INSTRUMENTATION AND CONTROL TUTORIAL 4 SYSTEM RESPONSE. On completion of this tutorial, you should be able to do the following.

Transcription

1 INSTRUMENTATION AND CONTROL TUTORIAL 4 SYSTEM RESPONSE Th tutral f nteret t any tudent tudyng cntrl ytem and n partcular the EC mdule D7 Cntrl Sytem Engneerng. On cmpletn f th tutral, yu huld be able t d the fllwng. Explan the utput repne f an ON/OFF cntrl ytem. Explan and defne the tandard mdel fr t and nd rder ytem. Explan and defne the tandard tme dependant nput t a ytem. Defne a lnear ytem and a lnear tme nvarant ytem. Explan and calculate the repne f a tandard t rder ytem t a tep change n the nput. Explan and calculate the repne f a tandard t (velcty) change n the nput. rder ytem t a ramp Explan and calculate the repne f a tandard t rder ytem t a nudal change n the nput. Explan hw t fnd the verall gan f a ytem. Ue partal fractn t lve repne. If yu are nt famlar wth ntrumentatn ued n cntrl engneerng, yu huld cmplete the tutral n Intrumentatn Sytem. In rder t cmplete the theretcal part f th tutral, yu mut be famlar wth bac mechancal and electrcal cence. Yu mut al be famlar wth the ue f tranfer functn and the Laplace Tranfrm (ee math tutral). Yu mut be able t cnvert plynmal nt partal fractn. D.J.DUNN

2 . INTRODUCTION A ytem may be defned a a et f cnnected thng. We are cncerned wth engneerng ytem and we may defne th mre precely a a et f cnnected element degned t prduce pecfed utput when a gven nput appled. Sytem may nclude analgue element r dgtal element. Frt we wll examne a pecal knd f mple ytem that ue ON/OFF cntrl.. ON/OFF CONTROL Cnder a ytem made up f a tank f lqud, a heater and a wtch. When the wtch perated, pwer uppled t the heater and the water get htter. The peratn f the wtch the nput actn and the temperature f the water the utput actn. The utput temperature related t the pwer by the fllwng frmula. Max Specfc Heat x Temperature Change mc POWER Fgure Tme t We may repreent ur ytem a a mple blck. Fgure Often t pble t repreent the rat f utput/input a a mple cntant and gve th cntant the ymbl G (t mght be thught f a gan but th wuld nt be trctly crrect). In th example we have the fllwng. Output t G Input P mc The ytem decrbed knwn a an OPEN LOOP SYSTEM becaue the gnal path de nt frm a cled lp and there n cntrl ver the temperature. When the wtch cled, the temperature wll keep rng wth tme. Nw cnder an mprvement t the ytem by ung a thermtatc cntrller and a thermcuple t meaure the utput. The dered temperature (nput) et n the cntrller and the actual temperature the utput f the ytem. The cntrller cmpare the meaured temperature wth the et temperature. If the water t cld, the heater turned n. If the temperature t ht t wtched ff. Suppe the temperature lw and crrect and the et temperature uddenly ncreaed. Th called a tep change. The thermtat wll turn the heater n and the temperature wll re. Smetme later, the temperature reache the crrect value. Fgure Ideally, the thermtat wuld wtch ff the heater at the prece mment that the crrect temperature reached. D.J.DUNN

