Dr. Seeler Department of Mechanical Engineering Fall 2009 Lafayette College ME 479: Control Systems and Mechatronics Design and Analysis

Transcription

1 Dr. Seeler Departmet of Mechaical Egieerig Fall 009 Lafayette College ME 479: Cotrol Systems ad Mechatroics Desig ad Aalysis Lab 0: Review of the First ad Secod Order Step Resposes The followig remarks are limited to physical systems that ca be described by liear ordiary differetial equatios with costat coefficiets. The mathematical of Classical Cotrol Theory depeds o superpositio. Ufortuately, superpositio does ot apply to o-liear systems (where doublig the iput does ot double the output) or systems with o-costat coefficiets (where the system's respose to idetical iputs depeds o whe they are applied). Cosequetly, cotrol of o-costat ad o-liear systems is graduate material. If the system is described by liear ordiary differetial equatios with costat coefficiets the superpositio applies ad the respose ca be broke dow ito two compoets: 1. The "atural" respose due to disturbig the system from its former state. The atural respose is the solutio of the homogeeous system equatio, as we will review. The atural respose is commoly but erroeously called the trasiet respose because it geerally, but eed't always, die away.. The "forced" respose due to the system followig the iput (the forcig fuctio). The forced respose may be either steady, whe the system is subjected to a step chage (a costat), or periodic, whe the system is forced or drive by a periodic iput. If a system is drive by a periodic forcig fuctio, such as a sie wave, the system will ultimately reach "steady-state" whe each cycle of its respose is idetical to the proceedig cycle. The shape ad magitude of the steady-state respose may be very differet tha the periodic iput, but the output will always have the same frequecy as the iput as the system tries to follow the forcig fuctio. First Order Step Resposes The step respose of a first order system is o-oscillatory. First order systems have oly oe idepedet eergy storage elemet. A first order system s step respose results from mootoic eergy trasfer across the system boudary ito or out of the oe idepedet eergy storage. It is useful to ormalize step resposes. The time axis is ormalized by dividig time by the time costat τ. The vertical axis ca be ormalized by dividig the output variable V(t) by the rage V. The respose of all first order systems subjected to a step iput ca be described by a two ormalized curves with the percet chage i the output variable plotted agaist time measured i uits of time costats. There are two fudametal first order step resposes are:

2 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 1. Decay to Zero: V(t) = Ve τ t where V= V0. Stable Growth from Zero: t V(t) = V(1 e τ ) where V= VSS Note that the growth ad decay curves are mirror images of oe aother. Other possible first order step resposes are described by the growth ad decay curves offset vertically by addig a costat to the growth ad decay fuctios.

3 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 3 3. Decay to a No-Zero Steady-State Value: t τ V(t) = Ve + V SS where V= V0 VSS 4. Stable Growth from No-Zero Value: t τ V(t) = V(1 e ) + V 0 where V = VSS V0 Note: Either V 0 or V SS or both ca be egative. We have a prejudice i favor of positive umbers, but a variable ca decay from or grow to a egative value. Example:

4 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 4 Secod Order Step Resposes Secod order step resposes have a wider rage of shapes that caot be reduced to a sigle ormalized curve. The two fudametal secod order step resposes are oscillatory, or uderdamped, ad o-oscillatory, or overdamped. Uderdamped ad Overdamped Secod Order Step Resposes The characteristic equatio of a uderdamped secod order system yields complex cojugate roots. The characteristic equatio of a overdamped secod order system has two real roots.

