EECE 301 Signals & Systems Prof. Mark Fowler1/7EECE 301 Signals & SystemsProf. Mark FowlerNote Set #31• C-T Systems: Laplace Transform… and System Response to an Input• Reading Assignment: Section 8.4 of Kamen and Heck2/7Ch. 1 IntroC-T Signal ModelFunctions on Real LineD-T Signal ModelFunctions on IntegersSystem PropertiesLTICausalEtcCh. 2 Diff EqsC-T System ModelDifferential EquationsD-T Signal ModelDifference EquationsZero-State ResponseZero-Input ResponseCharacteristic Eq.Ch. 2 ConvolutionC-T System ModelConvolution IntegralD-T System ModelConvolution SumCh. 3: CT Fourier SignalModelsFourier SeriesPeriodic SignalsFourier Transform (CTFT)Non-Periodic SignalsNew System ModelNew SignalModelsCh. 5: CT Fourier SystemModelsFrequency ResponseBased on Fourier TransformNew System ModelCh. 4: DT Fourier SignalModelsDTFT(for “Hand” Analysis)DFT & FFT(for Computer Analysis)New Signal ModelPowerful Analysis ToolCh. 6 & 8: Laplace Models for CTSignals & SystemsTransfer FunctionNew System ModelCh. 7: Z Trans.Models for DTSignals & SystemsTransfer FunctionNew SystemModelCh. 5: DT Fourier System ModelsFreq. Response for DTBased on DTFTNew System ModelCourse Flow DiagramThe arrows here show conceptual flow between ideas. Note the parallel structure between the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis).3/78.4: Response to Sinusoids and Arbitrary SignalsSinusoidal input: Before, we used FT methods to answer this question…but there we assumed the sinusoid extended infinitely in bothdirections:∞<<∞−+=ttAtx )cos()(0θωFor our studies of LT we have considered causal signals which are more practical!⎩⎨⎧<≥+=0,00),cos()(0tttAtxθωh(t)H(s))()cos()(0tutAtxθω+=?)(=ty8.2) (Table have we)()cos()(For 0tutAtxω=))(()(00202ωωωjsjsAssAssX−+=+=For ease, we’ll let θ= 0, but we can handle the case of θ≠0 using: and linearity[][])sin( )sin()cos( )cos()cos(000tAtAtAωθωθθω−=+4/7case) system ldimensiona-finite (for the )()()(Let sAsBsH =Assume system has no initial stored energy (i.e., no ICs) then we have:))()(()()()()(00ωωjsjssAsBAssXsHsY−+==Looks like what we got before using “double-sided”sinusoid. BUThere it starts at time t = 0…⎥⎦⎤⎢⎣⎡++−+=0000)()2/()()2/()()()(ωωωωγjsjHAjsjHAsAssYUse Partial Fraction Expansion Some Polynomial0))(cos()()()(000>∠++= tjHtjHAtytytωωωILT ILT… and we have this term… what does it look like?5/7)()()( that NotesAssYtγ=…and we know that the den. A(s) sets the behavior!yt(t) is “the transient response”0)),(cos()()(response"state-steady The"000≥∠+= tjHtjHAtyssωωωSo…And…- How fast it dies out depends on the real parts of the poles- The pole closest to jωaxis will “dominate” (it takes the longest to die out)- After enough time, all that is effectively left is:Note that A(s) is the system characteristic poly…. it sets the system poles.⇒ System poles determine the behavior of yt(t) If system is stable ⇒ the poles are in the LH plane ⇒ yt(t) consists of decaying terms (might also oscillate if poles are complex)6/70 2 4 6 8 10 12 14 16 18 20-6-4-20246t (sec)yt(t)0 2 4 6 8 10 12 14 16 18 20-6-4-20246t (sec)yss(t)0 2 4 6 8 10 12 14 16 18 20-6-4-20246t (sec)y(t)yss(t) Plots for Example 8.16: RC circuit with causal sinusoid appliedRC = 1 secondNote that the transient has completely died out by “5 time constants” (actually, even before that)7/7Arbitrary Inputs)()()(sAsBsH =)()()(sDsNsXXX=)()()()()(sDsAsCsBsY =If there are no common poles between X(s) & H(s): )()()()()(sDsFsAsEsYX+=)(sYt)(sYssE(s) & F(s) are “some polynomials”… they come from the math while factoringIf common poles, then Y(s) has repeated polesand you know how to modify for that caseSystem Den.Signal Den.)()()( tytytysst+=Will die out if system is stable Will persist if input persistsSo… if the system has poles in the “left-half plane” then the time-domain terms that arise from A(s) will
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