4. Itroductio Laplace trasform is a alterative to solve the differetial equatio by the complex frequecy domai ( s = σ + jω), istead of the usual time domai. The DE ca be easily trasformed ito a algebraic fuctio, i terms of the complex variables, s. The algebraic equatio ca the easily maipulated to produce solutio as fuctio s where the solutio ca be obtaied by referrig Laplace trasform pairs i the table. I some cases, partial fractio expasio techique may have to be employed to reduce the solutio i terms of complex variables to oes that coform to the preseted table. 4. Itroductio Advatages of LT : ability to give close approximatio to the system performace via graphical techique without solvig the system DE. : trasiet ad steady state solutio which cotribute to the total solutio of system DE ca be determied simultaeously.
4. Itroductio Laplace Trasform Table 4. Itroductio LT Table Time domai (t) s - domai (s) Result System performace Trasiet ad steady state solutio
4. Itroductio hard problem t y" y te y() y'() trasform L simple equatio ( s ) Y ( s ) Solve? solutio y y( t) cos t te t e t iverse trasform L algebraic operatio solutio Y Y ( s )( s ) 4. Uilateral Laplace Trasform (ULT) The LT of a fuctio of time x(t) is defied as : L{ x ( t )} X ( s ) t x ( t ) e st Bilateral LT whe t = - Uilateral LT whe t = The ULT is the LT of causal sigal ad sigals of all systems i practice are causal, i.e. the sigals starts at some fiite istat (origi). There are variatio i the defiitio of lower limits which could be +, ad - BUT the usage of ay lower limits make o differece to the LT as log the fuctio x(t) does ot ivolve a sigularity fuctio at t =. 3
Example Fid the LT for equatios x(t) below, for t, where A is a costat. a) x(t) = A cos ωt b) x(t) = A c) x(t) = Ae -at d) x(t) = At e) x(t) = A si ωt 4. Itroductio Laplace trasform is a alterative to solve the differetial equatio by the complex frequecy domai ( s = σ + jω), istead of the usual time domai. The DE ca be easily trasformed ito a algebraic fuctio, i terms of the complex variables, s. The algebraic equatio ca the easily maipulated to produce solutio as fuctio s where the solutio ca be obtaied by referrig Laplace trasform pairs i the table. I some cases, partial fractio expasio techique may have to be employed to reduce the solutio i terms of complex variables to oes that coform to the preseted table. 4
4. Properties Of Laplace Trasform a) Homogeeity b) Superpositio c) Liearity d) Shiftig I The Time Domai e) Shiftig I the S Domai f) Time Scalig g) Multiplicatio by t h) Differetiatio i The Time Domai i) Itegratio I time Domai j) Iitial Value Theorem k) Fial Value Theorem 4. Properties Of Laplace Trasform a) Homogeeity L { k x(t) } = k To prove : k x(t)e k x(t)e k st st 5
4. Properties Of Laplace Trasform b) Superpositio L { x (t) + x (t) } = X (s) + X (s) To prove : [ x (t) x (t) X ( s ) X ( s ) ] e st [ x (t) e ] st [ x (t) e st ] 4. Properties Of Laplace Trasform c) Liearity L { k x (t) + k x (t) } = k X (s) + k X (s) To prove : k [ k x (t) k x k X ( s ) k X (t) ( s ) ] e st st [ x (t) e ] k [ x (t) e st ] Example : Fid the LT of the sigal x(t) = (t + e -3t ) u(t) 6
4. Properties Of Laplace Trasform d) Shiftig I the t - domai L { x ( t ± t ) u ( t ± t ) } = e ± sto To prove : st [ x(t t ) u(t t ) ] e Let Τ t t, x(t) u(t) e it follows t s(t-t ) dt T - t ad dt Lim T = t t For t = ; T = + t = For t = ; T = + t = t e e x(t) e st st x(t) e s(t -t ) st dt Example : Fid the LT of the sigal x(t) = ½ t u( t- ) 4. Properties Of Laplace Trasform e) Shiftig I the s - domai L { e ± at x(t) } = X(s a) To prove : a t e x(t) X(s x(t) - a) Example : Fid the LT of the sigals : e e st ( s a ) t a) x(t) = e t r(t) u(t) b) x(t) = e -t cos 4(t) u(t) 7
4. Properties Of Laplace Trasform f) Time Scalig L { x(at) } = To prove : x(at) e st Let T at it follows, t T, a dt a a x a x(t) e s a s t a dt Example : Fid the LT of the sigals : a) x(t) = e -5t u(t) b) x(t) = u(t/4) c) x(t) = tu(t/3) 4. Properties Of Laplace Trasform g) Multiplicatio by t L { t x(t) } = () d ds To prove, let = : Let = d ds x(t)e st d [ x(t) e ds t x(t) e st st ] x(t)e The same coditio applied whe differetiatig with respect to s for times. t st x(t) e d d st [ x(t) e ] ds ds st 8
