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Unformatted text preview: ECE 147A FEEDBACK CONTROL SYSTEMS - THEORY AND DESIGN F02 Midterm Exam Closed book and notes. No calculators. Show all work. Name: Solution Problem 1: /10 Problem 2: /20 Problem 3: /20 Problem 4: /15 Total: /65 Miscellaneous:- b ± √ b 2- 4 ac 2 a , 1 . 8 ω n , 4 . 6 σ , exp(- πζ/ p 1- ζ 2 ) , d dx exp( f ( x )) = exp( f ( x )) d dx f ( x ) atan([0.5 2/3 1 2 5 10 15])*180/pi = [26.6 33.7 45 63.4 78.7 84.3 86.2] Problem 1 In a given, stable closed-loop unity feedback configuration, with unity gain prefilter, the controller has one pole at the origin and the plant has one pole at the origin. The loop has no zeros at the origin. Explain your answers to the following questions: 1. What is the system type for the transfer function from reference to tracking error? E ( s ) R ( s ) = 1 1 + L ( s ) , L ( s ) = 1 s 2 L ( s ) , < | L (0) | < ∞ lim s → L ( s ) = ∞ , lim s → sL ( s ) = ∞ , lim s → s 2 L ( s ) = L (0) , Hence Type II. By the way, notice that E ( s ) R ( s ) = s 2 s 2 + L ( s ) i.e., the transfer function has two zeros at the origin, hence it is Type II. 2. What is the system type for the transfer function from disturbance to output? L ( s ) = 1 s 2 L ( s ) , P ( s ) = 1 s P ( s ) , Y ( s ) D ( s ) = P ( s ) 1 + L ( s ) = s P ( s ) s 2 + L ( s ) i.e., one zero at the origin, so Type I. Name: Problem 2 A simplified, normalized model of an optical amplifier is given by the equation ˙ x =- x + e u [1- exp(4 x- 4)] + e d [1- exp(10 x- 4)] e y = exp(10 x- 4) . The output e y is the gain of the amplifier. The control input to the amplifier is e u . The disturbance input to the amplifier is e d . 1. Linearize the dynamics of the amplifier around the operating condition e y * = exp(2), e d * = 0 . 1. Find the transfer functions from e u- u * and e d- d * to e y- y * . (Estimate the numerical value of the open-loop poles(s) )....
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- Spring '10
- lim r, unity feedback configuration, lim L, 0.25 seconds