finding parameters by inverse problem in time domain
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mohammad hodaei
el 14 de Mayo de 2020
Comentada: Walter Roberson
el 17 de Mayo de 2020
Dear Walter,
First, Thank you for taking time and answering all of my questions.
I have a function in s-domain, TR. I used Durbin to bring it in time domain.
the function has a few parameters. I decidede to find one or a few parameters using inverse problem.
I have a file of another function which I need to match my function with it, then I can find my parameter.(MINIMIZATION)
My concern is to know how I use inverse problem here?
I haver to use Durbin, because they are in S-domain.
I decided to add a few lines at the end of my durbin regarding inverse problem(minimization) and start my durbin code with a function name [function y=f(x)].
attached, you can see my TR, f (include Durbin+ a few more line for inverse) and a mat file including the data of another function which I need to have minimization with it. Also, there is a mfile with name fint as it is just a function.
when I use optimization, I get "Index in position 1 is invalid. Array indices must be positive integers or logical values."
I went through your old respnse to others and removed that problem. Thank you.
Now, I have another problem and it is killing me.
it says Objective function is undefined at initial point. Fmincon cannot continue."
Please note that I use fmincon
If you know better function please let me know.
please find my files and see inside f.m
Again, thank you for your time and answers.
Best
M.H.
1 comentario
Walter Roberson
el 16 de Mayo de 2020
You did not include your f . I can guess that it invokes TR, but I cannot guess what value of s you might be passing in, or how the .mat is connected to this.
Respuesta aceptada
Más respuestas (3)
Walter Roberson
el 16 de Mayo de 2020
tthrough=8.8e-5:1e-9:tmax;
That becomes a vector of 12001 elements
trange=tthrough(1):(length(tthrough))/100:(length(tthrough));
That uses the 8.8e-5 as the first element, and increments by 12001/100 = 120.001 to 12001, for a total of 100 elements. But why ?? Why would use use the time value of 8.8e-5 as the first value, and otherwise be dealing with the length of the time vector??
length(trange)
for b=1:length(trange)
disp(b)
t=trange(b);
For the first iteration, t will be trange(1) which is the tthrough(1) which is 8.8e-5 .
For the second iteration, t will be 120.001 higher, about 120.010088 .
yt(b)=(2*exp(mu.*t)/T0)*(S+0.5*real(TR(x,mu)));
mu is close to 25000. Multiply that by the 120.010088 to get 3000102.19489026 . exp() of that is about 6.7*10^1302927 which overflows to inf.
??? Why are you building tthrough just to take its (known) first value and its length ???
I suspect you mean something more like
trange = linspace(8.8e-5, tmax, 100);
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mohammad hodaei
el 16 de Mayo de 2020
1 comentario
Walter Roberson
el 17 de Mayo de 2020
because I need to find fig 15 from the attached file
Figure 15 says nothing about incrementing by 120 at a time. The x axis for figure 15 goes up to about 10.2 * 10^(-5) not up to 12001 .
Fellah2006 is experimental point that I got from the figure
Figure 15 has two lines together on it, and you have mixed the data for the two lines together in Fellah2006, resulting in having multiple entries with the same time coordinate (the last 3 times are all the same for example.)
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