5 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 5 Poles of the Uderdamped ad Overdamped Secod Order Systems whose Step Resposes are Plotted Above. Although the oscillatory uderdamped respose requires more mathematics to describe, it is easier to characterize because its features are distictive. The o-oscillatory overdamped step respose rages from almost oscillatory critically damped case to a pair of widely spaced real poles which may be adequately described as a first order system cosistig of just the right most domiat pole. We will make extesive use of block diagrams i ME 478 ad ME 479. Recall that the lies i a block diagram represet the Laplace trasforms of time variables ad the blocks are operators. Block diagrams of the two fuctioal forms eeded to describe secod order systems are: Overdamped Secod Order System Uderdamped Secod Order System We will excite systems with step iputs of some magitude M, Laplace trasform is: u(t) = Mu (t) The s L { } L { } L { } 1 M u(t) = Mu s(t) = M u s(t) = M = s s Applyig this iput U(s) to the overdamped trasfer fuctio yields:

6 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 6 Y(s) M A M A U(s) = = = Y(s) U(s) s ( s a)( s a) + + s( s + a)( s + a) ad to the uderdamped trasfer fuctio yields: Y(s) M A M A U(s) = Y(s) = = U(s) s s + ζω s +ω ss+ ζω s+ω ( ) A abbreviated Laplace trasform table is attached. Trasform pair 17, below, is a overdamped, o-oscillatory secod order uit step respose. Trasform pair 4 is a uderdamped, oscillatory secod order uit step respose. No. f(t) F(s) at bt 1 1 ( be ae ) ab + a b ss ( + a)( s+ b) 4 1 ζω 1 e ( ) si ω 1 ζ t+φ 1 ζ ω 1 ω d 1 1 ta ta ζ φ= = ss+ ζω s+ ω σ ζ ( ) Sice these are uit step resposes ad our step iputs will have some magitude M other tha uity, we will eed to scale the resposes multiplyig by the magitude M of our step iput. We will also eed to scale the uit step resposes to correct the umerator term so that we have a exact match with the trasform pair i the table. I the overdamped case, we simply factor out the amplitude costat A: Y(s) Over Damped MA 1 = = M A ss ( + a)( s+ b) ss ( + a)( s+ b) I the uderdamped case, we multiply ad divide by the table: ω to match the Laplace trasform ω MA MA ω Y(s) = = ω ss ( + ζω s+ω ) ω ss ( + ζω s+ω)

7 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 7 Characterizig Secod Order Systems from their Step Resposes There are two ways to characterize a uderdamped secod order step respose. Oe characterizatio is used for system idetificatio, which meas to determie the system s trasfer fuctio. The dampig ratio ζ ad udamped atural frequecy ω are determied so that we ca establish the characteristic fuctio ad, hopefully, the etire trasfer fuctio of the system which produced the step respose. The other characterizatio of a uderdamped step respose is purely descriptive. I practice, you may fid occasioally that you eed ru a step respose test twice to get both the dyamic respose ad the steady-state respose data eeded. It ca be impossible to amplify a sigal o the oscilloscope sufficietly to see the detail of the dyamic respose without pushig part of the trace off the scree. Likewise, decreasig the time per divisio to stretch the time axis may make it impossible to catch the etire trasiet period. Ruig the same test more tha oce is always a good idea sice we eed to kow how repeatable the data are. Describig a Uderdamped Step Respose for a Performace Specificatio There is a stadard way to describe a uderdamped step respose which is used to specify the required performace of a system. If our purpose is to characterize a oscillatory step respose without determiig the system s trasfer fuctio, the we measure ad report time ad amplitude values of the respose at the poits show i the figure below. Note that settlig time is defied i terms of a evelope comprised of lies parallel to the steady-state value ad spaced at either +/- % or +/- 5% of the steady-state value. The settlig time is the time at whe the trace eters the settlig time evelope ad stays i it. Poits i a Performace Specificatio for a Oscillatory Step Respose

