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midterm06_sol - ECE 147A FEEDBACK CONTROL SYSTEMS THEORY...

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ECE 147A FEEDBACK CONTROL SYSTEMS - THEORY AND DESIGN F06 Midterm Exam Closed book and notes. No calculators. Show all work. Name: Solution Problem 1: /25 Problem 2: /35 Problem 3: /25 Total: /85 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 The dynamical equations for the height of a bead on a rotating hoop are given by ¨ θ = ω 2 sin( θ ) cos( θ ) - ν m ˙ θ - g R sin( θ ) e y = R [1 - cos( θ )] where the control input is the angular velocity of the hoop ω , the variable θ denotes the angle the bead makes relative to the vertical down position (see the figure), the coefficient ν is used to characterize friction, R is the radius of the hoop, m is the mass of the bead, g is the gravitational constant, and e y is the height of the bead above the floor. ω y ~ θ R Figure 1: The bead on a hoop 1. For each possible value of the height y * , find all of the possible input values ω * and state values that correspond to an operating point with the given height. (I’m not looking for numerical values here...) We take x 1 = θ , x 2 = ˙ θ , e u = ω and we get ˙ x 1 = x 2 =: f 1 ( x, e u ) ˙ x 2 = e u 2 cos( x 1 ) sin( x 1 ) - ν m x 2 - g R sin( x 1 ) = - ν m x 2 - sin( x 1 ) g R - e u 2 cos( x 1 ) =: f 2 ( x, e u ) e y = R [1 - cos( x 1 )] =: h ( x, e u ) . Operating points, which enforce f ( x * , ω * ) = 0 , must satisfy x * 2 = 0 and sin( x * 1 ) = 0 or g = R ( ω * ) 2 cos( x * 1 ) . Such points either have the form ( kπ, 0 , ω * ) where k is an arbitrary integer and ω * is an arbitrary real number, or else, as long as x * 1 is such that cos( x * 1 ) > 0 , ˆ x * 1 , 0 , ± s g R cos( x * 1 ) ! . It follows that the operating height y * must be in the interval [0 , R ) or else equal to 2 R . For y * = 0 or y * = 2 R , this corresponds to x * 1 = for some integer k and ω * can be anything. For y * (0 , R ) , x * 1 and ω * must satisfy cos( x * 1 ) = 1 - y * R , ω * = ± s g R (1 - y * /R ) = ± r g R - y * .
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Name: 2. Compute the transfer function from ω - ω * to e y - y * for any ω * that goes with y * = 0 . 5 R . Note that the formula for the inverse of a 2 × 2 matrix is given on the front page of the exam. y * = 0 . 5 R translates to cos( x * 1 ) = 0 . 5 and thus x * 1 = ± 60 or ± π/ 3 and the operating point is ± π/ 3 , 0 , ± q g 0 . 5 R · . We form the linearization matrices A = " 0 1 a 21 - ν m # , B = " 0 b 2 # , C = h c 1 0 i , D = 0 where a 21 = ( ω * ) 2 (cos( ± π/ 3) 2 - sin( ± π/ 3) 2 ) - g R cos( ± π/ 3) = g R 2 1
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