--- title: "ISI; duobinary with precoder for 0010110" subject: "Communication Systems" chapter: 6 key: "semester_6/communication_systems:isi-duobinary-precoder-0010110" source: https://www.bctengineer.com/q/semester_6/communication_systems/isi-duobinary-precoder-0010110 site: BCT Engineering --- # ISI; duobinary with precoder for 0010110 Communication Systems · Chapter 6: Baseband Digital Data Transmission - Asked in: 2073 Shrawan (CS II), 2+5 marks - Video: [Duobinary Encoder with Precoder Numerical Lecture 3|| Digital Communication|| Wireless Communication](https://www.youtube.com/watch?v=sG1_GlR3SB4) — Engineering Basics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/isi-duobinary-precoder-0010110 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6 ## Question What do you understand by intersymbol interference? Explain Duobinary coding technique with precoder and illustrate it using binary input sequence 0010110. ## Short answer ISI is the interference from overlapping neighboring pulses at sampling instants. Duobinary coding (a_k + a_{k-1}) transmits at Nyquist rate with half-cosine spectrum. Precoder d_k = b_k XOR d_{k-1} prevents error propagation; decode: |c_k| < 1 -> 1, else 0. Input 0010110 yields c_k = +2, +2, 0, -2, 0, 0, -2 and recovers the original bits. ## Answer #### Intersymbol interference **Intersymbol interference (ISI)** is the overlap of the tails of neighbouring pulses at the sampling instant of the present pulse. It arises because a band-limited channel spreads each pulse beyond its own bit interval. The sample at $t = kT_b$ becomes $$ y(kT_b) = \mu a_k + \mu\sum_{n\ne k} a_n\,p\big((k-n)T_b\big) + n(kT_b) $$ where the middle term is ISI. ISI closes the eye pattern and causes errors even without noise. #### Duobinary coding with precoder **Duobinary (class I partial response) signalling** deliberately adds a known amount of ISI from the previous bit so that data at rate $R_b$ can be sent in the minimum bandwidth $R_b/2$ with a realizable filter. $$ c_k = a_k + a_{k-1}, \qquad a_k = \pm 1 $$ $$ H(f) = 1 + e^{-j2\pi fT_b} = 2\cos(\pi fT_b)\,e^{-j\pi fT_b}, \quad |f| \le \frac{1}{2T_b} $$ The output has three levels (−2, 0, +2). **Need for precoder:** without it, the receiver decodes $\hat a_k = c_k - \hat a_{k-1}$, so one wrong decision spreads to all later bits (error propagation). **Precoder:** $d_k = b_k \oplus d_{k-1}$; level map $d_k = 1 \to +1$, $d_k = 0 \to -1$. Then $c_k = 0$ when $b_k = 1$ and $c_k = \pm 2$ when $b_k = 0$. **Decision rule:** $|c_k| < 1 \Rightarrow \hat b_k = 1$; $|c_k| > 1 \Rightarrow \hat b_k = 0$. ```text b_k -->(XOR)---+--> level map --> a_k --+------------>(+)--> [LPF, B = Rb/2] --> c_k ^ | d_k | ^ | v | | +--[Tb delay] +--[Tb delay]--+ ``` #### Illustration for 0010110 Initial precoder bit assumed $d_0 = 1$ ($a_0 = +1$). | k | 1 | 2 | 3 | 4 | 5 | 6 | 7 | |---|---|---|---|---|---|---|---| | $b_k$ | 0 | 0 | 1 | 0 | 1 | 1 | 0 | | $d_{k-1}$ | 1 | 1 | 1 | 0 | 0 | 1 | 0 | | $d_k = b_k \oplus d_{k-1}$ | 1 | 1 | 0 | 0 | 1 | 0 | 0 | | $a_k$ | +1 | +1 | −1 | −1 | +1 | −1 | −1 | | $a_{k-1}$ | +1 | +1 | +1 | −1 | −1 | +1 | −1 | | $c_k = a_k + a_{k-1}$ | +2 | +2 | 0 | −2 | 0 | 0 | −2 | | Decoded $\hat b_k$ | 0 | 0 | 1 | 0 | 1 | 1 | 0 | The decoded sequence **0010110** equals the input, and each bit is decided from its own sample only. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What causes intersymbol interference? A. Thermal noise added in the receiver B. A band-limited channel spreading pulses so tails overlap at neighbouring sampling instants C. Using too many quantization levels D. Wrong carrier frequency at the demodulator Answer: B. Band-limiting spreads each pulse beyond its bit interval, so tails fall on the sampling instants of other pulses. (Key point: Meaning of ISI) 2. In duobinary signalling c_k = a_k + a_(k-1) with a_k = ±1. What output levels occur? A. Two levels: -1 and +1 B. Three levels: -2, 0, +2 C. Four levels: -3, -1, +1, +3 D. Three levels: -1, 0, +1 Answer: B. Adding two values of ±1 gives -2, 0 or +2. (Key point: Duobinary equation and levels) 3. Why is a precoder d_k = b_k XOR d_(k-1) used in duobinary? A. To reduce the bandwidth below Rb/2 B. To remove the DC component C. To avoid error propagation by allowing each bit to be decided from its own sample D. To increase the number of levels Answer: C. Without precoding, each decision depends on the previous one (a_k = c_k - a_(k-1)), so one error spreads. Precoding removes this dependence. (Key point: Need for precoder) 4. With the precoder, what does a received sample c_k = 0 mean? A. b_k = 0 B. b_k = 1 C. The previous bit was in error D. d_k = 0 always Answer: B. c_k = 0 when b_k = 1, and c_k = ±2 when b_k = 0; decision: |c_k| < 1 gives 1. (Key point: Decision rule with precoder) ## Related in Communication Systems - [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; modified duobinary for 10110011](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aisi-modified-duobinary-10110011) — Communication Systems, chapter 6 - [ISI; solutions for zero ISI; duobinary with precoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aisi-zero-isi-duobinary-precoder) — 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× - [Correlative coding; minimizing error in duobinary](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acorrelative-coding-duobinary-error) — Communication Systems, chapter 6 - [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 ## Related in other subjects - [Line coding in PCM and the right technique](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aline-coding-pcm-solution) — Telecommunications, chapter 3