---
title: "Binary PAM error probability extended to M-ary"
subject: "Communication Systems"
chapter: 9
key: "semester_6/communication_systems:binary-pam-mary-error"
source: https://www.bctengineer.com/q/semester_6/communication_systems/binary-pam-mary-error
site: BCT Engineering
---

# Binary PAM error probability extended to M-ary

Communication Systems · Chapter 9: Noise in Communication Systems

- Asked in: 2069 Chaitra (CS II), 6+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/binary-pam-mary-error
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9

## Question

Derive the expression for error probability for binary PAM system and extend it to M-ary system.

## Short answer

For polar binary PAM, P_e= Q(A/σ)= Q(√(2E_b/N_0)). For M-ary PAM with levels ±A,±3A,…±(M−1)A, P_e=2(1−1/M) Q(√(6E_s/(( M^2−1)N_0)).

## Answer

#### 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 $0$.

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

```text
   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 $\log_2 M$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For polar binary PAM (levels +A and -A, threshold 0), when a 1 is sent an error occurs when
   A. n > A
   B. n < -A
   C. n < 0
   D. |n| < A
   Answer: B. y = A + n falls below 0 only if n < -A. (Key point: Binary PAM error event)

2. With a matched filter, the polar binary PAM error probability is
   A. Q(sqrt(E_b/N_0))
   B. Q(sqrt(2E_b/N_0))
   C. 2Q(sqrt(2E_b/N_0))
   D. Q(2E_b/N_0)
   Answer: B. P_e = Q(A/sigma) with A^2/sigma^2 = 2E_b/N_0. (Key point: Binary PAM error probability)

3. For M-ary PAM with levels +-A, +-3A, ..., +-(M-1)A, the average symbol energy is
   A. (M^2 - 1)A^2 / 3
   B. M A^2
   C. (M - 1)^2 A^2
   D. (M^2 - 1)A^2 / 6
   Answer: A. Averaging the squared levels gives E_s = (M^2 - 1)A^2/3, so A^2 = 3E_s/(M^2 - 1). (Key point: M-ary average symbol energy)

4. For fixed symbol energy E_s, increasing M causes
   A. lower P_e and lower bandwidth
   B. higher P_e but bandwidth reduced by log2 M
   C. higher P_e and higher bandwidth
   D. no change in P_e
   Answer: B. Levels get closer so error rises, but each symbol carries log2 M bits, reducing bandwidth by that factor. (Key point: M-ary trade-off)

## Related in Communication Systems

- [Error probability of binary ASK extended to M-ary](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aask-error-prob-mary) — Communication Systems, chapter 9
- [Binary baseband error probability; comparison with M-ary](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aerror-prob-binary-vs-mary) — Communication Systems, chapter 9, asked 3×
- [M-ary error probability; binary vs M-ary](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amary-error-compare-binary) — Communication Systems, chapter 9
- [Error probability of binary PAM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abinary-pam-error-probability) — Communication Systems, chapter 9
- [Error probability of M-ary system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amary-error-probability) — Communication Systems, chapter 9, asked 2×
- [Error probability in binary baseband system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aerror-probability-binary-baseband) — Communication Systems, chapter 9, asked 2×
