Nonlinear Control Systems
|
|
- Erin Allison
- 6 years ago
- Views:
Transcription
1 Nonlinear Control Systems António Pedro Aguiar 7. Feedback Linearization IST-DEEC PhD Course 1 1
2 Feedback Linearization Given a nonlinear system of the form ẋ = f(x) + G(x)u y = h(x) Does exist a state feedback control law and a change of variables u = α(x) + β(x)v z = T (x) that transforms the nonlinear system into a an equivalent linear system (ż = Az + Bv)?
3 Feedback Linearization Example: Consider the following system ẋ = Ax + Bγ(x)`u α(x) where γ(x) is nonsingular for all x in some domain D. Then, u = α(x) + β(x)v, with β(x) = γ 1 (x) yields ẋ = Ax + Bv If we would like to stabilize the system, we design v = Kx such that A BK is Hurwitz Therefore u = α(x) β(x)kx
4 Feedback Linearization Example: Consider now this example: ẋ 1 = a sin x ẋ = x 1 + u How can we do this? We cannot simply choose u to cancel the nonlinear term a sin x! However, if we first change the variables then ż 1 = z z 1 = x 1 z = a sin x = ẋ 1 ż = a cos x ẋ = a cos x ( x 1 + u) Therefore with u = x v, π/ < x < π/ a cos x we obtain the linear system ż 1 = z ż = v 4
5 Feedback Linearization A continuously differentiable map T (x) is a diffeormorphism if T 1 (x) is continuously differentiable. This is true if the Jacobian matrix T is nonsingular x D. T (x) is a global diffeormorphism if and only if T T (x) is proper, that is, lim x T (x) =. is nonsingular x Rn and Definition A nonlinear system ẋ = f(x) + G(x)u (1) where f : D R n and G : D R n p are sufficiently smooth on a domain D R n is said to be feedback linearizable (or input-state linearizable) if there exists a diffeomorphism T : D R n such tat D z = T (D) contains the origin and the change of variables z = T (x) transforms (1) into the form ż = Az + Bγ(x)`u α(x) 5
6 Feedback Linearization Suppose that we would like to solve the tracking problem for the system ẋ 1 = a sin x ẋ = x 1 + u y = x If we use state feedback linearization we obtain z 1 = x 1 z = a sin x = ẋ 1 u = x v a cos x ż 1 = z ż = v y = sin 1 (z /a) which is not good! Linearizing the state equation does not necessarily linearize the output equation. Notice however if we set u = x 1 + v we obtain ẋ = v y = x There is one catch: The linearizing feedback control law made x 1 unobservable from y. We have to make sure that x 1 whose dynamics are given by ẋ 1 = a sin x is well behaved. For example, if y = y d = cte x 1 (t) = x 1 () + ta sin y d. It is unbounded! 6
7 SISO system Input-Output Linearization ẋ = f(x) + g(x)u y = h(x) where f, g, h are sufficiently smooth in a domain D R n. The mappings f : D R n and g : D R n are called vector fields on D. Computing the first output derivative... ẏ = h ẋ = h [f(x) + g(x)u] =: L f h(x) + L gh(x)u In the sequel we will use the following notation: L f h(x) = h f(x) Lie Derivative of h with respect to f L gl f h(x) = (L f h) g(x) L f h(x) = h(x) L f h(x) = L f L f h(x) = (L f h) f(x) L k f h(x) = L f L k 1 f h(x) = (Lk 1 f h) f(x) 7
8 Input-Output Linearization ẏ = L f h(x) + L gh(x)u If L gh(x)u = then ẏ = L f h(x) (independent of u). Computing the second derivative... y () = (L f h) [f(x) + g(x)u] = L f h(x) + LgL f h(x)u If L gl f h(x)u = then ẏ () = L f h(x) (independent of u). Repeating this process, it follows that if L gl i 1 f h(x) =, i = 1,,..., ρ 1 L gl ρ 1 f h(x) then u does not appear in y, ẏ,..., y (ρ 1) and y (ρ) = L ρ f h(x) + LgL(ρ 1) f h(x)u 8
9 Input-Output Linearization Therefore, by setting y (ρ) = L ρ f u = h(x) + LgLρ 1 f h(x)u 1 L gl ρ 1 f h(x) [ Lρ f h(x) + v] the system is input-output linearizable and reduces to y (ρ) = v chain of ρ integrators Definition The nonlinear system ẋ = f(x) + g(x)u y = h(x) is said to have relative degree ρ, 1 ρ n, in the region D D if for all x D L gl i 1 f h(x) =, i = 1,,..., ρ 1 L gl ρ 1 f h(x) 9
10 Example 1: Van der Pol system Examples ẋ 1 = x ẋ = x 1 + ɛ(1 x 1 )x + u, ɛ > 1. y = x 1 Calculating the derivatives... ẏ = ẋ 1 = x ÿ = ẋ = x 1 + ɛ(1 x 1)x + u Thus the system has relative degree ρ = in R.. y = x Then ẏ = ẋ = x 1 + ɛ(1 x 1 )x + u In this case the system has relative degree ρ = 1 in R.. y = x 1 + x Then ẏ = x + x ( x 1 + ɛ(1 x 1 )x + u) In this case the system has relative degree ρ = 1 in D = {x R : x }. 1
11 Examples Example : ẋ 1 = x 1 ẋ = x + u y = x 1 Calculating the derivatives... ẏ = ẋ 1 = x 1 = y y (n) = y = x 1, n 1 The system does not have a well defined relative degree! Why? Because the output y(t) = x 1 (t) = e t x 1 () is independent of the input u. 11
12 Examples Example : where m < n and b m. H(s) = bmsm + b m 1 s m b s n + a n 1 s n a A state model of the system is the following ẋ = Ax + Bu y = Cx with A = : a a 1 a n 1 What is the relative degree ρ? 7 5 n n B = n 1 C = [ b b 1 b m ] 1 n 1
