How can i plot PI controller step respone

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Lina
Lina el 27 de Ag. de 2024
Editada: Sam Chak el 27 de Ag. de 2024
I used the following steps to plot the step response using (step) function:
-> num2_PI=conv(395.5*[1 1/8],[0 -0.0005854 -0.01121 0.0003176]);
den2_PI=[ 1 1.117 10.18 0.0005226 0]+num2_PI;
TF2_PI=tf(num2_PI,den2_PI);
step(TF2_PI)
the plot was :
So , how can I read the overshoot, final value and other readings from this plot
  2 comentarios
Sam Chak
Sam Chak el 27 de Ag. de 2024
You have successfully plotted the step response of an unknown system that doesn't look like a Pure PI Controller. Unfortunately, the response indicates instability and those characteristics {overshoot, final value, etc.} are undefined for unstable systems.
Mathieu NOE
Mathieu NOE el 27 de Ag. de 2024
hello
fyi, your "plant" is unstable in open loop and cannot be stabilized with this PI correction , neither with negative or positive feedback
% PI
num1 = 395.5*[1 1/8]; % P , I gains
den1 = [1 0];
C = tf(num1,den1);
% Plant
num2=[0 -0.0005854 -0.01121 0.0003176];
den2=[ 1 1.117 10.18 0.0005226 0];
G=tf(num2,den2);
step(G)
sys = feedback(G,C,-1) % standard negative feedback
sys = -0.0005854 s^3 - 0.01121 s^2 + 0.0003176 s ----------------------------------------------------------- s^5 + 1.117 s^4 + 9.948 s^3 - 4.462 s^2 - 0.4286 s + 0.0157 Continuous-time transfer function.
step(sys)
sys = feedback(G,C,+1)% positive feedback
sys = -0.0005854 s^3 - 0.01121 s^2 + 0.0003176 s ----------------------------------------------------------- s^5 + 1.117 s^4 + 10.41 s^3 + 4.463 s^2 + 0.4286 s - 0.0157 Continuous-time transfer function.
step(sys)

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Sam Chak
Sam Chak el 27 de Ag. de 2024
Editada: Sam Chak el 27 de Ag. de 2024
The 4th-order plant under consideration appears to be unstable and cannot be effectively stabilized using a low-order, simple 2-parameter PI controller. Additionally, the unstable plant also exhibits positive zeros, which can lead to initial undershoot in the system response.
In an effort to address these challenges, I have attempted to strategically place the poles and zeros of a matching 4th-order controller (on the feedback path) and the pre-filter, with the goal of limiting the undershoot percentage to less than 20% and ensuring the compensated system converges within 5 minutes. The MATLAB stepinfo() command can be utilized to analyze the step-response characteristics of the stabilized system.
%% Plant
Gp = tf([0, -0.0005854, -0.01121, 0.0003176], [1, 1.117, 10.18, 0.0005226, 0])
Gp = -0.0005854 s^2 - 0.01121 s + 0.0003176 ----------------------------------------- s^4 + 1.117 s^3 + 10.18 s^2 + 0.0005226 s Continuous-time transfer function.
%% Feedback Compensator
cz = [ 1.8729742918689 % zeros
-5.13296170895781e-05
-6.69459607654879e-09];
cp = [-3.27517464154048 + 0i % poles
2.1086089615566 + 0.506178267136755i
2.1086089615566 - 0.506178267136755i
0.0282900517606163 + 0i];
ck = -7456.33220301306; % gain
Gc = tf(zpk(cz, cp, ck))
Gc = -7456 s^3 + 1.397e04 s^2 + 0.7169 s + 4.799e-09 ----------------------------------------------- s^4 - 0.9703 s^3 - 9.083 s^2 + 15.66 s - 0.4357 Continuous-time transfer function.
%% Pre-filter
fz = []; % zeros
fp = [-19.1775896766893 % poles
-3.27517464154048];
fk = -6.9180823798271e-07; % gain
Gf = tf(zpk(fz, fp, fk))
Gf = -6.918e-07 --------------------- s^2 + 22.45 s + 62.81 Continuous-time transfer function.
%% Filtered Closed-loop
Gcl = feedback(Gp, Gc); % put Gc on the feedback path
Fcl = minreal(series(Gf, Gcl))
Fcl = 4.05e-10 s^4 - 1.731e-09 s^3 + 2.001e-09 s^2 - 1.091e-10 s + 1.524e-12 ----------------------------------------------------------------------------------------------------------------------- s^8 + 0.1467 s^7 + 0.01311 s^6 + 0.0007778 s^5 + 3.284e-05 s^4 + 9.877e-07 s^3 + 2.03e-08 s^2 + 2.561e-10 s + 1.524e-12 Continuous-time transfer function.
%% Step response
step(Fcl, 600), grid on
stepinfo(Fcl)
ans = struct with fields:
RiseTime: 68.8473 TransientTime: 287.5175 SettlingTime: 289.4285 SettlingMin: 0.9002 SettlingMax: 1.0118 Overshoot: 1.1755 Undershoot: 17.9067 Peak: 1.0118 PeakTime: 420.8477

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