2 Simulation exercise 2

Transcription

1 Smulaton exercse. Descrton of a dynamc system One of the man tasks n the course on CAD of automatc control system s to model, to smulate and to carry out control acton for some knd of dynamc system(s). One way or another, the dynamc system s descrbed as a mathematcal model that reresents a hyscal realty, wth a set of varables and a set of logcal and quanttatve relatonshs between them. Even though the model may make exlct assumtons that are known to be false (or ncomlete) n some detal, t has to carry out the essence of the hyscal rocess(es). Modern techncal systems are consdered to be comlex (comosed of many comonents) and cover mult-doman (electrcal, mechancal, thermal ). Over and over agan t s not trval to go from hyscal rocess to mathematcal model and comromse beng fast and accurate (what to model and what to exclude, nclude relevant hyscs, use arorate concets ). A model s always aroxmate and a good model s smle, yet catures the essentals! There can be dfferent methodologes how to gras the essentals or an effcent aroach to (buld a) model of a dynamc system of nterest. There mght be a few stes between establshng a hyscal understandng and defnng a mathematcal model: How does the system work? Understandng, defnton of urose, structurng nto subsystems and nteractons between them, once more understandng of hyscal realty Whch asects of the system are we nterested n and whch quanttes are mortant for descrbng the rocesses? Understandng, try to reflect logcal relatons, hyscal rncles, cause-and-effect, force-and-reacton relatonshs, controllablty, measurablty How do reflect the essence of the hyscal rocesses and how the quanttes of mortance nfluence each other? Organze varables and equatons, valdate and smlfy model o Mathematcal models are often based on balance equatons (nflow outflow = volume, energy etc change er tme unt, ower n ower out = accumulated energy er tme unt, sum of all currents enterng a node = 0) and consttutve relatons (statc relatons between quanttes Ohm s Law) o Focus modellng effort on dynamcs whose tme constants are relevant for the ntended urose of the model subsystems wth fast dynamcs aroxmated as statc relatons and varables that vary slowly are aroxmated by constants In the frst home assgnment a geometrc shae of a system.e. nductor was gven and the assgnment was to descrbe the system by a set of equatons that s also reflected from the block dagram. Fgure. A smle smulnk block dagram descrbng the nductor

2 ψ ψ = ( u R) dt = f = em (.) L The system s consdered to be nonlnear not only because of L = f ( B) f n B but also because of, whch s not ncluded n the revous exressons. There s assumed that L=const. Furthermore, when consderng knetcs of the nductor yoke (electromagnet yoke) then L=f(B,x) where x corresond to a dslacement of the yoke. As a matter of fact, the oosng nduced voltage can be ether a transformer voltage or/and a motonal voltage, whch s resectvely generated ether by the change of current or by the change of nductance generated by moton. d d d dl d dx dl e = ψ = L = L + = L + (.) dt dt dt dt dt dt dx At the moment, there s no further nvestgaton on the nonlnear system of the nductor or the electromagnet rather than to study a lnear tme nvarant (LTI) system. u S y=s(u) The system S s lnear f ( ku) ks( u) S ( u u ) = S( u ) + S( ) S = scalng + sueroston u A system s tme-nvarant f delayng the nut results n a delayed outut: y ( t τ ) = S( u( t τ )) Lnear tme-nvarant systems are easy to analyze Local stablty = global stablty: Egenvalues of the system matrx A (= oles of G(s)) n left half lane regardless of ntal ostons The transent resonse of a lnear system s comosed of the natural modes of the system: Sueroston, enough to know ste (or mulse) resonse Frequency analyss ossble: Snusodal nuts gve snusodal oututs The man tasks of ths home assgnments s to recall knowledge or become famlar wth Dfferent ways of descrbng a dynamc system Tme and frequency resonse characterzaton of a dynamc system Analyss of the automatc control system References to course lterature Automatc Control by Raul Naadel are gven n the arentheses {Eest keeles}. Each ste of solvng an assgnment corresonds to certan amount of credts gven n brackets (0). Use the and mrove the m-fle (Ch.7) to carry out your work

