MIT6_003S10_lec16

MIT6_003S10_lec16 - 6.003: Signals and Systems Fourier...

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6.003: Signals and Systems Fourier Transform April 6, 2010
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Mid-term Examination #2 Tomorrow, April 7, No recitations tomorrow. Coverage: Lectures 1–15 Recitations 1–15 Homeworks 1–8 Homework 8 will not collected or graded. Solutions are posted. 1 8 Closed book: 2 pages of notes ( 2 × 11 inches; front and back). Designed as 1-hour exam; two hours to complete. 7:30-9:30pm.
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Last Week: Fourier Series Representing periodic signals as sums of sinusoids . new representations for systems as filters . This week: generalize aperiodic signals.
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Fourier Transform An aperiodic signal can be thought of as periodic with infinite period. Let x ( t ) represent an aperiodic signal. x ( t ) t SS “Periodic extension”: x T ( t )= x ( t + kT ) k = −∞ x T ( t ) t T Then x ( t ) = lim x T ( t ) . T →∞
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Fourier Transform Represent x T ( t ) by its Fourier series. x T ( t ) t SS T T/ 2 2 π S 2 π sin 2 πkS 1 j kt 1 j T 2 sin ωS a k = x T ( t ) e T dt = e T dt = = T 2 T S πk T ω 2 sin ω ω = 0 = k 2 π T Ta k k ω ω 0 =2 π/T
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Fourier Transform Doubling period doubles # of harmonics in given frequency interval. x T ( t ) t SS T T/ 2 2 π S 2 π sin 2 πkS 1 j kt 1 j T 2 sin ωS a k = x T ( t ) e T dt = e T dt = = T 2 T S πk T ω 2 sin ω ω = 0 = k 2 π T Ta k k ω ω 0 =2 π/T
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Fourier Transform As T →∞ , discrete harmonic amplitudes a continuum E ( ω ) . x T ( t ) t SS T T/ 2 2 π S 2 π sin 2 πkS 1 j kt 1 j T 2 sin ωS a k = x T ( t ) e T dt = e T dt = = T 2 T S πk T ω 2 sin ω ω = 0 = k 2 π T Ta k k ω ω 0 =2 π/T lim k = lim 2 x ( t ) e jωt dt = 2 sin = E ( ω ) T →∞ T →∞ 2 ω
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Fourier Transform As T →∞ , synthesis sum integral. x T ( t ) t SS T Ta k 2 sin ωS ω = 0 = k 2 π ω T k ω ω 0 =2 π/T T/ 2 2 lim k = lim x ( t ) e jωt dt = sin = E ( ω ) T →∞ T →∞ 2 ω ± ± 1 1 π kt T 2 ω 0 E ( ω ) e j x ( t )= E ( ω ) e E ( ω ) e = 2 π 2 π −∞ T k = −∞ ² ³´ µ k = −∞ a k
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Fourier Transform Replacing E ( ω ) by X ( ) yields the Fourier transform relations. E ( ω )= X ( s ) | s = X ( ) Fourier transform X ( )= x ( t ) e jωt dt (“analysis” equation) −∞ 1 x ( t )= 2 π −∞ X ( ) e (“synthesis” equation)
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Fourier Transform Replacing E ( ω ) by X ( ) yields the Fourier transform relations. E ( ω )= X ( s ) | s = X ( ) Fourier transform X ( )= x ( t ) e jωt dt −∞ (“analysis” equation) x ( t )= 1 2 π −∞ X ( ) e (“synthesis” equation) Form is similar to that of Fourier series provides alternate view of signal.
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Relation between Fourier and Laplace Transforms If the Laplace transform of a signal exists and if the ROC includes the axis, then the Fourier transform is equal to the Laplace transform evaluated on the axis.
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MIT6_003S10_lec16 - 6.003: Signals and Systems Fourier...

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