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

Question

Muhammad 2022-12-24

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评论： Emmanouil Tzorakoleftherakis 2024-6-17

采纳的回答： Sam Chak

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

Sam Chak 2024-1-14

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Hi @Muhammad

I'm revisiting a control problem we encountered a year ago. Previously, we utilized the systune() method to establish PIDF gains, but the transient response exhibited significant oscillations, and the derivative term appeared unusual with an unexpectedly large time constant Tf. This time, I've discovered that a set of classical PID gains can fulfill the requirements: settling time (ts) less than 2.6, overshoot (os) of 8.83%, and a rise time (tr) of 0.561.

%% Original Plant

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

%% Target Reference

Gref= tf(64.9193300346851, [1, 7.93016824066555, 31.8718356679018, 64.9193300346851]);

%% Classical PID

kp = -0.733860787658412;

ki = 6.597481064703570;

kd = 0.100288660665148;

Gc = pid(kp, ki, kd);

%% Closed-loop system

Gcl = feedback(Gc*Gp, 1)

Gcl = 2.006 s^2 - 14.68 s + 131.9 --------------------------------- s^3 + 6.506 s^2 + 49.32 s + 131.9 Continuous-time transfer function.

Scl = stepinfo(Gcl)

Scl = struct with fields:

RiseTime: 0.5610 TransientTime: 2.0757 SettlingTime: 2.0757 SettlingMin: 0.9003 SettlingMax: 1.0883 Overshoot: 8.8301 Undershoot: 0 Peak: 1.0883 PeakTime: 0.9200

stepplot(Gref), hold on

stepplot(Gcl), grid on

stepplot(Gp), hold off

legend('Target System', 'PID Controller', 'Original Plant', 'location', 'east')

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Emmanouil Tzorakoleftherakis 2024-6-17

Hi @Sam Chak,

For completeness, can you maybe share how you reached to those pid gains so that the accepted answer is complete?

Thanks

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Answer 2

Sam Chak 2022-12-24

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编辑：Sam Chak 2022-12-24

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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

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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

在 MATLAB Online 中打开

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

Paul 2022-12-27

在 MATLAB Online 中打开

Hi Sam,

I haven't been looking at this problem too closely, still wasn't able to get systune to give me what I wanted, but I was getting closer. Anyway, there are several ways to insert an Analysis Point into a model. See the link from the doc page you linked for some examples, maybe other doc pages as well. Here's one way, using connect

G = tf(20,[1 4.5 64],'InputName','u','OutputName','y');
K = tunablePID('PID','PID');
K.InputName = 'e';
K.OutputName = 'u';
% Create the closed loop system, include an analysis point at the error
% signal
S = sumblk('e = r - y');
CL0 = connect(G,K,S,'r','y','e')
CL0 =

  Generalized continuous-time state-space model with 1 outputs, 1 inputs, 3 states, and the following blocks:
    AnalysisPoints_: Analysis point, 1 channels, 1 occurrences.
    PID: Tunable PID controller, 1 occurrences.

Type "ss(CL0)" to see the current value, "get(CL0)" to see all properties, and "CL0.Blocks" to interact with the blocks.

Then, in systune, the "location" input would 'e'

Sam Chak 2022-12-28

Thank you, @Paul. Will look into it.

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Answer 3

Sulaymon Eshkabilov 2022-12-24

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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 4

Paul 2022-12-24

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在 MATLAB Online 中打开

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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Sam Chak 2024-1-13

编辑：Sam Chak 2024-1-13

在 MATLAB Online 中打开

Hi @Hansi K,

Given the lack of a clear description of the rise time by the OP, the definition of rise time appears to be somewhat subjective. MATLAB typically measures rise time from 10% to 90%. In contrast, many control theory textbooks use a formula defining it as the time needed for the response to rise from 0% to 100% of its final value.

