CHAPTER 7: SECOND-ORDER CIRCUITS

EEE5: CI RCUI T THEORY CHAPTER 7: SECOND-ORDER CIRCUITS 7. Inroducion Thi chaper conider circui wih wo orage elemen. Known a econd-order circui becaue heir repone are decribed by differenial equaion ha conain econd derivaive. Example of econd-order circui are hown in figure 7. o 7.4. Figure 7. Figure 7. 7

EEE5: CI RCUI T THEORY Figure 7.3 Figure 7.4 7

EEE5: CI RCUI T THEORY 7. Finding Iniial and Final Value Objecive: Find v ( ), i(), dv(), di(), i( ), v( ) Two key poin: (a) The direcion of he curren i() and he polariy of volage v(). Figure 7.5 (b) Figure 7.6 The capacior volage i alway coninuou: v ( ) v( and he inducor curren i alway coninuou: i ( ) i( ) ) 73

EEE5: CI RCUI T THEORY Example: The wich in Figure 7.7 ha been cloed for a long ime. I i open a. Find i ( ), v( ), di( ), dv( ), i( ), v( ) Figure 7.7 The wich i cloed a long ime before, hu he circui ha reached dc eady ae a. The inducor ac like a hor circui. The capacior ac like an open circui. Figure 7.8 74

EEE5: CI RCUI T THEORY i( ) A 4 v( ) i( ) () 4 A he inducor curren and capacior volage canno change abruply, i( ) i( ) v( ) v( ) A 4 V V A, he wich i open and he equvalen can be drawn a: Since i C C ( ) i( ) dv ( dv Figure 7.9 A i, dv i C and ) C i C ( C ) C. V/ 75

EEE5: CI RCUI T THEORY Similarly, Since Thu, L di v, di / v L, applying KVL L L / i( ) vl ( ) 4 v( ) v L ( ) 8 4 di( ) v L ( L ).5 A/ For >, he circui undergoe ranience. Bu, he circui reache eady ae again. The inducor ac like a hor circui. The capacior ac like an open circui. Thu, Figure 7. i( ) A v( ) V 76

EEE5: CI RCUI T THEORY 7.3 The Source-Free Serie RLC Circui Conider he ource-free erie RLC circui in Figure 7.. Figure 7. The circui i being excied by he energy iniially ired in he capacior and inducor. V - he iniial capacior volage I - he iniial inducor curren Thu, a v ( ) i V C i ( ) I Applying KVL around he loop: di Ri L i C Differeniae wih repec o : 77

EEE5: CI RCUI T THEORY d i R L di i LC - he econd-order differenial equaion Le circui di() Ri ( ) L V di() ( RI V ) L i Ae - he exponenial form for order Thu, we obain A Ae e AR e L R L R A e LC LC or L LC Thi quadraic equaion i known a he characeriic equaion ince he roo of he equaion dicae he characer of i. The roo are: R L R L LC 78

EEE5: CI RCUI T THEORY or R L R L LC (7.) α α ω, α α ω where R, L ω LC α (7.) The roo, are called naural frequencie, meaured in neper per econd (Np/). - hey are aociaed wih he naural repone of he circui. ω i known a he reonan frequency or ricly a he undamped naural frequency, expreed in radian per econd (rad/). α i he neper frequency or he damping facor, expreed in neper per econd. poible oluion for i: i A e, i Ae 79

EEE5: CI RCUI T THEORY d i R di L i LC i a linear equaion any linear combinaion of he wo diinc oluion and i i alo a oluion for he equaion. Thu, ( ) i A e Ae where A and A are deermined from he iniia value () i and di ( ) From Equaion 7.: (i) If α > ω - overdamped cae. (ii) If α ω - criically damped cae. (iii) If α < ω - underdamped cae Overdamped cae: α > ω implie - C > 4L R. - boh roo are negaive and real. - The repone, i( ) A e A e (7.3) which decay and approache zero a increae a hown in Figure 7. i 8

EEE5: CI RCUI T THEORY Criically Damped Cae: - α ω implie - C α - The repone, i( ) Figure 7. R L 4L R α α A e Ae 3 3 A A A A e α where - Thi canno be he oluion becaue he wo iniial condiion canno be aified wih he ingle A. conan 3 - Le conider again: d i α ω R di L R / i LC - L, hu, 8

