Second-Order Circuits [相容模式]

1 2012/10/241 Second-Order Circuits Introduction Finding Initial and Final Values The Source-Free SeriesRLCC ircuit The Source-Free ParallelRLCC ircuit Step Response of a SeriesRLCC ircuit Step Response of a ParallelRLCC ircuit General Second-Order Circuits Duality ApplicationsIntroduction A Second-Order circuit is characterized by asecond- order differential equation. It consists of resistors and the equivalent oftwo energy storage Initial and Final Values v(0),i(0),dv(0)/dt,di(0)/dt,v( ), andi( ) Two key points: vandiare defined according to thepassive signconvention. Continuity properties: Capacitor voltage:(VS-like) Inductor current:(IS-like))0()0()0()0( LLCC iivvvi+_Example).

2 (,)((c),)0(,)0((b)),0(,)0((a)Find:Q vidtdvdtdivi V4)0()0(A2)0()0(V4)0(2)0(A22412)0( tanalysisdcApply(a):Solvviiivi2012/10/24 3 Cont dV12)(A0)(.0foranalysisdcApply(c):Sol vitCont :0findTo(b):Sol CidtdviiCidtdvidtdvCdtdvCCCC)()()()()( t= 0+2 A2012/10/244 Cont dt= 0+A/s0250000haveweThus0481200000412gives KVLapplying, (b):Sol .)()()()()()()(:)(LvdtdivvvivLvdtdivdtdi LdtdiLLCLLLL The Source-Free SeriesRLCC ircuit )4(1)0(0)0()0(gives(2)and(1) )0((3),solveTo(3)0(2)01givesKVLA pplying(1b)10(1a)0:conditionsinitialAssu med00022000 VRIL dtdiVdtdiLRidtdiLCidtdiLRdtididtCdtdiLRi VidtCvIitC 000221)0(0:conditionsInitial0 VRIL dtdiIiLCidtdiLRdtid2012/10/245 Cont d :Let1)0(0.)

3 ConditionsInitial022200022 LCsLRsLCsLRsAeeLCAseLAReAssAAeiVRIL dtdiIiLCidtdiLRdtidststststst LCLRssLCLRLRsLCLRLRs12where1221220202220 212221 CharacteristicequationNaturalfrequencies DampingfactorResonantfrequency(or undamped naturalfrequency) )(:solutiongeneralA,:)(ifsolutionsTwo212 12211212121 AAeAeAtieAieAisststststs LCLR ssss12where02:equationsticCharacteri0202 22021202 Three cases discussed Overdamped case (distinct real roots): > 0 Critically damped case (repeated real root): = 0 Underdamped case (complex-conjugate roots): < 02012/10/246 Overdamped Case ( > 0)tstseAeAtissRLCLCLR2121212)( ti(t)tse1tse2 Critically damped Case ( = 0) to!

4 Conditionsinitialosatisfy twtcan'constantSingle)(24 Let222321212 idtdiidtdidtdidtdidtideAeAeAtiRLssRLCttt ttttttteAtAtiAtAieAiedtdAiedtdieeAidtdie Affdtdfidtdif 21211111)(0 Let2012/10/247 Critically damped Case (Cont d)ti(t) 1te tte teAtAti 21)(Underdamped Case ( < 0) 2122112121212121)(2)(1220220222012ewhers incos)(sincossincossincos)()(where4 LetBBjABBAtAtAetitBBjtBBetjtBtjtBeeBeBee BeBtijsjsRLCddtddtddddttjtjttjtjddddddd 2012/10/248 Underdamped Case (Cont d) LR etAtAtitdd2,sincos)(21 ti(t)d 2te Finding The ConstantsA1,2)(1)0(or0)0(gives0atKVL2.)0 ( )0(and)0(needwe,anddetermineTo0000021 VRIL dtdiVRIdtdiLtIi/dtdiiAA 2012/10/249 Conclusions The concept ofdamping The gradual loss of the initial stored energy Due to the resistanceR Oscillatory response is possible.

5 The energy is transferred betweenLandC. Ringingdenotes the damped oscillation in theunderdamped case. With the same initial conditions, theoverdamped case has the longest settling underdamped case has the fastest decay.(If a constant 0is assumed.)ExampleFindi(t).t< 0t> 0(6+3)2012/10/2410 Example (Cont d)t< 0t> 0 tAtAetij , LR biviat, )( )(V6)0(6)0(,A16410)0()(219202210 )1(92)0()0(1)0(10:conditionsInitial21 AAvRiLdtdi)i(-The Source-Free ParallelRLCC ircuit 011becomesequationsticcharacterithe,)(Le t(3)01(2)01givesKCLA pplying(1b)0(1a))(10:conditionsinitialAs sumed222000 LCsRCsAetvLCvdtdvRCdtvddtdvCvdtLRvVvdttv LIistt LCRCs121where02022,1 2012/10/2411 Summary Overdamped case: > 0 Critically damped case: = 0 Underdamped case.

