Identify a node voltage at each of the nonreference nodes and a current with every element in the circuit Step 2 Write KCL connection constraints in terms of the element currents at the nonreference nodes Step 3 Use the element admittances and the f ID: 33046
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S-Domain Analysis s-Domain Circuit AnalysisTime domain DifferentialLaplace TransformInverse Transform Laplace Transform Transformed Kirchhoffs Laws in s-Domaint domains domain )(1ti)(4ti)(2ti)(3ti0)()()()(4321titititi0)()()()(4321sIsIsIsI0)()()(321tvtvtv0)()()(321sVsVsV )(2tv)(4tv)(1tv)(3tv)(5tv Signal Sources in s Domain L t domains domain _+)(tvS _+ )(ti()()(titvtvSVoltage Source: _+)(sV)(sVS _+ )(sIVoltage Source:circuit()()(sIsVsVS +_)(tv)(tiS )(ti()()(tvtitiSCurrent Source: )(sV )(sIS +_ )(sICurrent Source:circuit()()(sVsVsIS Time and s-Domain Element ModelsImpedance and Voltage Source for Initial ConditionsTime Domain L s-Domain +)(tvRR Resistor: +)(sVRR Resistor: +)(tvLL )(tiL tvLL)()( + _+)(sVLLs)0(L (sIL)0( ()(LLLV +)(tvCC )(tiC )0( (1)(0CtCCvdiCtv + _+)(sVC Cs1 svC)0( )(sIC s)(vsIVCCC0 (1)( Impedance and Voltage Source for Initial Conditions sIsVsZRRR)()()( 0)0(th ()()(LLLLiIsVsZ Impedance Z(s) Impedance of the three passive elements Time and s-Domain Element ModelsAdmittance and Current Source for Initial ConditionsTime Domain L s-Domain +)(tvRR )(tiR Resistor: +)(sVRR )(sIR Resistor: +)(tvLL )(tiL 0( (1)(0LtLLidvLti _+)(tvCC )(tiC tiCC)()( )(isVILLL0 (1)( _+)(sVLLs siL)0( )(sIL )0( ()(CCCI +)(sVC Cs1)0(C (sIC Admittance and Current Source for Initial sVsIsYRRR1)()()( 0)0(th )()()(LLLLiVsIsY Admittance Y(s) Admittance of the three passive elements Example: Solve for Current Waveform i(t) L + )(tuVAR)(tiL _+)(sVL)0(LLi)(sVR _+ sVAR)(sILs _+ 0)()(sVsVsVLRABy KVL:)()(sVR0()()(LLLisLsIsV )0()()(LAV LRsiLRsRVsRVLRsiLRssLVsILAALA)0()0()()( Inverse Transform:forced response Series Equivalence and Voltage Division ()()()()()()()(2211sVsZsZsVsVsZsZsVEQ)()()()()(2222sIsZsIsZsV)()()(21sZsZsZ 1 2 )(1sV )(2sV)(sV)(sI )(1sI )(2sI )(sV)(sI 21ZZZ Parallel Equivalence and Current Division ()()()()()()()(2211sIsYsYsIsIsYsYsIEQ)()()(22sVsYsI)()()(21sYsYsY )(sV)(sI 21YYY 1 2 )(sV)(sI )(2sI )(1sI )(1sV1EQZ _+_+Ls (2sV A B Z (1sV _+ A B Z (1sV1EQZ _+_+Ls (2sV A B Example: 11)()(21RLCsRCsZZEQ CssZsYEQ1)(1)(11 Find equivalent impedance at A and B at t = 0 )(1tvR C _+_+L (2tv A B (1sVR _+_+Ls (2sV A B (1sV1EQZ _+_+Ls (2sV A B ()()()(12112sVRRCLssVZsZsVEQ (1sVR _+_+Ls (2sV A B1EQZ General Techniques for s-Domain Circuit AnalysisNode Voltage Analysis (in s-domain)Use Kirchhoffs Current Law (KCL)Get equations of node voltagesUse current sources for initial conditionsVoltage source current source Mesh Current Analysis (in s-domain)Use Kirchhoffs Voltage Law (KVL)Get equations of currents in the meshUse voltage sources for initial conditionsCurrent source voltage source (Works only for Planar circuits) Transform the circuit into the s domain using current of the non-reference nodes and a current with every element in the circuitWrite KCL connection constraints in terms of the element currents at the non-reference nodesUse the element admittances and the fundamental property of node voltages to express the element currents in terms of the node voltagesSubstitute the device constraints from Step 3 into the KCL connection constraints from Step 2 and arrange the Example: Formulating Node-Voltage Equations L (tiSR C L t domain )(sISR s domain siL)0()0(C(sVA)(2sI )(1sI )(3sI )(sVB Reference Step 0:Transform the circuit into the s domain using current sources to represent capacitor and inductor initial conditionsStep 1:Identify N-1=2 node voltages and a current with each elementStep 2:Apply KCL at nodes A and B: Step 3:Express element equations in terms of node voltages ()()()(1 where)()()()()()(1)()()()(321sVsYsIRGsVsYsIsVsVVsVsYsIBBCAARBABAL Step 2:Apply KCL at nodes A and B: Step 3:Express element equations in terms of node voltages Step 4:Substitute eqns. in Step 3 into eqns. in Step 2 and collect common terms to yield node-voltage eqns. siVVisIsVVLCBALSBA)0()0()(1)(1B 0()()(1)(1A Circuit Determinant: Depends on circuit element parameters: L, C, G=1/R, not on driving force and initial conditionsSolve for node A using Cramers rule: Zero Statewhen initial condition are turned off Solve for node B using Cramers rule: Zero State Zero input Network FunctionsDriving-point function relates the voltage and current at a given pair of terminals called a port TransformInput Network Transfer function relates an input and response at different ports in the circuit )()((12sVsVsTV in thezero-state )(sV)(sI in thezero-state Input + V2V)(sTV ()(Transfer Current (12sIsIsTI )()((12sVsIsTY )()((12sIsVsTZ _+ V2I)(sTY In I2V)(sTZ I)(sTI In I Calculating Network Functions Z1 2 )(2sV)(1sV _+ 1 2 (1sI)(2sI )()()(21sZsZsZEQ Driving-point impedanceVoltage transfer function: )()()()()(12122sVsZsZsZsV )()()(21sYsYsY Driving-point admittanceVoltage transfer function: )()()()()(12122sYsYsYsYsI Impulse Response and Step Response T(s)(sX Input-output relationship in s-domainWhen input signal is an impulseImpulse response equals network functionH(s) = impulse response transformh(t) = impulse response waveform When input signal is a stepG(s) = step response transformg(t) = step response waveform(=) means equal almost everywhere,excludes those points at which g(t)has a discontinuity dthdhsgt)())(( ,)()(0