3 . Electrcal machne wth ermanent magnet exctaton The second smulaton exercse focus on a seed controlled electrcal machne n an electrc drve system. The frst task s to establsh a hyscal understandng of the electromechancal energy converson. An electrc motor s a machne whch converts electrcal energy nto mechancal energy. The oeraton of a most of electrcal motors s based on the rncle that when a current-carryng conductor s laced n a magnetc feld, t exerences a mechancal force whose drecton s gven by Flemng's Left-hand rule. I I E M M B ψ ω F jωψ I rotaton axs Fgure. a smlfed constructon of a ermanent magnet excted electrcal machne. We can assume that ermanent magnet exctaton gves a constant feld and the rotaton occurs n the drecton and seed ω. The moton of col s a subject of lnkng wth feld lnes and nducng feld E nsde the col. There are 4 waveforms n Fgure.3 that mght be ether flux or nduced voltage and your task s to select and motvate s flux and emf that corresond to the machne n Fgure.. The dmensons can be recalculated from the student dentfcaton number: n=[n n n 3 n 4 n 5 n 6 ] accordng to the formulaton n.7 (m-scrt). waveform, A (t) [-] A (t) relatve erod, T [-] waveform, A (t) [-] A (t) relatve erod, T [-] waveform, A 3 (t) [-] A 3 (t) relatve erod, T [-] waveform, A 4 (t) [-] A 4 (t) relatve erod, T [-] Fgure.3 Normalsed waveforms of a hyscal quantty A over a erod T.. Pck correct flux waveform () and motvate your choce (3).. Pck correct emf waveform () and motvate your choce (3).

4 .3 Frst order system Study the electromechancal resonse of a dc machne wthout consderng the knetcs of the system.e. the rotatonal seed Ω=0 rad/s. The SISO has voltage as an nut and electromagnetc torque as an outut. 3. Descrton of the system. Gve arametrc exressons and check t numercally by hel of Matlab. a. Block dagram {Struktuurskeem..4} reresentaton (and reducton). Draw structure block dagram accordng to Fgure. n smulnk and use In and Out blocks to secfy nut and outut and gve a name to the model e.g. AutRP mdl () b. Wrte (a sngle) dfferental equaton {Dfferentsaalvõrrand.}of the system () c. State varable model. Descrbe the system by state equaton {Olekuvõrrand.3}. (). Choose state varables. Exress tme dervatves of states n terms of states and nuts. Exress oututs n terms of states and nuts v. Use MATLAB command n work sace [A,B,C,D] = lnmod('autrp mdl') and comare the results d. Transfer functon {ülekandefunktsoon..} () 4. resonse of the system. Use MATLAB command n work sace [num,den] = lnmod('autrp mdl'); tf(num,den) and comare the results e. tme resonse durng t=0:e-4:0.05. ste resonse {ühkhüe.4.}. Use MATLAB command n work sace y=ste(num,den,t); lot the results and characterse the resonse (show the tme constant of the frst order ste resonse) (). ste resonse {ühkmulss.4.}. Use MATLAB command n work sace y=mulse(num,den,t); lot the results and characterse the resonse () f. frequency resonse. Bode dagram {Bode dagramm..3}. Use MATLAB command n work sace y=bode(num,den); lot the results and characterse the resonse (show the bandwdth of the frst order frequency resonse) (). Nyqust dagram {Nyqust dagramm 3...4}. Use MATLAB command n work sace y=nyqust(num,den); lot the results and characterse the resonse (show the tme constant of the frst order ste resonse) () 5. Estmate the stablty of the system g. Accordng to egenvalues of the system matrx A {tunnusvõrrand lahendd 3...}. Use MATLAB command n work sace eg(a) and characterse the results. () h. Stablty accordng to Nyqust and Bode dagrams {3...4 ja 3...5} ()

