Solutions. Problems of Chapter 1

Transcription

1 Solutions Problems of Chapter 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 in IR n n is closed. Therefore det(a is generically different from zero and A is generically invertible. 1. A square, symmetric, real matrix A of order n, that can be seen as a point in IR (n +n/, is positive definite if and only if x t A x > 0 for every x IR n. The trivial case n = 1 is obvious: positiveness is not a generic property for real numbers. The general case can be dealt with by recalling that A is positive definite if and only if the determinants of all its upper-left submatrices A i, i = 1,..., n, are positive. Then, consider the function D : IR (n +n/ IR n defined, for any square, symmetric real matrix A IR (n +n/, by D(A = (det(a 1,..., det(a n T and take e.g. the point P = (1,..., 1, 1 T IR n. Since the matrix A = , belongs to D 1 (P and D is a continuous function in the entries of A, the conterimage D 1 (V ɛ (P of any spherical neighborhood V ɛ (P of P of radius ɛ > 0 is a neighborhood of A. Taking 0 < ɛ 1, D 1 (V ɛ (P consists of non positive definite matrices, then the conclusion follows. 1.3 The controllability matrix [BABA B...A n 1 B]

2 has rank equal to n when at least one of its n n minors has non zero determinant. The same argument as in (1.1 proves that this is a generic property. { e 1/x, ifx < It can be shown that the function f(x = 0, ifx 0 admits continuous derivatives f (m (x of any order m in all points of IR and f (m (0 = 0 for all m. The proof, by induction, is mainly based on the fact that for any integer m lim x 0 x m e 1/x = 0 According to what precedes, the Taylor series of f at 0 is n=0 0 n! xn = 0 f(x unless x 0. Consequently f is not analytic at 0, since it does not admit a Taylor expansion in a neighborhood of x = Integration of one-forms (a The differential of the one-form v = (1 + cos(x + ydx + cos(x + ydy, is dv = ( sin(x+y dx dy +( sin(x+y dy dx = ( sin(x+y dx dy ( sin(x + y dx dy = 0. Then, v is closed and a function ϕ that, locally, satisfies the relation v = dϕ is ϕ = x + sin(x + y. (b The differential of the one-form v = x + y x 3 y dx + 1 xy dy is dv = x4 y 1 1 x 6 dx dy+ y x dy dx = y x dx dy y x dx dy = 0. Then, y v is a closed one-form and a function ϕ that, locally, satisfies the relation v = dϕ is ϕ = x+sin(x+y that, locally, satisfies the relation ϕ = x + y x y. x (c The differential of the one-form v = x + y dx + y x dy is + y dv = xy xy xy + xy (x + y dx dy + (x + y dy dx = (x + y dx dy = 0. Then, v is closed a function ϕ that, locally, satisfies the relation v = dϕ is ϕ = 1 log(x + y. (d The differential of the one-form v = y x + y dx + dv = x + y (x + y dx dy + x + y (x + y dy dx = = x + y ( x + y (x + y dx dy = 0. x x dy is + y

3 Then, v is closed and a function ϕ that, locally, ( satisfies the relation v = dϕ can be found, in this case, computing y x + y dx. We find ϕ = arctan( x y. (e The differential of the one-form x v = x + y + z dx + y x + y + z dy + z x + y dz is + z xy dv = (x + y + z dx dy + xz (x + y + z dx dz + xy (x + y + z dy dx+ yz + (x + y + z dy dz + xz (x + y + z dz dx + yz (x + y + z dz dy = xy + xy xz + xz yz + yz = (x + y + z dx dy + (x + y + z dx dz + (x + y + z dy dz = 0 Then, v is closed and a function ϕ that, locally, ( satisfies the relation x v = dϕ can be found, in this case, computing x + y + z dx. We find ϕ = 1 log(x + y + z. (f The differential of the one-form y + z v = (x y z dx + x (x y z dy + x dz is (x y z dv = x y z x y z x y z 3 dx dy + 3 dx dz + 3 dy dx+ (x y z (x y z (x y z x x y z + 3 dy dz + (x y z (x y z 3 dz dx + x 3 dz dy = (x y z x y z ( x y z x y z ( x y z = (x y z 3 dx dy + (x y z 3 dx dz+ x x + 3 dy dz = 0 (x y z Then, v is closed and a function ϕ that, locally, ( satisfies the relation y + z v = dϕ can be found, in this case, computing (x y z dx. We find ϕ = y + z x y z. 1.6 The one-form ω = ( x 3 cos(ydx + (xsin(ydy is not closed, since we have dω = (x 3 sin(y dx dy + sin(y dy dx 0 However,

