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

# SQNR of delta modulation

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2070 Chaitra (CS II), 6 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-derive
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Derive the expression for evaluating signal to quantization noise ratio (SQNR) for Delta modulation.

## Short answer

For a sinusoid at slope overload limit, A_max = Δf_s/(2πf_m). Signal power is Δ^2 f_s^2/(8π^2 f_m^2), granular noise is Δ^2/3, reduced by LPF to Δ^2 f_M/(3f_s). SQNR = (3/(8π^2))(f_s^3/(f_m^2 f_M)).

## Answer

In delta modulation the quantization noise has two parts: slope-overload noise and granular noise. SQNR is derived assuming the step $\Delta$ is just large enough to avoid slope overload, so only **granular noise** remains.

**1. Maximum signal amplitude (no slope overload)**
For $m(t) = A\cos\omega_mt$ the maximum slope is $2\pi f_mA$. The staircase can change by $\Delta$ per $T_s$, so we need
$$
2\pi f_mA \le \Delta f_s \;\Rightarrow\; A_{max} = \frac{\Delta f_s}{2\pi f_m}
$$

**2. Signal power**
$$
S = \frac{A_{max}^2}{2} = \frac{\Delta^2f_s^2}{8\pi^2f_m^2}
$$

**3. Granular noise power**
Without slope overload the error $e(t) = m(t) - m_q(t)$ lies in $(-\Delta, +\Delta)$ and is assumed uniform with pdf $1/(2\Delta)$:
$$
\overline{e^2} = \int_{-\Delta}^{\Delta}e^2\frac{1}{2\Delta}\,de = \frac{\Delta^2}{3}
$$

**4. Effect of the output LPF**
The error waveform changes at the sampling rate, so its power spectral density is taken as roughly flat from 0 to $f_s$. The receiver LPF passes only $0$ to $f_M$ (message bandwidth), so
$$
N_q = \frac{\Delta^2}{3}\cdot\frac{f_M}{f_s}
$$

**5. SQNR**
$$
\begin{aligned}
\text{SQNR} &= \frac{S}{N_q} = \frac{\Delta^2f_s^2}{8\pi^2f_m^2}\cdot\frac{3f_s}{\Delta^2f_M} \\
&= \frac{3}{8\pi^2}\,\frac{f_s^3}{f_m^2f_M}
\end{aligned}
$$
With $f_m = f_M = W$:
$$
\text{SQNR} = \frac{3}{8\pi^2}\left(\frac{f_s}{W}\right)^3 \approx 0.038\left(\frac{f_s}{W}\right)^3
$$

**Observations**
- SQNR is independent of $\Delta$ (when $\Delta$ is set at the overload limit).
- SQNR $\propto f_s^3$: doubling $f_s$ improves SQNR by $10\log_{10}8 = 9$ dB.
- Compared with PCM (6 dB per bit, i.e. per unit increase of bandwidth factor), DM needs a much higher sampling rate for good quality. Example: $f_s/W = 32$ gives $0.038 \times 32\,768 \approx 1245$, i.e. about 31 dB.

## Check yourself

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

1. For m(t) = A cos(wm t) the maximum amplitude without slope overload is:
   A. Delta fs / (2 pi fm)
   B. 2 pi fm / (Delta fs)
   C. Delta / fs
   D. Delta fm
   Answer: A. From 2 pi fm A <= Delta fs. (Key point: No overload condition)

2. With error uniform in (-Delta, +Delta), noise power is:
   A. Delta^2/12
   B. Delta^2/3
   C. Delta^2
   D. Delta/2
   Answer: B. Integral of e^2/(2 Delta) from -Delta to Delta equals Delta^2/3. (Key point: Granular noise power)

3. Why is the noise multiplied by fM/fs?
   A. Noise spectrum is flat from 0 to fs and the LPF passes only 0 to fM
   B. Noise is only at fM
   C. Sampling halves the noise
   D. Signal is attenuated
   Answer: A. Only the fraction fM/fs of the flat noise spectrum passes the receiver LPF. (Key point: Output LPF effect)

4. The resulting SQNR is:
   A. (3/8 pi^2)(fs/W)^3
   B. 1.5 (fs/W)
   C. (3/8 pi^2)(fs/W)
   D. 1.5 L^2
   Answer: A. SQNR = 3 fs^3/(8 pi^2 fm^2 fM). (Key point: Final expression)

5. Doubling fs in DM improves SQNR by about:
   A. 3 dB
   B. 6 dB
   C. 9 dB
   D. 18 dB
   Answer: C. SQNR is proportional to fs^3, so 10 log 8 = 9 dB. (Key point: Doubling fs)

## Close variants set in other exams

- [Quantization; SQNR for linear quantization](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aquantization-sqnr-linear) — Communication Systems, chapter 4, asked 2×

## Related in Communication Systems

- [SQNR of delta modulation for a single-tone signal](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-sqnr-single-tone) — Communication Systems, chapter 4
- [SQNR for uniform quantization in terms of levels and bits](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asqnr-uniform-levels-bits) — Communication Systems, chapter 4, asked 2×
- [Need for non-uniform quantization; SQNR of DM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anonuniform-quantization-dm-sqnr) — Communication Systems, chapter 4, asked 2×
- [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
- [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
- [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

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