No logro estabilizar mi sistema por PID

A pesar de aplicar de todo no se logra estabilizar

Answer 1

[Sam Chak]

Hi @Bethsy Sheyla

For learning purposes, I am providing a demo below to illustrate that an ideal Proportional-Derivative (PD) controller can be designed to stabilize the unstable 3rd-order system. The effectiveness of this approach stems from the fact that only the proportional and derivative terms in the denominator carry negative signs.

First, we need to construct a target closed-loop transfer function (Tcl) around the stable accelerative term, . Subsequently, we can calculate the desired PD gains from the stable proportional and derivative terms of Tcl.

%% Unstable 3rd-order system
Gp  = tf(-19.81, [1, 0.5, -392.4, -196.2])

Gp =
 
              -19.81
  -------------------------------
  s^3 + 0.5 s^2 - 392.4 s - 196.2
 
Continuous-time transfer function.

[num, den]  = tfdata(Gp, 'v')

num = 1×4

0 0 0 -19.8100

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den = 1×4

1.0000 0.5000 -392.4000 -196.2000

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%% Convert Gp to state-space model (Note that B(3) = num(4). It is an input signal amplifier gain)
sys = compreal(Gp);
sys = ss(sys.A', sys.C', sys.B', sys.D')

sys =
 
  A = 
          x1     x2     x3
   x1      0      1      0
   x2      0      0      1
   x3  196.2  392.4   -0.5
 
  B = 
           u1
   x1       0
   x2       0
   x3  -19.81
 
  C = 
       x1  x2  x3
   y1   1   0   0
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.

%% Target Closed-loop TF (This will give us a stable critically-damped response)
wn  = den(2)/3;                         % find natural freq using conservation of accelerative gain
Tcl = tf(wn^3, [1, 3*wn, 3*wn^2, wn^3]) % based on Hurwitz Polynomial for 3rd-order critically-damped

Tcl =
 
                0.00463
  -----------------------------------
  s^3 + 0.5 s^2 + 0.08333 s + 0.00463
 
Continuous-time transfer function.

figure(1)
step(Tcl), grid on

%% Calculate the proportional and derivative gains
% Solve 3*wn^2 = -392.4 + den(4)*Kd for Kd
Kd  = (23549/60)/num(4);    % Why divide by num(4)? Because den(4) amplifies desired control input signal

% Solve   wn^3 = -196.2 + den(4)*Kp for Kp
Kp  = (211901/1080)/num(4);

%% Ideal PD Controller (not realizable in practice, but can be fun for learning)
Gc  = pid(Kp, 0, Kd)

Gc =
 
             
  Kp + Kd * s
             

  with Kp = -9.9, Kd = -19.8
 
Continuous-time PD controller in parallel form.

%% Checking the coefficients of the numerator
Gap = series(Gc, Gp)    % whether -392.4 + 392.5 = 3*wn^2 and -196.2 + 196.2 = wn^3 (not zero!)

Gap =
 
          392.5 s + 196.2
  -------------------------------
  s^3 + 0.5 s^2 - 392.4 s - 196.2
 
Continuous-time transfer function.

%% If the above step is correct, then we can apply feedback to find the Closed-loop TF
Gcl = feedback(Gc*Gp, 1)

Gcl =
 
            392.5 s + 196.2
  -----------------------------------
  s^3 + 0.5 s^2 + 0.08333 s + 0.00463
 
Continuous-time transfer function.

figure(2)
step(Gcl), grid on

%% Since Gcl is stable and the zero of Gcl lies on the LHS of s-plane, it can be cancelled out using a Pre-filter
[num, den]  = tfdata(Gcl, 'v')

num = 1×4

0 0 392.4833 196.2046

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den = 1×4

1.0000 0.5000 0.0833 0.0046

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%% Pre-filter
Gf  = tf(den(4), num(3:4))

Gf =
 
      0.00463
  ---------------
  392.5 s + 196.2
 
Continuous-time transfer function.

%% Filtered Closed-loop
Fcl = minreal(series(Gf, Gcl))  % verify if Fcl == Tcl

Fcl =
 
                0.00463
  -----------------------------------
  s^3 + 0.5 s^2 + 0.08333 s + 0.00463
 
Continuous-time transfer function.

figure(3)
step(Fcl), grid on