5 .4 Second order system Include knematcs to the revously derved system. 6. Reeat revous exercses (3, 4, 5) wth the renewed system (0). 7. How behaves the system f the back emf feedback E=m*Ω s not ncluded, how ths can be exlaned n the sense of hyscal energy converson rocesses (5). 8. Include load momentum of nerta Jload=0.0 kgm and characterse system behavour (5). Note that you can nclude more than sngle outut when studyng the dynamcs of the system e.g. one outut for the mechancal seed, the other for the machne current..5 Seed control Prevously an electrc drve system has been derved that nut voltage u(t) nfluences the outut Ω(t). The am of the control system s to acheve a desred resonse of a system and not just to affect the outut smlar to dsturbances or load varatons. The am here s to carry out the seed control for the electrcal drve The control loos n electrcal drve are usually always connected n cascade Cascaded control concet does only work f the bandwdth ncreases from the outer to the nner loos. It s assumed that torque control acton s erformed erfectly wthout affectng the desgn of the seed control loo. The system resonse wth PI controller has a good steady state erformance e.g. t can acheve the zero steady-state error. The control acton of a PI-controller bases on the control error e(t), whch s the dfference between the reference y*(t) and the measured value y(t). e () t = y * () t y() t The control outut s roortonal to the error and to the ntegral of the error. In the leftmost exresson of u(t) the controller gan can be changed wthout changng the tme constant. u = + dt T () t e() t + e() t dt = e() t e() t The breakont n Bode dagram /T wll be unchanged (shown from the exressons of transfer functons). From the rghtmost exressons can be seen that the gan and the breakont wll be affected by both and. u + st st () s = e() s = + e() s st By assumng that the torque source s nfntely fast the transfer functon of the closed seed controlled system can be wrtten Ω * Ω () s () s ( + st ) = s JT + s T + the oles for the system are (.3) (.4) (.5) (.6)

6 oles = ± J 4J JT (.7) Ths exresson shows that the oles of the system can be laced arbtrarly wth and T. Practcally, f s too large then the assumton that the torque loo s nfntely fast s not vald. The realstc assumton for the torque resonse corresonds to a low ass flter wth a tme constant equal to the samlng erod n the torque control loo. () s = () s + stc T (.8) * T By consderng the realstc aroxmate resonse of the torque controlled electrcal machne, the oen loo transfer functon for the system can be wrtten W + st ω = (.9) st + st sj ( j ) c Your assgnment s to draw an oen loo block dagram and to study frequency resonse. Fgure.4 oen loo of a seed controlled electrcal drve 9. Draw the block dagram shown Fgure.4, select Tc=0.00 seconds, T>Tc, = and study the frequency resonse from Bode dagram (4). 0. Wrte on Bode dagram the breakng frequences /T, /Tc and the frequency ω0 = where the maxmum hase aears (4). T T c.6 Symmetrc otmum The PI controller arameters are selected accordng to symmetrc otmum crteron. An dea behnd that s to select the cross-over frequency n the ont where the hase margnal s the largest. Ths wll gve the best damng for the closed system. The condton W(jω) = at ω 0 gves the frst equaton: T = a T where a > as T > T c c (.0) By relacng Tc n eq.0 wth T accordng to eq. gves the next relaton a J = (.) T In accordance wth the closed seed loo Ω * Ω () s () s c ( + st ) = 3 s JT T + s JT + s T + the characterstc equaton can be wrtten (.)

7 3 T a s + s T + + = 0 sa (.3) a T where one root s T a s = ω 0 = (.4) Polynomal dvson of the characterstc olynomal (eg.4) gves T a s + s T + = 0 (.5) a a where the other roots can be calculated analytcally. We assume that they are comlex conjugates, where ( ξ ± ) s (.6),3 = ω0 ξ ξ = a (.7) s seen as the relatve damng for the oles to the system. The arameter T can now be selected to lace the oles n a sutable way. If the relatve damng s selected to ξ =, then a=.4. Ths gves drectly value on T as a functon of Tc. If we on the other hand assume that no comlex oles are to exst n the seed control system then ξ = and a=3. All three oles wll thus be equal to - ω 0. Your task s to choose PI seed controller arameters accordng to symmetrc otmum crteron.. Select PI controller arameters accordng to Tc=0.00 sec and a=3. Analyse the system wthout and wth load momentum of nerta Jload=0.0 kgm and characterse system behavour accordng to tme resonse and frequency resonse (8).. Use dfferent and T n the PI regulator and characterse system behavour and stablty accordng to tme resonse and frequency resonse (0). 3. Study the torque resonse wth the new nerta and estmate the motor current accordng to your motor (4). 4. Add a low ass flter nto the seed feedback as n ractcal cases t s consdered that the seed sensor s subjected to nose that needs to be fltered out to some extent. As you see the system becomes 4-th order system. In order to solve the roblem the flter tme constant s selected at least one order of magntude longer comared to torque loo. Thereafter the flter tme constant wll be bass to desgn PI seed controller accordng to symmetrc otmum. Snce the torque dynamcs s one order of magntude faster than the flter dynamcs t can be omtted. Reeat analyss accordng to onts and (0).

