Characteristics Equations, Overdamped-, Underdamped-, …

1 Kevin D. Donohue, University of Kentucky1 Characteristics Equations, overdamped -, underdamped -, and Critically Damped CircuitsKevin D. Donohue, University of Kentucky2In 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 Kentucky3 Find the differential equation for the circuit below in terms of vc and also terms of iLShow:vs(t)RLC+vc(t)_iL(t)cccscccsvLCdt dvLRdtvdLCtvvdtdvRCdtvdLCtv1)( )(2222++= ++= ++= ++=tLLLstLLLsdiLCiLRdtdiLtvdiCRidtdiLtv )(1)( )(1)(Kevin D.

2 Donohue, University of Kentucky4 Find the differential equation for the circuit below in terms of vc and also terms of iLShow:LLLsLLLsiLCdtdiRCdtidLCtiidtdiRLd tidLCti11)( )(2222++= ++= ++= ++=tcccstcccsdvLCvRCdtdvCtidvLvRdtdvCti )(11)( )(11)(is(t)RLC+vc(t)_iL(t)Kevin D. Donohue, University of Kentucky5 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.

3 Donohue, University of Kentucky6 Find characteristic equation from homogeneous equation: xadtdxadtxd21220++= Convert to polynomial by the following substitution: nnndtxds=2120asas++=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 .5-1-0 . s po ns ea 1*e xp(-t)+e xp(-2 t) for a 1 =[-3 :2 ] If roots are real and distinct ( ), natural solution becomes04221> aa)exp()exp()(2211tsatsatxn+=a1=-3a1=2 Kevin D. Donohue, University of .5-2-1 .5-1-0 . 1 *e xp (-t)+t*e xp (-t) fo r a 1 =[-3 :2 ] If roots are real and repeated ( ), natural solution becomes04221= aa)exp()exp()(1211tstatsatxn+=a1=-321ss= a1=2 Kevin D.

4 Donohue, University of Kentucky9 If roots are complex ( ), natural solution becomes:04221< .5-1-0 . (-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 Kentucky10 Find the unit step response forvcandiLfor 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()(tutttiL =