EEE5: CI RCUI T THEORY - Le, - Thu, d i di α α i d di di α i α αi f df di α f αi which i he oluion - So, di e α f A e αi di order differenial equaion wih α A e α e α αi A which can be wrien a: d ( α e i) A - Inergraing boh ide: α e i A A 8

EEE5: CI RCUI T THEORY or i α ( A A ) e - Hence, he naural repone of he criically damped circui i a um of wo erm: a negaive exponenial and a negaive exponenial muliplied by a linear erm: i ) α ( A A ) e ( (7.4) Figure 7.3 Underdamped Cae: α < ω implie C < - 4L / R - The roo can be wrien a: α ( ω α ) α α ( ω α ) α jω d jω d 83

EEE5: CI RCUI T THEORY where ω d ω α, which i called he damping frequency. - Boh ω and ω d are naural frequencie becaue hey help deermine he naural repone. ω i called he undamped naural frequency. - - ω d i called he damped naural frequency. - The naural repone i i( ) e A e α - Uing Euler ideniie, e jθ coθ - We ge, ( α jω ) ( α jω ) d A e ( j ω j ) d ωd A e A e j inθ, e jθ d coθ j inθ [ A ( coω j inω ) A ( coω j inω ) ] α i( ) e d d d [( A A ) coω j( A A ) inω] A and j( ) α i( ) e - Replacing conan ( A ) wih conan B and B, we ge ( B coω B inω ) α i( ) e d d A A (7.5) d 84

EEE5: CI RCUI T THEORY - Wih he preence of ine and coine funcion, i i clear ha he naural repone for hi cae i exponenially damped and ocillaory in naure. - The repone ha a ime conan of / α and a period of T π / ω d Figure 7.4 Concluion: (i) - The behaviour of uch nework i capured by he idea of damping, which i he gradual lo of he iniial ored energy. - The damping effec i due o he preence of reiance R. - The damping facor α deermine he rae a which he repone i damped. - If R, hen α and we have an LC circui wih LC a he undamped naural frequency. Since α < ω in hi 85

EEE5: CI RCUI T THEORY (ii) (iii) cae, he repone i no only undamped bu alo ocillaory. - The circui i aid o be lole becaue he diipaing or damping elemen (R) i aben. - By adjuing he value of R, he repone may be made undamped, overdamped, criically damped or underdamped. Ocillaory repone i poible due o he preence of he wo ype of orage elemen. - Having boh L and C allow he flow of energy back and forh beween he wo. - The damped ocillaion exhibied by he underdamped repone i known a ringing. - I em from he abiliy of he orage elemen L and C o ranfer energy back and forh beween hem. - I i difficul o differeniae beween he overdamped and criically damped repone. - he criically damped repone i borderline and decay he fae. - The overdamped ha he longe eling ime becaue i ake he longe ime o diipae he iniial ored energy. 86

EEE5: CI RCUI T THEORY - If we deire he fae repone wihou ocillaion or ringing, he criically damped circui i he righ choice. Example: In Figure 7.5, R 4 Ω, L 4H, C / 4F. Calculae he characeriic roo of he circui. I he naural repone overdamped, underdamped or criically damped. α The roo are R L 5, Figure 7.5 ω LC, α ± α ω 5 ± 5., 9.899 α > ω, he repone i overdamped. Since 87

EEE5: CI RCUI T THEORY 7.4 The Source-Free Parallel RLC Circui Parallel RLC circui find many pracical applicaion e.g. incommunicaion nework and filer deign. Conider he parallel RLC circui hown in Figure 7.6: Figure 7.6 Aume iniial inducor curren I and iniial capacior volage V. i() v() I V L v( ) Since he hree elemen are in parallel, hey have he ame volage v acro hem. According o paive ign conenion, he curren i enering each elemen 88

EEE5: CI RCUI T THEORY - he curren hrough each elemen i leaving he op node. Thu, applying KCL a he op node give v v R L C dv Taking he derivaive wih repec o and dividing by C reul in d v dv v RC LC Replace he fir derivaive by and he econd derivaive by. Thu, RC LC The roo of he characeriic equaion are or where RC, ± RC, α ± α ω LC (7.6), ω RC LC α (7.7) 89

