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Meeting Design Specs using Root Locus
ECE 461
Problem: Change the previous problem so that the 2% settling time is equal to 4 seconds.
In this case you want to place the closed-loop poles at a specific spot: s = -1.000 + 1.3685
Step 1: Design a compensator which makes the closed-loop system too fast.
See the previous solution
Step 2: Adjust the fast pole so that the root locus passes through -1 + j1.3685.
Let
(s+1)(s+2)
K(s) = k ⎛⎝ s(s+a) ⎞⎠
so
10k
⎞
GK = ⎛⎝ s(s+4)(s+5)(s+a)
⎠
We know the point s = -1.000 + 1.3685 should be on the root locus. This means the phase of GK must be 180
degrees at this point. Plugging it in:
10k
⎞
∠ ⎛⎝ s(s+4)(s+5)
⎠
= −169.56 0
s=−1+j1.3685
so, to make the angles add up to 180 degrees,
∠(s + a) = 10.4354 0
a=1+
1.3685
tan ⎛⎝ 10 4354 0 ⎞⎠
= 8.4305
and
(s+1)(s+2)
K(s) = k ⎛⎝ s(s+8.4305) ⎞⎠
10k
⎞
GK = ⎛⎝ s(s+4)(s+5)(s+8.4305)
⎠
Finding k:
10k
⎛
⎞
⎝ s(s+4)(s+5)(s+8.4305) ⎠ s=−1+j1.3685 = −1
k = 17.8517
and
(s+1)(s+2)
K(s) = 17.8517 ⎛⎝ s(s+8.4305) ⎞⎠
JSG
3
rev October 5, 2007