How can I fit an double exponential curve?

Hi! I have a problem fitting a curve. I have two curves, one voltage and the other current, and I need the Z=V/I impedance. What I need to do is: Find the fit for the voltage curve. And with the new fit curve I found new impedance.

I made a code that I got a good fit of the curve, however, when plotting the new impedance it doesn't come out anything similar to the "original" (curve without adjustments).

can anybody help me?

Here is my code and data.

load('500_CONSTANTE_2u_V.MAT');
load('500_CONSTANTE_2u_C.MAT');
load('t.MAt');
t=t(2:200000);
%Impedancia
Z_Trad_Primeira_500=tensao_at_500./corrente_at_500;
figure(1)
plot(t*10^3,Z_Trad_Primeira_500,'r.');hold on;
%FIT
fcn1 = @(b,t) b(1).*exp(b(2).*t) + b(3).*exp(b(4).*t);
[f1, fval] = fminsearch (@ (b) norm (tensao_at_500 - fcn1 (b, t)), [1; -1; -1; -1]);
figure(2)
plot(t, tensao_at_500, 'p')
hold on
plot(t, fcn1(f1,t), '-')
hold off
grid
%new Z
Znew=fcn1(f1,t)./corrente_at_500;
figure(5)
plot(t*10^3,Z_Trad_Primeira_500,'r.');hold on;
plot(t*10^3,Znew ,'g--');hold on;

[Sargondjani]

There could be two issues:

1. there is a problem with your code;
2. there is problem with your equations.

For an outsider it is hard to guess which of the two it is. However, since you minimize least squares you might get stuck in a local minimum. Did you try different starting value?

[Walter Roberson]

Fitting sum of exponential is always difficult. A lot of the time almost all of the value will go into one of the components and the coefficients of the other will go to near 0 to make the second contribution close to a linear offset.

[Star Strider]

Post the equation for the network (and if possible, code for it), noting if it is a oneport or a twoport, and if a twoport, the input and output nodes.

.

[Amanda Botelho Amaral]

Thank you so much for replying! I did the test all points, I couldn't find the error in the code. I actually modified the equation, got a better approximation.

fcn1 = @(b,t) b(1).*exp(b(2).*t) + b(3).*exp(b(4).*t);

Let me explain the reason for this analysis. As I mentioned, I want to analyze the impedance value on my system. I have a default system, which I add some variables to for analysis, so for every change I make I will have a new curve.

My study is now divided into two stages: 1) I have to check the differences between the curves and to check the influence of each parameter. (I used dtw in this case, if you know of another analysis I can do I appreciate it). 2) Try to get the standard system curve from the other curves. Which is this analysis I did in the code above. From the parameters of the equations I can check what the change in each one of them according to the variation of parameters.

Does it make sense what I'm thinking?

[Star Strider]

‘Does it make sense what I'm thinking?’

With all due respect, no it does not.

In order to fit the data to the circuit to be analysed in the time-domain correctly, the time-domain representation of the circuit must be used as the model. This is relatively straightforward with the Symbolic Math Toolbox — model it in the Laplace ‘s’ domain, simplify it, then invert it, simplify the inverted expression, the convert that to an anonymous function with matlabFunction and use that as the objective function to fit the data.

Then, the parameters will be estimated correctly, and the result will be meaningful.

.

[Amanda Botelho Amaral]

Thank you very much

I'm totally lost then!

What I was trying to do is more or less what they did in this article.

http://iclp.epfl.ch/rms/modules/request.php?module=oc_program&action=view.php&id=252&file=data/252.pdf

[Star Strider]

My pleasure!

How enlightening! (Pun intended.)

It seems that the objective is to identify a relatively simple electrical network. I would use the System Identification Toolbox (if available) to identify the system. (It will be necessary to experiment with the number of poles and zeros, however in all likelihood both should be not greater than 2.) After it is identified, use the compare function to see how appropriate the identification is, then use the impulse function to simulate it.

.

[Amanda Botelho Amaral]

I'll try to analyze the signal at frequency then. I have never used System Identification Toolbox again, thanks for the tip! You said I shouldn't have poles and zeros greater than 2? why?

