{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lec17-handout-6up

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

This preview shows pages 1–3. Sign up to view the full content.

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 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 1-hour 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 ) = 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 π T kt dt = 1 T S S e j 2 π T kt dt = sin 2 πkS T πk = 2 T sin ωS ω 2 sin ωS ω ω 0 = 2 π/T ω = 0 = 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 T/ 2 T/ 2 x T ( t ) e j 2 π T kt dt = 1 T S S e j 2 π T kt dt = sin 2 πkS T πk = 2 T sin ωS ω 2 sin ωS ω ω 0 = 2 π/T ω = 0 = k 2 π T Ta k k ω

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

View Full Document
6.003: Signals and Systems Lecture 17 November 10, 2009 2 Fourier Transform As T → ∞ , discrete harmonic amplitudes a continuum E ( ω ) . x T ( t ) t S S T a k = 1 T T/ 2 T/ 2 x T ( t ) e j 2 π T kt dt = 1 T S S e j 2 π T kt dt = sin 2 πkS T πk = 2 T sin ωS ω 2 sin ωS ω ω 0 = 2 π/T ω = 0 = k 2 π T Ta k k ω lim T →∞ Ta k = lim T →∞ 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 0 = 2 π/T ω = 0 = k 2 π T Ta k k ω lim T →∞ Ta k = lim T →∞ T/ 2 T/ 2 x ( t ) e jωt dt = 2 ω sin ωS = E ( ω ) x ( t ) = k = −∞ 1 T E ( ω ) a k e j 2 π T kt = k = −∞ ω 0 2 π E ( ω ) e jωt 1 2 π −∞ E ( ω ) e jωt Fourier Transform Replacing E ( ω ) by X ( ) yields the Fourier transform relations.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online