what is the maximum data rate for a 6kHz channel that transmits data using 4 levels (binary) ?(only give the value in bits/second, don't put the unit)
Question
what is the maximum data rate for a 6kHz channel that transmits data using 4 levels (binary) ?(only give the value in bits/second, don't put the unit)
Solution
The maximum data rate can be calculated using the Shannon Capacity formula, which is C = B log2 (1 + SNR), where C is the capacity, B is the bandwidth, and SNR is the signal-to-noise ratio. However, in this case, we are not given the SNR, so we will use the simpler formula for a noiseless channel, which is C = B log2 M, where M is the number of discrete signal or quantization levels.
Given that B = 6 kHz = 6000 Hz and M = 4 (since we are using 4 levels), we can substitute these values into the formula:
C = 6000 log2 4 = 6000 * 2 = 12000 bits/second.
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