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quiz2practicesol

# quiz2practicesol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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± ± ² ± ² ² ± ² ± ² ³ MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Aeronautics and Astronautics 16.060: Principles of Automatic Control Fall 2003 Quiz 2 Practice Problems SOLUTIONS State-Space 1. (a) The poles of the system are the eigenvalues of A : | sI A | =d e t s 1 = s ( s +3)+2 = ( s +1)( s +2) 2 s +3 s = 2 , 1 The poles are in the left half-plane, so the system is stable. (b) The transfer function matrix is given by the formula G ( s )= C [ sI A ] 1 B + D . s 1 ( sI A 2 s ⇒ | sI A | = s 2 s +2 1 ± s 1 [ sI A ] 1 = s 2 s 2 s Hence ± ² ²± ² ± 1 ² 1 0 1 ± s +3 1 0 = s 2 +3 s +2 G ( s s 01 s 2 s 2 s 1 s 2 +3 s +2 (c) The state transition matrix is given by Φ( t L 1 ( sI A ) 1 . From Problem 1(a) above we know that 2 1 1 s +2 s +1 s +2 [ sI A ] 1 = 1 s 1 = s +1 1 2 + 2 1 + 2 s 2 s 2 s s +1 s +2 s +1 s +2 Taking the inverse Laplace Transform: 2 t t e 2 t ² 2 e t e e Φ( t 2 t e t e 2 t 2 e t e (d) To ﬁnd the step response we use the following formula: t x =Φ( t ) x ( 0 )+ Φ( t τ ) B u ( τ ) 0 Since y = x in this case and u ( t )=0 for t< 0and u ( t )=1 t> 0, we have the following: 1

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± ² ± ² ³ ´ ³ ³ µ ´ · · ´ ´ ¸ ´ ´ ¸ ´ ¸ ´ ¸ t y ( t ) x ( 0 )+ Φ( t τ ) B u ( τ ) 0 2 t e t e 2 t ¸´ ¸ 2 e t e 1 = 2 t e t +2 e 2 t 2 e t e 1 t ´ ( t τ ) e 2( t τ ) ( t τ ) e 2( t τ ) ¸ 2 e e 0 + [1] 2( t τ ) 2( t τ ) 1
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quiz2practicesol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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