Feedback Linearization Lectures delivered at IIT-Kanpur, TEQIP program, September 2016.

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1 Feedback Linearization Lectures delivered at IIT-Kanpur, TEQIP program, September 216 Ravi N Banavar banavar@iitbacin September 24, 216 These notes are based on my readings o the two books Nonlinear Control Systems, A Isidori, 3rd Edition, Springer, 1992 Nonlinear Systems, H Khalil, Prentice Hall, 22 Consider a dynamical system o the orm ẋ = (x) + g(x)u x(t) U R n (1) where ( ), g( ) are smooth unctions and u takes values in a subset o R The vector (x) is called the drit vector ield and the vectors g i (x) are called the control vector ields Now let us examine the behaviour o the system in a region U R containing a point x in the state-space (x U) Associated with system (1) let us deine the output as y = h(x) The irst derivative o the output is expressed as y (1) = L h(x) + L g h(x)u where L and L g denote the Lie derivatives along the vector ields and g respectively and (1) denotes the irst derivative with respect to time Let us assume that L g h(x) is zero in U and urther, that L g L k h(x) = or k =, 1,, r 2(r n) x U and L g L r 1 h(x )

2 The ollowing two equalities are equivalent L g h(x) = L g L 1 h(x) = = L g L k h(x) = L g h(x) = L [,g] h(x) = = L ad k gh(x) = x U x U where ad k g = [,, [, g]]] (k-brackets) Proo: Let us assume that the irst equality is true The proo is based on an iteration From L g h(x) = and L g L h(x) = we have Similarly, L ad gh(x) = L [,g] h(x) = [L L g L g L ]h(x) = [L L g h(x) L g L h(x)] = L ad 2 gh(x) = L [,[,g]] h(x) = [L L [,g] L [,g] L ]h(x) = L L ad gh(x) L ad gl h(x) = L ad gl h(x) = [L L g L g L ]L h(x) = Based on the above arguments, it ollows that Going on this way, we have L ad k gl l h(x) = i il g L k+l h(x) = L g h(x) = L [,g] h(x) = = L ad k gh(x) = x U For a system with relative degree r at x, the r row vectors dh(x ), d(l h(x )),, d(l r 1 h(x )) are linearly independent Proo: ( Note that < dl j h(x), adi g(x) >= L ad i g(x)l j h(x) and L ad k gl l h(x ) k + l = r 1 ) Consider the product dh(x ) d(l h(x )) d(l r 1 h(x )) [ g(x ) ad g(x ) ad r 1 g(x ) ]

3 < dh(x ), ad r 1 g(x ) > < d(l = h(x )), ad r 1 g(x ) > < d(l r 1 h(x )), g(x ) > < d(l r 1 h(x )), ad r 1 g(x ) > In the irst row, < dh(x ), ad r 1 g(x ) > Consider the second row The elements are,,, < d(l h(x )), ad r 2 g(x ) >, < d(l h(x )), ad r 1 g(x ) > =,,, L ad r 2 g L h(x ), L ad r 1 g L h(x ) The last but one (r 1th) element o the row is non-zero Similarly the third row has its last but second (r 2th) element to be non-zero Thus the (r i + 1)th element o row i is non-zero and the matrix is lower triangular Hence the matrix has rank r From the result on the product o two linear transormations (or matrices) Rank(AB) = minimum{rank(a), Rank(B)} it ollows that the two matrices on the LHS have atleast rank r The purpose o the lemma is that it helps us to deine a new set o coordinates in a neighbourhood around x as z = Φ(x) = h(x) L h(x) L r 1 h(x) φ 1 (x) φ n r (x) where the φ i s are n r smooth unctions such that Φ( ) is a dieomorphism and the jacobian Φ(x) x x=x has ull rank The system evolution in the new set o coordinates is ż = d dt ż = d dt Φ(x) Φ(x) = ((x) + g(x)u) x Φ(x) Φ(x) = x x=φ 1 (z)((φ 1 (z)) + g(φ 1 (z))u)

