final07_sol - ECE 147A FEEDBACK CONTROL SYSTEMS THEORY AND...

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ECE 147A FEEDBACK CONTROL SYSTEMS - THEORY AND DESIGN F07 Final Exam Closed book and notes. No calculators. Show all work. Name: Solution Problem 1: /20 Problem 2: /20 Problem 3: /20 Problem 4: /20 Problem 5: /20 Total: /100 Miscellaneous: tan - 1 (1 / 3) 18 degrees, tan - 1 (1 / 2) 26 degrees, tan - 1 (1) = 45 degrees. tan - 1 (2) 63 degrees, tan - 1 (3) 72 degrees, tan - 1 (5) 78 degrees, tan - 1 (10) 84 degrees, tan - 1 (20) 87 degrees, sin(45 ) = 2 / 2 0 . 7, sin(50 ) 0 . 77, sin(55 ) 0 . 82, sin(60 ) = 3 / 2 0 . 87.
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Name: Problem 1 1. For the unity negative feedback configuration, describe the three properties that typically characterize a good choice for the function ω 7→ | C ( ω ) P ( ω ) | and explain why these three properties are important. (a) The values should be large for small ω . Since S ( s ) = 1 / (1 + C ( s ) P ( s )) , this ensures that the sensitivity transfer function is small for small ω . This guarantees good disturbance attenuation and reference tracking for low frequency disturbances and references since y = [1 - S ( s )] F ( s ) r + S ( s ) P ( s ) d . (b) The values should be small for large ω . This ensures that the sensitivity transfer funtion is close to 1 for large ω . This guarantees good measurement noise suppression since y = - [1 - S ( s )] n . (c) The slope near where the magnitude is equal to one should not be too steep. By the Bode gain/phase relationship, this says that the phase is not too negative, and by the Nyquist stability test, this says that the closed-loop will be stable. 2. What are the typical motivations for using a pole at the origin in a feedback controller? As long as the closed-loop is stable, a pole at the origin in the controller will guarantee that the output of the feedback system asymptotically tracks constant step references and asymptotically rejects constant disturbances, even if the parameters of the plant are not known to the controller exactly. Placing a pole at the origin in the controller makes the sensitivity transfer function equal to zero at ω = 0 . 3. What does the Bode integral say about the sensitivity transfer function for a stable loop that has relative degree two at least? What does this imply about amplifying and/or attenuating disturbances at the plant input? The Bode integral for the described situation says Z 0 ln | S ( ω ) | = 0 . Since ln( s ) > 0 for s > 1 and ln( s ) < 0 for s (0 , 1) , it follows that if there are values of ω where | S ( ω ) | < 1 , which corresponds to disturbance attenuation, then there are values of ω where | S ( ω ) | > 1 , which corresponds to disturbance amplification.
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Problem 2 Answer the following questions: 1. The loop transfer function L ( s ) with the Nyquist plot shown below has two unstable poles.
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