13 Examples ẏ = CAx + CBu If m = n 1 CB = b m ρ = 1 Otherwise, CB = y () = CA x + CABu Note that CA is obtained by shifting the elements of C one position to the right and CA i by shifting i positions. Therefore, CA i 1 B =, for i = 1,,... n m 1 CA n m 1 B = b m y (n m) = CA n m x + CA n m 1 Bu ρ = n m In this case the relative degree of the system is the difference between the degrees of the denominator and numerator polynomials of H(s). 1
14 Consider again the linear system given by the transfer function 8 < D(s) can be written as H(s) = N(s) D(s) with : deg D = n deg N = m < n ρ = n m D(s) = Q(s)N(s) + R(s) where the degree of the quotient deg Q = n m = ρ and the degree of the reminder deg R < m Thus H(s) = N(s) Q(s)N(s) + R(s) = Q(s) R(s) Q(s) N(s) and therefore we can conclude that H(s) can be represented as a negative feedback connection with 1/Q(s) in the forward path and R(s)/N(s) in the feedback path. 14
15 Note that the ρ-order transfer function 1/Q(s) has no zeros and can be realized by where ξ = (A c + B cλ T )ξ + B cb me y = C cξ ξ = ˆy ẏ... y (ρ 1) T R ρ and (A c, B c, C c) is a canonical form representation of a chain of ρ integrators: 1 1. A c = ρ ρ B c = ρ 1 B cλ T = , λ Rρ λ T C c = ˆ1 1 ρ 15
16 R(s) N(s) η = A η + B y w = C η The eigenvalues of A are the zeros of the polynomial N(s), which are the zeros of the transfer function H(s). Thus, the system H(s) can be realized by the state model Note that y = C cξ and η = A η + B C cξ ξ = A cξ + B c(λ T ξ b mc η + b mu) y = C cξ ξ = A cξ + B c(λ T ξ b mc η + b mu) ξ 1 = ξ ξ = ξ. ξ ρ = λ T ξ b mc η + b mu and therefore y (ρ) = λ T ξ b mc η + b mu 16
17 Thus, setting results in y (ρ) = λ T ξ b mc η + b mu u = 1 b m [ λ T ξ + b mc η + v] η = A η + B C cξ Internal dynamics: It is unobservable from the output y ξ = A cξ + B cv chain of integrators y = C cξ Suppose we would like to stabilize the output y at a constant reference r, that is, ξ ξ = (r,,..., ) T. Defining ζ = ξ ξ we obtain ζ = A cζ + B cv Therefore, setting v = Kζ = K(ξ ξ ) with (A c B ck) Hurwitz we obtain the closed-loop system η = A η + B C c(ξ + ζ) ζ = (A c B ck)ζ y = C cξ where the eigenvalues of A are the zeros of H(s). If H(s) is minmum phase (zeros in the open left-half complex plan) then A is Hurwitz. 17
18 Feedback Linearization Can we extend this result η = A η + B C cη ξ = A cξ + B c `λt ξ b mc η + b mu y = C cx for the nonlinear system (SISO) that is find a z = T (x), where» η z = ξ ẋ = f(x) + g(x)u y = h(x) = 6 4 φ 1 (x). φ n ρ (x) h(x). L ρ 1 f. h(x) such that T (x) is a diffeomorphism on D D and φ i g(x) =, for 1 i n ρ, x D. Note that Does exist such T (x)? 7 5 η = φ i ẋ = φ i f(x) + φ i g(x)u 18
19 Normal form Theorem (1.1) Consider the SISO system ẋ = f(x) + g(x)u y = h(x) and suppose that it has relative degree ρ n in D. Then, for every x D, there exists a such diffeomorphism T (x) on a neighborhood of x. Using this transformation we obtain the system re-written in normal form: η = f (η, ξ) ξ = A cξ + B cγ(x)[u α(x)] y = C cξ where ξ R ρ, η R n ρ and (A c, B c, C c) is the canonical form representation of a chain of integrators, and f (η, ξ) := φ f(x) x=t 1 (z) γ(x) = L gl ρ 1 f h(x) α(x) = Lρ f h(x) L gl ρ 1 f h(x) 19
20 η = f (η, ξ) ξ = A cξ + B cγ(x)[u α(x)] y = C cξ The external part can be linearized by u = α(x) + β(x)v with β(x) = γ 1 (x). The internal part is described by η = f (η, ξ) Setting ξ = result η = f (η, ) This is called the zero-dynamics Note that for the linear case we have η = A η, where the eigenvalues of A are the zeros of H(s).
21 Definition The system is said to be minimum phase if η = f (η, ) has an asymptotically stable equilibrium point in the domain of interest. The zero dynamics can be characterized in the original coordinates by notting that y(t) =, t ξ(t) = u(t) = α(x(t)) where the first implication is due to the fact that ξ = [y, ẏ,...] T and the second due to ξ = A ξ + B γ(x)[u α(x)]. Thus, when y(t) =, the solution of the state equation is confined to the set n o Z = x D : h(x) = L f h(x) =... = L ρ 1 f h(x) = and the input that is u = u (x) := α(x) x Z ẋ = f (x) := [f(x) + g(x)α(x)] x Z In the special case that ρ = n η does not exist. In that case the system has no zero dynamics and by default is said to be minimum phase. 1
22 Example 1 Example ẋ 1 = x ẋ = x 1 + ɛ(1 x 1)x + u y = x It is in the normal form (ξ = y, η = x 1 ) Zero-dynamics? ẋ 1 =, which does not have an asymptotic stable equilibrium point. Hence, the system is not minimum phase. Example ẋ 1 = x x 1 + x u ẋ = x ẋ = x x + u y = x What is the relative degree and the zero dynamics?
23 ẋ 1 = x x 1 + x u Computing the time-derivative... ẋ = x ẋ = x x + u y = x ẏ = ẋ = x ÿ = ẋ = x 1 x + u Thus, the relative degree is ρ =. Analyzing the zero-dynamics we have y = ẏ = ÿ = we have x = x = and from the last we have u = x 1 x =. Therefore, ẋ 1 = x 1 and the system is minimum phase.