EEE5: CI RCUI T THEORY There are hree poible oluion, depending on wheher α > ω, α ω, or α < ω. Overdamped Cae (D! Z ) α > ω when L > 4R C. The roo of he characeriic equaion are real and negaive The repone i v( ) A e A e (7.8) Criically Damped Cae (D Z ) For α ω, L 4R C. The roo are real and equal The repone i v( ) ( A A) e α (7.9) Underdamped Cae (D < Z ) When α < ω, L < 4R C. The roo are complex and may be expreed a S, -α ± jω d 9

EEE5: CI RCUI T THEORY Where ω d The repone i ω α α v( ) e ( A co ω A in ω ) (7.) The conan A and A in each cae can be deermined from he iniial condiion. We need v() and dv()/. The fir erm i known from: v ( ) V For econd erm i known by combining and a or i() v() v R V R I V L v L I C v( ) C dv() dv 9

EEE5: CI RCUI T THEORY dv ( ) ( V RI ) RC The volage waveform are imilar o hoe hown in Secion 7.3. Having found he capacior volage v() for he parallel RLC circui a hown above, we can readily obain oher circui quaniie uch a individual elemen curren. For example, he reior curren i i R v/r and he capacior volage i v C C dv/. Noice ha we fir found he inducor curren i() for he RLC erie circui, wherea we fir found he capacior volage v() for he parallel RLC circui. Example: In he parallel circui of Figure 7.7, find v() for >, auming v() 5V, i(), L H and C mf. Conider hee cae: R.93Ω, R 5Ω, and R 6.5Ω. CASE If R.93 Ω α RC x.93 x x x 3 6 9

EEE5: CI RCUI T THEORY ω LC 3 x x Since α > ω, he repone i overdamped. The roo of he characeriic equaion are, α ± α ω, 5 and he correponding repone i v( ) A e A e 5 We now apply he iniial condiion o ge A and A. From v ( ) 5 A A dv() v() Ri() RC dv() 5.93 x x v( ) A e A 5, dv A e A e 5 5 A, 6 -A 5A Thu, A.65 and A -5.65 and v().65e - e -5.65e 3-5 6 93

EEE5: CI RCUI T THEORY CASE When R 5Ω α RC 3 x 5 x While ω remain he ame. x Since α ω, he repone i criically damped. Hence, -, and v() (A - A)e To ge A and A, we apply he iniial condiion dv() From v ( ) 5 A v() Ri() RC 5 5 3 v() (A A )e x x -, dv ( A A A e ) A -A A Thu, A 5 and A 5 and v() (5 5)e - V 94

EEE5: CI RCUI T THEORY CASE 3 α When R 6.5 Ω RC 3 x 6.5 while ω remain he ame. x x A α < ω in hi cae, he repone i underdamped. The roo of he characeriic equaion are, α ± α ω 8 ± j6 Hence, v() (A co 6 A in 6)e -8 We now obain A and A, a v() 5 A dv() v() Ri() RC 6.5 5 3 x From v() (A co 6 A in 6)e -8, dv ( 8A 6A co 6 co 6) e 8A 8 in 6 A, 8-8A 6A Thu, A 5 and A. and v() (5 co 6 in 6) e -8 x 6A 8 in 6 8 95

EEE5: CI RCUI T THEORY Noe: by increaing he value of R, he degree of damping decreae and he repone differ. The repone for hoe hree cae: Figure 7.7 96

EEE5: CI RCUI T THEORY 7.5 Sep Repone of a Serie RLC Circui Reviion: he ep repone i obained by he udden applicaion of a dc ource. Conider he erie RLC circui hown in Figure 7.8. Figure 7.8 Applying KVL around he loop for >, Bu di L Ri v i C dv V Subiuing for i and rearranging erm, d v R L dv v LC V LC 97

EEE5: CI RCUI T THEORY v v v The oluion o he equaion ha wo componen: he ranien repone v () and he eady-ae repone v (); v( ) v ( ) v ( ) The ranien repone v () i he componen of he oal repone ha die ou wih ime. The form of he ranien repone i he ame a he form of he oluion obained in Secion 7.3. Therefore, he ranien repone v () for he overdamped, underdamped and criically damped cae are: ( ) ( ) ( ) A e ( Overdamped) α ( A A) e ( Criically damped) α ( A coω A inω ) e ( Underdamped ) A d e d The eady-ae repone i he final value of v(). In he circui in Figure 7.8 he final value of he capacior volage i he ame a he ource volage V. Hence, ( ) V v ( ) v 98

