---
title: "Bits per sample and information rate (4.5 kHz, 8 levels)"
subject: "Communication Systems"
chapter: 6
key: "semester_6/communication_systems:bits-info-rate-4-5khz-8-levels"
source: https://www.bctengineer.com/q/semester_6/communication_systems/bits-info-rate-4-5khz-8-levels
site: BCT Engineering
---

# Bits per sample and information rate (4.5 kHz, 8 levels)

Communication Systems · Chapter 6: Baseband Digital Data Transmission

- Asked in: 2069 Chaitra (CS II), 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/bits-info-rate-4-5khz-8-levels
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6

## Question

A signal of bandwidth 4.5 kHz is sampled at the double rate given by Nyquist, the signal is quantized in 8 levels, the probability of occurrence of the level are 0.1, 0.15, 0.15, 0.05, 0.2, 0.05, 0.18, 0.12. Find the minimum no of bits per sample and information rate.

## Short answer

Nyquist rate is 9 kHz, so sampling rate is 18 kHz. For 8 levels, minimum fixed-length bits per sample is 3. Information rate using entropy (2.862 bits/sample) yields 51.52 kbps.

## Answer

**Given:** $f_m = 4.5$ kHz; sampled at twice the Nyquist rate; 8 levels with $P = \{0.1, 0.15, 0.15, 0.05, 0.2, 0.05, 0.18, 0.12\}$ (sum = 1).

**Sampling rate:** Nyquist rate $= 2f_m = 9$ kHz, so

$$
f_s = 2 \times 9\ \text{kHz} = 18\ 000\ \text{samples/s}
$$

**Minimum number of bits per sample:** with a fixed-length binary code, 8 levels need

$$
n = \log_2 8 = 3\ \text{bits/sample}
$$

**Entropy** (the theoretical minimum average number of bits per sample with ideal source coding):

| Symbol | $p_i$ | $\log_2(1/p_i)$ | $p_i\log_2(1/p_i)$ |
|---|---|---|---|
| Q1 | 0.1 | 3.3219 | 0.3322 |
| Q2 | 0.15 | 2.7370 | 0.4105 |
| Q3 | 0.15 | 2.7370 | 0.4105 |
| Q4 | 0.05 | 4.3219 | 0.2161 |
| Q5 | 0.2 | 2.3219 | 0.4644 |
| Q6 | 0.05 | 4.3219 | 0.2161 |
| Q7 | 0.18 | 2.4739 | 0.4453 |
| Q8 | 0.12 | 3.0589 | 0.3671 |
| **Total** | 1 | | **2.8622** |

$$
H = 2.8622\ \text{bits/sample}
$$

**Information rate:**

$$
R = f_s H = 18\ 000 \times 2.86224 = 51\ 520.24\ \text{bits/s} \approx 51.52\ \text{kbps}
$$

(With 3-bit PCM words the line bit rate is $18\ 000 \times 3 = 54$ kbps. If "the double rate given by Nyquist" is read as the Nyquist rate itself, $f_s = 9$ kHz and $R \approx 25.76$ kbps.)

**Answer:** 3 bits/sample (entropy limit 2.862 bits/sample); information rate ≈ **51.52 kbps**.

## Check yourself

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

1. A 4.5 kHz signal is sampled at twice the Nyquist rate. What is fs?
   A. 9 kHz
   B. 18 kHz
   C. 4.5 kHz
   D. 36 kHz
   Answer: B. Nyquist rate = 2 x 4.5 = 9 kHz; twice that is 18 kHz. (Key point: Sampling rate)

2. How many bits per sample does a fixed-length code need for 8 levels?
   A. 2
   B. 3
   C. 4
   D. 8
   Answer: B. log2 8 = 3 bits. (Key point: Bits per sample)

3. With H = 2.8622 bits/sample and fs = 18 000, what is the information rate?
   A. 54 kbps
   B. 51.52 kbps
   C. 25.8 kbps
   D. 18 kbps
   Answer: B. R = fs x H = 18 000 x 2.8622 = 51.52 kbps (the 3-bit PCM line rate would be 54 kbps). (Key point: Information rate)

## Related in Communication Systems

- [Information rate of a quantized 100 Hz signal](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ainformation-rate-100hz-four-symbols) — Communication Systems, chapter 6
- [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
- [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
- [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: minimum channel bandwidth numerical](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-bandwidth-3khz-0-5pct) — Communication Systems, chapter 4
- [Entropy and information rate; Shannon limit as bandwidth → ∞](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aentropy-info-rate-shannon-limit) — Communication Systems, chapter 6

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