Lecture 7 Circuit analysis via Laplace transform

1 S. BoydEE102 Lecture7 CircuitanalysisviaLaplacetransform analysisof generalLRCcircuits impedanceand admittancedescriptions naturaland forcedresponse circuitanalysiswithimpedances naturalfrequenciesand stability7{1 CircuitanalysisexamplePSfragreplacements iuyLRinitialcurrent:i(0)KCL, KVL,and branchrelationsyield: u+Li0+y= 0,y=Ritake Laplacetransforms to get U+L(sI i(0)) +Y= 0; Y=RIsolveforYto getY=U+Li(0)1 +sL=R=11 +sL=RU+L1 +sL=Ri(0)Circuitanalysisvia Laplacetransform7{2in the timedomain:y(t) =1 TZt0e =Tu(t )d +Ri(0)e t=TwhereT=L=Rtwo termsiny(orY): rst termcorrespondsto solutionwithzeroinitialcondition rst termis convolutionof sourcewitha function secondtermcorrespondsto solutionwithzerosourcewe'll see theseare generalproperties..Circuitanalysisvia Laplacetransform7{3 Analysisof generalLRCcircuitsconsidera circuitwithnnodes andbbranches,containing independentsources linear elements(resistors, op-amps.)}}}

2 Inductors & capacitorsCircuitanalysisvia Laplacetransform7{4sucha circuitis described by threesets of equations: KCL:Ai(t) = 0(n 1equations) KVL:v(t) =ATe(t)(bequations) branchrelations(bequations)where A2R(n 1) bis the reducednode incidencematrix i2 Rbis the vector of branchcurrents v2 Rbis the vector of branchvoltages e2Rn 1is the vector of node potentialsCircuitanalysisvia Laplacetransform7{5 Branchrelations independentvoltagesource:vk(t) =uk(t) resistor:vk=Rik capacitor:ik=Cv0k inductor:vk=Li0k VCVS:vk=avj and so on (currentsource,VCCS,op-amp,.. )thus:circuitequationsare a set of2b+n 1(linear) algebraic and/ordi erentialequationsin2b+n 1variablesCircuitanalysisvia Laplacetransform7{6 Laplacetransformof circuitequationsmostof the equationsare the same, , KCL, KVLbecomeAI= 0,V=ATE independentsources, ,vk=ukbecomesVk=Uk linear staticbranchrelations, ,vk=RikbecomesVk=RIkthedi erentialequationsbecomealgebraicequation s: capacitor:Ik=sCVk Cvk(0) inductor.}}}

3 Vk=sLIk Lik(0)thus,in frequencydomain,circuitequationsare a set of2b+n 1(linear) algebraic equationsin2b+n 1variablesCircuitanalysisvia Laplacetransform7{7thus,LRCcircuitscan be solvedexactlylike staticcircuits, except all variablesare Laplacetransforms,not real numbers capacitors and inductors havebranchrelationsIk=sCVk Cvk(0),Vk=sLIk Lik(0)interpretation:an inductor is like a \resistance"sL, in serieswithanindependentvoltagesource Lik(0)a capacitor is like a \resistance"1=(sC), in parallelwithan independentcurrentsource Cvk(0) these\resistances"are calledimpedances thesesourcesare impulsesin the timedomainwhichset up the initialconditionsCircuitanalysisvia Laplacetransform7{8 Impedanceandadmittancecircuitelementor devicewithvoltagev, currentiPSfragreplacementsivthe relationV(s) =Z(s)I(s)is calledanimpedancedescriptionof thedevice Zis calledthe (s-domain)impedanceof the device in the timedomain,vandiare relatedby convolution.}}

4 V=z isimilarly,I(s) =Y(s)V(s)is calledanadmittancedescription(Y= 1=Z)Circuitanalysisvia Laplacetransform7{9 Examples a resistor has an impedanceR an inductorwithzeroinitialcurrenthas an impedanceZ(s) =sL(admittance1=(sL)) a capacitorwithzeroinitialvoltagehas an impedanceZ(s) = 1=(sC)(admittancesC) SSS analysiswithphasors: resistor:V=RI inductor:V= (j!L)I capacitor:V= (1=j!C)Is-domainand phasor impedanceagreefors=j!, but are not the sameCircuitanalysisvia Laplacetransform7{10we can express the branchrelationsasM(s)I(s) +N(s)V(s) =U(s) +Wwhere Uis the independentsources Wincludesthe termsassociatedwithinitialconditions MandNgive the impedanceor admittanceof the branchesfor example,if branch13 is an inductor,(sL)I13(s) + ( 1)V13(s) =Li13(0)(thisgivesthe 13throw ofM,N,U, andW)Circuitanalysisvia Laplacetransform7{11we can writecircuitequationsas one big matrixequation.}}}

