CHE302 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang

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1 CHE3 ECTURE V APACE TRANSFORM AND TRANSFER FUNCTION Profeor Dae Ryook Yang Fall Dept. of Chemical and Biological Engineering Korea Univerity CHE3 Proce Dynamic and Control Korea Univerity 5-

2 SOUTION OF INEAR ODE t -order linear ODE Integrating factor: [ xe ] = fte () atdt () atdt () High-order linear ODE with contant coeff. Mode: root of characteritic equation For ax + ax + ax = f (), t Depending on the root, mode are pt pt Ditinct root: ( e, e ) pt pt Double root: ( e, te ) α Imaginary root: ( t αt e co βte, in βt) dx For atx () f( x), I.F. exp( atdt () ) dt + = = () () xt () = [ fte () atdt dt+ Ce ] ap + ap+ a = a ( p p )( p p ) = Many other technique for different cae CHE3 Proce Dynamic and Control Korea Univerity 5- atdt Solution i a linear combination of mode and the coefficient are decided by the initial condition.

3 APACE TRANSFORM FOR INEAR ODE AND PDE aplace Tranform Not in time domain, rather in frequency domain Derivative and integral become ome operator. ODE i converted into algebraic equation PDE i converted into ODE in patial coordinate Need invere tranform to recover time-domain olution (D.E. calculation) u(t) ODE or PDE y(t) U() Tranfer Function Y() (Algebraic calculation) CHE3 Proce Dynamic and Control Korea Univerity 5-3

4 DEFINITION OF APACE TRANSFORM Definition t F() = f() t fte () dt F() i called aplace tranform of f(t). f(t) mut be piecewie continuou. F() contain no information on f(t) for t <. The pat information on f(t) (for t < ) i irrelevant. The i a complex variable called aplace tranform variable Invere aplace tranform and are linear. { } f() t = () - { F} CHE3 Proce Dynamic and Control Korea Univerity { + } = + af () t bf () t af() bf ()

5 APACE TRANSFORM OF FUNCTIONS Contant function, a t a t a a a = ae dt = e = = { } f(t) a t Step function, S(t) for t f() t = S( t) = for t < S( ) t t t e dt e = = = = { } f(t) t Exponential function, e -bt = = = + b + b ( + ) { e } e e dt e bt bt t b t f(t) b< b> t CHE3 Proce Dynamic and Control Korea Univerity 5-5

6 Trigonometric function Euler Identity: jωt e coωt+ jinωt jωt jωt jωt jωt = ( + ) inωt = ( e e ) coωt e e ω ω ω j j j jω + jω + ω jωt jωt { coωt} = e + e = + = jω + jω + ω j t j t { inωt} = e e = = j in(t) π ω t Rectangular pule, P(t) for t > tw f() t = P( t) = h for tw t for t < w tw t h t h tw { P( t) } = he dt = e = ( e ) t f(t) h t w t CHE3 Proce Dynamic and Control Korea Univerity 5-6

7 Impule function, δ () t for t > tw f() t = δ () t = lim / tw for tw t tw for t < tw t tw { δ () t } = lim e dt lim ( e ) tw = = t tw w t w f () t f () t 'Hopital' rule: lim = lim t gt () t g () t Ramp function, t f(t) /t w t w t { } = t t te dt f(t) t t e t e dt e dt t = = = t ( ) Integration by part: f ' gdt = f g f gdt Refer the Table 3. (Seborg et al.) for other function CHE3 Proce Dynamic and Control Korea Univerity 5-7

8 CHE3 Proce Dynamic and Control Korea Univerity 5-8

9 CHE3 Proce Dynamic and Control Korea Univerity 5-9

10 PROPERTIES OF APACE TRANSFORM Differentiation df t t t = f e dt fte () f ( e ) dt (by ibp...) dt = t = f e dt f() = F() f() d f t t t t f e dt f () t e f ( e ) dt f e dt f () = dt = = = F() f() f () = F () f() f () ( ) n d f ( n) t ( n ) t ( n ) t f e dt f() t e f ( e ) dt n = = dt n ( n ) t ( n ) d f ( n ) = f e dt f () = f () n dt = F f f f n n ( n ) ( n ) ( ) () () () CHE3 Proce Dynamic and Control Korea Univerity 5-

