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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 ζω n s + ω 2 n For unit step response, we need M p ≤ 25% ,t s ≤ . 1sec Solve for ζ : M p = e- πζ/ √ 1- ζ 2 ≤ . 25 ζ = s (log M p ) 2 π 2 + (log M p ) 2 ≥ . 4037 Solve for ω n : t s ’ 4 . 6 ζω n ≤ . 1 ω n ’ 114 Now solve for a and K with: 2 ζω n = 25 + a ω 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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