Challenge_9 - InvestiGATOR 1 EEL 3135 Dr. Fred J. Taylor,

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
EEL 3135 Dr. Fred J. Taylor, Professor Lesson #9 Synergy Challenge 09 The reconstructed signal can also be derived from the synthesis equation: T kt j k k e c t x / 2 ) ( π -∞ = = where c -8 = -0.043+0.033j, c -6 = -0.027+0.061j, c 0 =0.217+0j, c 6 = -0.027-0.061j, and c 8 = -0.043-0.033j (machine calculated). What is x(t)? Required Challenge response: What is the 6 th harmonic of x(t)? ------------------------------------------------ Laying down some key parameters, c 8 = -0.043-j0.033 = 0.0542 -152.5 ° ; c -8 = -0.043+j0.033=c 8 *=0.0542 152.5 ° c 6 = -0.027+j0.061 = 0.0663 113.9 ° , c -6 = -0.027-j0.061=c 6 *= 0.0663 -113.9 ° , c 0 =0.217+j0 = 0.217 Remember that it was posted that the frequency resolution of the machine computed Fourier series is given by f=135 Hz/256 harmonics = 0.527 Hz/harmonic. You are also told that you process T 0 =1.8975 seconds of data with an FFT for each displayed spectrum (i.e., 256 samples at T s =1/f s sec/sample). T s = 1/135, T 0 = 1.8975 The 0 th harmonic is 0 f= 0 Hz. The 6 th harmonic is at 6 f= (6)( 0.527)=3.162 Hz The 8 th harmonic is at 8 f= (8)( 0.527)=4.216 Hz
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: InvestiGATOR 1 EEL 3135 Dr. Fred J. Taylor, Professor------------------------------------------------------Direct substitution: ( 29 { } ( 29 { } ( 29 ( 29 ( 29 ( 29 ) . ) . ( cos( . ) . ) . ( cos( . . . . . . . . . . . . ) ( . / . / . / . / / . / . / . / . / / / / / 5 152 216 4 2 1084 9 113 162 3 2 1326 217 0542 0542 0663 0663 217 0542 0542 0663 0663 217 5 152 16 5 152 16 9 113 12 9 113 12 16 5 152 16 5 152 12 9 113 12 9 113 16 8 16 8 12 6 12 6 2-+ + + = = + = + + + = = + + + + + = = + + + + = = =---+-+------- = t t e e e e e e e e e e e e e a e a e a e a a e a t x o o o o o o o o o o o o o T t j T t j T t j T t j T t j j T t j j T t j j T t j j T t j T t j T t j T t j T kt j k k assuming the arithmetic is correct. The 6 th harmonic component is therefore: ) . ) . ( cos( . ) ( 9 113 162 3 2 1326 6 + = t t x 2...
View Full Document

This note was uploaded on 04/15/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

Page1 / 2

Challenge_9 - InvestiGATOR 1 EEL 3135 Dr. Fred J. Taylor,

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online