24 The single-input system Full-State Linearization ẋ = f(x) + g(x)u with f, g sufficiently smooth in a domain D R n is feedback linearizable if there exists a sufficiently smooth h : D R such that the system has relative degree n in a region D D. ẋ = f(x) + g(x)u y = h(x) This implies that the normal form reduces to Note that Thus which is equivalent to ż = A cz + B cγ(x)[u α(x)] y = C cz z = T (x) ż = T ẋ A cz + B cγ(x)[u α(x)] = T [f(x) + g(x)u] 4
25 Splinting in two we obtain Equation () is equivalent to T f(x) = AcT (x) Bcγ(x)α(x) () T g(x) = Bcγ(x) () T 1 f(x) = T (x) T f(x) = T (x). T n 1 f(x) = Tn(x) T n f(x) = α(x)γ(x) 5
26 and () is equivalent to Setting h(x) = T 1, we see that and T 1 g(x) = T g(x) =. T n 1 g(x) = T n g(x) = γ(x) T i+1 (x) = L f T i (x) = L i f h(x), i = 1,,..., n 1 L gl i 1 f h(x) =, i = 1,,..., n 1 L gl n 1 f Therefore we can conclude that if h(.) satisfies (4) the system is feedback linearizable. (4) 6
27 The existence of h(.) can be characterized by necessary and sufficient conditions on the vector fields f and g. First we need some terminology. Definition Given two vector fields f and g on D R n, the Lie Bracket [f, g] is the vector field [f, g](x) = g f f(x) g(x) Note that Adjoint representation [f, g] = [g, f] f = g = cte [f, g] = ad f g(x) = g(x) ad 1 f g(x) = [f, g](x) ad k f g(x) = [f, adk 1 f g](x), k 1 7
28 Then,» f(x) = Example 1 x, g(x) = sin x 1 x» x1 " [f, g](x) = g g1 g f 1 f(x) g(x) = 1 g g 1»»» x = 1 sin x 1 x»»» x1 = = x x 1 x 1 x 1 + x # " f1 f 1 f(x) 1 f f 1» 1 cos x 1 1 x1 # g(x) ad f g = [f, ad f g] = ad f g f f(x) ad f g(x)»»» 1 x = sin x 1 x cos x 1 1» x = 1 x x 1 + x sin x 1 x 1 cos x 1» x 1 x 1 + x 8
29 Example : f(x) = Ax and g(x) = g is a constant vector field. Then, ad f g(x) = [f, g](x) = g f f(x) g(x) = Ag ad f g(x) = [f, ad f g](x) = ad f g f f ad f g = A( Ag) = A g ad k f g = ( 1)k A k g Definition For vector fields f 1, f,..., f k on D R n, a distribution is a collection of all vector spaces (x) = span {f 1 (x),..., f k (x)}, where for each fixed x D, (x) is the subspace of R n spanned by the vectors f 1 (x),..., f k (x). The dimension of (x) is defined by which may depend on x. dim( (x)) = rank[f 1 (x), f (x),..., f k (x)] If f 1, f,..., f k are linearly independent, then dim( (x)) = k, x D. In this case, we say that is a nonsingular distribution on D. A distribution is involutive if g 1, g [g 1, g ]. If is a nonsingular distribution on D, generated by f 1,..., f k then it is involutive if and only if [f i, f j ], 1 i, j k 9
30 Example Let D = R, = span {f 1, f } and f 1 = 4 x 1 5, f = x dim( (x)) = rank[f 1, f ] =, x D is involutive? [f 1, f ] = f f 1 f 1 f = x x 5 = Checking that [f 1, f ] is the same to see if [f 1, f ] can be generated by f 1, f, that is if rank[f 1 (x), f (x), [f 1, f ](x)] =, x D. But rank Hence, is not involutive. 4 x 1 1 x 1 5 =, x D
31 Theorem 1. Theorem The system ẋ = f(x) + g(x)u, with x R n, u R is feedback linearizable if and only if there is a domain D D such that 1. The matrix G(x) = [g(x), ad f g(x),..., ad n 1 f g] has rank n x D.. The distribution D = span{g, ad f g(x),..., ad n f g} is involutive in D. Example» a sin x ẋ = f(x) + gu, f(x) = x 1 we have seen that ad f g = [f, g] = g f f» g = a cos x x 1», g = 1» 1» a cos x =» a cos x The matrix G = [g, ad f g] = has rank G =, cos x 1. The distribution D = span {g} is involutive. Thus, we can conclude that there exists a T (x) in D = x R : cos x that allow us to do feedback linearization. 1
32 Now we have to find h(x) that satisfies Thus, h(.) must be independent of x. L f h(x) = h f(x) = h L f h h g = h g = ; (L f h) g ; h() = h i» h h 1 = h = 1 h 1 h a sin x 1 1 i» a sin x x 1 = h a sin x 1 i» h a sin x 1 1 = h 1 a cos x h In conclusion, and h =. 1 Examples of such h(x) include h(x) = x 1 or h(x) = x 1 + x 1. Given h(x) we can now perform input-output linearization.
33 Example A single link manipulator with flexible points f(x) = 6 4 x a sin x 1 b(x 1 x ) x 4 c(x 1 x ) ẋ = f(x) + gu 7 5, g = 6 4 G = [g, ad f g, ad f g, ad f g] d 7 5 a, b, c, d > ad f g = [f, g] = g f f g = 6 4 ad f g = [f, ad f g] = (ad f g) f f ad f g 1 = 6 a cos x 1 b b c c 1 a cos x 1 b b 1 c c d 7 5 = 6 4 bd cd d 7 5 = 6 4 d 7 5
34 g, h() = 4 Thus, ad f g = [f, ad f g] = ad f g f f ad f g 1 = 6 a cos x 1 b b c c G = 6 4 d The distribution = span d bd cd bd cd n g, ad f g, ad f bd cd 7 5 = rank G = 4, x R4 bd cd 7 5 o is involutive since g, ad f g, ad f are constant vector fields ([g, ad f g] = ). Therefore, we can conclude that there exists an h(x) : R 4 R and a T (x) that make the system full state feedback linearizable. In particular, h(x) must satisfy L i 1 f h g =, i = 1,,. L f h
35 For i = 1, we have h g = h 4 = and so h(x) must be independent of x 4. L f h(x) = h From f(x) = h 1 x + h [ a sin x 1 b(x 1 x )] + h x 4. L f h g = L f h 4 = h = Thus, h(x) is independent of x. Thus, L f h(x) = h x + h [ a sin x 1 b(x 1 x )] 1. L f h(x) = L f L f h(x) = (L f h) f(x) = (L f h) x + (L f h) [ a sin x 1 1 b(x 1 x )] + (L f h) x 4 + (L f h) c(x 1 x ) 4 For i =, Thus h is independent of x. L f h g = `L f h = h = 5
36 Hence, L L f h(x) = L f L f h = f h L f(x) = f h 1 L + f L h x 4 + f h c(x 1 x ) 4 Also, from condition L f h(x) g L f h L x + f h [ a sin x 1 b(x 1 x )] 1 L f h Therefore, let h(x) = x 1. Then, the change of variables h 1 z 1 = h(x) = x 1 z = L f h(x) = x z = L f h(x) = a sin x 1 b(x 1 x ) z 4 = L f h(x) = a cos x 1 ẋ 1 bẋ 1 + bẋ = ax cos x 1 bx + bx 4 transforms the state equation into ż 1 = z ż = z ż = z 4 ż 4 = (a cos z 1 + b + c)z + a(z c) sin z 1 + bd u 6
37 Example - field controlled DC motor Consider the system with Computing the f(x) = 4 ad f g = [f, g] = 4 Rank of G? a cx θx ẋ = f(x) + gu ax 1 bx + k cx 1 x θx 1 x G = [g, ad f g, ad f g] = 5, g = 5, ad f g = [f, ad f g] = a cx θx a (a + b) cx (b a) θx θk a (a + b) cx (b a) θx θk 5 5 7
38 det G = cθ( k + bx )x. Hence G has rank for x k b and x. Let s check the distribution D = span{g, ad f g} ˆg, adf g = (ad f g) g = 4 c = 4 5 θ Hence D is involutive because [g, ad f g] D. Therefore, the conditions of Theorem 1. are satisfied in particular for the domain j D = x R : x > k ff b, x > h(.) =? 8
39 with f(x) = 4 ẋ = f(x) + gu ax 1 bx + k cx 1 x θx 1 x 5, g = Equilibrium points: x 1 =, x = k b, x = cte. Suppose we are interested in the desired operating point x = [ k b w ] T. Then, h(x) must satisfy (n = ) and h(x ) =. h g = ; (L f h) g = ; (L f h) g that is, h must be independent of x 1. L f h(x) =... h g = h = 1 9