4 dω ω = ( x 6 sin(y cos(y dx dy dx + x 4 sin(y dy dx dy+ x 3 sin(y cos(y dy dx dx + x sin(y dy dx dy = 0 Then, by theorem 1.15, ω is colinear to an exact form, i.e. there exist λ and ϕ in K such that λω = dϕ. The integrating factor λ must satisfy the relation d(λω = 0. Since [ d(λω = x 3 cos(y λ ] [ ] λ y + λx3 sin(y dx dy + xsin(y + λsin(y dy dx = [ x = x 3 cos(y λ y + λ x3 sin(y xsin(y λ ] x λ sin(y dx dy [ = x 3 cos(y λ ( ] dλ y + sin(y dx x + (1 x3 λ dx dy = 0 Assume that λ depend only on x, i.e. equivalent to λ y λ = 1 x exp (x3 /3 solves equation (1.51 and (1.50 = 0, then equation (1.50 is dλ dx x + (1 x3 λ = 0 (1.51 λω = ( x exp (x 3 /3 cos(ydx + (exp (x 3 /3sin(ydy is, by construction, a closed one-form. A function ( ϕ that, locally, satisfies the relation dϕ = λω can be found computing x cos(y e x3 /3 dx. We find ϕ = e x3 /3 cos(y. 1.7 Exterior differentiation dω = (cos(x + y dx dy + (x dy dx = [cos(x + y x] dx dy. (b dω = sin(z dx dz (c ω = (x dx dy, dω = 0. (d dω = dy dz dx = dy dx dz = dx dy dz (e dω = Since we have that dξ i0... dξ ij... dξ ik... dξ is = dξ i0... dξ ik... dξ ij... dξ is, if dξ ij = dξ ik for some index j and k, the conclusion follows. 1.9 Exterior product (a dx (sin(ydy (x dx + (y dy = = (sin(y dx dy ( x dx + y dy = = xsin(y dx dy dx + sin(y y dx dy dy = 0 by 1.8.

5 ( (b cos(xydx + (y 3 dy (z dx + y dz = = z cos(xy dx dx + y cos(xy dx dz + y 3 z dy dx + y 4 dy dz = = z y 3 dx dy + y cos(xy dx dz + y 4 dy dz (c (x dx + (x + y dy + (1 zdz (ydx xdz = = x dx dz y(x + y dx dy x(x + y dy dz y(1 z dx dz =. = y(x + y dx dy + ( x y + yz dx dz x(x + y dy dz (d (e x dx dy + x dy dz (e (dx dy (cos(x + ydy dz = cos(x + ydx dy dy dz = 0 by 1.8. Problems of Chapter.1 The ball and beam 1. Set x 1 = r, x = ṙ, u = α, and y = r. Then, ẋ 1 = x ( 1 J ẋ = R + m [ mx1 u mg sin u ] y = x 1 is the generalized realization which is sought.. Introduce the notation x 3 = u, and x 4 = u. Write the extended system ẋ 1 = x ( 1 J ẋ = R + m [ ] mx1 x 4 mg sin x 3 ẋ 3 = x 4 ẋ 4 = ü y = x 1 Compute H 1 = span{dx 1, dx, dx 3, dx 4 } H = span{dx 1, dx, dx 3 } H 3 = span{dx 1, ω} ( J where ω = R + m dx mx 1 x 4 dx 3. Since H 3 is not fully integrable, there does not exist any classical state-space realization.. The actuated ball and beam Set x 1 = r, x = ṙ, x 3 = α, x 4 = α, u = α, and y = r. Then, a classical realization is obviously obtained as