%% Parameters

zeta = 0.61131525436442019; % Damping ratio that gives OS 8.83

tr = 0.561; % Rise time from 0 to 100%

%% Natural frequency of a linear second order system

wn = ((1/sqrt(1 - zeta^2))*(pi - atan(sqrt(1 - zeta^2)/zeta)))/tr;

tau = 1/wn % this will be the tau value I'd probably take

tau = 0.1992

%% 2nd-order transfer function that yields Rise time 0.561 s & Overshoot 8.83%

G = tf(wn^2, [1 2*zeta*wn wn^2])

G = 25.2 -------------------- s^2 + 6.137 s + 25.2 Continuous-time transfer function.

S = stepinfo(G)

S = struct with fields:

RiseTime: 0.3752 TransientTime: 1.1880 SettlingTime: 1.1880 SettlingMin: 0.9087 SettlingMax: 1.0883 Overshoot: 8.8300 Undershoot: 0 Peak: 1.0883 PeakTime: 0.7954

step(G), grid on

xline(tr, '--', sprintf('Time from 0 to 100%%: %.3f sec', tr), 'color', '#7F7F7F', 'LabelVerticalAlignment', 'bottom')

Upon analyzing the issue, it is evident that the OP aims to slow down the fast-responsive plant and reduce the overshoot, hence the inclusion of a negative sign in the proportional gain in the auto PID tuning. However, this poses an issue in the pole placement design when incorporating the Prefilter.

Gp  = tf(20, [1 4.5 64])
Gp =
 
         20
  ----------------
  s^2 + 4.5 s + 64
 
Continuous-time transfer function.
stepinfo(Gp)
ans = struct with fields:
         RiseTime: 0.1627
    TransientTime: 1.7235
     SettlingTime: 1.7235
      SettlingMin: 0.2629
      SettlingMax: 0.4369
        Overshoot: 39.8224
       Undershoot: 0
             Peak: 0.4369
         PeakTime: 0.4093
[numP, denP] = tfdata(Gp, 'v');
a1  = denP(2);
a2  = denP(3);
b   = numP(3);
%% Target Compensated System that yields Rise time 0.561 s (10-90%) & Overshoot 8.83%
G   = tf(64.9193300346851, [1, 7.93016824066555, 31.8718356679018, 64.9193300346851])
G =
 
               64.92
  --------------------------------
  s^3 + 7.93 s^2 + 31.87 s + 64.92
 
Continuous-time transfer function.
[numG, denG] = tfdata(G, 'v');
d1  = denG(2);
d2  = denG(3);
d3  = denG(4);
%% Classical PID using pole placement (but it doesn't affect zero)
kp  = -(a2 - d2)/b;
ki  = d3/b;
kd  = -(a1 - d1)/b;
Gc  = pid(kp, ki, kd)
Gc =
 
             1          
  Kp + Ki * --- + Kd * s
             s          

  with Kp = -1.61, Ki = 3.25, Kd = 0.172
 
Continuous-time PID controller in parallel form.
%% Compensated system (with 2 zeros on the right-half plane)
Gcl = feedback(Gc*Gp, 1)
Gcl =
 
     3.43 s^2 - 32.13 s + 64.92
  --------------------------------
  s^3 + 7.93 s^2 + 31.87 s + 64.92
 
Continuous-time transfer function.

It is commonly taught that a Prefilter (Gf) can eliminate undesired zeros in the closed-loop system (Gcl). Nevertheless, some less-experienced students may unintentionally introduce unstable poles while designing the Prefilter. Mathematically, it is observed that Gcl's numerator is miraculously nullified by Gf's denominator, resulting in the Target Compensated System.

How would you design the controller instead?

%% Pre-filter

Gf = tf(b*ki, [b*kd b*kp b*ki])

Gf = 64.92 -------------------------- 3.43 s^2 - 32.13 s + 64.92 Continuous-time transfer function.