40 State feedback control Consider the system ẋ = f(x) + G(x)u and let z = T (x) = [T 1 (x), T (x)] T such that η = f (η, ξ) ξ = Aξ + Bγ(x)[u α(x)] Suppose that (A, B) is controllable, γ(x) is nonsingular x D, f (, ) = and f (η, ξ), α(x), γ(x) C 1. Goal: Design a state feedback control law to stabilize the origin z =. Setting u = α(x) + β(x)v, β(x) = γ 1 (x), we obtain the triangular system η = f (η, ξ) ξ = Aξ + Bv Let v = Kξ with (A BK) Hurwitz then we can conclude the following: 4
41 Lemma (1.1) State feedback control The origin z = is asymptotically stable (AS) if the origin of η = f (η, ) is AS (that is, if the system is minimum phase). Proof. By the converse theorem V 1 (η) : V 1 η f (η, ) α ( x ) in some neighborhood of η =, where α K. Let P = P T > be the solution of the Lyapunov equation P (A BK) + (A BK) T P = I. V = V 1 + k p ξ T P ξ, k > V V 1 η f (η, ξ) + k p ξ T P ξ ξt [P (A BK) + (A BK) T P ]ξ V 1 η f (η, ) + V 1 η [f (η, ξ) f (η, )] α ( η ) + k 1 ξ kk ξ, k 1, k > kξt ξ p ξ T P ξ where the last inequality follows from restricting the state to be in any bounded neighborhood of the origin and using the continuous differentiability property of V 1 and f. Thus, choosing k > k 1 /k ensures that V <, which follows that the origin is AS. 41
42 State feedback control Lemma (1.) The origin z = is GAS if η = f (η, ξ) is ISS. Note that if η = f (η, ) is GAS or GES does NOT imply ISS. But, if it is GES + globally Lipschitz then it is ISS. Otherwise, we have to prove ISS by further analysis. Example: η = η + η ξ ξ = v Zero dynamics: η = η η = is GES but η = η + η ξ is not ISS, e.g., if ξ(t) = 1 and η() then η(t), t, which implies that η grows unbounded. However with v = Kξ, K > we achieve AS. 4
43 To view this, let ν = ηξ, then ν = η ξ + ηξ = ηv ηξ + η ξ = Kηξ ηξ + η ξ = (1 + K)ν + υ = [(1 + K) ν]ν Thus, with ν() < 1 + K υ and therefore we can also conclude that T t : υ(t) 1, t T. Consider now V = 1/η. Then V = η η = η + η ξ = η (1 ηξ) = η (1 ν) <, t T Thus, η and note also that ξ = Kξ. Therefore, the control law v = Kξ can achieve semiglobal stabilization. 4
44 One may think that we can assign the eigenvalues of (A BK) to the left half-complex plane to make ξ = (A BK)ξ decay to zero arbitrarily fast. BUT this may have consequences: the zero-dynamics may go unstable! This is due to the peaking phenomenon. Example η = 1 (1 + ξ )η ξ 1 = ξ ξ = v» Setting v = Kξ = k ξ 1 kξ A BK = 1 k k and the eigenvalues are k, k. Note that» e (A BK)t (1 + kt)e kt te = kt k te kt (1 kt)e kt which shows that as k, ξ(t) will decay to zero arbitrary fast. However, the element (, 1) reaches a maximum value k/e at t = 1. There is a peak k of order k! Furthermore, the interaction of peaking with nonlinear growth could destabilize the system. 44
45 E.g., consider the initial conditions η() = η ξ 1 () = 1 ξ () = Then, and ξ (t) = k te kt In this case the solution η(t) is given by η (t) = which has a finite escape if η > 1. η = 1 (1 + ξ )η = 1 (1 k te kt )η η 1 + η [t + (1 + kt)e kt 1]. 45
46 Tracking η = f (η, ξ) ξ = A ξ + B γ(x)[u α(x)] y = C ξ Goal: Design u such that y asymptotically tracks a reference signal r(t). Assume that r(t), ṙ(t),..., r (ρ) are bounded and available on-line. Note that the reference signal could be the output of a pre-filter. Example: If ρ =, the pre-filter could be Then wn G(s) = s + ξw ns + w n ẏ 1 = y ẏ = w n y 1 ξw ny + w n y d r = y 1 Note that in this case ṙ = ẏ 1 = y and r = ẏ. Consider a system with relative degree ρ. Let r 6 r = 4. r (ρ 1) 7 6 5, and e = 4 ξ 1 r. ξ ρ r (ρ 1) 7 5 = ξ r 46
47 Then, the error system is given by η = f (η, e + r) ė = A ce + B c(γ(x)[u α(x)] r (ρ) ) Setting u = α(x) + β(x)[v r (ρ) ] with β = 1 it follows that γ(x) η = f (η, e + r) ė = A ce + B cv Thus, selecting v = Ke with (A c B ck) Hurwitz we can conclude that the states of the closed-loop system η = f (η, e + r) ė = (A B K)e are bounded if η = f (η, e + R) is ISS and that e as t. 47
Feedback Linearization
Feedback Linearization Peter Al Hokayem and Eduardo Gallestey May 14, 2015 1 Introduction Consider a class o single-input-single-output (SISO) nonlinear systems o the orm ẋ = (x) + g(x)u (1) y = h(x) (2)
More informationOutput Feedback and State Feedback. EL2620 Nonlinear Control. Nonlinear Observers. Nonlinear Controllers. ẋ = f(x,u), y = h(x)
Output Feedback and State Feedback EL2620 Nonlinear Control Lecture 10 Exact feedback linearization Input-output linearization Lyapunov-based control design methods ẋ = f(x,u) y = h(x) Output feedback:
More informationNonlinear Control. Nonlinear Control Lecture # 24 State Feedback Stabilization
Nonlinear Control Lecture # 24 State Feedback Stabilization Feedback Lineaization What information do we need to implement the control u = γ 1 (x)[ ψ(x) KT(x)]? What is the effect of uncertainty in ψ,
More informationNonlinear Control Systems
Nonlinear Control Systems António Pedro Aguiar pedro@isr.ist.utl.pt 5. Input-Output Stability DEEC PhD Course http://users.isr.ist.utl.pt/%7epedro/ncs2012/ 2012 1 Input-Output Stability y = Hu H denotes
More informationNonlinear Control Lecture 9: Feedback Linearization
Nonlinear Control Lecture 9: Feedback Linearization Farzaneh Abdollahi Department of Electrical Engineering Amirkabir University of Technology Fall 2011 Farzaneh Abdollahi Nonlinear Control Lecture 9 1/75
More informationCONTROL DESIGN FOR SET POINT TRACKING
Chapter 5 CONTROL DESIGN FOR SET POINT TRACKING In this chapter, we extend the pole placement, observer-based output feedback design to solve tracking problems. By tracking we mean that the output is commanded
More information1 The Observability Canonical Form
NONLINEAR OBSERVERS AND SEPARATION PRINCIPLE 1 The Observability Canonical Form In this Chapter we discuss the design of observers for nonlinear systems modelled by equations of the form ẋ = f(x, u) (1)
More informationIntroduction to Geometric Control
Introduction to Geometric Control Relative Degree Consider the square (no of inputs = no of outputs = p) affine control system ẋ = f(x) + g(x)u = f(x) + [ g (x),, g p (x) ] u () h (x) y = h(x) = (2) h
More information1 Relative degree and local normal forms
THE ZERO DYNAMICS OF A NONLINEAR SYSTEM 1 Relative degree and local normal orms The purpose o this Section is to show how single-input single-output nonlinear systems can be locally given, by means o a
More informationNonlinear Control Systems
Nonlinear Control Systems António Pedro Aguiar pedro@isr.ist.utl.pt 3. Fundamental properties IST-DEEC PhD Course http://users.isr.ist.utl.pt/%7epedro/ncs2012/ 2012 1 Example Consider the system ẋ = f
More informationFeedback Linearization Lectures delivered at IIT-Kanpur, TEQIP program, September 2016.
