Consider an ideal digital optical communications system using on-off keying (OOK). A constant optical power incident on a photodiode generates carrier pairs in independent random events. The probability of detecting L photons in a time period T obeys the Poisson distribution: where N is the expected number of photons in this period. Assume that the variation in numbers of generated carriers per time interval is the only source of noise. (i) Choose an optimum decision threshold for the receiver; (ii) (iii) Deduce the corresponding expression for bit error probability; Hence calculate the quantum limit for a bit error probability of Pe = 10. P P(L) = [8 marks] (b) Consider the situation in which the system in part (a) is operated under ASK, so that during transmission of a binary "0", the average number of incident photons in one bit period is No and for a binary "1" it is N, where N > No. If the decision threshold is set to a, it can be shown that the error probability is given by: = 1 NL e-N L! 2 _n=a+l NO n! -No e + n=0 N" n! The optimum value of a will be that which minimises the value of Pe such that AP = P(a) - P(a 1)~0 Use this information to obtain an expression for the optimum value of a in terms of No and N. [7 marks] (c) Using the expression from part (b), calculate a for a system operating at Br = 100 Mb/s and = 800 nm. Binary "zeros" and "ones" are represented by an optical power of 1 nW and 5 nW respectively. The photodetector has a quantum efficiency of n = 0.9 and a dark current of ip = 1.5 nA.