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Signal Theory

Detection: threshold, matched filter and error probability

Signal-to-noise ratio, deciding between two symbols, error probability, the matched filter with proof, BPSK and Eb/N0, error-correcting codes, Nyquist pulses and the eye diagram.

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A digital receiver must make a decision: having observed a signal disturbed by noise, which symbol was sent? It will be wrong now and then, and the design consists of making errors as rare as possible. In this lesson we see how the decision is made, which filter maximises the signal-to-noise ratio and how to compute the error probability of a binary transmission.

Signal-to-noise ratio

The quality of a received signal is measured by the signal-to-noise ratio, the ratio of signal power to noise power:

    \[ \mathrm{SNR} = \frac{P_S}{P_N},\qquad \mathrm{SNR}_{dB} = 10\log_{10}\frac{P_S}{P_N} \]

What matters is not how strong the signal is in absolute terms, but how strong it is compared with the noise. Amplifying does not help: it amplifies both, and adds the amplifier’s own noise.

Deciding between two symbols

In an antipodal binary transmission +A is sent for bit 1 and -A for bit 0. The receiver observes r = \pm A + n, with n Gaussian with zero mean and standard deviation \sigma.

Binary detection in Gaussian noise: the two densities of the received sample and the threshold
The two curves are the densities of the received value when −A is sent (symbol 0) and when +A is sent (symbol 1). The threshold is halfway, at zero. The coloured tails beyond the threshold are the error probabilities: a 1 mistaken for a 0 and vice versa.

If the two bits are equally likely, the best rule (the maximum likelihood decision, which follows from Bayes’ theorem) is to choose the symbol whose density is higher at the observed value: for two equal, symmetric Gaussians, the threshold is halfway, at zero. Decide 1 if r \gt 0, 0 otherwise.

Error probability. An error occurs when noise pushes the sample beyond the threshold: with +A sent, when n \lt -A. By the symmetry of the Gaussian,

    \[ P_e = P(n \gt A) = Q\!\left(\frac{A}{\sigma}\right) \]

with the Q function from the probability lesson. It depends only on the ratio of signal amplitude to noise amplitude.

The matched filter

The real signal is not a single number but a pulse s(t) of duration T, buried in white noise. Before sampling and deciding, the receiver passes it through a filter h(t). Which filter maximises the SNR at the sampling instant T?

Theorem (matched filter). With white noise of density N_0/2, the SNR at instant T is maximised by

    \[ h(t) = s(T - t) \]

and the maximum value is \mathrm{SNR}_{max} = \frac{2E}{N_0}, where E = \int s^2\,dt is the pulse energy.

Proof. At instant T the useful output signal is y_s(T) = \int h(\tau)s(T - \tau)\,d\tau, the noise power is \frac{N_0}{2}\int h^2\,d\tau (lesson on random processes). So

    \[ \mathrm{SNR} = \frac{\left(\int h(\tau)\,s(T - \tau)\,d\tau\right)^2}{\frac{N_0}{2}\int h^2(\tau)\,d\tau} \le \frac{\int h^2\,d\tau\int s^2(T - \tau)\,d\tau}{\frac{N_0}{2}\int h^2\,d\tau} = \frac{2E}{N_0} \]

by the Cauchy-Schwarz inequality. Equality holds when h(\tau) is proportional to s(T - \tau). ∎

The matched filter is correlation with the expected pulse (lesson on correlation): it measures how much the received signal resembles what it expects. The surprising result is that the maximum SNR depends only on the pulse energy, not on its shape: for detection, different pulses with the same energy are equivalent.

Bit error probability

Combining the two results for antipodal transmission (BPSK) with energy per bit E_b:

    \[ P_e = Q\!\left(\sqrt{\frac{2E_b}{N_0}}\right) \]

E_b/N_0P_e
4 dB1.25\cdot 10^{-2}
7 dB7.7\cdot 10^{-4}
9.6 dB1\cdot 10^{-5}

The probability falls exponentially: a few more decibels take you from one error every hundred bits to one every hundred thousand. The ratio E_b/N_0, energy per bit divided by the noise density, is the right way to compare systems running at different speeds.

Errors in practice

  • Error-correcting codes. By adding suitably computed redundancy bits, the receiver corrects errors. Modern codes (LDPC, polar, turbo) achieve the same error probability with several decibels less, and make it possible to approach Shannon’s theoretical limit.
  • Intersymbol interference. If the channel spreads the pulses, each symbol “smears” its neighbours. Nyquist pulses (such as the raised cosine) are designed to be zero at the sampling instants of the other symbols, and allow 2B symbols per second in a bandwidth B. On an oscilloscope, overlaying segments of the received signal gives the eye diagram: the wider the eye is open, the easier the decision.
  • Synchronisation. All of this assumes knowing when to sample: the receiver must recover the symbol timing from the signal itself, one of the most delicate tasks in a modem.

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