--- title: "ISI; solutions for zero ISI; duobinary with precoder" subject: "Communication Systems" chapter: 6 key: "semester_6/communication_systems:isi-zero-isi-duobinary-precoder" source: https://www.bctengineer.com/q/semester_6/communication_systems/isi-zero-isi-duobinary-precoder site: BCT Engineering --- # ISI; solutions for zero ISI; duobinary with precoder Communication Systems · Chapter 6: Baseband Digital Data Transmission - Asked in: 2074 Chaitra (CS II), 2+6 marks - Video: [DC#54 Precoded Duobinary signalling for controlled ISI || EC Academy](https://www.youtube.com/watch?v=qkSMnruB808) — EC Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/isi-zero-isi-duobinary-precoder - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6 ## Question What is ISI? State two solutions for zero ISI. Explain duo-binary encoding with the use of precoder. ## Short answer ISI is pulse overlap causing errors. Two zero-ISI solutions: Nyquist pulse shaping and correlative coding. Duobinary encoding uses c_k = a_k + a_{k-1} with a precoder; decision: |c_k|>1 gives 0. ## Answer #### ISI **Intersymbol interference (ISI)** is the overlap of pulse tails from neighbouring symbols at the sampling instant of the present symbol, caused by the limited bandwidth and dispersion of the channel. It causes decision errors even without noise. #### Two solutions for zero ISI 1. **Nyquist pulse shaping:** use an overall pulse with $p(nT_b) = 0$ for $n \ne 0$, e.g. the raised cosine pulse (bandwidth $\frac{R_b}{2}(1+\alpha)$). 2. **Correlative (partial response) coding:** e.g. duobinary signalling, which adds a known, controlled ISI that the receiver removes, allowing the minimum bandwidth $R_b/2$ with practical filters. (Adaptive equalization is a third common method.) #### Duobinary encoding with precoder **Duobinary signalling:** each output sample is the sum of the present and previous binary levels: $$ c_k = a_k + a_{k-1}, \qquad a_k = \pm 1 $$ The equivalent filter is a delay-and-add followed by an ideal low-pass filter: $$ H(f) = 2\cos(\pi fT_b)\,e^{-j\pi fT_b}, \quad |f| \le \frac{1}{2T_b} $$ $$ h(t) = \frac{T_b^2\,\sin(\pi t/T_b)}{\pi t\,(T_b - t)} $$ $H(f)$ falls smoothly to zero at $f = 1/(2T_b)$, so it is realizable. The output has three levels: $-2, 0, +2$. **Problem:** without precoding, decoding uses $\hat a_k = c_k - \hat a_{k-1}$, so one error propagates to later bits. **Precoder:** form $d_k = b_k \oplus d_{k-1}$ (modulo-2) before level mapping ($d = 1 \to +1$, $d = 0 \to -1$). Then - $c_k = \pm 2$ when $d_k = d_{k-1}$, i.e. $b_k = 0$; - $c_k = 0$ when $d_k \ne d_{k-1}$, i.e. $b_k = 1$. Decision rule: $|c_k| < 1 \Rightarrow \hat b_k = 1$, $|c_k| > 1 \Rightarrow \hat b_k = 0$. Each bit is found from its own sample, so there is no error propagation. ```text b_k -->(XOR)---+--> level map --> a_k --+------------>(+)--> [ideal LPF] --> c_k ^ | d_k | ^ | v | | +--[Tb delay] +--[Tb delay]--+ ``` **Example:** input $b_k = 1101001$, initial $d_0 = 1$. | k | 1 | 2 | 3 | 4 | 5 | 6 | 7 | |---|---|---|---|---|---|---|---| | $b_k$ | 1 | 1 | 0 | 1 | 0 | 0 | 1 | | $d_k$ | 0 | 1 | 1 | 0 | 0 | 0 | 1 | | $a_k$ | −1 | +1 | +1 | −1 | −1 | −1 | +1 | | $c_k = a_k + a_{k-1}$ | 0 | 0 | +2 | 0 | −2 | −2 | 0 | | $\hat b_k$ | 1 | 1 | 0 | 1 | 0 | 0 | 1 | ($a_0 = +1$.) The decoded data equal the input. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which pair are solutions for zero ISI? A. Nyquist pulse shaping and correlative coding B. Increasing noise and lowering bandwidth C. Parity bits and CRC D. Amplitude clipping and filtering Answer: A. Nyquist pulse shaping (e.g. raised cosine) gives zeros at other sampling instants; correlative coding adds controlled, removable ISI. (Key point: Zero-ISI solutions) 2. Duobinary output is: A. c_k = a_k - a_(k-2) B. c_k = a_k + a_(k-1) C. c_k = a_k a_(k-1) D. c_k = a_k - a_(k-1) Answer: B. It sums present and previous levels, giving three levels -2, 0, +2. (Key point: Duobinary rule) 3. Why is a precoder used with duobinary? A. To increase bandwidth B. To avoid error propagation in decoding C. To add DC D. To create four levels Answer: B. Without it decoding uses a_k = c_k - a_(k-1), so one error propagates; precoding allows each bit to be decoded from its own sample. (Key point: Need for precoder) 4. With precoding d_k = b_k XOR d_(k-1), the decision is: A. |c_k| < 1 gives 1; |c_k| > 1 gives 0 B. |c_k| > 1 gives 1 C. c_k > 0 gives 1 D. c_k = 0 gives 0 Answer: A. c_k = 0 when d_k differs from d_(k-1) (b = 1); |c_k| = 2 when equal (b = 0). (Key point: Decision rule) ## Related in Communication Systems - [Nyquist criteria for zero ISI; duobinary difficulties](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anyquist-zero-isi-duobinary-difficulties) — Communication Systems, chapter 6 - [ISI; duobinary with precoder for 0010110](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aisi-duobinary-precoder-0010110) — Communication Systems, chapter 6 - [Reducing ISI; modified duobinary with precoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aisi-reduce-modified-duobinary) — Communication Systems, chapter 6 - [ISI; Nyquist criteria; duobinary with example](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aisi-nyquist-duobinary-example) — Communication Systems, chapter 6, asked 2× - [ISI; modified duobinary for 10110011](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aisi-modified-duobinary-10110011) — Communication Systems, chapter 6 - [ISI in baseband systems; ideal and practical solutions](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aisi-ideal-practical-solutions) — Communication Systems, chapter 6, asked 2×