--- title: "Error probability in a binary communication system" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:error-probability-binary-system" source: https://www.bctengineer.com/q/semester_6/communication_systems/error-probability-binary-system site: BCT Engineering --- # Error probability in a binary communication system Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2081 Bhadra (CS II), 7 marks - Video: [Probability Of Error - Binary Communication - GATE 2018](https://www.youtube.com/watch?v=h0eFOj0x0LQ) — Sri Hari B H - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/error-probability-binary-system - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Evaluate the error probability in a binary communication system with appropriate expression. ## Short answer For equiprobable binary signaling in AWGN, the error probability is P_e = (1/2) erfc( sqrt(E_d/(4N0)) ), where E_d = ∫(s1−s2)^2 dt is the difference energy. For polar/PSK, P_e = (1/2) erfc( sqrt(E_b/N0) ); for unipolar/ASK or coherent FSK, P_e = (1/2) erfc( sqrt(E_b/(2N0)) ). The optimum threshold is midway between noise-free outputs. ## Answer **Model.** In a binary system, symbol 1 is sent as $s_1(t)$ and 0 as $s_2(t)$, each lasting $T_b$, with equal probability. Channel adds white Gaussian noise $w(t)$ of PSD $N_0/2$. The receiver filter output is sampled at $t=T_b$: ```text s_i+w +--------+ y(t) sample y +---------+ ------>| filter |------>-/ ------->| y > lam?|--> 1/0 +--------+ t=Tb +---------+ ``` The sample is $y=a_i+n$, where $a_1, a_2$ are the noise-free outputs ($a_1>a_2$) and $n$ is Gaussian with zero mean and variance $\sigma^2$. Conditional pdfs: $$ f(y\mid1)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_1)^2/2\sigma^2},\qquad f(y\mid0)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2} $$ ```text f(y|0) f(y|1) .--. .--. / \ | / \ / \ | / \ / ...\...|.../... \ --------a2---lam---a1----------> y Pe0 = area right of lam under f(y|0) Pe1 = area left of lam under f(y|1) ``` **Decision rule:** choose 1 if $y>\lambda$, else 0. For equal priors the optimum threshold is midway, $\lambda=\frac{a_1+a_2}{2}$. **Error when 0 is sent** ($y>\lambda$): with $u=(y-a_2)/\sqrt2\sigma$, $$ \begin{aligned} P_{e0} &= \int_{\lambda}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y-a_2)^2/2\sigma^2}dy=\frac{1}{\sqrt{\pi}}\int_{(\lambda-a_2)/\sqrt2\sigma}^{\infty}e^{-u^2}du \\ &= \frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right) \end{aligned} $$ By symmetry $P_{e1}$ (1 sent, $y<\lambda$) is the same. The average error probability is $$ P_e=\tfrac12P_{e0}+\tfrac12P_{e1}=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\,\sigma}\right)=Q\left(\frac{a_1-a_2}{2\sigma}\right) $$ where $\operatorname{erfc}(x)=\frac{2}{\sqrt\pi}\int_x^{\infty}e^{-u^2}du$ and $Q(x)=\frac12\operatorname{erfc}(x/\sqrt2)$. **With a matched filter** (matched to $s_1-s_2$), the maximum of $\frac{(a_1-a_2)^2}{\sigma^2}$ is $\frac{2E_d}{N_0}$, where $E_d=\int_0^{T_b}[s_1(t)-s_2(t)]^2dt$. Therefore $$ P_e=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_d}{4N_0}}\right) $$ $P_e$ depends only on the **energy of the difference signal** relative to $N_0$, not on pulse shape. Making the two signals as different as possible (antipodal) gives the lowest $P_e$. **Special cases:** | Scheme | $s_1(t)$, $s_2(t)$ | $E_d$ | $P_e$ | |---|---|---|---| | Unipolar / ASK (OOK) | $A\cos\omega_ct$, $0$ | $2E_b$ | $\frac12\operatorname{erfc}\sqrt{E_b/2N_0}$ | | Polar / PSK | $\pm A\cos\omega_ct$ | $4E_b$ | $\frac12\operatorname{erfc}\sqrt{E_b/N_0}$ | | Coherent FSK | $A\cos\omega_1t$, $A\cos\omega_2t$ | $2E_b$ | $\frac12\operatorname{erfc}\sqrt{E_b/2N_0}$ | ($E_b$ = average energy per bit.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For equally likely bits, where is the optimum threshold? A. At a1 B. Midway, (a1 + a2)/2 C. At zero always D. At a1 - a2 Answer: B. Symmetric Gaussian pdfs cross midway. (Key point: Decision threshold) 2. What is the error probability in terms of a1, a2 and sigma? A. (1/2) erfc( (a1 - a2) / (2 sqrt2 sigma) ) B. erfc( (a1 - a2) / sigma ) C. (1/2) erfc( (a1 + a2) / (2 sigma) ) D. exp(-(a1 - a2)/sigma) Answer: A. Integrating the Gaussian tail beyond the midpoint gives this. (Key point: Pe expression) 3. With a matched filter, what does Pe depend on? A. Pulse shape only B. Peak amplitude only C. Energy E_d of the difference signal relative to N0 D. Bandwidth only Answer: C. Pe = (1/2) erfc sqrt(E_d/4N0). (Key point: Matched filter Pe) 4. Which signalling gives the lowest Pe for the same Eb? A. Unipolar / ASK B. Coherent FSK C. All are equal D. Antipodal (polar / PSK) Answer: D. Polar has E_d = 4Eb versus 2Eb for ASK and FSK. 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