---
title: "Error probability of coherent ASK"
subject: "Communication Systems"
chapter: 9
key: "semester_6/communication_systems:ask-coherent-error-probability"
source: https://www.bctengineer.com/q/semester_6/communication_systems/ask-coherent-error-probability
site: BCT Engineering
---

# Error probability of coherent ASK

Communication Systems · Chapter 9: Noise in Communication Systems

- Asked in: 2074 Asoj (CS II), 6 marks; 2069 Chaitra (CS II), 6 marks (asked 2 times)
- Video: [L 12 | Probability of Error of ASK | Amplitude Shift Key | Digital Communication Performance |](https://www.youtube.com/watch?v=PQ4NSkIDepE) — Dopamine
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/ask-coherent-error-probability
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9

## Question

Derive the expression of error probability for coherent detection of Amplitude Shift Keying (ASK).

## Short answer

For coherent ASK, error probability is P_e = ½ erfc(√(E_b/(2N_0))) = Q(√(E_b/N_0)), where E_b is the average energy per bit. The optimum threshold is midway; ASK needs 3 dB more power than BPSK for same performance.

## Answer

In **ASK** (on–off keying) a carrier is switched on for bit 1 and off for bit 0. Coherent detection uses a correlator with a locally generated carrier of the same frequency and phase.

**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 $E_1$: $P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}$.)

**Remarks:** coherent ASK needs 3 dB more average power than BPSK for the same $P_e$, and the threshold depends on received amplitude, so ASK is sensitive to fading. Example: for $E_b/N_0=10$ (10 dB), $P_e=Q(\sqrt{10})=Q(3.162)\approx7.8\times10^{-4}$.

## Check yourself

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

1. In on-off keying ASK, how are bits represented?
   A. Bit 1 as A cos(wc t), bit 0 as no signal
   B. Bit 1 as +A cos(wc t), bit 0 as -A cos(wc t)
   C. Bit 1 at frequency f1, bit 0 at f2
   D. Bit 1 as a phase of 90 degrees
   Answer: A. OOK switches the carrier on for 1 and off for 0, so s2(t) = 0. (Key point: ASK signal set)

2. What are the noise-free correlator outputs a1 and a2?
   A. a1 = A Tb and a2 = -A Tb
   B. a1 = A Tb/2 and a2 = -A Tb/2
   C. a1 = A Tb/2 and a2 = 0
   D. a1 = A^2 Tb/2 and a2 = 0
   Answer: C. Integrating A cos^2(wc t) over Tb gives A Tb/2; for a 0 the output is zero. (Key point: Correlator outputs)

3. What is the variance of the correlator noise?
   A. N0 Tb/2
   B. N0 Tb/4
   C. N0 Tb
   D. N0/4
   Answer: B. (N0/2) times the integral of cos^2 over Tb (= Tb/2) gives N0 Tb/4. (Key point: Noise variance)

4. What is the coherent ASK error probability in terms of average Eb?
   A. (1/2) erfc( sqrt(Eb / N0) )
   B. (1/2) erfc( sqrt(2 Eb / N0) )
   C. (1/2) exp(-Eb / 2N0)
   D. (1/2) erfc( sqrt(Eb / 2N0) )
   Answer: D. With Eb = A^2 Tb/4 (half the bits are zero), the argument becomes sqrt(Eb/2N0), 3 dB worse than BPSK. (Key point: Final ASK Pe)

## Related in Communication Systems

- [AM vs FM; error probability of coherent ASK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-fm-compare-ask-error) — Communication Systems, chapter 9, asked 2×
- [Threshold effect in FM; error probability of coherent PSK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Athreshold-effect-psk-error) — Communication Systems, chapter 9
- [Need for shift keying; ASK vs FSK vs PSK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashift-keying-ask-fsk-psk) — Communication Systems, chapter 7
- [AM vs FM efficiency and complexity; coherent ASK error](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-fm-bw-efficiency-ask-error) — Communication Systems, chapter 9, asked 2×
- [Coherent ASK and PSK error; ASK needs double power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aask-psk-error-double-power) — Communication Systems, chapter 9
- [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

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