How can I fit a second Fourier component to a polar histogram?
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I would like to fit a second Fourier series function: E(x) = (1/2*pi)*(1+A1*cos(2*xdata-A2))
to
theta = atan2(rand(100000,1)-0.5,2*(rand(100000,1)-0.5)); polarhistogram(theta,25);
Regards, Eric
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回答 (1 件)
David Goodmanson
2017 年 5 月 24 日
Hi Eric, see how this works. I added an adjustable tilt angle to the random data to test the fit. The code compares the fit to the unnormalized histogram, with its total of 1e5 points. To go to the normalized expression you have, then A1 = B1/c(n0) and A2 = -B2.
npts = 1e5;
n = 25; % should be odd
tilt = pi/4;
theta = atan2(rand(npts,1)-0.5,2*(rand(npts,1)-0.5)) + tilt;
theta = mod(theta+pi,2*pi)-pi;
h = polarhistogram(theta,n);
% start fit
val = h.Values;
c = fftshift(fft(ifftshift(val)))/n; % fourier coefficients
% n0 is index for constant term. c(n0) = npts/n = average bin value
n0 = (n+1)/2;
B1 = 2*abs(c(n0+2));
B2 = angle(c(n0+2));
theta1 = (h.BinEdges(1:end-1) + h.BinEdges(2:end))/2; % bin centers
E = c(n0) + B1.*cos(2*theta1 + B2);
hold on
polarplot(theta1,E,'-o')
hold off
3 件のコメント
David Goodmanson
2017 年 5 月 26 日
Hi Eric, what version of matlab are you using? This seems a bit odd since your original question contains that function.
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