Lecture : Feedback Linearization
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1 ecture : Feedbac inearization Niola Misovic, dipl ing and Pro Zoran Vuic June 29 Summary: This document ollows the lectures on eedbac linearization tought at the University o Zagreb, Faculty o Electrical Engineering and Computing Motivation The dynamic equation that describes the water level h(t) in a tan with an opening at the bottom (cross section is a) is shown with () The assumption is that the tan has the same cross section A throughout the whole height Input water low is denoted with q(t) ḣ (t) A q (t) a A 2gh(t) () The aim is not to linearize the system, which would give approximate dynamics around the operating point, but we want to design a controller such that it linearizes the system completely, in the whole operating region We can resolve this by introducing a new control input v such that The system dynamics can now be written as q(t) a 2gh(t) + Av(t) ḣ(t) v(t) Since the system is completely linear, one can design a controller so that the system is guided to the desired value, eg by choosing v(t) K h des (t) h (t) the control closed loop dierential equation becomes ḣ (t) K h des (t) h (t) Kḣ (t) + h (t) h des (t) h(s) h des (s) K s+ or h (t) h des (t) as t The ollowing part o this lecture will describe how to ormalize the described approach, what to do with unstable dynamics, etc
2 2 Introduction In this part a short description o ie algebra on vector ields is described, together with an idea o relative degree in nonlinear systems 2 ie Algebra Deinition (ie derivatives) et h : R n R be a smooth scalar unction and : R n R be a smooth vector ield on R n, then the ie derivative o h with respect to is a scalar unction deined by h h The symbol is oten written as δ maing the deinition as ollows: δx h δh The ie δx derivative is interpreted as dierentiation o unction h in the direction o the vector Example 2 (ie derivative) I a homogenous nonlinear system is described as ẋ (x) y h(x) then the derivatives o the output can be represented using the ie derivatives: ẏ xẋ h h ÿ h ẋ 2 x h Deinition 3 (ie bracets) et and g be two vector ields on R n The ie bracet o and g is a third vector deined by, g g g The ie bracet, g is commonly written as ad g where ad stands or adjoint Repeated ie bracets can be deined recursively by Some properties o ie bracets: ad o g g ad i g, ad i g or i, 2, bilinearity sew comutativity α + α 2 2, g α, g + α 2 2, g, α g + α 2 g 2 α, g + α 2, g 2, g g, 2
3 Jacobi identity ad gh g h g h Example 4 (ie bracets) et ẋ (x) + g(x)u with two vector ields and g deined by 2x + ax 2 + sin x g(x) x 2 cosx cos(2x ) The ie bracet can be computed as 2x + ax, g 2 + sin(x ) 2 + cosx a 2 sin(x ) x 2 cosx x 2 sin x cosx cos(2x ) a cos(2x ) cosx cos(2x ) 2 sin(2x )( 2x + ax 2 + sin x ) 22 Relative Degree A SISO nonlinear system is given in a orm (2) ẋ (x) + g(x)u y h(x) I we observe the output derivatives we obtain the ollowing: ẏ xẋ h h (x) + g(x)u x h x (x) + h }{{} x g(x) u }{{} h gh h + ( g h) u et s assume that the irst derivative o the output is not inluenced by the input signal u, ie g h and ẏ h Then the next derivative o the output is ÿ h Induction leads us to the ollowing: d dtẏ d dt ( h) ( h) x ẋ ( + gu) x ( h) + g ( h) u 2 h + g ( h) u }{{} ẏ h ÿ 2 h y (r) r h + g ( r h ) u et s assume that the r th derivation o the output sees the input signal u Then the value r is called the relative degree o the system 3 (2)
