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2 Pages

### project5

Course: MA 531, Fall 2008
School: N.C. State
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Word Count: 454

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531 Math Project 5 Due Thursday, November 13 (1) Consider the system x = Ax + Bu y = Cx + Du where 3 6 4 6 10 A= 9 7 7 9 2/3 1/3 B = 1/3 2/3 1/3 1/3 C= 123 336 D = 0. (a) Show that the system is not controllable and use a QR decomposition to nd a Kalman controllable canonical form. (b) Show that the system is not observable and use a QR decomposition to nd a Kalman observable canonical form. (2)...

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531 Math Project 5 Due Thursday, November 13 (1) Consider the system x = Ax + Bu y = Cx + Du where 3 6 4 6 10 A= 9 7 7 9 2/3 1/3 B = 1/3 2/3 1/3 1/3 C= 123 336 D = 0. (a) Show that the system is not controllable and use a QR decomposition to nd a Kalman controllable canonical form. (b) Show that the system is not observable and use a QR decomposition to nd a Kalman observable canonical form. (2) Consider the system x = Ax + Bu y = Cx . In the next chapter, we will study feedback systems in which the control is specied by u(t) = F x(t) where F is chosen to attain or enhance stability margins or achieve some other goal. Note that the resulting system has the form x = (A F B)x y = Cx . One way to choose F is through a technique termed pole or eigenvalue placement which is facilitated by the companion controller form (further details can be found on pages 330-335 of our text). To illustrate, consider a system 0 1 0 0 0 0 0 0 1 0 . . . . .. z + . u . . . . z= . . . . . . . where the i s are the coecients of the characteristic polynomial 0 0 0 1 0 1 2 n1 0 1 (s) = |sI A| = sn + n1 sn1 + + 1 s + 0 . We now consider a feedback control of the form u(t) = f0 z1 f1 z2 fn1 zn = Fc z . The closed loop system will now have the form x= 0 0 . . . 1 0 . . . 0 1 . . . .. . 0 0 . . . 0 0 0 1 (0 + f0 ) (1 + f1 ) (2 + f2 ) (n1 fn1 ) x. Note that the of signs the fi are opposite those in the book. This is to accommodate the more common convention of considering A BF , and yields the same nal result. Now, if the desired closed loop eigenvalues are specied by sd , sd , , sd , then the desired characteristic n 12 equation will be d (s) = (s sd )(s sd ) (s sd ) n 1 2 = sn + d sn1 + + d s + d n1 1 0 from which it follows that 0 + f0 = d f0 = d 0 0 0 1 + f1 = d f1 = d 1 1 1 . . . n1 + fn1 = d fn1 = d n1 n1 n1 Given specied eigenvalue locations, this provides an algorithm for computing the feedback gain F . Note that if the system is not in companion form, previously discussed similarity transforms can be applied rst to obtain this form. (a) Consider the system 0 1 0 0 0 1 z + 0 u. z= 0 1 2 5 10 Determine the eigenvalues for the system and discuss its stability. Now nd a fe...

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