MIT16_30F10_lec08

MIT16_30F10_lec08 - Topic #8 16.30/31 Feedback Control...

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Topic #8 16.30/31 Feedback Control Systems State-Space Systems System Zeros Transfer Function Matrices for MIMO systems
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± ± Fall 2010 16.30/31 8–1 Zeros in State Space Models Roots of transfer function numerator called the system zeros . Need to develop a similar way of deFning/computing them using a state space model. Zero: generalized frequency s 0 for which the system can have a non-zero input u ( t ) = u 0 e s 0 t , but exactly zero output y ( t ) 0 t Note that there is a speciFc initial condition associated with this response x 0 , so the state response is of the form x ( t ) = x 0 e s 0 t u ( t ) = u 0 e s 0 t x ( t ) = x 0 e s 0 t y ( t ) 0 Given x ˙ = A x + B u , substitute the above to get: x 0 s 0 e s 0 t = A x 0 e s 0 t + B u 0 e s 0 t ² s 0 I A B ³ x 0 = 0 u 0 Also have that y = C x + D u = 0 which gives: C x 0 e s 0 t + D u 0 e s 0 t = 0 ² C D ³ x 0 = 0 u 0 So we must Fnd the s 0 that solves: ± ± s 0 I A B x 0 = 0 C D u 0 Is a generalized eigenvalue problem that can be solved in MATLAB using eig.m or tzero.m 1 1 MATLAB is a trademark of the Mathworks Inc. October 17, 2010
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± ² ³ ± ± Fall 2010 16.30/31 8–2 There is a zero at the frequency s 0 if there exists a non-trivial solution of ± s 0 I A B det = 0 C D Compare with equation on page 6– ?? x 0 Key Point: Zeros have both direction u 0 and frequency s 0 Just as we would associate a direction (eigenvector) with each pole (frequency λ i ) s +2 Example: G ( s ) = s 2 +7 s +12 7 12 1 ´ µ A = B = C = 1 2 D = 0 1 0 0 ± s 0 + 7 12 1 det s 0 I A B = det 1 s 0 0 C D 1 2 0 = ( s 0 + 7)(0) + 1(2) + 1( s 0 ) = s 0 + 2 = 0 so there is clearly a zero at s 0 = 2 , as we expected. For the directions, solve: ⎤⎡ s 0 + 7 12 1 x 01 5 12 1 x 01 1 s 0 0 x 02 = 1 2 0 ⎦⎣ x 02 = 0? 1 2 0 u 0 1 2 0 u 0 s 0 = 2 gives x 01 = 2 x 02 and u 0 = 2 x 02 so that with x 02 = 1 x 0 = 2 and u = 2 e 2 t 1 October 17, 2010
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Fall 2010 16.30/31 8–3 Further observations: apply the specifed control input in the ±re- quency domain, so that Y 1 ( s ) = G ( s ) U ( s ) where u = 2 e 2 t , so that U ( s ) = 2 s +2 s + 2 2 2 Y 1 ( s ) = = s 2 + 7 s + 12 · s + 2 s 2 + 7 s + 12 Say that s = 2 is a blocking zero or a transmission zero. The response Y 1 ( s ) is clearly non-zero, but it does not contain a component at the input ±requency s = 2 . That input has been “blocked”. Note that the output response le±t in Y 1 ( s ) is a very special ±orm it corresponds to the (negative the) response you would see ±rom ± T the system with u ( t ) = 0 and x 0 = 2 1 Y 2 ( s ) = C ( sI A ² ) 1 x 0 ³ 1 ² ³ ± s + 7 12 2 = 1 2 1 s 1 ² ³² ³ ± s 12 2 1 = 1 2 1 s + 7 1 s 2 + 7 s + 12 2 = s 2 + 7 s + 12 So then the total output is Y ( s ) = Y 1 ( s )+ Y 2 ( s ) showing that Y ( s ) = 0 y ( t ) = 0 ,
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MIT16_30F10_lec08 - Topic #8 16.30/31 Feedback Control...

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