Prashant Kumar answered . 2024-05-04 01:01:18

s = tf('s');
G = 1.5531/(s^2+10.0938*s);
G.u = 'V'; %input (voltage)
G.y = 'X'; % ouput (x position)
% PID Controller Transfer Function
H = pid(120,10,40,1e-3)

H =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = 120, Ki = 10, Kd = 40, Tf = 0.001
 
Continuous-time PIDF controller in parallel form.

H.u = 'e'; % input (error)
H.y = 'V'; % output (voltage)
% Combined Transfer Function for R to X (X/R)
Sum = sumblk('e = R - X'); % solve for e,rror in terms of R and X
%XR = connect(H,G,Sum,'R','X');
% Solve for step response
%[x1, t1] = step(XR,1.5);
% Combined Transfer Function for R to V  and X 
clsys = connect(H,G,Sum,'R',{'V' 'X'});
zpk(clsys)

ans =
 
  From input "R" to output...
        40120 s (s+0.08579) (s+2.905) (s+10.09)
   V:  -----------------------------------------
       (s+0.08621) (s+2.584) (s+74.75) (s+932.7)
 
              62310 (s+2.905) (s+0.08579)
   X:  -----------------------------------------
       (s+0.08621) (s+2.584) (s+74.75) (s+932.7)
 
Continuous-time zero/pole/gain model.

% call step twice because the X repsonse takes longer than the V response
figure
step(clsys('X','R'))

 

 

 

figure
step(clsys('V','R'))

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