# Linear Control Systems

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1 Chapter 3 Linear Control Systems Topics : 1. Controllability 2. Observability 3. Linear Feedback 4. Realization Theory Copyright c Claudiu C. Remsing, 26. All rights reserved. 7

2 C.C. Remsing 71 Intuitively, a control system should be designed so that the input u( ) controls all the states; and also so that all states can be observed from the output y( ). The concepts of (complete) controllability and observability formalize these ideas. Another two fundamental concepts of control theory feedback and realization are introduced. Using (linear) feedback it is possible to exert a considerable influence on the behaviour of a (linear) control system.

3 72 AM3.2 - Linear Control 3.1 Controllability An essential first step in dealing with many control problems is to determine whether a desired objective can be achieved by manipulating the chosen control variables. If not, then either the objective will have to be modified or control will have to be applied in some different fashion. We shall discuss the general property of being able to transfer (or steer) a control system from any given state to any other by means of a suitable choice of control functions Definition. The linear control system Σ defined by ẋ = A(t)x + B(t)u(t) (3.1) where A(t) R m m and B(t) R m l, is said to be completely controllable (c.c.) if for any, any initial state x( ) = x, and any given final state x f, there exist a finite time t 1 > and a control u :, t 1 R l such that x(t 1 ) = x f. Note : (1) The qualifying term completely implies that the definition holds for all x and x f, and several other types of controllability can be defined. (2) The control u( ) is assumed piecewise-continuous in the interval, t Example. Consider the control system described by ẋ 1 = a 1 x 1 + a 2 x 2 + u(t) ẋ 2 = x 2. Clearly, by inspection, this is not completely controllable (c.c.) since u( ) has no influence on x 2, which is entirely determined by the second equation and x 2 ( ). We have x f = Φ(t 1, ) x + Φ(, τ)b(τ)u(τ) dτ

4 C.C. Remsing 73 or = Φ(t 1, ) x Φ(, t 1 )x f + Φ(, τ)b(τ)u(τ) dτ. Since Φ(t 1, ) is nonsingular it follows that if u( ) transfers x to x f, it also transfers x Φ(, t 1 )x f to the origin in the same time interval. Since x and x f are arbitrary, it therefore follows that in the controllability definition the given final state can be taken to be the zero vector without loss of generality. Note : For time-invariant control systems in the controllability definition the initial time can be set equal to zero. The Kalman rank condition For linear time-invariant control systems a general algebraic criterion (for complete controllability) can be derived Theorem. The linear time-invariant control system ẋ = Ax + Bu(t) (3.2) (or the pair (A, B) ) is c.c. if and only if the (Kalman) controllability matrix has rank m. Proof : C = C(A, B) : = B AB A 2 B... A m 1 B R m ml ( ) We suppose the system is c.c. and wish to prove that rank (C) = m. This is done by assuming rank(c) < m, which leads to a contradiction. Then there exists a constant row m-vector q such that In the expression qb =, qab =,..., qa m 1 B =. x(t) = exp(ta) x + t exp( τa)bu(τ) dτ

5 74 AM3.2 - Linear Control for the solution of (3.2) subject to x() = x, set t = t 1, x(t 1 ) = to obtain (since exp(t 1 A) is nonsingular) x = exp( τ A)Bu(τ) dτ. Now, exp( τa) can be expressed as some polynomial r(a) in A having degree at most m 1, so we get x = (r I m + r 1 A + + r m 1 A m 1 )Bu(τ) dτ. Multiplying this relation on the left by q gives qx =. Since the system is c.c., this must hold for any vector x, which implies q =, contradiction. ( ) We asume rank(c) = m, and wish to show that for any x there is a function u :, t 1 R l, which when substituted into t x(t) = exp(ta) x + exp( τa)bu(τ) dτ produces Consider the symmetric matrix W c : = x(t 1 ) =. exp( τa)bb T exp( τa T ) dτ. (3.3) One can show that W c is nonsingular. Indeed, consider the quadratic form associated to W c α T W c α = = ψ(τ)ψ T (τ) dτ ψ(τ) 2 e dτ where α R m 1 is an arbitrary column vector and ψ(τ) : = α T exp ( τa)b. It is clear that W c is positive semi-definite, and will be singular only if there

