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Unformatted text preview: Flight Control Systems Homework #6 Solutions Dr. Bishop  Fall 2007 Reading Assignment: Modern Control Systems: Chapter 6 1. Exercises: E6.24 2. Problems: P6.14 3. Advanced Problems: AP6.1 4. Design Problems: DP6.6 5 . Computer Problems: CP6.5. E6.24 The closedloop transfer function is T ( s ) = 10 2 s 2 + ( K 20) s + 10 10 K . Therefore, the characteristic equation is 2 s 2 + ( K 20) s + 10 10 K = 0 . The Routh array is s 2 2 1010K s 1 K 20 s o 10 10 K We see that the system is stable for any value of K > 20 and any K < 1. Therefore, the system is unstable for all K > 0 since the gain K cannot be simultaneously greater than 20 and less than 1. P6.14 (a) The Routh array is s 3 1 5 s 2 5 6 s 1 3.8 s o 6 Examining the first column of the Routh array, we see no sign changes. So, the system is stable. (b) The roots of the system are s 1 = . 3246 and s 2 , 3 = 2 . 3377 ± 3 . 6080 j ....
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This note was uploaded on 09/01/2010 for the course ASE 270L taught by Professor Dr.bishop during the Spring '10 term at University of Texas.
 Spring '10
 Dr.Bishop

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