6.16 Digital communication in the presence of noise

When we incorporate additive noise into our channel model, so that $r(t)=\alpha s_i(t)+n(t)$ , errors can creep in. If the transmitter sent bit 0 using a BPSK signal set , the integrators' outputs in the matched filter receiver would be:

It is the quantities containing the noise terms that cause errors in the receiver's decision-making process. Because theyinvolve noise, the values of these integrals are random quantities drawn from some probability distribution that varyerratically from bit interval to bit interval. Because the noise has zero average value and has an equal amount of power in allfrequency bands, the values of the integrals will hover about zero. What is important is how much they vary. If the noise issuch that its integral term is more negative than $\alpha A^{2}T$ , then the receiver will make an error, deciding that the transmitted zero-valued bit was indeed a one. Theprobability that this situation occurs depends on three factors:

• Signal Set Choice — The difference between the signal-dependent terms in the integrators'outputs (equations [link] ) defines how large the noise term must be for an incorrect receiverdecision to result. What affects the probability of such errors occurring is the energy in the difference of the received signals incomparison to the noise term's variability. The signal-difference energy equals $$\int_0^T (s_1(t)-s_0(t))^2\,d t$$ For our BPSK baseband signal set, the difference-signal-energy term is $4\alpha ^{2}A^{4}T^2$ .
• Variability of the Noise Term — We quantify variability by the spectral height of the white noise $\frac{N_0}{2}$ added by the channel.
• Probability Distribution of the Noise Term — The value of the noise terms relative to the signal terms and the probability of their occurrencedirectly affect the likelihood that a receiver error will occur. For the white noise we have been considering, theunderlying distributions are Gaussian. Deriving the following expression for the probability the receiver makes an error on any bit transmission is complicated but can be found at [link] and [link] .
  $p_e=Q(\sqrt{\frac{\int_0^T (s_1(t)-s_0(t))^2\,d t}{2N_0}})=Q(\sqrt{\frac{2\alpha ^{2}A^{2}T}{N_0}})\text{ for the BPSK case}$
  Here $Q(·)$ is the integral $Q(x)=\frac{1}{\sqrt{2\pi }}\int \,d \alpha$∞ x α 2 2 .This integral has no closed form expression, but it can be accurately computed. As [link] illustrates, $Q(·)$ is a decreasing, very nonlinear function.

The term $A^{2}T$ equals the energy expended by the transmitter in sending the bit; we label this term $E_b$ . We arrive at a concise expression for the probability the matched filter receiver makes a bit-receptionerror.