Tuning a PID controller to desired Overshoot, settling time and rise time

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Muhammad 2022-12-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1883952-tuning-a-pid-controller-to-desired-overshoot-settling-time-and-rise-time

评论： Sam Chak ，2022-12-28

The uncompensated plant transfer function in this case is:

as shown in the simulink model below:

I want to tune the PID so that the Rise time becomes 0.561 s, settling time becomes 2.6 s and overshoot narrows to 8.83%. However the PID tuner app is very inconvenient in this regard as it only varies the response time and transient behaviour, it is impossible to achieve the desired settling time and overshoot with such a PID tuner.

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Answer 1

Sulaymon Eshkabilov 2022-12-24

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https://ww2.mathworks.cn/matlabcentral/answers/1883952-tuning-a-pid-controller-to-desired-overshoot-settling-time-and-rise-time#answer_1135717

Simply double-click on PID block and its parameters window opens. Click on "Tune" button as shown in the attached screen shot - Step 1 - PID_tune_step1.jpg.

Then follow step 2 as shown in the attached file: PID_tune_step2.jpg

Alternative way is using pidtune() in MATLAB.

Good luck

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Answer 2

Paul 2022-12-24

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The CST offers some tools to do this. Start from this doc page. Or from this doc page if you want to use Simulink.

Here is an example of the command line workflow. The result I got is not that good, so you'll have to make it work better to get the result you want

Define the plant model

G = tf(20,[1 4.5 64],'InputName','u','OutputName','y');

Define a tunable PID compensator

K = tunablePID('PID','PID');
K.InputName = 'e';
K.OutputName = 'u';

Create the closed loop system

S = sumblk('e = r - y');
CL0 = connect(G,K,S,'r','y');

Define a second order model response that meets the requirements for rise time, settling time, and overshoot.

Rtrack = TuningGoal.StepTracking('r','y',.30,8.8);
stepinfo(Rtrack.ReferenceModel)
ans = struct with fields:
         RiseTime: 0.5654
    TransientTime: 1.7893
     SettlingTime: 1.7893
      SettlingMin: 0.9077
      SettlingMax: 1.0880
        Overshoot: 8.7988
       Undershoot: 0
             Peak: 1.0880
         PeakTime: 1.1967

Use systune to design the PID gains

CL = systune(CL0,Rtrack);
Final: Soft = 3.09, Hard = -Inf, Iterations = 44

The PID controller is:

showTunable(CL)
PID =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = -0.101, Ki = 6.45, Kd = -23.9, Tf = 1.37e+04
 
Name: PID
Continuous-time PIDF controller in parallel form.

Those negative gains look peculiar!

The closed-loop step response is

stepplot(CL)

stepinfo(CL)

ans = struct with fields:

RiseTime: 0.4233 TransientTime: 2.0735 SettlingTime: 2.0735 SettlingMin: 0.8039 SettlingMax: 1.0237 Overshoot: 2.3675 Undershoot: 0.0210 Peak: 1.0237 PeakTime: 1.5024

Presumably there are other Tuning Goals that can be applied to get the desired response.

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Answer 3

Sam Chak 2022-12-24

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https://ww2.mathworks.cn/matlabcentral/answers/1883952-tuning-a-pid-controller-to-desired-overshoot-settling-time-and-rise-time#answer_1135792

编辑：Sam Chak 2022-12-24

Hi @Muhammad

This is the Root Locus compensator design approach, first attempt.

s   = tf('s');
Gp  = 20/(s^2 + 4.5*s + 64)
Gp =
 
         20
  ----------------
  s^2 + 4.5 s + 64
 
Continuous-time transfer function.
controlSystemDesigner('rlocus', Gp)

The app will show root locus of

.

Fig. 1: Right-click the white area and add a new design requirement.

Fig. 2: Select Percent overshoot and insert the design value.

Fig. 3: An exclusion region is indicated by the yellow shaded area, separated from the white area by the thick black lines.

Fig. 4: Add a second design requirement, select Settling time and insert the design value.

Fig. 5: A second exclusion region overlaps with the first exclusion region. A new thick black line can also be seen.

Fig. 6: Right-click the white area, click on Edit compensator, and then add an Integrator, because the Plant is a Type-0 system.

Fig. 7: Right-click the white area, click on Edit compensator, and then add a Real Pole, placing it relatively far away from the Plant's poles

Fig. 8: Right-click the white area, click on Edit compensator, and then add the Complex Zeros to cancel out the Plant's stable Complex poles. Natural frequency

, and Damping ratio

.

Note: This doesn't work if the Plant is unstable.

Fig. 9: The "magenta dot markers" can be seen. The Plant's stable poles are cancelled out by the Compensator's zeros. Next, the design task is to drag one of the magenta markers (on the real axis) along the root locus until a response plot stays outside of the associated exclusion regions.

