P only control model----PID controlor

最新推荐文章于 2024-07-29 11:34:56 发布

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本文通过MATLAB代码实例，详细分析了PID控制器中比例系数Kp对输出的影响，包括静差变化、过调现象及稳定性，通过不同Kp值的比较，直观展示了PID控制器参数调整对系统响应的影响。

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如图PID是什么不解释了。

对于仅仅只有P控制的模型

%% 
% Code writer：EOF
% Time ： 2013.09.23 In XTU
% 
% This code just used for understand how PID work .So you may not to be
% intertwind with this code how to work!

% Matlab language is not like the C language which is strict to detail of a
% description of  problems

% What I want to say is that you should understand what is PID and then
% program it
clear all
clc

syms s
hold on
%% abstract the object which is controled into a function 1/(s^2+3*s+1)
num = 1;
den = sym2poly(s^2+3*s+1);
G = tf(num,den);

%% Set up the coefficient
%--------------------------
H = 1;%THE TARGET VALUE!!
%--------------------------
% PID系数
Kp = 2;
Ki = 0;
Kd = 0;
%% Set up the coefficient controlor
C = pid(Kp,Ki,Kd);% set up a PID controlor
T = feedback(C*G,H);%backforward loop

step(T)

Kp = 4;% coefficient of  PID
Ki = 0;
Kd = 0;

C = pid(Kp,Ki,Kd);

T = feedback(C*G,H);

step(T)
%% repeat simply
Kp = 6;
Ki = 0;
Kd = 0;

C = pid(Kp,Ki,Kd);

T = feedback(C*G,H);

step(T)

Kp = 40;
Ki = 0;
Kd = 0;

C = pid(Kp,Ki,Kd);

T = feedback(C*G,H);

step(T)
Kp = 400;
Ki = 0;
Kd = 0;

C = pid(Kp,Ki,Kd);

T = feedback(C*G,H);

step(T)

legend('Kp = 2','Kp = 4','Kp = 6','Kp = 40','Kp=400');

我们假定控制对象抽象成一个函数