Gclf= minreal(Gf*Gcl)

Gclf = 64.92 -------------------------------- s^3 + 7.93 s^2 + 31.87 s + 64.92 Continuous-time transfer function.

stepinfo(Gclf)

ans = struct with fields:

RiseTime: 0.5610 TransientTime: 1.6437 SettlingTime: 1.6437 SettlingMin: 0.9152 SettlingMax: 1.0883 Overshoot: 8.8333 Undershoot: 0 Peak: 1.0883 PeakTime: 1.2146

%% Wishful step response

step(Gclf), grid on

Sam Chak 2024-1-13

在 MATLAB Online 中打开

@Hansi K, the systune() command is employed to identify the tunable parameters in the PID controller, optimizing them to best meet the requirements of the target step response that yields Rise time 0.561 and Overshoot 8.83%.

%% Original Plant

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

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

Gc0 = tunablePID('Gc', 'PID');

%% Initial Closed-loop system (untuned)

CL0 = feedback(Gc0*Gp, 1);

CL0.InputName = 'r';

CL0.OutputName = 'y';

%% Reference system for target step response (Rise time 0.561 & Overshoot 8.83%)

Gref= tf(64.9193300346851, [1, 7.93016824066555, 31.8718356679018, 64.9193300346851]);

%% Tuning goal

Req = TuningGoal.StepTracking('r', 'y', Gref);

stepinfo(Req.ReferenceModel)

ans = struct with fields:

RiseTime: 0.5610 TransientTime: 1.6437 SettlingTime: 1.6437 SettlingMin: 0.9152 SettlingMax: 1.0883 Overshoot: 8.8333 Undershoot: 0 Peak: 1.0883 PeakTime: 1.2146

%% Tune CL0 according to tuning goals specified in Req

CL = systune(CL0, Req);

Final: Soft = 2.82, Hard = -Inf, Iterations = 48

%% Display the values of all tunable Control parameters

showTunable(CL)

Gc = 1 s Kp + Ki * --- + Kd * -------- s Tf*s+1 with Kp = -0.437, Ki = 5.23, Kd = -27.2, Tf = 2.67e+04 Name: Gc Continuous-time PIDF controller in parallel form.

%% Show performances of tuned Closed-loop system

stepinfo(CL)

ans = struct with fields:

RiseTime: 0.4516 TransientTime: 2.2666 SettlingTime: 2.2692 SettlingMin: 0.8034 SettlingMax: 1.0123 Overshoot: 1.2300 Undershoot: 2.4392 Peak: 1.0123 PeakTime: 1.6758

stepplot(Gref), hold on

stepplot(CL), grid on

stepplot(Gp), hold off

legend('Target System', 'Compensated', 'Original Plant', 'location', 'east')

Sam Chak 2024-1-15

在 MATLAB Online 中打开

Hi @Hansi K

I'm not entirely certain about deriving the formula for

when the rise time isn't measured from 0 to 100% of its final value. However, @Paul has found the inequality

quite useful. If the Rise time <

requirement is loosely defined, this formula provides

for the exact

measured from 0 to 100%:

.

In simpler terms, it ensures that

will yield the Rise time <

when measured from 10% to 90% or 5% to 95%. The Wendepunkt-based rise time seems to approximate the rise time measured from 10% to 90%.

The control problem presented by the OP is somewhat unique. In my opinion, the systune() tool proves to be more versatile than the pidtune() tool, especially when the Control Designer can effectively express the Reference system for a desired step response.

The pidtune() is better suited for cases, where the 0 dB gain crossover frequency (

) and target phase margin (pm) are precisely specified. Nonetheless, a challenge arises as not many users are familiar with the process of converting desired rise time (

) or settling time (

) into

and target phase margin. Maybe there are, but I don't remember seeing related examples in the documentation.

Here is an example illustrating how the powerful systune() tool can design the PIDF controller to compensate for a slower plant, yielding an output that precisely mirrors the Reference system for a desired step response.