6 C.C. Remsing 75 exists an ᾱ such that ᾱ T W c ᾱ =. However, in this case, it follows (using the properties of the norm) that ψ(τ), τ t 1. Hence, we have ( ) ᾱ T I m τa + τ2 2! A2 τ3 3! A3 + B =, τ t 1 from which it follows that ᾱ T B =, ᾱ T AB =, ᾱ T A 2 B =, This implies that ᾱ T C =. Since by assumption C has rank m, it follows that such a nonzero vector ᾱ cannot exist, so W c is nonsingular. Now, if we choose as the control vector then substitution into (3.3) gives u(t) = B T exp( ta T )W 1 c x, t, t 1 x(t 1 ) = exp(t 1 A) x exp( τa)bb T exp( τa T ) dτ (Wc 1 x ) = exp(t 1 A) x W c Wc 1 x = as required Corollary. If rank(b) = r, then the condition in Theorem reduces to rank B AB... A m r B = m. Proof : Define the matrix C k : = B AB A k B, k =, 1, 2... If rank (C j ) = rank(c j+1 ) it follows that all the columns of A j+1 B must be linearly dependent on those of C j. This then implies that all the columns of A j+2 B, A j+3 B,... must also be linearly dependent on those of C j, so that rank(c j ) = rank(c j+1 ) = rank(c j+2 ) =

7 76 AM3.2 - Linear Control Hence the rank of C k increases by at least one when the index k is increased by one, until the maximum value of rank (C k ) is attained when k = j. Since rank (C ) = rank(b) = r and rank(c k ) m it follows that r+j m, giving j m r as required Example. Consider the linear control system Σ described by ẋ = x + u(t). 1 1 The (Kalman) controllability matrix is 1 2 C = C Σ = 1 which has rank 2, so the control system Σ is c.c. Note : When l = 1, B reduces to a column vector b and Theorem can be restated as : A linear control system in the form ẋ = Ax + bu(t) can be transformed into the canonical form ẇ = Cw + du(t) if and only if it is c.c. Controllability criterion We now give a general criterion for (complete) controllability of control systems (time-invariant or time-varying) as well as an explicit expression for a control vector which carry out a required alteration of states Theorem. The linear control system Σ defined by ẋ = A(t)x + B(t)u(t)

8 C.C. Remsing 77 is c.c. if and only if the symmetric matrix, called the controllability Gramian, W c (, t 1 ) : = is nonsingular. In this case the control Φ(, τ)b(τ)b T (τ)φ T (, τ) dτ R m m (3.4) u (t) = B T (t)φ T (, t)w c (, t 1 ) 1 x Φ(, t 1 )x f, t, t 1 transfers x( ) = x to x(t 1 ) = x f. Proof : ( ) Sufficiency. If W c (, t 1 ) is assumed nonsingular, then the control defined by u (t) = B T (t)φ T (, t)w c (, t 1 ) 1 x Φ(, t 1 )x f, t, t 1 exists. Now, substitution of the above expression into the solution t x(t) = Φ(t, ) x + Φ(, τ)b(τ)u(τ) dτ of ẋ = A(t)x + B(t)u(t) gives x(t 1 ) = Φ(t 1, ) x + Φ(, τ)b(τ)( B T (τ)φ T (, τ)w c (, t 1 ) 1 x Φ(, t 1 )x f ) dτ = Φ(t 1, ) x W c (, t 1 )W c (, t 1 ) 1 x Φ(, t 1 )x f = Φ(t 1, ) x x + Φ(, t 1 )x f = Φ(t 1, )Φ(, t 1 )x f = x f. ( ) Necessity. We need to show that if Σ is c.c., then W c (, t 1 ) is nonsingular. First, notice that if α R m 1 is an arbitrary column vector, then from