8 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 8 Uderdamped System Idetificatio System idetificatio works backwards from the observed respose of a system to the fuctioal form of the system that created the respose. If the step repose is oscillatory, tha we kow that there is a secod order factor i the characteristic fuctio of the form: s + ζω s+ ω Cosequetly, we eed to determie the dampig ratio ζ ad ideal, udamped atural frequecy ω. We ca determie the damped period from the observed step respose. T = t t d peak + 1 peak The damped frequecy is calculated from the damped period: π ω d = T d The dampig ratio ζ is determied usig the log-decremet formula: Log Decremet Formula ζ = 1 x l 1 1 x 1 x 4π + l 1 1 x Recall that the peak amplitude terms x 1 ad x are relative to the steady-state value of the step respose. The damped frequecy ad the dampig ratio are the used calculate the ideal udamped atural frequecy: ω = ω d 1 ζ Example: A secod order system s respose to a step iput with a amplitude of 0 is show i the trace below. Determie the trasfer fuctio.

9 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 9 Measurig the time ad amplitude of the first ad secod peaks: Time (sec) Amplitude Amplitude Mius Steady-State Value Peak Peak Td = = 0.40sec π rad π rad rad ω d = = = 15.7 T 0.4sec sec d Usig the form: l 1. ζ= = π + l 1. ω 15.7 rad rad 16 1 ζ 1 0. sec sec d ω = = = ( ) ( )( ) ( ) s + ζω s +ω = s s + 16 = s + 6.4s + 56 ( ) Y s M ω = K s s + ζω s+ ω

10 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 10 where M is the magitude of the step iput ad K is a yet to be determied gai, so far we kow: = 0 56 s s + 6.4s + 56 ( ) K Y s We determie the gai of the trasfer fuctio from the fial value of the respose. From the Laplace trasform pair, we see that the steady-state value of the time domai uit step respose is 1 because the real expoetial t approaches zero as time icreases. 4 e ζω ( ) 1 ω ζ +φ 1 ζ ζω 1 e t si 1 t 1 ω d 1 1 ζ φ= ta = ta σ ζ ω ss s ( + ζω +ω) Our data has a steady-state value of 15 for a step iput with a magitude of 0. Hece: ad the trasfer fuctio is: ( ) ( ) 15 = 0K K = 0.75 ( ) ω ( )( ) Output s Y s = = K = = Iput s M s + ζω s +ω s + 6.4s + 56 s + 6.4s + 56 s Overdamped System Idetificatio A overdamped step respose cosists of a expoetial decay from a positive value ad a expoetial decay from a egative value superimposed o a step. A overdamped step respose is more difficult to characterize tha a uderdamped step respose because its features are less distict. There are two iterative methods. The older method uses a family of ormalized respose curves. This method is ow obsolete as a techique for derivig a trasfer fuctio, but it retais some usefuless i providig iitial values for the computer-based iteratio. Step Respose Plots Method The output variable axis of secod order step respose plots is the observed or actual respose y(t) ormalized (or divided by) the steady-state value y ss (t). Cosequetly, the ormalized steady-state value of the all curves is 1. The time axis is ormalized by multiply time by the udamped atural frequecy ω of the system: t Normalized = t ω

