PID root locus

Question

I have open loop function 31.75/(s^3+3*s^2+1.25s) I am trying to design PID controller for this using given percentage overshoot n settling time. I m using root locus graphical technique to do it. I have design PD controller and PI controller separatleti I want to know how can i desing PID controller andits transfrom function .I also want to know how can i determine the PId compensator gain.Please provide any related information as i m new in it.
Dost

Answer 1

Hi
Thanks for reply.well for the time being using LAbview or matlab is not an issue infact I am desinging P dcontroller and calcualting the parameter of Kd and Kp.open loop transfer function for my problem is : 31.75/s(s+2.5)(s+0.5) .
you can see there is an integrator in the open loop and theortically it should nt give any steady state error the close loop function obtained from this openloop with feedback is:
5.286
----------------------------       the step respons of the close loop function shows a unbounded system ;any comments on it????? should we include 31.75 the numerator of the open l
s^3 + 3 s^2 + 1.25 s + 31.72
loop in the calcualtion of the close loop?
i developed aPD compensator using graphical root locus techniques using the following time doima specification overshoot 10% settling time 10 secod.the desired pole from this specification is -2.-+2,76i which gives from gemoetry the position of the required zero as -1.24 using this zero does not give the required specification so i used z=-.3 (any comments on it)I am also perplexed about the calculation of the Kd :from graphical  root locus methos i have used following formula
Kd= product of vector lenght of finte pole to  desire pole/(productor of vector length of zero to desired pole * 31.75)
 the obtained PD compensator is :
 0.2866 s + 0.08599 and coreesponding open loop P Dcompensator is :
 9.09 s + 2.727
s^3 + 3 s^2 + 1.25 s
the close loop is obtained as
     1.515 s + 0.4545
s^3 + 3 s^2 + 10.34 s + 2.727
this gives settling time 3 second overshoot is 16 % but steady state is .167 i.e about 99 % steady state error.
Looking forrwad for ur reply.
Dost