---
title: "Bit error probability for binary ASK"
subject: "Communication Systems"
chapter: 9
key: "semester_6/communication_systems:binary-ask-ber"
source: https://www.bctengineer.com/q/semester_6/communication_systems/binary-ask-ber
site: BCT Engineering
---

# Bit error probability for binary ASK

Communication Systems · Chapter 9: Noise in Communication Systems

- Asked in: 2071 Chaitra (CS II), 6 marks
- 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/binary-ask-ber
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9

## Question

With necessary assumption, derive the expression for bit error probability for binary ASK system.

## Short answer

Assuming coherent detection, AWGN, equally likely symbols, and known carrier phase. The bit error probability for binary ASK is P_e = Q(√(E_b/N_0)), where E_b = A^2 T_b/4 is the average energy per bit. This requires 3 dB more energy than coherent BPSK.

## Answer

Binary ASK (on-off keying) sends a carrier burst for "1" and nothing for "0". With coherent (matched-filter/correlator) detection in AWGN, the bit error probability is $P_e = Q\left(\sqrt{E_b/N_0}\right)$, where $E_b$ is the average energy per bit.

**Assumptions**

1. Signals over $0 \le t \le T_b$:
$$
s_1(t) = A\cos\omega_c t \ \ (\text{"1"}), \qquad s_0(t) = 0 \ \ (\text{"0"})
$$
with $f_c$ an integer multiple of $1/T_b$.
2. Channel adds white Gaussian noise $n(t)$ of two-sided PSD $N_0/2$.
3. Symbols are equally likely; carrier phase and bit timing are known (coherent detection).

**Receiver**

```text
 r(t) --->(X)--->[ integrate 0..Tb ]--> y --> [ y > Vth ? ] --> 1/0
           ^                                   sample at Tb
     cos(wc t)  (local coherent carrier)
```

**Correlator output**

$$
y = \int_0^{T_b} r(t)\cos\omega_c t\,dt
$$

- For "1": signal part $= \int_0^{T_b}A\cos^2\omega_c t\,dt = \dfrac{AT_b}{2}$
- For "0": signal part $= 0$

**Noise at the output**: $n_o = \int_0^{T_b} n(t)\cos\omega_c t\,dt$ is Gaussian, zero mean, with variance

$$
\sigma^2 = \frac{N_0}{2}\int_0^{T_b}\cos^2\omega_c t\,dt = \frac{N_0T_b}{4}
$$

**Threshold.** For equal priors and equal variances, the optimum threshold is midway: $V_{th} = AT_b/4$.

**Error probabilities**

$$
\begin{aligned}
P(e|0) &= P\left(n_o > \tfrac{AT_b}{4}\right) = Q\!\left(\frac{AT_b/4}{\sigma}\right) \\
P(e|1) &= P\left(n_o < -\tfrac{AT_b}{4}\right) = Q\!\left(\frac{AT_b/4}{\sigma}\right)
\end{aligned}
$$

$$
P_e = \tfrac12 P(e|0) + \tfrac12 P(e|1) = Q\!\left(\frac{AT_b/4}{\sqrt{N_0T_b/4}}\right) = Q\!\left(\sqrt{\frac{A^2T_b}{4N_0}}\right)
$$

**In terms of energy.** Energy of the "1" pulse $E_1 = A^2T_b/2$; the "0" has zero energy, so average energy per bit $E_b = A^2T_b/4$. Hence

$$
P_e = Q\!\left(\sqrt{\frac{E_b}{N_0}}\right) = Q\!\left(\sqrt{\frac{E_1}{2N_0}}\right) = \frac12\,\text{erfc}\!\left(\sqrt{\frac{E_b}{2N_0}}\right)
$$

**Remarks**

- Coherent BPSK gives $Q(\sqrt{2E_b/N_0})$, so ASK needs **3 dB more** average energy for the same $P_e$.
- Non-coherent (envelope) ASK gives approximately $P_e \approx \tfrac12 e^{-E_b/2N_0}$, slightly worse but simpler.

## Check yourself

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

1. In coherent binary ASK, what is the correlator output signal part for a transmitted 1 and the noise variance?
   A. A T_b / 2 and N_0 T_b / 4
   B. A T_b and N_0 T_b / 2
   C. A T_b / 2 and N_0 T_b / 2
   D. A T_b / 4 and N_0 T_b / 4
   Answer: A. The integral of A cos^2 is A T_b/2, and the noise variance is (N_0/2)(T_b/2) = N_0 T_b/4. (Key point: Correlator signal and noise)

2. For equally likely symbols, the optimum decision threshold for binary ASK is
   A. 0
   B. A T_b / 2
   C. A T_b / 4
   D. A T_b
   Answer: C. The threshold is midway between the signal values 0 and A T_b/2, i.e. A T_b/4. (Key point: Optimum threshold)

3. The 1 pulse has energy A^2 T_b/2 and the 0 has zero. What is the average energy per bit E_b?
   A. A^2 T_b / 2
   B. A^2 T_b / 4
   C. A^2 T_b
   D. A^2 T_b / 8
   Answer: B. With equal priors E_b = (1/2)(A^2 T_b/2) + (1/2)(0) = A^2 T_b/4. (Key point: Average energy per bit)

4. Coherent ASK has P_e = Q(sqrt(E_b/N_0)) while BPSK has Q(sqrt(2E_b/N_0)). So for the same P_e, ASK needs
   A. 3 dB more average energy
   B. 3 dB less average energy
   C. 6 dB more average energy
   D. the same energy
   Answer: A. The argument is smaller by a factor sqrt(2), so ASK needs twice (3 dB) the average energy. (Key point: ASK versus BPSK)

## 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
- [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×
- [Error probability in a binary communication system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aerror-probability-binary-system) — Communication Systems, chapter 9
- [Error probability of coherent ASK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aask-coherent-error-probability) — Communication Systems, chapter 9, asked 2×
- [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×
- [Bit error probability of coherent BFSK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abfsk-coherent-ber) — Communication Systems, chapter 9

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