TRANSFORM AND ITS APPLICATION

LAPLACE TRANSFORM AND ITS APPLICATION IN CIRCUIT ANALYSIS C.T. Pan. Definition of the Laplace Tranform. Ueful Laplace Tranform Pair.3 Circuit Analyi in S Domain.4 The Tranfer Function and the Convolution Integral C.T. Pan

.5 The Tranfer Function and the Steady tate Sinuoidal Repone.6 The Impule Function in Circuit Analyi C.T. Pan 3. Definition of the Laplace Tranform Pierre Simon Laplace (749-87) : A French atronomer and mathematician Firt preented the Laplace tranform and it application to differential equation in 979. C.T. Pan 4

. Definition of the Laplace Tranform Definition: [ ] = = 0 t L f() t F () f() te dt = σ jω a complex variable The Laplace tranform i an integral tranformation of a function f(t) from the time domain into the complex frequency domain, F(). C.T. Pan 5. Definition of the Laplace Tranform One-ided (unilateral) Laplace tranform Two-ided (bilateral) Laplace tranform [ ] = = π j j σ ω σ jω L F () f() t Fe () d t Look-up table,an eaier way for circuit application f() t F () C.T. Pan 6

. Definition of the Laplace Tranform Similar to the application of phaor tranform to olve the teady tate AC circuit, Laplace tranform can be ued to tranform the time domain circuit into S domain circuit to implify the olution of integral differential equation to the manipulation of a et of algebraic equation. C.T. Pan 7. Ueful Laplace Tranform Pair Function f(t), t> F() impule tep ramp exponential δ () t ut () t 0 S S at e S a ine inωt S ω ω C.T. Pan 8

. Ueful Laplace Tranform Pair Function f(t), t> F() 0 coine damped ramp damped ine e S S ω ( S a) coωt at te at inωt ω ( ) S a ω damped coine e at coωt S a ( ) S a ω C.T. Pan 9. Ueful Laplace Tranform Pair d L f ( t) = F ( ) f (0 ) dt t F ( ) L f ( ) d = - τ τ 0 S L f ( t a ) u ( t a ) = a e F ( ), a > 0 [ ] at L e f ( t) = F ( a ) L [ f ( at) ] = F ( ), a > 0 a a f ( t) = F ( ) lim t 0 lim [ ] [ ] lim lim f ( t) = F ( ) t 0 C.T. Pan 0

. Ueful Laplace Tranform Pair Example Ue the Laplace tranform to olve the differential equation. dv dv 6 8v= ut () dt dt v(0) = v '(0) = Take Laplace tranfrom [ ] V() v(0) v'(0) 6 V() v(0) 8 V() = C.T. Pan. Ueful Laplace Tranform Pair [ ] V() v(0) v'(0) 6 V() v(0) 8 V() = ( 6 8) V() = 4 4 4 V() = = ( 6 8) ( )( 4) t 4t vt () = ( e e ) ut () 4 C.T. Pan

.3 Circuit Analyi in S Domain () KCL, i () t = 0, for any node. n k Take Laplace Tranform n I k () = 0, for any node. () KVL, v () t = 0, for any loop. m k Take Laplace Tranform Vk () = 0, for any loop. m C.T. Pan 3.3 Circuit Analyi in S Domain (3) Circuit Component Model reitor v () t = Ri () t R R V () = RI () R R I () = GV () R R C.T. Pan 4

.3 Circuit Analyi in S Domain inductor dil vl() t = L dt t il() t = il(0 ) v ( ) 0 L d L τ τ Li L (0 ) V () = LI () Li (0 ) L L L VL() il(0 ) I () = L i L (0) C.T. Pan 5.3 Circuit Analyi in S Domain capacitor dvc ic = C dt t vc() t = vc(0 ) i ( ) 0 C d C τ τ C v C (0 ) I () = CV () Cv (0 ) C C c vc(0 ) VC() = IC() C C Cv C (0 ) C.T. Pan 6

.3 Circuit Analyi in S Domain Coupling inductor i M i v L L v - - d i = L M dt d i = M L dt d i dt d i dt V ()=L SI ()-L i (0)MSI ()-Mi (0) V ()= MSI ()- Mi (0)L SI ()-L i (0) v v C.T. Pan 7.3 Circuit Analyi in S Domain For zero initial condition impedance @ V() = Z () IS ( ) admittance @ IS ( ) = Y() = V() Z () V() = ZI () () ohm' lawindomain C.T. Pan 8

