---
title: "PCM: minimum channel bandwidth numerical"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:pcm-bandwidth-3khz-0-5pct"
source: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-bandwidth-3khz-0-5pct
site: BCT Engineering
---

# PCM: minimum channel bandwidth numerical

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2081 Baisakh (CS II), 2+5 marks
- Video: [EE 370, 73- Minimum Bandwidth for Binary PCM](https://www.youtube.com/watch?v=tdU9eN-CTnc) — Wessam Mesbah
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-bandwidth-3khz-0-5pct
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Define Pulse Code Modulation (PCM). A band-limited signal m(t) with a bandwidth of 3 kHz is sampled at a rate equal to twice the Nyquist rate. The step size used during the quantization process is 0.5% of the peak amplitude mp. Assuming binary encoding, determine the minimum bandwidth of the channel to transmit the encoded binary signal.

## Short answer

Sampling at 12 kHz, L=400 requires 9 bits/sample, bit rate 108 kbps. Minimum Nyquist bandwidth for binary transmission is half the bit rate, so 54 kHz.

## Answer

#### Pulse code modulation (PCM)

**PCM** is a method of converting an analog signal into a digital (binary) signal by three steps: **sampling** at or above the Nyquist rate, **quantizing** each sample to one of $L$ levels, and **encoding** each level into an $n$-bit binary code word ($L = 2^n$). The binary pulses are then transmitted.

#### Numerical

**Given:** $W = 3$ kHz, sampling at twice the Nyquist rate, step size $\Delta = 0.5\%$ of $m_p$, binary coding.

**Step 1: Sampling rate**

$$
f_N = 2W = 6\ \text{kHz}, \qquad f_s = 2 \times 6 = 12\ \text{kHz}
$$

**Step 2: Number of levels.** The signal swings from $-m_p$ to $+m_p$, so

$$
\Delta = \frac{2m_p}{L} = 0.005\,m_p \;\Rightarrow\; L = \frac{2}{0.005} = 400
$$

**Step 3: Bits per sample.** $n$ must be an integer with $2^n \ge 400$:

$$
n = \lceil\log_2 400\rceil = \lceil 8.64\rceil = 9 \quad (2^9 = 512)
$$

**Step 4: Bit rate**

$$
R_b = n f_s = 9 \times 12000 = 108\ \text{kbps}
$$

**Step 5: Minimum channel bandwidth.** A binary signal of rate $R_b$ needs at least $R_b/2$ Hz (Nyquist bandwidth for ISI-free transmission):

$$
B_{min} = \frac{R_b}{2} = \frac{108}{2} = 54\ \text{kHz}
$$

**Answer:** minimum channel bandwidth = 54 kHz (9 bits/sample, 108 kbps).

## Check yourself

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

1. What are the three steps of PCM?
   A. Sampling, quantizing, encoding
   B. Encoding, mixing, sampling
   C. Filtering, modulating, detecting
   D. Sampling, filtering, amplifying
   Answer: A. Each sample becomes an n-bit word. (Key point: PCM steps)

2. With step 0.5% of mp and range 2mp, L = ?
   A. 100
   B. 400
   C. 200
   D. 512
   Answer: B. L = 2/0.005 = 400. (Key point: Levels)

3. How many bits per sample?
   A. 10
   B. 400
   C. 9
   D. 8
   Answer: C. 2^8 = 256 < 400 <= 512 = 2^9. (Key point: Bits)

4. What is the minimum channel bandwidth?
   A. 12 kHz
   B. 54 kHz
   C. 108 kHz
   D. 27 kHz
   Answer: B. Rb = 9 x 12 kHz = 108 kbps; B = Rb/2 = 54 kHz. (Key point: Bandwidth)

## Related in Communication Systems

- [Non-uniform quantization; PCM numerical (4 kHz, ±2.4 V, 64 kHz)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anonuniform-companding-pcm-4khz-64khz) — Communication Systems, chapter 4
- [TDM of 24 voice signals: PAM bandwidth and PCM sampling rate](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Atdm-24-voice-pam-pcm) — Communication Systems, chapter 5
- [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
- [Uniform vs non-uniform quantization; 16-level PCM numerical](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Auniform-nonuniform-16-level-pcm) — Communication Systems, chapter 4
- [Entropy and information rate of 6-level PCM signal](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aentropy-info-rate-5khz-six-levels) — Communication Systems, chapter 6
- [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 other subjects

- [Band-limited signals; aliasing; Nyquist rate](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Abandlimited-aliasing-nyquist) — Signals and Systems, chapter 4
- [Aliasing in a band-limited signal](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aaliasing-bandlimited-w) — Signals and Systems, chapter 4
- [Minimum sampling rate (sampling theorem)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aminimum-sampling-rate) — Digital Control System, chapter 1
- [Frequency of a signal sampled at 300 Hz](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampled-signal-frequency-300) — Signals and Systems, chapter 4