---
title: "Delta modulator numerical (3 kHz, 10 kHz sampling)"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:dm-3khz-10khz-numerical"
source: https://www.bctengineer.com/q/semester_6/communication_systems/dm-3khz-10khz-numerical
site: BCT Engineering
---

# Delta modulator numerical (3 kHz, 10 kHz sampling)

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2072 Kartik (CS II), 2+3+1 marks
- Video: [Examples of Delta Modulation | Solved Problems of Delta Modulation | Digital Communication](https://www.youtube.com/watch?v=mH9HsunLmBc) — 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-3khz-10khz-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

A Delta modulator is used to encode speech signal band-limited to 3 kHz with sampling frequency 10 kHz. For maximum signal amplitude of Amax = 1, find:
i) Minimum step size to avoid slope overloading.
ii) Assuming the speech signal to be sinusoidal, find Signal to quantization noise ratio
iii) Determine the minimum transmission bandwidth.

## Short answer

Minimum step size to avoid slope overload: Δ_min = 2πfm Amax / fs = 2π×3000×1/10000 = 1.885 V. For sinusoidal input, SQNR = (3/(8π²))(fs³/(fm² fM)) = 1.407 (1.48 dB). Minimum transmission bandwidth BT = Rb/2 = 5 kHz.

## Answer

**Given:** $f_m = 3$ kHz, $f_s = 10$ kHz, $A_{max} = 1$. Assume output LPF bandwidth $f_M = f_m = 3$ kHz.

#### i) Minimum step size to avoid slope overload
The staircase slope $\Delta f_s$ must be at least the maximum signal slope $2\pi f_mA$:
$$
\begin{aligned}
\Delta_{min} &= \frac{2\pi f_mA_{max}}{f_s} \\
&= \frac{2\pi \times 3000 \times 1}{10\,000} = 1.885\ \text{V}
\end{aligned}
$$

#### ii) SQNR for a sinusoidal signal
Signal power at maximum amplitude $S = A^2/2 = \dfrac{\Delta^2f_s^2}{8\pi^2f_m^2}$; granular noise after the LPF $N_q = \dfrac{\Delta^2}{3}\dfrac{f_M}{f_s}$:
$$
\begin{aligned}
\text{SQNR} &= \frac{3}{8\pi^2}\,\frac{f_s^3}{f_m^2f_M} = \frac{3}{8\pi^2}\left(\frac{10}{3}\right)^3 \\
&= 0.0380 \times 37.04 = 1.407 \\
\text{SQNR}_{dB} &= 10\log_{10}1.407 = 1.48\ \text{dB}
\end{aligned}
$$
This very low value shows that $f_s = 10$ kHz is far too low for DM; DM needs $f_s$ many times the Nyquist rate.

#### iii) Minimum transmission bandwidth
One bit per sample: $R_b = f_s = 10$ kbps.
$$
B_{T,min} = \frac{R_b}{2} = 5\ \text{kHz}
$$

**Answer: $\Delta_{min} = 1.885$ V; SQNR = 1.407 (1.48 dB); $B_T = 5$ kHz** (10 kHz if $B_T = R_b$ is taken).

## Check yourself

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

1. To avoid slope overload the step size must satisfy:
   A. Delta >= 2 pi fm A / fs
   B. Delta <= 2 pi fm A / fs
   C. Delta = fs / (2 pi fm A)
   D. Delta >= fs A
   Answer: A. The staircase slope Delta fs must be at least the maximum signal slope 2 pi fm A. (Key point: Slope overload condition)

2. For fm = 3 kHz, fs = 10 kHz, A = 1, minimum Delta is:
   A. 0.3 V
   B. 1.885 V
   C. 6.28 V
   D. 0.53 V
   Answer: B. Delta = 2 pi x 3000 / 10000 = 1.885 V. (Key point: Minimum step)

3. Which formula gives SQNR of DM for a sinusoid?
   A. 3/(8 pi^2) x fs^3 / (fm^2 fM)
   B. 1.5 L^2
   C. Delta^2/12
   D. 3/(8 pi^2) x fs / fm
   Answer: A. Signal power Delta^2 fs^2/(8 pi^2 fm^2) divided by granular noise (Delta^2/3)(fM/fs). (Key point: DM SQNR)

4. Why is the SQNR here only about 1.48 dB?
   A. Delta is too large
   B. fs = 10 kHz is too low for DM, which needs fs far above the Nyquist rate
   C. Amplitude is too small
   D. A quantizer is missing
   Answer: B. SQNR scales as (fs/fm)^3, and fs/fm is only 3.33. (Key point: Result)

5. One bit per sample at fs = 10 kHz gives a minimum bandwidth of:
   A. 10 kHz
   B. 20 kHz
   C. 3 kHz
   D. 5 kHz
   Answer: D. Rb = fs = 10 kbps and BT = Rb/2 = 5 kHz. (Key point: DM bandwidth)

## Close variants set in other exams

- [Companding; delta modulator numerical (5 kHz, 200 kHz)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acompanding-dm-5khz-200khz) — Communication Systems, chapter 4, asked 2×

## Related in Communication Systems

- [PCM: minimum channel bandwidth numerical](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-bandwidth-3khz-0-5pct) — 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