[Star Strider]

‘You said I shouldn't have poles and zeros greater than 2? why?’

Because it is likely a simple exponential.

Consider —

syms s t 
F(s) = 1/(s + 1)

F(s) = 

figure
fplot(F,[0 10])
title('Transfer Function')
xlabel('j\omega')
ylabel('H(s+j\omega)')

f(t) = ilaplace(F,s,t)

f(t) = 

figure
fplot(f,[0 10])
xlabel('t')
ylabel('h(t)')

This (considerably simplified) transfer function has one pole at s=0 and a zero at ∞. The identified system will determine if there are other poles and zeros, in which case it will be necessary to invert the Laplace transform of the identified s-domain result in order to determine what they are. Use the Symbolic Math Toolbox to invert them.

.

[Amanda Botelho Amaral]

wow

I'm getting to know a toolbox that I didn't even know existed.

I'm new to matlab so everything for me is new! but I believe that now I can do the analysis

Thank you very much, I will try to do as suggested!

Thank you so much! Your tips were very valuable

[Star Strider]

‘I'm getting to know a toolbox that I didn't even know existed.’

For a good time, look at the left CONTENTS bar in the Help Center documentation!

My pleasure!

If you require any further help with the System Identification Toolbox, I will post my replies as an Answer. I have a reasonably amount of experience with it.

.

[Amanda Botelho Amaral]

Thank you so much once again!

Certainly many appeared! I'll try it myself, if I don't work I'll post it here again!

And thanks for understanding the English errors! Another thing I'm learning too :)

[Star Strider]

My pleasure!

I will be here!

No worries!

[Amanda Botelho Amaral]

hi Star Strider, how are you?

As mentioned I want the impedance value of my system to compare the influence of some parameters on my circuit. So with the circuit polynomial I can analyze the terms of the equation that have changed with my parameters.

I have as output, for example, these two curves. The yellow one is my original system and the purple one with parameter change. I would like to evaluate the influence of my parameter on my system and if it is possible, knowing only the purple curve, return to the yellow curve.

That's why I thought about analyzing it by polynomials, but I'm not getting good results...

Once again...

Does my logic make sense?

[Mathieu NOE]

hello Amanda

welcome back

as far as I understandthe publication you are refering to, the attempt is to do a simple two parameters exponential fit once for the voltage signal, one for the current signal

this is done here :

load('tensao_at_500.MAT');
load('corrente_at_500.MAT');
load('t.MAt');

figure(1)
plot(t,tensao_at_500,'b',t,corrente_at_500,'r');
title(' Voltage and current');
xlabel('time (s)');
ylabel('Amplitude');

% FIT voltage signal
% use only decaying portion of signal for the fit
[val,indm] = max(tensao_at_500);
tensao_fit = tensao_at_500(indm:end);
t_fit = t(indm:end);

fcn1 = @(b,t) b(1).*exp(b(2).*t);
[f1, fval] = fminsearch (@ (b) norm (tensao_fit - fcn1 (b, t_fit)), [val; -1e3;]);
figure(2)
plot(t, tensao_at_500, '*',t_fit, fcn1(f1,t_fit), '-')
title(' Voltage');
legend('voltage','exp fit');
xlabel('time (s)');
ylabel('Amplitude');
grid on

% FIT current signal
% use only decaying portion of signal for the fit
[val,indm] = max(corrente_at_500);
corrente_fit = corrente_at_500(indm:end);
t_fit = t(indm:end);

fcn2 = @(b,t) b(1).*exp(b(2).*t);
[f2, fval] = fminsearch (@ (b) norm (corrente_fit - fcn2 (b, t_fit)), [val; -1e3;]);
figure(3)
plot(t, corrente_at_500, '*',t_fit, fcn2(f2,t_fit), '-')
title(' Current');
legend('current','exp fit');
xlabel('time (s)');
ylabel('Amplitude');
grid on

[Amanda Botelho Amaral]

"use only the decaying portion of the signal for adjustment"

That's just what I don't think! Thank you very much!