EECE 301 Signals & Systems Prof. Mark FowlerEECE 301 Signals & SystemsProf. Mark FowlerDiscussion #6• Fourier Transform ExamplesFT ExamplesExample:Find FT of x(t) given below: )(txAA−-22tSolution:Note:)(2tApA)1()1()( :So22−−+=tAptAptx{}{}{})1()1()( :y LinearitUse22−−+=tpAtpAtx FFF{}ωωjePtp )()1( :Shift Time Use22=+F{}ωωjePtp−=− )()1(22F[]ωωωωjjeeAPXSo−−= )()(...2⎥⎦⎤⎢⎣⎡−=−jeejAPjj22)(2ωωω)()sin(22ωωPjA=⎟⎠⎞⎜⎝⎛=πωωsinc2)( :Table From2P⎟⎠⎞⎜⎝⎛=πωωωsinc)sin(4)( jAXExampleFind FT of x(t) B2−2)(txtSolution #111− 1)(2tptNote:()2)()()(22Btptptx ∗=Verify it!!Using Convolution Property:)(2)(22ωωPBX =From Table:⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛=πωπωωsinc222sinc2)(2P⎟⎠⎞⎜⎝⎛=πωω2sinc2)(... BXSoSolution #2get to)( Takedttdxdttdx )(2B2B−2−2tTake another derivative to get:dttdx )(22BB−2−2t2BNow by Linearity and & time shift:{})(tδF{}{}{})2(2)()2(2)(2−+−+=⎭⎬⎫⎩⎨⎧tBtBtBdttdxδδδFFFF())(sin21)2cos(21121222ωωωωBBeeBjj−=−=⎥⎦⎤⎢⎣⎡+−=−Euler! ⇒ cos(ω2)Now by derivative property:()[]22222222sin2)sin(2)(sin21)()(1)()()(2⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⋅⎟⎠⎞⎜⎝⎛=⎥⎦⎤⎢⎣⎡=−−=⇒⎭⎬⎫⎩⎨⎧−=⇒=⎭⎬⎫⎩⎨⎧−=πωππωπωωωωωωωωωωBBBXdttxdXXjdttxdFF321⎟⎠⎞⎜⎝⎛=πωω2sinc2)( BX Same Result!!Semi-real-world example-We want to wirelessly send binary data -Suppose FCC has allocated the band centered at 50MHz with bandwidth of 20kHzof=[]MHz01.5099.49ResponseFreq.2=⇒=oofπω1f=2f=Suppose…“FCC Rule”: No more than 2% of energy outside band112 fπω=222 fπω=-Suppose you wish to use a positive, rect. pulse to send a “1” and a negative, rect. pulse to send a “0”AA−)(txTtClearly, the smaller you make T the faster you can send data: ↓T = ↑Data RateQ: How fast can you send data while adhering to “FCC Rule”?Rough SolutionBuild x(t) from shifted rect. pulses: ∑=−=NnoTntnTtpAtx0)cos()()(ω0on 1forAAn±=By linearity of FT, modulation prop., & time shift property:∑=−⎥⎦⎤⎢⎣⎡−++=NnjnTototnePPAX02)()()(ωωωωωωModulation prop. Delay prop.To get a rough ideal of the answer: ⇒ Look at energy of onepulse ⇒ We seek to ensure no more than 2% of energy of itselflies outside the band. (In EECE 377/477) you’ll learn more precise ways to do this)Total energy of pulse:)cos()( ttpoTω)cos()( ttpotω∫−±=2/2/22)(cos)(TTodttAEω⎥⎦⎤⎢⎣⎡+⎟⎠⎞⎜⎝⎛=⎥⎦⎤⎢⎣⎡+==⇒±=⇒±===⇒=−−∫2212sin41221)2sin(41)(cos221122222222TTAExxAdxxAETxTtdxdtxttxletoooTToTToooooooooωωωωωωωωωωωωωSo 98% of this must be in band: 0.98EOkay…set that result aside. Look at FT of one pulse (Note: since delay does not affect the magnitude of the FT it does not affect energy) {}⎥⎦⎤⎢⎣⎡−++==Δ2)()()cos()()(oToToTPPAttApPωωωωωωF{}⎥⎦⎤⎢⎣⎡−++=≡2)()()cos()()(oToToTPPAttApPωωωωωωF2)(oTPωω+oω−12ωω−− Band2)(oTPωω−oω21ωωBand()()[]2224)(oToTPPAPωωωωω−++=() ()()()oToToToTPPPAPAωωωωωωωω−++−++=81442222(Ignore because it is small)Now:∫∫−−+++=212122)()(ωωωωωωωωdPdPEbandParseval!! versa) viceand0)( where0)((≠+≈−oToTPPωωωωBut by even symmetry of magnitude of FT these two terms are the same![]ωωωωωωωωωωωdPPAdPEoToTBand∫∫−++≈=⇒2121)()(42)(222220≈[]21,over ωωω∈∫∫ΔΔ−Δ=−Δ−=−⎟⎠⎞⎜⎝⎛==ΔΔωωωωωωωωωπωζζdTTAdPAT2sinc2)(2222220201Just re-write again in ω… who cares…ζ…ω… both just dummy variables!020220101101 because0 because0let ωωωωζωωωωωωζωωωζωωζ><−=→=<<−=→==⇒−= ddNow…ωωωωωπωωωdTTTAETTTBand∫ΔΔ−⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛222222sin222sin2sincNasty Integral!! Note: We want Eband/E = 0.98⇒A2in top and bottom cancels!⇒Ignore A (A = 1)Then… resort to “numerical integration”TforTTEoooo offunction a as 105022212sin412 Compute.16××=⎥⎦⎤⎢⎣⎡+⎟⎠⎞⎜⎝⎛=πωωωω Matlab)use - nintegratio numerical using( Compute.2bandE .Plot .3 TvsEEband1.98minTTSmallest T value you can use!Sets max rate at which you can send
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