9 78 AM3.2 - Linear Control (3.4) since W = W c (, t 1 ) is symmetric we can construct the quadratic form α T Wα = = θ T (τ, )θ(τ, ) dτ θ 2 e dτ where θ(τ, ) : = B T (τ)φ T (, τ)α, so that W c (, t 1 ) is positive semi-definite. Suppose that there exists some ᾱ such that ᾱ T Wᾱ =. Then we get (for θ = θ when α = ᾱ) θ 2 e dτ = which in turn implies (using the properties of the norm) that θ(τ, ), τ t 1. However, by assumption Σ is c.c. so there exists a control v( ) making x(t 1 ) = if x( ) = ᾱ. Hence Therefore ᾱ = Φ(, τ)b(τ)v(τ) dτ. ᾱ 2 e = ᾱ T ᾱ = = v T (τ)b T (τ)φ T (, τ)ᾱdτ v T (τ) θ(τ, ) dτ = which contradicts the assumption that ᾱ. Hence W c (, t 1 ) is positive definite and is therefore nonsingular Example. The control system is ẋ = x + u(t) Observe that λ = is an eigenvalue of A, and b = is a corresponding 1 eigenvector, so the controllability rank condition does not hold. However, A is

10 C.C. Remsing 79 1 similar to its companion matrix. Using the matrix T = 1 1 before (see Example 2.5.1) and w = Tx we have the system 1 1 ẇ = w + u. 3 computed Differentiation of the w 1 equation and substitution produces a second-order ODE for w 1 : ẅ 1 + 3ẇ 1 = 3u + u. One integration produces a first-order ODE ẇ 1 + 3w 1 = 3 u(τ) dτ + u which shows that the action of arbitrary inputs u( ) affects the dynamics in only a one-dimensional space. The original x equations might lead us to think that u( ) can fully affect x 1 and x 2, but notice that the w 2 equation says that u( ) has no effect on the dynamics of the difference x 1 x 2 = w 2. Only when the initial condition for w involves w 2 () = can u( ) be used to control a trajectory. That is, the inputs completely control only the states that lie in the subspace span b Ab = span {b} = span Solutions starting with x 1 () = x 2 () satisfy x 1 (t) = x 2 (t) = t 1 1 u(τ) dτ + x 1 (). One can steer along the line x 1 = x 2 from any initial point to any final point x 1 (t 1 ) = x 2 (t 1 ) at any finite time t 1 by appropriate choice of u( ). On the other hand, if the initial condition lies off the line x 1 = x 2, then the difference w 2 = x 1 x 2 decays exponentially so there is no chance of steering to an arbitrarily given final state in finite time..

11 8 AM3.2 - Linear Control Note : The control function u ( ) which transfers the system from x = x( ) to x f = x(t 1 ) requires calculation of the state transition matrix Φ(, ) and the controllability Gramian W c (, τ ). However, this is not too dificult for linear timeinvariant control systems, although rather tedious. Of course, there will in general be many other suitable control vectors which achieve the same result Proposition. If u( ) is any other control taking x = x( ) to x f = x(t 1 ), then u(τ) 2 e dτ > u (τ) 2 e dτ. Proof : Since both u and u satisfy x f = Φ(t 1, ) x + Φ(, τ)b(τ)u(τ) dτ we obtain after subtraction = Φ(, τ)b(τ) u(τ) u (τ) dτ. Multiplication of this equation on the left by x Φ(, t 1 )x f T W c (, t 1 ) 1 T gives and thus Therefore (u ) T (τ) u (τ) u(τ) dτ = u (τ) 2 e dτ = (u ) T (τ)u(τ) dτ. < t u (τ) u(τ) 2 e dτ = = = u (τ) u(τ) T u (τ) u(τ) dτ ( u(τ) 2 e + u (τ) 2 e 2(u ) T (τ)u(τ) ) dτ ( u(τ) 2 e u (τ) 2 ) e dτ

12 C.C. Remsing 81 and so u(τ) 2 e dτ = ( u (τ) 2 e + u (τ) u(τ) 2 ) t1 e dτ > u (τ) 2 e dτ as required. Note : This result can be interpreted as showing that the control u (t) = B T (t)φ T (, t)w c (, t 1 ) 1 x Φ(, t 1 )x f is optimal, in the sense that it minimizes the integral ( u(τ) 2 e dτ = u 2 1 (τ) + u 2 2(τ) + + u 2 l(τ) ) dτ over the set of all (admissible) controls which transfer x = x( ) to x f = x(t 1 ), and this integral can be thought of as a measure of control energy involved. Algebraic equivalence and decomposition of control systems We now indicate a further aspect of controllability. Let P( ) be a matrixvalued mapping which is continuous and such that P(t) is nonsingular for all t. (The continuous maping P :, ) GL(m, R) is a path in the general linear group GL (m, R).) Then the system Σ obtained from Σ by the transformation x = P(t)x is said to be algebraically equivalent to Σ Proposition. If Φ(t, ) is the state transition matrix for Σ, then is the state transition matrix for Σ. P(t)Φ(t, )P 1 ( ) = Φ(t, ) Proof : We recall that Φ(t, ) is the unique matrix-valued mapping satisfying Φ(t, ) = A(t)Φ(t, ), Φ(, ) = I m

