If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20MHz, what is the maximum capacity of the channel?

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

C = W*log_2 (1 + P/N)

- Plug the values in:

C = (20*10^6) * log_2 (1 + 250*10^-3/10)

C = (20*10^6) * log_2 (25001)

C = (20*10^6) * 14.6096

C = 292 Mbps