A digital communication system uses N=15 identical routes. The modulation used is 2-PPM. Calculate the value of $$ \varepsilon_b/N_0 $$ on the exit from the first route so that a final error probability not greater than $$10^-3$$ is granted in two cases: regenerative repeaters and non-regenerative repeater. In the case of regenerative repeaters calculate also the bit-rate, $$R_b$$knowing the noise level $$N_0=10^-10 W/Hz$$ and the transmitter power of $$P_u = 2mW$$
I already solved almost all the exercise, I just don't know how to calculate the bit rate though, could i use the following formula: $$ P_u = \varepsilon_b/T_b = \varepsilon_b * R_b <=> R_b = P_u/varepsilon_b = P_u/(N_0 * 10db)$$ where I used the value that I found in the regenerative case of $$ ( \varepsilon_b ) / N_0 = 10dB $$
