Consider a noiseless channel with a bandwidth of 5000Hz transmitting a signal with two signal levels. The maximum bit rate is:

Correct option is B

The maximum bit rate of a noiseless channel can be calculated using the Nyquist Theorem for a noiseless channel. The theorem states that the maximum bit rate $$R_{\text{max}}$$​ can be given by:
​$$R_{\text{max}} = 2 \times B \times \log_2(L)$$​​
Where:
• B is the bandwidth of the channel in Hz.
• L is the number of signal levels.
Given:
• The bandwidth B = 5000 Hz.
• The signal has L = 2 signal levels (since the signal is binary).
Substituting these values into the formula:
​$$R_{\text{max}} = 2 \times 5000 \times \log_2(2)$$​​
Since $$\log_2(2) = 1$$​, we get:
​$$R_{\text{max}} = 2 \times 5000 \times 1 = 10000 \, \text{bps}$$​​
Information Booster:
1. Nyquist Theorem:
o The Nyquist theorem is used to determine the maximum bit rate of a noiseless channel.
o It states that the maximum bit rate is proportional to the bandwidth and the number of signal levels.
2. Bandwidth and Signal Levels:
o Bandwidth determines how much data can be transmitted per second.
o Signal levels represent the distinct states in which data can be encoded. For binary signals, there are 2 signal levels (0 and 1).
3. Logarithmic Calculation:
o The term $$\log_2(L)$$​ calculates the number of bits that can be transmitted per signal level. For 2 levels, $$\log_2(2) = 1$$​, meaning each signal carries one bit of information.
4. Formula Impact:
o The formula shows that with higher bandwidth or more signal levels, the bit rate increases.
o The noiseless condition implies there is no error, allowing for maximum data transmission.
Additional Knowledge:
• Maximum Bit Rate Calculation: If the number of signal levels increases, the bit rate increases. For example, with 4 signal levels, L = 4, and the formula would become $$2 \times B \times \log_2(4) = 2 \times B \times 2$$​.
• Noise and Channel Capacity: In a real-world channel, noise limits the maximum bit rate, which is why the Shannon-Hartley theorem is used for noisy channels. However, for a noiseless channel, the Nyquist theorem provides the ideal maximum bit rate.