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Why DPCM over PCM; DPCM transmitter and receiver

Short answer

DPCM preferred over PCM because correlated samples have redundancy; DPCM quantizes prediction error giving 4-11 dB SQNR gain or lower bit rate. Transmitter predicts sample, quantizes error, encodes, and feeds reconstructed sample to predictor. Receiver decodes error, uses identical predictor, sums to reconstruct signal.

Explain why DPCM is preferred over PCM? Explain the working principle of DPCM with necessary transmitter and receiver.

Why DPCM is preferred over PCM

Speech and video samples taken at or above the Nyquist rate are highly correlated: adjacent samples differ only slightly. PCM codes every sample fully, so it sends much redundant information.

DPCM predicts each sample from past samples and quantizes only the difference (prediction error). Since the error has a much smaller range than the signal:

  • For the same number of bits, the step size is smaller, so SQNR improves by the prediction gain Gp=σm2/σe2G_p = \sigma_m^2/\sigma_e^2 (typically 4–11 dB for speech).
  • For the same SQNR, fewer bits per sample are needed, so bit rate and bandwidth are lower (e.g. 32 kbps ADPCM vs 64 kbps PCM).

DPCM transmitter

m[n]-->(+)--e[n]-->[Quantizer]--+--eq[n]->[Encoder]->
        ^ -                     |
        |                       v
      m^[n]                    (+)<------+
        |                       | mq[n]  |
        +----[Predictor]<-------+        |
                  |                      |
                  +----------------------+
  1. Prediction error: e[n]=m[n]−m^[n]e[n] = m[n] - \hat{m}[n].
  2. Quantized error: eq[n]=e[n]+q[n]e_q[n] = e[n] + q[n], which is encoded and transmitted.
  3. The predictor input is the reconstructed sample mq[n]=m^[n]+eq[n]m_q[n] = \hat{m}[n] + e_q[n].
  4. Then mq[n]=m^[n]+e[n]+q[n]=m[n]+q[n]m_q[n] = \hat{m}[n] + e[n] + q[n] = m[n] + q[n]: the quantized sample differs from the original only by the quantization error of e[n]e[n], so errors do not accumulate.
  5. The predictor (a linear FIR filter, m^[n]=∑wkmq[n−k]\hat{m}[n] = \sum w_k m_q[n-k]) works on mqm_q, the same signal available at the receiver.

DPCM receiver

in->[Decoder]--eq[n]-->(+)--+--> mq[n] -->[LPF]--> m(t)
                       ^    |
                  m^[n]|    |
                       +--[Predictor]

The decoder recovers eq[n]e_q[n]; the same predictor as in the transmitter gives m^[n]\hat{m}[n]; their sum gives mq[n]=m[n]+q[n]m_q[n] = m[n] + q[n]. A low-pass filter reconstructs the analog signal.

Note: delta modulation is a special case of DPCM with a 1-bit quantizer and a one-sample delay as the predictor.

✓ Answer checked by Claude Opus 5.5 (high effort)

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