Fig. 10: The "magenta dot markers" are dragged to the breakaway point (My preference).

Fig. 11: This can also be directly adjusted through entering a design value to Compensator gain.

Fig. 12: Check the Step response if the performance is satisfactory.

Fig. 13: If satisfactory, then export the designed Compensator to Workspace. With the design values, they can be entered in the Transfer function block or zpk block in Simulink.

C   = zpk(1.8*(s^2 + 4.5*s + 64)/(s*(s + 12)))
C =
 
  1.8 (s^2 + 4.5s + 64)
  ---------------------
        s (s+12)
 
Continuous-time zero/pole/gain model.
Gcl = tf(feedback(C*Gp, 1))
Gcl =
 
           36 s^2 + 162 s + 2304
  ---------------------------------------
  s^4 + 16.5 s^3 + 154 s^2 + 930 s + 2304
 
Continuous-time transfer function.
% Can check in MATLAB whether the design requirements are satisfied.
S   = stepinfo(Gcl)
S = struct with fields:
         RiseTime: 0.5599
    TransientTime: 0.9725
     SettlingTime: 0.9725
      SettlingMin: 0.9055
      SettlingMax: 1.0000
        Overshoot: 0
       Undershoot: 0
             Peak: 1.0000
         PeakTime: 2.1337

If satisfactory, find the equivalent PID gains via algebraic calculations.

% This step is unnecessary, because the Compensator works just as good!
kp   = -0.125;
ki   =  9.6;
kd   =  77/480;
Tf   =  1/12;
Cpid = pid(kp, ki, kd, Tf)
Gpid =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = -0.125, Ki = 9.6, Kd = 0.16, Tf = 0.0833
 
Continuous-time PIDF controller in parallel form.
% Proof
zpk(Cpid)
ans =
 
  1.8 (s^2 + 4.5s + 64)
  ---------------------
        s (s+12)
 
Continuous-time zero/pole/gain model.

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Sam Chak 2022-12-25

Hi @Muhammad

If you want to stay at the borderline design requirements, then pick these PID gain values. You should also try out and learn @Paul's tuning method. But you need to set tau

, instead of

.

Gp = tf(20, [1 4.5 64])

Gp = 20 ---------------- s^2 + 4.5 s + 64 Continuous-time transfer function.

% PID Controller (manual via

kp = -1.522560;

ki = 8.787510;

kd = 0.508165;

N = 4.105485;

Tf = 1/N;

Gc = pid(kp, ki, kd, Tf)

Gc = 1 s Kp + Ki * --- + Kd * -------- s Tf*s+1 with Kp = -1.52, Ki = 8.79, Kd = 0.508, Tf = 0.244 Continuous-time PIDF controller in parallel form.

Gcl = feedback(Gc*Gp, 1)

Gcl = 11.27 s^2 + 50.73 s + 721.5 --------------------------------------------- s^4 + 8.605 s^3 + 93.75 s^2 + 313.5 s + 721.5 Continuous-time transfer function.

% Desired Rise time 0.561 sec takes precedence over the desired Settling time 2.6 sec

% Desired Percent overshoot 8.83% is also met.

S = stepinfo(Gcl)

S = struct with fields:

RiseTime: 0.5609 TransientTime: 1.7760 SettlingTime: 1.7760 SettlingMin: 0.9087 SettlingMax: 1.0883 Overshoot: 8.8278 Undershoot: 0 Peak: 1.0883 PeakTime: 1.1890

step(Gcl, 5)

In Simulink, the following P, I, D, N values are entered into the PID controller block.

zpk(Gc)
ans =
 
  0.5637 (s^2 + 4.5s + 64)
  ------------------------
        s (s+4.105)
 
Continuous-time zero/pole/gain model.
p = pole(minreal(Gcl, 1e-4))
p = 
  -2.0527 + 2.6571i
  -2.0527 - 2.6571i

The root locus of the compensated system looks like this. As expected, because the compensator only meets the borderline design requirement, the loci of the closed-loop poles (

marked by the magenta squares) are situated at the boundary lines separating the white and yellow shaded areas.

Sam Chak 2022-12-26

编辑：Sam Chak 2022-12-27

Hi @Paul,

Edit: I'm learning Tuning Goals from your example. Let's say the desired closed-loop transfer function is known.

G = tf(11.27405828, [1 4.10548454 11.27405828])
G =
 
          11.27
  ---------------------
  s^2 + 4.105 s + 11.27
 
Continuous-time transfer function.
pole(G)
ans = 
  -2.0527 + 2.6571i
  -2.0527 - 2.6571i

How should I insert location of the desired poles in this syntax?

