6.003: Signals and Systems Fourier Transform November 3, 2011 1 Last - PowerPoint PPT Presentation

6.003: Signals and Systems Fourier Transform November 3, 2011 1

Last Time: Fourier Series Representing periodic signals as sums of sinusoids . → new representations for systems as filters . Today: generalize for aperiodic signals. 2

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 ∞ 0 “Periodic extension”: x T ( t ) = x ( t + kT ) k = −∞ x T ( t ) t − S S T Then x ( t ) = lim x T ( t ) . T →∞ 3

Fourier Transform Represent x T ( t ) by its Fourier series. x T ( t ) t − S S T T kt dt = sin 2 πkS T/ 2 S a k = 1 = 2 sin ωS T kt dt = 1 − j 2 π − j 2 π T x T ( t ) e e T T πk T ω − T/ 2 − S Ta k ω = kω 0 = k 2 π 2 sin ωS ω T k ω ω 0 = 2 π/T 4

Fourier Transform Doubling period doubles # of harmonics in given frequency interval. x T ( t ) t − S S T T kt dt = sin 2 πkS T/ 2 S a k = 1 = 2 sin ωS T kt dt = 1 − j 2 π − j 2 π T x T ( t ) e e T T πk T ω − T/ 2 − S Ta k ω = kω 0 = k 2 π 2 sin ωS ω T k ω ω 0 = 2 π/T 5

Fourier Transform As T → ∞ , discrete harmonic amplitudes → a continuum E ( ω ) . x T ( t ) t − S S T T kt dt = sin 2 πkS T/ 2 S a k = 1 = 2 sin ωS T kt dt = 1 − j 2 π − j 2 π T x T ( t ) e e T T πk T ω − T/ 2 − S Ta k ω = kω 0 = k 2 π 2 sin ωS ω T k ω ω 0 = 2 π/T T / 2 − jωt dt = 2 T →∞ T a k = lim lim x ( t ) e ω sin ωS = E ( ω ) T →∞ − T / 2 6

Fourier Transform As T → ∞ , synthesis sum → integral. x T ( t ) t − S S T Ta k ω = kω 0 = k 2 π 2 sin ωS ω T k ω ω 0 = 2 π/T T / 2 − jωt dt = 2 T →∞ T a k = lim lim x ( t ) e ω sin ωS = E ( ω ) T →∞ − T / 2 ∞ ∞ ∞ 1 0 ω 0 1 2 π 0 j T kt = jωt → jωt dω x ( t ) = E ( ω ) e E ( ω ) e E ( ω ) e T 2 π 2 π −∞ k = −∞ k = −∞ � �� � a k 7

Fourier Transform Replacing E ( ω ) by X ( jω ) yields the Fourier transform relations. E ( ω ) = X ( jω ) Fourier transform ∞ − jωt dt X ( jω )= x ( t ) e (“analysis” equation) −∞ ∞ 1 jωt dω x ( t )= X ( jω ) e (“synthesis” equation) 2 π −∞ Form is similar to that of Fourier series → provides alternate view of signal. 8

Relation between Fourier and Laplace Transforms If the Laplace transform of a signal exists and if the ROC includes the jω axis, then the Fourier transform is equal to the Laplace transform evaluated on the jω axis. Laplace transform: ∞ − st dt X ( s ) = x ( t ) e −∞ Fourier transform: ∞ − jωt dt = X ( s ) | s = jω X ( jω ) = x ( t ) e −∞ 9

Relation between Fourier and Laplace Transforms Fourier transform “inherits” properties of Laplace transform. Property x ( t ) X ( s ) X ( jω ) Linearity ax 1 ( t ) + bx 2 ( t ) aX 1 ( s ) + bX 2 ( s ) aX 1 ( jω ) + bX 2 ( jω ) − st 0 X ( s ) − jωt 0 X ( jω ) Time shift x ( t − t 0 ) e e � � 1 � � s 1 jω Time scale x ( at ) X X | a | a | a | a dx ( t ) Differentiation sX ( s ) jωX ( jω ) dt d 1 d Multiply by t tx ( t ) − X ( s ) − X ( jω ) ds j dω Convolution x 1 ( t ) ∗ x 2 ( t ) X 1 ( s ) × X 2 ( s ) X 1 ( jω ) × X 2 ( jω ) 10

