3.2.1 PPM signaling on the noisy photon counting channel

The capacity is determined by Eq .(1) and the probability of correct uncoded PPM detection $p$ from the photon counter. Let Pois$(i,n) = n^i e^{-n} / i!$ denote the probability that a Poisson random variable with mean $n$ takes on the value $i$, $i = 0,1,\ldots$. The probability that an uncoded $M$-PPM symbol is detected correctly is given by

$$p =$$ Pr(uncoded PPM symbol is detected correctly)

$$= {1 \over M} \left(\parbox{1in}{\raggedright 0 photons detected in all slots}\... ...ggedright less than $i$\ photons detected in all $M-1$\ nonsignal slots}\right)$$

$$\quad + \sum_{i=1}^\infty \sum_{r=1}^{M-1} {1 \over r+1} {M-1 \choose r} \left(\parbox{0.8in}{\raggedright $i$\ photons detected in signal... ...{\raggedright less than $i$photons detected in $M-1-r$\ nonsignal slots}\right)$$

$$= {e^{-(\eta\bar n_s+M\eta\bar n_b)} \over M} + \sum_{i=1}^\infty (i,\eta\bar n_s+\eta\bar n_b)\left[\sum_{m=0}^{i-1} \mbox{Pois}(m,\eta\bar n_b)\right]^{M-1}$$

$$\quad + \sum_{i=1}^\infty \sum_{r=1}^{M-1} {1 \over r+1} {M-1 \choose r} (i,\eta\bar n_s+\eta\bar n_b) \left(\mbox{Pois}(i,\eta\bar n_b)\right)^r \left[\sum_{m=0}^{i-1} \mbox{Pois}(m,\eta\bar n_b)\right]^{M-1-r}.$$

After some straightforward algebra, this simplifies to [Gag95,Che92]:

$$p = {1 \over M} e^{-(\eta\bar n_s+M\eta\bar n_b)} + \sum_{k=1}^\i... ...}^{k-1} \mbox{Pois}(m,\eta\bar n_b)\right)^{M-1} {((1+a)^M-1) \over aM}\right],$$ (6)

where $a =$   Pois$(i,\eta\bar n_b) / \sum_{m=0}^{i-1}$   Pois$(m,\eta\bar n_b)$. Eq. (6) may be plugged into Eq. (1) to compute the capacity of PPM when an ideal photon counter is used.