1817_07basic math for process control

7/28/2019 1817_07Basic Math for process control

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7Laplace Transforms

HistoryThe origin of Laplace transforms dates back to the era of a Bri t ish civi lengineer named Ol iver Heavis ide , who l ived f rom 1850 to 1925. Ana c c o m p l i sh e d m a t h e m a t i c i a n , H e a v i s i d e w a s e x p e r i m e n t i n g w i t h w h a tcame to be cal led mathemat ical operators . In operat ional calculus , forexample , the le t ter p might replace the operat ion of taking a der ivat ive , sotha t if x = f(t), the n px w as e qu iva len t to

dx/dtOl iver Heav i s ide ' s con temporar i es r id icu led h i s work , no t so muchbecause of the opera to r s them se lves , bu t because he ac tua l ly m ov ed oper a tors around in a lgebraic expressions as i f they were ordinary terms. Thisd id no t de te r h im, how ever , because as fa r a s he w as concerned , them e t h o d w o r k e d .Eventual ly , resul ts prevai led , and f rom Ol iver Heavis ide 's beginnings,ma them at ic i ans d eve lo ped a se t of opera t iona l t ransfo rms tha t came to bek n o w n a s t h e Laplace transforms. Just as the use of logar i thms can reducemu l t ip l i ca t ion and d iv i s ion to add i t ion an d sub t r ac t ion , Lap lace tr ansforms, where they can be appl ied , can reduce the problem of solving a dif-ferent ia l equat ion to one of solving an a lgebraic equat ion.In the analysis of control systems, process var iables vary wi th t ime. Theobserved behavior may be descr ibed by a d i f ferent ia l equat ion, which hast ime as the independent var iable . In such cases , what i s required i s anexpre ssion x = f (t ), w hic h descr ibes the beh avio r of the de pe nd en t var iable

93


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94 Basic Math for Process Control

on a t ime bas is , an d w hic h is c lear of any der iv at ives . Laplace t ransform sis a m athem at ica l t echnique th ro ug h whic h th is m ay be ach ieved .

In Laplace t ransforms, t ime is replaced as the independent var iable by anew var iab le , wh ich has been des igna ted s. Th i s has been do ne by m ul t i plying the function f( t) by e - s t , and then in tegra t ing the p roduct be tweenthe lower a nd u p pe r l imi ts of zero an d infinity . The resul t ing integralappea r s a s

The value s that s can acquire m us t necessar i ly be l imi ted to those that w i l lcause the integral to assu m e som e finite value . The type s of di fferent ia lequat ions wi th which this text deals are essent ia l ly l inear , which meansthat s wi l l ha ve a real valu e an d w i l l be greater th an zero. Most funct ionsf(t) that occur in control sys tems eng inee r ing are Laplace t ransform able,s ince w i th s real an d p os i t ive, e - s t decreases rapidly as t approaches inf ini ty , w hic h m ore th an co m pen sates for an increas ing va lue of f (t ).Inasmuch as the domain of the integral is from zero to infinity, al l act ivitys tar ts a t t = 0 an d pr oce ed s in the pos i t ive di rect ion. Ne gat iv e valu es of thave no mean i ng .There are tw o fur ther res t r ic t ions on the appl icat ion of Laplace t ransforms.1. The function f( t) must be s ingle valued, that is , any value of tg rea te r than zero prod uce s on ly one va lue of the de pe nd entvar iab le .2. M ore mu st be kn ow n a bo ut the relat ion x = f(t) than jus t i tsdifferential equation. Specifically, the value of f(t) at t = 0 must beknown if the differential equation is of the first order. If the

differential equation is of the second order, then the value of thefirs t deriv ativ e of f(t), tha t is , the rate of ch an ge of the de pe nd en tvar iab le mu st be kn ow n at t = 0 as wel l .The Lap lace trans form of a function f(t), de no te d L {f(t)} in m ath em atic alshor thand, i s def ined as

L{f(t)} = 0f(t)e - s td t an d is gene ral ly de no ted F(s) .If these con dit io ns are sat isfied, th en th e function f(t) is Laplace transform-able.


