Chapter 5Basic IdeaSlide 3Example – Rectangular SignalFourier Transform of Unit Impulse FunctionFourier Series PropertiesFourier Series Properties - LinearitySlide 8Fourier Series Properties - Time ScalingFourier Series Properties - Duality or SymmetrySlide 11Slide 12Fourier Series Properties - ConvolutionSlide 14Slide 15Fourier Series Properties - Frequency ShiftingFourier Series Properties - Time DifferentiationMore…Schaums’ Outlines ProblemsChapter 5The Fourier TransformBasic Idea•We covered the Fourier Transform which to represent periodic signals•We assumed periodic continuous signals •We used Fourier Series to represent periodic continuous time signals in terms of their harmonic frequency components (Ck). •We want to extend this discussion to find the frequency spectra of a given signalBasic Ideanotes•The Fourier Transform is a method for representing signals and systems in the frequency domain •We start by assuming the period of the signal is T= INF•All physically realizable signals have Fourier Transform•For aperiodic signals Fourier Transform pairs is described asRemember: Fourier Transforms of f(t)Inverse Fourier Transforms of F()Example – Rectangular Signal •Compute the Fourier Transform of an aperiodic rectangular pulse of T seconds evenly distributed about t=0. •Remember this the same rectangular signal as we worked before but with T0 infinity!All physically realizable signals have Fourier Transforms )2/(sin)/()2/(sin)()()2/()2/()(TcTVTtVrectthusTcTVdtetfFTtVuTtVutftj T/2-T/2VnotesFourier Transform of Unit Impulse Function)(22)(21)(21)()}({)()(001000 tjtjtjtjethusededeFtfFFExample: AtthusAFtAtfifAedtettAdtetfFttAtftjtjtj )()()0()(:)()()()()(000Plot magnitude and phase of f(t)Fourier Series PropertiesMake sure how to use these properties!Fourier Series Properties - Linearity)(22)(22)(2:Re}{2}{2)()}({)(2/)cos()(000000000 BBemembereBeBFtfeeBtBtftjtjtjtjtjFind F(w)Fourier Series Properties - Linearity)(22)(22)(2:Re}{2}{2)()}({)(2/)cos()(000000000 BBemembereBeBFtfeeBtBtftjtjtjtjtjDue to linearityFourier Series Properties - Time Scaling)4/(sin)2()4/(sin))(21(1)()/(||/1)()(1)()()()2/(sin)()/()(:Re)/2()(111121111TcTTcVTVGThenaFawGatfVtgwGtgThusTcVTFTtVrecttfmemberTtrecttga   Due to Time Scaling PropertyRemember: sinc(0)=1;sinc(2pi)=0=sinc(pi)rect(t/T)rect(t/(T/2))Fourier Series Properties - Duality or Symmetry  ttffwhereftFftFFtfif)()(:)(2)}({)(2)()()(:Example: Find the time-domain waveform for)2/()(sin)(sin2)()2/(2)/(2)(2)(___sin)2/()()()(:_)2/(sin)/(:Re)()()(BArectBtcBABtcBAtFThusBArectTArectftFfindDualitygUBArectBAuBAuFhaveWeTcTVTtVrectmemberBAuBAuF  Remember we had:Arect(/2B)Refer to FTPTable FTP: Fourier Transfer PairFourier Series Properties - Duality or SymmetryExample: find the frequency responseOf y(t)Fourier Series Properties - Duality or SymmetryExample: find the frequency responseOf y(t)We know Using Fourier Transform PairsUsing duality  ttffwhereftFftFFtfif)()(:)(2)}({)(2)()()(:Fourier Series Properties - ConvolutionProofProofFourier Series Properties - Convolution)2/(sin)/()2/(sin)()()1()1()(TcTVTtVrectthusTcTVdtetfFtututftj Example:Find the Fourier Transform of x(t)=sinc2(t))2/()(sin)2/()(sin)2/()(sin:RerecttcrecttcBArectBtcBAmember  X1() X2()In this case we have B=1, A=1 Refer to Schaum’sProb. 2.6Fourier Series Properties - ConvolutionExample:Find the Fourier Transform of x(t)= sinc2(t) sinc(t)Refer to Schaum’sProb. 2.6We need to find the convolution of a rect and a triangle function:Fourier Series Properties - Frequency Shifting)800(5)12 00(5)1200(5)800(5)1000200(5)1000200(5)100020 0(5)1000200(5)()200cos(5)200cos(5)(3100010003GetettgtjtjExample:Find the Fourier Transform of g3(t) if g1(t)=2cos(200t), g2(t)=2cos(1000t); g3(t)=g1(t).g2(t) ; that is [G3(w)])()(00 XetxtjRemember: cosa . cosb=1/2[cos(a+b)+cos(a-b)]Fourier Series Properties - Time Differentiation   dttfdtetfFNoteGFFjdftgAlsoFjdttdftgFtftjt)()()0()()()0()(1)()()(/)()()()(Example:1/)()sgn(1)()(2/)()(/)()()()sgn()(  dttdftthusttdttdftdttdgtutgttfMore…•Read your notes for applications of Fourier Transform. •Read about Power Spectral Density •Read about Bode PlotsSchaums’ Outlines Problems•Schaum’s Outlines: –Do problems 5.16-5.43–Do problems 5.4, 5.5, 5.6. 5.7, 5.8, 5.9, 5.10, 5.14•Do problems in the