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EECE 301 Signals & Systems Prof. Mark FowlerEffect of Changing oEffect of Changing o1/30EECE 301 Signals & SystemsProf. Mark FowlerNote Set #30• C-T Systems: Laplace Transform… and System Stability• Reading Assignment: Section 8.1 of Kamen and Heck2/30Ch. 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).3/30Ch. 8: System analysis and design using the transfer functionWe have seen that the system transfer function H(s) plays an important role in the analysis of a system’s output for a given input. e.g. for the zero-state case:{})()()()()()(1sHsXtysHsXsY-L=→=Much insight can be gained by looking at H(s) and understanding how its structure will affect the form of y(t). Section 8.1: First we’ll look at how H(s) can tell us about a system’s “stability”Section 8.4:Then we’ll see how the form of H(s) can tell us about how the system output should behaveSection 8.5:Then we’ll see how to design an H(s) to give the desired frequency response.4/30Roughly speaking: A “stable” system is one whose output does not keep getting bigger and bigger in response to an input that does not keep getting bigger. We can state this mathematically and then use our math models (e.g. h(t) or H(s)) to determine if a system will be stable. Mathematical checks for stability:The following are given without proof. They are all equivalent checks so you only need to test for one of them.1.2. All poles are in the “open left-half of the s-plane”3. Routh-Hurwitz test (section 8.2, we’ll skip it)"Integrable Absolutely" )(0∫∞∞<dtthSection 8.1 System StabilityMath definition of stability“Bounded-Input, Bounded-Output” (BIBO) stability:A system is said to be BIBO stable if for any bounded input: |x(t)| ≤ C1< ∞∀t ≥ 0 (x(t)=0, t<0)the output remains bounded: |y(t)| ≤ C2< ∞ For SOME C1& C25/30can only happen if |h(t)| decays fast enough. This is hard to check!!! ∫∞∞<0)( dtthDiscussion of Stability ChecksSo the system having this impulse response is not BIBO stable… it is unstable…that means that there is a bounded input that will (eventually) drive the system’s output to infinity.Note: Real H/W will “encounter problems” long before the output “goes to infinity”!!! )()sin()(0tutthω=Consider a system withThis is not absolutely integrable… as you integrate its absolute valuefrom 0 to ∞the value of the integral keeps growing without bound…For this system, if you put in )()sin()(0tuttxω=the system output is driven to infinity…6/30We can verify this claim using convolution:…h(τ)τx(t – τ)τtx(t – τ) h(τ)τ…To get y(t)… Integrate… gives area of the humps… as t grows you get more and more humps… output grows without bound!!!Just look at tvalues where these line up like thisThus, the upper “envelope” of the output grows by the area of one of those humps each time we increase t by one-half the period of the sinusoid…So… the upper envelope grows linearly with time… it grows without bound!!!)()()()sin()(0thtxuh==ττωτ7/30Computing the complete convolution gives something like this:We just analyzed what happens to these pointsWe just analyzed what happens to these pointsWe just analyzed what happens to these pointsWe just analyzed what happens to these pointsWe just analyzed what happens to these pointsUpper EnvelopeDoing numerical calculations like this are helpful… but they don’t PROVE that something happens… the plot by itself does not show what happens after 0.5 seconds!! As far as we know… anything could happen out there!!That is the value of mathematical analysis (based on sound models, of course!)8/30So… we can use the absolute integrability check … but it is not that easy to do in general. It also doesn’t tell us much about why a system is stable.Well… if the impulse response can be used to check for stability… it should be no surprise that the transfer function can also be used!!!A term for each real poleA term for each complex pole pair)(...)()(1thththp++=After the Inverse LT we get:)(...)(......)(101110111sHsHasasasbsbsbsbsHpNNNMMMM++=++++++++=−−−−Consider:After Partial Fraction ExpansionFrom our understanding of partial fraction expansion we know what each of these terms can look like:poles repeated- )()(poledistinct )()(ktuetcthtuecthtpkiitpiiii==poles repeated- )()tcos()(poledistinct )()cos()(iiktuetcthtutecthtpkiiiitiiiiθωθωσ+=+=For Real PoleFor Complex Pole Pair9/30So what do these look like?For the distinct pole cases:)()cos()()( tuθtωecthecthiitσiitpiiii+==This decays if pi< 0This decays if σi< 0We can capture the condition for decay with: “the real-part of the pole is negative”Although we won’t prove it here… it can be shown that this decay is fast enough to ensure “absolute integrability”… and thus stability. )()tcos()()()(iituetcthtuetcthtpkiitpkiiiiθω+==For the repeated pole cases:We’ve got a race!!! The tkterms are “going up” and the exponentials are “going down” for poles whose real parts are negative… WHO WINS???We’ve got a race!!! The tkterms are “going up” and the exponentials are “going down” for poles whose real parts are negative… WHO WINS???The decaying exponential wins!!!10/30So we have argued that… a system is stable if all its poles have negative real parts{}ipi∀<,0ReStability condition (not proved here!)An Nthorder system (with N


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