Previously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system

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1 Advnced contol nd utomtion Extensions to bckstepping contolle designs Tcking Obseves (nonline dmping) Peviously Lst lectue we looked t designing nonline contolles using the bckstepping technique fo the egultion poblem We will now conside how to djust the bckstepping design fo tcking contolles, the thn set point egultion Finlly we will look t obseves in nonline systems, nd how they diffe fom line systems Best to use n exmple bsed ppoch Tcking using bckstepping Suppose we conside the genel system = f( x) + x = f ( x,..., x ) + x i i i i i+ = f ( x,..., x ) + bu n n n y = x How do we get y to tck some efeence signl y? Fist mke the ssumption tht the fist n deivtives of y e vilble, bounded nd continuous, Now since we wnt x y the thn 0, we should set the fist vitul stte to be z = x y We will need to modify the highe vitul sttes similly fom the egultion cse, ie fo i =..n z x y () i i = i i The poblem is now effectively egultion poblem fo the vitul sttes 3 Tcking exmple Lets go bck to the pendulum exmple = x = sin x x y = x Solution: Lets stt by defining the two vitul sttes z = x y z = ( ) = z + + y z = x We cn now define clf fo the fist vitul stte V = z = z z = z z + The fist stte epesents line system! So lets choose the vitul contol = kz 4

2 We cn now define clf fo the totl system V = k z + z z + z z = kz + z z + z z x y So lets choose the ctul contol u = kz z+ sin x+ x + y k z kz = sin x + x + y k + k z + k z = ( ) z = sin x x k z y ( sin ) = k z + z z x x + k z k z y Compe to feedbck lineistion contol lw fom lectues go! Homewok: show the two contol lws e identicl Expected since no good nonlineites in system, ie we just used bckstepping to lineise system nywy! Plotting the eo s function of time fo diffeent gins: 5 Tcking exmple Lets conside n bity nonline exmple (ok I couldn t find n esy el-wold one to do concisely on this slide) = x + x = x y = x Solution: Agin stt by defining the two vitul sttes z = x y z = x z = y = x + z + + y ( ) We cn now define clf fo the fist vitul stte V = z Cnt do nything bout this yet, keep it until next clf = zz = z x+ z+ + y So lets choose the vitul contol = ( kz x ) 6 We cn now define clf fo the totl system = kz+ zz+ zz = kz + z( z+ z) Whee z = x z = ( x ) ( kz x y) y = ( x ) ( ( k+ x) + ky y) y 3 = ( x ) ( ( k+ x)( x + x) + ky ) y So in ode to mke = kz kz 3 Lets choose u = k ( ( z z x + k+ x x + x) + ky ) + y Even fo eltively simple looking nonline system we cn end up with fily complex looking contol lw! Obseves in nonline systems Conside the following system 4 = x+ x + xx = kx k =k =0. Ref = cos(t) Eo 0 exponentilly fst! x (0) = x (0) =0. If we let = x x 7 then = k Looks good so f 8 We know how to design contolle vi the bckstepping ppoch tht will povide stbilistion of the system But wht hppens if we cnt mesue the stte x? In the line systems we looked t lst semeste the nswe ws esy Replicte the unknown stte equtions fo n obseve Use the obseve in the contol lw insted of the ctul stte So: An obseve in ou nonline system would look like: x = kx

