6.003: Signals and SystemsRelations among Fourier RepresentationsNovember 19, 2009Fourier RepresentationsWe’ve seen a variety of Fourier representations:• CT Fourier series• CT Fourier transform• DT Fourier seriesOne more today: DT Fourier transform... and relations among all four representations.DT Fourier transformRepresenting aperiodic DT signals as sums of complex exponentials.DT Fourier transformX(ejΩ)=∞Xn=−∞x[n]e−jΩn(“analysis” equation)x[n]=12πZ<2π>X(ejΩ)ejΩndΩ (“synthesis” equation)Comparison to DT Fourier SeriesFrom periodic to aperiodic.DT Fourier Seriesak= ak+N=1NXn=<N>x[n]e−jΩ0kn; Ω0=2πN(“analysis” equation)x[n]= x[n + N] =Xk=<N>akejΩ0kn(“synthesis” equation)DT Fourier transformX(ejΩ)=∞Xn=−∞x[n]e−jΩn(“analysis” equation)x[n]=12πZ<2π>X(ejΩ)ejΩndΩ (“synthesis” equation)Comparison to DT Fourier SeriesSum over an infinite number of time samples instead of N.DT Fourier Seriesak= ak+N=1NXn=<N>x[n]e−jΩ0kn; Ω0=2πN(“analysis” equation)x[n]= x[n + N] =Xk=<N>akejΩ0kn(“synthesis” equation)DT Fourier transformX(ejΩ)=∞Xn=−∞x[n]e−jΩn(“analysis” equation)x[n]=12πZ<2π>X(ejΩ)ejΩndΩ (“synthesis” equation)Comparison to DT Fourier SeriesSum over an infinite number of frequency components (P→R).DT Fourier Seriesak= ak+N=1NXn=<N>x[n]e−jΩ0kn; Ω0=2πN(“analysis” equation)x[n]= x[n + N] =Xk=<N>akejΩ0kn(“synthesis” equation)DT Fourier transformX(ejΩ)=∞Xn=−∞x[n]e−jΩn(“analysis” equation)x[n]=12πZ<2π>X(ejΩ)ejΩndΩ (“synthesis” equation)DT Fourier Series and TransformExample of series.Let x[n] represent the following periodic DT signal.nx[n] = x[n + 8]−1 1−8 8ak=1NXn=<N>x[n]e−jΩ0kn; Ω0=2πN=181 +12e−j2π8k+12ej2π8k=1 + cosπk48kak−1 1−8 814DT Fourier Series and TransformExample of transform.Let x[n] represent the aperiodic base of the previous signal.nx[n]−1 1H(ejΩ) =∞Xn=−∞x[n]e−jΩn=1 +12e−jΩ+12ejΩ= 1 + cos ΩΩX(ejΩ)2π−2π2DT Fourier Series and TransformSimilarities.DT Fourier serieskak−1 1−8 814DT Fourier transformΩX(ejΩ)2π−2π2Relations among Fourier RepresentationsDifferent Fourier representations are related because they apply tosignals that are related.DTFS (discrete-time Fourier series): periodic DTDTFT (discrete-time Fourier transform): aperiodic DTCTFS (continuous-time Fourier series): periodic CTCTFT (continuous-time Fourier transform): aperiodic CTperiodic DTDTFSaperiodic DTDTFTperiodic CTCTFSaperiodic CTCTFTN → ∞periodic extensionT → ∞periodic extensioninterpolate sample interpolate sampleRelation between Fourier Series and TransformA periodic signal can be represented by a Fourier series or by anequivalent Fourier transform.x(t) = x(t + T) =∞Xk=−∞akejω0kt; ω0=2πTBecause the Fourier transform of ejω0ktis 2πδ(ω − kω0),X(jω) =∞Xk=−∞2πakδ(ω − kω0).This expression shows the relation between the Fourier Series andFourier transform for a periodic signal.Relation between Fourier Series and TransformA periodic signal can be represented by a Fourier series or by anequivalent Fourier transform.x(t) = x(t + T ) =∞Xk=−∞akejω0ktωFourier Transform0ω02πa02πa12πa−12πa22πa−22πa32πa−32πa42πa−4kFourier