5 24A000I ATM(s)N(s)03524I(s)V(s)E(s)35=2400U(s) +W35 Circuitanalysisvia Laplacetransform7{12hence,24I(s)V(s)E(s) 35=24A000I ATM(s)N(s)035 12400U(s) +W35in the timedomain,24i(t)v(t)e(t)35=L 10B@24A000I ATM(s)N(s)035 12400U(s) +W351CA this givesaexplicitsolutionof the Circuit theseequationsareidenticalto thosefor a linear staticcircuit(exceptinsteadof real numbers we haveLaplacetransforms, ,complex-valuedfunctionsofs) hence,muchof whatyou know extendsto this caseCircuitanalysisvia Laplacetransform7{13 Naturalandforcedresponselet'sexpress solutionas24i(t)v(t)e(t)35=L 10B@24A000I ATM(s)N(s)035 12400U(s)351CA+L 10B@24A000I ATM(s)N(s)035 12400W351 CAthuscircuitresponseis equalto: thenaturalresponse, , solutionwithindependentsourceso , plus theforcedresponse, , solutionwithzeroinitialconditionsCircuit analysisvia Laplacetransform7{14 the forcedresponseis linear inU(s), , the independentsourcesignals the naturalresponseis linear inW, , the inductor & capacitor initialconditionsCircuitanalysisvia Laplacetransform7{15 Backto theexamplePSfragreplacementsiuyLRinitial current:i(0)naturalresponse:set sourceto zero,get LR circuitwithsolutionynat(t) =Ri(0)e t=T; T=L=Rforcedresponse:assumezeroinitialcur rent,replaceinductor withimpedanceZ=sL:Circuitanalysisvia Laplacetransform7{16 PSfragreplacementsUYfrcZ=sLRby voltagedividerrule (for impedances),Yfrc=URR+sL(as if theyweresimpleresistors!)}}}}}

6 Soyfrc=L 1(R=(R+sL)) u, ,yfrc(t) =1 TZt0e =Tu(t )d all together,the voltageisy(t) =ynat(t) +yfrc(t)(sameas before)Circuitanalysisvia Laplacetransform7{17 Circuitanalysiswithimpedancesfor a circuitwith linear staticelements(resistors, op-amps,dependentsources,.. ) independentsources elementsdescribed by impedances(inductors & capacitorswithzeroinitialconditions, .. )we can manipulate Laplacetransforms of voltages,currents impedancesas if theywere (real,constant)voltages,currents,and resistances,respectivelyCircuitanalysisv ia Laplacetransform7{18reason:theyboth satisfythe sameequationsexamples: series,parallelcombinations voltage& currentdividerrules Thevenin,Nortonequivalents nodal analysisCircuitanalysisvia Laplacetransform7{19example:PSfragreplac ementsIinVin3 4H2F1 let's nd inputimpedance, ,Zin=Vin=Iinby series/parallelcombinationrules,Zin= 1=(2s) + (1k4s) + 3 =12s+4s1 + 4s+ 3we haveVin(s) = 12s+4s1 + 4s+ 3 Iin(s)providedthe capacitor & inductor havezeroinitialconditionsCircuitanalysis via Laplacetransform7{20example.}}}}

7 Nodal analysisPSfragreplacementsIinE1E23 1 4 5H2 Fnodal equationsareGE=Isrcwhere Isrcis totalof currentsources owinginto nodes Giiis sumof admittancestied to nodei Gijis minusthe sumof all admittancesbetween nodesiandjCircuitanalysisvia Laplacetransform7{21for this examplewe have: 1 + 2s+13 (2s+13) (2s+13)13+ 2s+14+15s E1(s)E2(s) = Iin(s)0 (whichwe couldsolve.. )Circuitanalysisvia Laplacetransform7{22example:Theveninequi valentPSfragreplacements1 e t2H1 ABvoltagesourceis1s 1s+1=1s(s+1)ins-domainTheveninvoltageis open-circuitvoltage, ,Vth=1s(s+ 1)11 + 2sTheveninimpedanceis impedancelookinginto terminalswithsourceo , ,Zth= 1k2s=2s1 + 2sCircuitanalysisvia Laplacetransform7{23 Theveninequivalentcircuitis:PSfragreplac ementsVth(s) =1s(s+ 1)(1+ 2s)Zth(s) =2s1 + 2sABCircuitanalysisvia Laplacetransform7{24 Naturalfrequenciesandstabilitywe say a circuitisstableif its naturalresponsedecays ( , convergestozeroast!)}}}}

8 1) for all initialconditionsin this casethe Circuit \forgets"its initialconditionsastincreases;thenatural responsecontributesless and less to the solutionastincreases, ,y(t)!yfrc(t)ast!1circuitis stablewhenpoles of the naturalresponse,callednaturalfrequencies , havenegativereal parttheseare givenby the zerosofdet24A000I ATM(s)N(s)035 Circuitanalysisvia Laplacetransform7{25}