hw4soln_F2009 - MAE170 Homework 4 Solution Fall 2009 E6.4...

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Unformatted text preview: MAE170 Homework 4 Solution Fall 2009 E6.4 Define G c = 2 s +1 , G p = k s ( s +2) and G = ( G c- 1) G p = k (1- s ) s 3 +3 s 2 +2 s The closed-loop transfer function is T ( s ) = G 1 + G = k (1- s ) s 3 + 3 s 2 + (2- k ) s + k (1) Therefore, the characteristic polynomial is s 3 + 3 s 2 + (2- k ) s + k (2) The corresponding Routh Array is given by s 3 1 2- k s 2 3 k s 1 b s k (3) where b =- 1 3 ( k- 3(2- k )) = 2- 4 k 3 (4) For stability we require k > 0 and b > 0. Thus the range of k for stability is < k < 1 . 5 1 E6.11 The closed-loop characteristic polynomial is s 4 + 6 s 3 + 2 s 2 + s + 3 (5) The system is unstable so the steady-state error does not exist. The poles are s 1 =- 5 . 6614, s 2 =- . 9002 and s 3 , 4 = 0 . 2808 . 714 j We can also show that the system is unstable by writing the corresponding Routh Array s 4 1 2 3 s 3 6 1 0 s 2 11 6 3 0 s 1-8.82 0 s 3 (6) Not all the coefficients in the 1 st column of the Routh array have the same sign (all positive or all negative), therefore the system is unstable . The number of sign changes in the first column is two, therefore it is predicted that there are two unstable poles (with positive real parts), which consistent with the answer above....
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hw4soln_F2009 - MAE170 Homework 4 Solution Fall 2009 E6.4...

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