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
More informationA nemlineáris rendszer- és irányításelmélet alapjai Relative degree and Zero dynamics (Lie-derivatives)
A nemlineáris rendszer- és irányításelmélet alapjai Relative degree and Zero dynamics (Lie-derivatives) Hangos Katalin BME Analízis Tanszék Rendszer- és Irányításelméleti Kutató Laboratórium MTA Számítástechnikai
More informationDisturbance Decoupling Problem
DISC Systems and Control Theory of Nonlinear Systems, 21 1 Disturbance Decoupling Problem Lecture 4 Nonlinear Dynamical Control Systems, Chapter 7 The disturbance decoupling problem is a typical example
More information1. Find the solution of the following uncontrolled linear system. 2 α 1 1
Appendix B Revision Problems 1. Find the solution of the following uncontrolled linear system 0 1 1 ẋ = x, x(0) =. 2 3 1 Class test, August 1998 2. Given the linear system described by 2 α 1 1 ẋ = x +
More informationChap. 3. Controlled Systems, Controllability
Chap. 3. Controlled Systems, Controllability 1. Controllability of Linear Systems 1.1. Kalman s Criterion Consider the linear system ẋ = Ax + Bu where x R n : state vector and u R m : input vector. A :
More informationIntroduction to Nonlinear Control Lecture # 3 Time-Varying and Perturbed Systems
p. 1/5 Introduction to Nonlinear Control Lecture # 3 Time-Varying and Perturbed Systems p. 2/5 Time-varying Systems ẋ = f(t, x) f(t, x) is piecewise continuous in t and locally Lipschitz in x for all t
More informationEL 625 Lecture 10. Pole Placement and Observer Design. ẋ = Ax (1)
EL 625 Lecture 0 EL 625 Lecture 0 Pole Placement and Observer Design Pole Placement Consider the system ẋ Ax () The solution to this system is x(t) e At x(0) (2) If the eigenvalues of A all lie in the
More informationStabilization and Passivity-Based Control
DISC Systems and Control Theory of Nonlinear Systems, 2010 1 Stabilization and Passivity-Based Control Lecture 8 Nonlinear Dynamical Control Systems, Chapter 10, plus handout from R. Sepulchre, Constructive
More informationHigh-Gain Observers in Nonlinear Feedback Control. Lecture # 3 Regulation
High-Gain Observers in Nonlinear Feedback Control Lecture # 3 Regulation High-Gain ObserversinNonlinear Feedback ControlLecture # 3Regulation p. 1/5 Internal Model Principle d r Servo- Stabilizing u y
More informationSolution of Additional Exercises for Chapter 4
1 1. (1) Try V (x) = 1 (x 1 + x ). Solution of Additional Exercises for Chapter 4 V (x) = x 1 ( x 1 + x ) x = x 1 x + x 1 x In the neighborhood of the origin, the term (x 1 + x ) dominates. Hence, the
More informationModeling and Analysis of Dynamic Systems
Modeling and Analysis of Dynamic Systems Dr. Guillaume Ducard Fall 2017 Institute for Dynamic Systems and Control ETH Zurich, Switzerland G. Ducard c 1 / 57 Outline 1 Lecture 13: Linear System - Stability
More informationControl engineering sample exam paper - Model answers
Question Control engineering sample exam paper - Model answers a) By a direct computation we obtain x() =, x(2) =, x(3) =, x(4) = = x(). This trajectory is sketched in Figure (left). Note that A 2 = I
More informationTopic # /31 Feedback Control Systems. Analysis of Nonlinear Systems Lyapunov Stability Analysis
Topic # 16.30/31 Feedback Control Systems Analysis of Nonlinear Systems Lyapunov Stability Analysis Fall 010 16.30/31 Lyapunov Stability Analysis Very general method to prove (or disprove) stability of
More informationNonlinear Control. Nonlinear Control Lecture # 25 State Feedback Stabilization
Nonlinear Control Lecture # 25 State Feedback Stabilization Backstepping η = f a (η)+g a (η)ξ ξ = f b (η,ξ)+g b (η,ξ)u, g b 0, η R n, ξ, u R Stabilize the origin using state feedback View ξ as virtual
More informationNonlinear Control Lecture # 36 Tracking & Regulation. Nonlinear Control
Nonlinear Control Lecture # 36 Tracking & Regulation Normal form: η = f 0 (η,ξ) ξ i = ξ i+1, for 1 i ρ 1 ξ ρ = a(η,ξ)+b(η,ξ)u y = ξ 1 η D η R n ρ, ξ = col(ξ 1,...,ξ ρ ) D ξ R ρ Tracking Problem: Design
More informationMCE693/793: Analysis and Control of Nonlinear Systems
MCE693/793: Analysis and Control of Nonlinear Systems Input-Output and Input-State Linearization Zero Dynamics of Nonlinear Systems Hanz Richter Mechanical Engineering Department Cleveland State University
More informationNonlinear Control. Nonlinear Control Lecture # 8 Time Varying and Perturbed Systems
Nonlinear Control Lecture # 8 Time Varying and Perturbed Systems Time-varying Systems ẋ = f(t,x) f(t,x) is piecewise continuous in t and locally Lipschitz in x for all t 0 and all x D, (0 D). The origin
More informationProblem List MATH 5173 Spring, 2014
Problem List MATH 5173 Spring, 2014 The notation p/n means the problem with number n on page p of Perko. 1. 5/3 [Due Wednesday, January 15] 2. 6/5 and describe the relationship of the phase portraits [Due
More informationConverse Lyapunov theorem and Input-to-State Stability
Converse Lyapunov theorem and Input-to-State Stability April 6, 2014 1 Converse Lyapunov theorem In the previous lecture, we have discussed few examples of nonlinear control systems and stability concepts
More informationCDS Solutions to Final Exam
CDS 22 - Solutions to Final Exam Instructor: Danielle C Tarraf Fall 27 Problem (a) We will compute the H 2 norm of G using state-space methods (see Section 26 in DFT) We begin by finding a minimal state-space
More informationNonlinear Systems and Control Lecture # 12 Converse Lyapunov Functions & Time Varying Systems. p. 1/1
Nonlinear Systems and Control Lecture # 12 Converse Lyapunov Functions & Time Varying Systems p. 1/1 p. 2/1 Converse Lyapunov Theorem Exponential Stability Let x = 0 be an exponentially stable equilibrium
More informationIntro. Computer Control Systems: F8
Intro. Computer Control Systems: F8 Properties of state-space descriptions and feedback Dave Zachariah Dept. Information Technology, Div. Systems and Control 1 / 22 dave.zachariah@it.uu.se F7: Quiz! 2
More informationLMI Methods in Optimal and Robust Control
