RF-Toolbox: timeresp with time delay although delay is zero

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Lukas Preiss
Lukas Preiss 2019-2-26
回答： Lukas Preiss 2019-5-7
I am using the function timeresp of the RF-Toolbox in a loop with 500 iterations. I observed that the signal is delayed every loop by one index which is equal to a delay of 1 ts. I checked the input and the delay of the rationalfit transfer function is 0. This is because although delay is set to 0, y(1) is set to 0:
y(1:delaynum) = 0;
Input Output 1. loop Output 2. loop
9999.99595344380 0 0
9999.99595344820 299.634476573046 0
9999.99595345260 478.100830944963 8.97808558815310
To avoid this, I changed the code and removed this section.
But the signal is still delayed, because D is 0 in my case and X(1) is 0 by definition, the first time step of the response y(1) will always be 0. So I got the same output as before. Then I changed the code again, so that I shifted the response signal y_new(1)=y_old(2) … y_new(end-1)=y_old(end) and y_new(end)=y_old(end). I still encounter the problem of a delayed signal.
y(1:end-1,1)=yold(2:end,1);
y(end,1)=yold(end,1);
Input Output 1. loop Output 2. loop Output 10. loop
9999.99595319807 299.634476565683 8.97808558793248 1.94682740271114e-10
9999.99595320246 478.100830933214 19.6730244510586 1.23828384323514e-09
9999.99595320678 856.857068263674 45.5557442730978 5.93943329969475e-09
I do not expect a delay caused by my used rationalfit transfer function. I would only expect a lower rise time of my signal after 500 iteration. The Output of the 10. loop is greater than 1 after 62 time steps and greater than 1000 after 195 time steps with an input of 10 000. This still looks like a delay. This does not seem reasonable to me.
How can I overcome this problem that my undelayed signal is delayed?
Thank you in advance for your help.
Kind regards
Lukas
%%https://de.mathworks.com/help/rf/ug/timeresp.html
%%[...]
%%Check the property TF
checkproperty(h);
%%Get the property
poles = h.A;
c = h.C;
d = h.D;
delay = h.Delay;
%%Calculate output signal
f = exp(poles*ts);
g = (f - 1)./poles;
x=zeros(length(poles), 1);
n = length(u);
y=zeros(n, 1);
yold=zeros(n, 1); %New
for k=1:n
yold(k) = sum(c.*x) + d*u(k); %d*u(k) will always be 0
x = f.*x + g*u(k);
%X(1)=0 -> X(2)=g*u(k) -> X(3)=f*g*u(k)+g*u(k)
end
y(1:end-1,1)=yold(2:end,1); %New
y(end,1)=yold(end,1); %New
delaynum = 0 ; % round(delay/ts); %Delay is set to 0!
%y(delaynum+1:delaynum+n) = y;
%y(1:delaynum) = 0;
%y = real(y(:));
t = double(0:(delaynum+n-1))'*ts;
end

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采纳的回答

Lukas Preiss
Lukas Preiss 2019-5-7
Increasing the temporal resolution does not effect the solution, because the artificial delay is caused by the non linear group delay of the transfer function. What I observed was the self cumulative effect of the dispersion. A work around is to add a threshold. I chose 1/1000 of the input and then the results are fine.

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