optimum detector for digital signals in AWGN; in practice it is built as a correlator.

RC low-pass filter as approximate matched filter

The exact matched filter for a rectangular pulse is an integrate-and-dump circuit, which is not always convenient. A simple single-pole RC low-pass filter can approximate it; we find the best bandwidth and the loss.

Input pulse: s(t)=As(t)=A for 0≤t≤T0\le t\le T (energy E=A2TE=A^2T), white noise PSD N0/2N_0/2.

Filter: H(f)=11+jf/fcH(f)=\dfrac{1}{1+jf/f_c}, fc=12πRCf_c=\dfrac{1}{2\pi RC} (3 dB bandwidth, variable).

Signal output. The step response of the RC filter rises as A(1−e−t/RC)A(1-e^{-t/RC}); it is maximum at the end of the pulse, t=Tt=T:

so(T)=A(1−e−T/RC)=A(1−e−2πfcT)s_o(T)=A\left(1-e^{-T/RC}\right)=A\left(1-e^{-2\pi f_cT}\right)
 s(t)                 s_o(t)
 A +-----+          A |      ..-.  peak at t=T
   |     |            |    .'    '.
   |     |            |  .'        '..
 --+-----+--> t     --+-'-----+-------'--> t
   0     T            0       T

Noise output power:

No=N02∫−∞∞df1+(f/fc)2=N02⋅πfcN_o=\frac{N_0}{2}\int_{-\infty}^{\infty}\frac{df}{1+(f/f_c)^2}=\frac{N_0}{2}\cdot\pi f_c

Output SNR:

SNRo=A2(1−e−2πfcT)2πN0fc/2=2A2TN0⋅2(1−e−x)2x,x=2πfcT\text{SNR}_o=\frac{A^2\left(1-e^{-2\pi f_cT}\right)^2}{\pi N_0f_c/2}=\frac{2A^2T}{N_0}\cdot\frac{2\left(1-e^{-x}\right)^2}{x},\qquad x=2\pi f_cT

Compared with the matched filter (SNRmax=2E/N0=2A2T/N0\text{SNR}_{max}=2E/N_0=2A^2T/N_0):

ρ(x)=SNRoSNRmax=2(1−e−x)2x\rho(x)=\frac{\text{SNR}_o}{\text{SNR}_{max}}=\frac{2(1-e^{-x})^2}{x}

Effect of variable bandwidth:

  • fcf_c too small: the filter cannot follow the pulse, so(T)s_o(T) is small, SNR low.
  • fcf_c too large: signal fully passes but much extra noise enters, SNR low.
  • An optimum lies between. Setting dρ/dx=0d\rho/dx=0 gives (1+2x)e−x=1(1+2x)e^{-x}=1, solved numerically: x≈1.256x\approx1.256.
fcT=1.2562π≈0.2,ρmax=2(1−e−1.256)21.256≈0.815f_cT=\frac{1.256}{2\pi}\approx0.2,\qquad \rho_{max}=\frac{2(1-e^{-1.256})^2}{1.256}\approx0.815

Result: the best RC filter has fc≈0.2Tf_c\approx\dfrac{0.2}{T} (i.e. RC≈0.8TRC\approx0.8T) and gives an SNR only 10log⁡100.815≈−0.8910\log_{10}0.815\approx-0.89 dB, i.e. less than 1 dB below the ideal matched filter. Hence a simple RC filter is a good practical substitute for the matched filter of a rectangular pulse.

✓ Answer checked by Claude Opus 5.5 (high effort)

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