LMI Methods in Optimal and Robust Control Matthew M. Peet Arizona State University Lecture 15: Nonlinear Systems and Lyapunov Functions Overview Our next goal is to extend LMI s and optimization to nonlinear
More informationLecture 9 Nonlinear Control Design
Lecture 9 Nonlinear Control Design Exact-linearization Lyapunov-based design Lab 2 Adaptive control Sliding modes control Literature: [Khalil, ch.s 13, 14.1,14.2] and [Glad-Ljung,ch.17] Course Outline
More informationEN Nonlinear Control and Planning in Robotics Lecture 3: Stability February 4, 2015
EN530.678 Nonlinear Control and Planning in Robotics Lecture 3: Stability February 4, 2015 Prof: Marin Kobilarov 0.1 Model prerequisites Consider ẋ = f(t, x). We will make the following basic assumptions
More informationREGULATION OF NONLINEAR SYSTEMS USING CONDITIONAL INTEGRATORS. Abhyudai Singh
REGULATION OF NONLINEAR SYSTEMS USING CONDITIONAL INTEGRATORS By Abhyudai Singh A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE
More informationNonlinear Control Lecture # 14 Tracking & Regulation. Nonlinear Control
Nonlinear Control Lecture # 14 Tracking & Regulation Normal form: η = f 0 (η,ξ) ξ i = ξ i+1, for 1 i ρ 1 ξ ρ = a(η,ξ)+b(η,ξ)u y = ξ 1 η D η R n ρ, ξ = col(ξ 1,...,ξ ρ ) D ξ R ρ Tracking Problem: Design
More informationControl Systems. Internal Stability - LTI systems. L. Lanari
Control Systems Internal Stability - LTI systems L. Lanari outline LTI systems: definitions conditions South stability criterion equilibrium points Nonlinear systems: equilibrium points examples stable
More informationModule 03 Linear Systems Theory: Necessary Background
Module 03 Linear Systems Theory: Necessary Background Ahmad F. Taha EE 5243: Introduction to Cyber-Physical Systems Email: ahmad.taha@utsa.edu Webpage: http://engineering.utsa.edu/ taha/index.html September
More informationProf. Krstic Nonlinear Systems MAE281A Homework set 1 Linearization & phase portrait
Prof. Krstic Nonlinear Systems MAE28A Homework set Linearization & phase portrait. For each of the following systems, find all equilibrium points and determine the type of each isolated equilibrium. Use
More informationZeros and zero dynamics
CHAPTER 4 Zeros and zero dynamics 41 Zero dynamics for SISO systems Consider a linear system defined by a strictly proper scalar transfer function that does not have any common zero and pole: g(s) =α p(s)
More informationInput-to-state stability and interconnected Systems
10th Elgersburg School Day 1 Input-to-state stability and interconnected Systems Sergey Dashkovskiy Universität Würzburg Elgersburg, March 5, 2018 1/20 Introduction Consider Solution: ẋ := dx dt = ax,
More informationLyapunov Stability Theory
Lyapunov Stability Theory Peter Al Hokayem and Eduardo Gallestey March 16, 2015 1 Introduction In this lecture we consider the stability of equilibrium points of autonomous nonlinear systems, both in continuous
More informationState Regulator. Advanced Control. design of controllers using pole placement and LQ design rules
Advanced Control State Regulator Scope design of controllers using pole placement and LQ design rules Keywords pole placement, optimal control, LQ regulator, weighting matrixes Prerequisites Contact state
More informationNonlinear Control. Nonlinear Control Lecture # 6 Passivity and Input-Output Stability
Nonlinear Control Lecture # 6 Passivity and Input-Output Stability Passivity: Memoryless Functions y y y u u u (a) (b) (c) Passive Passive Not passive y = h(t,u), h [0, ] Vector case: y = h(t,u), h T =
More informationIntroduction to Nonlinear Control Lecture # 4 Passivity
p. 1/6 Introduction to Nonlinear Control Lecture # 4 Passivity È p. 2/6 Memoryless Functions ¹ y È Ý Ù È È È È u (b) µ power inflow = uy Resistor is passive if uy 0 p. 3/6 y y y u u u (a) (b) (c) Passive
More information6 Linear Equation. 6.1 Equation with constant coefficients
6 Linear Equation 6.1 Equation with constant coefficients Consider the equation ẋ = Ax, x R n. This equating has n independent solutions. If the eigenvalues are distinct then the solutions are c k e λ
More informationState Feedback and State Estimators Linear System Theory and Design, Chapter 8.
1 Linear System Theory and Design, http://zitompul.wordpress.com 2 0 1 4 2 Homework 7: State Estimators (a) For the same system as discussed in previous slides, design another closed-loop state estimator,
More information2.10 Saddles, Nodes, Foci and Centers
2.10 Saddles, Nodes, Foci and Centers In Section 1.5, a linear system (1 where x R 2 was said to have a saddle, node, focus or center at the origin if its phase portrait was linearly equivalent to one
More informationTTK4150 Nonlinear Control Systems Solution 6 Part 2
TTK4150 Nonlinear Control Systems Solution 6 Part 2 Department of Engineering Cybernetics Norwegian University of Science and Technology Fall 2003 Solution 1 Thesystemisgivenby φ = R (φ) ω and J 1 ω 1
More informationLecture 4. Chapter 4: Lyapunov Stability. Eugenio Schuster. Mechanical Engineering and Mechanics Lehigh University.
Lecture 4 Chapter 4: Lyapunov Stability Eugenio Schuster schuster@lehigh.edu Mechanical Engineering and Mechanics Lehigh University Lecture 4 p. 1/86 Autonomous Systems Consider the autonomous system ẋ
More informationA Light Weight Rotary Double Pendulum: Maximizing the Domain of Attraction
A Light Weight Rotary Double Pendulum: Maximizing the Domain of Attraction R. W. Brockett* and Hongyi Li* Engineering and Applied Sciences Harvard University Cambridge, MA 38, USA {brockett, hongyi}@hrl.harvard.edu
More informationEntrance Exam, Differential Equations April, (Solve exactly 6 out of the 8 problems) y + 2y + y cos(x 2 y) = 0, y(0) = 2, y (0) = 4.
Entrance Exam, Differential Equations April, 7 (Solve exactly 6 out of the 8 problems). Consider the following initial value problem: { y + y + y cos(x y) =, y() = y. Find all the values y such that the
More informationLecture 8. Chapter 5: Input-Output Stability Chapter 6: Passivity Chapter 14: Passivity-Based Control. Eugenio Schuster.
