--- title: "Binary baseband error probability; comparison with M-ary" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:error-prob-binary-vs-mary" source: https://www.bctengineer.com/q/semester_6/communication_systems/error-prob-binary-vs-mary site: BCT Engineering --- # Binary baseband error probability; comparison with M-ary Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2076 Asoj (CS II), 8 marks - Video: [ESE 471 A General Formula for Prob Symbol Error in M ary PAM](https://www.youtube.com/watch?v=6o-uUbFNicQ) — Neal Patwari - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/error-prob-binary-vs-mary - 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 and compare it with M-ary system. ## Short answer Binary baseband error probability is P_e = 1/2 erfc((a1-a2)/(2√2 σ)) = Q((a1-a2)/(2σ)); with matched filter, P_e = 1/2 erfc(√(E_d/(4N_0))) where E_d is difference signal energy. M-ary systems save bandwidth by factor log2 M but require more power for same P_e, trading power for bandwidth with P_e ≈ (1-1/M) erfc(A/(√2 σ)) for PAM. ## Answer #### Error probability in binary baseband system **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 $P_e$. For polar NRZ ($\pm A$ at the sampler): $P_e=\frac12\operatorname{erfc}\left(\frac{A}{\sqrt2\sigma}\right)$. #### Comparison with M-ary system For M-ary PAM with levels $\pm A,\pm3A,\dots,\pm(M-1)A$ (spacing $2A$), inner levels can err on both sides and outer levels on one side, giving $$ P_e=\left(1-\frac1M\right)\operatorname{erfc}\left(\frac{A}{\sqrt2\,\sigma}\right),\qquad P_{av}=\frac{(M^2-1)A^2}{3} $$ which reduces to the binary result for $M=2$. | Point | Binary ($M=2$) | M-ary ($M=2^k$) | |---|---|---| | Bits per symbol | 1 | $k=\log_2M$ | | Bandwidth for same bit rate | $R_b/2$ (Nyquist) | $R_b/(2\log_2M)$, i.e. $k$ times less | | Level spacing for same peak power | Large | Smaller, by factor $M-1$ | | $P_e$ for same average power | Lowest | Higher; needs about $(M^2-1)/3$ times more power | | Thresholds | One | $M-1$ | | Receiver complexity | Simple | More complex | | Typical use | Noisy, power-limited links | Band-limited links (e.g. telephone modems) | So M-ary signalling **trades power for bandwidth**: it saves bandwidth but needs more signal power for the same error probability. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which gives the binary baseband Pe with matched filter, in terms of difference-signal energy Ed? A. (1/2) erfc sqrt(Ed/4N0) B. (1/2) erfc sqrt(Ed/N0) C. Q( sqrt(Ed/4N0) ) D. erfc sqrt(2Ed/N0) Answer: A. Maximum (a1-a2)^2/sigma^2 = 2Ed/N0 gives (1/2) erfc sqrt(Ed/4N0). (Key point: Binary Pe) 2. For M-ary PAM with level spacing 2A, what is the symbol error probability? A. (1/M) erfc( A/(sqrt2 sigma) ) B. (1 - 1/M) erfc( A/(sqrt2 sigma) ) C. (1/2) erfc( A/(sqrt2 sigma) ) for all M D. M erfc( A/(sqrt2 sigma) ) Answer: B. Inner levels err on both sides, outer on one side; averaging gives (1 - 1/M) erfc, which equals the binary result at M = 2. (Key point: M-ary PAM Pe) 3. For the same bit rate, how does M-ary bandwidth compare with binary? A. log2 M times greater B. The same C. M times less D. log2 M times less Answer: D. Each symbol carries k = log2 M bits, so the symbol rate and bandwidth fall by k. (Key point: Bandwidth trade-off) 4. For the same Pe and average power, what happens as M increases? A. Spacing shrinks, so about (M^2 - 1)/3 times more power is needed B. Power requirement falls C. Power stays the same D. Only one threshold is needed Answer: A. P_av = (M^2-1)A^2/3, so level spacing shrinks and more power is needed: M-ary trades power for bandwidth. (Key point: Power penalty) ## Close variants set in other exams - [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 of M-ary system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amary-error-probability) — Communication Systems, chapter 9, asked 2× ## 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 - [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 - [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 - [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× - [Short note: M-ary baseband communication system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashort-note-mary-baseband) — Communication Systems, chapter 6, asked 3×