---
title: "Error probability of binary ASK extended to M-ary"
subject: "Communication Systems"
chapter: 9
key: "semester_6/communication_systems:ask-error-prob-mary"
source: https://www.bctengineer.com/q/semester_6/communication_systems/ask-error-prob-mary
site: BCT Engineering
---

# Error probability of binary ASK extended to M-ary

Communication Systems · Chapter 9: Noise in Communication Systems

- Asked in: 2073 Shrawan (CS II), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/ask-error-prob-mary
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9

## Question

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

## Short answer

For binary ASK, error probability is (1/2)erfc(√(E_b/(2N_0))) = Q(√(E_b/N_0)). For M-ary ASK, P_e = 2(1-1/M) Q( √[3E_s/((M-1)(2M-1)N_0)] ).

## Answer

#### Binary ASK

**ASK (on–off keying) signals** over $0\le t\le T_b$:

$$
s_1(t)=A\cos\omega_ct\ \ (\text{bit }1),\qquad s_2(t)=0\ \ (\text{bit }0)
$$

**Coherent receiver (correlator = matched filter):**

```text
 r(t) -->(x)--> integrate --> sample --> y > lam ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (local carrier, in phase)
```

Correlator output at $t=T_b$, with $r(t)=s_i(t)+w(t)$:

$$
y=\int_0^{T_b}r(t)\cos\omega_ct\,dt=a_i+n
$$

- Bit 1: $a_1=\int_0^{T_b}A\cos^2\omega_ct\,dt=\frac{AT_b}{2}$ (for $f_c$ an integer multiple of $1/T_b$).
- Bit 0: $a_2=0$.
- Noise: $n=\int_0^{T_b}w(t)\cos\omega_ct\,dt$ is Gaussian, mean 0, variance

$$
\sigma^2=\int_0^{T_b}\!\!\int_0^{T_b}\frac{N_0}{2}\delta(t-u)\cos\omega_ct\cos\omega_cu\,dt\,du=\frac{N_0}{2}\cdot\frac{T_b}{2}=\frac{N_0T_b}{4}
$$

Threshold midway: $\lambda=\frac{AT_b}{4}$. Using the binary result $P_e=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\sigma}\right)$:

$$
\begin{aligned}
\frac{a_1-a_2}{2\sqrt2\,\sigma} &= \frac{AT_b/2}{2\sqrt2\sqrt{N_0T_b/4}}=\frac{A\sqrt{T_b}}{2\sqrt{2N_0}}=\sqrt{\frac{A^2T_b}{8N_0}}
\end{aligned}
$$

Energy of a "1" pulse is $E_1=\frac{A^2T_b}{2}$; the average energy per bit (half the bits are 0) is $E_b=\frac{E_1}{2}=\frac{A^2T_b}{4}$. Hence $\frac{A^2T_b}{8N_0}=\frac{E_b}{2N_0}$ and

$$
P_e(\text{ASK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)
$$

(In terms of peak energy $E_1$: $P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}$.)

#### Extension to M-ary ASK

In M-ary ASK the carrier takes $M$ amplitudes $A_i=iA$, $i=0,1,\dots,M-1$, each symbol of duration $T_s$ carrying $\log_2M$ bits. The same coherent correlator gives

$$
y=a_i+n,\qquad a_i=\frac{iAT_s}{2},\qquad \sigma^2=\frac{N_0T_s}{4}
$$

Levels are spaced $\Delta=\frac{AT_s}{2}$ apart, and thresholds are placed midway. The two end levels can be mistaken on one side only; the $M-2$ inner levels on both sides:

$$
\begin{aligned}
P_e &= \frac1M\left[(M-2)\cdot2Q\left(\frac{\Delta}{2\sigma}\right)+2\cdot Q\left(\frac{\Delta}{2\sigma}\right)\right]=2\left(1-\frac1M\right)Q\left(\frac{\Delta}{2\sigma}\right) \\
\frac{\Delta}{2\sigma} &= \frac{AT_s/2}{2\sqrt{N_0T_s/4}}=\sqrt{\frac{A^2T_s}{4N_0}}
\end{aligned}
$$

Average symbol energy: $E_i=\frac{i^2A^2T_s}{2}$, so

$$
E_s=\frac{A^2T_s}{2M}\sum_{i=0}^{M-1}i^2=\frac{A^2T_s(M-1)(2M-1)}{12}
$$

Substituting $A^2T_s=\frac{12E_s}{(M-1)(2M-1)}$:

$$
P_e(\text{M-ASK})=2\left(1-\frac1M\right)Q\left(\sqrt{\frac{3E_s}{(M-1)(2M-1)N_0}}\right)
$$

**Check $M=2$:** $P_e=Q\left(\sqrt{E_s/N_0}\right)$ with $E_s=E_b$, the binary ASK result. For larger $M$ the levels crowd together for the same average energy, so $P_e$ grows, while the bandwidth per bit falls by $\log_2M$.

## Check yourself

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

1. In coherent binary ASK with correlator output y = a_i + n, what are the signal levels?
   A. a1 = AT_b, a2 = 0
   B. a1 = AT_b/2, a2 = 0, with threshold AT_b/4
   C. a1 = AT_b/2, a2 = -AT_b/2
   D. a1 = A^2 T_b/2, a2 = 0, with threshold A^2 T_b/2
   Answer: B. Integrating A cos^2(wc t) over T_b gives AT_b/2 for bit 1 and 0 for bit 0; the midway threshold is AT_b/4. (Key point: Binary ASK correlator outputs)

2. For white noise of PSD N0/2 and a correlator with cos(wc t) over T_b, the noise variance is:
   A. N0 T_b / 4
   B. N0 T_b / 2
   C. N0 / 2
   D. N0 T_b
   Answer: A. Variance = (N0/2) times the integral of cos^2 = (N0/2)(T_b/2) = N0 T_b/4. (Key point: Noise variance of the correlator)

3. The average energy per bit of ASK is E_b = A^2 T_b/4. The resulting error probability is:
   A. (1/2) erfc( sqrt(E_b/N0) )
   B. (1/2) erfc( sqrt(E_b/2N0) ) = Q( sqrt(E_b/N0) )
   C. (1/2) erfc( sqrt(2E_b/N0) )
   D. (1/2) exp(-E_b/2N0)
   Answer: B. The erfc argument sqrt(A^2T_b/8N0) equals sqrt(E_b/2N0), which gives Q(sqrt(E_b/N0)). (Key point: Binary ASK error probability)

4. For M-ary ASK with equally spaced levels and midway thresholds, the symbol error probability has the factor:
   A. (1 - 1/M)
   B. 2(1 - 1/M)
   C. M - 1
   D. 2/M
   Answer: B. The 2 end levels err on one side and the M-2 inner levels on both, giving (1/M)[2(M-2)+2]Q = 2(1 - 1/M)Q. (Key point: M-ary ASK error probability)

## Related in Communication Systems

- [Binary PAM error probability extended to M-ary](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abinary-pam-mary-error) — 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
- [Bit error probability for binary ASK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abinary-ask-ber) — 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×
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