PID controller behaves strangely for larger Ts values

Question

Wynand 2024-10-17

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2160480-pid-controller-behaves-strangely-for-larger-ts-values

编辑： Sam Chak 2024-10-17

在 MATLAB Online 中打开

I am designing a PID controller for a rudimentary elevator system

The elevator has to take more than 5s to arrive at the next floor but no more than 6s

The transfer function is as follows:

syms Ki Kd Kp s 
M = 1019;
B = 100;
K = 19875;
num=[1];
den=[M B K];
G = tf(num, den)
G =
 
             1
  ------------------------
  1019 s^2 + 100 s + 19875
 
Continuous-time transfer function.
Zeta = 1;
Ts = 5;
Wn = 0.8;

Using the PID function

GC = (Kp*s + Ki + Kd*s^2)/s

The close loop function is found to be:

T = (Kp*s + Ki + Kd*s^2)/(M*s^3 + B*s^2 + K*s + Kp*s + Ki + Kd*s^2)

Comparing this with the ITAE ideal tranfer function for a third order system yields the following equations

ITAE_tf3 = vpa(Wn^3/(s^3 + (Wn)*s^2 + (Wn^2)*s + Wn^3))
Kp = (2.15)*(M)*(Wn^2)-(K)
Kp = -1.8473e+04
Ki = M*(Wn^3)
Ki = 521.7280
Kd = (1.75)*(M)*(Wn)-B
Kd = 1.3266e+03

However the large Kp value drives the system downwards when it should be going up and the large integrator value also causes some issues

Changing zeta and ts gives me a results that is almost what I need

overshoot = 10/100;
Zeta = (-log(overshoot))/sqrt(pi^2+(log(overshoot)^2));
% Zeta = 1
Ts = 2.4;;
Wn = 4/(Zeta*Ts);

But of course the Ts value is too high.

Any help would be greatly appreciated

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Answer 1

Sam Chak 2024-10-17

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2160480-pid-controller-behaves-strangely-for-larger-ts-values#answer_1533665

编辑：Sam Chak 2024-10-17

在 MATLAB Online 中打开

Hi @Wynand

There are a few things for which you can find detailed explanations in control theory books. In short, the ideal ITAE third-order transfer function is different from your actual PID-controlled third-order transfer function.

Blunder: Two coefficients are missing in your version of the ITAE third-order transfer function.

While your derived PID-controlled third-order transfer function is correct, to make an apples-to-apples comparison, we must divide both the numerator and the denominator by M.

%% Plant TF

M = 1019;

B = 100;

K = 19875;

num = [1];

den = [M B K];

Gp = tf(num, den)

Gp = 1 ------------------------ 1019 s^2 + 100 s + 19875 Continuous-time transfer function.

%% Ideal TF

wn = 3; % Tune this parameter only, DO NOT individually tune P-I-D gains (Always use single parameter tuning!)

Gi = tf(wn^3, [1, 1.781*wn, 2.171*wn^2, wn^3])

Gi = 27 ------------------------------ s^3 + 5.343 s^2 + 19.54 s + 27 Continuous-time transfer function.

figure(1)

step(Gi, 5), grid on

title('Step response of Ideal TF')

%% Pole placement

ki = M*wn^3;

kp = 2.171*M*wn^2 - K;

kd = 1.781*M*wn - B;

Gc = pid(kp, ki, kd)

Gc = 1 Kp + Ki * --- + Kd * s s with Kp = 35.2, Ki = 2.75e+04, Kd = 5.34e+03 Continuous-time PID controller in parallel form.

%% Closed-loop TF

Gcl = minreal(feedback(Gc*Gp, 1))

Gcl = 5.245 s^2 + 0.03458 s + 27 ------------------------------ s^3 + 5.343 s^2 + 19.54 s + 27 Continuous-time transfer function.

zero(Gcl) % to be cancelled out later

ans =

-0.0033 + 2.2689i -0.0033 - 2.2689i

figure(2)

step(Gcl, 5), grid on

title('Step response of Closed-loop TF')

%% Design of Prefilter, Gf

[num, den] = tfdata(Gcl, 'v');

Gf = tf(den(end), num)

Gf = 27 -------------------------- 5.245 s^2 + 0.03458 s + 27 Continuous-time transfer function.

%% Filtered Closed-loop TF

Fcl = minreal(series(Gf, Gcl))

Fcl = 27 ------------------------------ s^3 + 5.343 s^2 + 19.54 s + 27 Continuous-time transfer function.

figure(3)

step(Fcl, 5), grid on

title('Step response of Filtered Closed-loop TF')

My preferred control architecture: Simple yet accurate!

%% My PID controller

Ts = 5.5; % Desired Settling Time (single tuning parameter)

[Gc, Gh] = chakpid(Gp, Ts) % Gc in forward path, Gh in feedback path

Gc = 1 Kp + Ki * --- + Kd * s s with Kp = 1.03e+03, Ki = 367, Kd = 725 Continuous-time PID controller in parallel form. Gh = 2.862 s^2 - 25.29 s + 0.5059 ---------------------------- s^2 + 1.423 s + 0.5059 Continuous-time transfer function.

%% Closed-loop System

Gcl = minreal(feedback(Gc*Gp, Gh))

Gcl = 0.7113 s^4 + 2.024 s^3 + 2.159 s^2 + 1.024 s + 0.1821 --------------------------------------------------------- s^5 + 3.556 s^4 + 5.059 s^3 + 3.598 s^2 + 1.28 s + 0.1821 Continuous-time transfer function.

S = stepinfo(Gcl) % Performances

S = struct with fields:

RiseTime: 3.0891 TransientTime: 5.5000 SettlingTime: 5.5000 SettlingMin: 0.9002 SettlingMax: 0.9995 Overshoot: 0 Undershoot: 0 Peak: 0.9995 PeakTime: 10.7900

figure(4)

step(Gcl, round(3*Ts, 0)), grid on, grid minor

%% Double compensation control scheme

function [C, H] = chakpid(P, Ts)

% Gp is a 2nd-order Plant

% Ts is the desired settling time

[numP, denP] = tfdata(minreal(P), 'v');

a1 = denP(2);

a2 = denP(3);

b = numP(3);

Tc = -log(0.02)/Ts; % Desired time constant

%% The formulas

k1 = (3*(b^2)*((Tc/b)^2) - a2)/b;

k2 = (b^2)*((Tc/b)^3);

k3 = (3*b*(Tc/b) - a1)/b;

k4 = 2*b*((Tc/b)^2); % P-gain of PID controller

k5 = k2; % I-gain of PID controller

k6 = Tc/b; % D-gain of PID controller

C = pid(k4, k5, k6); % Classical PID controller

H = minreal(tf([k3 k1 k2], [k6 k4 k5])); % Compensator at the Sensor path

end

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PID controller behaves strangely for larger Ts values

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PID controller behaves strangely for larger Ts values

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