Differential part of PID and its low pass filter- time domain

Question

I programmed PID in MATLAB:

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

And tested it using a random low pass filter as the plant:

function r = getResponse(t)
r = 1 - exp(-5*t);
end

The test code:

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')

Note that the parameters are set to kp=0;ki=1;kd=1;. I'm only testing the differential part here. The result is very wrong:

Notice the Y-axis is scaled by 10^307. It gets too big that after ~1.6s the PID value exceeds the range of double precision and therefore, the curve stops.

I have ensured that both P and I parts work well enough (see this question I asked a while ago).

From the curve for the D component (see figure below), one can clearly see that it starts to oscillate heavy from the very beginning; its value reaches >50k after the 5th timestamp at 0.04s:

I'm almost certain I must have made a mistake in implementing the low pass filter, but I also noticed that even with the low pass filter removed, the differential values still behave similarly.

---

To have some sort of reference and comparison, I also made a Simulink simulation of the same system, using the exact same PID gains (i.e. kp=0;ki=1;kd=1;). Below is the block diagram (left), figure for input and output (top right figure) and figure for PID values (bottom right)

Note that there is no top/lower limit in the gain blocks and the initial inputs/outputs are set to zeros.

These PID gains are nowhere near optimised but they give completely different results in the simulation and coded PID.

Therefore the big question is am I doing something wrong here? Why is there a difference between the two results?

Answer 1

The implementation of the low pass filter is not correct. The difference equation of a low pass filter is as shown:

The call of the getResponse function could look like this:

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

Thus getResponse is equivalent to alpha

function r = getResponse(wc,Ts)
    r = 1 - exp(-wc*Ts);
end

Further you have to modify the lowPass function in the PID class.

    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