3 In realty th cannt happen a heat wll g n beng put ut after t wtched ff and the thermtat ha t detect that the temperature t ht befre t wtche ff. Cnequently, the temperature re abve the crrect value befre beng wtched ff and then the lqud tart t cl. It wll cl t jut belw the crrect value befre the thermtat detect the errr and wtche the heater back n. The lqud heated agan and the prce cntnue ndefntely prducng a temperature tme graph a hwn. Th graph called the SYSTEM TIME RESPONSE GRAPH. Fgure 4 The blck dagram f th ytem hw that the thermcuple turn the gnal path nt a cled lp t a CLOSED LOOP SYSTEM. Fgure 5 The prblem wth all ON/OFF cntrl that the utput mut cycle between an OFF value and an ON value whch are jut abve and belw the SET value. If yu brng the ON and OFF level cler tgether the ytem mut wtch n and ff mre quckly. It mpble t get prece cntrl wth th methd. Th adequate fr many ytem uch a central heatng where the exact temperature nt mprtant. Sytem that need precn cntrl (uch a the mvement f a machne tl) requre a mre phtcated methd f regulatn and thee are what we mut tudy n detal.. RESPONSE OF CONTINUOUS CONTROL SYSTEMS We are gng t dcu hw t et abut lvng the way a gven ytem behave wth repect t tme. It huld be nted, hwever, that tme nt alway the man varable but we wll aume t. We culd, fr example, be dcung hw a rbt manpulatr mve n repne t an nput r hw a valve n a ppe lne mve n repne t an nput.. STANDARD MODELS The frt tep t derve a mathematcal mdel alng the lne hwn n the prevu tutral. In the tutral t wa hwn that the fllwng mdel appled t everal dfferent ytem that were analgue f each ther. Thee are the tandard frt rder and ecnd rder equatn. Standard t rder equatn. = T d /dt +...() Standard nd rder equatn. = T d /dt + T d /dt +...() T a tme cntant and a dampng rat. Thee wll be dcued n greater detal later. T and are example f ytem parameter r cntant. D.J.DUNN

4 efre yu can lve hw the utput change wth tme, yu have t decde hw the nput change wth tme. Fr example, yu mght make a udden change t the nput r yu mght be changng t ver a perd f tme. Let lk at th next.. STANDARD INPUTS Standard nput are uually lted n the fllwng rder.. AN IMPULSE Th an ntantaneu change n latng fr zer length f tme and returnng t the ntal value. Th mtly appled t dgtal ytem where ntantaneu value are ampled by dgtal t analgue cnverter. It al wdely ued a a tandard nput t a ytem t cmpare the repne f dfferent ytem. Fgure 6. A STEP CHANGE Th an ntantaneu change n the nput whch then reman at the new value. = H at all value f tme after t=. H the change r heght f the tep. Fgure 7. A RAMP OR VELOCITY CHANGE Th when the nput change at a cntant rate. It al called a velcty nput. d /dt = c r = ct c the rate f change (velcty). Fgure 8 v. A PARAOLIC r ACCELERATION CHANGE Th when d /dt = a r = at / Th al knwn an acceleratn nce the rate f rate f change a cntant a. Fgure 9 D.J.DUNN 4

5 v. A SINUSOIDAL CHANGE In th cae the nput change wth tme nudally. = A n(t + ) A the ampltude and the phae angle. Fgure v. AN EXPONENTIAL CHANGE Th when the nput change expnentally wth tme. = A( - e -at ) There are everal expnental frm e.g. repreentng grwth and decay. Fgure Once an equatn ha been frmed t repreent hw the nput t a ytem change wth tme, the man tak t deduce hw the utput change wth tme that yu can ee hw accurate and hw fat the repne wll be. It th apect whch we wll dcu n detal and n partcular the ue f LAPLACE TRANSFORMS t help u d t. efre th, hwever, let' lk at me mre defntn and termnlgy ued t decrbe type f ytem.. LINEAR SYSTEM A prperty f a lnear ytem that f SUPERPOSITION. Th mean that f ne nput prduce a gven utput and anther nput prduce a ecnd utput, then bth nput tgether wll prduce the um f the tw utput. A gd example f th the deflectn f a prng. If a frce F appled the reultng extenn x. If a frce F appled the reultng extenn x. If a frce F + F appled, the reultng extenn x + x. Anther prperty f a lnear ytem that f HOMOGENEITY. Th mean that f an nput prduce a gven utput, then multplyng the nput wll multply the utput by the ame fgure. In the cae f the prng f a frce F prduce deflectn x then a frce f nf prduce a deflectn nx..4 LINEAR TIME-INVARIANT SYSTEMS Cnder the mathematcal mdel fr a dfferental preure flw meter. Q = C (p) ½. C a cntant fr the meter and the pwer f ½ a parameter, bth f whch defne the equatn. Q and p are varable. If the cntant and parameter d nt change ver the perd f tme beng tuded, the ytem a lnear tme-nvarant ytem. T and n equatn () and () are example f cntant fr uch ytem. Fr the mment we need t tudy hw uch equatn are lved rather than hw they are derved. D.J.DUNN 5