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Chapter 7 - Laplace Transforms 95

Example 1: Step ChangeI t i s w or t hw hi le to wo rk ou t the Laplace transfo rm for one par t icu lar funct ion f( t) , s ince th is funct ion be com es im po r ta nt in the s tud y of t ransferfunctions. Fortunately, i t is the easiest one to evaluate.In the process of tes t ing control system components to determine thei rs ta t ic and dynamic proper t ies , the input that i s used most of ten i s a s tepinp ut , that i s , a su d de n jum p at t equ als zero f rom a zero s ignal to a constant s ignal of f in i te valu e . N ote that th is conforms to the requ irem ent thatthe in i tia l va lue of the funct ion be zero . If the inp ut s ign al su dd en lyas su m es a f inite va lue C at t = 0, th en f(t) = C is the func tion to be tr an sformed. The Laplace t ransform for the s tep change wi l l then be ,

Transforms of DerivativesThe fo l lowing shor thand symbols a r e genera l ly used when work ing wi thLaplace t ransforms.The f ini te value that x = f( t) assumes at t = 0 is designated x 0. In otherw o r d s , x0 = f(0).The valu e ofdx/dtat t = 0 is des ign ate d (dx/dt)0.W i thou t go ing th rou gh the ma them at ica l chores invo lved , we hav e the fo llowing r e l a t ionsh ips .1. The Laplace transform of the function f( t) is designated F(s) .2. The Laplace t ransfo rm of the fi rs t der iva t ive dx/dtwill be sF(s) - x0.

3 . The Laplace t ransform of the second der ivat ived2x/dt2wil l be

S 2 F(s) - sx 0 - (dx/dt)0.These facts wil l be required for the solution of differential equations usingLap lace t r ansfo rms .


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Example 2: T ime ConstantFigure 7-1 descr ibes the type of dy na m ic respo nse tha t i s typical of m an yelements that appear in control sys tems. The pr imary character is t ic of th isco m po ne nt i s that a t any poin t on i ts respo nse c urve , the ra te of cha nge ofthe depe nd en t var i ab le w i th t ime is p rop or t iona l to the d i st ance r em ain ingfor i t to at tain i ts ul t imate value. In this case, the dependent variable is y,an d in resp ons e to a s tep chan ge inp ut , y beg ins to cha ng e from z ero toeven tua l ly r each a new va lue Y.

Figure 7 - 1 . Behavior of a Time Constant Element.

At an y poin t in the respon se of the var iable y , the dis tance re m ainin g is(Y - y ) . De scribing th e be ha vio r as a differential eq ua tion ,

dy/dt (Y- y), ord y / d t=1 / T ( Y - y)w he re T is a cons tan t . In th is expre ss ion, the uni ts tur n ou t to be m orerealis tic if 1/T is cho sen as the cons tan t rath er th an T.Rear rang ing , T d y / d t + y = Y .Then , app ly ing the Lap lace tr ans format ion t e rm by t e rm ,

T{sF(s) - y 0 }+ F ( s ) = Y/s.


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Since y 0 = 0, the express ion bec om es

A table of Laplace t ransform s revea ls that the t ransform

F(s) = 1/s(Ts+1)originates from

-1/T tf(t ) = 1 - e .Acc ordingly, the solution to the differential eq ua tion ,

S ince the expon ent of e m us t be dim ens io nless , an d t is in time un i ts , Tmust also be in units of t ime. T is , in fact , the t ime constant of the componen t .

