midterm05_sol

midterm05_sol - ECE 147A FEEDBACK CONTROL SYSTEMS THEORY...

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ECE 147A FEEDBACK CONTROL SYSTEMS - THEORY AND DESIGN F05 Midterm Exam Closed book and notes. No calculators. Show all work. Name: Solution Problem 1: /30 Problem 2: /30 Problem 3: /35 Total: /95 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] - 1 dB 0 . 89, - 3 dB 0 . 71, - 5 dB 0 . 56 - 10 dB 0 . 32 - 15 dB 0 . 18 - 30 dB 0 . 0316.
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Name: Problem 1 For a particular experimental set-up, the equations of motion for a ball rolling on a beam are given by ¨ r = - sin( θ ) - 0 . 5 ˙ r + r ˙ θ 2 ¨ θ = 1 3 r 2 + 2 h - 6 r ˙ r ˙ θ - 30 r cos( θ ) - 2 ˙ θ + τ i where r is the ball position, ˙ r is the ball velocity, θ is the beam angle, and ˙ θ is the beam angular velocity. The output is the ball position and the input is the torque τ applied to the beam. 1. True or False: For each ball position r * there exists an input τ * such that r * is a component of a valid operating point. Explain. True. One can take the state vector to be x = ( x 1 , x 2 , x 3 , x 4 ) = ( r, ˙ r, θ, ˙ θ ) . To be an operating point, x * must have x * 2 = ˙ r * = 0 and x * 4 = ˙ θ * = 0 . In turn, this implies that sin( x * 3 ) = sin( θ * ) = 0 and τ * = 30 r * cos( θ * ) , which is either 30 r * or - 30 r * . Henceforth, we will assume that the angle θ stays in the range ± 90 degrees, so that the ball does not fall oF of the beam. In that case, θ * = 0 and τ * = 30 r * . 2. Compute the matrices A , B , C and D that would be used to generate the transfer function from τ - τ * to r - r * for an operating point corresponding to r * = 2. Write the formula for the transfer function in terms of A , B , C , and D . (You do not need to compute the transfer function). How many poles will the transfer function have? We have f ( x, τ ) = x 2 - sin( x 3 ) - 0 . 5 x 2 + x 1 x 2 4 x 4 1 3 x 2 1 + 2 [ - 6 x 1 x 2 x 4 - 30 x 1 cos( x 3 ) - 2 x 4 + τ ] h ( x, τ ) = x 1 A = ∂f ( x, τ ) ∂x | ( x = x * = τ * ) = 0 1 0 0 ( x * 4 ) 2 - 0 . 5 - cos( x * 3 ) 2 x * 1 x * 4 0 0 0 1 - 6 x * 2 x * 4 - 30 cos( x * 3 ) 3( x * 1 ) 2 +2 + 6 x * 1 (6 x * 1 x * 2 x * 4 +30 x * 1 cos( x * 3 )+2 x * 4 - τ * ) (3( x * 1 ) 2 +2) 2 - 6 x * 1 x * 4 3( x * 1 ) 2 +2 30 x * 1 sin( x * 3 ) 3( x * 1 ) 2 +2 - 2 - 6 x * 1 x * 2 3( x * 1 ) 2 +2 = 0 1 0 0 0 - 0 . 5 - 1 0 0 0 0 1 - 30 14 0 0 - 2 14
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Name: B = ∂f ( x, τ ) ∂τ | ( x = x * = τ * ) = 0 0 0 1 3( x * 1 ) 2 + 2 = 0 0 0 1 14 C = ∂h ( x, τ ) ∂x | ( x = x * = τ * ) = h 1 0 0 0 i D = ∂h ( x, τ ) ∂τ | ( x = x * = τ * ) = 0 .
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midterm05_sol - ECE 147A FEEDBACK CONTROL SYSTEMS THEORY...

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