# lag compensation design assumptions?

I have a plant, Gp = 100/s(1+s/100)^2
I need to compensate this plant using lag compensation to give a phase margin of 45 and a Kv 0...
So I know that my compensator will be of form:
Gc = Kc(1+s/b)/(1+s/a).
I know Kc = 2.
a < b < c < 100, where c = crossover frequency in rad/s and a, b are respective frequencies at which the lag compensator comes into affect. I solve a system of 3 equations from:
|GcGp|= 1 @ c
phase GcGp @c = -3pi/4 (using Taylor series arctan approximations)
and
d/dw[phase] @ c = 0.
Solving the 3 equations, I came up with nonsense results-- ie values for a,b, and c that violated the assumed relationships between a,b,c,100. So, I know the assumption is wrong--but I do not understand how one knows what the 'right' assumption is regarding the lag compensation parameters and the cross over frequency and the double pole at 100.
Yes, this is a homework -- and I'm not looking for a direct answer--but rather a hint in knowing--or rather how to know what the relationships of a,b, c and 100 are? Or can this only be done by trial and error? Tim Wescott has been quite helpful in the past with giving me clues--and hoping he or someone else here will again. My thanks in advance to any/all.
My next best guess is a < c < b < 100-- but I REALLY don't want to solve 3 more nasty simultaneuos equations again only to discover the assumption is bad. How do you determine a 'best' assumption? My original assumption was modeled after an in class example the teacher worked.
Thanks,
Bo
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Bo wrote:

Your instructor uses different notation than I do, so I'm not sure what the 'Kv' means. Illuminate me, and maybe I'll be more help.
Since you're using approximations for the phase contributions, it may be a good idea to get things close with a Bode plot -- you can use nearly any math package, including pencil and paper. Once you find a combination that's close, _then_ use your approximations to figure out the 'exact' answer.
--

Tim Wescott
Wescott Design Services
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Tim is right, except I would suggest always sketching a Bode Plot. Walking through the mathematics can lead to many dead ends/unrealistic solutions. Whereas the Bode plot is a really good way to get in a realistic neighborhood. Typically you can do a sketch in a short period of time.
Ray
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

You need to calcuate the closed loop transfer function (CLTF) T(s)=Gp(s)*Gc(s)/(1+Gp(s)*Gc(s)) I do this symbollically using Mathcad. Then one must find the a and b that meet your requirements. I use a minimization package the minimizes the error between my CLTF and the desired response or Bode plot.
Yes, the minimization function uses a lot of trial and error but it is very smart about picking values for a and b.
What is Kv?
Peter Nachtwey
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
napisa�(a):
(...)

I'm almost sure that:
1/Kv = e [velocity steady error] (when input is a ramp signal) describes ability to follow the ramp signal
e(t)lim't->0' = lim 's->0' s*R*1/(1+openloopTF) where R=1/s^2 is input e(t)lim't->0' = lim 's->0' s/s^2 * 1/(1+openloopTF) e(t)lim't->0' = lim 's->0'      1/(s*(1+openloopTF)) e(t)lim't->0' = lim 's->0' 1/(s+s*openloopTF)
e=1/Kv Kv = lim 's->0' (s*openloopTF)
I hope it looks human readible.
--
Mikolaj

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
napisa�(a):

(...)
Not necessarily. There are very simple procedures based on open-loop bode plot. Of course you can then design phase margin only and overshoot will be the sweet mistery without calculating CLTF.
assumption: LAG (PI) form C=(1+aTs)/(1+Ts) and a<1
1. design proportional gain to achieve steady error (gain=2 as Bo said)     and plot Bode of Gp*gain 2. find frequency 'w' where openloop has needed phase margin plus about 6deg.     (about 36rad/sek in this case) 3. what is required attenuation to achieve this phase margin     (gain of openloop at 36rad/sek, in this case you need to attenuate by 14dB     so it is 0.2, this is your 'a' coeff.) 4. calculate zero (1/aT) of the controller, it should be at     freq. 10 times higher then the pole so 1/aT/w     (T=1.1 in your case) END
For more details look over internet.
--
Mikolaj

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

The method our prof has outlined, while an estimate because of arctan approximations, is a 'more' exact method than guessing how much additional phase one needs for the CLTF. We actually design for max (d/dw = 0) of the arctan equation of the open loop transfer function equation. It is interesting that this procedure gives pretty good phase margin results--+/- 3 degrees--but that the % overshoot is 30% higher than the approximations would suggest.

Turns out I made math errors that were giving me the incongruent answers/assumptions. Solving the system of 3 non-linear equations has proven to be the hardest thing to do by hand. Using MATLAB, it's a piece of cake--but no matlab/laptops allowed on midterms :(
Your 10 times higher statement in step 4-- proved to be true--although that generality has yet to show up in class lectures...is this always true?
Thanks again to all who replied.
Bo
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Thanks to all who responded.
Yes, Mikolaj had the meaning of Kv correct-- ie the velocity constant. I'll followup to his post...
Bo
= -3pi/4 (using Taylor series arctan approximations)

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
= -3pi/4 (using Taylor series arctan approximations)

Yes-- but trying to solve by hand--without a lot of trial and error--is my objective. Solving with matlab is easy :)

Velocity constant
Bo