{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol99f2

Sol99f2

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online