11 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 11 ωd π rad which yields a dimesioless value sice ω = ad ω d =. T d is the damped 1 ζ Td period ad carries uits of time. Radias are the dimesioless ratio of arc legth divided by radius. The dampig ratio ζ is also dimesioless. Although scalig time i terms of frequecy makes sese for a oscillatory system, it is a misomer for the step respose of a overdamped system. The step respose has o udamped atural frequecy ω sice it does t oscillate. Recall the geometric costructio i the s-plae used to express s i terms of the ideal, udamped atural frequecy ad the dampig ratio z: s + bs+ c= s + ζω s+ω Note that the distace from the origi to a poit i the left-half of the s-plae is the atural frequecy ω. Whe s =σ, i.e. s is o the real axis, the σ =ω, IMPORTANT. whe s = σ, σ ζω sice the dampig ratio ζ may be greater tha oe. Recall that for overdamped secod order systems ζ 1. Although the Step Respose Method is ow obsolete, you should be familiar with it sice it ca be exted to other applicatio ad allows you to use published data. The method is a follows: Pick a poit o the actual respose ad determie its amplitude y(t) ad time t. Normalize the amplitude y(t) by dividig it by the steady-state value. Refer to the respose plots. Estimate (or guess) a value of the dampig ratio ζ which characterizes the actual system ad fid the respose curve closest to that value. Usig the step respose curve which correspods to the estimated dampig ratio ζ, fid the ormalized time ω t which correspods to the ormalized amplitude. You ca ow calculate a value of udamped atural frequecy ω. Now test your values. Pick a differet poit o the ormalized step respose curve ad determie its ormalized amplitude ad ormalized time. Retur to the actual overdamped respose. Multiply the ormalized amplitude by the steady-state value. Divide the ormalized time by ω. It the resultig poit is reasoably close to the actual overdamped respose curve, the you have a valid estimate of ζ. If it is t, refie your estimate ad try agai. If this method seems very approximate, it is. It is typical of how egieerig aalyses were performed i the bad-old-days before calculators ad persoal computers.

12 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 1 Computer-Based, Iterative Method A computer-based, iterative method is more practical tha usig ormalized respose curves. The first step is to idetify, if possible, whether the respose is a uderdamped step respose, ad described by the Laplace trasform pair No. 4, or a overdamped step respose described by Laplace trasform pair No. 17. If the dampig ratio ζ is ot close to 1, the it is easy to distiguish betwee the two types of resposes, sice very uderdamped systems are very oscillatory ad very overdamped systems creep towards steady-state. If ζ is ear 1, the it is difficult to distiguish betwee uderdamped ad overdamped systems visually. If you ca t tell which model to use, the guess. If you fid it impossible to iterate to a reasoable fit to the data, the try the other model. Havig picked a step respose model, the ext step is to choose iitial values of the parameters to begi the iteratio. It ca be very time cosumig to blidly iterate to fit three ukows. It usually saves a sigificat amout of time to use a approximate techique, such as iterpolatig from Respose Curves, to provide the iitial values for the iteratio. Example: Say you have the followig respose to a step iput of 0: Respose to a Step Iput of Magitude 0 This looks overdamped, so we will use: Y(s) Over Damped MA 1 = = M A ss ( + a)( s+ b) ss ( + a)( s+ b) The time domai respose is the uit step respose of trasform pair No. 17 scaled by the product M A: MA 1 y(t) = 1 be ae ab + a b at bt ( ) where M = 0

13 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 13 The faster secodary pole is resposible for the smaller curve at the origi. The slower domiat pole is resposible for the rest of the dyamic respose. Estimate the domiat pole by eglectig the secodary pole ad calculatig a time costat as if the trace were a first order growth. The precisio is improved slightly by straighteig the curve (fittig a straight lie segmet to it) so as to remove the curve at the origi. Estimate of the Domiat Pole by Modelig the Respose as a First Order Growth The time costat τ is the egative iverse of first order pole. 1 τ = p This ca be see by comparig the first order step respose fuctio V(t) = V(1 e τ ) with the MA 1 at bt time domai respose y(t) = 1 ( be ae ) ab + a b. Watch out for sig errors!! They are VERY commo. Note that the two poles are s = -a ad s = -b: t Y(s) Over Damped MA = ss a s b ( + )( + ) Now, does it make a differece whether we use s = -a or s = -b as the domiat pole i this expressio? Fortuately, o. Let s call the domiat pole -a: 1 1 a = = = 3. τ 0.31