6 .5. LAPLACE TRANSFORMS In rder t lve a a functn f tme we ue a mathematcal tranfrmatn knwn a LAPLACE TRANSFORMS. Th change the equatn frm a dfferental equatn nt an algebrac equatn whch can then be manpulated and cnverted back nt a dfferental equatn by nvere tranfrm. Yu huld tudy the tutral n th n the math ectn unle yu are already famlar wth t. Frt we replace d/dt wth the ymbl fr example d /dt becme and d /dt becme. Yu may g n t apprecate that n t wn becme and that the ntegral f becme - and n. Cnductng th change n equatn () and () prduce the algebrac equatn a fllw. and = T + = (T + )...() = T + T + = ( T + T + )...(4) It nw pble t wrte the equatn a a rat and when th dne, the equatn r mdel called a TRANSFER FUNCTION. Equatn and 4 becme and T T T The tranfer functn are nw functn f f() ntead f functn f tme f(t). efre lutn f the tranfer functn can prceed, we mut change the nput frm a functn f tme f(t) nt a functn f f(). Th where the man tranfrmatn ccur. Yu d nt have t tranfrm the functn mathematcally but ntead yu wll ue tandard table. Here are tw example f hw t dne (agan yu wll fnd detal n the math ectn) WORKED EXAMPLE N. Fnd the Laplace tranfrm L H when H a cntant (a tep change) SOLUTION L t -t t e H e f t dt e H dt H H f() H H LH f() Fr a unt tep H = and the Laplace tranfrm / D.J.DUNN 6

7 WORKED EXAMPLE N.. Fnd the Laplace tranfrm f e -at SOLUTION L L -at e f t dt ( a)t - at e e L e L e - at - at a f() a t -at ( a)t e e dt e a a dt D.J.DUNN 7

8 TALE OF COMMON LAPLACE TRANSFORMS Nte that the ue f the letter fr cntant arbtrary and that ften the lutn fund by nterchangng a wth /T and. H and k are arbtrary cntant. Tme dman f(t) Frequency dman f() Decrptn e -at f(t) dt f( + a) t Unt Impule H H Step H 4 ct c Ramp 5 H(t - T) T e H 6 T - e 7 k e -at k a 8 kt e -at k ( a) 9 K(e -at - e -bt ) k(b a) ( a)( b) 9 k n (t) k k t n(t) kω k c (t) k e -at n (t) k e -at c (t) 4 t k e T 5 k t 6 k(-ct) T t e ( k ω ω kω ) ( a) a ( a) ka ( a) ( ka kω a) ω ω a = /T ( ω ) k ω c φ nφ 7 k n (t + ) 8 ω at a e (c bt n bt) a b b a b Delayed Step Rectangular pule Expnental Snudal C nudal Damped nudal Damped c nudal Expnental grwth Standard ecnd rder ytem tep repne D.J.DUNN 8

