20095ee141_1_EE141_hw2_sol

20095ee141_1_EE141_hw2_sol - EE141 Principles of Feedback...

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Unformatted text preview: EE141 Principles of Feedback Control (Fall 2009) Solutions to Homework 2 Problem 3.19 (a) Figure 1: Block diagram for 3.19(a) Y R = G 2 + G 1 1 + G 1 (b) Figure 2: Block diagram for 3.19(b) Reduce the original diagram to: Figure 3: Reduced block diagram for 3.19(b) Y R = G 7 + G 1 1 + G 1 G 2 G 3 G 4 1 + G 4 G 5 G 6 = G 7 + G 1 G 3 G 4 G 6 (1 + G 1 G 2 )(1 + G 4 G 5 ) (c) 1 Figure 4: Block diagram for 3.19(c) Reduce the original diagram to: Figure 5: Reduced block diagram for 3.19(c) Y R = G 7 + ( G 6 + G 1 G 2 1 + G 2 G 3 ) ( G 4 1 + G 4 ) G 5 = G 7 + G 4 G 5 G 6 1 + G 4 + G 1 G 2 G 3 G 4 G 5 (1 + G 2 )(1 + G 4 ) Problem 3.24 The transfer function is: Y ( s ) R ( s ) = 100 K s 2 + (25 + a ) s + 25 a + 100 K = 100 K s 2 + 2 n s + 2 n For unit step response, we need M p 25% ,t s . 1sec Solve for : M p = e- / 1- 2 . 25 = s (log M p ) 2 2 + (log M p ) 2 . 4037 Solve for n : t s 4 . 6 n . 1 n 114 Now solve for a and K with: 2 n = 25 + a 2 n = 25 a + 100 K we get a 67 ,K 113 . 2 Here is step response of the system using MATLAB. We can find that the two specifications are satisfied....
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20095ee141_1_EE141_hw2_sol - EE141 Principles of Feedback...

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