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lab2_v3

# lab2_v3 - University of California Santa Barbara Department...

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University of California, Santa Barbara Department of Electrical & Computer Engineering ECE 147b: Digital Control Lab 2: Linear Flexible Joint Overview The linear flexible joint system builds on the motor cart system used in the previous lab. Here, the motor cart is coupled to a second cart (which we will refer to as the load cart) via a spring. The second cart is equipped with an optical encoder, making position measurements possible. Below is a free-body diagram describing this system. Modeling x 1 x 2 m 2 m 1 F Figure 1: Free Body Diagram of Linear Flexible Joint The variables for this system are: F (input force to the motor cart in Newtons), k (spring stiffness in Newtons/meter), and m 1 and m 2 (mass of the motor and load carts, respectively, in kilograms). Letting x 1 and x 2 be the position of the motor and load carts respectively, we can write the following equations describing the system behavior: F - k ( x 1 - x 2 ) = m 1 ¨ x 1 (1) k ( x 1 - x 2 ) = m 2 ¨ x 2 (2) The conversion from force to input voltage is given by F = K m K g R m r V - K 2 m K 2 g R m r 2 ˙ x 1 . (3)

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2 ECE147b: Lab #2 Taking Laplace transforms of Equations 1, 2, and 3 and substituting Equation 3 into Equation 1 yields: X 2 ( s )( m 2 s 2 + k ) = kX 1 ( s ) X 1 ( s )( m 1 s 2 + K 2 m K 2 g R m r 2 s + k ) = K m K g R m r V ( s ) + kX 2 ( s ) By substitution, we can generate the three transfer functions X 1 ( s ) V ( s ) , X 2 ( s ) V ( s ) , and X 2 ( s ) X 1 ( s ) . Two of the transfer functions, with numerical values substituted, are provided for ease of reference. X 2 ( s ) X 1 ( s ) = 61 . 2 s 2 + 61 . 2 (4) X 1 ( s ) V ( s ) = 2 . 97 s 2 + 61 . 2 s 4 + 13 . 24 s 3 + 127 . 15 s 2 + 810 . 37 s (5) We can now close loops around the position of either cart or perform a cascade control design. Sample Design Suppose we close a simple proportional gain loop around the system P ( s ) = X 1 ( s ) V ( s ) . By examining the root locus plot for P ( s ), we pick a gain K = 10 to push the poles as far to the left as we can. This places the continuous-time closed-loop pole locations at s = - 3 . 29 ± j 9 . 33 and - 3 . 33 ± j 3 . 60. The zero-order hold plant equivalent, with sampling period T s = 0 . 05s, is P ( z ) = 2 . 964 × 10 - 3 z 3 + 1 . 047 z 2 + 0 . 4821 z + 0 . 8016 z 4 - 3 . 249 z 3 + 4 . 086 z 2 - 2 . 353 z + 0 . 5158 . With K = 10, the discrete-time closed-loop poles are at z = 0 . 983 ± j 0 . 046 and 0 . 984 ± j 0 . 018, yielding a stable system. Suppose, by looking at the step response, we decided that the rise time is to slow. We can add a lead compensator to improve the response. Using the compensator C ( s ) = K c s + 12 s + 25 we can again plot the root locus of C ( s ) P ( s ) and select K c = 35 to push the closed-loop poles to the left in the s -plane. The continuous-time closed-loop poles are at s = - 3 . 86 ± j 7 . 69, - 4 . 88 ± j 5 . 07, and - 20 . 76. Discretizing this controller via the bilinear transformation yields C ( z ) = 28 z - 0 . 5386 z - 0 . 2308 .
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