14 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 14 Now estimate the secod pole. A useful rule-of-thumb is that if the secodary pole is a order of magitude or larger tha the domiat pole, the its effects (other tha the amplitude scalig) ca be eglected. (This rule-of-thumb oly holds for systems without zeros.) This gives us a upper boud for -b: b 10( a) b ( 10)( 3.) = 3 We ow have eough values to make a first estimate of the amplitude term A usig the steadystate value of the step respose, y(ss) = 5.0: ( ) MA 1 at bt y(ss) = lim{ y(t) } = 1+ b e 0 ae t ab 0 = a b MA ab A = y(ss)a b M ( 5.0)( 3.)( 3) A= = 6 0 Now we are ready to begi iteratig our model usig Mathcad. It is very helpful to plot poits from the observed respose o the same graph as the Mathcad model. Mathcad will plot fuctios ad vectors o the same graph. The iitial ad fial values ad three to five additioal poits from the observed respose will give us sufficiet accuracy:

15 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 15 First Iteratio Model Not bad. I may cases, this model would be adequate ad there would be o eed for the iteratio. However, it takes little time to iterate to a model which is more precise. The techique is to shift the poles aloe the real axis of the s-plae to adjust their speed or time costats. Values of σ closer to the imagiary axis have larger time costats. Cosequetly, movig a pole to the right slows the pole. Values of σ further from the imagiary axis toward egative ifiity are have smaller time costats. Cosequetly, movig a pole to the left makes it faster. Ispectig the fit of the model to the data, we see that the faster pole (s = -b) is a bit too fast sice the model plots to the left (at earlier time) of the data. Recall that we used a upper boud estimate for s = -b. Referrig to the s-plae, we will slow s = -b dow by movig it to the right. Let s try splittig the iterval betwee s = -3 ad s = -3 ad use s = -18 = -b. We eed to recalculate the amplitude term A usig b = 18: ( )( ) ( )( )( ) y(ss) a b A= = = 14 M 0

16 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 16 Secod Iteratio Model We have a good fit at the begiig of the respose. Now we eed to adjust the positio of the domiat (right most) pole s = -a. The model plots to the right (later i time) of the data. We eed to speed up the domiat pole by movig to the left o the s-plae. Let s try doublig its magitude from a = -3. to a = -6. Agai, recalculatig the amplitude term: ( )( ) ( )( )( ) y(ss) a b A= = = 7 M 0

17 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 17 Third Iteratio Model What happeed to our good fit at the begiig of the respose? The domiat pole is ow too fast ad its respose is overlappig with the faster pole. We moved from s= 3. to s= 6. Let s try s= 5. Recalculatig A: ( )( ) ( )( )( ) y(ss) a b A = = =.5 M 0

18 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 18 Fourth Iteratio Model We improved the fit ad the begiig ad pick up three poits at the ed of the respose. We eed to slow dow the faster pole to pick up poits at the begiig. Our two poles are s = -a = -5 ad s = -b = -18. Let s move b half way toward a ad use s = -b = -1. Recalculatig A: ( )( ) ( )( )( ) y(ss) a b A= = = 15 M 0

19 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 19 Fifth Iteratio Model The faster pole still looks too fast sice we are to the left of the data poits (at earlier time) except for the origi, which, of course, does t cout. Let s agai reduce the spacig betwee the poles by half ad move the faster pole from s = -1 to s = The amplitude factor is ow: y(ss) ( a)( b) ( 5.0)( 5)( 8.5) A = = = 10.6 M 0

20 ME 479 Lab 0: Review of First ad Secod Order Step Resposes page 0 Fial Model We have a good curve fit for the step respose. IMPORTANT: Remember to divide the Laplace trasform of the output Y(s): Y(s) Over Damped MA = ss a s b ( + )( + ) by the Laplace trasform of the iput: M U(s) = s Y s to yield the trasfer fuctio G( s) Us : ( ) ( ) MA Output(s) Y(s) ss ( + a)( s+ b) A 10.6 = = = = G(s) Iput(s) U(s) M ( s + a)( s + b) ( s + 5)( s + 8.5) s Commo errors are to either (1) to forget to divide the Laplace trasform of the respose by the Laplace trasform of the step iput or () to divide by the Laplace trasform of a uit step whe the actual iput was a differet magitude.