.3 Circuit Analyi in S Domain The elegance of uing the Laplace tranform in circuit analyi lie in the automatic incluion of the initial condition in the tranformation proce, thu providing a complete (tranient and teady tate) olution. C.T. Pan 9.3 Circuit Analyi in S Domain Circuit analyi in domain Step : Tranform the time domain circuit into -domain circuit. Step : Solve the -domain circuit. e.g. Nodal analyi or meh analyi. Step 3 : Tranform the olution back into time domain. C.T. Pan 0

.3 Circuit Analyi in S Domain Example Find v o (t) given v o (0)=5V S-domain equivalent circuit 0 0 C.T. Pan.3 Circuit Analyi in S Domain 0 0 Nodal analyi 0 Vo () Vo() Vo() 0.5 = 0 0 0/ 5 35 0 5 Vo() = = ( )( ) t t v () t = (0e 5 e ) ut () V o C.T. Pan

.4 The Tranfer Function and the Convolution Integral Given a linear circuit N in domain a hown below Tranfer function H() i defined a H () = Y() X() C.T. Pan 3.4 The Tranfer Function and the Convolution Integral If Y() = V (), X() = V() ; then H() = voltage gain o i If Y() = I (), X() = I () ; then H() = current gain o i If Y() = V(), X() = I () ; then H() = impedance If Y() = I (), X() = V() ; then H() = admittance C.T. Pan 4

.4 The Tranfer Function and the Convolution Integral Given the tranfer funtion H() and input X(), then Y()=H()X() If the input i δ(t), then X()= and Y()=H() Hence, the phyical meaning of H() i in fact the Laplace tranform of the impule repone of the correponding circuit. C.T. Pan 5.4 The Tranfer Function and the Convolution Integral Y()=H()X(), in -domain yt () = ht ( τ) x( τ) dτ @ ht () xt () in time domain Geometrical interpretation of finding the convolution integral value at t=t k i baed on : C.T. Pan 6

.4 The Tranfer Function and the Convolution Integral ()Approximating the input function by uing a erie of impule function. ()Shifting property of linear ytem input x(t) outputy(t) x(t-τ) output y(t- τ) (3)Superpoition theorem for linear ytem (4)Definition of integral : finding the area C.T. Pan 7.4 The Tranfer Function and the Convolution Integral ()Input x(τ) i approximated uing impule function, x(τ)=0, for τ<0 x( τ ) x(0) 0 x() τ 0 τ τ τ τ 3 4 τ5 τ τ intant τ, x( τ ) value, area 0 τ = Vτ, x( τ ), x( τ ) Vτ τ = Vτ, x( τ ), x( τ ) Vτ τ = 3 Vτ, x( τ ), x( τ ) Vτ 3 3 3 C.T. Pan 8 k k τ = 0, x(0), x(0) Vτ x( τ) f δτ ( ) fδτ ( τ ) f δτ ( τ ) L 0 @ xk ( Vτ) Vτδ( τ kvτ) k= 0

.4 The Tranfer Function and the Convolution Integral () Ue the linearity property input output( repone) x(0) Vτδ( τ) x(0) Vτh( τ) x( τ ) Vτδ( τ τ ) x( τ ) Vτh( τ τ ) x( τ ) Vτδ( τ τ ) x( τ ) Vτh( τ τ ) M M up to τ =t k C.T. Pan 9.4 The Tranfer Function and the Convolution Integral (3) Ue uperpoition theorem to find the total approximate repone $ yt ( k) = xk ( Vτ) Vτ ht ( k kvτ) k= 0 tk n= integer[ ] Vτ n C.T. Pan 30

.4 The Tranfer Function and the Convolution Integral (4) Take the limit, Δτ dτ, $ yt ( ) yt ( ) k k yt ( k) = x( τ) ht ( k τ) dτ Due to cauality principle, h(t-τ)=0 for t t k and x(τ)=0 for τ<0 t k = ht ( k τ) x ( τ) d τ 0 C.T. Pan 3.4 The Tranfer Function and the Convolution Integral Example: t Given xt () = ut (), ht () = e ut (), find y(4), yt () = ht ( τ) x( τ) dτ y(4) = h(4 τ) x( τ) dτ C.T. Pan 3

.4 The Tranfer Function and the Convolution Integral Step. xx ( τ ).0 0.8 0.6 0.4 0. 4 4 0 4 h ( τ ).0 τ 4 4 0 4 C.T. Pan 33 0.8 0.6 0.4 0. τ.4 The Tranfer Function and the Convolution Integral Step. Shift to t k =4 h (4 4τ ).0 0.8 0.6 0.4 0. 4 4 0 4 C.T. Pan 34 τ

