我在MATLAB中编写PID:
classdef PID < handle
properties
Kp = 0
Ki = 0
Kd = 0
SetPoint = 1
Dt = 0.01
end
properties (Access = private)
IState = 0
PreErr = 0
end
methods
function obj = PID(Kp, Ki, Kd, SetPoint, Dt)
if nargin == 0
return;
end
obj.Kp = Kp;
obj.Ki = Ki;
obj.Kd = Kd;
obj.SetPoint = SetPoint;
obj.Dt = Dt;
end
function output = update(obj, measuredValue, t)
err = obj.SetPoint - measuredValue;
P = obj.getP(err);
I = obj.getI(err);
val = lowPass(obj,t);
D = obj.getD(err*val);
output = P + I + D;
end
function val = getP(obj, err)
val = obj.Kp*err;
end
function val = getI(obj, err)
obj.IState = obj.IState + err * obj.Dt;
val = obj.Ki * obj.IState;
end
function val = getD(obj, err)
val = obj.Kd * (err - obj.PreErr) / obj.Dt;
obj.PreErr = err;
end
function val = lowPass(obj,t)
N = 10;
val = 1-exp(-N*t);
end
end
end
并使用随机低通滤波器作为工厂进行测试:
function r = getResponse(t)
r = 1 - exp(-5*t);
end
测试代码:
sr = 1e2; % sampling rate 100Hz
st = 10; % sampling time 10s
ss = st*sr+1; % sample size
t = 0:1/sr:st; % time
input = ones(1,ss)*100;
output = zeros(1,ss);
measured = 0;
pid = PID(0,1,1,input(1),t(2)-t(1));
for i = 2:ss
rPID(i) = pid.update(measured, t(i));
output(i) = rPID(i)*getResponse(t(i));
measured = output(i);
end
figure
plot(t,output)
hold on;
plot(t,input)
plot(t,rPID)
legend('Output','Input','PID')
请注意,参数设置为kp=0;ki=1;kd=1;
。我只在这里测试差分部分。结果非常错误:
请注意,Y轴按10 ^ 307缩放。它变得太大,在约1.6s后PID值超出双精度范围,因此曲线停止。
我确保P和I部分都能很好地工作(参见this question I asked a while ago)。
从D分量的曲线(见下图),人们可以清楚地看到它从一开始就开始变重;在第5个时间戳为0.04秒后,其值达到> 50k:
我几乎可以肯定我在实现低通滤波器时犯了一个错误,但我也注意到即使删除了低通滤波器,差值仍然表现相似。
为了进行某种参考和比较,我还使用完全相同的PID增益(即kp=0;ki=1;kd=1;
)对同一系统进行了Simulink仿真。下面是方框图(左),输入和输出图(右上图)和PID值图(右下)
请注意,增益块中没有上限/下限,初始输入/输出设置为零。
这些PID增益远未达到优化,但它们在模拟和编码PID中给出了完全不同的结果。
因此,最重要的问题是我在这里做错了吗?为什么这两个结果有区别?
低通滤波器的实现不正确。低通滤波器的差分方程如下所示:
getResponse函数的调用可能如下所示:
pid = PID(0,1,1,input(1),t(2)-t(1));
for i = 2:ss
rPID(i) = pid.update(measured, t(i));
alpha = getResponse(0.25,0.01);
output(i) = rPID(i)*alpha+(1-alpha)*output(i-1);
measured = output(i);
end
因此getResponse等同于alpha
function r = getResponse(wc,Ts)
r = 1 - exp(-wc*Ts);
end
此外,您必须修改PID类中的lowPass函数。
function output = update(obj, measuredValue)
err = obj.SetPoint - measuredValue;
P = obj.getP(err);
I = obj.getI(err);
val = lowPass(obj,err,0.1,0.01);
D = obj.getD(val);
output = P + I + D;
end
% ...
function val = lowPass(obj,err,wc,Ts)
alpha = getResponse(wc,Ts);
output = err*alpha+(1-alpha)*obj.output_1;
obj.output_1 = output;
val = output;
end