11 If f () = f () = f () = = f (n-) () =, Initial condition effect are vanihed. It i very convenient to ue deviation variable o that all the effect of initial condition vanih. df = dt d f = n d f n = Tranform of linear differential equation. yt () Y (), ut () U () F() F CHE3 Proce Dynamic and Control Korea Univerity 5- dt dt () n F () dy() t Y() (if y() = ) dt dy() t τ = yt () + Ku() t ( y() = ) ( τ+ ) Y() = KU( ) dt T T T() v ( Tw T) τhv = + + ( τh+ ) T () = T w() t z τ z H

12 Integration { ( ξ) ξ} = ( ( ξ) ξ) t t t f d f d e dt t e t t F () = f( ξ) dξ + f e dt = (by ibp...) Time delay (Tranlation in time) Derivative of aplace tranform d bt () db t da t () () eibniz rule: f( ) d f(()) bt f( at ()) dt τ τ = at () dt dt f t f t t + θ in t () ( θ)s( θ) t ( τ + θ) { } f ( t θ)s( t θ) = f ( t θ) e dt = f ( τ) e dτ (let τ = t θ) θ θ τ θ = e f( τ) e dτ = e F () df() d t d t t = f e dt f e dt ( t f ) e dt t f () t d d = = = d CHE3 Proce Dynamic and Control Korea Univerity 5- f(t) θ [ ] t

13 Final value theorem From the T of differentiation, a approache to zero lim df t e dt lim [ F () f () ] = dt df dt = f ( ) f () = lim F () f () f ( ) = lim F () dt imitation: f ( ) ha to exit. If it diverge or ocillate, thi theorem i not valid. Initial value theorem From the T of differentiation, a approache to infinity df t lim e dt lim [ F() f ()] = dt df lim e t dt lim F () f () f () lim F () = = = dt CHE3 Proce Dynamic and Control Korea Univerity 5-3

14 EXAMPE ON APACE TRANSFORM () f(t) 3.5 t for t< 3 for t < 6 f() t = for 6 t 6 t for t < f() t =.5tS( t).5( t )S( t ) 3S( t 6) () { ()} ( F = f t = e ) e 5 For F() =, find f() and f( ). Uing the initial and final value theorem f() = lim F() = lim = f( ) = lim F() = lim = 5 5 But the final value theorem i not valid becaue 5 lim f() t = lime t = t t CHE3 Proce Dynamic and Control Korea Univerity 5-4

15 EXAMPE ON APACE TRANSFORM () What i the final value of the following ytem? x + x + x= in t; x() = x () = + ( + )( + + ) ( )=lim x = X () + X() + X = x()= ( + )( + + ) Actually, x( ) cannot be defined due to in t term. Find the aplace tranform for? df() From = [ t f() t ] d d ω ω t in ωt = d ω = + ( + ω ) [ ] (in t ωt) CHE3 Proce Dynamic and Control Korea Univerity 5-5

16 INVERSE APACE TRANSFORM Ued to recover the olution in time domain - { F } () = f () t From the table By partial fraction expanion By inverion uing contour integral { } t f() t = F() = efd () π j C - Partial fraction expanion After the partial fraction expanion, it require to know ome imple formula of invere aplace tranform uch a θ ( n )! e,,,, etc. n ( τ+ ) ( + b) + ω τ + ζτ + CHE3 Proce Dynamic and Control Korea Univerity 5-6

17 PARTIA FRACTION EXPANSION F () N () N () α α n D () ( + p ) ( + p ) ( + p ) ( + p ) = = = + + n n Cae I: All p i are ditinct and real By a root-finding technique, find all root (time-conuming) Find the coefficient for each fraction Comparion of the coefficient after multiplying the denominator Replace ome value for and olve linear algebraic equation Ue of Heaviide expanion Invere T: Multiply both ide by a factor, (+p i ), and replace with p i. α i N() = ( + pi) D () CHE3 Proce Dynamic and Control Korea Univerity 5-7 = p f t e e e pt pt n () = α + α + + α i n pt

18 Cae II: Some root are repeated F () N() N() b + + b α α = = = = + + D () ( + p) ( + p) ( + p) ( + p) r r r r r r Each repeated factor have to be eparated firt. Same method a Cae I can be applied. Heaviide expanion for repeated factor αr i = + = i! d D () () i d N() () ( ) r p i ( i,, r ) = p Invere T α f () t = α e + α te + + t e ( r )! pt pt r r pt CHE3 Proce Dynamic and Control Korea Univerity 5-8