.4 The Tranfer Function and the Convolution Integral Step3. Find the product h(4-τ)x(τ) C.T. Pan 35.4 The Tranfer Function and the Convolution Integral Step4. Find the integral (area) 4 4 (4 τ) 4 τ 4 τ 4 y(4) = e dτ = e edτ= e e Step5. Check 0 0 e e = e 4 4 4 = ( ) ( ) Y () = HX () () = = a a a a t yt () = ( e ), a= a 4 y(4) = ( e ) C.T. Pan 36 0

.5 The Tranfer Function and the Steady State Sinuoidal Repone From definition of tranfer function H() = Y() X() Y() = H() X() Aume input X(t)=Aco(wtФ) and H() i given, then one can get the teady tate olution without needing a eparate phaor analyi. C.T. Pan 37.5 The Tranfer Function and the Steady State Sinuoidal Repone proof : X() t = AcoφcoωtAinφin ωt A(coφin φω) X() = ω Y() = H() X() * K K = jω jω other term dueto pole under teady tate : * K K Y () = jω jω HA () ( coφωin φ) K = jω =ω j = H( jω) Ae C.T. Pan 38 jφ

.5 The Tranfer Function and the Steady State Sinuoidal Repone Let H( jω ) = H( jω) e jθ( ω) and takeinvere Laplacetranform Then y () t = A H( jω) co[ ω t φ θ( ω)] ie.. PXt ( ()) = A φ then Py ( ()) t = A H( jω) φ θ( ω) C.T. Pan 39.5 The Tranfer Function and the Steady State Sinuoidal Repone Example : The tranfer function H() of the circuit given below i known. Find the teady tate olution of V o (t) for the given V g (t). C.T. Pan 40

.5 The Tranfer Function and the Steady State Sinuoidal Repone 000( 5000) H() = 6 6000 5 0 Vg( t) = 0co(5000t 30 ) V C.T. Pan 4.5 The Tranfer Function and the Steady State Sinuoidal Repone Solution: Let = jω= j5000 000( j5000 5000) Evaluate H(5000) j = 5 0 j5000(6000) 5 0 = 45 6 6 6 Then Vo () t = 0 co(5000t 30 45 ) 6 = 0 co(5000t5 ) V C.T. Pan 4

.5 The Tranfer Function and the Steady State Sinuoidal Repone In theory, the relationhip between H() and H(jw) provide a link between the time domain and the frequency domain. In ome cae, we can determine H(jw) experimentally and then contruct H() from the data. C.T. Pan 43.5 The Tranfer Function and the Steady State Sinuoidal Repone Example: Find the impule repone of the following circuit. v in _ R C v o _ C.T. Pan 44

.5 The Tranfer Function and the Steady State Sinuoidal Repone (a) Time domain olution dvo RC vo = δ () t dt At t = 0, v (0 ) = 0 o t δ () t At t = 0, vo (0 ) = dt C 0 R = V RC δ () t v () t o C.T. Pan 45.5 The Tranfer Function and the Steady State Sinuoidal Repone For t > 0, δ() t = 0 R C vo () t, v(0 ) = _ RC dv RC dt o v = 0 t RC vo () t = e ut () RC o C.T. Pan 46

.5 The Tranfer Function and the Steady State Sinuoidal Repone (b) -domain olution Find the tranfer function Vo() H() = V () in zero IC.. Tranform into -domain circuit V () in R C V () o _ C.T. Pan 47.5 The Tranfer Function and the Steady State Sinuoidal Repone Vo() H() = = C = V () in R RC C - ht () = L [ H()] V () in R C V () o _ = e RC t RC ut () Same anwer. C.T. Pan 48

.6 The Impule Function in Circuit Analyi Example : Impule voltage ource excitation it () V δ () t O i(0 ) = 0 C.T. Pan 49.6 The Impule Function in Circuit Analyi (a) Time domain olution At t = 0, i(0 ) = 0 t i(0 ) = V ( ) 0 Oδ x dx L VO = ( A) L V Oδ () t i(0 ) The impule voltage ource ha tored energy, L( ) i(0 ), in the inductor a an initial current in an infiniteimal moment. C.T. Pan 50

.6 The Impule Function in Circuit Analyi For t > 0, δ() t = 0 di L Ri = 0, natural repone dt VO i(0 ) = A L V L L R O t / τ it () = e ut (), τ = it () Note that the impule ource jut build up an initial inductor current but doe not contribute to any forced repone. C.T. Pan 5.6 The Impule Function in Circuit Analyi (b) -domain olution VO VO / L I () = = R L R L VO t/ τ it () = e ut () L V O Same anwer but much eaier. R I () L C.T. Pan 5

