---
title: "Companding; DPCM with derivation and block diagram"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:companding-dpcm"
source: https://www.bctengineer.com/q/semester_6/communication_systems/companding-dpcm
site: BCT Engineering
---

# Companding; DPCM with derivation and block diagram

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2075 Chaitra (CS II), 2+4 marks
- Video: [COMPANDING AND DPCM](https://www.youtube.com/watch?v=wETjxivRPZU) — Prof. Barapate's Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/companding-dpcm
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

What do you mean by companding? Briefly explain the operation of Differential PCM (DPCM) along with derivation and block diagram.

## Short answer

Companding uses compressor and expander for non-uniform quantization. DPCM transmits prediction error instead of sample, reducing bit rate. Block diagram includes predictor, quantizer, encoder, decoder.

## Answer

#### Companding
**Companding** (compressing + expanding) is a method of achieving non-uniform quantization. A compressor at the transmitter boosts weak signal amplitudes relative to strong ones, the result is uniformly quantized, and an expander at the receiver applies the inverse law. Weak signals thus get small effective steps and strong signals large steps, keeping SQNR almost constant over a wide range of speech levels. Standard laws are the µ-law ($\mu = 255$) and A-law ($A = 87.6$).

#### Differential PCM (DPCM)
Speech/video samples are strongly correlated, so DPCM transmits the **difference between the current sample and its predicted value** rather than the sample itself. The difference has a smaller range, so fewer bits are needed for the same SQNR.

```text
Transmitter
m[k]-->(+)--d[k]-->[Quantizer]--dq[k]-->[Encoder]--> out
        ^-                       |
   m^[k]|                        v
        +---[Predictor]<--(+)<---+
                           ^ m^[k]
Receiver
in-->[Decoder]--dq[k]-->(+)--mq[k]--> LPF --> m(t)
                         ^    |
                         +-[Predictor]
```

**Operation and derivation**
1. Prediction error: $d[k] = m[k] - \hat m[k]$, where $\hat m[k]$ is predicted from past reconstructed samples.
2. Quantized error: $d_q[k] = d[k] + q[k]$, with $q[k]$ the quantization error.
3. Predictor input:
$$
\begin{aligned}
m_q[k] &= \hat m[k] + d_q[k] = \hat m[k] + d[k] + q[k] \\
&= m[k] + q[k]
\end{aligned}
$$
So the reconstructed sample differs from the true one only by the quantization error of $d[k]$; errors do not accumulate.
4. The receiver has the same predictor, adds $d_q[k]$ to its prediction and gets $m_q[k]$, then low-pass filters it.
5. SNR gain: since $d[k]$ is quantized instead of $m[k]$,
$$
\text{SNR}_{DPCM} = \frac{\overline{m^2}}{\overline{d^2}}\cdot\frac{\overline{d^2}}{\overline{q^2}} = G_p \times \text{SNR}_q
$$
where $G_p = \overline{m^2}/\overline{d^2}$ is the prediction gain (4–11 dB for speech), equivalent to saving 1–2 bits/sample.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In companding, the receiver does what?
   A. Applies the inverse (expander) characteristic of the compressor
   B. Applies a second compressor
   C. Doubles the step size
   D. Samples at a higher rate
   Answer: A. The expander undoes the compression so the overall signal is restored. (Key point: Companding)

2. In DPCM the signal that is quantized is:
   A. The sample m[k] itself
   B. The prediction error d[k] = m[k] - predicted value
   C. The sampling clock
   D. The reconstructed signal
   Answer: B. DPCM quantizes the difference between sample and prediction. (Key point: Prediction error)

3. Why do quantization errors not accumulate in DPCM?
   A. The predictor uses reconstructed samples, so mq[k] = m[k] + q[k]
   B. The quantizer has infinite levels
   C. The receiver has no predictor
   D. The sampling rate is very high
   Answer: A. Feeding back the reconstructed value means only the current q[k] remains as error. (Key point: No error accumulation)

4. The SNR improvement of DPCM over PCM equals:
   A. Gp = mean m^2 / mean d^2
   B. Gp = mean d^2 / mean m^2
   C. L^2
   D. 6 dB always
   Answer: A. SNR_DPCM = Gp x SNR_q, where Gp is the signal-to-prediction-error power ratio (4-11 dB for speech). (Key point: Prediction gain)

## Related in Communication Systems

- [PCM, DPCM and DM comparison; SQNR of PCM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-dpcm-dm-compare-sqnr) — Communication Systems, chapter 4
- [PCM block diagram; T1 signaling rate and frame](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-block-t1-rate-frame) — Communication Systems, chapter 5
- [Companding; T1 hierarchy of TDM-PCM telephony](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acompanding-t1) — Communication Systems, chapter 5
- [Why DPCM over PCM; DPCM transmitter and receiver](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adpcm-preferred-transmitter-receiver) — Communication Systems, chapter 4, asked 2×
- [DPCM vs delta modulation; E1 hierarchy](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adpcm-vs-dm-e1) — Communication Systems, chapter 4
- [Aperture effect, DPCM and DM slope overload numerical](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aaperture-effect-dpcm-dm-slope) — Communication Systems, chapter 4

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