9 4. THE STANDARD ST ORDER SYSTEM 4. RESPONSE TO AN IMPULSE AND STEP Any ytem wth an mpule appled t the nput wth have an nput f () =. It fllw that the utput n the tme dman wll mply be the nvere Laplace tranfrm f the tranfer functn. WORKED EXAMPLE N. Cmpare the tme repne f a ytem wth the tranfer functn unt tep nput. t a unt mpule and a T SOLUTION UNIT IMPULSE () ubttute () = fr an mpule nput. T () (t) nvere Laplace tranfrm f T /T rearrange nt a recgnab le tranfrm and /T UNIT STEP () ubttute () = / fr an tep nput. T (t) nvere Laplace tranfrm f (T) /T rearrange nt a recgnab le tranfrm and ( /T) The dagram hw the tw repne. (t) (/T)e (t) e t/t t/t Fgure D.J.DUNN 9

10 WORKED EXAMPLE N.4 A R- C crcut hwn n whch R = and C = 5 F. The vltage V uddenly changed frm t Vlt. Determne the tme cntant and hw lng t take V t reach 9.99V. SOLUTION Fgure The mdel fr the R C crcut hwn wa derved n tutral and hwn t be V = TdV /dt + V Frt replace dv /dt by V V = T V + V Next rearrange nt a tranfer functn G() = V /V = /(T + ) Th may be repreented dagrammatcally a hwn. Fgure 4 Next change the nput frm a functn f tme nt a functn f by makng a Laplace tranfrmatn. Fr a tep nput V (t) = H the tranfrm V () =H/. In th cae H = Vlt. V H V T T Next manpulate the equatn nt a recgnable Laplace tranfrm a fllw. H/T V /T Lkng n the table f tranfrm we ee Ka/(+a) the tranfrm f K( - e -at ). Th mean that f we put a = /T and K = H the lutn fr V V = H( - e -t/t ) The tme cntant T = RC = x 5 x -6 =. ecnd Put V = 9.99 V and H = V t e.999 e e t.. ln t. t t.7 D.J.DUNN

11 4.. FULL ANALYSIS OF THE RESPONSE Fr any tandard frt rder ytem wth a tep nput, the utput repne = H( e -t/t ). If we plt bth nput and utput agant tme we get the reult hwn. 4.. FIRST ORDER TIME CONSTANT Fgure 5 The tme cntant T ha a real meanng. Cnder the repne graph. The rate f change f d dt H e T At the tart f the change t =, the ntal rate f change t T d H e dt T The ntal gradent (rate f change) H/T. If the ntal gradent prjected a hwn, t wll ntercept at tme t = T. Th lead t a defntn f T a the tme t wuld take the utput t match the nput f t kept changng at the ntal rate. Anther way f defnng the tme cntant T a fllw. At tme t = T the value f the utput = H{ - e - } =.6H In ther wrd, the utput wll reach 6.% f the dered value after T ecnd. When t = 4T the utput = H( - e -t/t ) = H( - e -4 ) =.999H. Th a cle t the crrect value a we are lkely t want we nrmally aume that the ytem ha reached the crrect value after t = 4T ecnd. H T D.J.DUNN

12 4. RESPONSE OF A STANDARD ST ORDER SYSTEM TO A RAMP INPUT A ramp r velcty nput ccur when (t) = ct and ()= c/ () T () c c(/t) () T (T ) ( /T) If we replace c wth k and /T wth a we have ca () ( a) and we can fnd th n the table f tranfrm. The nvere tranfrm gve the lutn (t) = c {t - T( - e-t/t )} Fgure 6 Plttng the utput agant tme prduce the reult hwn. The equatn may be wrtten a: (t) = c t ct( - e-t/t )} At large value f tme t the term ( e -t/t ) becme neglgbly mall and the utput becme: (t) = c {t - T()} = c(t - T) The errr becme e (t) = - = ct ct + ct e (t) = ct In ther wrd, a cntant errr ct reult after an ntal tranent tage and th the teady tate errr. WORKED EXAMPLE N.5 A ptn cntrl ytem ha a tranfer functn G() = /(. + ). The nput changed at a cntant rate f 5 degree/ frm the zer ptn. Calculate the errr after.4 ecnd and the teady tate errr. SOLUTION Cmparng parameter t apparent that T =. and c = 5 deg/. After.4 ecnd = 5 x.4 = degree. = c t ct( - e -t/t )} = 5 x.4 5 x. x ( e -.4/. ) =.85 =.5 degree. e = -.5 =.865 degree The teady tate errr ct where T =. and c = 5 degree/. e = 5 x. = degree. D.J.DUNN

