---
title: "Companding; delta modulator numerical (5 kHz, 200 kHz)"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:companding-dm-5khz-200khz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/companding-dm-5khz-200khz
site: BCT Engineering
---

# Companding; delta modulator numerical (5 kHz, 200 kHz)

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2079 Bhadra (CS II)
- Video: [COMPANDING](https://www.youtube.com/watch?v=rYnbbkVyIEY) — GATE ACHIEVERS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/companding-dm-5khz-200khz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

What is companding? A Delta modulator is used to encode speech signal band-limited to 5 kHz with a sampling frequency of 200 kHz. For maximum signal amplitude of Amax = 1, find
a) Minimum step size to avoid slope overloading
b) Assuming the speech signal to be sinusoidal, find signal to quantization noise ratio
c) Determine the minimum transmission bandwidth

## Short answer

Companding compresses weak signals before uniform quantization to maintain constant SQNR. For DM: minimum step size 0.157 V, SQNR 2431.7 (33.86 dB), minimum bandwidth 100 kHz.

## Answer

#### Companding
**Companding** = **comp**ressing + exp**anding**. At the transmitter the signal is passed through a compressor that amplifies weak signals more than strong ones; it is then uniformly quantized; at the receiver an expander applies the inverse characteristic. The overall effect is non-uniform quantization: small steps for weak signals and large steps for strong signals, so SQNR stays nearly constant over a wide dynamic range. Standard laws: **µ-law** ($\mu = 255$, North America/Japan) and **A-law** ($A = 87.6$, Europe/India/Nepal).

#### Given
$f_m = 5$ kHz, $f_s = 200$ kHz, $A_{max} = 1$, $T_s = 1/f_s$.

#### a) Minimum step size to avoid slope overload
The staircase can rise at most $\Delta/T_s = \Delta f_s$. For $m(t) = A\sin 2\pi f_m t$ the maximum slope is $2\pi f_m A$. No slope overload requires
$$
\begin{aligned}
\Delta f_s &\ge 2\pi f_m A_{max} \\
\Delta_{min} &= \frac{2\pi f_m A_{max}}{f_s} = \frac{2\pi \times 5000 \times 1}{200\,000} = 0.157\ \text{V}
\end{aligned}
$$

#### b) SQNR for a sinusoidal signal
- Signal power at the limit of slope overload: $S = A^2/2$ with $A = \Delta f_s/(2\pi f_m)$, so $S = \dfrac{\Delta^2 f_s^2}{8\pi^2 f_m^2}$.
- Granular noise is uniform in $(-\Delta, \Delta)$: power $\Delta^2/3$, spread evenly over $0$ to $f_s$; the output LPF (bandwidth $f_M = f_m$) passes the fraction $f_M/f_s$:
$N_q = \dfrac{\Delta^2}{3}\dfrac{f_M}{f_s}$.

$$
\begin{aligned}
\text{SQNR} &= \frac{3}{8\pi^2}\,\frac{f_s^3}{f_m^2 f_M} = \frac{3}{8\pi^2}\left(\frac{f_s}{f_m}\right)^3 \\
&= 0.0380 \times \left(\frac{200}{5}\right)^3 = 0.0380 \times 64\,000 = 2431.7 \\
\text{SQNR}_{dB} &= 10\log_{10}2431.7 = 33.86\ \text{dB}
\end{aligned}
$$

#### c) Minimum transmission bandwidth
DM sends one bit per sample, so $R_b = f_s = 200$ kbps. The minimum (Nyquist) bandwidth for binary data is $R_b/2$:
$$
B_{T,min} = \frac{R_b}{2} = \frac{200}{2} = 100\ \text{kHz}
$$

**Answer: $\Delta_{min} = 0.157$ V; SQNR = 2431.7 (33.86 dB); $B_T = 100$ kHz** (if rectangular pulses with $B_T = R_b$ are assumed, 200 kHz).

## Check yourself

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

1. What is the effect of companding?
   A. Uniform quantization with larger steps
   B. Non-uniform quantization: small steps for weak signals and large steps for strong ones
   C. Increasing the sampling rate
   D. Removing the aperture effect
   Answer: B. A compressor, a uniform quantizer and an expander together give non-uniform quantization with nearly constant SQNR. (Key point: Companding)

2. To avoid slope overload in a delta modulator with sinusoid amplitude A, we need:
   A. Delta x fs >= 2 pi fm A
   B. Delta >= fs/(2 pi fm)
   C. Delta x fm >= 2 pi fs A
   D. Delta <= 2 pi fm A / fs
   Answer: A. The staircase can rise at most Delta x fs per second, which must be at least the maximum signal slope 2 pi fm A. (Key point: Slope overload condition)

3. With fm = 5 kHz, fs = 200 kHz and Amax = 1, the minimum step size is about:
   A. 0.0157 V
   B. 0.157 V
   C. 1.57 V
   D. 0.314 V
   Answer: B. Delta = 2 pi x 5000 x 1 / 200000 = 0.157 V. (Key point: Minimum step size)

4. DM sends 1 bit per sample at fs = 200 kHz. The minimum (Nyquist) transmission bandwidth is:
   A. 200 kHz
   B. 100 kHz
   C. 5 kHz
   D. 400 kHz
   Answer: B. Rb = 200 kbps and the minimum bandwidth is Rb/2 = 100 kHz. (Key point: Bandwidth)

## Close variants set in other exams

- [Delta modulator numerical (3 kHz, 10 kHz sampling)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-3khz-10khz-numerical) — Communication Systems, chapter 4, asked 2×