Homework 8 Solution

Homework 8 Solution - MEEN 260 Introduction to Engineering...

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Unformatted text preview: MEEN 260 Introduction to Engineering Experimentation Homework 8: Sampling Theory and the Nyquist Frequency Solution Assigned: Thursday, 26 Mar. 2009 Due: Thursday, 2 Apr. 2009, 5:00pm Learning Objectives: After completing this homework assignment, you should be able to: 1) 2) 3) 4) Determine the Nyquist Frequency based on a sampling frequency Determine if a harmonic signal will be aliased, and the aliased (apparent) frequency of the reconstructed signal Determine the digital signal after sampling an analog signal, including the digital frequencies Reconstruct an analog signal from a digital signal Homework Problems: Problem 1) Nyquist Frequncy a) What sampling frequency produces a Nyquist Frequency of 256 Hz? b) A waveform contains harmonics of 75 Hz. Which of these can be accurately measured assuming the sampling frequency from part a)? (definition: a harmonic is based on integer multipliers of a fundamental frequency) c) An engineer wants to measure vibration effects in a motorcycle engine that redlines at 9500 rpm. Assuming this is also the limit of the frequencies present, what is the minimum sampling rate should be used for the monitoring equipment? Problem 1 Solution a) Nyquist Frequency 2 ; therefore, fs=512Hz b) Harmonics of 75Hz are 75,150,225,300,375, and so on. Those frequencies below the Nyquist Frequency will be accurately measured. Therefore, the frequencies 75Hz,150Hz,225Hz will be kept. c) According to sampling theory, the sampling frequency must be at least twice the highest frequency that is to be recorded. The highest engine frequency is given by: 1 1 158.33 158 9500 1 60 Therefore, the sampling frequency should be about 316-317 Hz Problem 2) Aliasing Assume that a 6000 Hz cosine wave is sampled at a frequency of 4000 Hz. a) What apparent frequency is measured? b) Draw a full cycle of the apparent signal and the original input signal on the same graph. Why does this happen? (Visit http://www.dsptutor.freeuk.com/aliasing/AliasingDemo.html for an example graph) Problem 2 Solution Aliasing a) In general, when a sinusoid of frequency f is sampled with frequency f s , the resulting samples are indistinguishable from those of another sinusoid of frequency | | for any integer N being the actual signal frequency. Therefore, |6000 4000| 2000 b) Figure shown below: Problem 3) Digital Signals and Signal Reconstruction Given the time domain signal: x (t ) = 4 cos(16π t − ) + 5sin(4π t + ) 4 3 π π a) Find the discrete time signal for sampling frequency 6 Hz. What are the digital frequencies present? (remember to put all digital frequencies into the principal range) b) Reconstruct the analog signal, assuming a sampling frequency of 6 Hz c) Which elements of the original signal were aliased, if any? Problem 3 Solution Digital Signals and Signal Reconstruction 8 and 2. a) From the CT signal, If sampling frequency, 6 , the digital frequencies are Outside principal range Inside principal range F1 should be placed in principal range: The final discrete time signal is given by: 1 4 cos 2 4 3 b) From If sampling frequency 6, 6 6 The reconstructed signal is given by: 4 cos 4 4 5sin 4 3 2 2 1 5sin 2 1 3 3 Problem 4) Aliasing and Signal Reconstruction Consider the following continuous-time signal xa (t ) = 2 cos(1000π t ) − 9sin(600π t ) + 3sin(400π t ) + sin(500π t ). Suppose this original signal, xa(t) is sampled at an unknown frequency, Fs Hz to obtain the discrete-time signal x[n], which is then reconstructed assuming the same sampling frequency, Fs, to obtain the following continuous-time signal: xr (t ) = 2 cos(600π t ) − 9sin(600π t ) + 3sin(400π t ) + sin(500π t ). Determine the sampling frequency, Fs used in sampling. Justify your answer. Problem 4 Solution Aliasing and Signal Reconstruction By examining the original signal and the reconstructed signals, we see that only the 500Hz signal was aliased. Therefore we know that 600 1000 and we need only consider the term 2cos(600πt). Applying 500 500 300 The absolute value leads to two outcomes: 300 And 300 500 800 800 . 800 500 200 100 200 |500 | 500 The second answer fits our range, and is correct: ...
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This note was uploaded on 07/28/2010 for the course MEEN 260 taught by Professor Langari during the Fall '08 term at Texas A&M.

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