4 Question: Why do we have the requirements - 1 that Φ be a dieomorphism? 2 that the Jacobian have ull rank at x? What would happen i these were not satisied? Let us now deine the dynamics o the system in this new set o coordinates ż 1 = d dt h(x) = L h(x) = z 2 ż 2 = d dt L h(x) = L 2 h(x) = z 3 and ż r = d dt Lr 1 h(x) = L r h(x) + L g L r 1 h(x)u = b(z) + a(z)u where a(z) = L g L r 1 h(φ 1 (z)) and b(z) = L r h(φ 1 (z)) I the φ i s are chosen such that L g φ i (x) = (this is called the normal orm), and d dt φ i(x) = L φ i (x) = q i (z) then the system in the new coordinates looks like ż 1 ż r ż r+1 ż n By deining a new control input v as z 2 b(z) + a(z)u = q r+1 (z) q n (z) v = b(z) + a(z)u we notice that the sub-system rom v to z is a linear system By means o the eedback u = 1 (v b(z)) a(z) this is state-eedback the nonlinear system has been partially linearized as ż 1 ż r = 1 z 1 z r + 1 v ż r+1 ż n = q r+1 (z) q n (z) I the relative degree r = n (order o the system), then the original nonlinear system can be ully eedback linearized

5 Zero dynamics Deining ξ = z 1 z 2 z r η = z r+1 z n the system can be written as ż 1 ż r = z 2 b(ξ, η) + a(ξ, η)u η = q(ξ, η) Suppose the output y = z 1 is identically zero Then it ollows that z 1 (t) z 2 (t) = z r (t) ( ξ ) and ż r = b(, η) + a(, η)u The input history u(t) is then given by b(, η(t)) u(t) = a(, η(t)) where η(t) is the solution o the dierential equation η = q(, η) η() = η The dynamics o the above equation are called the zero dynamics o the system Zero dynamics o a linear system Consider the transer unction G(s) = b + b 1 s + s n r a + a 1 s + s n The state-space realization in a canonical orm is 1 A = B = a a 1 a n 1 1 [ C = ] b b 1 b n r 1 1

6 Following a similar procedure as beore z 1 = y = b x 1 + b 1 x b n r 1 x n r + x n r+1 z 2 = ż 1 = b x 2 + b 1 x b n r 1 x n r+1 + x n r+2 z r = ż r 1 = b x r + b 1 x r b n r 1 x n 1 + x n To ensure that the rank o the jacobian is n, choose the other n r coordinates as z r+1 = x 1 z r+2 = x 2 z n = x n r (Check on the rank o the jacobian) As beore, partition the new set o variables into ξ = (z 1,, z r ) and η = (z r+1,, z n ) The dynamics are ż 1 = z 2 ż r 1 = z r ż r = Rξ + Sη + u η = P ξ + Qη I we identically zero the output (ξ ), then η = Qη where Q = 1 b b 1 b n r 1 The eigen values o Q correspond to the zeros o the transer unction! Remark 1 Please note that zero dynamics is a concept related to identically zeroing the output with a particular control The reason we are concerned about the hidden part (the η part) is that we do not wish that these variables get unbounded when the output is identically zero Hence we insist on asymptotic stability o this subsystem So whatever the initial conditon o η, this variable then decays to zero As seen rom above, the same concepts apply to a linear system Though one can draw comparisons to observability/detectability, please remember that here we are using an input We are not studying an output due to initial conditions alone In the situation o the output identically being zero, as we have seen, the ollowing identities hold y(t) = h(x(t)) y (1) (t) = L h(x(t)) y (r 1) (t) = L r 1 h(x(t))

7 and y (r) (t) = L r h(x(t)) + L g L r 1 h(x(t))u(t) (Note: Each independent algebraic constraint on the system dynamics reduces the dimension o the maniold on which the system trajectories evolve by 1) We recall that the r row vectors dh(x ), d(l h(x )),, d(l r 1 h(x )) are linearly independent and this holds in a region around x (since the unctions are smooth) and hence the zero-dynamics trajectory lies in an n r dimensional submaniold (Z)o the original n-dimensional maniold The control input required to realize the output zeroing is and we have u (x(t)) = Lr h(x(t)) L g L r 1 h(x(t)) h(x) dh(x) (L h(x)) d(l h(x)) [(x) + g(x)u (x)] = d(l r 1 L r 1 h(x) h(x)) For all x Z, the output is zero and hence dh(x) d(l h(x)) [(x) + g(x)u (x)] = d(l r 1 h(x)) which implies that the vector ield [(x) + g(x)u (x)] is tangential to Z The system trajectories or a zero output thus evolve on Z - the n r dimensional maniold Local asymptotic stability Zero dynamics are helpul in determining the local asymptotic stability o a system around a critical point (when the linearization contains eigen values on the imaginary axis which are not controllable) Proposition 1 Consider the system (1) with an equilibrium x = and where the relative degree is r Without loss o generality assume that (ξ, η) = (, ) is the equilibrium point in the transormed coordinates Suppose the equilibrium η = o the zero dynamics o the system is locally asymptotically stable and the eedback linearized subsystem is stable, then the system with the eedback control law is locally asymptotically stable at (ξ, η) = (, ) Proo:

