MIT6_003S10_lec16_handout

MIT6_003S10_lec16_handout - 6.003: Signals and Systems...

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± ± ± ± 6.003: Signals and Systems Lecture 16 April 6, 2010 6.003: Signals and Systems Mid-term Examination #2 Fourier Transform Tomorrow, April 7, No recitations tomorrow. 7:30-9:30pm. Coverage: Lectures 1–15 Recitations 1–15 Homeworks 1–8 Homework 8 will not collected or graded. Solutions are posted. Closed book: 2 pages of notes ( 8 1 2 × 11 inches; front and back). Designed as 1-hour exam; two hours to complete. April 6, 2010 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 ) SS “Periodic extension”: x T ( t )= x ( t + kT ) k = −∞ x T ( t ) t t T Then x ( t ) = lim x T ( t ) . T →∞ π T π T Fourier Transform Represent x T ( t ) by its Fourier series. x T ( t ) t T 1 T/ 2 2 1 S 2 sin 2 πkS 2 sin ωS a k = x T ( t ) e j kt dt = e j kt dt = T = T 2 T S πk T ω Ta k 2 π 2 sin ω ω = 0 = k T k ω ω 0 =2 π/T π T π T Fourier Transform Doubling period doubles # of harmonics in given frequency interval. x T ( t ) t T 1 2 2 1 S 2 sin 2 πkS 2 sin a k = x T ( t ) e j kt dt = e j kt dt = T = T 2 T S T ω k 2 π 2 sin ω ω = 0 = k T k ω ω 0 π/T 1
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± 6.003: Signals and Systems Lecture 16 April 6, 2010 Fourier Transform As T →∞ , discrete harmonic amplitudes a continuum E ( ω ) . x T ( t ) t SS T T/ 2 2 S 2 sin 2 πkS a k = 1 ² x T ( t ) e j kt dt = 1 ² e j kt dt = T = 2 sin ωS T 2 T S πk T ω Ta k 2 sin ω = 0 = k 2 π ω T k ω ω 0 =2 π/T ² 2 2 lim k = lim x ( t ) e jωt dt = sin = E ( ω ) T →∞ T →∞ 2 ω π T π T Fourier Transform As T , synthesis sum integral. x T ( t ) t T 2 sin ω ω 0 π/T ω = 0 = k 2 π T k k ω lim T →∞ k = lim T →∞ ² 2 2 x ( t ) e dt = 2 ω sin = E ( ω ) x ( t )= ³ k = −∞ 1 T E ( ω ) ´ µ¶ · a k e j 2 π T kt = ³ k = −∞ ω 0 2 π E ( ω ) e 1 2 π ² −∞ E ( ω ) e Fourier Transform Replacing E ( ω ) by X ( ) yields the Fourier transform relations. E ( ω X ( s ) | s = X ( ) Fourier transform ² X ( )= x ( t ) e dt (“analysis” equation) −∞ x ( t )= 1 2 π ² −∞ X ( ) e (“synthesis” equation) Form is similar to that of Fourier series provides alternate view of signal.
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MIT6_003S10_lec16_handout - 6.003: Signals and Systems...

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