13 4. RESPONSE OF A STANDARD ST ORDER SYSTEM TO A SINUSOIDAL INPUT The tandard frt rder ytem tranfer functn In the teady tate = and () k T k () k t fllw that k the teady tate ytem gan. Nte that t mut be + n the bttm lne t make th. T a tme cntant. Fr a nudal nput let (t) = A n(t) ω The Laplace tranfrm () ω k ω The Output () (T ) (T ) (ω ) The nvere tranfrm gve the utput a a functn f tme. Plttng the nput and utput gve a typcal reult lke th wth A= and K=). Fgure 7 The plt hw that the utput quckly ettle dwn nt teady tate repne wth a nudal frm but wth dfferent ampltude and laggng the nput wth a phae hft. The tranent tage nly lat fr ne cycle. A mre detaled analy f the teady tate repne cvered n the next tutral. D.J.DUNN

14 SELF ASSESSMENT EXERCISE N.. Shw the dervatn f the tranfer functn fr prng and damper ytem hwn. Gven that the dampng ceffcent k d. and the prng tffne k 4 kn/m, determne the tme cntant fr the ytem. (Anwer 7.5 ) If a frce f F = N uddenly appled, calculate the value f x after T ecnd. (Anwer 6 mm) Fgure 8. A blck f metal ha a ma f.5 kg, pecfc heat capacty 46 J/kg K and temperature f = C. It drpped nt a large tank f l at = C and t fund that the temperature f the blck take 6 mnute t reach 9 C. Aume that the temperature f the blck change by the law () (T ) Shw that the temperature f the blck change wth tme by the law = + ( - )(-e -t/t ) Determne the tme cntant T and hence the thermal retance between the blck and the l. (Yu huld ee the tutral fr detal) (Anwer R=.45 K/W). A hydraulc mtr ha a nmnal dplacement f k m /radan. The peed cntrlled by a mple valve uch that the preure t the mtr k x where x the nput ptn f the valve. The mtr ha a mment f nerta J kg m and a dampng ceffcent f k Nm /radan. Gven that the trque develped by the mtr k p, hw that the pen lp tranfer functn ω relatng utput t nput x gven by kmx T J k k T and km k k The nput gven a tep change. Sketch the repne f the utput. Determne the % change n the utput at t = T and t = 4T. (6.% and 99.9%) Shw n the ketch the affect f ncreang the mment f nerta. 4. A ptn cntrl ytem ha a tranfer functn G() = /( + ). The nput changed at a cntant rate f 4 mm/ frm the zer ptn. Calculate the errr after ecnd and the teady tate errr. (.6 mm and mm) D.J.DUNN 4

15 4. GAIN OF t ORDER SYSTEMS DC GAIN :- th an electrcal term and the gan when there n varatn f the nput n the tme dman. When an nput cntant all the dervatve value f the tranfer functn are zer the D.C. the magntude f G() when =. Hence the gan f the tandard t rder cled lp tranfer functn () unty. T When the tranfer functn f the frm K (), the D.C. gan K. T A The D.C. gan nt alway bvu at frt glance e dervatn. Cnder the cae () T A/ Put = and the D.C. gan A/. We culd rearrange the tranfer functn t () (T/) Th nw n the frm () K T T a new tme cntant reultng frm the change. In the teady tate = K the teady tate gan f the ytem. D.C. GAIN = A/ T = T/ The D.C. gan al affect the magntude f a dynamc repne The repne t a tep change H wll hence ettle dwn at KH. The repne t a ramp nput wll nt ettle dwn at a fxed value but (t) = Kc {t - T( - e-t/t )}. The affect f ntrducng gan hwn n the repne dagram. Fgure 9 In the cae f the tep nput = H, the utput reache a value f K H. In the cae f the ramp nput = ct, the utput pae and exceed the nput mre and mre a tme prgree. D.J.DUNN 5

