Binary PAM error probability extended to M-ary

Binary PAM error probability

Assumptions: polar binary PAM; the sampled receiver output is $+A$ for "1" and $-A$ for "0"; additive Gaussian noise $n$ of zero mean and variance $\sigma^2$; equal priors, so the threshold is $0$.

$$y = \pm A + n, \qquad f_N(n) = \frac{1}{\sqrt{2\pi}\sigma}e^{-n^2/2\sigma^2}$$

   f(y|0)            f(y|1)
     /\                /\
    /  \     |        /  \
 __/    \____|_______/    \__ y
    -A       0  threshold +A
          shaded tails = errors

Error when "1" is sent: $y < 0 \Rightarrow n < -A$:

$$P(e|1) = \int_{-\infty}^{-A}\frac{1}{\sqrt{2\pi}\sigma}e^{-n^2/2\sigma^2}dn = Q\!\left(\frac{A}{\sigma}\right)$$

Error when "0" is sent: $n > A$, so by symmetry $P(e|0) = Q(A/\sigma)$.

$$P_e = \tfrac12P(e|1) + \tfrac12P(e|0) = Q\!\left(\frac{A}{\sigma}\right) = \frac12\text{erfc}\!\left(\frac{A}{\sqrt2\sigma}\right)$$

With a matched filter, $A^2/\sigma^2 = 2E_b/N_0$, so

$$P_e = Q\!\left(\sqrt{\frac{2E_b}{N_0}}\right)$$

(For unipolar on-off PAM with levels $0$ and $A$, and threshold $A/2$: $P_e = Q(A/2\sigma)$.)

Extension to M-ary PAM

Levels $\pm A, \pm3A, \dots, \pm(M-1)A$ (spacing $2A$), with thresholds midway between adjacent levels.

• Each of the $M-2$ inner levels is in error if $|n| > A$: probability $2Q(A/\sigma)$.
• Each of the 2 outer levels can be in error on one side only: probability $Q(A/\sigma)$.

$$P_e = \frac{(M-2)\,2Q(A/\sigma) + 2Q(A/\sigma)}{M} = 2\left(1-\frac1M\right)Q\!\left(\frac{A}{\sigma}\right)$$

The average symbol energy is $E_s = (M^2-1)A^2/3$, so with a matched filter ($\sigma^2 = N_0/2$):

$$P_e = 2\left(1-\frac1M\right)Q\!\left(\sqrt{\frac{6E_s}{(M^2-1)N_0}}\right)$$

For $M=2$ this reduces to the binary result. For a fixed $E_s$, a larger $M$ gives a larger $P_e$, but for a given bit rate the required bandwidth falls by a factor of $\log_2 M$.