Final answer:
The maximum transmission rate in a system with 256 equiprobable symbols, 4 kHz bandwidth, and SNR of 31, calculated using the Shannon-Hartley theorem, is 20,000 symbols per second.
Step-by-step explanation:
The student is asking about the maximum transmission rate of symbols per second in a communication system that uses a source of 256 equiprobable symbols, a channel with a 4 kHz bandwidth, and an SNR of 31. This involves using the Shannon-Hartley theorem, which is defined in the fields of information theory and telecommunications. This theorem states that the maximum data rate (C) that can be achieved over a noisy channel with a certain bandwidth (B) and signal-to-noise ratio (SNR) is given by the equation C = B log2(1 + SNR).
In this case, the bandwidth is 4 kHz and the SNR is 31. Converting the SNR to the same base used in the logarithm (base 2), we need to calculate the logarithm of (1 + 31). First, let's find the log base 2 of 32, which is 5. Then we can apply the theorem:
C = 4 kHz * log2(1 + 31) = 4 kHz * 5 = 20k symbols
Therefore, the maximum rate at which symbols can be transmitted over the AWGN channel with an arbitrarily low probability of error is 20,000 symbols per second.