19 Cae III: Some root are complex F () N () c+ c α ( + b) + βω = = = D () d d ( b) ω Each repeated factor have to be eparated firt. Then, α ( + b) + βω ( + b) = α + β ω ( + b) + ω ( + b) + ω ( + b) + ω Invere T where b = d /, ω = d d /4 ( ) α = c, β = c αb / ω bt f () t = α e coωt+ β e inωt bt CHE3 Proce Dynamic and Control Korea Univerity 5-9

20 EXAMPES ON INVERSE APACE TRANSFORM ( + 5) A B C D F () = = (ditinct) ( + )( + )( + 3) Multiply each factor and inert the zero value ( + 5) B C D = A A 5/ = = ( + )( + )( + 3) 3 = ( + 5) A ( + ) C ( + ) D ( + ) = + B+ + B= ( + )( + 3) 3 = + + = ( + 5) A ( + ) B ( + ) D ( + ) = + + C+ C = 3/ ( + )( + 3) 3 = + + = ( + 5) A ( + 3) B ( + 3) C ( + 3) = D D= /3 ( + )( + ) = = f() t = { F() } = e + e e t t 3t CHE3 Proce Dynamic and Control Korea Univerity 5-

21 A + B+ C D F () = = + (repeated) 3 3 ( + )( + ) ( + ) ( + ) = ( A + B+ C)( + ) + D ( + ) 3 = ( A D ) (A B 3 D ) (B C 3 D ) ( C D) A= D, A+ B+ 3D =, B+ C + 3D =, C+ D= A=, B =, C =, D = Ue of Heaviide expanion + + α α α3 = + + ( + ) ( + ) ( + ) ( + ) ( ) ( i = ): α = + + = = () i α d N () () ( ) r r i= p i ( i,, r ) + = i! d D () = p d ( i= ): α = ( + + ) =! d = d ( i= ): α = ( + + ) =! d = - t t t t f() t = { F() } = e te + te e CHE3 Proce Dynamic and Control Korea Univerity 5-

22 ( + ) A ( + ) + B C+ D () (complex) ( ) ( + ) + F = = + + = A ( + ) + B + ( C+ D)( ) 3 = ( A+ C ) + (A+ B+ 4 C+ D ) + (5C+ 4 D ) + 5D A= C, A+ B+ 4C+ D=, 5C+ 4D =, 5D = A= /5, B= 7/5, C = /5, D = /5 A ( + ) + B ( + ) 7 B = ( ) 5 ( ) 5 ( ) C+ D = = { } = t t f() t F() e cot e in t t CHE3 Proce Dynamic and Control Korea Univerity 5-

23 + e A B F () = = + ( + e ) (Time delay) (4+ )(3+ ) A= /(3 + ) = 4, B = /(4 + ) = 3 = /4 = /3 f() t = () = + { F} 4 3 4e 3e ( ) t/4 t/3 ( t )/4 ( t )/3 = e e + e e t S( ) It i a brain teaer!!! But you have to live with it. Hang on!!! You are half way there. CHE3 Proce Dynamic and Control Korea Univerity 5-3

24 SOVING ODE BY APACE TRANSFORM Procedure. Given linear ODE with initial condition,. Take aplace tranform and olve for output 3. Invere aplace tranform dy dt + = = Example: Solve for 5 4y ; y() dy 5 +{ 4y} = { } 5( Y() y()) + 4 Y () = dt 5 + (5+ 4) Y() = + 5 Y() = (5+ 4).5.5 yt () = { Y ()} = + =.5+.5e t CHE3 Proce Dynamic and Control Korea Univerity 5-4

25 TRANSFER FUNCTION () Definition An algebraic expreion for the dynamic relation between the input and output of the proce model dy y = u; y() = U () Tranfer Y () dt Function, G() et y = y and u = u 4 Y ().5 (5+ 4) Y () = U () = = = G ( ) U () How to find tranfer function. Find the equilibrium point. If the ytem i nonlinear, then linearize around equil. point 3. Introduce deviation variable 4. Take aplace tranform and olve for output 5. Do the Invere aplace tranform and recover the original variable from deviation variable CHE3 Proce Dynamic and Control Korea Univerity 5-5