4 Deinition 5 (Relative degree) The nonlinear system has relative degree r at the point x i ẋ (x) + g(x)u y h(x) g h(x) or all x in the neighbourhood o x and all < r g r h(x ) In other words, we should dierentiate the output until the input appears in the equation The derivative to which we have come is the degree o the system Note: Relative degree is deined only i there exists an output rom the system It also depends on the chosen output equation In linear systems the relative degree is the pole excess o the system, ie the dierence between the number o poles and the number o inite zeros o the transer unction o the systems I the relative degree o the linear system is less then n, that means that there exist inite zeros (which may be unstable) Since transer unctions do not exist in nonlinear systems, the act that the relative degree o the nonlinear system is smaller then the order o the system, means that there are some internal dynamics which cannot be seen rom the output Example 6 (Relative degree o the Van der Pool oscillator) Calculate the relative degree o the van der Pool oscillator given with x ẋ (x) + g(x)u 2 2ωζ ( µx 2 ) x 2 ω 2 + u x y h (x) x For starters, let s dierentiate the output g h g h h x g h x This one does not depend on the input!! h g h h g x x g 2 2ωζ ( µx 2 )x 2 ω 2 x g x 2 x g This one depends on the input so the relative degree is r 2 (we had to mae 2 calculations, ie we had to dierentiate the output twice to get to the input signal) 4
5 I the output were deined as: then the ollowing calculations would tae place y h (x) sin (x 2 ), g h gh h x g h x cos (x ) h cos (x ) so the relative degree is at all points apart rom x π (2 + ) At these points, relative degree is 2 not deined From this example it is obvious that the relative degree depends on the chosen output, just as the number o zeros in the linear system depend on the chosen output 3 Input/Output inearization We have shown beore that the nonlinear system (2) can be transormed to a orm y h ẏ h ÿ 2 h y (r) r h + ( g r h ) u I we introduce α (x) r h and β (x) ( g r h ) then y (r) α (x) + β (x) u Since the input signal can be whatever we want it to be, let s choose it in such a way that the nonlinearities in the system are compensated or, ie y (r) α (x) + β (x) u v rom where it ollows that v α (x) + β (x) u u α (x) + v β(x) Graphically, this is what happened: Another way o thining is that we have deined a new state space vector, which consists o the derivatives o the output: z y h z 2 ẏ h z r y (r ) r h z r+ y (r) α + βu v 5
6 So the system can be represented in a state space orm as ż z ż 2 z 2 + v ż r z r z y z 2 which is in act a cascade o r integrators Now we can design any type o controller or the r integrator linear system The most appropriate one would be a state controller What was the transormation between the two state spaces, x and z? z Φ (x) Algorithm or input/output linearization: z r ϕ (x) ϕ 2 (x) ϕ r (x) Find the relative degree r o the nonlinear system g h(x) h h r g r h(x ) h 2 Mae the r state transormations z Φ (x) ϕ (x) ϕ 2 (x) ϕ r (x) 3 Deine the new input so that the states are linearized h h r h α (x) r h β (x) g ( r h ) u α (x) + v β (x) 6
7 Example 7 (r n) Find the relative degree ẋ g h gh h x g h x g h g h h x g x + x 2 2 x x 2 y h(x) x 3 h x 3 e x 2 e x 2 + e x 2 e x 2 u x + x 2 2 x x 2 g x x 2 g e x 2 e x 2 g 2 h h g x g x x 2 g x + x 2 2 x x 2 g x x 2 2 2x 2 e x 2 e x 2 e x 2 ( + 2x 2 ) Relative degree is 3 i + 2x 2, it is the same as the order o the nonlinear system Mae the r state transormations z ϕ (x) h h x 3 z 2 ϕ 2 (x) h x + x 2 2 x x 2 z 3 ϕ 3 (x) 2 h h Deine the new input so that the states are linearized α (x) 3 h 2 x x 2 2 β (x) g ( 2 h ) g x x 2 2 x x 2 2x2 2x2 7 x + x 2 2 x x 2 x + x 2 2 x x 2 e x 2 e x 2 x x 2 2 ( ) 2x 2 x + x 2 2 ( + 2x 2 )e x 2
8 u β (x) α (x) + v 2x 2 (x + x 2 2 ) ( + 2x 2 ) e x 2 ( + 2x 2 ) e x 2 v By introducing this control, we have obtained a system with the ollowing dynamics ż z ż 2 z 2 + v ż 3 z 3 y z z 2 z 3 and now we can design whatever type o control we desire et s tae a loo at an example where relative degree is smaller than the order o the system (r n) Example 8 (r < n THIS EXAMPE IS WRONG REGARDING STABIITY - NEEDS REVI- SION) ẋ x 2 x 2 ẋ 2 3x 2 + u y h(x) 2x x 2 The state-space orm is then ẋ x 2 x 2 + ẋ 2 3x 2 y h(x) 2x x 2 u Find the relative degree g h gh h x g h x h 2 Relative degree is That means that we have state transormation Notice that we have a second order system!! Mae the r state transormations The second state stays the same z ϕ (x) h h 2x x 2 Deine the new input so that the states are linearized α (x) h h x 2 2x 2 x 2 3x 2 8
