Nyquist Formula Calculates 24 Kbps | Generated by AI
Question: According to the Nyquist formula, if the channel bandwidth H = 3 KHz and each signal symbol can take N = 16 discrete values, what is the maximum data transmission rate of the channel?
Answer:
The Nyquist formula determines the maximum data transmission rate in a noiseless channel. Here’s the solution:
Given Information:
- Channel bandwidth (H) = 3 KHz
- Number of discrete signal levels (N) = 16
Solution Process:
According to the Nyquist theorem, the maximum symbol rate (baud rate) in a noiseless channel is:
Maximum symbol rate = 2 × H
Therefore:
- Maximum symbol rate = 2 × 3 KHz = 6 kilobaud (symbols per second)
The relationship between data transmission rate and symbol rate is:
Data transmission rate = Symbol rate × log₂(N)
Where log₂(N) represents the number of bits per symbol.
Since N = 16:
- log₂(16) = 4 bits per symbol
Therefore: Maximum data transmission rate = 6 × 4 = 24 kbps
Explanation:
The Nyquist theorem establishes that the maximum symbol rate is twice the bandwidth. With 16 possible discrete values per symbol, each symbol can encode log₂(16) = 4 bits of information. Multiplying the maximum symbol rate by the bits per symbol gives us the maximum data transmission rate.
Correct Answer: B (24 Kbps)
Your answer of C (48 Kbps) appears to have multiplied the bandwidth by 16 directly without properly applying the Nyquist theorem’s 2× factor, or incorrectly calculated log₂(16) as 8 instead of 4.