Zero values of a fitted curve
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Hi to everybody, i have a time history dataset that i have fitted with a fourier series to find its peaks. Now i am in need of the zero crossing values of the curve, and as output the corresponding values of time of these zero crossing. Is there a function that does it automatically? I have tried with fnzeros but seems not to work for fitted curves but only for functions, am i wrong? Maybe i'm missing some statement in the code. Waiting for someone's reply i would thank in advance those willing to help me :D
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Star Strider
2015-9-25
Without actually seeing your data and the analysis you did, I cannot provide a definitive response.
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arich82
2015-9-25
There's a paper by Prof. Boyd from the University of Michigan that appears to address this:
"Computing the zeros, maxima and inflection points of Chebyshev, Legendre and Fourier series: solving transcendental equations by spectral interpolation and polynomial rootfinding" http://link.springer.com/article/10.1007%2Fs10665-006-9087-5#page-1
A summary seems to be given here: http://math.stackexchange.com/questions/370996/roots-of-a-finite-fourier-series
I think this does what you want (not that the roots of the fourier series are given in the variable t):
% fourier coefficients
% (dummy data)
a = [1, 1, 1, 1];
b = [0, 0, 1, 0];
% Note:
% a(1) corresponds to a_0 in the reference
% a(2:end) corresponds to a(1:N) in the reference;
N = numel(a) - 1;
h = [a(end:-1:2) + 1i*b(end:-1:2), 2*a(1), (a(2:end) - 1i*b(2:end))];
B = diag(ones(2*N - 1, 1), 1);
B(end, :) = -h(1:end - 1)/(a(end) - 1i*b(end));
z = eig(B);
t = angle(z);
% note: duplicate t's are imgaginary roots
% eliminate duplicates within tolerance
tol = 10*eps;
t = sort(t);
ind = find(diff(t)./abs(t(2:end)) < tol);
ind = [ind; ind + 1];
t(ind(:)) = [];
% plot results
% f = ugly inline fourier series anonymous function
f = @(x, a, b) sum(bsxfun(@times, a(:).', cos(x(:)*(0:numel(a) - 1))) + bsxfun(@times, b(:).', sin(x(:)*(0:numel(b)-1))), 2);
x = linspace(-pi, pi, 128 + 1);
hf = figure;
ha = axes;
axis([-pi, pi, -sum(a+b)-0.5, sum(a+b)+0.5]);
hold(ha, 'on');
plot(x, f(x, a, b));
plot(t, f(t, a, b), 'o');
plot([-pi, pi], [0, 0], '--c');
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/178506/image.jpeg)
I probably won't be around much in the next two days, but feel free to leave a comment, and I'll try to get back to you.
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更多回答(2 个)
Matt J
2015-9-25
编辑:Matt J
2015-9-25
fnzeros only appears to work with spline fits, which makes sense. That's the only way the zeros will have an analytical closed form solution. However, one option would be to sample your Fourier series fit at a finely spaced selection of points. Then, you can fit a spline to those points and use fnzeros on the spline fit to approximate the zeros of the Fourier fit.
If needed, you can refine further by using fzero() on the Fourier fit to do a numerical search. The result of fnzeros from above will give you a list of the initial seed points or search intervals that you would loop over, applying fzero() to each one.
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