6 ẋ 1 = x ( 1 J ẋ = R + m [ ] mx1 x 4 mg sin x 3 ẋ 3 = x 4 ẋ 4 = u y = x 1. The pendulum on a cart Theorem.16 can not be applied directly as the input-output equation is not given. To cope with the auxiliary variable r, it is necessary to derive the socalled inverse dynamic model, i.e. to write explicitly r and θ. From the direct dynamic model (M + m r + bṙ + ml θ cos θ ml θ sin θ = F (I + ml θ + mgl sin θ = ml r cos θ the elimination of θ is obtained by multiplying the first equation by (I + ml and the second equation by ( ml cos θ. The resulting summation yields A r + b(i + ml ṙ ml(i + ml θ sin θ m gl sin θ cos θ = (I + ml F where A = (I + ml (M + m m l cos θ. Thus, r = b(i + ml ṙ + ml(i + ml θ sin θ + m gl sin θ cos θ + (I + ml F (I + ml (M + m m l cos θ The elimination of r in the direct dynamic model is obtained in a similar vein by multiplying the first equation by ml cos θ and the second equation by (M + m: B θ + bmlṙ cos θ m l sin θ cos θ = (M + mmgl sin θ + ml cos θf where B = m l cos θ (M + m(i + ml. Thus, θ = bmlṙ cos θ + m l sin θ cos θ + (M + mmgl sin θ + ml cos θf m l cos θ (M + m(i + ml Set x 1 = r, x = ṙ, x 3 = θ, and x 4 = θ, then direct inspection of the inverse dynamic model yields the following classical realization: x b(i+ml x +ml(i+ml x 4 sin x 3+m gl sin x 3 cos x 3 (I+ml ẋ = (M+m m l cos x 3 x 4 bmlx cos x 3+m l sin x 3 cos x 3+(M+mmgl sin x 3 m l cos x 3 (M+m(I+ml 0 (I+ml + (I+ml (M+m m l cos x 3 0 F ml cos x 3 m l cos x 3 (M+m(I+ml y = x 1

7 Problems of Chapter The actuated ball and beam The dynamics were obtained as ẋ 1 = x ( 1 J ẋ = R + m [ ] mx1 x 4 mg sin x 3 ẋ 3 = x 4 ẋ 4 = u Compute H = span{dx 1, dx, dx 3 } Obviously, dx 1 H 3 since its time-derivative is in H. Let ω be a general vector of H 3, thus ω H and ω = α 1 dx 1 + α dx + α 3 dx 3 where α i K, for i = 1,, 3. The problem consists in finding all solutions (α 1, α, α 3 such that ω H. Compute: ω = i α i dx i + i α i dẋ i Since i α idx i H for any α i, write ω = + i α idẋ i ( 1 J = + α 1 dx + α m R + m d [ ] x 1 x 4 g sin x 3 + α3 dx 4 ( 1 J = + [α m R + m x 1 x 4 + α 3 ]dx 4 ( 1 J Thus, ω H if and only if α m R + m x 1 x 4 + α 3 = 0. A nonzero solution of the latter is and α = 1 α 3 = m ( 1 J R + m x 1 x 4 ( 1 J H 3 = span{dx 1, dx m R + m x 1 x 4 dx 3 } To compute H 4, denote again ω as a general vector of H 3 : ( ] 1 J ω = α 1 dx 1 + α [dx m R + m x 1 x 4 dx 3