8 .7 M-fle Ths s the m-fle that you should use n your home assgnment n order to get the ntal data for an electrcal machne. Take an advantage of the scrt to carry out effcently all the assgnments! % Home assgnment on CAD of Automatc Control Systems % machne descrton % the ermanent magnet machne s assumed to have rotor dameter of D, % stator dameter of D and machne length l. Magnetc ga s D/4, wndng % heght D/4 and fll factor over the crcular cross-secton s 0.5. % suly voltage V, ga flux densty s assumed 0.8T and allowed current % densty 3A/mm wthn the thermal lmt. Mass densty of the machne s % 7000 kg/m3 rjuta sa oma martkl number jättes numbrte vahele tühkud % arametersaton n = [ ]; % student dentfcaton number mu0 = 4**e-7; D = (n(6) + 0)*e-3; l = (n(5) + 40)*e-3; Un = ; n = 4000; Ph = 0.5**D*l * 0.8; [Vs] % magnetc ermeablty n vacuum % rotor dameter [m] % length of machne [m] % suly voltage [V] % nomnal seed [rm] % maxmum magnetc Bga=0.8T N = 0.5*((D-D/4)^-(D/)^)* * 0.5*3e+6; % magnetomotve Jm=3e+6A/m [Aturns] dph= Ph/(/); Tn = N*dPh; Pn = Tn*n/60**; N = cel(un/(dph*n/60**)); In = N/N; m = dph*n; M = *D^*l * 7000; % dph/dtheta [V/(rad/s)] % electromagnetc torque [Nm] % electromagnetc ower [W] % number of turns [turns] % rated current [A] % machne constant [Nm/A=Vs/rad] % mass of the machne [kg] R =.4e-8**(l+.6*D)/(0.5*((D-D/4)^-(D/)^)*)*N^; % resstance [Ohm] L = /(/mu0*d/4/(0.5**d*l))*n^; % armature nductance [H]

9 J = /4*7000*l**(D/)^*((D/)^); % rotor momentum of nerta [kg*m^] Dm = J/0.*0; % mechancal damng %J=J+0.0; Tc=0.00; % torque resonse tme, sec Tflt=0.05; % seed flter tme constant, sec %Tflt=Tc; % PI seed controller arameters a=3; % otmum control Tw=a^*Tflt; % ntegral tme constant w=j*a/tw; % roortonal gan w=/tw; % ntegral gan % Smulatons %oen('autrp_a_mdl.mdl') [A,B,C,D] = lnmod('autrp mdl'); [num,den]=lnmod('autrp mdl'); t=0:e-4:0.05; fgure(); clf; scrsz = get(0,'screensze'); set(gcf,'poston',[5 scrsz(4)/0 scrsz(3)-0 scrsz(4)/],'color',[ ],'Name','Tme Doman'); % ste resonse y=ste(num,den,t); %y=y./(ones(length(y),)*max(y)); sublot(,,); hold on; grd on; lot(t,y); ttle(['ste resonse']); xlabel('tme, t [sec]'); ylabel('outut'); % mulse resonse y=mulse(num,den,t); sublot(,,); hold on; grd on; lot(t,y) ttle(['mulse resonse']); xlabel('tme, t [sec]'); ylabel('outut'); fgure(66); nyqust(num,den) fgure(67); bode(num,den)