Lecture 8 Chapter 5: Input-Output Stability Chapter 6: Passivity Chapter 14: Passivity-Based Control Eugenio Schuster schuster@lehigh.edu Mechanical Engineering and Mechanics Lehigh University Lecture
More informationHigh-Gain Observers in Nonlinear Feedback Control. Lecture # 2 Separation Principle
High-Gain Observers in Nonlinear Feedback Control Lecture # 2 Separation Principle High-Gain ObserversinNonlinear Feedback ControlLecture # 2Separation Principle p. 1/4 The Class of Systems ẋ = Ax + Bφ(x,
More informationSolution of Linear State-space Systems
Solution of Linear State-space Systems Homogeneous (u=0) LTV systems first Theorem (Peano-Baker series) The unique solution to x(t) = (t, )x 0 where The matrix function is given by is called the state
More informationEN Nonlinear Control and Planning in Robotics Lecture 10: Lyapunov Redesign and Robust Backstepping April 6, 2015
EN530.678 Nonlinear Control and Planning in Robotics Lecture 10: Lyapunov Redesign and Robust Backstepping April 6, 2015 Prof: Marin Kobilarov 1 Uncertainty and Lyapunov Redesign Consider the system [1]
More informationControl Systems Design
ELEC4410 Control Systems Design Lecture 18: State Feedback Tracking and State Estimation Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 18:
More information5. Observer-based Controller Design
EE635 - Control System Theory 5. Observer-based Controller Design Jitkomut Songsiri state feedback pole-placement design regulation and tracking state observer feedback observer design LQR and LQG 5-1
More informationMultivariable MRAC with State Feedback for Output Tracking
29 American Control Conference Hyatt Regency Riverfront, St. Louis, MO, USA June 1-12, 29 WeA18.5 Multivariable MRAC with State Feedback for Output Tracking Jiaxing Guo, Yu Liu and Gang Tao Department
More informationRobust Semiglobal Nonlinear Output Regulation The case of systems in triangular form
Robust Semiglobal Nonlinear Output Regulation The case of systems in triangular form Andrea Serrani Department of Electrical and Computer Engineering Collaborative Center for Control Sciences The Ohio
More informationNonlinear Control Lecture # 1 Introduction. Nonlinear Control
Nonlinear Control Lecture # 1 Introduction Nonlinear State Model ẋ 1 = f 1 (t,x 1,...,x n,u 1,...,u m ) ẋ 2 = f 2 (t,x 1,...,x n,u 1,...,u m ).. ẋ n = f n (t,x 1,...,x n,u 1,...,u m ) ẋ i denotes the derivative
More informationTRACKING AND DISTURBANCE REJECTION
TRACKING AND DISTURBANCE REJECTION Sadegh Bolouki Lecture slides for ECE 515 University of Illinois, Urbana-Champaign Fall 2016 S. Bolouki (UIUC) 1 / 13 General objective: The output to track a reference
More informationCDS Solutions to the Midterm Exam
CDS 22 - Solutions to the Midterm Exam Instructor: Danielle C. Tarraf November 6, 27 Problem (a) Recall that the H norm of a transfer function is time-delay invariant. Hence: ( ) Ĝ(s) = s + a = sup /2
More informationSolutions. Problems of Chapter 1
Solutions Problems of Chapter 1 1.1 A real square matrix A IR n n is invertible if and only if its determinant is non zero. The determinant of A is a polynomial in the entries a ij of A, whose set of zeros
More informationMCE/EEC 647/747: Robot Dynamics and Control. Lecture 8: Basic Lyapunov Stability Theory
MCE/EEC 647/747: Robot Dynamics and Control Lecture 8: Basic Lyapunov Stability Theory Reading: SHV Appendix Mechanical Engineering Hanz Richter, PhD MCE503 p.1/17 Stability in the sense of Lyapunov A
More informationInput to state Stability
Input to state Stability Mini course, Universität Stuttgart, November 2004 Lars Grüne, Mathematisches Institut, Universität Bayreuth Part IV: Applications ISS Consider with solutions ϕ(t, x, w) ẋ(t) =
More informationLinearization problem. The simplest example
Linear Systems Lecture 3 1 problem Consider a non-linear time-invariant system of the form ( ẋ(t f x(t u(t y(t g ( x(t u(t (1 such that x R n u R m y R p and Slide 1 A: f(xu f(xu g(xu and g(xu exist and
More informationHigh-Gain Observers in Nonlinear Feedback Control
High-Gain Observers in Nonlinear Feedback Control Lecture # 1 Introduction & Stabilization High-Gain ObserversinNonlinear Feedback ControlLecture # 1Introduction & Stabilization p. 1/4 Brief History Linear
More informationChapter III. Stability of Linear Systems
1 Chapter III Stability of Linear Systems 1. Stability and state transition matrix 2. Time-varying (non-autonomous) systems 3. Time-invariant systems 1 STABILITY AND STATE TRANSITION MATRIX 2 In this chapter,
More informationSYSTEMTEORI - ÖVNING 5: FEEDBACK, POLE ASSIGNMENT AND OBSERVER
SYSTEMTEORI - ÖVNING 5: FEEDBACK, POLE ASSIGNMENT AND OBSERVER Exercise 54 Consider the system: ẍ aẋ bx u where u is the input and x the output signal (a): Determine a state space realization (b): Is the
More informationControl Systems. Frequency domain analysis. L. Lanari
Control Systems m i l e r p r a in r e v y n is o Frequency domain analysis L. Lanari outline introduce the Laplace unilateral transform define its properties show its advantages in turning ODEs to algebraic
More informationState Feedback and State Estimators Linear System Theory and Design, Chapter 8.
1 Linear System Theory and Design, http://zitompul.wordpress.com 2 0 1 4 State Estimator In previous section, we have discussed the state feedback, based on the assumption that all state variables are
More informationNonlinear Control Lecture 7: Passivity
Nonlinear Control Lecture 7: Passivity Farzaneh Abdollahi Department of Electrical Engineering Amirkabir University of Technology Fall 2011 Farzaneh Abdollahi Nonlinear Control Lecture 7 1/26 Passivity
More informationState will have dimension 5. One possible choice is given by y and its derivatives up to y (4)
A Exercise State will have dimension 5. One possible choice is given by y and its derivatives up to y (4 x T (t [ y(t y ( (t y (2 (t y (3 (t y (4 (t ] T With this choice we obtain A B C [ ] D 2 3 4 To
More informationEEE582 Homework Problems
EEE582 Homework Problems HW. Write a state-space realization of the linearized model for the cruise control system around speeds v = 4 (Section.3, http://tsakalis.faculty.asu.edu/notes/models.pdf). Use
More informationPassivity-based Stabilization of Non-Compact Sets
Passivity-based Stabilization of Non-Compact Sets Mohamed I. El-Hawwary and Manfredi Maggiore Abstract We investigate the stabilization of closed sets for passive nonlinear systems which are contained
More informationMathematical Systems Theory: Advanced Course Exercise Session 5. 1 Accessibility of a nonlinear system
Mathematical Systems Theory: dvanced Course Exercise Session 5 1 ccessibility of a nonlinear system Consider an affine nonlinear control system: [ ẋ = f(x)+g(x)u, x() = x, G(x) = g 1 (x) g m (x) ], where
More informationControl Systems. Laplace domain analysis
Control Systems Laplace domain analysis L. Lanari outline introduce the Laplace unilateral transform define its properties show its advantages in turning ODEs to algebraic equations define an Input/Output
More informationControl design with guaranteed ultimate bound for feedback linearizable systems
Control design with guaranteed ultimate bound for feedback linearizable systems Ernesto Kofman, Fernando Fontenla Hernan Haimovich María M. Seron CONICET; Depto. de Electrónica, Fac. de Cs. Exactas, Ing.