Example 3: PendulumFigure 7-2 sh ow s a pe n d ul um of leng th L, w hic h osci lla tes abou t it s fixedcenter O. At any t ime t , the angle that the shaf t of the pe nd u lu m m ak esw ith the ver t ical i s 6 . Onc e the free en d of the pe nd u lu m ha s be en raisedso that the pendulum s tar ts f rom a pos i t ion 0, t he pe nd ul um wi l l beg in toswing under the inf luence of gravi ty .If w e can assum e that the m ass of the p en d ul u m is m , an d tha t the ma ss ismainly concentrated at the center of gravity of the bob at the free end ofthe pe nd ulu m , then the force of g rav i ty wi l l be mg d ow nw ard f rom i tscenter of gravity.The path BP is perpendicular to the shaf t of the pendulum and is theins tan tan eous pa th a long whic h the bo b is m oving . The ang le APB is ac tu al ly equa l to the angle 6 . Therefore the com po ne nt of m g in the di rect ionBP is eq ual to

m g x the cos ine of the angle that BP m ake s wi t h the v er t ical= m g x cos (90 - ) = mg (cos 90 cos + s in 90 s in )

= mg ( 0 + s in ) = m g s in .


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Figure 7-2 . A Simple Pendulum .

If the pe nd ul u m is m ad e to swin g so that the angle 0 i s kep t smal l , then thevalu e of sin 9 bec om es vi r tual ly the sam e as the value of 6. Consequ ent ly ,the com po ne nt of the gravi ta t ion al force m g in the di rect ion of m ot io n ofthe bob w i ll hav e a mag ni tu de m g0. Ho wev er , if w e agree tha t va lues ofthe an gle 9, an d of forces, are posit iv e to the r ig ht s ide of the vert ical an dnegative to the lef t , then the tangential force exerted on the bob will be( -m g ) .The force -m g 9 wi l l be equ al to the ma ss m of the bo b mu l t ip l ied by i t saccelerat ion along the path BP. If d is tances a lon g the arc are deno ted bythe var iable z , then

- mg = md2z/dt2.Ho w ever , we are not so m uc h interes ted in the pos i t ion of the weig ht in it st rack , w i th t im e, as w e are in the angle that the shaft m ake s wi th the ver t i cal.


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Chapter 7 - Laplace Transform s 99

Th e basic relat ion (arc) z = (radiu s) L x 9 can be differentiated twic e, wh ic hgives

W ha t rema ins now is the solut ion of th is d i f ferentia l eq ua t ion us in gLaplace t ransforms.Ina sm uc h as there i s a second o rder d er ivat iv e invo lved , it i s necessa ry to

d 0kn ow the s tar t in g ( t = 0) va lue s of 8 a nd of .S ince the m ot io n of the pe n du lu m is cycl ical , the s tar t ing p oint ca n be chosen at any point in the cycle . From the viewpoint of knowing the requireds tar t ing poi nt value s , the bes t poin t from w hic h to s tar t is sh ow n in Figure7-2 . With the p en d ul um in th is pos i t ion, the in i t ia l value of 0 wi l l be w ha teve r angle is give n to the pe n d u lu m to s tar t it off. Th is wil l be the ev en tua lam p l i t u d e of t h e p e n d u l u m an d can b e d e s ig n a t ed 0.

d 9The angula r ve loc ity of the pen du lu m , , wi l l be zero at th i s po in t .Apply ing the Lap lace t r ans format ions t e rm by t e rm,

The table of Laplace t ransform s sho ws tha t the t ransform s/s2+ 2

origin ates from the function f( t) = cos t . In this case, 2 = g/L

and the funct ion sought i s consequent ly


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This result is not really as relevant as the period of oscil lat ion of the pendu lum . The t e rm in the cos ine expres sion has to hav e un i t s of s econ ds - 1so that the argu m en t cot i s d im ens io nless . The term is actual ly the a ng ular veloci ty of the shaf t, wi th un i ts in rad ian s per second .

