The function f(t) = 3 cos 5001Tt + 5 cos 8001Tt is sampled | StudySoup

Textbook Solutions for Introduction to Engineering Experimentation

Chapter 5 Problem 5.20

Question

The function \(f(t)=3 \cos 500 \pi t+5 \cos 800 \pi t\) is sampled at 400 samples per second starting at t = 0.00025 s. What false alias frequencies would you expect in the output?

Solution

Step 1 of 4

Consider the following signal:

\(f\left( t \right) = 3\cos \left( {500\pi t} \right) + 5\cos \left( {800\pi t} \right)\)

The sampling frequency is,

\({f_s} = 400\;{\rm{samples/s}}\)

Calculate the sampling time interval as,

\({T_s} = \frac{1}{{{f_s}}}\)

\({T_s} = \frac{1}{{400}}\)

\({T_s} = 0.0025\;{\rm{s}}\)

Consider that the starting time of the sampled signal is 0.00025 s.

Though the sampling starts at 0.00025 s, the sampling period, which is the time between the samples, does not change. Hence, the sampling frequency does not change.

\({f_s} = 400\;{\rm{samples/s}}\)

Step 2 of 4

The maximum frequency component in a message signal is,

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full solution

Title Introduction to Engineering Experimentation 3 
Author Anthony J. Wheeler, Ahmad R. Ganji
ISBN 9780131742765

The function f(t) = 3 cos 5001Tt + 5 cos 8001Tt is sampled

Chapter 5 textbook questions

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