Relation between Fourier and Laplace Transforms There are also important differences. − t Compare Fourier and Laplace transforms of x ( t ) = e u ( t ) . x ( t ) t Laplace transform ∞ ∞ 1 − t − st dt = − ( s +1) t dt = X ( s ) = e u ( t ) e e ; Re ( s ) > − 1 1 + s 0 −∞ a complex­valued function of complex domain. Fourier transform ∞ ∞ 1 − t − jωt dt = − ( jω +1) t dt = X ( jω ) = e u ( t ) e e 1 + jω 0 −∞ a complex­valued function of real domain. 11

Laplace Transform The Laplace transform maps a function of time t to a complex­valued function of complex­valued domain s . x ( t ) t � � 1 � � | X ( s ) | = � � 1 + s � � 10 Magnitude 0 1 I m 1 0 a 0 g i -1 n Real(s) -1 a r y ( s ) 12

Fourier Transform The Fourier transform maps a function of time t to a complex­valued function of real­valued domain ω . x ( t ) t � � 1 � � � � � = � X ( j ) ω � � 1 + jω � � ω 0 1 Frequency plots provide intuition that is difficult to otherwise obtain. 13

Check Yourself Find the Fourier transform of the following square pulse. x 1 ( t ) 1 t − 1 1 1. X 1 ( jω ) = 1 2. X 1 ( jω ) = 1 � − ω � ω − e e ω sin ω ω 3. X 1 ( jω ) = 2 4. X 1 ( jω ) = 2 � − ω � ω − e e ω sin ω ω 5. none of the above 14

Fourier Transform Compare the Laplace and Fourier transforms of a square pulse. x 1 ( t ) 1 t − 1 1 Laplace transform: 1 � 1 − st e = 1 � � � − st dt = − s s − e − s X 1 ( s ) = e e [function of s = σ + jω ] � s � − 1 − 1 Fourier transform 1 � 1 − jωt e 2 sin ω � − jωt dt = X 1 ( jω ) = e = [function of ω ] � − jω ω � − 1 − 1 15

Check Yourself Find the Fourier transform of the following square pulse. 4 x 1 ( t ) 1 t − 1 1 1. X 1 ( jω ) = 1 2. X 1 ( jω ) = 1 � � � � ω − e − ω ω e ω sin ω 3. X 1 ( jω ) = 2 4. X 1 ( jω ) = 2 � � � � ω − e − ω ω e ω sin ω 5. none of the above 16

Laplace Transform Laplace transform: complex­valued function of complex domain. x 1 ( t ) 1 t − 1 1 � � 1 � s ( e s − e − s ) � | X ( s ) | = � � � � 30 20 10 0 5 5 0 0 -5 -5 17

Fourier Transform The Fourier transform is a function of real domain: frequency ω . Time representation: x 1 ( t ) 1 t − 1 1 Frequency representation: X 1 ( jω ) = 2 sin ω ω 2 ω π 18

Check Yourself Signal x 2 ( t ) and its Fourier transform X 2 ( jω ) are shown below. x 2 ( t ) X 2 ( jω ) 1 b ω t − 2 2 ω 0 Which is true? 1. b = 2 and ω 0 = π/ 2 2. b = 2 and ω 0 = 2 π 3. b = 4 and ω 0 = π/ 2 4. b = 4 and ω 0 = 2 π 5. none of the above 19

Check Yourself Find the Fourier transform. 2 � � 2 − jωt e 2 sin 2 ω 4 sin 2 ω � � − jωt dt = X 2 ( jω ) = e = = � � − jω − 2 ω 2 ω � � − 2 4 ω π/ 2 20

