CHBE3 ECTURE V APACE TRANSFORM AND TRANSFER FUNCTION Profeor Dae Ryook Yang Spring 8 Dept. of Chemical and Biological Engineering 5-
Road Map of the ecture V aplace Tranform and Tranfer function Definition of aplace tranform Propertie of aplace tranform Invere aplace tranform Definition of tranfer function How to get the tranfer function Propertie of tranfer function + - Controller Actuator PROCESS Senor ecture IV to VII 5-
SOUTION OF INEAR ODE t -order linear ODE Integrating factor: [ xe ] f ( t) e 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 ) Imaginary root: ( t t e co t, e in t) dx For atx ( ) f ( x), I.F. exp( a( t) dt) dt () () xt () [ f() te atdt dtce ] ap apa a( p p)( p p) Many other technique for different cae atdt Solution i a linear combination of mode and the coefficient are decided by the initial condition. 5-3
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) 5-4
DEFINITION OF APACE TRANSFORM Definition 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. t F() f () t f () te dt - F f() t ( ) - af () t bf () t af () bf () 5-5
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 t t S( 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 5-6
Trigonometric function jt co Euler Identity: e t jint jt jt jt jt int e e cot e e j j j j j jt jt cot e + e j j j t j t int e e j in(t) t Rectangular pule, P(t) for t tw f( t) P( t) h for tw t for t tw t t P( ) t w t he dt e e h t w h f(t) h t w t 5-7
Impule function, for t tw f() t () t lim/ tw for tw t tw for t tw t () lim lim t w t e dt e tw t tw w tw f () t f() t 'Hopital' rule: lim lim t gt () t g() t Ramp function, t () 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 5-8
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PROPERTIES OF APACE TRANSFORM Differentiation df t t t f e dt f ( te ) 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 n n ( n) ( n) () () () () F f f f 5-
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 Tranform of linear differential equation. yt () Y (), ut () U() dt n d f dt n F() () F n F () dy() t Y ( ) (if y() ) dt dy() t yt () Kut () ( y() ) ( ) Y( ) KU( ) dt T T T( ) v ( Tw T) Hv ( H) T ( ) T w( ) t z z H 5-
Integration t t ( ) ( ) t f d f d e dt t e t t F( ) f ( ) d f e dt (by ib.. p.) Time delay (Tranlation in time) Derivative of aplace tranform d bt () db t da t f d f b t f a t dt at () dt dt () () eibniz rule: ( ) ( ( )) ( ( )) f t f t t in t () ( )S( ) t ( ) f( t )S( t) f( t) e dt f( ) e d (let t) e f( ) e d e F( ) f(t) t df() d t d t t f e dt f e dt ( t f ) e dt t f ( t) d d d 5-3
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 lim t e dt lim F( ) f() dt df lim e t dt lim F ( ) f () f () lim F ( ) dt 5-4
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( t6).5 3 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) lim e t t t 5-5
EXAMPE ON APACE TRANSFORM () What i the final value of the following ytem? xxxin 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 tint d ( ) (in t t) 5-6
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 Partial fraction expanion t f() t F() e F() d j C - After the partial fraction expanion, it require to know ome imple formula of invere aplace tranform uch a ( n)! e,,,, etc. b n ( ) ( ) 5-7
PARTIA FRACTION EXPANSION F () N () N () 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 Multiply both ide by a factor, (+p i ), and replace with p i. N () i ( pi) D () Invere T: p f t e e e pt pt n () i n p t 5-8
