6.003: Signals and SystemsFourier TransformApril 6, 2010Mid-term Examination #2Tomorrow, April 7, 7:30-9:30pm, 34-101.No recitations tomorrow.Coverage: Lectures 1–15Recitations 1–15Homeworks 1–8Homework 8 will not collected or graded. Solutions are posted.Closed book: 2 pages of notes (812× 11 inches; front and back).Designed as 1-hour exam; two hours to complete.Last Week: Fourier SeriesRepresenting periodic signals as sums of sinusoids.→ new representations for systems as filters.This week: generalize for aperiodic signals.Fourier TransformAn 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”: xT(t) =∞Xk=−∞x(t + kT )xT(t)t−S STThen x(t) = limT →∞xT(t).Fourier TransformRepresent xT(t) by its Fourier series.xT(t)t−S STak=1TZT/2−T/2xT(t)e−j2πTktdt =1TZS−Se−j2πTktdt =sin2πkSTπk=2Tsin ωSω2 sin ωSωω0= 2π/Tω = kω0= k2πTT akkωFourier TransformDoubling period doubles # of harmonics in given frequency interval.xT(t)t−S STak=1TZT/2−T/2xT(t)e−j2πTktdt =1TZS−Se−j2πTktdt =sin2πkSTπk=2Tsin ωSω2 sin ωSωω0= 2π/Tω = kω0= k2πTT akkωFourier TransformAs T → ∞, discrete harmonic amplitudes → a continuum E(ω).xT(t)t−S STak=1TZT/2−T/2xT(t)e−j2πTktdt =1TZS−Se−j2πTktdt =sin2πkSTπk=2Tsin ωSω2 sin ωSωω0= 2π/Tω = kω0= k2πTT akkωlimT →∞T ak= limT →∞ZT/2−T/2x(t)e−jωtdt =2ωsin ωS = E(ω)Fourier TransformAs T → ∞, synthesis sum → integral.xT(t)t−S ST2 sin ωSωω0= 2π/Tω = kω0= k2πTT akkωlimT →∞T ak= limT →∞ZT/2−T/2x(t)e−jωtdt =2ωsin ωS = E(ω)x(t) =∞Xk=−∞1TE(ω)| {z }akej2πTkt=∞Xk=−∞ω02πE(ω)ejωt→12πZ∞−∞E(ω)ejωtdωFourier TransformReplacing E(ω) by X(jω) yields the Fourier transform relations.E(ω) = X(s)|s=jω≡ X(jω)Fourier transformX(jω)=Z∞−∞x(t)e−jωtdt (“analysis” equation)x(t)=12πZ∞−∞X(jω)ejωtdω (“synthesis” equation)Fourier TransformReplacing E(ω) by X(jω) yields the Fourier transform relations.E(ω) = X(s)|s=jω≡ X(jω)Fourier transformX(jω)=Z∞−∞x(t)e−jωtdt (“analysis” equation)x(t)=12πZ∞−∞X(jω)ejωtdω (“synthesis” equation)Form is similar to that of Fourier series→ provides alternate view of signal.Relation between Fourier and Laplace TransformsIf the Laplace transform of a signal exists and if the ROC includes thejω axis, then the Fourier transform is equal to the Laplace transformevaluated on the jω axis.Laplace transform:X(s) =Z∞−∞x(t)e−stdtFourier transform:X(jω) =Z∞−∞x(t)e−jωtdt = H(s)|s=jωRelation between Fourier and Laplace TransformsFourier transform “inherits” properties of Laplace transform.Property x(t) X(s) X(jω)Linearity ax1(t) + bx2(t) aX1(s) + bX2(s) aX1(jω) + bX2(jω)Time shift x(t − t0) e−st0X(s) e−jωt0X(jω)Time scale x(at)1|a|Xsa1|a|XjωaDifferentiationdx(t)dtsX(s) jωX(jω)Multiply by t tx(t) −ddsX(s) −1jddωX(jω)Convolution x1(t) ∗ x2(t) X1(s) × X2(s) X1(jω) × X2(jω)Relation between Fourier and Laplace TransformsThere are also important differences.Compare Fourier and Laplace transforms of x(t) = e−tu(t).x(t)tLaplace transformX(s) =Z∞−∞e−tu(t)e−stdt =Z∞0e−(s+1)tdt =11 + s; Re(s) > −1a complex-valued function of complex domain.Fourier transformX(jω) =Z∞−∞e−tu(t)e−jωtdt =Z∞0e−(jω+1)tdt =11 + jωa complex-valued function of real domain.Laplace TransformThe Laplace transform maps a function of time t to a complex-valuedfunction of complex-valued domain s.x(t)t-101-101010Real(s)Imaginary(s)Magnitude|X(s)| =11 + sFourier TransformThe Fourier transform maps a function of time t to a complex-valuedfunction of real-valued domain ω.x(t)t0 1X(j )=11 + jωωωFourier TransformThe Fourier transform maps a function of time t to a complex-valuedfunction of real-valued domain ω.x(t)t0 1X(j )=11 + jωωωFrequency plots provide intuition that is difficult to otherwise obtain.Check YourselfFind the Fourier transform of the following square pulse.−1 1x1(t)1t1. X1(jω) =1ωeω− e−ω2. X1(jω) =1ωsin ω3. X1(jω) =2ωeω− e−ω4. X1(jω) =2ωsin ω5. none of the aboveFourier TransformCompare the Laplace and Fourier transforms of a square pulse.−1 1x1(t)1tLaplace transform:X1(s) =Z1−1e−stdt =e−st−s1−1=1ses− e−s[function of s = σ + jω]Fourier transformX1(jω) =Z1−1e−jωtdt =e−jωt−jω1−1=2 sin ωω[function of ω]Check YourselfFind the Fourier transform of the following square pulse. 4−1 1x1(t)1t1. X1(jω) =1ωeω− e−ω2. X1(jω) =1ωsin ω3. X1(jω) =2ωeω− e−ω4. X1(jω) =2ωsin ω5. none of the aboveLaplace TransformLaplace transform: complex-valued function of complex domain.−1 1x1(t)1t-505-5050102030|X(s)| =1s(es− e−s)Fourier TransformThe Fourier transform is a function of real domain: frequency ω.Time representation:−1 1x1(t)1tFrequency representation:2πX1(jω) =2 sin ωωωCheck YourselfSignal x2(t) and its Fourier transform X2(jω) are shown below.−2 2x2(t)1tbω0X2(jω)ωWhich is true?1. b = 2 and ω0= π/22. b = 2 and ω0= 2π3. b = 4 and ω0= π/24. b = 4 and ω0= 2π5. none of the aboveCheck YourselfFind the Fourier transform.X2(jω) =Z2−2e−jωtdt =e−jωt−jω2−2=2 sin 2ωω=4 sin 2ω2ω4π/2ωCheck YourselfSignal x2(t) and its Fourier transform X2(jω) are shown below.−2 2x2(t)1tbω0X2(jω)ωWhich is true? 31. b = 2 and ω0= π/22. b = 2 and ω0= 2π3. b = 4 and ω0= π/24. b = 4 and ω0= 2π5. none of the aboveFourier TransformsStretching time compresses frequency.−1 1x1(t)1t2πX1(jω) =2 sin ωωω−2 2x2(t)1t4π/2X2(jω) =4 sin 2ω2ωωCheck YourselfStretching time compresses frequency.Find a general scaling rule.Let x2(t) = x1(at).If time is stretched in going from x1to x2, is a > 1 or a < 1?Check YourselfStretching time compresses frequency.Find a general scaling rule.Let x2(t) = x1(at).If time is stretched in going from x1to x2, is a > 1 or a < 1?x2(2) = x1(1)x2(t) = x1(at)Therefore a = 1/2, or more generally, a < 1.Check YourselfStretching time compresses frequency.Find a general scaling rule.Let x2(t) = x1(at).If time is stretched in going from x1to x2, is a > 1 or a < 1?a < 1Fourier TransformsFind a general
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