13 82 AM3.2 - Linear Control and is nonsingular. Clearly, Φ(, ) = I m ; differentiation of x = P(t)x gives x = Px + Pẋ = (Ṗ + PA)x + PBu = (Ṗ + PA)P 1 x + PBu. We need to show that Φ is the state transition matrix for x = ( P + PA)P 1 x + PBu. We have Φ(t, ) = d P(t)Φ(t, t )P 1 ( ) dt = P(t)Φ(t, )P 1 ( + P(t) Φ(t, )P 1 ( ) ( ) = P(t) + P(t)A(t) P 1 (t) P(t)Φ(t, )P 1 ( ) ( ) = P(t) + P(t)A(t) P (t) 1 Φ(t, t ) Proposition. If Σ is c.c., then so is Σ. Proof : The system matrices for Σ are so the controllability matrix for Σ is W = = Ã = ( P + PA)P 1 and B = PB Φ(t, τ) B(τ) B T (τ) Φ T (, τ) dτ P( )Φ(, τ)p 1 (τ)p(τ)b(τ)b T (τ)p T (τ) ( P 1 (τ) ) T Φ T (, τ)p T ( ) dτ = P( )W c (, t 1 )P T ( ).

14 C.C. Remsing 83 Thus the matrix W = W c (, t 1 ) is nonsingular since the matrices W c (, t 1 ) and P( ) each have rank m. The following important result on system decomposition then holds : Theorem. When the linear control system Σ is time-invariant then if the controllability matrix C Σ has rank m 1 < m there exists a control system, algebraically equivalent to Σ, having the form ẋ(1) A1 A 2 x(1) B1 = + u(t) ẋ (2) A 3 x (2) y = C 1 C 2 x where x (1) and x (2) have orders m 1 and m m 1, respectively, and (A 1, B 1 ) is c.c. We shall postpone the proof of this until a later section (see the proof of Theorem 3.4.5) where an explicit formula for the transformation matrix will also be given. Note : It is clear that the vector x (2) is completely unaffected by the control u( ). Thus the state space has been divided into two parts, one being c.c. and the other uncontrollable. 3.2 Observability Closely linked to the idea of controllability is that of observability, which in general terms means that it is possible to determine the state of a system by measuring only the output Definition. The linear control system (with outputs) Σ described by ẋ = A(t)x + B(t)u(t) (3.5) y = C(t)x

15 84 AM3.2 - Linear Control is said to be completely observable (c.o.) if for any and any initial state x( ) = x, there exists a finite time t 1 > such that knowledge of u( ) and y( ) for t, t 1 suffices to determine x uniquely. Note : There is in fact no loss of generality in assuming u( ) is identically zero throughout the interval. Indeed, for any input u :, t 1 R l and initial state x, we have Defining we get t y(t) C(t)Φ(t, τ)b(τ)u(τ)dτ = C(t)Φ(t, )x. t ŷ(t) : = y(t) C(t)Φ(t, τ)b(τ)u(τ)dτ ŷ(t) = C(t)Φ(t, )x. Thus a linear control system is c.o. if and only if knowledge of the output ŷ( ) with zero input on the interval, t 1 allows the initial state x to be determined Example. Consider the linear control system described by ẋ 1 = a 1 x 1 + b 1 u(t) ẋ 2 = a 2 x 2 + b 2 u(t) y = x 1. The first equation shows that x 1 ( ) (= y( )) is completely determined by u( ) and x 1 ( ). Thus it is impossible to determine x 2 ( ) by measuring the output, so the system is not completely observable (c.o.) Theorem. The linear control system Σ is c.o. if and only if the symmetric matrix, called the observability Gramian, W o (, t 1 ) : = is nonsingular. Φ T (τ, )C T (τ)c(τ)φ(τ, ) dτ R m m (3.6)

