Transient Response for Second-Order CircuitsSecond-Order Circuits:ExampleSlide 4Solving Second-Order Systems:Natural SolutionsOverdampedCritically DampedUnderdampedSlide 10Kevin D. Donohue, University of Kentucky1Transient Response for Second-Order Circuits Characteristics Equations, Overdamped-, Underdamped-, and Critically Damped CircuitsKevin D. Donohue, University of Kentucky2Second-Order Circuits:In previous work, circuits were limited to one energy storage element, which resulted in first-order differential equations. Now, a second independent energy storage element will be added to the circuits to result in second order differential equations:xadtdxadtxdty2122)( Kevin D. Donohue, University of Kentucky3ExampleFind the differential equation for the circuit below in terms of vc and also terms of iLShow: vs(t)RLC+vc(t)_iL(t)cccscccsvLCdtdvLRdtvdLCtvvdtdvRCdtvdLCtv1)( )(2222tLLLstLLLsdiLCiLRdtdiLtvdiCRidtdiLtv)(1)( )(1)(Kevin D. Donohue, University of Kentucky4ExampleFind the differential equation for the circuit below in terms of vc and also terms of iLShow: LLLsLLLsiLCdtdiRCdtidLCtiidtdiRLdtidLCti11)( )(2222tcccstcccsdvLCvRCdtdvCtidvLvRdtdvCti)(11)( )(11)( is(t)R LC+vc(t)_iL(t)Kevin D. Donohue, University of Kentucky5Solving Second-Order Systems:The method for determining the forced solution is the same for both first and second order circuits. The new aspects in solving a second order circuit are the possible forms of natural solutions and the requirement for two independent initial conditions to resolve the unknown coefficients.In general the natural response of a second-order system will be of the form:)exp()exp()(2211tsKtstKtxmKevin D. Donohue, University of Kentucky6Natural SolutionsFind characteristic equation from homogeneous equation: xadtdxadtxd21220 Convert to polynomial by the following substitution: nnndtxds 2120 asas to obtainBased on the roots of the characteristic equation, the natural solution will take on one of three particular forms. Roots given by: 2422112,1aaasKevin D. Donohue, University of Kentucky70 0.5 1 1.5 2 2.5 3 3.5 4-2-1.5-1-0.500.511.522.53timeresponsea1*exp(-t)+exp(-2t) for a1 =[-3:2]OverdampedIf roots are real and distinct ( ), natural solution becomes04221 aa)exp()exp()(2211tsatsatxna1=-3a1=2Kevin D. Donohue, University of Kentucky80 0.5 1 1.5 2 2.5 3 3.5 4-3-2.5-2-1.5-1-0.500.511.52timeresponsea1*exp(-t)+t*exp(-t) for a1 =[-3:2] Critically DampedIf roots are real and repeated ( ), natural solution becomes04221 aa)exp()exp()(1211tstatsatxna1=-321ss a1=2Kevin D. Donohue, University of Kentucky9UnderdampedIf roots are complex ( ), natural solution becomes:04221 aa0 0.5 1 1.5 2 2.5 3 3.5 4-2-1.5-1-0.500.511.522.5timeresponseexp(-t).*(cos(6*pi*t)+2*sin(6*pi*t))  js 2,1 ) sin() cos()exp()(21tctcttxn) cos() exp()( ttAtxnor2221ccA 121tancc)sin( )cos(21AcAc Kevin D. Donohue, University of Kentucky10ExampleFind the unit step response for vc and iL for the circuit below when:vs(t)RLC+vc(t)_iL(t)a) R=16, L=2H, C=1/24 Fb) R=10, L=1/4H, C=1/100 Fc) R=2, L=1/3H, C=1/6 FShow:)()2exp(23)6exp(211)( tutttvc )()6exp()2exp(81)( tutttiL )()20exp(20)20exp(1)( tuttttvc  )()3sin()3cos()3exp(1)( tuttttvca)b)c) )()20exp(4)( tuttiL )()3sin()3exp()(