8 Compute ω, and denote terms which belong to H 3 by ( 1 J ω = + α 1 dx + α [m R + m d [ ] x 1 x 4 g sin x 3 ( 1 ( ] 1 J J m R + m (x x 4 + x 1 udx 3 m R + m x 1 x 4 dx 4 ( 1 ( 1 J J = + α 1 dx + α [ mg R + m cos x 3 m R + m (x x 4 + x 1 u] dx 3 ( 1 ( ] 1 J J +α [m R + m x 1 x 4 m R + m x 1 x 4 dx 4 ( 1 ( 1 J J = + α 1 dx + α [ mg R + m cos x 3 m R + m (x x 4 + x 1 u] dx 3 ω H 3 if, for instance, ( J α 1 = mg α = m R + m ( 1 J R + m x 1 x 4 1 ( J cos x 3 m 1 R + m (x x 4 + x 1 u Thus, H 4 = span { [ ( J mg R + m [ + m ( J R + m 1 ( J cos x 3 m 1 R + m (x x 4 + x 1 u] dx 1 ] [ 1 ( ] 1 J x 1 x 4 dx m R + m x 1 x 4 dx 3 } Since dimh 4 = 1, either H 5 = H 4 or H 5 = 0. By time derivation of the one-form in H 4, one checks that H 5 = 0 and the system is thus accessible. 3. The hopping robot The dynamics (3.19 of the hopping robot are x x 1 x 6 ẋ = x x 6 x x 6 x /m 0 0 1/J mx 1 0 ( u1 Its controllability indices are derived from the dimensions of the H k spaces. Compute H = span{dx 1, dx 3, dx 5, Jdx 4 + mx 1dx 6 } u

9 To compute H 3, consider a general vector ω H : ω = α 1 dx 1 + α dx 3 + α 3 dx 5 + α 4 (Jdx 4 + mx 1dx 6 Compute ω, and denote terms which belong to H by ω = + α 1 dx + α dx 4 + α 3 dx 6 + α 4 [mx 1 x dx 6 mx 1d( x x 6 x 1 = + α 1 dx + α dx 4 + α 3 dx 6 + α 4 [mx 1 x dx 6 mx 1 x 6 dx mx 1 x dx 6 ] = + α 1 dx + α dx 4 + α 3 dx 6 mx 1 x 6 α 4 dx = + (α 1 mx 1 x 6 α 4 dx + α dx 4 + α 3 dx 6 The constraint ω H has to independent solutions: α 1 = 0, α 4 = 0, α = J, and α 3 = mx 1 α 1 = mx 1 x 6, α 4 = 1, α = 0, and α 3 = 0 and thus, H 3 = span{jdx 3 + mx 1dx 5, mx 1 x 6 dx 1 + Jdx 4 + mx 1dx 6 } = span{jdx 3 + mx 1dx 5, d(jx 4 + mx 1x 6 } As it has been noted in Section 3.8, d(jx 4 + mx 1x 6 H ; thus, either H 4 = H 3 or H 4 = H. The time derivative of Jdx 3 + mx 1dx 5 equals Jdx 4 + mx 1 x dx 5 + mx 1dx 6 which does not belong to H 3, and consequently H 4 = H. Finally, the two controllability indices are {, 3}. 3.3 Linear systems Consider the linear system ẋ = Ax + Bu: H = B where B denotes the left kernel of matrix B. More generally, H k = [B, AB,..., A k B] Due to Caley-Hamilton Theorem, H = [B, AB,..., A n 1 B]. By duality, H = 0 rank[b, AB,..., A n 1 B] = n ] Problems of Chapter Given a system Σ of the form { ẋ(t = f(x(t + g(x(tu(t Σ = y(t = h(x(t