.6 The Impule Function in Circuit Analyi Example : Impule current ource excitation I δ () t o R C vt (), v(0 ) = 0 _ C.T. Pan 53.6 The Impule Function in Circuit Analyi (a) Time domain olution At t = 0, v(0 ) = 0 v(0) = 0, hort circuit I δ () t t v(0 ) = I ( ) 0 o x dx C δ Io = C The impule current ource ha tored energy, o C( ) v(0 ) R, in the capacitor a an initial voltage in an infiniteimal moment. C.T. Pan 54

.6 The Impule Function in Circuit Analyi For t > 0, δ () t = 0, open circuit dv v C = 0, natural repone dt R Io v(0 ) = C t I o τ vt () = e ut (), τ = RC C vt () Note that the impule current jut build up an initial capacitor voltage but doe not contribute to any forced repone. C.T. Pan 55.6 The Impule Function in Circuit Analyi (b) -domain olution Tranform into -domain circuit I o R C C.T. Pan 56 V _ () R Io/ C V( ) = I C o = R C RC t I o τ vt () = e ut () C Same anwer but much eaier.

.6 The Impule Function in Circuit Analyi Example 3: Impule caued by witching operation The witch i cloed at t=0 in the following circuit. Notethat v(0 ) v (0 ) C.T. Pan 57.6 The Impule Function in Circuit Analyi Tranform into -domain Vo I () = @ Vo Ce C C C C V C V C Ce= V = = C C C C C v t o e o, () it () = VC o eδ() t C () = V C C C.T. Pan 58 o

.6 The Impule Function in Circuit Analyi At t=0, a finite charge of C i tranferred to C intantaneouly. Note that, a the witch i cloed, the voltage acro C doe not jump to V o of C but to it final value of the two paralleled capacitor. C.T. Pan 59.6 The Impule Function in Circuit Analyi C C = =, > 0 C C C = =, > 0 C C =, > 0 Note Q C V Vo t Q C V Vo t Q Q C Vo t Alo att = Q = CV Q =, 0, o, 0 Q Q CV o = Conervationof charge. C.T. Pan 60

.6 The Impule Function in Circuit Analyi If we conider charged capacitor a voltage ource, then we hould not connect two capacitor with unequal voltage in parallel. Due to violation of KVL, an impule will occur which may damage the component. C.T. Pan 6.6 The Impule Function in Circuit Analyi Example 4: Impule caued by witching operation The witch i opened at t = 0 in the following circuit. 00V 0Ω 3H i L t = 0 i L 5Ω H vo _ At t = 0, teady tate olution 00V i (0 ) = = 0 A 0Ω i (0 ) = 0 A C.T. Pan 6

.6 The Impule Function in Circuit Analyi For t > 0, the S-domain S circuit i 00 0Ω 3 I () L 30 L 5Ω V () o (00/ ) 30 4 I () = = 5 5 5 60 0 Vo () = I () ( 5) = 5 it = e ut A 5t () (4 ) ()...( ) vt () = δ () t (60 0 e ) ut ()...( B) 5t C.T. Pan 63.6 The Impule Function in Circuit Analyi 0Ω 3 30 00 I () L L 5Ω V () o Note t = 0, i (0 ) = 0 A, i (0 ) = 0 A L L t = 0, from ( A), i (0 ) = 6 A, i (0 ) = 6 A L L Alo, from ( B), there exit δ() t at v (). t o C.T. Pan 64

.6 The Impule Function in Circuit Analyi Thu, if we conider an inductor current a a current ource, then two inductor with unequal current hould not be connected in erie. Due to violation of KCL, it will reult in impule voltage which may damage the component. 00 C.T. Pan 65 0Ω 3 I() L 30 L 5Ω V () o SUMMARY Objective : Know the component model in -domain. Objective : Be able to tranform a time domain circuit into the -domain circuit. Objective 3 : Know how to analyze the -domain circuit and tranform the olution back to time domain. C.T. Pan 66

SUMMARY Objective 4 : Undertand the ignificance of tranfer function and be able to calculate the tranfer function from the -domain circuit. Objective 5 : Know the geometrical interpretation of convolution integral and be able to calculate the integral. C.T. Pan 67 SUMMARY Objective 6 : Know the relation between the phaor olution technique for finding inuoidal teady tate olution and the -domain olution technique. Objective 7 : Know how to ue -domain olution technique to olve a circuit containing impule ource or a witching circuit which may reult in impule function. C.T. Pan 68

SUMMARY Chapter problem : 3.3 3.0 3.7 3.36 3.57 3.85 3.88 C.T. Pan 69