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midterm07_sol

# midterm07_sol - ECE 147A FEEDBACK CONTROL SYSTEMS THEORY...

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ECE 147A FEEDBACK CONTROL SYSTEMS - THEORY AND DESIGN F07 Midterm Exam Closed book and notes. No calculators. Show all work. Name: Solution Problem 1: /25 Problem 2: /25 Problem 3: /25 Problem 4: /25 Total: /100 Miscellaneous: - b ± b 2 - 4 ac 2 a , 1 . 8 ω n , 4 . 6 σ , exp ˆ - πζ p 1 - ζ 2 ! , atan([0.1 0.5 2/3 1 2 5 10])*180/pi = [5.7 26.6 33.7 45 63.4 78.7 84.3] acos([0.5 0.7071 0.866])*180/pi =[60 45 30] asin([0.5 0.7071 0.866])*180/pi =[30 45 60] - 1 dB 0 . 89, - 3 dB 0 . 71, - 5 dB 0 . 56 - 10 dB 0 . 32 - 15 dB 0 . 18 - 30 dB 0 . 0316. " m 11 m 12 m 21 m 22 # - 1 = " m 22 - m 12 - m 21 m 11 # m 11 m 22 - m 21 m 12

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Name: Problem 1 A nonlinear control system is modeled by the second-order differential equation ¨ z = g 1 ( z ) + g 2 ( ˙ z ) + e u e y = g 3 ( z ) where g 1 ( z ) = z 2 , g 2 ( ˙ z ) = - 3 ˙ z ( ˙ z + 1) 2 and g 3 ( z ) = z 3 . 1. Find an operating point corresponding to a constant output equal to 8. Is there one or more than one such operating point? We start by picking the state vector x = " z ˙ z # giving us the equations ˙ x 1 = x 2 ˙ x 2 = g 1 ( x 1 ) + g 2 ( x 2 ) + e u e y = x 3 1 . Clearly, any operating point must have x * 2 = 0 and we note that g 2 (0) = 0 . Also, if e y = x 3 1 = 8 then we must have x * 1 = 3 8 = 2 . If x * 1 = 2 then g 1 ( x * 1 ) = 2 2 = 4 so that we must have u * = - 4 . There are no other values that give an operating point with e y = 8 . The operating point is (2 , 0 , - 4) . 2. Compute the transfer function from e u - u * to e y - 8 where u * is an input that makes the output a constant equal to 8. (Note that the formula for the inverse of a 2 × 2 matrix is given on the front page of the exam.) The transfer function will be C ( sI - A ) - 1 B + D where A = " 0 1 2 x * 1 - 3( x * 2 + 1) 2 - 6 x * 2 ( x * 2 + 1) # = " 0 1 4 - 3 # and B = " 0 1 # , C = h 3 x 2 1 0 i = h 12 0 i , D = 0 . Using the formula on the front of the exam, we get ( sI - A ) - 1 = " s + 3 1 4 s # s 2 + 3 s - 4 , C ( sI - A ) - 1 B + D = 12 s 2 + 3 s - 4 .
Name: 3. Is the transfer function stable or unstable? The transfer function is unstable since the coefficients of the denominator polynomial are not all of the same sign. 4. Suppose a controller C ( s ) has been designed to control the system near the operating point determined above, using the negative unity feedback configuration with the reference prefilter F ( s ). Draw the corresponding block diagram implementation of the entire control system, including the nonlinear model of the system that is being controlled. The implementation is shown below.

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Name: Problem 2 Consider a unity negative feedback control problem with P ( s ) = 2( s - 1)( s - 30) s ( s + 1)( s + 30) . The goal is to design a lead controller to stabilize the system and make all three closed-loop poles have real part less than or equal to - 1. You may use the controller zero to cancel a stable plant pole if you find that helpful.
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