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§1-3 Solution of a Dynamical Equation 1. Solution of a homogeneous differential equation x A (t )x (1 - 50) • A preliminary theorem for x A (t )x f (t ), x (t 0 ) x 0 R n x , f R n , A (t ) (a ij (t ))n n (A1) Preliminary Theorem: Let every element of A and f be continuous over (, +). Then for any t0(, +) and any constant vector x0, (A1) has a solution x(t) defined over (, +), such that x (t 0 ) x 0 (A 2) Furthermore, the solution satisfying (A2) is unique. Corollary 1: If (t) is a solution to the differential equation x A (t )x and for some t0, (t0)=0, then Y (t ) 0, t Proof:It is clear that x(t)0, t is one of the solutions of the above equation. On the other hand, since (t0)=0 and (t) with the initial condition is unique, we have Y (t ) x (t ) 0, t Q.E.D Theorem 2: The set of all solutions of x A (t )x (1-50) forms an n-dimensional vector space over the field of real numbers. Outline of the proof. a) All the solutions of (1-50) form a linear space; b) The dimension of the linear space is n. More specifically, 1) (1-50) has n linearly independent solutions; 2) Any solution can be expressed as the linear combination of the n solutions. Proof: a) All the solutions form a linear space. Let 1 and 2 be two arbitrary solutions of (1-50). Then, for any real numbers 1 and 2, we have d (a 1Y 1 a 2 Y 2 ) a 1Y 1 a 2 Y 2 a 1A (t )Y 1 a 2 A (t )Y 2 dt A (t )(a 1Y 1 a 2 Y 2 ) b) The dimension of the space is n. 1). Let e1, e2,…, en be n linearly independent constant vectors, and i(t), i=1, 2,…, n, are the solutions of (150) with i(t0)=ei, i=1, 2,…, n We then prove that i(t), i=1, 2,…, n, are linearly independent over (, +). The proof is by contradiction. If there exists a t0 (, +), such that i(t), i=1, 2,…, n are linearly dependent. Then, there exists an n×1 nonzero vector , such that [Y 1 (t ') Y 2 (t ') a 1Y 1 (t ') a 2 Y 2 (t ') Y n (t ')]a 0 a n Y n (t ') 0 Note that (t)0, t is a solution of (1-50) and X (t ) : a 1Y 1 (t ) a 2 Y 2 (t ) a n Y n (t ) is also a solution of (1-50) with X (t ') 0 Hence, the uniqueness of the solution implies that Y (t ) X (t ) a 1Y 1 (t ) a 2 Y 2 (t ) a n Y n (t ) 0 t In particular, for t=t0, [Y 1 (t 0 ) Y 2 (t 0 ) [e1 e 2 Y n (t 0 )]a e n ]a 0 which implies that the vectors e1, e2,…, en are linearly dependent, a contradiction. Hence i(t), i=1, 2,…, n, are linearly independent over (, +). 2) To prove that any solution can be expressed as the linear combination of the n solutions. That is, all the solutions of (1-50) form an ndimensional vector space. Let (t) be an arbitrary solution of (1-50) with (t0)=e Then, e can be uniquely expressed as e a1e 1 a 2e 2 n a n e n a ie i i 1 n i a Y (is t )also the solution of (1-50) with Since i i 1 n n i 1 i 1 i i a Y ( t ) a e i i e 0 The uniqueness of the solution implies that Y (t ) n i a Y i (t ) i 1 Q.E.D 2. Fundamental matrix and state transition matrix Fundamental matrix Definition: The matrix [Y 1 (t ) Y 2 (t ) Y n (t )] : Y (t ), t (, ) which is formed by n linearly independent solutions of (1-50) is said to be a fundamental matrix of (1-50). Some properties for a fundamental matrix: Y A (t )Y Y (t 0 ) E (1 - 51) where E is a nonsingular constant matrix. Theorem 3 The fundamental matrix of equation of (1-50) is nonsingular for every t over (, +). . Proof: By using Corollary 1 directly. Corollary 1: If (t) is a solution to the differential equation x A (t )x and for some t0, (t0)=0, then Y (t ) 0, t Theorem 4: Let 1 and 2 be two fundamental matrixes of (1-50). Then, there exists