EEE5: CI RCUI T THEORY v( ) v( ) v( ) Thu, he complee oluion for he overdamped, and criically damped cae are: V V V A e ( Overdamped) α ( A A) e ( Criically damped) α ( A coω A inω ) e ( Underdamped ) A d e (7.) The value of he conan A and A are obained from he iniial condiion: v() and dv()/. Noe: v and i are repecively, he volage acro he capacior and he curren hrough he inducor. Therefore, he Eq. 7. only applie for finding v. Bu once he capacior volage v C v i known we can deermine i C dv/, which i he ame curren hrough he capacior, inducor and reior. Hence, he volage acro he reior i v R ir, while he inducor volage i v L L di/. Alernaively, he complee repone for any variable x() can be found direcly, becaue i ha he general from x( ) x ( ) x ( ) Where he x x ( ) i he final value and x () i he ranien repone. The final value i found a in Secion 7.. d 99

EEE5: CI RCUI T THEORY Example For he circui in Figure 7.9, find v() and i() for >. Conider hee cae: R 5 Ω. Figure 7.9 For <, he wich i cloed. The capacior behave like an open circui while he inducor ac like a hor circui. The iniial curren hrough he inducor i 4 i( ) 4A 5 And he iniial volage acro he capacior i he ame a he volage acro he -Ω reior; ha i, v( ) i() 4V For >, he wich i opened, o he -Ω reior diconneced. Wha remain i he erie RLC circui wih he volage ource. The characeriic roo are deermined a follow.

EEE5: CI RCUI T THEORY R 5 α.5 L ω LC.5, α ± α ω, 4 α > ω Since, we have he overdamped naural repone. The oal repone i herefore v( ) v ( A e 4 A e ) where v i he eady-ae repone. I i he final value of he capacior volage. In Figure 7.8 v f 4 V. Thu, v( ) 4 ( A e 4 A e ) Find A and A uing he iniial condiion or v ( ) 4 4 A A A A The curren hrough he inducor canno change abruply and i he ame curren hrough he capacior a becaue he inducor and capacior are now in erie.

EEE5: CI RCUI T THEORY Hence, dv() i ( ) C 4 dv() 4 4 6 C.5 v( ) 4 A e Ae dv A e A e 4 4, From ( 4 ) A, Thu, and dv() 6 A 4 A -64/3 and A 4/3. v( ) 4 4 3 A ( 4 6e e )V ince he inducor and capacior are in erie for >, he inducor curren i he ame a he capacior curren. Hence, i ( ) C dv

EEE5: CI RCUI T THEORY Therefore, i( ) 4 ( 4 4e e )A 3 Noe ha i() 4 A, a expeced 3

EEE5: CI RCUI T THEORY 7.6 Sep Repone of a Parallel RLC Circui Conider he parallel RLC circui hown in Figure 7.. Figure 7. Objecive: Find i due o a udden applicaion of a dc curren. Applying KCL a he op node for >, Bu v dv i C R v di L I Subiuing for v and dividing by LC, d i RC di i LC I LC 4

EEE5: CI RCUI T THEORY The complee oluion coni of he ranien repone i () and he eady-ae repone i ; i() i () i () The eady-ae repone i he final value of i. In he circui in Figure 7., he final value of he curren hrough he inducor i he ame a he ource curren I, Thu, i() I A e Overdamped A e i() I (A A )e -α Criically damped i() I (A co ω d A in ω d )e -α Underdamped The conan A and A in each cae can be deermined from he iniial condiion for i and di/. Fir, find he inducor curren i. Once he inducor curren i L i i known, we can find v L di/, which i he ame volage acro inducor, capacior and reior. Hence, he curren hrough he reior i i R v/r, while he capacior curren i i C C dv/. Alernaively, he complee repone for any variable x() may be found direcly, uing 5

EEE5: CI RCUI T THEORY x() x () x () where x and x are i final value and ranien repone, repecively. Example In he circui in Figure 7. find i() and i R () for >. Figure 7. For <, he wich i open and he circui i pariioned ino wo independen ubcircui. The 4-A curren flow hrough he inducor, o ha i() 4 A Since 3u(-) 3 when < and when >, he volage ource i operaive for < under conideraion. The capacior ac like an open circui and he volage acro i i he ame a he volage acro he -Ω reior conneced in parallel wih i. 6

