Q43.Marks: +2.0UGC NET Paper 2: Computer Sc 6th Jan 2025 Shift 1
Consider a noiseless channel with a bandwidth of 5000Hz transmitting a signal with two signal levels. The maximum bit rate is
1.2500 bps
2.10000 bps✓ Correct
3.5000 bps
4.20000 bps
Solution
The correct answer is 10000 bps.
Key Points
According to the Nyquist formula for a noiseless channel, the maximum bit rate can be calculated as:
Bit Rate = 2 * Bandwidth * log2(L)
Where L is the number of signal levels.
In this case, the bandwidth is 5000 Hz and there are 2 signal levels.
Substituting the values, we get:
Bit Rate = 2 * 5000 * log2(2)
log2(2) is 1.
Therefore, Bit Rate = 2 * 5000 * 1 = 10000 bps
Hence, the maximum bit rate is 10000 bps.
Additional Information
The Nyquist theorem is fundamental in digital communications for determining the maximum data rate of a noiseless channel.
This theorem is particularly useful in understanding the limitations of signal transmission in various communication systems.
Practical implementations may achieve lower bit rates due to various factors such as noise and signal attenuation.
The number of signal levels (L) directly influences the bit rate; higher levels can increase the bit rate, but also increase complexity and potential error rates.