3 We hve n exponentilly conveging estimto eo (s in line systems) so lets use the estimted vlue fo ou vitul contol design (s in line systems!) Fist step, lets ewite the st stte eqution in tems of 4 the obseve: = x+ x + xx + x (*) x = kx = k To pply the bckstepping pocedue to (*), sensible design choice fo the clf, vitul stte nd contol might be V = x ( x) = x z = x ( x) = x + x And to find the contol, we ugment the clf z = kx + x x + x 4 + x x + x V V z We cnt do nything bout the stte estimtion eo (nd its conveging nywy) so choose u = cz x 3 + kx x x + x 4 + x x = + 3 ( 4 = x + z x kx + x x+ x + xx + x ) 9 And hence the closed loop system equtions become This tem my cuse = x+ xz + x 3 3 divegence if x lge nd z = cz x + x positive = k Lets conside the cse whee z=0 (ie the vitul contol is exctly equl to the contol we wnt to give), the system equtions educe to x = kx () t = (0) e kt = x + x = x + x e (0) kt If we mke the chnge of vibles w = /x, the nd eqution becomes kt w = = w (0) e x Solving this eqution fo w: (0) t (0) kt wt () = w(0) e + e k + k + x (0)( k + ) x () t = k + x (0) (0) e t + x (0) (0) e [ ] kt 0 x () t = x (0)( k + ) [ + ] t + k x (0) x (0) e x (0) (0) e This hs seious implictions if k + < x(0) (0) In this cse we will hve x (t) eching infinity in finite mount of time!!! The system is clely not stbilised! Whee did it ll go wong? The ssumption tht we could use n obseve in exctly the sme wy we did fo line system is not lwys sfe ssumption fo nonline system! So does men we cnt use obseves t ll in nonline system? Not vey pcticl! Thee is solution to this poblem nonline dmping kt Nonline dmping Suppose we hve system in the following fom = f( x) + g( x) u+ h( x) Uncetinty Now suppose the system without uncetinty = f( x) + g( x) u is stbilisble though the use of feedbck contol u = (x). Ie thee exists clf V(x) fo the bove system with =0 (x) ensues (x)/dt < 0 Fo the uncetin system, modifying the stte feedbck contol in the mnne Often efeed u = ( x) k g( x) h( x) k > 0 to s xs(x) guntees globl boundedness of x(t) nd convegence to set ound the oigin (whose size is ntully dependent on the mximum vlue of the uncetinty) This is clled Nonline Dmping

4 Nonline dmping poof We hve system in the following fom = f( x) + g( x) ( u+ h( x) ) And thee exists clf, V, fo the system = f( x) + g( x) ie ( f ( x ) + g ( x ) ) W ( x ) u = ( x) k g( x) h( x) Fo the uncetin system, tking the deivtive of the clf gives = ( f gu) gh + + = ( f + g ) k h + gh Using esult fom bove: W( x) k h + gh Using uppe bound on : W( x) k h + g h Completing the sque: = W( x) k g h + k 4k Convegence gunteed W( x) + 4k when RHS < 0 3 Nonline dmping with vitul contols The nonline dmping theoem is eqully pplicble to systems with vitul contols In this cse, insted of using the stbilising vitul contol we pedict fo system without uncetinty, we should modify it to contin nonline dmping tem Lets we evisit the fist eqution (*) of ou pevious exmple 4 = x + x + x x + x The vitul contol we hd initilly chosen ws ( x ) = x Lets now dd on nonline dmping tem V = x x = g( x ) = x Choose ( x ) = x k x hx ( ) = 3 4 ( x ) = x k x 3 Using the sme vitul stte ppoch z = x ( x) we get the fist closed loop system eqution becomes 5 This tem is now = x kx + xz + x d outweighed by the nd z = x ( x ) = kx x nonline dmping tem! The deivtive of the clf, V=0.5x, thus becomes = x = x kx + x z+ x When we ugment this with the second stte V 0.5 g h = V + z 4 3 d 4 d ( = x k x + z x kx +. u x+ x + xx) x If thee ws no uncetinty in the system, we could choose 3 d 4 ( x, z) = k z x + kx + ( x + x + xx) Wont expnd this tem hee due to spce esons 5 3 d 4 ( x, z) = kz x + kx + ( x+ x + xx) Howeve the pesence of the uncetinty mens we need to include nonline dmping tem. We hve z dz = d g = k g h = kz x dz d h= x Hence the ctul stbilising contol fo the uncetin system is given by u = ( x, z) k g h dz 3 d 4 d = kz x + kx + ( x + x + xx ) kz x Cn you see why contol enginees like line systems? 6

5 A futhe step (possible) Summy A genel ule of thumb fo Lypunov stbility poblems is we ty to include nything in the clf tht we wnt to mke 0 So we could even tke the pevious exmple poblem one step futhe Becuse we wnt the estimto eo to ppoch 0, long with the stte, x, nd the vitul stte, z, it would mke sense to include it in futhe ugmented clf This won t chnge the contol lw we hve developed But it would llow fo moe ccute bounds of pefomnce to be scetined We will leve tht fo nothe couse 7 We consideed two bckstepping poblems tody which incese the nge of el-wold poblems to which we cn pply ou newly lent contol technique. Tcking Esy, fo the n th vitul stte just subtct the n th deivtive of the desied output fom the stte eqution we would use fo egultion Uncetin systems We sw ignoing n uncetinty cn hve dmtic consequences in nonline system This is pticully impotnt fo systems equiing obseves In line ppoches the uncetinty is genelly ignoed We looked t solution to this poblem, which ws to dd nonline dmping tem to the contol Incesed the complexity of the contol! 8