16 C.C. Remsing 85 Proof : ( ) Sufficiency. Assuming u(t), t, t 1, we have y(t) = C(t)Φ(t, )x. Multiplying this relation on the left by Φ T (t, )C T (t) and integrating produces Φ T (τ, )C T (τ)y(τ) dτ = W o (, t 1 )x so that if W o (, t 1 ) is nonsingular, the initial state is so Σ is c.o. x = W o (, t 1 ) 1 Φ T (τ, )C T (τ)y(τ) dτ ( ) Necessity. We now assume that Σ is c.o. and prove that W = W o (, t 1 ) is nonsingular. First, if α R m 1 is an arbitrary column vector, α T Wα = (C(τ)Φ(τ, )α) T C(τ)Φ(τ, )α dτ so W o (, t 1 ) is positive semi-definite. Next, suppose there exists an ᾱ such that ᾱ T Wᾱ =. It then follows that C(τ)Φ(τ, )ᾱ, τ t 1. This implies that when x = ᾱ the output is identically zero throughout the time interval, so that x cannot be determined in this case from the knowledge of y( ). This contradicts the assumption that Σ is c.o., hence W o (, t 1 ) is positive definite, and therefore nonsingular. Note : Since the observability of Σ is independent of B, we may refer to the observability of the pair (A, C). Duality Theorem. The linear control system (with outputs) Σ defined by ẋ = A(t)x + B(t)u(t) y = C(t)x

17 86 AM3.2 - Linear Control is c.c. if and only if the dual system Σ defined by ẋ = A T (t)x + C T (t)u(t) y = B T (t)x is c.o.; and conversely. Proof : We can see that if Φ(t, ) is the state transition matrix for the system Σ, then Φ T (, t) is the state transition matrix for the dual system Σ. Indeed, differentiate I m = Φ(t, )Φ(t, ) 1 to get = d dt I m = Φ(t, )Φ(t, ) 1 + Φ(t, ) Φ(, t) = A(t)Φ(t, )Φ(t, ) 1 + Φ(t, ) Φ(, t) = A(t) + Φ(t, ) Φ(, t). This implies Φ(, t) = Φ(, t)a(t) or Φ T (, t) = A T (t)φ T (, t). Furthermore, the controllability matrix W Σ c (, t 1 ) = Φ(, τ)b(τ)b T (τ)φ T (, τ) dτ (associated with Σ ) is identical to the observability matrix W Σ o (, t 1 ) (associated with Σ ). Conversely, the observability matrix W Σ o (, t 1 ) = Φ T (τ, )C T (τ)c(τ)φ(τ, ) dτ (associated with Σ ) is identical to the controllability matrix Wc Σ (, t 1 ) (associated with Σ ).

18 C.C. Remsing 87 Note : This duality theorem is extremely useful, since it enables us to deduce immediately from a controllability result the corresponding one on observability (and conversely). For example, to obtain the observability criterion for the time-invariant case, we simply apply Theorem to Σ to obtain the following result Theorem. The linear time-invariant control system ẋ = Ax + Bu(t) y = Cx (3.7) (or the pair (A, C)) is c.o. if and only if the (Kalman) observability matrix O = O(A, C) : = has rank m. C CA CA 2. CA m 1 R mn m Example. Consider the linear control system Σ described by ẋ = x + u(t) 1 1 y = x 1. The (Kalman) observability matrix is O = O Σ = which has rank 2. Thus the control system Σ is c.o.

19 88 AM3.2 - Linear Control In the single-output case (i.e. n = 1), if u( ) = and y( ) is known in the form γ 1 e λ 1t + γ 2 e λ 2t + + γ m e λmt assuming that all the eigenvalues λ i of A are distinct, then x can be obtained more easily than by using x = W o (, t 1 ) 1 Φ T (τ, )C T (τ)y(τ) dτ. For suppose that = and consider the solution of ẋ = Ax in the spectral form, namely We have x(t) = (v 1 x())e λ 1t w 1 + (v 2 x())e λ 2t w (v m x())e λmt w m. y(t) = (v 1 x())(cw 1 )e λ 1t + (v 2 x())(cw 2 )e λ 2t + + (v m x())(cw m )e λmt and equating coefficients of the exponential terms gives v i x() = γ i (i = 1, 2,..., m). cw i This represents m linear equations for the m unknown components of x() in terms of γ i, v i and w i (i = 1, 2,..., m). Again, in the single-output case, C reduces to a row matrix c and Theorem can be restated as : A linear system (with outputs) in the form ẋ = Ax y = cx can be transformed into the canonical form v = Ev if and only if it is c.o. y = fv

