--- title: "Nyquist criteria for zero ISI; duobinary difficulties" subject: "Communication Systems" chapter: 6 key: "semester_6/communication_systems:nyquist-zero-isi-duobinary-difficulties" source: https://www.bctengineer.com/q/semester_6/communication_systems/nyquist-zero-isi-duobinary-difficulties site: BCT Engineering --- # Nyquist criteria for zero ISI; duobinary difficulties Communication Systems · Chapter 6: Baseband Digital Data Transmission - Asked in: 2071 Chaitra (CS II), 3+6 marks - Video: [Unit 2 L4.1 | Nyquist Criterion for Distortion less Base Band Transmission | Digital communication](https://www.youtube.com/watch?v=za-fgUWRnmQ) — Techno Tutorials ( e-Learning) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/nyquist-zero-isi-duobinary-difficulties - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6 ## Question State Nyquist Criteria for zero ISI in both time and frequency domain. What are two major difficulties with duo binary encoding method and explain how can they be solved? ## Short answer Time domain: pulse satisfies p(nTb)=0 for n≠0. Frequency domain: sum of shifted spectra equals constant. Duobinary has error propagation, solved by precoding d_k = b_k ⊕ d_{k-1}. Its non-zero DC is overcome by modified duobinary using c_k = a_k − a_{k-2}, which has a spectral null at DC. ## Answer #### Nyquist criterion for zero ISI Let $p(t)$ be the overall pulse (transmit filter, channel, receive filter), normalized so $p(0) = 1$, and $T_b$ the bit period. **Time domain:** the pulse must be zero at every sampling instant except its own: $$ p(nT_b) = \begin{cases} 1, & n = 0 \\ 0, & n \ne 0 \end{cases} $$ **Frequency domain:** the spectrum and its copies shifted by multiples of $R_b = 1/T_b$ must add to a constant: $$ \sum_{n=-\infty}^{\infty} P\left(f - \frac{n}{T_b}\right) = T_b $$ The minimum-bandwidth solution is $P(f) = T_b\,\text{rect}(fT_b)$ (bandwidth $R_b/2$, pulse $\text{sinc}(t/T_b)$); the practical solution is the raised cosine spectrum with bandwidth $\frac{R_b}{2}(1 + \alpha)$. #### Two major difficulties with duobinary encoding and their solutions Duobinary sends $c_k = a_k + a_{k-1}$ with $H(f) = 2\cos(\pi fT_b)e^{-j\pi fT_b}$. **1. Error propagation** The receiver estimates $\hat a_k = c_k - \hat a_{k-1}$. If one decision is wrong, the wrong value is used for the next decision, and the error carries forward. **Solution: precoding.** Before level mapping form $$ d_k = b_k \oplus d_{k-1} $$ Then $c_k = 0$ if $b_k = 1$ and $c_k = \pm 2$ if $b_k = 0$, so the detector decides each bit from its own sample: $|c_k| < 1 \Rightarrow 1$, otherwise 0. No previous decision is needed, so errors do not propagate. Example ($d_0 = 1$): $b = 0\,1\,1$ gives $d = 1\,0\,1$, $a = +1, -1, +1$, $c = +2, 0, 0$, decoded $0\,1\,1$. **2. Non-zero DC (low-frequency) content** $|H(f)| = 2\cos(\pi fT_b)$ is maximum at $f = 0$, so the duobinary signal has a strong DC component. It cannot pass through AC-coupled (transformer or capacitor) lines, and it is unsuitable for single-sideband transmission, which needs a spectral null near zero frequency. **Solution: modified duobinary signalling.** Correlate symbols two bits apart: $$ c_k = a_k - a_{k-2}, \qquad H(f) = 2j\sin(2\pi fT_b)\,e^{-j2\pi fT_b} $$ $H(0) = 0$, so there is no DC component. It is used with the precoder $d_k = b_k \oplus d_{k-2}$ to avoid error propagation. (Another drawback: three output levels instead of two need more signal power for the same error rate; this is accepted as the price for minimum bandwidth.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the Nyquist criterion require in the time domain for zero ISI? A. p(nTb) = 1 for all n B. p(nTb) = 1 for n = 0 and 0 for n not equal to 0 C. p(t) must be zero for all t D. p(t) must be a rectangular pulse Answer: B. The pulse must be zero at all other sampling instants. (Key point: Time-domain Nyquist criterion) 2. What is the frequency-domain Nyquist condition? A. P(f) must be zero outside Rb B. The sum of P(f - n/Tb) over all n equals a constant (Tb) C. P(f) must be infinite at f = 0 D. P(f) must be a cosine Answer: B. Shifted copies of the spectrum by multiples of Rb must add to a constant. (Key point: Frequency-domain criterion) 3. What problem does precoding solve in duobinary? A. Non-zero DC content B. Error propagation C. Excess bandwidth D. Too many levels Answer: B. Precoding lets each bit be decoded from its own sample without using the previous decision. (Key point: Precoding solution) 4. How does modified duobinary remove the DC component? A. c_k = a_k + a_(k-1) B. c_k = a_k - a_(k-2), giving H(0) = 0 C. By raising the roll-off factor D. By adding a DC blocking capacitor only Answer: B. Correlating symbols two bits apart gives H(f) proportional to sin(2 pi f Tb), zero at f = 0. 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