PI controller does not have the effect I would expect from the calculations
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I am designing a PI controller for the following plant
K = 11041666;
B = 100;
M = 1019;
num = [K];
den = [M B K];
sys = tf(num,den);
I need to design for 10% overshoot and a settling time of 5 seconds
The closed loop transfer function is as follows:
syms k m b s kp ki
G = k/(m*s^2+b*s+k);
Gc = (kp*s+ki)/s;
T = (kp*s*k+ki*k)/(m*s^3+b*s^2+k*s*(1+kp)+ki*k)
These are the calculations for Kp and Ki
overshoot = 10/100;
Zeta = (-log(overshoot))/sqrt(pi^2+(log(overshoot)^2));
Wn = 5/(4*Zeta);
Ki = (Wn^3*M)/K;
Kp = ((2.15*Wn^2*M)/K)-1;
However with these values this is the response I get
The block diagram
The controller internals
The step input
Yellow if the target and blue is the system responseI am making a mistake somewhere but I have been unable to find it. Any assistance would be greatly appreciated
2 Comments
Any reason why you are not using a PID? Your system is lightly damped. I doubt you can achieve both stability and performance with a PI controller. You would need a damping term in your controller.
K = 11041666;
B = 100;
M = 1019;
num = [K];
den = [M B K];
sys = tf(num,den);
damp(sys)
The formulas for the overshoot etc are valid if the system is stable already. They do not tell if a system is stable or not. Also I am not sure about the formulas you use to get Kp and Ki from Wn,M and K values.
Wynand
on 16 Oct 2024
Accepted Answer
Hi @Wynand
Please see my explanations below.
%% Plant
K = 11041666;
B = 100;
M = 1019;
num = [K];
den = [M B K];
Plant = tf(num, den)
Step Response of the Uncontrolled Plant:
The response is highly oscillatory, and the settling time takes roughly 100 seconds (which is way too long). No real elevator behaves in this manner. This suggests that the modeling is incorrect, an issue that should be addressed by the person who posted the unverified solution in the original thread or by posing a new question to modeling experts.
figure(1)
step(Plant), grid on, grid minor
Controller Design:
The damping ratio has been calculated correctly; however, the natural frequency of the desired closed-loop system has not been. It seems that you may have attempted to use an approximation formula but forgot to verify its correct application. The precise value (not an approximation) of ω is 1.18516590360739 rad/s. Additionally, a 1st-order PI controller is insufficient to provide adequate compensation for a generic 2nd-order plant. Therefore, a 2nd-order PID controller should be employed.
%% Kontroler (Polish)
overshoot = 10/100;
Zeta = (-log(overshoot))/sqrt(pi^2+(log(overshoot)^2))
Wn = 3.9/(5*Zeta)
Wn = 5/(4*Zeta) % super precise value is 1.18516590360739 rad/s
Ki = (Wn^3*M)/K
Kp = ((2.15*Wn^2*M)/K)-1
Kontroler = pid(Kp, Ki)
Analysis of the Closed-Loop System:
After designing the controller, the closed-loop poles should be checked to ensure that all real parts are negative. Clearly, at least one pole has a positive real part, as indicated by the response, which explodes, as shown in the figure below.
%% Closed-loop system
ClosedLoop = minreal(feedback(Kontroler*Plant, 1))
pole(ClosedLoop) % check for positive real part
figure(2)
step(ClosedLoop), grid on, grid minor
Suggested Control Solution:
If you wish to continue pursuing a control solution for this questionable elevator model, consider using the following PID control configuration. No overshoot is always preferable for the passengers' experience.
%% PID controller
Ts = 5; % Desired Settling Time
[Gc, Gh] = chakpid(Plant, Ts) % Gc in forward path, Gh in feedback path
%% Closed-loop System
Gcl = minreal(feedback(Gc*Plant, Gh))
S = stepinfo(Gcl) % Performances
figure(3)
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
4 Comments
Thank you for your time.
I have followed another approach to the design but the respone time is way too fast.
From first principles the values for the parameters were found to be:
I determined the closed loop transfer function to be:
syms Kp Ki Kd s m k b Wn
T = (Kp +Ki/s+s*Kd)/(m*s^2+b*s+k+Kp+Ki/s+s*Kd)
Then using the ideal 3rd order transfer function equation I calculated the PID parameters:
ITF = s^3+1.75*Wn*s^2+2.15*Wn^2*s+Wn^3;
K = 12421874;
M = 1019;
B = 100;
Wn = sqrt(K/M)
Kd = ((1.75)*(Wn*M)-B)
Ki = (M*(Wn^3))
Kp = ((2.15*(Wn^2)*M)-K)
Implementing that into my controller results in the following graph
By manually tuning a gain (0.01) placed before the controller, I can get to the results I want, but I prefer to not have to tune it by hand at all. I also see small oscillations that are still present in the position response that I would like to remove completely. Any tips?
Hi @Wynand
Your design methodology is somewhat correct from a stabilization perspective, as you have only placed the closed-loop poles according to the Hurwitz polynomial of your ideal transfer function. Therefore, stabilization is guaranteed. However, the transient response is not guaranteed due to the non-ideal nature of the numerator in the actual closed-loop transfer function.
If you compare the characteristic polynomial of the actual closed-loop transfer function with that of the ideal transfer function, it appears that your computed PID gain has not successfully placed the closed-loop poles at the desired locations, possibly due to incorrect formulas.
Nonetheless, I cannot replicate your result with a gain factor of 0.01. A gain factor of 
appears to provide a satisfactory response (under 5 seconds), albeit rather chattering.
appears to provide a satisfactory response (under 5 seconds), albeit rather chattering.It is important to understand that Control Theory indicates that a high value of ω leads to a very short settling time, and vice versa, provided that the system is stable.
K = 12421874;
M = 1019;
B = 100;
num = [K];
den = [M B K];
Gp = tf(num, den);
Wn = sqrt(K/M) % No control theory books will tell you to use this formula
%% Ideal Transfer Function
s = tf('s');
ITF = Wn^3/(s^3 + 1.75*Wn*s^2 + 2.15*Wn^2*s + Wn^3)
pole(ITF) % locations of desired closed-loop poles
figure(1)
step(ITF), grid on, grid minor
Kd = ((1.75)*(Wn*M)-B);
Ki = (M*(Wn^3));
Kp = ((2.15*(Wn^2)*M)-K);
Gc = pid(Kp, Ki, Kd)
%% Closed-loop System (without scaling factor)
Gcl = minreal(feedback(Gc*Gp, 1))
pole(Gcl) % locations of actual closed-loop poles
%% Closed-loop System (with scaling factor)
Gcl = minreal(feedback(1e-9*Gc*Gp, 1))
S = stepinfo(Gcl)
figure(2)
step(Gcl), grid on, grid minor
Wynand
on 16 Oct 2024
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