---
title: "PCM 7-bit at 50 Mbps: maximum message bandwidth"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:pcm-7bit-50mbps-bandwidth"
source: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-7bit-50mbps-bandwidth
site: BCT Engineering
---

# PCM 7-bit at 50 Mbps: maximum message bandwidth

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2073 Shrawan (CS II), 3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-7bit-50mbps-bandwidth
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

A PCM system uses a uniform quantizer followed by a 7 bit binary encoder. The bit rate of the system is equal to 50×10⁶ bits/sec. What is the maximum message signal bandwidth for which the system operates satisfactorily?

## Short answer

Sampling rate fs = Rb/n = 50×10⁶/7 ≈ 7.143 MHz. Maximum message bandwidth for Nyquist sampling is W_max = fs/2 ≈ 3.57 MHz.

## Answer

**Given:** $n = 7$ bits/sample, $R_b = 50 \times 10^6$ bps, uniform quantizer.

For satisfactory operation the sampling rate must be at least the Nyquist rate, $f_s \ge 2W$.

1. Bit rate relation:
$$
R_b = nf_s \;\Rightarrow\; f_s = \frac{R_b}{n} = \frac{50 \times 10^6}{7} = 7.143\ \text{MHz}
$$
2. Maximum message bandwidth (sampling exactly at Nyquist rate):
$$
W_{max} = \frac{f_s}{2} = \frac{R_b}{2n} = \frac{50 \times 10^6}{14} = 3.571\ \text{MHz}
$$

**Answer: Maximum message bandwidth ≈ 3.57 MHz.**

## Check yourself

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

1. Which relation gives the sampling rate from the bit rate?
   A. fs = n x Rb
   B. fs = Rb / n
   C. fs = Rb / 2n
   D. fs = Rb - n
   Answer: B. Rb = n fs, so fs = Rb/n = 50x10^6/7 = 7.143 MHz. (Key point: Bit rate relation)

2. What condition on fs gives satisfactory operation, and hence the maximum message bandwidth?
   A. fs >= 2W, so Wmax = fs/2
   B. fs >= W, so Wmax = fs
   C. fs >= 4W, so Wmax = fs/4
   D. fs <= 2W, so Wmax = fs/2
   Answer: A. Sampling must be at least the Nyquist rate 2W, so the largest W is fs/2. (Key point: Nyquist condition)

3. What is the maximum message bandwidth for the 7-bit, 50 Mbps PCM system?
   A. 7.14 MHz
   B. 3.57 MHz
   C. 1.79 MHz
   D. 25 MHz
   Answer: B. Wmax = Rb/(2n) = 50x10^6/14 = 3.571 MHz. (Key point: Final result)

## Related in Communication Systems

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