To become more mathematically correct, Circuit equations are Algebraic Differential equations. from KVL, KCL from the constitutive relationship

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1 Laplace Tranform (Lin & DeCarlo: Ch 3) ENSC30 Elecric Circui II The Laplace ranform i an inegral ranformaion. I ranform: f ( ) F( ) ime variable complex variable From Euler > Lagrange > Laplace. Hence, differenial equaion can be ranformed o algebraic equaion. Digreion: he circui we analyze are decribed by ODE. To become more mahemaically correc, Circui equaion are Algebraic Differenial equaion ADE from KVL, KCL from he coniuive relaionhip Laplace Tranform: ODE linear algebra equaion linear or, more formally: algebraic differenial equaion ime domain algebraic equaion frequency domain Once he yem of algebraic equaion i olved (here are many known mehod and algorihm o do o), we have o ue Invere Laplace Tranform and ranform he oluion X ( ) x( ) capial cae mall cae, ialic x Laplace ranform grealy implifie he proce and enable u o find oluion o linear circui more eaily han olving differenial equaion in he ime domain. Solving equaion in he domain provide addiional inigh ino he circui behaviour. - frequency repone of he circui: ofen ued in he engineering approach o deign circui Example: Filer - -

2 ENSC30 Elecric Circui II H() ranfer funcion frequency range hi i an ideal high pa filer A more realiic cae: lope depend on he filer deign - Chebylev - ellipic We ofen deire cerain yem performance on he frequency domain. Hence, dealing wih circui repone in he domain provide imporan inigh ino he yem behaviour. Wha i wrong wih wriing and olving DE (ODE, ADE)? Procedure: Sep : wrie DE - Two equaion of he order - One equaion of he nd order Sep : - -

3 ENSC30 Elecric Circui II wrie he characeriic equaion and find characeriic frequencie (naural frequencie) of he circui Sep 3: Predic he oluion baed on he forcing funcion and find he conan Sep 4: Deermine conan baed on he circui iniial condiion and circui equaion Thi work very well for imple circui. Recall our example from la week: order: RL, RC: find conan k in x( ) ke + x ( ) p find k wa eay depend on he driving funcion nd order: RLC: find k and k x( ) k e + k e + x p ( ) we had o ue: x (0-): V c x (0-): i L and alo: dx d dx d are no known They had o be found from he DE and x (0-) and x (0-). Hence, even in hi imple cae, finding conan required ome work and hough. For more complicaed circui, he procedure fail becaue i ge raher complicaed o find he conan! - 3 -

4 See: example 3. in Lin & DeCarlo: pp496-7 Laplace Tranform inpu ignal circui or a yem: ENSC30 Elecric Circui II conrol yem mechanical yem MEMS oupu ignal volage curren volage curren Laplace Tranform of he inpu ignal Laplace Tranformed circui Laplace Tranform of he oupu ignal Equivalenly: Inpu ignal DE oupu ignal LT of inpu ignal LT of DE LT of oupu ignal Tranform: a funcion ha decribe he behaviour of he yem (we will learn more abou i in Ch. 6 of Lin & DeCarlo) x() h() y() Time domain: y() x()*h(), where h(): impule repone of he circui convoluion X() Laplace Tranformaion Y() H() X() H() Y() imple muliplicaion - 4 -

5 X() L[x()] Y() L[y()] H() L[h()], where L refer o he Laplace Tranform ENSC30 Elecric Circui II H(): ranfer funcion (noe: i can be found direcly from he circui afer he circui i ranformed by he Laplace ranform). Hence, L H() X() L - x() X() Y() y() L: Laplace Tranform x ( ) e d L - : Invere Laplace Tranform X ( ) e d ωj definiion Noe: convoluion in ime domain i equivalen o muliplicaion in Laplace domain, which i good becaue convoluion i difficul o apply.) Baic ignal:, 0 u ( ) uni ep funcion 0, < 0 L[u()] U() By definiion: L[f()] f ( ) e d one-ided Laplace ranform, σ + jω & j - 5 -

6 ENSC30 Elecric Circui II U ( ) e U ( ) e d 0 U ( ) U ( ) : Laplace ranform of u(): uni ep funcion Dicuion:. One-ided: 0 f ( ) e d no -: if we ue -, hen we alk abou wo-ided L.T.. Why 0-? To accoun for iniial condiion in a circui. Noe ha we know: v c (0-) i L (0-) 3. Wha if f() 0 for < 0? In circui analyi: uually f() 0 for < Wha ype of funcion have Laplace Tranform? No every funcion! Exponenially bounded funcion have Laplace Tranform (hough no a neceary condiion). The inegral f ( ) e d hould exi. Conider: f ( ) e u( ) F ( ) e e d ( F( ) e ) d F σ jω ( ) e e d σ + jω Euler conan jinω A σ, e area under e σ - 6 -

7 ENSC30 Elecric Circui II Noe ha hi inegral doe no exi for any σ. 5. Why i σ imporan? Conider he uni ep funcion: Bu only if σ > 0 L[u() u e ( ) e d ( σ + jω ) d ( σ + jω ) e σ + jω 0 σ + jω σ + jω e ( σ + jω ) e σ (coω j inω) a e σ 0 if σ > 0 RoC: Region of Convergence for u() σ > 0 6. I L.T. valid only wihin he RoC? No. There i analyical calculaion mehod ha permi exenion o he enire S plane. Thi i he reaon ha we uually do no menion Region of Convergence when dealing wih one-ided Laplace Tranform

8 ENSC30 Elecric Circui II Properie: L.T. i linear (nice!) Recall: L[u()] L[a f () + a f ()] a L[f ()] + a L[f ()] caling wih a conan uperpoiion (boh have lineariy) Oher ignal o remember: - 8 -