20 C.C. Remsing 89 Decomposition of control systems By duality, the result corresponding to Theorem is : Theorem. When the linear control system Σ is time-invariant then if the observability matrix O Σ has rank m 1 < m there exists a control system, algebraically equivalent to Σ, having the form ẋ(1) A1 x(1) = + A 2 A 3 ẋ (2) x (2) B1 B 2 u(t) y = C 1 x (1) where x (1) and x (2) have orders m 1 and m m 1, respectively and (A 1, C 1 ) is c.o. We close this section with a decomposition result which effectively combines together Theorems and to show that a linear time-invariant control system can split up into four mutually exclusive parts, respectively c.c. but unobservable c.c. and c.o. uncontrollable and unobservable c.o. but uncontrollable Theorem. When the linear control system Σ is time-invariant it is algebraically equivalent to ẋ (1) A 11 A 12 A 13 A 14 ẋ (2) A 22 A 24 = ẋ (3) A 33 A 34 A 44 ẋ (4) x (1) x (2) x (3) x (4) y = C 2 x (2) + C 4 x (4) where the subscripts refer to the stated classification. B 1 B 2 + u(t)

21 9 AM3.2 - Linear Control 3.3 Linear Feedback Consider a linear control system Σ defined by ẋ = Ax + Bu(t) (3.8) where A R m m and B R m l. Suppose that we apply a (linear) feedback, that is each control variable is a linear combination of the state variables, so that u(t) = Kx(t) where K R l m is a feedback matrix. The resulting closed loop system is ẋ = (A + BK)x. (3.9) The pole-shifting theorem We ask the question whether it is possible to exert some influence on the behaviour of the closed loop system and, if so, to what extent. A somewhat surprising result, called the Spectrum Assignment Theorem, says in essence that for almost any linear control system Σ it is possible to obtain arbitrary eigenvalues for the matrix A + BK (and hence arbitrary asymptotic behaviour) using suitable feedback laws (matrices) K, subject only to the obvious constraint that complex eigenvalues must appear in pairs. Almost any means that this will be true for (completely) controllable systems. Note : This theorem is most often referred to as the Pole-Shifting Theorem, a terminology that is due to the fact that the eigenvalues of A+BK are also the poles of the (complex) function z 1 det (zi n A BK) This function appears often in classical control design. The Pole-Shifting Theorem is central to linear control systems theory and is itself the starting point for more interesting analysis. Once we know that arbitrary sets of eigenvalues can be assigned, it becomes of interest to compare the performance

22 C.C. Remsing 91 of different such sets. Also, one may ask what happens when certain entries of K are restricted to vanish, which corresponds to constraints on what can be implemented Theorem. Let Λ = {θ 1, θ 2,..., θ m } be an arbitrary set of m complex numbers (appearing in conjugate pairs). If the linear control system Σ is c.c., then there exists a matrix K R l m such that the eigenvalues of A + BK are the set Λ. Proof (when l = 1) : Since ẋ = Ax + Bu(t) is c.c., it follows that there exists a (linear) transformation w = Tx such that the given system is transformed into ẇ = Cw + du(t) where C = k m k m 1 k m 2... k 1, d =. 1. The feedback control u = kw, where k : = k m k m 1... k 1 produces the closed loop matrix C + dk, which has the same companion form as C but with last row γ m γ m 1 γ 1, where k i = k i γ i, i = 1, 2,..., m. (3.1) Since C + dk = T(A + bkt)t 1 it follows that the desired matrix is K = kt

