2.2.2 Bits per second

Instead of using an enormous value of $M$ and transmitting one symbol, we would be better off transmitting two $(M/2)$-PPM symbols in the same amount of time (assuming $M >> T_d/T_s$), because there is a potential for $2\log_2 (M/2)$ bits received, as opposed to only $\log_2 M$ bits. Neglecting dead time, the capacity of the errorless channel is $\log_2 M/M$ bits per slot, which is maximized when $M=3$. (The noninteger maximum occurs when $M=e$.)

The optimum value of $M$ may be much higher than three when the required dead time is taken into account. On an error-free channel using $M$-PPM, a slot time of $T_s$ and a laser dead time of $T_d$, the capacity in bits per second is

$$C = {\log_2 M \over MT_s + T_d}$$   bits/second$$.$$

$M$ may be chosen to maximize this equation. For the laser used in this report, $T_s = 3.125\times 10^{-8}$ seconds and $T_d = 4.32\times 10^{-4}$ seconds, and an errorless channel capacity is optimized when $M=2082$. For channels that produce errors, more complicated expressions of capacity result (shown later in Eq.s (1), (7), (11), and (16)) and a different optimal value of $M$ emerges.