9 β (x) g ( h ) g h u α (x) + v β (x) Now the system o nonlinear equations is transormed to ż v ẋ 2 3x 2 + u 3x 2 2x 2 x 2 3x 2 v 2x 2 x 2 v et s observe the zero dynamics This is done so that the output is ept and the behavior o the system is observed I the output is zero y 2x x 2 x 2 2x which means that ż v ẋ 2 2x 3 2 v what is a stable system This means that the zero dynamics are stable et s tae a loo at another example where the zero dynamics are not stable Example 9 (r < n, unstable zero dynamics) ẋ x 2 + x 3 + u ẋ 2 u y h(x) x The state space orm is then ẋ ẋ 2 y h(x) x x2 + x 3 + u Find the relative degree g h g h h x g h x h Relative degree is That means that we have state transormation Notice that we have a second order system!! Mae the r state stransormations z ϕ (x) h h x The second state stays the same Deine the new input so that the states are linearized α (x) h h x x2 + x 3 x 2 + x 3 9
10 β (x) g ( h ) g h u β (x) α (x) + v x 2 x 3 + v Now the system o nonlinear equations is transormed as ż v ẋ 2 x 2 + x 3 v Not only is it not linearized, but let s say that we want to design a proportional controller so that the output goes to, ie v K p (y y des ) K p y The closed loop system is then ż v K p z ẋ 2 x 2 + z 3 + K p z I y has to go to, that means that z has to go to zero I that one goes to zero, that means that x 2 is governed by ẋ 2 x 2 which is unstable!!!!! Obviously we have some internal dynamics which are unstable The reason why we didn t notice these internal dynamics is the act that the relative degree is smaller than the degree o the nonlinear system There is something else, some other states that we orgot about - n r states The internal dynamics might not be unstable, maybe it would be enough or us to linearize the r states o the system (which are based on the output) and the rest would have been stable But we cannot be sure Bare in mind that we perormed only the transormation to the r states The question is: can we derive such a transormation so that the relative degree o the system is n? This way we could include all o the dynamics in the design procedure What should this transormation loo lie? What are the constraints? Is this transormation even possible? What i the relative degree o the system is not equal to the degree o the nonlinear system, ie r n? This means that we have the ollowing situation h (x) z z r ξ r+ ξ n 4 Input/State inearization Φ (x) Ψ (x) r h (x) ψ r+ (x) ψ n (x) et s rephrase the problem I a nonlinear system is given with ẋ (x) + g(x)u Find a transormation ξ Ψ (x) so that the system in the new coordinates can be linearized and so that it is controllable can be linearized This will be possible i we ind an output y λ (x) such that the system has relative degree n at some x How to ind a suitable λ(x)? Just as in the case when the output was available, this unction has to satisy the ollowing properties:
11 g λ (x) g λ (x) g n 2 λ (x) (all the terms next to the input u have to disappear) g n λ (x ) It can be shown by the use o the Jacobi identity that these conditions can be rewritten as λ (x) g λ (x) ad g λ (x) ad n 2 g λ (x) ad n g or λ (x) g ad g ad n 2 g λ (x) ad n g These equations are much more appropriate since the unnown unction λ(x) has been separated rom the rest o the calculations I we manage to ind this unction, the state transormation will ollow as ξ ξ n Ψ (x) ψ (x) ψ n (x) λ (x) n λ (x) How do we now that there even exists a transormation such that the relative degree is n? It is the same as asing: how do we now that the previously stated partial dierential equations have a solution? The Frobenius theorem tells us i the solution exists Theorem (Frobenius) The nonlinear system with (x) and g(x) being smooth vector ields, is input-state linearized i and only i there exists a region Ω such that the ollowing conditions hold: the vector ields {g, ad g,, ad n g} are linearly independent in Ω (the matrix has to have ull ran) 2 the set {g, ad g,, ad n 2 g} is involutive in Ω, Note: The irst condition can be interpreted as controllability condition or the nonlinear system The involutivity condition is less intuitive It is trivially satisied or linear systems, but not generally satisied in the nonlinear case
