--- title: "Delta modulation encoder/decoder; PCM vs DM" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:dm-encoder-decoder-pcm-vs-dm" source: https://www.bctengineer.com/q/semester_6/communication_systems/dm-encoder-decoder-pcm-vs-dm site: BCT Engineering --- # Delta modulation encoder/decoder; PCM vs DM Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2075 Chaitra (CS II), 6+1 marks - Video: [Delta Modulation (Basics, Block Diagram, Working, Waveforms, Applications, Pros & Cons) Explained](https://www.youtube.com/watch?v=vzkLq0tcmhw) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/dm-encoder-decoder-pcm-vs-dm - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question Explain the Delta Modulation encoder and decoder with its derivations and diagram. Compare between PCM and DM? ## Short answer Delta Modulation uses a 1-bit quantizer with fixed step Δ. The encoder compares input sample with previous staircase approximation, outputting +Δ or -Δ. The decoder accumulates these steps and low-pass filters to reconstruct. Slope overload occurs when signal slope exceeds Δfs; granular noise appears for flat signals. PCM uses n bits/sample at Nyquist rate with quantization noise; DM uses 1 bit at high rate with slope overload and granular noise. SQNR improves 6 dB/bit for PCM, 9 dB per fs doubling for DM. ## Answer #### Delta modulation (DM) DM is a 1-bit DPCM: the signal is sampled much faster than the Nyquist rate and only **one bit** per sample is sent, telling whether the signal is above or below the previous staircase approximation. The staircase then steps up or down by a fixed $\Delta$. **Encoder** ```text m(t)->[Sampler]-m[k]->(+)-e[k]->[1-bit Q]-+-> eq[k] ^- | (+D/-D) mq[k-1] | v +--[Delay Ts]<--(+)< mq[k] ^ | mq[k-1] ``` **Decoder** ```text eq[k] -->(+)--mq[k]--> [LPF] --> m(t) ^ | +-[Delay Ts] (accumulator) ``` **Equations** $$ \begin{aligned} e[k] &= m[k] - m_q[k-1] \\ e_q[k] &= \Delta\,\text{sgn}(e[k]) = \pm\Delta \\ m_q[k] &= m_q[k-1] + e_q[k] = \sum_{i=0}^{k} e_q[i] \end{aligned} $$ The feedback loop (delay + adder) is an **accumulator (integrator)**, so the transmitter tracks the input with a staircase. The decoder is the same accumulator followed by a LPF that smooths the staircase. **Errors** - **Slope overload:** if the signal slope exceeds $\Delta f_s$ the staircase cannot follow. For $m(t) = A\sin 2\pi f_m t$ we need $$ \Delta f_s \ge 2\pi f_m A \;\Rightarrow\; A_{max} = \frac{\Delta f_s}{2\pi f_m} $$ - **Granular noise:** for a flat signal the staircase hunts $\pm\Delta$; error is uniform in $(-\Delta, \Delta)$ with power $\Delta^2/3$. After the LPF of bandwidth $f_M$: $$ N_q = \frac{\Delta^2}{3}\frac{f_M}{f_s}, \qquad \text{SQNR}_{max} = \frac{3}{8\pi^2}\frac{f_s^3}{f_m^2 f_M} $$ So SQNR rises 9 dB for every doubling of $f_s$. #### PCM vs DM | Point | PCM | DM | |---|---|---| | Bits/sample | $n$ (e.g. 8) | 1 | | Sampling rate | Near Nyquist | Much higher | | Errors | Quantization noise | Slope overload + granular | | SQNR improves by | 6 dB per bit | 9 dB per doubling of $f_s$ | | Bandwidth | $nW$ | $f_s/2$, high since $f_s \gg 2W$ (lower than PCM only at low SQNR) | | Complexity | ADC, framing needed | Very simple | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the DM transmitter, the feedback loop (delay plus adder) acts as: A. A differentiator B. An accumulator (integrator) forming a staircase C. A compressor D. A mixer Answer: B. mq[k] = mq[k-1] + eq[k] is a running sum, the integrator. (Key point: DM loop) 2. Slope overload in DM occurs when: A. The signal slope exceeds Delta x fs B. The signal is constant C. Delta is very small and fs very large D. The signal is zero Answer: A. The staircase can rise at most Delta fs; a steeper input cannot be followed. (Key point: Slope overload) 3. Granular noise in DM appears when: A. The signal is steep B. The signal is nearly flat and the staircase hunts by +/-Delta C. fs is below Nyquist D. The LPF is missing Answer: B. For a flat input the staircase oscillates by Delta, giving noise power Delta^2/3 before filtering. (Key point: Granular noise) 4. In DM, SQNR rises by how much for every doubling of fs? A. 3 dB B. 6 dB C. 9 dB D. 12 dB Answer: C. SQNR is proportional to fs^3, and 10 log10(8) = 9 dB. 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