---
title: "Symbol rate, entropy and information rate (6 symbols per 10 μs)"
subject: "Communication Systems"
chapter: 6
key: "semester_6/communication_systems:symbol-rate-six-symbols-10us"
source: https://www.bctengineer.com/q/semester_6/communication_systems/symbol-rate-six-symbols-10us
site: BCT Engineering
---

# Symbol rate, entropy and information rate (6 symbols per 10 μs)

Communication Systems · Chapter 6: Baseband Digital Data Transmission

- Asked in: 2070 Asar (CS II), 6 marks
- Video: [Calculate Source Entropy and Information Rate | Communication system.](https://www.youtube.com/watch?v=pi1XA8j5dcY) — ELECTRICAL HUB
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/symbol-rate-six-symbols-10us
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6

## Question

A discrete source emits one of 6 possible symbols per 10 μs in statistically independent manner. The symbol probabilities are 1/4, 1/4, 1/4, 1/8, 1/16 and 1/16 respectively. Calculate symbol rate, entropy and information rate.

## Short answer

Symbol rate r=1/(10 μs)=100 ksym/s. Entropy H = Σ p_i log2(1/p_i) = 2.375 bits/symbol. Information rate R = rH = 237.5 kbps.

## Answer

**Given:** one symbol per $T_s = 10\ \mu$s; $P = \{\tfrac14, \tfrac14, \tfrac14, \tfrac18, \tfrac1{16}, \tfrac1{16}\}$ (sum = 1).

#### Symbol rate

$$
r = \frac{1}{T_s} = \frac{1}{10\times10^{-6}} = 10^5\ \text{symbols/s} = 100\ \text{kbaud}
$$

#### Entropy

| Symbol | $p_i$ | $\log_2(1/p_i)$ | $p_i\log_2(1/p_i)$ |
|---|---|---|---|
| x1 | 0.25 | 2.0000 | 0.5000 |
| x2 | 0.25 | 2.0000 | 0.5000 |
| x3 | 0.25 | 2.0000 | 0.5000 |
| x4 | 0.125 | 3.0000 | 0.3750 |
| x5 | 0.0625 | 4.0000 | 0.2500 |
| x6 | 0.0625 | 4.0000 | 0.2500 |
| **Total** | 1 | | **2.3750** |

$$
\begin{aligned}
H &= 3\left(\tfrac14\log_2 4\right) + \tfrac18\log_2 8 + 2\left(\tfrac1{16}\log_2 16\right) \\
&= 3(0.5) + 0.375 + 2(0.25) \\
&= 1.5 + 0.375 + 0.5 = 2.375\ \text{bits/symbol}
\end{aligned}
$$

(Maximum possible for 6 symbols: $\log_2 6 = 2.585$ bits/symbol.)

#### Information rate

$$
R = rH = 10^5 \times 2.375 = 2.375\times10^5\ \text{bits/s} = 237.5\ \text{kbps}
$$

**Answer:** $r = 10^5$ symbols/s, $H = 2.375$ bits/symbol, $R = 237.5$ kbps.

## Check yourself

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

1. One symbol is emitted every 10 microseconds. What is the symbol rate?
   A. 10^4 symbols/s
   B. 10^5 symbols/s
   C. 10^6 symbols/s
   D. 10 symbols/s
   Answer: B. r = 1/(10 x 10^-6) = 10^5 symbols/s. (Key point: Symbol rate)

2. For P = {1/4,1/4,1/4,1/8,1/16,1/16}, what is the entropy?
   A. 2.375 bits/symbol
   B. 2.585 bits/symbol
   C. 2.0 bits/symbol
   D. 1.875 bits/symbol
   Answer: A. H = 3(0.5) + 0.375 + 2(0.25) = 2.375 bits/symbol. (Key point: Entropy)

3. What is the information rate?
   A. 100 kbps
   B. 237.5 kbps
   C. 258.5 kbps
   D. 23.75 kbps
   Answer: B. R = rH = 10^5 x 2.375 = 237.5 kbps. (Key point: Information rate)

## Related in Communication Systems

- [Message vs information; 5 symbols per 10 μs](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amessage-info-five-symbols-10us) — Communication Systems, chapter 6
- [Importance of source coding; Huffman for eight symbols at 1000 symbols/s](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asource-coding-huffman-eight-1000) — 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
- [Source vs channel coding; Huffman for five symbols](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asource-channel-huffman-five) — Communication Systems, chapter 6
- [Huffman code for seven-symbol source](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ahuffman-seven-symbols) — Communication Systems, chapter 6
- [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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- [Numerical: max symbol rate without equalizer](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Asymbol-period-power-delay-profile) — Wireless Communication, chapter 3