6 < 0tstseAeAtv2121)( tetAAtv 21)( tAtAetvjsddtdd sincos)(where212202,1 Finding The ConstantsA1,2 RCRIV dtdvdtdvCIRVtVv/dtdvvAA)()0(or0)0(gives0atKCL2.)0( )0(and)0(needwe,anddetermineTo0000021 2012/10/2412 Comparisons LRIV dtdiIiLCLRs00002022,1)0()0(:conditionsInitial12where SeriesRLCC ircuit ParallelRLCC ircuit RCRIV dtdvVvLCRCs00002022,1)0()0(:conditionsInitial121where Example 1tt,eAeAtv sLC, RC R50221202210)(50, :1 Case Findv(t) for t > (0) = 5 V,i(0) = 0 Consider three cases:R= R= 5 R= )0()0()0(5)0(:conditionsInitial21 AARCR ivdtdvv2012/10/2413 Example 1 (Cont d) t,etAAtv sLC, RC R1021202210)(1010110215:2 Case 505100)0()0()0(5)0(.

7 ConditionsInitial21 AARCR ivdtdvv t,etAtAtvj sLC, RC R8212022106sin6cos)( :3 Case )0()0()0(5)0(:conditionsInitial21 AARCR ivdtdvvExample 1 (Cont d)2012/10/2414 Example 2t< 0t> 0 Findv(t).Getx(0).Getx( ),dx(0)/dt,s1,2, A1, 2 (Cont d)tt,eAeAtv sLC RC 14628541202210)(146,854354150021 t> 0 )0()0()0(50503040)0(25)40(503050)0( < 02012/10/2415 Step Response of A theasformsamethehas(2)(2)But(1),0forKVLA pplying22 LCVLC vdtdvLRdtvddtdvCiVvdtdiLRitSS responsestate-steadythe:responsetransien tthe:where)()()(sstsstvvtvtvtvCharacteri stic theasSame01becomesequationsticcharacteri The0,Let02''2'2'22 LCsLRsLCvdtdvLRdtvdVvvLCVvdtdvLRdtvdSS20 12/10/2416 Summary.

8 0(and)0(fromobtainedarewheresincos)()()( 0)(where)()()(2,121212121/dtdvvAetAtAetA AeAeAtvVvvvtvtvtvtddttststSsstsst (Overdamped)(Critically damped)(Underdamped)ExampleFindv(t),i(t) for t > three cases:R= 5 R= 4 R=1 t< 0t> 0 Getx(0).Getx( ),dx(0)/dt,s1,2, A1, 1:R= 5 dtdvCtieAeAvtv sLC LR ttss, )()(4, )1(25242120221012( )24 VInitial conditions:24(0)4 A ,(0) 1 (0)4 V5 1(0)(0)4(0)1664 34 3ssvvividvdviCdtdtCAA t< 0t> 0 Case 2:R= 4 dtdvCtietAAvtv sLC LR tss, )()(2212)1(242221210 )0()0()0( )0(1)0(, )0(:conditionsInitialV24)(21 AACdtdvdtdvCiivivvss2012/10/2418 Case 3:R= 1 dtdvCtietAtAvtvjsLC LR tss, )( )( )1( )0()0()0(V12)0(1)0(,A121124)0(.)

9 ConditionsInitial24)(21 AACdtdvdtdvCiivivvssExample (Cont d)2012/10/2419 Step Response of A theasformsamethehas(2)(2)1 But(1),0forKCLA pplying22 LCILC idtdiRCdtiddtdiLvIdtdvCiRvtSS responsestate-steadythe:responsetransien tthe:where)()()(sstsstiitititiCharacteri stic theasSame011becomesequationsticcharacter iThe01,Let012''2'2'22 LCsRCsLCidtdiRCdtidIiiLCIidtdiRCdtidSS20 12/10/2420 Summary .)0(and)0(fromobtainedarewheresincos)()( )(0)(where)()()(2,121212121/dtdiiAetAtAe tAAeAeAtiIitiititititddttststSsstsst (Overdamped)(Critically damped)(Underdamped)General Second-Order Circuits Steps required to determine the step response.

10 Determinex(0),dx(0)/dt, andx( ). Find the transient responsext(t). Apply KCL and KVL to obtain the differential equation. Determine the characteristic roots (s1,2). Obtainxt(t) with two unknown constants (A1,2). Obtain the steady-state responsexss(t) =x( ). Usex(t) =xt(t) +xss(t)to determineA1,2from thetwo initial conditionsx(0)anddx(0) ,ifor t > < 0t> 0 Getx(0).Get x( ),dx(0)/dt,s1,2, A1, (Cont d)t> 0t< 0 V4)(2)(A22412)(:forvaluesFinal(1c)V/s12)0()0(ivitCidtdvCA6)0(2)0()0()0(,)0(nodeatKCLA pplying(1b)0)0()0((1a)V12)0()0(:conditionsInitial CCiviitaiivv2012/10/2422 Example (Cont d)t> 0065.