How Approximate Model FOPDT

73 次查看(过去 30 天)
LORIS IACOBAN
LORIS IACOBAN 2024-3-13
回答: Sam Chak 2024-3-18
Hi, I have a problem.
My G(s)=7/(s^2+7), so is a 2 order undumped. I need to approximate it in a FOPDT , How i can? Thanks to everyone.

回答(2 个)

akshatsood
akshatsood 2024-3-18
编辑:akshatsood 2024-3-18
To approximate a second-order undamped system with a First Order Plus Dead Time (FOPDT) model, you shall aim to find an equivalent system that captures the essential dynamics of the original system with a simpler representation. The FOPDT model is given by: where K is the process gain, is time constant and, θ is dead time.
As stated in the question, your system is repereseted as: . This is a second-order system without damping as does not have a term (where ξ is the damping ratio and is the natural frequency). The natural frequency is in your case.
You can consider the following steps to approximate to FOPDT
Identify Key Characteristics - For the original system, identify characteristics like the time to reach first peak, settling time, or steady-state gain. For a second-order undamped system, the steady-state gain is the coefficient of the transfer function, which is 7 in this case.
Approximate Parameters
Gain (K) : The gain of the FOPDT model should match the steady-state gain of the original system. Thus, .
Time Constant (τ) : The time constant should represent the time it takes for the system to reach a significant portion of its final value. For a second-order undamped system, you might use the time it takes to reach the first peak or some empirical methods for estimation.
Dead Time (θ): This could represent the delay before the system starts responding. For systems without an explicit delay, this might be approximated as a fraction of time constant or based on system's response characteristics.
Approximation Method
A generic approach aimed towards approximating involves comparing step response of original system to that of FOPDT model and adjusting τ and θ to match key features of the response, such as the time to reach first peak or the half-rise time. However, for a purely undamped second-order system, there is no perfect match because the original system oscillates indefinitely while the FOPDT system does not.
Given lack of a standard method for directly converting a second-order undamped system to FOPDT, consider using empirical methods or judgment based on the system's step response, you might consider the following :
  1. Estimate τ based on when the system reaches a significant portion of its steady-state value for the first time.
  2. θ is based on when the system output first starts to rise, but for an undamped second-order system (as in your case), this is effectively zero.
As a footnote, this approach is highly approximate and relies on matching key characteristics rather than a direct mathematical transformation. For precise transformation, consider using simulations to iteratively adjust τ and θ until the FOPDT model's response closely aligns with that of the original system.
I hope this helps.

Sam Chak
Sam Chak 2024-3-18
Considering that the open-loop model lacks the characteristic S-shaped process reaction curve and exhibits oscillations, you can employ whatever suitable manipulation technique to close the loop and transform it into a First-Order Plus Time Delay (FOPDT) model. Once that is achieved, you can utilize the Ziegler-Nichols Tuning method with the help of Dr. Yi Cao's 'ReactionCurve()' code, which can be downloaded from the File Exchange using the following link:
Gp = tf(7, [1 0 7]) % Process model, Gp
Gp = 7 ------- s^2 + 7 Continuous-time transfer function.
Ts = 1.5; % desired Settling Time
kp =-0.25;
ki = 1.94464386024797;
kd = 0.309945826765284;
N = 7.77857544099188; % bandwidth of derivative filter
Gc = pid(kp, ki, kd, 1/N) % PID controller (for manipulation purposes)
Gc = 1 s Kp + Ki * --- + Kd * -------- s Tf*s+1 with Kp = -0.25, Ki = 1.94, Kd = 0.31, Tf = 0.129 Continuous-time PIDF controller in parallel form.
Gcl = feedback(Gc*Gp, 1); % Closed-loop system, Gcl (that looks like FOPDT)
S = stepinfo(Gcl);
figure(1)
step(Gp, 2*Ts), hold on
step(Gcl, 2*Ts), grid on, hold off
xline(S.SettlingTime, '--', sprintf('Settling Time: %.4f sec', S.SettlingTime), 'color', '#265EF5', 'LabelVerticalAlignment', 'bottom')
legend('Open-loop Gp', 'Closed-loop Gcl', '')
figure(2)
[y, t] = step(Gcl, 3);
[model, controller] = ReactionCurve(t, y);
T = feedback(Gcl*controller.PID, 1);
step(T, 3), grid on, hold off
function [model, controller] = ReactionCurve(t, y, u)
...
end

类别

Help CenterFile Exchange 中查找有关 PID Controller Tuning 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by