%% Original Plant (2nd-order system with ts ≈ 8 and os ≈ 16%)

Gp = tf(1, [1 1 1])

Gp = 1 ----------- s^2 + s + 1 Continuous-time transfer function.

stepinfo(Gp)

ans = struct with fields:

RiseTime: 1.6390 TransientTime: 8.0759 SettlingTime: 8.0759 SettlingMin: 0.9315 SettlingMax: 1.1629 Overshoot: 16.2929 Undershoot: 0 Peak: 1.1629 PeakTime: 3.5920

%% Set tunable parameters in the Compensator Gc

Gc0 = tunablePID('Gc', 'PID');

%% Initial Closed-loop system (untuned)

CL0 = feedback(Gc0*Gp, 1);

CL0.InputName = 'r';

CL0.OutputName = 'y';

%% Reference system for target step response (ts < 1 & No Overshoot)

wn = 6; % this value gives Settling time under 1 sec

zeta = 1; % damping ratio 1 gives zero overshoot

Gref = tf(wn^2, [1, 2*zeta*wn, wn^2]) % I want the output to look like this!

Gref = 36 --------------- s^2 + 12 s + 36 Continuous-time transfer function.

%% Tuning goal

Req = TuningGoal.StepTracking('r', 'y', Gref);

stepinfo(Req.ReferenceModel)

ans = struct with fields:

RiseTime: 0.5597 TransientTime: 0.9724 SettlingTime: 0.9724 SettlingMin: 0.9019 SettlingMax: 0.9999 Overshoot: 0 Undershoot: 0 Peak: 0.9999 PeakTime: 1.9879

%% Tune CL0 according to tuning goals specified in Req

CL = systune(CL0, Req);

Final: Soft = 7.96e-13, Hard = -Inf, Iterations = 96

CLsys = getIOTransfer(CL, 'r', 'y');

Gcl = tf(CLsys)

Gcl = From input "r" to output "y": 36 s^2 + 36 s + 36 --------------------------------- s^4 + 13 s^3 + 49 s^2 + 48 s + 36 Continuous-time transfer function.

%% Display the values of all tunable Control parameters

showTunable(CL)

Gc = 1 s Kp + Ki * --- + Kd * -------- s Tf*s+1 with Kp = 2.75, Ki = 3, Kd = 2.77, Tf = 0.0833 Name: Gc Continuous-time PIDF controller in parallel form.

%% Show performances of tuned Closed-loop system

stepinfo(CL)

ans = struct with fields:

RiseTime: 0.5597 TransientTime: 0.9724 SettlingTime: 0.9724 SettlingMin: 0.9019 SettlingMax: 0.9999 Overshoot: 0 Undershoot: 0 Peak: 0.9999 PeakTime: 1.9879

%% Plot result and comparison

subplot(211)

stepplot(Gref, 12), grid on

ylim([0 1.2])

legend('Target System', 'location', 'East')

subplot(212)

stepplot(CL), hold on, grid on

ylim([0 1.2])

stepplot(Gp), hold off

legend('PID Controller', 'Original Plant', 'location', 'East')

Paul 2024-1-15

在 MATLAB Online 中打开

IIRC, the approxiation tr ~ 1.8/wn applies for "intermediate" values of zeta. In this problem, the specified overshoot was 8%, which implies a zeta of around 0.62 (which sounds like it's intermediate), which happens to correspond very closes to wn = 1.8/tr.

Here's the plot (assuming wn = 5, the plot should be agnostic to the value of wn)

zeta = 0.1:.1:0.9; wn = 5;

for ii = 1:numel(zeta)

S = stepinfo(tf(wn^2,[1 2*zeta(ii)*wn wn^2]));

tr(ii) = S.RiseTime;

Mp(ii) = S.Peak - 1;

end

figure

plot(zeta,tr*wn,zeta,Mp)

xlabel('zeta')

yline([0.08 1.8])

xline(0.62)

legend('10-90 Rise Time*wn','%OverShoot/100','Location','NorthWest')

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