---
title: "SQNR of delta modulation for a single-tone signal"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:dm-sqnr-single-tone"
source: https://www.bctengineer.com/q/semester_6/communication_systems/dm-sqnr-single-tone
site: BCT Engineering
---

# SQNR of delta modulation for a single-tone signal

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2081 Baisakh (CS II), 7 marks
- Video: [SQNR of Delta Modulation | Designing of Delta Modulation System| SQNR Derivation of Delta Modulation](https://www.youtube.com/watch?v=y9gpu_5YuvQ) — 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-sqnr-single-tone
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Derive an expression for the SQNR (in dB) of the delta modulation for a single-tone analog signal.

## Short answer

For a single tone at max amplitude without slope overload, SQNR = 3 fs³/(8π² fm² W). In dB: 30log10 fs – 20log10 fm – 10log10 W – 14.2. It is independent of Δ and improves 9 dB per doubling of fs.

## Answer

**Setup.** Delta modulation uses step size $\Delta$, sampling rate $f_s = 1/T_s$ and a reconstruction LPF of bandwidth $W$ at the receiver. The message is a single tone:

$$
m(t) = A_m\cos 2\pi f_m t
$$

#### Step 1: Maximum amplitude without slope overload

The staircase can change by at most $\Delta$ in $T_s$, so the maximum slope it can follow is $\Delta/T_s = \Delta f_s$. The maximum slope of the tone is

$$
\left|\frac{dm}{dt}\right|_{max} = 2\pi f_m A_m
$$

No slope overload requires $2\pi f_m A_m \le \Delta f_s$, so

$$
A_{m,max} = \frac{\Delta f_s}{2\pi f_m}
$$

#### Step 2: Signal power

$$
S = \frac{A_{m,max}^2}{2} = \frac{\Delta^2 f_s^2}{8\pi^2 f_m^2}
$$

#### Step 3: Granular (quantization) noise power

Without slope overload, the error $q(t) = m(t) - m_q(t)$ stays within $\pm\Delta$. Assuming it is uniformly distributed in $(-\Delta, \Delta)$:

$$
E[q^2] = \int_{-\Delta}^{\Delta} q^2\frac{1}{2\Delta}\,dq = \frac{1}{2\Delta}\cdot\frac{2\Delta^3}{3} = \frac{\Delta^2}{3}
$$

This noise power is spread approximately uniformly over $0$ to $f_s$. The LPF passes only $0$ to $W$:

$$
N_q = \frac{\Delta^2}{3}\cdot\frac{W}{f_s}
$$

#### Step 4: SQNR

$$
\begin{aligned}
\text{SQNR} &= \frac{S}{N_q} = \frac{\Delta^2 f_s^2}{8\pi^2 f_m^2}\cdot\frac{3 f_s}{\Delta^2 W} \\
&= \frac{3 f_s^3}{8\pi^2 f_m^2 W}
\end{aligned}
$$

#### Step 5: In dB

Since $10\log_{10}\frac{3}{8\pi^2} = -14.2$ dB,

$$
\text{SQNR}_{dB} = 30\log_{10} f_s - 20\log_{10} f_m - 10\log_{10} W - 14.2\ \text{dB}
$$

**Observations:**

- SQNR $\propto f_s^3$: doubling the sampling rate improves SQNR by **9 dB** (compared with 6 dB per extra bit in PCM).
- SQNR does not depend on $\Delta$ (at the maximum amplitude), because the allowed amplitude and noise both scale with $\Delta$.

**Example:** $f_s = 64$ kHz, $f_m = 1$ kHz, $W = 4$ kHz:

$$
\text{SQNR} = \frac{3(64000)^3}{8\pi^2(1000)^2(4000)} = 2490 \approx 33.96\ \text{dB}
$$

## Check yourself

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

1. What is the maximum tone amplitude without slope overload?
   A. 2 pi Delta fs/fm
   B. Delta fm/(2 pi fs)
   C. Delta/fs
   D. Delta fs/(2 pi fm)
   Answer: D. Max slope 2 pi fm Am <= Delta fs. (Key point: Overload)

2. What is the signal power at max amplitude?
   A. Delta fs/(2 pi fm)
   B. Delta^2 fs^2/(8 pi^2 fm^2)
   C. Delta^2/3
   D. Delta^2 fs^2/(4 pi^2 fm^2)
   Answer: B. S = Am^2/2. (Key point: Signal power)

3. What is the output noise power?
   A. Delta^2/12
   B. (Delta^2/3)(W/fs)
   C. (Delta^2/12)(W/fs)
   D. Delta^2/3
   Answer: B. Error uniform in +-Delta. (Key point: Noise)

4. What is the SQNR?
   A. 3 fs^3/(8 pi^2 fm^2 W)
   B. 3 fs^2/(8 pi^2 fm W)
   C. 3 fs/(8 pi^2 fm W)
   D. fs^3/(fm^2 W)
   Answer: A. S/Nq simplified; independent of Delta. (Key point: SQNR)

## Related in Communication Systems

- [DSB-FC single tone expression and modulation index](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbfc-single-tone-index) — Communication Systems, chapter 2
- [SQNR of delta modulation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-sqnr-derive) — Communication Systems, chapter 4, asked 2×
- [Double-tone AM expression, bandwidth and modulation index](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adouble-tone-am-bw-index) — Communication Systems, chapter 2
- [Angle modulation; single tone FM time and frequency domain](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-single-tone-fm-time-freq) — Communication Systems, chapter 3
- [Need for modulation; single tone AM expressions](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amodulation-need-am-single-tone) — Communication Systems, chapter 2, asked 2×
- [Single-tone FM in terms of Bessel coefficients](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asingle-tone-fm-bessel) — Communication Systems, chapter 3, asked 5×

## Related in other subjects

- [Single tone FM parameters](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afm-single-tone-numeric) — Data Communication, chapter 5
- [Digital vs analog modulation; GMSK](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Adigital-modulation-gmsk) — Wireless Communication, chapter 4
- [Analog vs digital signal; convert A2.64H](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aanalog-digital-convert-a264h) — Digital Logic, chapter 1
- [Single-pulse PWM: modulation index and Fourier series](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-pulse-pwm-fourier) — Power Electronics, chapter 5