--- title: "Correlative coding; minimizing error in duobinary" subject: "Communication Systems" chapter: 6 key: "semester_6/communication_systems:correlative-coding-duobinary-error" source: https://www.bctengineer.com/q/semester_6/communication_systems/correlative-coding-duobinary-error site: BCT Engineering --- # Correlative coding; minimizing error in duobinary Communication Systems · Chapter 6: Baseband Digital Data Transmission - Asked in: 2073 Chaitra (CS II), 6+2 marks - Video: [DC#53 Duobinary signalling for controlled ISI || EC Academy](https://www.youtube.com/watch?v=bL3lmoH9a-c) — 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/correlative-coding-duobinary-error - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6 ## Question Explain any one type of correlative coding technique with its impulse response and transfer function. Mention how can we minimize error in duo-binary coding. ## Short answer Duobinary signaling uses c_k = a_k + a_{k-1}. Transfer function H(f)=2cos(πfTb)e^{-jπfTb} for |f|≤1/(2Tb); impulse response is sum of two sinc pulses. Error propagation is minimized by precoding d_k = b_k XOR d_{k-1}, enabling independent bit decisions using threshold |c_k|=1. ## Answer **Correlative (partial response) coding** introduces a controlled, known amount of ISI between successive symbols so that the data can be sent at rate $R_b$ through the minimum Nyquist bandwidth $R_b/2$ using realizable filters. The receiver removes the known ISI. The most common type is **duobinary signalling**. #### Duobinary signalling "Duobinary" means doubling the transmission capacity of a straight binary system. The binary input $b_k$ is mapped to $a_k = \pm 1$ and passed through a delay-and-add circuit followed by an ideal low-pass filter: $$ c_k = a_k + a_{k-1} $$ ```text a_k ---+------------>(+)--> [ideal LPF ] --> c(t) | ^ [|f|<1/2Tb ] +--> [delay Tb]+ ``` **Transfer function:** the delay-and-add has $1 + e^{-j2\pi fT_b}$; with the ideal Nyquist filter $H_N(f) = 1$ for $|f| \le 1/(2T_b)$: $$ \begin{aligned} H_I(f) &= \left(1 + e^{-j2\pi fT_b}\right)H_N(f) \\ &= \left(e^{j\pi fT_b} + e^{-j\pi fT_b}\right)e^{-j\pi fT_b} \\ &= 2\cos(\pi fT_b)\,e^{-j\pi fT_b}, \quad |f| \le \frac{1}{2T_b} \end{aligned} $$ and zero elsewhere. The magnitude is a half cosine that falls smoothly to zero at $f = 1/(2T_b)$, so it is much easier to approximate than the brick-wall filter. **Impulse response:** sum of two sinc pulses one bit apart: $$ h_I(t) = \frac{\sin(\pi t/T_b)}{\pi t/T_b} + \frac{\sin[\pi(t-T_b)/T_b]}{\pi(t-T_b)/T_b} = \frac{T_b^2\,\sin(\pi t/T_b)}{\pi t\,(T_b - t)} $$ $h_I(t)$ equals 1 at $t = 0$ and $t = T_b$ and is zero at all other sampling instants; its tails decay as $1/t^2$, so it is less sensitive to timing errors. **Output levels:** $c_k \in \{-2, 0, +2\}$. Detection without precoding: $\hat a_k = c_k - \hat a_{k-1}$. ```text |H(f)| 2 +---. | '. | '. | '. 0 +-----------+----> f 0 1/(2Tb) ``` #### Minimizing error in duobinary coding The main error source is **error propagation**: since each decision uses the previous decision, one error causes further errors. It is minimized by **precoding**: $d_k = b_k \oplus d_{k-1}$ at the transmitter. Then $c_k = 0$ means $b_k = 1$ and $c_k = \pm 2$ means $b_k = 0$, so each bit is decoded from its own sample (threshold $|c_k| = 1$) and an error does not spread. Proper thresholds and noise filtering also help. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the aim of correlative (partial response) coding? A. To eliminate all ISI by making tails zero B. To introduce a controlled, known ISI so data at Rb can be sent in bandwidth Rb/2 with realizable filters C. To add error-correcting parity bits D. To increase bandwidth for better noise immunity Answer: B. The known ISI is removed at the receiver, allowing minimum Nyquist bandwidth Rb/2 with a gradual (cosine) filter. (Key point: Purpose of correlative coding) 2. What is the magnitude of the duobinary transfer function for |f| <= 1/(2Tb)? A. 2 cos(pi f Tb) B. 2 sin(pi f Tb) C. 1 (flat) D. 2 sin(2 pi f Tb) Answer: A. H(f) = (1 + e^(-j2 pi f Tb)) H_N(f) = 2 cos(pi f Tb) e^(-j pi f Tb), a half cosine. (Key point: Duobinary transfer function) 3. The duobinary impulse response is the sum of two sinc pulses. How are they placed? A. Both at t = 0 B. One bit period Tb apart C. Two bit periods apart D. One at 0 and one at Tb/2 Answer: B. h(t) = sinc(t/Tb) + sinc((t-Tb)/Tb); it equals 1 at t = 0 and t = Tb and tails decay as 1/t^2. (Key point: Duobinary impulse response) 4. How is error propagation in duobinary minimized? A. Using a wider channel B. Precoding the data at the transmitter with d_k = b_k XOR d_(k-1) C. Increasing the number of levels D. Removing the low-pass filter Answer: B. Precoding lets each bit be decoded from its own sample, so a wrong decision does not carry forward. 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