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lec17-handout-6up - 6.003: Signals and Systems Lecture 17...

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Unformatted text preview: 6.003: Signals and Systems Lecture 17 November 10, 2009 1 6.003: Signals and Systems Fourier Transform November 10, 2009 Mid-term Examination #3 Wednesday, November 18, 7:30-9:30pm, Walker Memorial (this exam is after drop date). No recitations on the day of the exam. Coverage: cumulative with more emphasis on recent material lectures 118 homeworks 110 Homework 10 will not collected or graded. Solutions will be posted. Closed book: 3 page of notes ( 8 1 2 11 inches; front and back). Designed as 1-hour exam; two hours to complete. Review sessions during open office hours. Conflict? Contact freeman@mit.edu by Friday, November 13, 2009. Last Week: Fourier Series Representing periodic signals as sums of sinusoids . new representations for systems as filters . This week: generalize for aperiodic signals. Fourier Transform An aperiodic signal can be thought of as periodic with infinite period. Let x ( t ) represent an aperiodic signal. x ( t ) t S S Periodic extension: x T ( t ) = k = x ( t + kT ) x T ( t ) t S S T Then x ( t ) = lim T x T ( t ) . Fourier Transform Represent x T ( t ) by its Fourier series. x T ( t ) t S S T a k = 1 T T/ 2 T/ 2 x T ( t ) e j 2 T kt dt = 1 T S S e j 2 T kt dt = sin 2 kS T k = 2 T sin S 2 sin S = 2 /T = k = k 2 T T a k k Fourier Transform Doubling period doubles # of harmonics in given frequency interval. x T ( t ) t S S T a k = 1 T T/ 2 T/ 2 x T ( t ) e j 2 T kt dt = 1 T S S e j 2 T kt dt = sin 2 kS T k = 2 T sin S 2 sin S = 2 /T = k = k 2 T T a k k 6.003: Signals and Systems Lecture 17 November 10, 2009 2 Fourier Transform As T , discrete harmonic amplitudes a continuum E ( ) . x T ( t ) t S S T a k = 1 T T/ 2 T/ 2 x T ( t ) e j 2 T kt dt = 1 T S S e j 2 T kt dt = sin 2 kS T k = 2 T sin S 2 sin S = 2 /T = k = k 2 T T a k k lim T T a k = lim T T/ 2 T/ 2 x ( t ) e jt dt = 2 sin S = E ( ) Fourier Transform As T , synthesis sum integral. x T ( t ) t S S T 2 sin S = 2 /T = k = k 2 T T a k k lim T T a k = lim T T/ 2 T/ 2 x ( t ) e jt dt = 2 sin S = E ( ) x ( t ) = k = 1 T E ( ) a k e j 2 T kt = k = 2 E ( ) e jt 1 2 E ( ) e jt d Fourier Transform Replacing E ( ) by X ( j ) yields the Fourier transform relations....
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This note was uploaded on 08/24/2011 for the course EECS 6.003 taught by Professor Dennism.freeman during the Spring '11 term at MIT.

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lec17-handout-6up - 6.003: Signals and Systems Lecture 17...

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