fitting with a rectangular pulse function
18 次查看(过去 30 天)
显示 更早的评论
Hi, I would like to fit a signal pulse with a rectangular pulse function. this is my script, a rectangular pulse signal should be fitted with the same function:
%%%%%%%%%%
clear all
xdata=[1:1000];
a=100;
b=30;
c=12;
ydata=squarepulse(xdata,a,b,c);
plot(xdata,ydata)
par0=[ 100
2
7];
lb=[xdata(1)
1
5];
ub=[xdata(end)
50
20];
options = optimoptions('lsqcurvefit','tolx',1e-10,'tolfun',1e-10,'maxfunevals',5000000)
parOut=lsqcurvefit(@squarefun,par0,xdata,ydata,lb,ub,options)
yfit=squarefun(parOut,xdata);
plot(xdata,ydata,xdata,yfit,'r')
%%%%%%%%
and this is the script for build the rectangular pulse signal and used to fit:
%%%%%%%%%%%%%
function y = squarefun(par,xx)
y=xx*0;
%y=par(3)*exp(-(xx-par(1)).^2/par(2)^2);
for i=1:length(xx)
if xx(i)<=par(1)
y(i)=heaviside(xx(i)-par(1)+par(2))*par(3);
else
y(i)=heaviside(-xx(i)+par(1)+par(2))*par(3);
end
end
end
%%%%%%%%%%%%
the problem is that the parameters par(1) and par(2) (x position of the signal and its width) do not change, only the par(3) is correctly found. If I substitute the rectangular function with a gaussian the minimisation works good. Thank you regards Andrea
4 个评论
Jim Joy
2017-9-5
Thank you for clarifying.
I have played around with this a little bit, and I see the issue that you're dealing with. It likely has to do with the fact that the gradients of step-functions are either 0 or infinity.
What is the end goal of your calculation? Are you trying to measure pulse height and width?
Best,
Jim
采纳的回答
Jim Joy
2017-9-6
I would recommend using an edge detection method other than least-squares fitting. Since your signal is a perfect step function, you can use the "diff" function to determine where there are jumps in the signal. For example, consider the code below using your definition of 'ydata':
>> yDiff = diff(ydata);
>> idx = find(yDiff)
idx =
69 70 129 130
>> ydata(71)
ans =
12
Here, we can see that the square pulse rises around the index of 70, and falls around 130. Looking at a point interior to the pulse, you can see that the height of the pulse is about 12. Truth be told, this approach uses some a priori knowledge about the signal, and it will become more difficult with real data (where noise will almost certainly be an issue). In these cases, I would recommend either preconditioning the data by taking the convolution with an averaging filter, taking a numerical derivative that involves more points (see http://people.math.sfu.ca/~cbm/aands/page_883.htm for examples), or some combination of both.
The MATLAB function "findchangepts" (https://www.mathworks.com/help/signal/ref/findchangepts.html) is also useful for this type of analysis. For example, the code below shows that the most abrupt changes in the data occur at indices 70 and 131:
>> idx = findchangepts(ydata,'MaxNumChanges',2)
idx =
70 131
In addition, you may find some approaches based on wavelet transforms to be useful. See the link below for some examples:
https://www.mathworks.com/help/wavelet/examples/detecting-discontinuities-and-breakdown-points.html
Best Regards,
Jim
0 个评论
更多回答(0 个)
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!