a n×n nonsingular constant matrix C, such that 1(t)=2(t)C State transition matrix Definition 9: Let (t) be any fundamental matrix. Then F t , t 0 Y (t )Y (t 0 ) 1 is said to be the state transition matrix of (1-50), where t, t0(, +). Some important properties of state transition matrix: 1). F (t , t ) = I 2 ). F - 1 (t , t0 ) = Y (t0 )Y - 1 (t ) = F (t 0 , t ) 3 ). F (t 2 , t0 ) = F (t 2 , t 1 )F (t 1 , t 0 ) 4) d F (t , t 0 ) A (t )F (t , t 0 ) dt F (t 0 , t 0 ) I (1 52) Under the condition x(t0)=x0, the solution of (1-50) is x (t ) t ,t 0 x 0 (1-53) Hence, (t, t0) can be considered as a linear transformation which maps the state at t0 to the state x(t) at time t. In fact, x(t) can always be expressed as x (t ) Y(t ) , 0 In particular, x (t 0 ) Y(t 0 )a a Y1 (t 0 )x (t 0 ) from which we can obtain the conclusion. Example: Prove that the state transition matrix is unique, i.e. the state transition matrix is regardless of the choice of fundamental matrix. 3. Non-homogeneous equation Solutions of linear time-variant dynamical equations x A (t )x B (t )u (1 49) The solution to the state equation can be obtained by using the method of variation of parameters. Let x (t ) F t ,t 0 x(t ) Then, we have (s 1) Theorem 1-5 The solution of the state equation is given by x (t ) F t , t 0 x 0 F (t , t )B (t )u (t )d t t t0 where F t , t 0 x 0 is called zero-input response, and t t0 F (t , t )B (t )u (t )d t is called zero-state response. (1 54) Relationship between I/O and state space descriptions Corollary 1-5 The output of the dynamical equation (1-34) is y (t ) C (t )F t , t 0 x 0 t C (t )F (t , t )B (t )u (t )d t D (t )u (t ) t0 (1 57) If x(t0)=0, the impulse response matrix is t t : G (t , t ) C (t ) (t , t )B (t ) D (t )d(t t ) t t : G (t , t ) 0 Solutions of a linear time-invariant dynamical equation Consider the following linear time-invariant dynamical equation x A x Bu y Cx Du (1 60) where A, B, C and D are n×n, n×p, q×n and q×p real constant matrices. From the corresponding homogeneous equation, we have Fundamental matrix: eAt State transition matrix: F t , t 0 e A t (e A t 0 )1 e A (t t 0 ) F t t 0 x (t ) e y (t ) Ce A (t t 0 ) t x 0 e A (t t )B u (t )d t t0 A (t t 0 ) t x 0 Ce A (t t )B u (t )d t Du (t ) t0 Usually, we assume that t0=0, t x (t ) e x (0) e A (t t )Bu (t )d t At 0 t (1 63) y (t ) Ce x (0) Ce A (t t )Bu (t )d t Du (t ) At 0 (1 64) The corresponding impulse response matrix is t t G (t t ) Ce t t A (t t ) B D d(t t ) G (t , t ) 0 Usually we write it as G (t ) Ce B D d(t ), t 0 At (1 65) The corresponding transfer function matrix of the above equation is G( s) C ( sI A)1 B D which is a rational function matrix. (1 66) Equivalent dynamical equations a) Time-invariant case Definition 1-10: An LTI dynamical equation x A x Bu y Cx Du (1 67) is said to be equivalent to the dynamical equation (A, B, C, D) if and only if there exists a nonsingular matrix P, such that A PAP 1 , B PB, C CP 1 , D D where P is said to be an equivalence transfor-mation. Equivalence transformation implies that the choice of the state is not unique; different methods of analysis often lead to different choices of the state. Definition: Two time-invariant dynamical systems are said to be zero-state equivalent if and only if they have the same impulse response matrix or the same transfer function matrix. Theorem: Two equivalent LTI systems are zero-state equivalent. Example: Consider the two systems x 3x 2u and 1 4 1 x x u , y 1 1 x 4 1 1 which are not equivalent dynamical equations, but are zero-state equivalent.