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Unformatted text preview: 6.003: Signals and Systems Fourier Transform November 10, 2009 Midterm Examination #3 Wednesday, November 18, 7:309:30pm, 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 1–18 homeworks 1–10 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 1hour exam; two hours to complete. Review sessions during open office hours. Conflict? Contact [email protected] 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 ) = ∞ Ø 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 Ú T/ 2 − T/ 2 x T ( t ) e − j 2 π T kt dt = 1 T Ú S − S e − j 2 π T kt dt = sin 2 πkS T πk = 2 T sin ωS ω 2 sin ωS ω ω = 2 π/T ω = kω = k 2 π T Ta k k ω Fourier Transform Doubling period doubles # of harmonics in given frequency interval. x T ( t ) t − S S T a k = 1 T Ú T/ 2 − T/ 2 x T ( t ) e − j 2 π T kt dt = 1 T Ú S − S e − j 2 π T kt dt = sin 2 πkS T πk = 2 T sin ωS ω 2 sin ωS ω ω = 2 π/T ω = kω = k 2 π T Ta k k ω Fourier Transform As T → ∞ , discrete harmonic amplitudes → a continuum E ( ω ) . x T ( t ) t − S S T a k = 1 T Ú T/ 2 − T/ 2 x T ( t ) e − j 2 π T kt dt = 1 T Ú S − S e − j 2 π T kt dt = sin 2 πkS T πk = 2 T sin ωS ω 2 sin ωS ω ω = 2 π/T ω = kω = k 2 π T Ta k k ω lim T →∞ Ta k = lim T →∞ Ú T/ 2 − T/ 2 x ( t ) e − jωt dt = 2 ω sin ωS = E ( ω ) Fourier Transform As T → ∞ , synthesis sum → integral. x T ( t ) t − S S T 2 sin ωS ω ω = 2 π/T ω = kω = k 2 π T Ta k k ω lim T →∞ Ta k = lim T →∞ Ú T/ 2 − T/ 2 x ( t ) e − jωt dt = 2 ω sin ωS = E ( ω ) x ( t ) = ∞ Ø k = −∞ 1 T E ( ω ) ü ûú ý a k e j 2 π T kt = ∞ Ø k = −∞ ω 2 π E ( ω ) e jωt → 1 2 π Ú ∞ −∞ E ( ω ) e jωt dω Fourier Transform Replacing E ( ω ) by X ( jω ) yields the Fourier transform relations. E ( ω ) = X ( s )  s = jω ≡ X ( jω ) Fourier transform X ( jω )= Ú ∞ −∞ x ( t ) e − jωt dt (“analysis” equation) x ( t )= 1 2 π Ú ∞ −∞ X ( jω ) e jωt dω (“synthesis” equation) Relation between Fourier and Laplace Transforms If the Laplace transform of a signal exists and if the ROC includes the...
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 Spring '11
 DennisM.Freeman
 Complex number

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