Assume a discrete signal of 8 voltage levels has been transmitted through a noisy channel with bandwidth of 3500 Hz. The SNR is 2500 Watts and the power of the noise is 2 Watts.
A. What is the power of the signal?
B. What should be the minimum sampling rate?
C. What is the capacity of this channel?
given SNR = 2500
and noise power = 2W
SNR = signal power / Noise power
2500 = signal power / 2W
signal power = 5000W = 5 KW
In order not to have spectral aliasing of the discrete signal, you need to have a sampling rate that is 2 times the bandwidth. So the relationship is simple F = 2B
so the minimum sampling frequency is 2*3500 = 7000hz =
7KHZ
Reference
https://www.physicsforums.com/threads/relation-between-bandwidth-and-ssamplig-rate.645585/
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