23 92 AM3.2 - Linear Control the entries k i (i = 1, 2,..., m) being given by (3.1). In this equation k i (i = 1, 2,..., m) are the coefficients in the characteristic polynomial of A ; that is, det (λi m A) = λ m + k 1 λ m k m and γ i (i = 1, 2,..., m) are obtained by equating coefficients of λ in λ m + γ 1 λ m γ m (λ θ 1 )(λ θ 2 ) (λ θ m ). Note : The solution of (the closed loop system) ẋ = (A + BK)x depends on the eigenvalues of A + BK, so provided the control system Σ is c.c., the theorem tells us that using linear feedback it is possible to exert a considerable influence on the time behaviour of the closed loop system by suitably choosing the numbers θ 1, θ 2,..., θ m Corollary. If the linear time-invariant control system ẋ = Ax + Bu(t) y = cx is c.o., then there exists a matrix L R m 1 such that the eigenvalues of A + Lc are the set Λ. This result can be deduced from Theorem using the Duality Theorem Example. Consider the linear control system ẋ = x + u(t)

24 C.C. Remsing 93 The characteristic equation of A is char A (λ) λ 2 3λ + 14 = which has roots 3±i Suppose we wish the eigenvalues of the closed loop system to be 1 and 2, so that the characteristic polynomial is We have k 1 λ 2 + 3λ + 2. = k 1 γ 1 = 3 3 = 6 k 2 = k 2 γ 2 = 14 2 = 12. Hence K = kt = = It is easy to verify that A + bk = does have the desired eigenvalues Lemma. If the linear control system Σ defined by ẋ = Ax + Bu(t) is c.c. and B = b 1 b 2 b l with b i, i = 1, 2,..., l, then there exist matrices K i R l m, i = 1, 2,..., l such that the systems ẋ = (A + BK i )x + b i u(t) are c.c. Proof : For convenience consider the case i = 1. Since the matrix C = B AB A 2 B... A m 1 B

25 94 AM3.2 - Linear Control has full rank, it is possible to select from its columns at least one set of m vectors which are linearly independent. Define an m m matrix M by choosing such a set as follows : M = b 1 Ab 1... A r1 1 b 1 b 2 Ab 2... A r2 1 b 2... where r i is the smallest integer such that A r i b i is linearly dependent on all the preceding vectors, the process continuing until m columns of U are taken. Define an l m matrix N having its r th 1 column equal to e 2, the second column of I l, its (r 1 + r 2 ) th column equal to e 3, its (r 1 + r 2 + r 3 ) th column equal to e 4 and so on, all its other columns being zero. It is then not difficult to show that the desired matrix in the statement of the Lemma is K 1 = NM 1. Proof of Theorem when l > 1 : Let K 1 be the matrix in the proof of Lemma and define an l m matrix K having as its first row some vector k, and all its other rows zero. Then the control u = (K 1 + K )x leads to the closed loop system ẋ = (A + BK 1 )x + BK x = (A + BK 1 )x + b 1 kx where b 1 is the first column of B. Since the system ẋ = (A + BK 1 )x + b 1 u is c.c., it now follows from the proof of the theorem when l = 1, that k can be chosen so that the eigenvalues of A + BK 1 + b 1 k are the set Λ, so the desired feedback control is indeed u = (K 1 + K )x.

26 C.C. Remsing 95 deduce If y = Cx is the output vector, then again by duality we can immediately Corollary. If the linear control system ẋ = Ax + Bu(t) y = Cx is c.o., then there exists a matrix L R m n such that the eigenvalues of A + LC are the set Λ. Algorithm for constructing a feedback matrix The following method gives a practical way of constructing the feedback matrix K. Let all the eigenvalues λ 1, λ 2,..., λ m of A be distinct and let W = w 1 w 2... w m where w i is an eigenvector corresponding to the eigenvalue λ i. With linear feedback u = Kx, suppose that the eigenvalues of A and A BK are ordered so that those of A BK are to be µ 1, µ 2,..., µ r, λ r+1,..., λ m (r m). Then provided the linear system Σ is c.c., a suitable matrix is K = fg W where W consists of the first r rows of W 1, and α1 α 2 α r g = β 1 β 2 β r r (λ i µ j ) j=1 if r > 1 α i = r (λ i λ j ) β = j=1 j i λ 1 µ 1 if r = 1 T β 1 β 2... β r = WBf