EEE5: CI RCUI T THEORY By volage diviion, he iniial capacior volage i v( ) (3) 5V, For >, he wich i cloed and we have a parallel RLC circui wih a curren ource. The volage ource i off or hor-circuied. The wo -Ω reior are now in parallel. They are combined o give R Ω. The characeriic roo are deermined a follow: α ω RC LC x x 8 x α ± α ω x 8 x 3 3 6.5 ± 6.5 ± 6.5.5 39.65 5.78 6.5 or -.978, -.58 Since α > ω, we have he overdamped cae. Hence, i() I A e -.978 A e -.58 where I 4 i he final value of i(). Now ue he iniial condiion o deermine A and A. 7

EEE5: CI RCUI T THEORY A, i() 4 4 A A A -A Taking he derivaive of i() in i() I A e -.978 A e -.58 di.978..978a e A e 58. 58 o ha a, Bu di() L di().978a. 58A di() v() 5 5 L 5.75 Thu, i R ( ).75 (.978.58)A A.655, A -.655 The complee oluion a i() 4.655 (e -.58 e -.978 ) A From i(), we obain v() L di / and v( ) L di.785e.978.34e.58 A 8

EEE5: CI RCUI T THEORY 7.7 General Second-Order Circui Given a econd-order circui, we deermine i ep repone x() (which may be volage or curren) by aking he following four ep:. Fir, deermine he iniial condiion x() and dx()/ and he final value x( ) a dicued in Secion 7... Find he ranien repone x () by applying KCL and KVL. Once a econd-order differenial equaion i obained, deermine i characeriic roo. Depending on wheher he repone i overdamped, criically damped, or underdamped, we obain x () wih wo unknown conan a we did in he previou ecion. 3. Obain he forced repone a x f ( ) x( ) where x( ) i he final value of x, obained in Sep. 4. The oal repone i now found a he um of he ranien repone and eady-ae repone x( ) x ( ) x ( ) 9

EEE5: CI RCUI T THEORY Finally deermine he conan aociaed wih he ranien repone by impoing he iniial condiion x() and dx()/, deermined in ep. Example: Find he complee repone v and hen i for > in he circui of Figure 7.. Figure 7. Fir find he iniial and final value. A >, he circui i a eady ae. The wich i open, he equivalen circui i hown in Figure 7.3. Figure 7.3

EEE5: CI RCUI T THEORY From he figure, v ( ) V i( ) A >, he wich i cloed, he equivalen circui i in Figure 7.4. Figure 7.4 By he coninuiy of capacior volage and inducor curren, v ( ) v( ) V i( ) i( ) To ge dv > ( ) /, ue C dv/ i c or dv/ i C /C. Applying KCL a node a in Figure 7.4, v( ) i( ) ic ( ) ic ( ) ic ( ) 6A

EEE5: CI RCUI T THEORY Hence dv( ) 6.5 V / The final value are obained when he inducor i replaced by a hor circui and he capacior by an open circui in Figure 7.4, giving i 4 4 ( ) A v( ) i( ) V Nex, obain he naural repone for >. By urning off he -V volage ource, we have he circui in Figure 7.5. Figure 7.5 Applying KCL a node a in Figure 7.5 give i v dv

EEE5: CI RCUI T THEORY Applying KVL o he lef meh reul in Thu, or di 4 i v dv dv d v v v d v dv 5 6v From hi, we obain he characeriic equaion a 5 6 Wih roo - and -3. Thu, he naural repone i v n ( ) Ae Be 3 where A and B are unknown conan o be deermined laer. The forced repone i v f ( ) 4 ( ) v The complee repone i v( ) v n v f 4 Ae Be 3 We now deermine A and B uing he iniial value. We know ha v(), hu a : 3

EEE5: CI RCUI T THEORY 4 8 4 B A B A Taking he derivaive of v in f n Be Ae v v v 3 4 ) ( Be Ae dv 3 3 From V dv /.5 6 ) (, a : 3 3 B A B A Thu, 4, B A o ha,, 4 4 ) ( 3 > V e e v From v, we can obain oher quaniie of inere (refer o Figure 7.4): e e e e dv v i 3 3 6 6, 4 6 3 > A e e