Cae II: Some root are repeated F () N () N () b b 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 r i i! d D( ) () i d N( ) () ( ) r p ( i,, r ) i p Invere T f () t e te t e ( r )! pt pt r r pt 5-9
Cae III: Some root are complex F () N () cc ( b) D () d d ( b) Each repeated factor have to be eparated firt. Then, ( b) ( b) ( b) ( b) ( b) Invere T where bd /, d d / 4 c, c b / bt f () t e cot e int bt 5-
EXAMPES ON INVERSE APACE TRANSFORM ( 5) A B C D F ( ) (ditinct) ( )( )( 3) 3 Multiply each factor and inert the zero value ( 5) B C D A A5/6 ( )( )( 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 ( )( ) 3 3 5 3 f() t F( ) e e e 6 3 - t t 3 t 5-
A B C D F ( ) (repeated) 3 3 ( ) ( ) ( ) ( ) ( A B C)( ) D( ) 3 3 ( ) ( 3 ) ( 3 ) ( ) A D AB D BC D CD AD, AB3D, BC3D, CD A, B, C, D Ue of Heaviide expanion r i ( ) ( ) ( ) ( ) 3 3 3 ( i ): 3 d ( i ):! d d ( i ):! d f() t F( ) e te t e e - t t t t () i d N( ) () ( ) r (,, ) i p i r i! d D( ) p 5-
( ) A( ) B CD ( ) (complex) ( 45) ( ) F A B CD ( ) ( )( 4 5) 3 ( A C ) (AB4 CD ) (5C4 D ) 5D AC, AB4CD, 5C4D, 5D A/5, B7/5, C /5, D/5 A ( ) B ( ) 7 B ( ) 5 ( ) 5 ( ) C D 5 5-7 () () t co t f t F e t e int t 5 5 5 5 5-3
e A B F () ( e ) (Time delay) (4)(3) 4 3 A /(3 ) 4, B /(4 ) 3 /4 /3 4 3 4e 3e 4 3 4 3 f() t F() t/4 t/3 ( t)/4 ( t)/3 e e e e S( t) 5-4
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 (54) Y() 5 Y() (5 4).5.5 yt () Y().5.5e 5 4.8 t 5-5
TRANSFER FUNCTION () Definition An algebraic expreion for the dynamic relation between the input and output of the proce model dy 5 4 y u; y() U() dt et y y and u u4 Y ().5 (5 4) Y() U() G() U ( ) 54.5 Y () Tranfer Function, G() 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 5-6
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() G3() 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()= 5-7
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 () (t) G () 5-8
EXAMPE Horizontal cylindrical torage tank (Ex4.7) dm dv qi q dt dt h dv dh V( h) wi( h) dh wi ( h) dt dt wi ( h)/ R ( Rh) ( Rh) h dh dh w i qi q ( qi q) (Nonlinear ODE) dt dt ( D h) h Equilibrium point: ( qi, q, h) ( qi q) /( ( Dh) h) (if qi q, h can be any value in h D.) inearization: dh f f f f ( hq,, q) ( hh) ( qq) ( qq) dt h q q i i i ( hq, i, q) i ( hq, i, q) ( hq, i, q) q i q R w i h 5-9
f ( qi q) ( qi q) h h ( D h) h ( hq, i, q) f f, q ( Dh) h q ( Dh) h i ( hq, i, q) ( hq, i, q) H () kqi () kq() H () and Q (): - k k H() and Q i (): et thi term be k Tranfer function between (integrating) Tranfer function between (integrating) h If i near or D, k become very large and i around D/, 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 D/. (le change in gain) inearized model would be valid in very narrow range near. h 5-3
PROPERTIES OF TRANSFER FUNCTION Additive property Y Y Y () () () G X G X () () () () X () X () G () G () Y () Y () + + Y() Multiplicative property X () X () X 3 () G () G () X3() G() X() G() G() X() G() G() X() 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) 5-3
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 () Ci () C () ( V / q) C () ( V / q) C() C() C() Ci() C () Ci() ( V / q) ( V/ q) C C i C C V V q 5-3
Pule input P 5.5 C i () ( e ) c i P 5.5 t Equivalent impule input C i ( ) (5.5) ( t).5 Pule repone v. Impule repone 5 C C e () P P.5 () i () ( ) 5 e c t e.5 ( ) P t/ ( ) 5( ) e t.5 C () Ci () () c ( t.5)/ 5( )S(.5).65e t / 5-33
5 C C e ( )(.5 ) ( )(.5 ) P P.5 ( ) i ( ) ( ) 5 4.5 e.5 c t e e P t/ t/.5 () (5 5 ) C ( ) Ci ( ) ()(.5).5 ()(.5) 5 3.75.5 c.5e.5e t/ t/.5.5 ( ) ( t.5)/ ( t.5)/.5 (5 e 5 e )S( t.5) 5-34