10 O is the limit of the observability filtration, 0 O 0 O 1 O... O k..., where O k := X (Y k + U and Y (k = span{dy (k, k 0}. 4. As seen above, O is spanned by the differential of dy (k, k 0, when the contribution of du (j, j 0, has been discarded, namely by y(k x. See, for instance the following example. Consider the system ẋ 1 = x + x 3 u ẋ = x 1 ẋ 3 = x y = x 1 for which and Therefore we have ẏ = x + x 3 u y ( = x 1 + x u + x 3 u y (3 = x + (x 1 + x 3 u + x u + x 3 ü,... dy = dx 1 dẏ = dx + udx 3 + x 3 du dÿ = dx 1 + udx + udx 3 + x 1 du + x 3 d u dy (3 = udx 1 + (1 + udx + (u + üdx 3 + (x 1 + x 3 du+ +(x d u + x 3 dü,... O 0 = {dx 1 } O 1 = span{dx 1, dx + udx 3 } O = span{dx 1, dx + udx 3, udx + udx 3 } and O is spanned by the columns of the matrix u 1 u u and its dimension coincides with rank K [ (y, ẏ,..., y (n 1 x ]. 4.3 Given the linear system { ẋ = Ax + Bu y = Cx

11 we have and Then, and ẏ = CAx + CBu ÿ = CA x + CABu + CB u,... dẏ = CAdx + CBdu dÿ = CA dx + CABdu + CBd u,... Y 0 = span{cdx} Y 1 = span{cdx, CAdx, CBdu} Y = span{cdx, CAdx, CA dx, CABdu, CBd u},... O 0 = span{dx} span{cdx} = span{cdx} O 1 = span{dx} span{cdx, CAdx, CBdu} = span{cdx, CAdx} O = span{dx} span{cdx, CAdx, CA dx, CABdu, CBd u} = = span{cdx, CAdx, CA dx},... and O = span{cdx, CAdx, CA dx,..., CA n 1 dx} Problems of Chapter The observability assumption does not play any role here. By contradiction, assume that the relative degree of the output is infinite, then dh H, thus dh = 0, which stands in contradiction with the assumption that h(x is nonconstant Compute ẏ 1 = u 1 ; thus, the order n 1 of the zero at infinity of output y 1 is n 1 = 1. In a similar vein, ẏ = x 3 u 1 and n = 1.. The list of orders of the zeros at infinity may be obtained applying the inversion, or structure, algorithm: Step 1: ẏ 1 = u 1 ẏ = x 3 ẏ 1 The system thus has one zero at infinity of order 1. Step : ÿ = x 3 ÿ 1 + x 4 ẏ 1 Step k, for 3 k n : Step n-1: y (k = x 3 y (k x k+ ẏ k 1

12 y (n 1 = x 3 y (n ẏ n 1 1 u The system thus has one zero at infinity of order n 1. The list of orders of the zeros at infinity is {1, n 1}. 3. The structure at infinity of the system is defined by one zero at infinity of order 1 and one zero at infinity of order n The rank of the systems equals, the total number of zeros at infinity. 5.3 Since the system has inputs and 1 output, its rank is at most 1 and the system can not be left-invertible. By inversion, the first time derivative of y depends on the input and the system is thus right-invertible: A right-inverse is obtained as ẏ = x u 1 + u η 1 = ẏ η = u 1 u = ẏ η u Since the system has 1 input and outputs, its rank is at most 1 and the system can not be right-invertible. The first time-derivative of y 1 depends on u, the system is thus left-invertible: ẏ 1 = x u A (dynamics free left-inverse system as u = ẏ 1 /y An alternative solution is provided by the computation of the time-derivative of y as ÿ = u which represents obviously a dynamics free left-inverse. 5.5 Compute ẏ = x + u The reduced order inverse system is obtained as η = η + ẏ u = η + ẏ Constraining y 0 in the above dynamic yields the zero dynamic η = η

