 BPSK, there are two ways an error can happen: The demodulator thinks you sent s((), but in fact you sent o(t), or the demodulator thinks you sent (t), but in fact you sent S1 (t). Writing this statistically, we have P(e)=P(output s,(t)sends(t)).P(send = So(t)) + P(output so(t)|send s,(t)) P(send (t)) (5.68) We can simplify this equation by making a simple realization: the symbols are equally likely, which means the P(sent S1 (t)) = P(sent o(t)) = 0.5. Plugging this information into our P(E) equation leads us to 0.5 [P(output (t) send So(t)) + P(output so(t)|send s,(t))] (5.69) Now, let's see if we can figure out another value for P (output (t) I sent s,(t)). Take a look at Figure 5.32(a). There, we sent (t), which corresponds to sending S11. The decision device picks up This decision device makes a decision according to the rule "Choose the symbol which the received r1 is closest to." Look at what happens in Figure 5.32(b), where N1 is bigger than . In that case, the r1 will be closest to S01, and the demodulator will decide S01 (that is, so(t)). If N1 is bigger than A/2, the demodulator will make an error. So, the P (output so(t) I sent s,(t)) is equal to p(n,>A[%). Chapter Five sent this 01(t) S11=-AV(T/2) So1=AV(T/2) r1=S11+n1 sent this 01(t) S11=-AV(T/2) S01=AV(T/2) r1=S11+n1 n1>AV(T/2) Figure 5.32 Sent and received signals in BPSK Similarly, we can show that P (output sq(t) sent so(t)) is equal to p (n,>AV%). Putting this information to good use, we plug it into our P(E) equation, which leads us to (5.70) The good news here is that the equation just gets simpler. One of the neat proper- ties of a Gaussian random variable is that, if X is a Gaussian random variable, p (x>A) and p (x<-A) are the same. Putting this information to use in the P(E) equation leads us to (5.71) (5.72) Getting It from Here to There: Modulators and Demodulators Now, we just make a simple substitution, namely to let U = no/ Substituting this into the integration brings us to (5.73) (5.74) where the Q(x) function is just shorthand nota