Exam ple 4: Sec ond Order L inear Di fferent ia l Eq uat ions wi th Con stantCoef f ic ientsExa mp le 2 in Ch ap ter 6 (Dif ferentia l E quat ions) d eal t wi th the osci l la t ionsof a weight suspended on a spr ing. The resul t ing di f ferent ia l equat ion was

d2 y/dt2 + A dy/dt + B y = 0w her e A and B a re cons tan t s .In the sp r in g / m as s p rob le m w ork ed as Exam ple 2 in Ch apte r 6 , A s toodfor r / m , whi le B s tood for k / m . The k term w as the con s tant of the spr in g,w hic h induc es the osci l la t ions . The r term w as the dam pi ng ef fect, w hic htends to reduce or e l iminate the osci l la t ions , whi le the m term was themass of the weight .Ho we ver , i t i s des i rable tha t the dif ferentia l e qua t ion rep resen ts mo re realw or ld sy s tems than s im ply tha t of a we igh t dan g l ing on a spr ing . Consequent ly , the B term in the equat io n wi l l no w be rega rded as the "dr iv ingforce" be hin d the osci l la t ions , an d the A term as the "d am pi ng effect."W hat these two para m eter s s t an d for shou ld be kep t in m ind . The r e la tivem ag ni t ud es of the pa ram ete rs A an d B wi l l ha ve a s ignif icant imp act onthe behavior of the sys tem.The differential equation is l inear with constant coeff icients , and furtherm ore, the r ight s ide is zero. Accordingly, the resolut ion of th is equ at ion bys tand ard t ech n iques i s r e la t ive ly s t r a igh t fo rwa rd , as w as sho w n in Chap ter 6 . W ha t shou ld n ow be at te m pte d is the resolut ion of th is d i fferent ia lequa t ion by us ing Lap lace t r ans forms .

Ap ply in g the Lap lace t r ans forms t e rm by t e rm ,


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To continue, i t is necessary to declare the ini t ial condit ions. Since there isno w ay o f kn ow ing w ha t the ve loc ity of the we igh t wi ll be wh en thewe igh t i s m ovin g , we a re ob l iged to des igna te t = 0 a t some po in t w herethe velocity of the weight is zero. This means at one end or the other of i tst ravel . If the w eig ht i s osci l la ting up an d d ow n across the y = 0 base l ine ,the n w e can specify tha t i t star ts o ut at t = 0 w ith z ero velocity from a po sit ion Y belo w th e base l ine .With these in i t ia l condi t ions, the t ransform equat ion becomes

{s 2F(s) - sY - 0} + {AsF(s) - AY} + BF(s) = 0.Co llect ing F(s) an d Y term s: F(s)[s2 + As + B] - Y[s + A ] = 0

The proc ed ure n ow cal ls for consul t ing a table of Laplace t ransform s toobtain the funct ion f ( t) , w hic h has the expression F(s) de term ine d ab ove asi t s t ransform . Unfor tunate ly , the t ransform

s A /s 2 + A s + Bdoes not appear anywhere in the table . I s the conclusion, therefore , thatth i s p rob le m canno t be reso lved us in g the Lap lace t ransfo rm techn ique?In fact, Laplace t ransfo rms can pro du ce the answ er , bu t here again , the dif-ference between success and fa i lure i s the required mathemat ical exper i ence . The den om ina to r of the t r ansfo rm can be r ea r r ang ed to comp le te thesquare of the f i r s t two terms.

There are now three possibil i t ies to be considered. The f irst is the case inw h i c hA 2 = B .4

This would imply that the dr iv ing force and the damping ef fect balanceeach other off. WithB - A2/4 = 0


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t he t r ansfo rm b ecom es

which can be r ea r r anged as

A n imp or tan t fact of m athem at ics , one wh ich i s no t d i sc losed in m an ytex ts on Lap lace tr ansfo rms , n ow emerges : If a Laplace transform, which can-not be inverted a s it stands, can be expressed as the sum of two parts, each ofwhich is capable of being inverted, then the solution will be the sum of theinverted parts.From the table of Laplace t ransforms, the inverse t ransforms ( in mathemat i cs shor thand , L -1 ) for this problem are

Since the solution does not contain ei ther a sine or cosine function, the system is no t going to oscil la te . Fu r the rm ore , th is system h ad as it s basis th at

A2/4 = B .