13 Note that the given system is linear, its transfer function is (s + 1/s, and its transmission zero equals 1 which is consistent with the results obtained from the state equations. 5.6 Compute ẏ = x u The reduced order inverse system is obtained as η = η ẏ u = η ẏ Constraining y 0 in the above dynamic yields the zero dynamic η = η Note that the given system is linear, its transfer function is (s 1/s, and its transmission zero equals +1 which is consistent with the results obtained from the state equations. 5.7 Compute ẏ = x + x u The reduced order inverse system is obtained as η = η 3 + ηẏ u = η + ẏ/η Constraining y 0 in the above dynamic yields the zero dynamic η = η 3 Problem of Chapter The generalized transformations are obtained from the inversion algorithm: y 1 = x 1 =: ζ 1 y = x =: ζ ẏ 1 = cos x 3 u 1 =: v 1 ẏ = sin x 3 u 1 =: ζ 3 ÿ = sin x 3 u 1 + cos x 3 u 1 u =: v

14 They yield the canonical form ζ 1 = v 1 ζ = ζ 3 ζ 3 = v y 1 = ζ 1 y = ζ The generalized transformation has the following inverse transformation x 1 = ζ 1 x = ζ x 3 = arctan(ζ 3 /v 1 u 1 = v 1 / cos[arctan(ζ 3 /v 1 ] u = v (ζ 3 v 1 v 1 + ζ 3 v1 Problems of Chapter The unicycle Both relative degrees are 1 since ẏ 1 = u 1 cos x 3 ẏ = u 1 sin x 3 The decoupling matrix is then obtained as [ cos x3 ] 0 sin x 3 0 and its rank equals 1. The condition (7.3 of Theorem 7.3 is not fulfilled. 7. The unicycle continued Both relative degrees are since ÿ 1 = v 1 cos x 3 v x 4 sin x 3 ÿ = v 1 sin x 3 + v x 4 cos x 3 The decoupling matrix is then obtained as [ ] cos x3 x 4 sin x 3 sin x 3 x 4 cos x 3 and its rank equals. The condition (7.3 of Theorem 7.3 is fulfilled.

15 Problem of Chapter The unicycle A dynamic state feedback is derived from the inversion equations. ẏ 1 = u 1 cos x 3 ÿ = ÿ 1 tan x 3 + ẏ 1 (1 + tan x 3 u Set η = u 1 cos x 3, ÿ = v, and ÿ 1 = v 1 which yields the definition of the following dynamic state feedback η = v 1 u 1 = η/ cos x 3 u = v v 1 tan x 3 η(1+tan x 3 A quasi-static state feedback that solves the noninteracting control problem is obtained by setting w 1 = ẏ 1, and w = ÿ : { u1 = w 1 / cos x 3 u = w ẇ 1 tan x 3 w 1 (1+tan x 3 Problem of Chapter The ball and beam Recall the considered state equations ẋ 1 = x ( 1 J ẋ = R + m [ ] mx1 x 4 mg sin x 3 ẋ 3 = x 4 ẋ 4 = u 1. The H k spaces have been computed in Problem 3.1. Obviously, H 3 is not fully integrable; thus, the dynamics are not fully linearizable.. H 4 is not fully integrable either, so, there is no function of the state which has relative degree 4. The exact one-form dx 1 can be picked in H 3 that has relative degree 3. Set z 1 = x 1 z = x ( 1 J z 3 = mr + 1 [ ] x1 x 4 g sin x 3 z 4 = x ( 4 1 J v = mr + 1 [ x x 4 gx 4 cos x 3 + x 1 x 4 u ]

16 Solving the last equation in u, yields the state feedback [ ( ] J u = gx 4 cos x 3 x x 4 + mr + 1 v /x 1 x 4 The closed-loop systems displays the subsystem of order 3: ż 1 = z ż = z 3 ż 3 = v Problem of Chapter Virus dynamics Recall the virus dynamics model from Section.10. T = s δt βt v T = βt v µt v = kt cv y 1 = T y = v This model is linear up to the single input-output injection βt v. Considering the input β, on defines the static output feedback and the closed loop reads that is fully linear. β = w/y 1 y T = s δt w T = w µt v = kt cv y 1 = T y = v

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