Setting a Specific Formula for PID Controller in Simulink (Matlab)

I need to add a PID Controller to a Simulink Project with formula 1/(s+4.54) but I couldn't find the proper constants for the formula.

I tried looking at the different types of PID crack Controller libraries. I also tried finding the constants by solving the equation but couldn't.

Answer 1

[Sam Chak]

Hi @nora

Use the Transfer Funtion block instead.

https://www.mathworks.com/help/simulink/slref/transferfcn.html

[dorrin]

Hi @Sam Chak Sorry I'm very new to this. Are they the same? Can we always use them interchangeably?

[Sam Chak]

Hi @dorrin (Is @nora your original account?)

The ideal output-feedback PID controller has a transfer function of the form:

.

However, your given transfer function is low-pass filter.

Thus, they are not the same.

[dorrin]

I posted the question on Stackoverflow and saw it here today (Maybe it's automatically reposted?).

Is it possible to use the PID Tuning feature to get the function ? This is the project I have to build:

[Sam Chak]

Hi @dorrin & @nora

The step response looks okay, no overshoot and settles within 6 seconds. Are the performance requirements satisfied?

% Plant
Gp   = tf(30, [1 30 0])

Gp =
 
      30
  ----------
  s^2 + 30 s
 
Continuous-time transfer function.

% Compensator
Gc   = tf(4, [1 4.54])

Gc =
 
     4
  --------
  s + 4.54
 
Continuous-time transfer function.

% Pre-filter
Gf   = tf(1, [1 1])

Gf =
 
    1
  -----
  s + 1
 
Continuous-time transfer function.

% Closed-loop system
Gcl  = feedback(Gc*Gp, 1)

Gcl =
 
                120
  -------------------------------
  s^3 + 34.54 s^2 + 136.2 s + 120
 
Continuous-time transfer function.

% Closed-loop system with Prefilter
Gclf = Gf*Gcl

Gclf =
 
                      120
  -------------------------------------------
  s^4 + 35.54 s^3 + 170.7 s^2 + 256.2 s + 120
 
Continuous-time transfer function.

% Plot Step response
step(Gclf, 10), grid on

stepinfo(Gclf)

ans = struct with fields:
         RiseTime: 3.1225
    TransientTime: 5.6431
     SettlingTime: 5.6431
      SettlingMin: 0.9034
      SettlingMax: 0.9989
        Overshoot: 0
       Undershoot: 0
             Peak: 0.9989
         PeakTime: 8.7038

Answer 2

[Sam Chak]

Hi @nora & @dorrin

If you intend to use a PID controller to make the system behave similarly to the designed compensator, you'll need to make some slight modifications to the control loop configuration. The system under the PID controller should have the same settling time.

% Plant
Gp   = tf(30, [1 30 0])

Gp =
 
      30
  ----------
  s^2 + 30 s
 
Continuous-time transfer function.

Gpp  = feedback(Gp, 1);

% PID Controller
kp   =  0.266920333340203;
ki   =  0.516920333340203;
kd   = -0.111860942402338;
Tf   =  0.483633519278621;
Gc   = pid(kp, ki, kd, Tf)

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

  with Kp = 0.267, Ki = 0.517, Kd = -0.112, Tf = 0.484
 
Continuous-time PIDF controller in parallel form.

% Closed-loop system
Gcl  = feedback(Gc*Gpp, 1)

Gcl =
 
          1.069 s^2 + 32.06 s + 32.06
  -------------------------------------------
  s^4 + 32.07 s^3 + 93.1 s^2 + 94.1 s + 32.06
 
Continuous-time transfer function.

% Plot Step response
step(Gcl, 10), grid on

stepinfo(Gcl)

ans = struct with fields:
         RiseTime: 3.2484
    TransientTime: 5.6431
     SettlingTime: 5.6431
      SettlingMin: 0.9019
      SettlingMax: 0.9999
        Overshoot: 0
       Undershoot: 0
             Peak: 0.9999
         PeakTime: 11.5370