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### ece604 lecture 09

Course: ECE 604, Spring 2010
School: UMass (Amherst)
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Word Count: 820

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9 ECE Lecture 604 State Variable Analysis Doug Looze Oct. 6, 2009 Linear Systems Douglas Looze 1 Announcements PS3 available Due Thursday Oct. 6, 2009 Linear Systems Douglas Looze 2 Last Time Characterizations of Controllability Observability In terms of Grammians Linear independence of vector time functions Oct. 6, 2009 Linear Systems Douglas Looze 3 Conditions Equivalent (A(t), B(t))...

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9 ECE Lecture 604 State Variable Analysis Doug Looze Oct. 6, 2009 Linear Systems Douglas Looze 1 Announcements PS3 available Due Thursday Oct. 6, 2009 Linear Systems Douglas Looze 2 Last Time Characterizations of Controllability Observability In terms of Grammians Linear independence of vector time functions Oct. 6, 2009 Linear Systems Douglas Looze 3 Conditions Equivalent (A(t), B(t)) controllable Independent rows of ( t , ) B ( ) Grammian invertible n n matrix Full rank W t0 , t f Oct. 6, 2009 ( ) tf = ( t0 , t ) B ( t ) BT ( t ) T ( t0 , t ) dt t0 Linear Systems Douglas Looze 4 Equivalent (A(t), C(t)) observable Independent columns of C ( t ) ( t , t0 ) Grammian invertible n n matrix Full rank M t0 , t f Oct. 6, 2009 ( ) tf T = ( t , t0 ) CT ( t ) C ( t ) ( t , t0 ) dt t0 Linear Systems Douglas Looze 5 Exercise Consider the linear, time-varying system 0 0 d x( t) = x( t) + b( t) u( t) dt 0 1 Specify whether the system is controllable on any interval for the following input matrices. If it is controllable on some interval, determine such an interval 1 a) b ( t ) = 1 Oct. 6, 2009 1 b) b ( t ) = t e 1 c) b ( t ) = -t e 6 Linear Systems Douglas Looze The transition matrix for this system is: 0 0 0 1 t 1 = e At = e 0 t 0 e The controllability kernel matrix is: 0 1 1 1 ( t0 , t ) B ( t ) = = t0 - t 1 t0 - t 0 e e Since the rows are linearly independent on any interval, the system is (C) on all intervals Oct. 6, 2009 Linear Systems Douglas Looze 7 The transition matrix for this system is: 0 0 0 1 t 1 = e At = e 0 t 0 e The controllability kernel matrix is: 0 1 1 1 ( t0 , t ) B ( t ) = = t0 - t t t0 0 e e e Since the rows are not linearly independent on any interval, the system is not (C) on any interval Oct. 6, 2009 Linear Systems Douglas Looze 8 The transition matrix for this system is: 0 0 0 1 t 1 = e At = e 0 t 0 e The controllability kernel matrix is: 0 1 1 1 ( t0 , t ) B ( t ) = = t0 - t -t t0 - 2t 0 e e e Since the rows are linearly independent on any interval, the system is (C) on all intervals Oct. 6, 2009 Linear Systems Douglas Looze 9 Today Characterization of linear independence for controllability and observability Exercises Reading Rugh, Ch. 9 p. 144146, 149 Oct. 6, 2009 Linear Systems Douglas Looze 10 Alternative Tests Controllability is a range space concept Observability is a null space concept Define Then Oct. 6, 2009 Linear Systems Douglas Looze 11 Question Is there a test for linear independence of a finite number of vector time that functions is Finite-dimensional Non-integral Independent of the transition matrix Yes If the vectors are sufficiently differentiable Oct. 6, 2009 Linear Systems Douglas Looze 12 Controllability Define K0 ( t ) = B ( t ) d K j ( t ) = - A ( t ) K j -1 ( t ) + K j -1 ( t ) dt ( t , ) B ( ) j Claim ( t , ) K j ( ) = j Proof (induction): Check true for j = 0 ( t , ) K 0 ( ) = ( t , ) B ( ) Oct. 6, 2009 Linear Systems Douglas Looze 13 Assume for j Check true for j + 1 ( t , ) K j ( ) j = [ ( t , ) B ( ) ] j Oct. 6, 2009 Linear Systems Douglas Looze 14 Theorem: Suppose that, for some t, B(t) is q times differentiable and A(t) is (q1) times differentiable. Then the linear system is controllable over any interval containing t if rank K 0 ( t ) L K q ( t ) = n Note: by the previous theorem K j ( t) = Oct. 6, 2009 ( t , ) B ( ) j t = Linear Systems Douglas Looze 15 j Proof: Suppose the rank test holds but there is an interval [t0,tf] containing t for which the system is not controllable. Then W(t0,tf) is not invertible Oct. 6, 2009 Linear Systems Douglas Looze 16 T x a ( t0 , ) B ( ) = 0 = T xb K 0 ( ) Oct. 6, 2009 Linear Systems Douglas Looze 17 Remarks Test for linear independence of rows of ( t0 , t ) B ( t ) Same proof can be used for any matrix V(t) rank V ( t ) d dq V( t) L V( t) = n dt dt q Can determine Kj(t) recursively Note strong conclusion: any interval containing t Sufficient only Oct. 6, 2009 Linear Systems Douglas Looze 18 Recall Test for independence of rows of a matrix V(t) d dq rank V ( t ) V( t) L V( t) = n q dt dt Can use transpose to present an equivalent test for independence of columns Recursion formula for derivatives Oct. 6, 2009 Linear Systems Douglas Looze 19 Define Observability L j ( t) = L0 ( t ) = C ( t ) j t C ( t ) ( t , ) j Theorem: Suppose that, for some t, C(t) is q times differentiable and A(t) is (q1) time differentiable. Then the linear system is observable over any interval containing t if L0 ( t ) rank M = n Lq ( t ) Linear Systems Douglas Looze d L j ( t ) = L j -1 ( t ) A ( t ) + L j -1 ( t ) dt t = Oct. 6, 2009 20 Remarks Test for linear independence of columns of C ( t ) ( t , t0 ) Can determine Lj(t) recursively Note strong conclusion: any interval containing t Sufficient only Oct. 6, 2009 Linear Systems Douglas Looze 21 Summary Rank tests for (C) and (O) Exercises Oct. 6, 2009 Linear Systems Douglas Looze 22 Next Time Basis change Duality LTI systems Reading Rugh, Ch. 9 p. 146149, 150152 Oct. 6, 2009 Linear Systems Douglas Looze 23
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