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Delta modulation encoder/decoder; PCM vs 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.

Explain the Delta Modulation encoder and decoder with its derivations and diagram. Compare between PCM and DM?

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 Δ\Delta.

Encoder

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

eq[k] -->(+)--mq[k]--> [LPF] --> m(t)
          ^       |
          +-[Delay Ts]   (accumulator)

Equations

e[k]=m[k]−mq[k−1]eq[k]=Δ sgn(e[k])=±Δmq[k]=mq[k−1]+eq[k]=∑i=0keq[i]\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 Δfs\Delta f_s the staircase cannot follow. For m(t)=Asin⁡2πfmtm(t) = A\sin 2\pi f_m t we need
Δfs≥2πfmA  ⇒  Amax=Δfs2πfm\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 ±Δ\pm\Delta; error is uniform in (−Δ,Δ)(-\Delta, \Delta) with power Δ2/3\Delta^2/3. After the LPF of bandwidth fMf_M:
Nq=Δ23fMfs,SQNRmax=38π2fs3fm2fMN_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 fsf_s.

PCM vs DM

PointPCMDM
Bits/samplenn (e.g. 8)1
Sampling rateNear NyquistMuch higher
ErrorsQuantization noiseSlope overload + granular
SQNR improves by6 dB per bit9 dB per doubling of fsf_s
BandwidthnWnWfs/2f_s/2, high since fs≫2Wf_s \gg 2W (lower than PCM only at low SQNR)
ComplexityADC, framing neededVery simple

✓ Answer checked by Claude Opus 5.5 (high effort)

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