Req = TuningGoal.Poles(location, mindecay, mindamping, maxfreq)

https://www.mathworks.com/help/control/ref/tuninggoal.poles-class.html#btjqa2o-1-location

*************************************************

Regarding the derivative filter I learned that If I design an ideal PID (without derivative filter)

,

then the filter time constant in the practical PID (with derivative filter)

,

should be selected to satisfy this condition

.

To be honest, I don't treat PID (with derivative filter) as the typical P + I + D thing, but more as a special form of compensator, which has an equivalent transfer function of

something that is similar to what is shown here:

https://www.mathworks.com/help/slcontrol/ug/when-tuning-the-pid-controller-the-d-gain-has-a-different-sign-from-the-i-gain.html

With a 2nd-order Compensator and a 2nd-order Plant, the closed-loop transfer function should be a 4th-order system without Pole–Zero cancellation. If applying the conventional design approach taught in textbooks, then we should select 4th-order system that gives the desired overshoot

.

G1  = tf(1, [1 2.6784513 3.7935254 2.6784513 1])
G1 =
 
                      1
  -----------------------------------------
  s^4 + 2.678 s^3 + 3.794 s^2 + 2.678 s + 1
 
Continuous-time transfer function.
S1  = stepinfo(G1)
S1 = struct with fields:
         RiseTime: 2.6359
    TransientTime: 8.2598
     SettlingTime: 8.2598
      SettlingMin: 0.9155
      SettlingMax: 1.0883
        Overshoot: 8.8279
       Undershoot: 0
             Peak: 1.0883
         PeakTime: 6.0531

Rise Time takes precedence over the Settling Time. The desired Rise Time

sec can be achieved by selecting a proper frequency

.

wn  = 4.6984048;
G2  = tf(wn^4, [1 2.6784513*wn 3.7935254*wn^2 2.6784513*wn^3 wn^4]);
S2  = stepinfo(G2)
S2 = struct with fields:
         RiseTime: 0.5610
    TransientTime: 1.7580
     SettlingTime: 1.7580
      SettlingMin: 0.9155
      SettlingMax: 1.0883
        Overshoot: 8.8279
       Undershoot: 0
             Peak: 1.0883
         PeakTime: 1.2883

However, if

is selected, the compensated system will have zeros on the right-half s-plane and large undershoot. Skipping the tuning

process,

is selected instead.

wn  = 7.5089051;
G3  = tf(wn^4, [1 2.6784513*wn 3.7935254*wn^2 2.6784513*wn^3 wn^4])
G3 =
 
                     3179
  -------------------------------------------
  s^4 + 20.11 s^3 + 213.9 s^2 + 1134 s + 3179
 
Continuous-time transfer function.
S3  = stepinfo(G3)
S3 = struct with fields:
         RiseTime: 0.3510
    TransientTime: 1.1000
     SettlingTime: 1.1000
      SettlingMin: 0.9155
      SettlingMax: 1.0883
        Overshoot: 8.8279
       Undershoot: 0
             Peak: 1.0883
         PeakTime: 0.8061

Applying the Pole Placement method (manual algebra calculation as I'm not good at setting TuningGoal and using systune) based on

and converting the 2nd-order compensator to PID form:

% Practical PID controller 
kp  = -0.2203631;
ki  = 10.1815615;
kd  = 0.26916585;
Tf  = 0.06405211;
Gc  = pid(kp, ki, kd, Tf)
Gc =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = -0.22, Ki = 10.2, Kd = 0.269, Tf = 0.0641
 
Continuous-time PIDF controller in parallel form.
% Closed-loop system
s   = tf('s');
Gp  = 20/(s^2 + 4.5*s + 64);    % Plant
Gcl = minreal(feedback(Gc*Gp, 1))
Gcl =
 
          79.64 s^2 + 134.8 s + 3179
  -------------------------------------------
  s^4 + 20.11 s^3 + 213.9 s^2 + 1134 s + 3179
 
Continuous-time transfer function.

Designing a Prefilter to cancel out the zeros on the left-half s-plane.

% Prefilter

Gf = tf(39.919688, [1 1.6929442 39.919688])

Gf = 39.92 --------------------- s^2 + 1.693 s + 39.92 Continuous-time transfer function.

% Step Response

Gclf = minreal(Gf*Gcl, 1e-4) % Should behave almost similar to G3

Gclf = 3179 ------------------------------------------- s^4 + 20.11 s^3 + 213.9 s^2 + 1134 s + 3179 Continuous-time transfer function.

step(Gf*Gcl, 3)

stepinfo(Gf*Gcl)

ans = struct with fields:

RiseTime: 0.3510 TransientTime: 1.1000 SettlingTime: 1.1000 SettlingMin: 0.9153 SettlingMax: 1.0883 Overshoot: 8.8279 Undershoot: 0 Peak: 1.0883 PeakTime: 0.8060

Sam Chak 2022-12-28

Thank you, @Paul. Will look into it.

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Tuning a PID controller to desired Overshoot, settling time and rise time

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