16 WORKED EXAMPLE N.6 The tranfer functn fr a mple cntrl ytem Slve and tabulate the value f SOLUTION Rearrange the tranfer functn. ().5. agant tme fr a tep nput f unt. () /. (.5/.) 5.5 Frm th we ee the gan 5 and the tme cntant.5. Fr a tep nput f unt the tranfrm 5() gve () = / Hence () (.5 ) The nvere tranfrm = ( - e -t/.5 ) = ( - e -.4t ) Plttng the reult we get the repne hwn belw. WORKED EXAMPLE N.7 Fgure The ame ytem ubjected t a ramp nput f mm/. Plt the utput agant tme and determne the teady tate errr. SOLUTION 5 () Fr a ramp nput = ct and c = mm/. () = c/ hence (.5 ) 5c () (.5 ) The nvere tranfrm f th () 5ct t - - e Fgure D.J.DUNN 6

17 SELF ASSESSMENT EXERCISE NO.. Fnd the D.C. gan and tme cntant fr the fllwng tranfer functn.. G() (4 and.4)..5. G()..5. ( and.5). G ( ) ( and ) 6 v. G ( ) 8 4 (4 and ). The utput peed f a mtr ( rad/) related t the angle f the nput enr ( radan) by the ω km tranfer functn () Tm Where k m = 5 - and T m = 4 Determne the D.C. gan and tme cntant f the ytem. (7.5 - and ecnd) D.J.DUNN 7

18 5. USING PARTIAL FRACTIONS TO SOLVE RESPONSES The repne t mt type f nput can be lved by the ue f partal fractn. Here are tw example ung the tandard frt rder tranfer functn. /T Frt Order Sytem G() T /T Put /T = p and () 5. UNIT STEP INPUT D.J.DUNN 8 p A unt tep nput n Laplace frm () = / Change nt partal fractn Slve A and A n the nrmal way fr partal fractn. The numeratr mut equate A p Put = -p and A A p p A = - Nw put = p p A Subttute back nt the lutn p p () p A A () p A A p p p A p p p p A and th true fr any value f A = + () tp tp Nw cnduct an nvere Laplace tranfrm. (t) e e Put p = /T (t) e And th the lutn fund by ther methd n prevu tutral. 5. UNIT RAMP Partal fractn Equate the numeratr and frt let p = - put and () () p p A ( t/t p p p p p () p p p p p put - p and Next put = and ubttute fr and t lve and fnd = -/p - pt (t) - t e Let /T = p and the lutn p p t T - - (t) -T t Te (t) t T e Th the lutn fund by ther methd prevuly. t T p p p p p) p

19 Let extend th t a rd rder ytem r hgher. UNIT STEP () = / Change nt partal fractn Invere Laplace Tranfrm () A, A, A and A etc mut then be lved. G() p p p K A () () A p p p A p -pt A e K A p A e -pt A p A e -pt UNIT RAMP () = / () p p p Change nt partal fractn Invere Laplace Tranfrm,, and etc mut then be lved. () (t) t K p -p t e p -p t e 4 p e -p t 4 WORKED EXAMPLE N.8 A ytem ha a tranfer functn nput. () Determne the tme repne t a unt tep ( )( ) SOLUTION () ( )( ) ( )( ) Cnvert t partal fractn t () Invere tranfrm (t) e ( ) e t Fgure D.J.DUNN 9

20 SELF ASSESSMENT EXERCISE NO.. A ytem ha a tranfer functn tep nput. Anwer (t) 6 e t. A ytem ha a tranfer functn t t ramp nput. (t) e e 4 4 () Determne the utput tme repne t a unt ( )( ) t e () Determne the utput tme repne t a unt ( )( ) t D.J.DUNN