---
title: "PCM vs DM SQNR for 6cos(5000πt) with 128 levels"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:pcm-dm-6cos5000pit-128"
source: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-dm-6cos5000pit-128
site: BCT Engineering
---

# PCM vs DM SQNR for 6cos(5000πt) with 128 levels

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2071 Shrawan (CS II), 2+3+2 marks
- Video: [PCM Part 2 Bandwidth, SQNR](https://www.youtube.com/watch?v=BRVoIkiRCV8) — Knowledge Sharing for Communication
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-dm-6cos5000pit-128
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

A message signal x(t) = 6cos(5000πt) is quantized in 128 levels using Nyquist sampling rate:
a) Find SQNR of the PCM signal
b) Find the sampling frequency required when same signal uses delta modulation for same SQNR
c) If the system uses DM using Nyquist sampling rate, find SQNR degradation in DM as compared to PCM.

## Short answer

SQNR for PCM with 128 levels is 43.91 dB. DM requires sampling frequency 216.2 kHz for same SQNR. DM at Nyquist rate is 49.08 dB worse than PCM.

## Answer

**Given:** $x(t) = 6\cos(5000\pi t)$, so $A = 6$ V, $f_m = 2500$ Hz. $L = 128$. Nyquist rate $f_s = 2f_m = 5$ kHz.

#### a) SQNR of PCM
$n = \log_2 128 = 7$ bits. For a full-load sinusoid ($m_p = 6$ V):
$$
\begin{aligned}
\Delta &= \frac{2 \times 6}{128} = 0.09375\ \text{V}, \quad N_q = \frac{\Delta^2}{12} \\
\text{SQNR} &= \frac{A^2/2}{\Delta^2/12} = 1.5L^2 = 1.5 \times 128^2 = 24\,576 \\
\text{SQNR}_{dB} &= 10\log_{10}24\,576 = 43.91\ \text{dB}
\end{aligned}
$$

#### b) Sampling frequency for DM with the same SQNR
For DM with a sinusoid at the slope-overload limit and output filter $f_M = f_m$:
$$
\text{SQNR}_{DM} = \frac{3}{8\pi^2}\left(\frac{f_s}{f_m}\right)^3
$$
Set equal to 24 576:
$$
\begin{aligned}
\left(\frac{f_s}{f_m}\right)^3 &= \frac{24\,576 \times 8\pi^2}{3} = 6.468 \times 10^5 \\
\frac{f_s}{f_m} &= 86.48 \\
f_s &= 86.48 \times 2500 = 216.2\ \text{kHz}
\end{aligned}
$$

#### c) SQNR degradation of DM at the Nyquist rate
With $f_s = 5$ kHz, $f_s/f_m = 2$:
$$
\begin{aligned}
\text{SQNR}_{DM} &= \frac{3}{8\pi^2}(2)^3 = \frac{3}{\pi^2} = 0.304 \\
\text{SQNR}_{DM,dB} &= -5.17\ \text{dB} \\
\text{Degradation} &= 43.91 - (-5.17) = 49.08\ \text{dB}
\end{aligned}
$$

**Answer: (a) 43.91 dB; (b) $f_s \approx 216.2$ kHz; (c) DM is about 49.08 dB worse than PCM.**

## Check yourself

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

1. For x(t) = 6cos(5000 pi t), the Nyquist sampling rate is:
   A. 2.5 kHz
   B. 5 kHz
   C. 10 kHz
   D. 5000 kHz
   Answer: B. fm = 2500 Hz, so fs = 2fm = 5 kHz. (Key point: Message frequency)

2. SQNR of PCM with 128 levels and full-load sinusoid is:
   A. 42.1 dB
   B. 43.91 dB
   C. 49.9 dB
   D. 38.4 dB
   Answer: B. SQNR = 1.5 x 128^2 = 24576, i.e. 43.91 dB (1.76 + 6.02 x 7). (Key point: PCM SQNR)

3. For DM with the same SQNR, the formula to solve is:
   A. SQNR = 1.5 L^2
   B. SQNR = 3/(8 pi^2) (fs/fm)^3
   C. SQNR = (fs/fm)
   D. SQNR = 6n
   Answer: B. Setting 3/(8 pi^2)(fs/fm)^3 = 24576 gives fs/fm = 86.5 and fs about 216 kHz. (Key point: DM sampling rate)

4. At the Nyquist rate DM SQNR is only 0.304 (-5.17 dB). The degradation vs PCM is:
   A. About 49 dB
   B. About 6 dB
   C. About 38 dB
   D. About 5 dB
   Answer: A. 43.91 - (-5.17) = 49.08 dB. (Key point: DM at Nyquist)

## Related in Communication Systems

- [Delta modulation encoder/decoder; PCM vs DM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-encoder-decoder-pcm-vs-dm) — Communication Systems, chapter 4
- [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
- [PCM bit rate for 4.5 MHz TV signal with 1024 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-tv-4-5mhz-1024) — Communication Systems, chapter 4
- [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
- [E1 digital hierarchy; SQNR in uniformly quantized PCM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-hierarchy-sqnr-uniform-pcm) — Communication Systems, chapter 5, asked 2×
- [SQNR of uniformly quantized PCM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asqnr-uniform-pcm-bits) — Communication Systems, chapter 4

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- [Sampling theorem; Nyquist of 1 + cos + sin](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampling-theorem-nyquist-1-cos-sin) — Signals and Systems, chapter 4, asked 5×
- [Prove sampling theorem; Nyquist of product](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aprove-sampling-theorem-nyquist) — Signals and Systems, chapter 4, asked 2×
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