# Using Fourier Analysis to fit function to data

I have 24 values for $Y$ and corresponding 24 values for $t$. The Y values are measured experimentally, while t has values $t=1,2,\dots 24$.

I want to find the relationship between Y and t as an equation using Fourier analysis.

I wrote the following MATLAB code:

Y=[10.6534
9.6646
8.7137
8.2863
8.2863
8.7137
9.0000
9.5726
11.0000
12.7137
13.4274
13.2863
13.0000
12.7137
12.5726
13.5726
15.7137
17.4274
18.0000
18.0000
17.4274
15.7137
14.0297
12.4345];

ts=1; % step
t=1:ts:24; % the period is 24
f=[-length(t)/2:length(t)/2-1]/(length(t)*ts); % computing frequency interval
M=abs(fftshift(fft(Y)));
figure;plot(f,M,'LineWidth',1.5);
grid % plot of harmonic components
figure; plot(t,Y,'LineWidth',1.5);
grid % plot of original data Y
figure; bar(f,M); grid % plot of harmonic components as bar shape


the results of the bar figure was and now is:

Now, I want to find the equation for these harmonic components which represent the data. After that I want to draw the original data Y with the data found from the fitting function and the two curves should be close to each other.

Should I use cos or sin or -sin or -cos? In other words, what is the rule to represent these harmonics as a function: $Y = f ( t )$ ?

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Depends a bit on the algorithm used by Matlab. Can you not read off the code to see what they use? I don't have Matlab, so I can't check. – Raskolnikov Mar 26 '11 at 22:38
It is not a matter of Matlab program, it is a concept problem. How to get a function of known data Y=f(t) using fourier analysis. From reading some books and papers, the plot of points is analysed by fourier series, then, from its harmonic components, an equation could be written and built in some approach which is I am trying to find. – user4700 Mar 26 '11 at 23:31
What do the harmonic components represent? Amplitudes? Then what are the phases? If you don't know that, you can't retrieve the function. – Raskolnikov Mar 26 '11 at 23:52

You can dispatch this problem with one line of MATLAB, f=fit(t',Y,'fourier8'), the result of which is:

f =

General model Fourier8:
f(x) =
a0 + a1*cos(x*w) + b1*sin(x*w) +
a2*cos(2*x*w) + b2*sin(2*x*w) + a3*cos(3*x*w) + b3*sin(3*x*w) +
a4*cos(4*x*w) + b4*sin(4*x*w) + a5*cos(5*x*w) + b5*sin(5*x*w) +
a6*cos(6*x*w) + b6*sin(6*x*w) + a7*cos(7*x*w) + b7*sin(7*x*w) +
a8*cos(8*x*w) + b8*sin(8*x*w)
Coefficients (with 95% confidence bounds):
a0 =       4.889  (-4.318, 14.09)
a1 =      -13.85  (-26.55, -1.152)
b1 =       6.884  (-5.952, 19.72)
a2 =      -2.135  (-2.839, -1.431)
b2 =          13  (-2.486, 28.48)
a3 =       6.205  (-2.355, 14.76)
b3 =        8.53  (0.1128, 16.95)
a4 =       6.537  (-2.725, 15.8)
b4 =       3.137  (2.424, 3.851)
a5 =       5.048  (0.8798, 9.216)
b5 =      -1.603  (-6.124, 2.918)
a6 =        1.68  (1.289, 2.071)
b6 =      -2.345  (-5.843, 1.153)
a7 =      -0.158  (-1.539, 1.223)
b7 =      -1.358  (-2.395, -0.3211)
a8 =     -0.5341  (-1.093, 0.02453)
b8 =     -0.3162  (-0.6587, 0.02635)
w =      0.1994  (0.19, 0.2087)


plot(t,f(t) looks like this:

The answer to your question then is: you usually use both sine and cosine terms. The exception is if the signal is odd or even, in which case you use just sines and cosines, respectively.

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