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Re: How to fit a sine wave to under-sampled data
Posted:
Jan 7, 2013 9:20 AM
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"Greg Heath" <heath@alumni.brown.edu> wrote in message <kbiqn8$d78$1@newscl01ah.mathworks.com>...
> "Scott Rauscher" wrote in message <kbg0a3$afm$1@newscl01ah.mathworks.com>...
> > I've poured through the forums several times and have almost found a good solution.
> >
> > I want to fit a sine wave to some under-sampled constant-frequency/phase/amplitude data. I have hundreds of data files so I need to automate it. I have used SineFit in the file exchange, and it works well most of the time, but has a tendency not to converge. I have also tried sinfapm, but the phase it gives never lines up.
> >
> > What I'm trying to do is best illustrated in this picture (red is the data, blue is the sine wave I want to find): http://s2.postimage.org/98qvgbnu1/matcode.jpg
> >
> > Does anyone have suggestions on where to start for a full-proof way of doing this?
>
> Not enough info.
>
> Do you get a dominant frequency with the dft ? If so, you can estimate A and B
> from the linear problem
>
> xi = A*cos(w*ti) + B*sin(w*ti); % w and ti known
>
> Hope this helps.
>
> Greg
Thanks for the response Greg. I'm been feeling under the weather, but let me elaborate with some code:
x = data;
N = length(x);
n = 2:N-1;
xs = x(n-1)+x(n+1);
C = xs'*x(n)/(x(n)'*x(n))/2;
frequency = acos(C)*SF*.5/pi; %Gives a good estimate
% xi = A*cos(w*ti) + B*sin(w*ti); % w and ti known
i1 = 2;
i2 = 15;
x1 = TX_i(i1);
x2 = TX_i(i2);
c1 = cos(2*pi*frequency*t_section(i1));
c2 = cos(2*pi*frequency*t_section(i2));
s1 = sin(2*pi*frequency*t_section(i1));
s2 = sin(2*pi*frequency*t_section(i2));
B = (x2 - (x1*c2/c1))/(s2 - (c2*s1/c1));
A = (x1 - B*s1)/c1;
t = linspace(t_section(i1),t_section(i2),100);
fit = A*cos(2*pi*frequency*t + pi) + B*sin(2*pi*frequency*t + pi);
plot(t_section,TX_i(astart:aend),'r',t,fit,'b')
output: http://s1.postimage.org/6yqy4ikun/untitled.jpg
So that method works well (thank you), but I'm having the same problem as before in that the phase shift is still slightly off. I suppose this is a complication of limited resolution/sampling frequency, but since the shift is visible I should be able to mathematically figure it out, right?
Any suggestions?
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