Check Yourself Signal x 2 ( t ) and its Fourier transform X 2 ( jω ) are shown below. x 2 ( t ) X 2 ( jω ) 1 b ω t − 2 2 ω 0 Which is true? 3 1. b = 2 and ω 0 = π/ 2 2. b = 2 and ω 0 = 2 π 3. b = 4 and ω 0 = π/ 2 4. b = 4 and ω 0 = 2 π 5. none of the above 21

Fourier Transforms Stretching time compresses frequency. X 1 ( jω ) = 2 sin ω x 1 ( t ) ω 1 2 ω t − 1 1 π X 2 ( jω ) = 4 sin 2 ω 2 ω 4 x 2 ( t ) 1 ω t − 2 2 π/ 2 22

Check Yourself Stretching time compresses frequency. Find a general scaling rule. Let x 2 ( t ) = x 1 ( at ) . If time is stretched in going from x 1 to x 2 , is a > 1 or a < 1 ? 23

Check Yourself Stretching time compresses frequency. Find a general scaling rule. Let x 2 ( t ) = x 1 ( at ) . If time is stretched in going from x 1 to x 2 , is a > 1 or a < 1 ? x 2 (2) = x 1 (1) x 2 ( t ) = x 1 ( at ) Therefore a = 1 / 2 , or more generally, a < 1 . 24

Check Yourself Stretching time compresses frequency. Find a general scaling rule. Let x 2 ( t ) = x 1 ( at ) . If time is stretched in going from x 1 to x 2 , is a > 1 or a < 1 ? a < 1 25

Fourier Transforms Find a general scaling rule. Let x 2 ( t ) = x 1 ( at ) . ∞ ∞ − jωt dt = − jωt dt X 2 ( jω ) = x 2 ( t ) e x 1 ( at ) e −∞ −∞ Let τ = at ( a > 0 ). � � � � ∞ − jωτ/a 1 1 jω X 2 ( jω ) = x 1 ( τ ) e dτ = X 1 a a a −∞ If a < 0 the sign of dτ would change along with the limits of integra- tion. In general, � � � � 1 jω x 1 ( at ) ↔ X 1 . | a | a If time is stretched ( a < 1 ) then frequency is compressed and ampli- tude increases (preserving area). 26

Moments The value of X ( jω ) at ω = 0 is the integral of x ( t ) over time t . ∞ ∞ ∞ − jωt dt = x ( t ) e j 0 t dt = X ( jω ) | ω =0 = x ( t ) e x ( t ) dt −∞ −∞ −∞ X 1 ( jω ) = 2 sin ω x 1 ( t ) ω 1 area = 2 2 ω t − 1 1 π 27

Moments The value of x (0) is the integral of X ( jω ) divided by 2 π . ∞ ∞ 1 1 jωt dω = x (0) = X ( jω ) e X ( jω ) dω 2 π −∞ 2 π −∞ X 1 ( jω ) = 2 sin ω x 1 ( t ) ω area 1 = 1 2 2 π + + + + ω t − − − − − 1 1 π 28

Moments The value of x (0) is the integral of X ( jω ) divided by 2 π . ∞ ∞ 1 1 jωt dω = x (0) = X ( jω ) e X ( jω ) dω 2 π −∞ 2 π −∞ X 1 ( jω ) = 2 sin ω x 1 ( t ) ω area 1 = 1 2 2 π + + + + ω t − − − − − 1 1 π equal areas ! 2 ω π 29

Stretching to the Limit Stretching time compresses frequency and increases amplitude (preserving area). X 1 ( jω ) = 2 sin ω x 1 ( t ) ω 1 2 ω t − 1 1 π 4 1 ω t − 2 2 π 1 2 π ω t New way to think about an impulse! 30

Fourier Transform One of the most useful features of the Fourier transform (and Fourier series) is the simple “inverse” Fourier transform. ∞ − jωt dt X ( jω )= x ( t ) e (Fourier transform) −∞ ∞ 1 jωt dω x ( t )= X ( jω ) e (“inverse” Fourier transform) 2 π −∞ 31