--- title: "Error probability in binary baseband system" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:error-probability-binary-baseband" source: https://www.bctengineer.com/q/semester_6/communication_systems/error-probability-binary-baseband site: BCT Engineering --- # Error probability in binary baseband system Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2080 Bhadra (CS II), 8 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-baseband - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Derive the expression for evaluating error probability in Binary baseband system. ## Short answer For binary baseband signals s1(t), s2(t) over T_b with AWGN, the matched filter output sample y = a_i + n has Gaussian noise variance σ². Optimal threshold λ = (a1+a2)/2 gives P_e = ½erfc((a1-a2)/(2√2σ)) = Q((a1-a2)/(2σ)). With matched filter, P_e = ½erfc(√(E_d/4N_0)) where E_d = ∫(s1-s2)²dt is the difference signal energy. ## Answer In a binary baseband system, bits are sent as pulses (e.g. polar NRZ $\pm A$) over a channel adding white Gaussian noise. The receiver samples the filtered signal once per bit and compares it with a threshold; an error occurs when noise pushes the sample across the threshold. **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 $P_e$. **Polar NRZ example.** Levels $a_1=+A$, $a_2=-A$, threshold 0: $P_e=\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right)$; with matched filter, $P_e=\frac12\operatorname{erfc}\sqrt{E_b/N_0}$. Unipolar ($A$, 0) needs twice the average power for the same $P_e$. ## 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 lambda? A. At a1 B. At a2 C. At zero always D. Midway, (a1 + a2)/2 Answer: D. With equal priors and equal Gaussian noise the pdfs cross midway between the two outputs. (Key point: Optimum threshold) 2. What is the average error probability for the binary baseband system? A. Q( (a1 - a2)/(2 sigma) ) = (1/2) erfc( (a1 - a2)/(2 sqrt2 sigma) ) B. Q( (a1 - a2)/sigma ) C. erfc( (a1 - a2)/(2 sigma) ) D. (1/2) erfc( (a1 - a2)/sigma ) Answer: A. The error area lies (a1-a2)/2 from the mean, i.e. (a1-a2)/(2 sigma) standard deviations, giving Q(.). (Key point: General Pe expression) 3. With a matched filter, Pe depends on which quantity? A. The pulse shape of s1(t) B. The peak amplitude only C. Ed = integral of [s1 - s2]^2, relative to N0: Pe = (1/2) erfc sqrt(Ed/4N0) D. The bandwidth of the filter Answer: C. Maximum (a1-a2)^2/sigma^2 = 2Ed/N0, so Pe depends only on the difference-signal energy, not pulse shape. (Key point: Matched filter result) 4. Which statement is correct? A. Unipolar (A, 0) needs half the power of polar for the same Pe B. Antipodal signals give the lowest Pe; unipolar needs twice the average power of polar C. Orthogonal signals give the lowest Pe D. Pe is independent of the signal levels Answer: B. Maximising Ed means antipodal signals; unipolar has half the distance energy per unit average power. (Key point: Best signalling and unipolar) ## Close variants set in other exams - [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× ## Related in Communication Systems - [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 - [Binary error probability; threshold effect in FM and mitigation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aerror-prob-threshold-effect) — Communication Systems, chapter 9, asked 2× - [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 - [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 - [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