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).(,)((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)(.

2 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:conditionsInitial022200022 LCsLRsLCsLRsAeeLCAseLAReAssAAeiVRIL dtdiIiLCidtdiLRdtidststststst LCLRssLCLRLRsLCLRLRs12where1221220202220 212221 CharacteristicequationNaturalfrequencies DampingfactorResonantfrequency(or undamped naturalfrequency) )(:solutiongeneralA,:)(ifsolutionsTwo212 12211212121 AAeAeAtieAieAisststststs LCLR ssss12where02.

3 EquationsticCharacteri020222021202 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!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.

4 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. 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: < 0tstseAeAtv2121)( tetAAtv 21)( tAtAetvjsddtdd sincos)(where212202,1 Finding The ConstantsA1,2 RCRIV dtdvdtdvCIRVtVv/dtdvvAA)()0(or0)0(gives0 atKCL2.)

5 0( )0(and)0(needwe,anddetermineTo0000021 2012/10/2412 Comparisons LRIV dtdiIiLCLRs00002022,1)0()0(:conditionsIn itial12where SeriesRLCC ircuit ParallelRLCC ircuit RCRIV dtdvVvLCRCs00002022,1)0()0(:conditionsIn itial121where 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(: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.)

6 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(:condition sInitial24)(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.)

7 Where)()()(sstsstiitititiCharacteristic 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. 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).

8 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:equationsticCharacteri(4)2465122121 22gives(3)into(2)ngSubstituti(3)1214give smeshleftthetoKVLA pplying(2)212givesnodeatKCLA pplying22222 ssvdtdvdtvdvdtvddtdvdtdvvvdtdiidtdvvia(2 )usingbyobtainbecan)(8,12obtainwe(1c)and (1a)From)(4)(where)()(3,2213221tiAAeAeAt vvvtvvtvstttsstss Duality Duality means the same characterizing equations withdual quantities sourceCurrent sourceNodeMeshSeries pathParallel pathOpen circuitShort circuitKVLKCLT heveninNortonTable for dual pairs2012/10/2423A Case Study0:KVL)()()(212211 nnnvvvifvifvifv 0:KCL)()()(212211 nnniiivfivfivfi.

9 +v1-+v2-+ +v_Element Transformations)(Voltage)(Current)ce(Cap acitan)e(Inductanc)ce(ConductanSSSSSSIIv IiVViVvLdtdvLidtdiLvCdtdiCvdtdvCiRRviRiv 2012/10/2424 Example 1 SeriesRLCC ircuit ParallelRLCC ircuit012 LCsLRs012 LCsLRsExample 22012/10/2425 Application: Smoothing CircuitsOutputfrom aD/Aconverterv0vs