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Solutions2HW10

# Solutions2HW10 - 10.9 Solution a The full order observer is...

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10.9 Solution: a. The full order observer is Ly Bu x LC A x + + - = ˆ ) ( ˆ Let the observer poles are {-10,-11,-12,-13,-14}, compare the coefficients of ) 4 , 3 , 2 , 1 , 0 ( = i i λ in both sides of ) 14 )( 13 )( 12 )( 11 )( 10 ( ) det( + + + + + = + - λ λ λ λ λ λ LC A I we have T 577.9009] - 83.4986, 499.8865, 133.8085, - 38.4366, [ = L Without the feedback, the augmented system is u B B x x LC A LC A x x + - = ˆ 0 ˆ = x x C C y y ˆ 0 0 ˆ Assume , ] 0 , 0 , 0 , 0 , 0 [ ) 0 ( T x = T x ] 1 , 1 , 1 , 1 , 1 [ ) 0 ( ˆ = , and 2 = u , the simulation results are shown as follows.
For reduced order observer, let

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= 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 1 3 W By the transformation ] , [ , , 2 1 1 2 1 22 21 12 11 1 C C CW C B B WB B A A A A WAW A = = = = = = - - where 0000 . 3 11 = A , - - = 0 0000 . 5 0 0 0000 . 5 0 0 0 0 3333 . 0 0000 . 1 6667 . 0 0000 . 1 0 0000 . 2 0000 . 1 22 A ] 0000 . 6 , 0000 . 3 , 0000 . 14 , 0000 . 13 [ 12 - = A , T A ] 0 , 6667 . 0 , 0 [ 21 = , 5 1 = B , T B ] 1 , 1 , 1 , 1 [ 2 = 1 1 = C , T C ] 0 , 0 , 0 , 0 [ 2 = . The reduced observer poles are eigenvalues of
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Solutions2HW10 - 10.9 Solution a The full order observer is...

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