Series0 1a0a1a−1a2a−2a3a−3a4a−4↔Relations among Fourier RepresentationsStart with an aperiodic CT signal. Determine its Fourier transform.Convert the signal so that it can be represented by alternate Fourierrepresentations and compare.periodic DTDTFSaperiodic DTDTFTperiodic CTCTFSaperiodic CTCTFTN → ∞periodic extensionT → ∞periodic extensioninterpolate sample interpolate sampleStart with the CT Fourier TransformDetermine the Fourier transform of the following signal.tx(t)−1 0 11Could calculate Fourier transform from the definition.X(jω) =Z∞−∞x(t)ejωtdtEasier to calculate x(t) by convolution of two square pulses:ty(t)−12121ty(t)−12121∗Start with the CT Fourier TransformIf the transform of y(t) issin(ω/2)ωty(t)−12121ωY (jω)12π−2π↔then the transform of x(t) = (y ∗ y)(t) is X(jω) = Y (jω) × Y (jω).tx(t)−1 11ωX(jω)12π−2π↔Relation between Fourier Transform and SeriesWhat is the effect of making a signal periodic in time?Find Fourier transform of periodic extension of x(t) to period T = 4.tz(t) =∞Xk=−∞x(t + 4k)−1 114−4Could calculate Z(jω) for the definition ... ugly.Relation between Fourier Transform and SeriesEasier to calculate z(t) by convolving x(t) with an impulse train.tz(t) =∞Xk=−∞x(t + 4k)−1 114−4z(t) =∞Xk=−∞x(t + 4k) = (x ∗ p)(t)wherep(t) =∞Xk=−∞δ(t + 4k)ThenZ(jω) = X(jω) × P (jω)We already know P (jω): it’s also an impulse train!Relation between Fourier Transform and SeriesConvolving in time corresponds to multiply in frequency.π/2ωZ(jω)π2−π2ωX(jω)12π−2πωP (jω)π2−π2π2Relation between Fourier Transform and SeriesThe Fourier transform of a periodically extended function is a dis-crete function of frequency ω.tz(t) =∞Xk=−∞x(t + 4k)−1 114−4π/2ωZ(jω)π2−π2Relation between Fourier Transform and SeriesThe weight (area) of each impulse in the Fourier transform of aperiodically extended function is 2π times the corresponding Fourierseries coefficient.π/2ωZ(jω)π2−π21/4kak1−1Relation between Fourier Transform and SeriesThe effect of periodic extension of x(t) to z(t) is to sample thefrequency representation.ωX(jω)12π−2ππ/2ωZ(jω)π2−π21/4kak1−1Relation between Fourier Transform and SeriesPeriodic extension of a CT signal produces a discrete function offrequency.Periodic extension= convolving with impulse train in time= multiplying by impulse train in frequency→ sampling in frequencyperiodic DTDTFSaperiodic DTDTFTperiodic CTCTFSaperiodic CTCTFTN → ∞periodic extensionT → ∞(sampling in frequency)periodic extensioninterpolate sample interpolate sampleRelations between CT and DT transformsSampling a CT signal generates a DT signal.x[n] = x(nT )tx(t)−1 0 11Take T =12.nx[n]−1 1What is the effect on the frequency representation?Relations between CT and DT transformsWe can generate a signal with the same shape by multiplying x(t) byan impulse train with T =12.xp(t) = x(t) × p(t) where p(t) =∞Xk=−∞δ(t + kT )tx(t)−1 0 11txp(t)−1 1Relations between CT and DT transformsMultiplying x(t) by an impulse train in time is equivalent to
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