8 Integrable distributions Consider a non-singular smooth distribution W = span{ 1,, r } o rank r(< n) in an open set U R n This implies that at every point p U, the r vectors { 1 (p),, r (p)} span an r-dimensional subspace W (p) o R n From basic vector space theory, the annihilator o W (p), denoted by W o (p), orms an (n r) dimensional subspace o R n So i y W o (p), then [y, i (p)] = i = 1,, r Now, as a consequence, we can ind (n r) smooth unctions {α 1,, α n r } such that [α j, i ](x) = i = 1,, r, j = 1,, n r and x U So, when can we say that the distribution W is integrable? I the annihilator co-vector unctions α i s can be expressed as gradients o smooth unctions λ i s, then the distribution is said completely integrable in U So i there exist (n r) unctions λ i ( ) : U R such that dλ 1 (x) [ ] 1 (x) r (x) = dλ 1 (x) holds, then the distribution W is said to be integrable The distribution W is said to be involutive i all Lie brackets lie in the distribution Mathematically expressed or all order Lie brackets and all i, j [ [ i, j ]]] W When can we completely eedback linearize? Question: At a point x, when can we ind a scalar unction λ(x) such that the system has relative degree n at x? Answer: The ollowing two conditions are both necessary and suicient to be able to eedback linearize the system around x 1 The matrix [ has rank n g(x ) ad g(x ) ad (n 2) g(x ) ad (n 1) g(x ) ]

9 2 The distribution D = {g(x), ad g (x),, ad(n 2) g(x)} is involutive around x Based on our previous study, the necessary conditions are L g λ(x) = L g L λ(x) = = L g L (n 2) λ(x) = x U(x U) L g L (n 1) λ(x ) The irst set o conditions are (as we have seen) equivalent to L g λ(x) = L ad g λ(x) = = L ad (n 2) g λ(x) = x U(x U) (2) and the second non-triviality condition is equivalent to L ad (n 1) g λ(x ) Recall lemma These conditions imply that the vectors g(x ), ad g (x ),, ad (n 1) g(x ) (3) are linear independent Now the condition (3) also implies that the distribution D = span{g(x), ad g (x),, ad(n 2) g(x)} (4) is nonsingular and (n 1) dimensional around x The conditions (2) can be written as dλ(x)[g(x), ad g (x),, ad(n 2) g(x)] = This implies that dλ( ) is a basis or the one-dimensional codistribution that annihilates D By Frobenius theorem theorem, the distribution D is involutive An example: A DC motor Complete eedback linearization The governing dierential equations o a DC motor in which the rotor voltage (V r ) is kept constant and the stator voltage (V s ) is used as a control input are given by L s di s dt + RI s = V s The electrical dynamics on the stator side

10 where L r di s dt + RI r = V r E J dω dt + F Ω = T L s, L r are stator and rotor inductances, R s, R r are stator and rotor resistances, E is the back EMF in the rotor, The electrical dynamics on the rotor side The mechanical dynamics o the rotor J, Ω, F, T are the rotor inertia, angular speed o the rotor, viscous riction and torque developed at the rotor shat The back EMF, the torque and the lux are urther given by the relations Choosing the state variables as E = KΦΩ T = KΦI r Φ = L s I s x 1 = I s x 2 = I r x 3 = Ω and the input as V s, the drit and control vector ields are (x) = Rs L s x 1 Rr L r x 2 + Vr L r KLs L r x 1 x 3 F J x 3 + KLs J x 1x 2 g(x) = 1 L s Question 1: Can we ully eedback linearize this system around an arbitrary x? Let us check whether the two conditions are satisied Condition 1: Is Rank{g(x ) [, g](x ) [, [, g]](x )} = 3? Answer: Yes, provided x 2 or x 3 Sounds alright since these variables correspond to the rotor current and rotor speed Condition 2: Is the distribution{g(x) [, g](x)} integrable around x? Answer: Let us try to ind a β : U R such that grad(β)[g(x) [, g](x)] = i e [ β x 1 Yes, one possible solution is β x 2 ] β x 3 [g(x) [, g](x)] = ( ) β(x) = L r x Jx 2 3

11 Zero dynamics A natural output in a DC motor is the speed o the rotor Suppose we wish to regulate this speed around a desired value Ω d How would we choose an appropriate y in our problem? I we wish to regulate the output to Ω d, a natural choice would be y = h(x) = Ω Ω d = x 3 x 3d But with this output are we able to ully eedback linearize the system? I not, what happens?