4. Properties Of Laplace Trasform g) Multiplicatio by t (Cotiue) Therefore : L { t x(t) } = () d ds Example : Fid the LT of the sigals x(t) = te -t u(t) 4. Properties Of Laplace Trasform h) Differetiatio I the Time Domai d x(t) L { } s s X() s dx() - s 3 dx () -... - s - dx () - To prove : Let = dx(t) dx(t) L{ } e Let u e -st du - se -st st dx(t) dv v x(t) udv uv dx(t) L{ } x(t)e vdu x() s -st x(t) se st 9
4. Properties Of Laplace Trasform h) Differetiatio I the Time Domai (Cotiue) Let = d x(t) d x(t) L{ } e From the, the eq.ca be simplified as dg(t) L{ } s(l{g(t)}) g() dx(t) s(l{ }) g() s ( s - X() ) - X () s - sx() - X () st dx(t) g(t) 4. Properties Of Laplace Trasform h) Differetiatio I the Time Domai (Cotiue) Let = 3 3 3 d x(t) d x(t) L{ } e 3 3 st d x(t) From the, the eq.ca be simplified as h(t) dh(t) L{ } s(l{h(t)}) h() d x(t) s(l{ }) X () s ( s - sx() - X () ) X () 3 s - s X() - sx () X ()
4. Properties Of Laplace Trasform h) Differetiatio I the Time Domai (Cotiue) Example 4 d x(t) Fid the Laplace trasform for = L{ } 4 4. Properties Of Laplace Trasform i) Itegratio I The Time Domai L { x(t) } = To prove : [ Let x(t) ] e u du x(t) st x(t) dv e -st -st e v - s udv uv vdu -st e x(t) s g() s s s x(t)e st Example : Fid the LT of the sigals
4. Properties Of Laplace Trasform j) Iitial Value Theorem x( + ) = x( ) lim s, provided limits exist s To prove : dx(t) L{ } s x( ) Limitig s o both sides, LHS lim s dx(t) e RHS lim[s x( )] s st Equatigboth sides x( ) lims If x(t)is cotiousat t, the x( ) x( ) ad thus: x( ) lims s s 4. Properties Of Laplace Trasform k) Fial Value Theorem lim x(t) lim s t s Limitig s o both sides, dx(t) st LHS lim e s, provided limits exist To prove, cosider the LT st order DE dx(t) L{ } s x( ) dx(t) limx(t) x( ) t x(t) RHS lim[s x( )] s Equatigboth sides lim x( t) lim s s s
4. Properties Of Laplace Trasform Example Determie the fial value ad iitial value of the system with the followig LT, usig the LT theorem. The prove that the value are correct, usig the iverse Laplace Trasform s G(s) (s 4s)(s) 4.3 Iverse Laplace Trasform The Iverse Laplace trasform is give by : - L {} x(t) However, the equatio is ot commoly used to derive the ILT sice the itegratio is i the complex plae. j σ jω I most cases, the Laplace trasform exists i the form of : G(s) N(s) D(s) Value of s that set N(s) to zero, e.g. z, z,.. z m kow as zeros Value of s that set D(s) to zero, e.g. p, p,.. p kow as poles. σ jω e st ds K(s z)(s z )... (s z m ) (s p )(s p )... (s p ) 3
4.3 Iverse Laplace Trasform If the degree of polyomial of N(s) is strictly less tha the degree of polyomial of D(s) [ m < ], the ILT ca be obtaied by applyig partial fractio expasio techique. Otherwise log divisio must be applied before the ILT executio. Summary m < m > = Iverse Laplace by partial fractio = Log divisio ad iverse Laplace by partial fractio 4.4 Partial fractio expasio method Expasio i partial fractio is a techique to reduce proper ratioal fuctios as G(s) N(s) D(s) fractio decompositio of G(s) I other words : a ks k or p q (s b) (s as b) irreducible. K(s z )(s z )... (s z m ) (s p )(s p )... (s p ) B(s) G(s) G (s) G(s)... G i(s) A(s) whereg (s) has oe of the followig forms i ito a sum of simple terms (partial where p, q commoly some o - egativeiteger while polyomial(s as b) is 4
4.4 Partial fractio expasio method There are three differet cases, depedig o the type of factor for A(s) a) G(s) cotais oly differet poles b) G(s) cotais of repeated poles c) G(s) cotais of complex poles 4.4 Partial fractio expasio method a) G(s) cotais oly differet poles If polyomial G(s) cotais oly liear o-repeated factors, it ca be writte as follows : B(s) a a a G(s)... A(s) s p s p s p where a are the residues of poles. Values of a i ca be obtaied by multiplyig both sides of the equatio above by (s+p i ) G(s) B(s) (s A(s) a a a pi ) (s pi) (s pi)... s p s p s p (s p ) 5