More informationx 3y 2z = 6 1.2) 2x 4y 3z = 8 3x + 6y + 8z = 5 x + 3y 2z + 5t = 4 1.5) 2x + 8y z + 9t = 9 3x + 5y 12z + 17t = 7
Linear Algebra and its Applications-Lab 1 1) Use Gaussian elimination to solve the following systems x 1 + x 2 2x 3 + 4x 4 = 5 1.1) 2x 1 + 2x 2 3x 3 + x 4 = 3 3x 1 + 3x 2 4x 3 2x 4 = 1 x + y + 2z = 4 1.4)
More informationMathematics for Control Theory
Mathematics for Control Theory Geometric Concepts in Control Involutivity and Frobenius Theorem Exact Linearization Hanz Richter Mechanical Engineering Department Cleveland State University Reading materials
More informationNonlinear Observers. Jaime A. Moreno. Eléctrica y Computación Instituto de Ingeniería Universidad Nacional Autónoma de México
Nonlinear Observers Jaime A. Moreno JMorenoP@ii.unam.mx Eléctrica y Computación Instituto de Ingeniería Universidad Nacional Autónoma de México XVI Congreso Latinoamericano de Control Automático October
More information4F3 - Predictive Control
4F3 Predictive Control - Discrete-time systems p. 1/30 4F3 - Predictive Control Discrete-time State Space Control Theory For reference only Jan Maciejowski jmm@eng.cam.ac.uk 4F3 Predictive Control - Discrete-time
More information1 Continuous-time Systems
Observability Completely controllable systems can be restructured by means of state feedback to have many desirable properties. But what if the state is not available for feedback? What if only the output
More informationRobust Control 2 Controllability, Observability & Transfer Functions
Robust Control 2 Controllability, Observability & Transfer Functions Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /26/24 Outline Reachable Controllability Distinguishable
More informationIN [1], an Approximate Dynamic Inversion (ADI) control
1 On Approximate Dynamic Inversion Justin Teo and Jonathan P How Technical Report ACL09 01 Aerospace Controls Laboratory Department of Aeronautics and Astronautics Massachusetts Institute of Technology
More informationStability, Pole Placement, Observers and Stabilization
Stability, Pole Placement, Observers and Stabilization 1 1, The Netherlands DISC Course Mathematical Models of Systems Outline 1 Stability of autonomous systems 2 The pole placement problem 3 Stabilization
More information1 Controllability and Observability
1 Controllability and Observability 1.1 Linear Time-Invariant (LTI) Systems State-space: Dimensions: Notation Transfer function: ẋ = Ax+Bu, x() = x, y = Cx+Du. x R n, u R m, y R p. Note that H(s) is always
More informationNonlinear Control Lecture 5: Stability Analysis II
Nonlinear Control Lecture 5: Stability Analysis II Farzaneh Abdollahi Department of Electrical Engineering Amirkabir University of Technology Fall 2010 Farzaneh Abdollahi Nonlinear Control Lecture 5 1/41
More informationNonlinear Control Systems 1. - Introduction to Nonlinear Systems
Nonlinear Control Systems 1. - Introduction to Nonlinear Systems Dept. of Electrical Engineering Department of Electrical Engineering University of Notre Dame, USA EE658-1 Nonlinear Control Systems 1.
More informationEE Control Systems LECTURE 9
Updated: Sunday, February, 999 EE - Control Systems LECTURE 9 Copyright FL Lewis 998 All rights reserved STABILITY OF LINEAR SYSTEMS We discuss the stability of input/output systems and of state-space
More informationBIFURCATION PHENOMENA Lecture 1: Qualitative theory of planar ODEs
BIFURCATION PHENOMENA Lecture 1: Qualitative theory of planar ODEs Yuri A. Kuznetsov August, 2010 Contents 1. Solutions and orbits. 2. Equilibria. 3. Periodic orbits and limit cycles. 4. Homoclinic orbits.
More informationGeorgia Institute of Technology Nonlinear Controls Theory Primer ME 6402
Georgia Institute of Technology Nonlinear Controls Theory Primer ME 640 Ajeya Karajgikar April 6, 011 Definition Stability (Lyapunov): The equilibrium state x = 0 is said to be stable if, for any R > 0,
More informationCDS 101/110a: Lecture 2.1 Dynamic Behavior
CDS 11/11a: Lecture.1 Dynamic Behavior Richard M. Murray 6 October 8 Goals: Learn to use phase portraits to visualize behavior of dynamical systems Understand different types of stability for an equilibrium
More informationECEN 605 LINEAR SYSTEMS. Lecture 7 Solution of State Equations 1/77
1/77 ECEN 605 LINEAR SYSTEMS Lecture 7 Solution of State Equations Solution of State Space Equations Recall from the previous Lecture note, for a system: ẋ(t) = A x(t) + B u(t) y(t) = C x(t) + D u(t),
More informationControllability, Observability, Full State Feedback, Observer Based Control
Multivariable Control Lecture 4 Controllability, Observability, Full State Feedback, Observer Based Control John T. Wen September 13, 24 Ref: 3.2-3.4 of Text Controllability ẋ = Ax + Bu; x() = x. At time
More informationObservability. Dynamic Systems. Lecture 2 Observability. Observability, continuous time: Observability, discrete time: = h (2) (x, u, u)
Observability Dynamic Systems Lecture 2 Observability Continuous time model: Discrete time model: ẋ(t) = f (x(t), u(t)), y(t) = h(x(t), u(t)) x(t + 1) = f (x(t), u(t)), y(t) = h(x(t)) Reglerteknik, ISY,
More informationAn Approach of Robust Iterative Learning Control for Uncertain Systems
,,, 323 E-mail: mxsun@zjut.edu.cn :, Lyapunov( ),,.,,,.,,. :,,, An Approach of Robust Iterative Learning Control for Uncertain Systems Mingxuan Sun, Chaonan Jiang, Yanwei Li College of Information Engineering,
More informationPoincaré Map, Floquet Theory, and Stability of Periodic Orbits
Poincaré Map, Floquet Theory, and Stability of Periodic Orbits CDS140A Lecturer: W.S. Koon Fall, 2006 1 Poincaré Maps Definition (Poincaré Map): Consider ẋ = f(x) with periodic solution x(t). Construct
More information