How to fix a Phase-Lead Compensator?

Discussion in 'Homework Help' started by woodmark75, Dec 16, 2014.

Dec 11, 2014
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I've just finnished trying to design a Phase-Lead compensator so that the step response fits the design requiements below:
• Ts<4 secs (+/- 5% settling value output)
• P.O.<16%
• compensator has a zero at -1
I used the 'angle criteion' method i.e. summing up the angles from the poles & zeros to a reference point & derived a compensator = (s+1.58)/(s+73.51). My root locus has been modified & the step output of the system now has P.O.<16%. But the Ts is still around 30 secs. How do I reduce this to within the spec?

System specifications:

2. t_n_k AAC Fanatic!

Mar 6, 2009
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784
Frankly, you seem to be very confused. You have a clearly stated point that the compensator is specified as having a zero at -1 and yet your "design" comes up with a zero at -1.58. What are you doing?

Dec 11, 2014
36
0
Edit: to first post the 1.58 should NOT be a zero. It is just the figure that the compensator pole must contribute. Used in the derivation of the actual pole s = -73.51. Sorry for the confusion....

4. t_n_k AAC Fanatic!

Mar 6, 2009
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784
So the the lead compensator isn't (s+1.58)/(s+73.51)? Or is it???
If it is, what are the dominant roots you are seeking to fix on the root locus? What did you determine to be the value of K1 as the other requirement to satisfy this condition?
Again, if this is the lead network design there are certain other considerations one should take into account.

For instance, with the compensator function form

$\text{Gc=\frac{1}{\rho}\frac{(s+Zc)}{(s+Pc)}}$

the ratio

$\text{\rho=\frac{Z_c}{P_c}}$

is of importance.
Generally it is considered good design to keep the ratio $\text{\rho}$ to a value at least greater than 0.05 (with 0.1 being rather more desirable). Of course the ratio will by definition be less than unity for a Lead Compensator.

$\text{\rho=\frac{1.58}{73.51}=0.021}$