ECE535_10S_hw4

ECE535_10S_hw4 - u(0 = 10 and develop a series compensator D z that achieves u kT = u(3 T,k ≥ 3 y kT = 1,k ≥ 3(the control from part(a achieved

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ECE 535 DISCRETE TIME SYSTEMS SPRING 2010 HOMEWORK ASSIGNMENT #4 Due: 15 March 2010 1. Text, problem 7.20. 2. Consider the following sampled data system. - ‰… ±» + - D ( z ) - ZOH - s + 1 s 2 + 4 s + 8 - @ @ @ ¡ ¡“ 6 - T r ( kT ) u ( kT ) u ( t ) y ( t ) (a) Find D ( z ) as a deadbeat control for sampling periods T 1 = 1 . 0 and T 2 = 0 . 5. For each control design, plot the resulting responses of u ( t ) and y ( t ) when r ( kT ) is a unit step. Be sure to show the intersample behavior. (b) Suppose that at each sampling instant k , the input magnitude must satisfy | u ( kT ) | ≤ 10. Your deadbeat control from (a) for T 2 should violate this con- tstraint. To modify the control law in order to satisfy the constraint, reformulate the deadbeat control problem as follows. Force
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Unformatted text preview: u (0) = 10, and develop a series compensator D ( z ) that achieves u ( kT ) = u (3 T ) ,k ≥ 3 y ( kT ) = 1 ,k ≥ 3 (the control from part (a) achieved the above result for k ≥ 2). To do this, you’ll set Y ( z ) U ( z ) = ˆ b 1 z + b 2 z 2 + a 1 z + a 2 !ˆ z-p z-p ! . Pick p so that the sum of the numerator coefficients is 1 / 10, and proceed with a deadbeat control for this plant. Explain why this should achieve the desired result, and compute D ( z ). Again plot u ( t ) and y ( t ) with a unit step reference, but now with the now D ( z ). Is the input magnitude constraint met?...
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This document was uploaded on 02/08/2012.

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