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Unformatted text preview: Flight Control Systems Homework #4 Solutions Dr. Bishop  Fall 2007 Reading Assignment: Modern Control Systems: Chapter 4 12. Exercises: E4.6, E4.10 3. Problems: P4.7 4. Advanced Problems: AP4.7 5. Design Problems: DP4.3 6. Computer Problems: CP4.10 E4.6 The closedloop transfer function is T ( s ) = 5( s + 2) s 2 + 15 s + 10 . The step response is shown in Figure 1. Time (sec. ) Amplitude Step Response 1 2 3 4 5 6 7 8 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y(t)/A Figure 1: Step response. E4.10 (a) The closedloop transfer function is T ( s ) = G c ( s ) G ( s ) 1 + G c ( s ) G ( s ) H ( s ) = 100 K 1 ( s + 5) s 2 + 105 s + (500 + 100 K 1 K 2 ) . 1 The steadystate tracking error is E ( s ) = R ( s ) Y ( s ) = bracketleftbigg 1 G c ( s ) G ( s )(1 H ( s )) 1 + G c ( s ) G ( s ) H ( s ) bracketrightbigg R ( s ) = s 2 + (105 100 K 1 ) s + 500 100 K 1 (5 K 2 ) s 2 + 105 s + 500 + 100 K 1 K 2 · 1 s and lim s → sE ( s ) = 5 K 1 (5 K 2 ) 5 + K 1 K 2 ....
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 Spring '10
 Dr.Bishop
 closedloop transfer function, Modern Control Systems, Flight Control Systems

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