---
title: "Information rate of a quantized 100 Hz signal"
subject: "Communication Systems"
chapter: 6
key: "semester_6/communication_systems:information-rate-100hz-four-symbols"
source: https://www.bctengineer.com/q/semester_6/communication_systems/information-rate-100hz-four-symbols
site: BCT Engineering
---

# Information rate of a quantized 100 Hz signal

Communication Systems · Chapter 6: Baseband Digital Data Transmission

- Asked in: 2076 Asoj (CS II), 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/information-rate-100hz-four-symbols
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6

## Question

An analog baseband signal, band limited to 100 Hz, is sampled at the Nyquist rate. The samples are quantized into four message symbols that occur independently with probabilities p1 = p4 = 0.125 and p2 = p3. Determine the information rate (bits/sec) of the message source signal.

## Short answer

Nyquist sampling gives symbol rate r = 200 symbols/s. Solving p2 = p3 = 0.375, entropy H = 1.8113 bits/symbol. Information rate R = r H = 362.26 bits/s.

## Answer

**Given:** $f_m = 100$ Hz, sampled at Nyquist rate; four symbols with $p_1 = p_4 = 0.125$ and $p_2 = p_3$.

**Step 1: find $p_2$ and $p_3$.** Probabilities add to 1:

$$
\begin{aligned}
p_1 + p_2 + p_3 + p_4 &= 1 \\
0.125 + 2p_2 + 0.125 &= 1 \;\Rightarrow\; p_2 = p_3 = 0.375
\end{aligned}
$$

**Step 2: symbol rate.** Nyquist sampling rate:

$$
r = f_s = 2f_m = 2 \times 100 = 200\ \text{samples (symbols)/s}
$$

**Step 3: entropy.**

$$
\begin{aligned}
H &= 2(0.125)\log_2\frac{1}{0.125} + 2(0.375)\log_2\frac{1}{0.375} \\
&= 2(0.125)(3) + 2(0.375)(1.4150) \\
&= 0.75 + 1.0613 = 1.8113\ \text{bits/symbol}
\end{aligned}
$$

**Step 4: information rate.**

$$
R = rH = 200 \times 1.8113 = 362.26\ \text{bits/s}
$$

Note: with 2 bits/sample (fixed coding) the bit rate would be 400 bits/s, so the source information rate is less than the coded bit rate.

**Answer:** information rate $R \approx$ **362.26 bits/s** ($H = 1.8113$ bits/symbol, $r = 200$ symbols/s).

## Check yourself

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

1. With p1 = p4 = 0.125 and p2 = p3, what is p2?
   A. 0.25
   B. 0.375
   C. 0.5
   D. 0.75
   Answer: B. 0.125 + 2 p2 + 0.125 = 1 gives p2 = 0.375. (Key point: Find p2 and p3)

2. A signal band limited to 100 Hz is sampled at the Nyquist rate. The symbol rate r is:
   A. 100 symbols/s
   B. 200 symbols/s
   C. 400 symbols/s
   D. 50 symbols/s
   Answer: B. Nyquist rate is 2 fm = 200 samples/s, each sample being one symbol. (Key point: Symbol rate)

3. The entropy of this source is about:
   A. 2 bits/symbol
   B. 1.81 bits/symbol
   C. 1.5 bits/symbol
   D. 3 bits/symbol
   Answer: B. H = 2(0.125)(3) + 2(0.375)(1.415) = 0.75 + 1.0613 = 1.8113 bits/symbol. (Key point: Entropy)

4. The information rate R is about:
   A. 400 bits/s
   B. 362.26 bits/s
   C. 181 bits/s
   D. 200 bits/s
   Answer: B. R = rH = 200 x 1.8113 = 362.26 bits/s, below the 400 bits/s of 2-bit fixed coding. (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
- [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
- [Reconstruction from samples taken at Nyquist rate](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Areconstruction-nyquist-samples) — Communication Systems, chapter 4
- [USB-SSB expression for a random band limited message](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ausb-ssb-random-message) — Communication Systems, chapter 2
- [Symbol rate, entropy and information rate (6 symbols per 10 μs)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asymbol-rate-six-symbols-10us) — Communication Systems, chapter 6
- [FM numerical: 10 kHz baseband, 100 MHz carrier, 75 kHz deviation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-10khz-100mhz-75khz) — Communication Systems, chapter 3

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