Note that i fA2/4w ere greater tha n B, the n the dam pi ng ef fect that A represe n ts w o u l d b e e v e n m o r e d o m i n a n t .


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Chapter 7 - Laplace Transform s 103

Consequent ly , for any value ofA2/4 hat is equal to or greater than B, thesys tem wil l no t osci l la te .

The Oscillatory CaseThe rem ainin g poss ib i l i ty therefore i s the one in w hic h B is greater thanA 2/4 ,and the dr iving force is dominant over the damping ef fect .

To s impl ify the var io us exp ress ions , le t A / 2 = a , an d le t

Since B is gre ater th an A 2 / 4 , 2 will be a posit ive quanti ty. Later i t wil l beseen that us ing 2 ins tead of wi l l m ak e it eas ier to inver t the t ransform.


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From the table of Laplace t ransform s:

Consis tency of Resul tsAn as tu te observer wo uld no t i ce tha t the so lu t ion ob ta ined th rou gh theuse of Laplace t ransfo rms is no t the same as that obtain ed in Cha pte r 6 forthe second order l inear differential equation with constant coeff icients ,w h i c h w a s

x = X e t

s in ( t + ? )) ( wh ere a w as subs t i tu ted fo r -A /2 ) .In th is express ion , x i s a general var iab le , not nece ssar i ly the displa cem entof a m ass o n a spr ing . X wi l l be the or iginal am pl i tud e of the osci l la tions ,wh erea s M , w he re i t ap pe ars , i s the valu e of x a t t = 0.W he n the dif ferentia l eq uat io n wa s solved us ing the Laplace t ransform s, itwas necessary to specify two init ial condit ions, and this fact must be recogn ized w he n com par ing the two ap par ent l y dif ferent resul ts . Specif ically,th is m ea ns that the in i tia l con di t ions , w hic h w ere specif ied w he n us ing theLaplace t ransforms, should be appl ied to the solut ion obtained inChapte r 6 .The ini t ia l condi t ions were:1. W he n t = 0, x = M.2. W he n t = 0 ,dx/dt=0.A pp ly in g the init ial cond it ion (1) to the so lutio n x = X e t s i n ( t + ? ):

M = X x 1 x s i n (0 + ?), from which X = M/sin?.


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To ap pl y init ial co nd it ion s (2) , i t wil l be ne cess ary t o take the d eriv ativ e ofx with respect to t for f(t) = X e t s in ( t + ?) . The rules for tak ing theder ivat ive of a product , and of a funct ion wi thin a funct ion, are needed(refer to Chapter 2) .

which is the solut ion obtained through the use of Laplace t ransforms.


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Table of Laplace TransformsA table containing some of the more common Laplace t ransforms is contain ed in Table 7-1.Table 7 - 1 . Short Table of Laplace Transforms

Function x = f(t) Lapla ce Transform F(s) = L {f(t)}( t > 0 )C (cons tan t )tt2

tn

e - t

1/T e -t/T

l - e -t/Tf ( t - L )sin tcos t

e- t sin t

e- t cos t

d x / d td 2 x / d t 2

C/s1/t2

2/s3

n!/sn+ 1

1/s +

1/Ts + 11/s(Ts + 1)

e - L s F (s)

/s2 + 2s/s2 + 2

/(s + )2 + 2

s +a/(s+ a) 2 + 2

s F(s) - x0s2 F(s) - s x 0 - ( d x / d t ) 0

x0 a n d ( d x / d t ) 0 are the va lues of x an d d x /d t w he n t = 0 .

1817_07basic math for process control

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