12 Deinition (Involutivity) A linear independent set o vectors,, m is involutive i i, j m α ij (x) (x), ie i you tae ie bracets you don t generate new vectors The input-state linearization can be perormed through the ollowing steps: Construct the vector ields g, ad g,, ad n g or the given system 2 Chec whether the controllability and involutivity conditions are satisied 3 I both are satisied, ind the output unction λ(x) (the output unction which will lead to the input-output linearization o the relative degree n) by using λ (x) g ad g ad n 2 g 4 Compute the state transormation ξ ξ n Ψ (x) and the input transormation with λ (x)ad n g ψ (x) ψ n (x) α (x) n λ(x) ( β (x) g n λ(x) ) u α (x) + v β(x) λ (x) n λ (x) Example 2 Consider a mechanism given by the dynamics which represents a single lin lexible joint robot Its equations o motion is derived as I q + Mg sin q + (q q 2 ) J q 2 (q q 2 ) u Because nonlinearities (due to gravitational torques) appear in the irst equation, while the control input u enters only in the second equation, there is no easy way to design a large range controller q x 2 x q q 2, Mg sin x l l (x x 3 ) x 4, g q 2 (x J x 3 ) J 2
13 ad g, g g g x x Mg cos x I I I J J ad 2 g, ad g ad g ad x x g Mg cos x I I I J J ad 3 g, ad 2 g ad2 g x x ad2 g Mg cos x I I I J J g ad g ad 2 g ad 3 g J J JI J 2 IJ IJ J J 2 J J 2 J JI J 2 It has ran 4 Furthermore, since the above vector ields are constant, they orm an involutive set Thereore the system is input-state linearizable et us ind the state-transormation z Φ(x) and the input transormation u α(x)+β(x)v so that input-state linearization is achieved λ (x) g x λ (x) ad g x λ (x) ad 2 g x λ (x) ad 3 g x x 3 x 3 x 3 3 x 3 x 4 g x 4 x 4 x 4 x 4 J IJ J 2 IJ J 2 JI J 2 x 3 x
14 It is obvious that lambda should be a unction o x only Thereore, we choose a unction which is a dieomorphism, eg λ (x) x The other states are obtained rom the ollowing calculations: The input transormation is then α(x) 4 λ (x) z 4 z 2 λ z x 2 z 3 2 λ z 2 Mg sin x I (x I x 3 ) z 4 3 λ z 3 Mg x I 2 cosx (x I 2 x 4 ) Mg I x 2 sin x Mg I cos x I I Mg x 2 I 2 sin x ( )( Mg cosx I Mg I I x 2 Mg sin x I (x I x 3 ) x 4 (x J x 3 ) sin x + (x I x 3 ) ) + 2 (x IJ x 3 ) β(x) g 3 λ (x) g z 4 Mg I x 2 sin x Mg I cosx I I with u (v α) β We end up with the ollowing set o linear equations ż z 2 ż 2 z 3 ż 3 z 4 ż 4 v thus completing the input/output linearization Example 3 ẋ x 3 ( + x 2 ) x x 2 ( + x ) + + x 2 u x 3 J IJ First to see i it is possible to ind a unction λ so that the system can be input/state linearized We have to ind g ad g ad 2 g 4
15 ad g g x x x 3 ( + x 2 ) x x 2 ( + x ) x x 2 ( + x ) ( + x 2 ) ( + x ) x 3 + x 2 x 2 + x + x 2 x 3 x ( + x ) ( + 2x 2 ) ad 2 g, ad g (ad g) x x g) x 3 ( + x 2 ) x ( + 2x 2 ) 2 ( + x ) x 2 ( + x ) x 3 + x 2 x x 2 + x ( + x )( + 2x 2 ) x 3 ( + x 2 ) x 3 ( + x 2 )( + 2x 2 ) 2x ( + x ) x x 3 + ( + x 2 )( + x )( + 2x 2 ) x 3 ( + x 2 ) x 3 ( + x 2 )( + 2x 2 ) 3x ( + x ) et s observe the necessary conditions around the equilibrium point x g ad g ad 2 g x x x 3 ( + x 2 ) ( + x ) ( + 2x 2 ) x ( + x ) The ran is ull, so the irst condition is satisied The second condition is involutivity g, ad g ad g g g ad x x g ( + 2x 2 ) 2 ( + x ) x ( + x )( + 2x 2 ) 2 ( + x ) ( + x 2 ) x 3 ( + x ) ( + 2x 2 ) 5 + x 2 x 3 x ( + x )( + 2x 2 )
16 The matrix g ad g g, ad g has ran 2 which means that the vectors that or it are linearly dependant That means that the third column is not a new vector but a linear combination rom the irst two columns This means that the desired distribution is involutive The desired unction λ(x) is then calculated rom and λ (x) g ad g x x 3 ad n 2 + x 2 x 3 g x ( + x ) ( + 2x 2 ) λ (x) ad n g This can be done with λ(x) x which is a dieomorphism 6
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