4.4 Partial fractio expasio method a) G(s) cotais oly differet poles (Cotiue) Thus : B(s) a i (s pi ) A(s) sp i By referrig Laplace trasform table : L a a e - i pit i (s pi ) 4.4 Partial fractio expasio method a) G(s) cotais oly differet poles (Cotiue) Example : Fid f(t) for F(s) give below : i) F(s) s (s 3)(s 5) ii) 4 3 s 3s 5s s 4 F(s) s 3s 6
4.4 Partial fractio expasio method b b) G(s) cotais of repeated poles If polyomial G(s) cotais oly liear o-repeated factors, it ca be writte as follows : B(s) A(s) B(s) b G(s) A(s) (s r r p ) b (s r- r p ) b a a3 a...... sp sp sp sp The multiple poles of p of multiplicity r ca be obtaied i the followig way : ad so forth r r (sp) sp i b d B(s) ds A(s) r r - (sp) sp i The other roots are assumed distict ad should be treated as Case b d k! ds k r r (s p k ) 3 B(s) A(s) sp i 4.4 Partial fractio expasio method b) G(s) cotais of repeated poles (Cotiue) Example : Fid f(t) for F(s) give below : i) F(s) s 5s -4 3 (s 3) ii) F(s) s s 3 3 (s ) 7
4.4 Partial fractio expasio method c) G(s) cotais of complex poles If polyomial G(s) cotais oly liear o-repeated factors, it ca be writte as follows : B(s) ks k a a a G(s)... A(s) s asb sp sp sp The simplest way to determie the coefficiets is to equate the coefficiets of differet power of s or by substitutig values of s that make the factors zero oe at a time 4.4 Partial fractio expasio method c) G(s) cotais of complex poles (Cotiue) Example : Fid f(t) for F(s) give below : i) 3s F(s) s s 5 ii) (s ) F(s) s(s )(s s ) 8
4.5 Applicatio of Laplace Trasform There are may applicatios of LT i solvig problems i system aalysis ad desig as : a) Solutio of DE b) Determiatio of fial value ad iitial value c) Determiatio of trasfer fuctio d) Stability i the s-domai 4.5 Applicatio of Laplace Trasform a) Solutio of DE DE modelig the cotiuous time liear time ivariat systems, are solved usig the differetiatio property of Laplace Trasform. The procedure is outlied as follows : a) Apply LT to both sides of the DE ad substitute the appropriate iitial coditios to obtai algebraic equatio i terms of Y(s) b) Solve the algebraic equatio for Y(s) or Y(s) as a sum of Y zi (s) ad Y zs (s) c) Apply ILT to obtai y(t) 9
4.5 Applicatio of Laplace Trasform Example Cosider the followig d order LT differetial equatio. '' ' y (t) 7y (t) y(t) x(t) x(t) is the iput ad y(t) is the iput. Give Iitial coditios y() =, y () = ad x(t) = u(t) Usig the LT method, determie : i) The zero iput respose, y zi (t) ii) The zero state respose, y zs (t) iii) The total respose, y(t) 4.5 Applicatio of Laplace Trasform b) Determiatio of Fial Value ad Iitial Value Determie the iitial ad fial value for a system with the followig LT, usig the LT Theorems. The prove that the values are correct,usig the iverse LT s s 3s 3s
4.5 Applicatio of Laplace Trasform c) Determiatio of Trasfer Fuctio Trasfer fuctio is defied as the LT ratio of the output to the iput with all iitial values set zeros Y(s) T.F (all iitial value are zeros) 4.5 Applicatio of Laplace Trasform Example Determie the trasfer fuctio for the followig D.E : '' ' y (t) y (t) y(t) 5x(t) ; y() ', y () 3
4.5 Applicatio of Laplace Trasform Example Determie the trasfer fuctio for the circuit as show i the followig diagram : Vi Vo 4.5 Applicatio of Laplace Trasform d) Stability i the s - domai Stability of the system (trasfer fuctio) is determied from the locatio of POLES i the s - plaes j(imagiary) LHP RHP σ (real) s - plae
4.5 Applicatio of Laplace Trasform d) Stability i the s - domai Arragemet of the stability as per their priority : a) Stable : if ad oly the poles of the TF exist i the LHP or poles are egative b) Critically : if oe pole or more exist o j axis or pole is stable c) Ustable : if oe pole or more exist i the RHP or poles are positive. 4.5 Applicatio of Laplace Trasform Example The TF, G(s) of a system is give below. Sketch the zero-pole plot ad determie the stability of the system. a) s 4 s 3s s 4s 8 b) Y(s) (s ) s s 4s 3 3