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Unformatted text preview: Solution to Midterm 2 for EE3530
1. The ultimate period is approximately: Pu = 1:4. Hence
K = 0:6Ku = 0:6 11:74 = 7:044;
D s TI = 0:5Pu = 0:7; TD = Pu =8 = 0:175 1 which gives the PID compensator: =
KTD s s 2 + T1 s + T 1T
D D I 2 = 1:2327s + 5:7143s + 8:1633 : s 2. For PID compensator, setting TD = 1=2 and 1=TD TI = 10 yields
2 4
DGs = KTD s2 + T1D s + TD1TI
s 3 5 400:2s + 1 s2 + 2s + 10s + 8s + 3 2 20 !n = ssK 0:2s + 1 s2s:5K3 = ss + 2
! + 8s + 3 + n For the given speci cations, p i ! = 2:5K = 1:8 1:8; 3 ii = 2! = p 1:5 p 2j log0:2j 2 = 0:4559: + j log0:2j 2:5K n n tr Solving i and ii yields 1:82 K 1:5 2 =2:5 = 4:3302: 1:2960 = 2:5 0:4559 Because we did approximation in eliminating an extra zero and pole, it may have adverse e ect on both rise time if K = 1:2960, and on overshot if K = 4:3302, we choose the middle value K = 1:2960 + 4:3302=2 = 2:8131 for safty. Hence the PID compensator is given by
D s...
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This note was uploaded on 02/03/2012 for the course EE 3530 taught by Professor Chen during the Fall '07 term at LSU.
 Fall '07
 Chen

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