26 TRANSFER FUNCTION () Benefit Once TF i known, the output repone to variou given input can be obtained eaily. { } { } { } { } yt Y GU G U () = () = () () () () Interconnected ytem can be analyzed eaily. By block diagram algebra X + - G G3 G Y Y() G( G ) ( ) = X() + G( G ) ( G ) 3( ) Eay to analyze the qualitative behavior of a proce, uch a tability, peed of repone, ocillation, etc. By inpecting Pole and Zero Pole: all atifying D()= Zero: all atifying N()= CHE3 Proce Dynamic and Control Korea Univerity 5-6

27 TRANSFER FUNCTION (3) Steady-tate Gain: The ratio between ultimate change in input and output ouput ( y( ) y()) Gain= K= = input ( u ( ) u ()) For a unit tep change in input, the gain i the change in output Gain may not be definable: for example, integrating procee and procee with utaining ocillation in output From the final value theorem, unit tep change in input with zero initial condition give y( ) K = = lim Y() = lim G() = lim G () The tranfer function itelf i an impule repone of the ytem Y() = GU () () = G () δ(t) = G () { } CHE3 Proce Dynamic and Control Korea Univerity 5-7

28 EXAMPE Horizontal cylindrical torage tank (Ex4.7) dm dv = ρ = ρqi ρq dt dt h dv dh V( h) = wi( hdh ) wi = ( h) dt dt wi ( h)/ = R + ( R h) = ( R hh ) dh dh w i = qi q = ( qi q) (Nonlinear ODE) dt dt ( D hh ) Equilibrium point: ( qi, qh, ) = ( qi q) /( ( D hh ) ) (if qi = q, h can be any value in h D.) inearization: dh f f f = f( hq,, q) = ( h h) + ( q q ) + ( q q) dt h q q i i i ( h, q, q) i ( h, q, q) ( h, q, q) i i i q i q R w i h CHE3 Proce Dynamic and Control Korea Univerity 5-8

29 f = ( qi q) = ( qi = q) h h ( D hh ) ( h, q, q) i f f =, = q ( D hh ) q ( D hh ) ( hqq,, ) i ( h, q, q) i H () = kq i () kq () Tranfer function between Tranfer function between CHE3 Proce Dynamic and Control Korea Univerity 5-9 i H () and Q (): - k k H () and Q i ( ): et thi term be k (integrating) (integrating) h h h / If i near or D, k become very large and i around, k become minimum. The model could be quite different depending on the operating condition ued for the linearization. The bet uitable range for the linearization in thi cae i around h /. (le change in gain) inearized model would be valid in very narrow range near

30 PROPERTIES OF TRANSFER FUNCTION Additive property Y Y Y () = () + () = G() X() + G() X() X () X () G () G () Y () Y () + + Y() Multiplicative property X3 = G X () () () [ ] = G() G() X() = G() G() X() X () X () X 3 () G () G () Phyical realizability In a tranfer function, the order of numerator(m) i greater than that of denominator(n): called phyically unrealizable The order of derivative for the input i higher than that of output. (require future input value for current output) CHE3 Proce Dynamic and Control Korea Univerity 5-3

31 EXAMPES ON TWO TANK SYSTEM Two tank in erie (Ex3.7) No reaction dc V qc qci dt + = dc V + qc = qc dt Initial condition: c ()= c ()= kg mol/m 3 (Ue deviation var.) Parameter: V /q= min., V /q=.5 min. Tranfer function C i C C () Ci () = C () ( V / q ) + C () = ( V / q ) + C () C () C () C = i() C () C = i() ( V / q ) + ( V/ q ) + ( )( ) CHE3 Proce Dynamic and Control Korea Univerity 5-3 C V V q

32 Pule input P 5.5 C i () = ( e ) c i P 5.5 t Equivalent impule input { δ } C δ i () = (5.5) () t =.5 Pule repone v. Impule repone 5 C C e + (+ ) P P.5 () = i () = ( ) 5 = + c t = e.5 ( e ) P t/ () 5( ).5 C δ δ () = C i () = + (+ ) c =.65e δ ( t.5)/ 5( e )S( t.5) t / CHE3 Proce Dynamic and Control Korea Univerity 5-3

33 5 C C e ( + )(.5 + ) ( + )(.5 + ) P P.5 ( ) = i ( ) = ( ) = c t = e + e P t/ t/.5 () (5 5 ) C δ δ ( ) = C i () (+ )(.5+ ).5 = (+ )(.5+ ) = +.5+ c =.5e.5e δ t/ t/.5.5 ( e ) + ( t.5)/ ( t.5)/.5 (5 e 5 e )S( t.5) CHE3 Proce Dynamic and Control Korea Univerity 5-33