Question

Assume a discrete signal of 8 voltage levels has been transmitted through a noisy channel with...

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?

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Answer #1

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/

By Shannon

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