27 96 AM3.2 - Linear Control f being any column l-vector such that all β i Example. Consider the linear system 1 2 ẋ = x u(t). We have Suppose that λ 1 = 1, λ 2 = 2 and 1 1 W =, W = µ 1 = 3, µ 2 = 4, so W = W 1. We have α 1 = 6, α 2 = 2 and β = WBf gives β1 β 2 = f = 5f 1 3f 1. Hence we can take f 1 = 1, which results in g = Finally, the desired feedback matrix is K = W 1 = Example. Consider now the linear control system ẋ = x + u(t) We now obtain β1 β 2 5f1 + 2f 2 = 3f 1 f 2

28 C.C. Remsing 97 so that f 1 = 1, f 2 = gives K = However f 1 = 1, f 2 = 1 gives β 1 = 3, β 2 = 2 so that g = 2 1 and from K = fg W we now have 1 K = W 1 = Realization Theory The realization problem may be viewed as guessing the equations of motion (i.e. state equations) of a control system from its input/output behaviour or, if one prefers, setting up a physical model which explains the experimental data. Consider the linear control system (with outputs) Σ described by ẋ = Ax + Bu(t) y = Cx (3.11) where A R m m, B R m l and C R n m. Taking Laplace transforms of (3.11) and assuming zero initial conditions gives sx(s) = Ax(s) + Bu(s) and after rearrangement x(s) = (si m A) 1 Bu(s).

29 98 AM3.2 - Linear Control The Laplace transform of the output is y(s) = Cx(s) and thus where the n l matrix y(s) = C (si m A) 1 Bu(s) = G(s)u(s) G(s) : = C (si m A) 1 B (3.12) is called the transfer function matrix since it relates the Laplace transform of the output vector to that of the input vector. Exercise 41 Evaluate (the Laplace transform of the exponential) L e at (s) : = and then show that (for A R m m ) : e st e at dt L exp(ta) (s) = (si m A) 1. Using relation (si m A) 1 = sm 1 I m + s m 2 B 1 + s m 3 B B m 1 char A (s) (3.13) where the k i and B i are determined successively by B 1 = A + k 1 I m, B i = AB i 1 + k i I m ; i = 2, 3,..., m 1 k 1 = tr(a), k i = 1 i tr(ab i 1) ; i = 2, 3,..., m the expression (3.12) becomes G(s) = sm 1 G + s m 2 G G m 1 χ(s) = H(s) χ(s)

30 where χ(s) = char A (s) and G k = C.C. Remsing 99 g (k) ij R n l, k =, 1, 2,..., m 1. The n l matrix H(s) is called a polynomial matrix, since each of its entries is itself a polynomial; that is, h ij = s m 1 g () ij + s m 2 g (1) ij + + g (m 1) ij. Note : The formulas above, used mainly for theoretical rather than computational purposes, constitute Leverrier s algorithm Example. Consider the electrically-heated oven described in section 1.3, and suppose that the values of the constants are such that the state equations are ẋ = x + 1 u(t). Suppose that the output is provided by a thermocouple in the jacket measuring the jacket (excess) temperature, i.e. y = 1 x. The expression (3.12) gives s G(s) = 1 1 s = s + 1 s 2 + 3s using (si 2 A) 1 = 1 char A (s) adj(si 2 A). Realizations In practice it often happens that the mathematical description of a (linear time-invariant) control system in terms of differential equations is not known, but G(s) can be determined from experimental measurements or other considerations. It is then useful to find a system in our usual state space form to which G( ) corresponds.

31 1 AM3.2 - Linear Control In formal terms, given an n l matrix G(s), whose elements are rational functions of s, we wish to find (constant) matrices A, B, C having dimensions m m, m l and n m, respectively, such that G(s) = C(sI m A) 1 B and the system equations will then be ẋ = Ax + Bu(t) y = Cx. The triple (A, B, C) is termed a realization of G( ) of order m, and is not, of course, unique. Amongst all such realizations some will include matrices A having least dimensions these are called minimal realizations, since the corresponding systems involve the smallest possible number of state variables. Note : Since each element in (si m A) 1 = adj (si m A) det(si m A) has the degree of the numerator less than that of the denominator, it follows that lim C(sI m A) 1 B = s and we shall assume that any given G(s) also has this property, G( ) then being termed strictly proper Example. Consider the scalar transfer function g(s) = 2s + 7 s 2 5s + 6 It is easy to verify that one realization of g( ) is 1 A =, b =, c = It is also easy to verify that a quite different triple is 2 1 A =, b =, c =

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