问题
I have 24 values for Y and corresponding 24 values for the Y values are measured experimentally,
while t has values : t=[1,2,3........24]
I want to find the relationship between Y and t as an equation using Fourier analysis,
what I have tried and done is:
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 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 another way, what is the rule to represent these harmonics as a function: Y = f (t) ?
回答1:
An example done with your data and Mathematica using Discrete sine transform. Hope you can extrapolate to Matlab:
n = 24;
xg = N[Range[n]]/n
fg = l (*your list *)
fp = ListPlot[Transpose[{xg, fg}], PlotRange -> All] (*points plot*)
coef = FourierDST[fg, 1]/Sqrt[n/2]; (*Fourier transform*)
Show[fp, Plot[Sum[coef[[r]]*Sin[Pi r x], {r, n - 1}], {x, -1, 1},
PlotRange -> All]]
The coefficients are:
{16.6411, -4.00062, 5.31557, -1.38863, 2.89762, 0.898562,
1.54402, -0.116046, 1.54847, 0.136079, 1.16729, 0.156489,
0.787476, -0.0879736, 0.747845, 0.00903859, 0.515012, 0.021791,
0.35001, 0.0159676, 0.215619, 0.0122281, 0.0943376, -0.00150218}
More detailed view:
Edit
However, as an even function seems to be better, I made also a discrete fourier cosine transform of type 3, which works much better:
In this case the coefficients are:
{14.7384, -8.93197, 4.56404, -2.85262, 2.42847, -0.249488,
0.565181,-0.848594, 0.958699, -0.468337, 0.660136, -0.317903,
0.390689,-0.457621, 0.427875, -0.260669, 0.278931, -0.166846,
0.18547, -0.102438, 0.111731, -0.0425396, 0.0484102, -0.00559378}
And the plotting of coeffs and function are obtained by:
coef = FourierDCT[fg, 3]/Sqrt[n];(*Fourier transform*)
f[x_]:= Sum[coef[[r]]*Cos[Pi (r - 1/2) x], {r, n - 1}]
You'll have to experiment a little ...
回答2:
Depends on what MATLAB gave you back. It's either sine and cosine or a complex exponential.
Most FFT algorithms that I know of usually demand that the number of data points be an integer power of two. The closest one for your data set is 32, so you should pad it out with zeros.
回答3:
Thanks for your help.
I found the solution I was aiming to get but for some reason everything is shifted by 1
Here is the code:
ts = 1; % time step
t = [1:ts:24];
fs = 1/ts; % frequency step
f=[-length(t)/2:length(t)/2-1]/(length(t)*ts); % frequency formula
%data
P=[10.7083
9.7003
8.9780
8.4531
8.1653
8.2633
8.8795
9.9850
11.3289
12.5172
13.2012
13.2720
12.9435
12.6647
12.8940
13.8516
15.3819
17.0033
18.1227
18.3039
17.4531
15.8322
13.9056
12.1154];
plot(t,P,'LineWidth',1.5);grid
xlabel('time (hours)');ylabel('Power (MW)')
title('Power Profile for 2nd Feb, 1998')
% fourier transform analysis
P1 = fft(P)/length(t);
P2=fftshift(P1);
amp=abs(P2); % amplitude
phi = angle(P2); % phase angle
figure
subplot(211),stem(f,amp,'LineWidth',1.5),grid
xlabel('frequency (Hz)');ylabel('amplitude (MW)')
subplot(212),stem(f,phi,'LineWidth',1.5),grid
xlabel('frequency (Hz)');ylabel('phase angle (rad)')
% NOW, I WILL CONSTRUCT THE MODEL FROM THE FIGURE
% THE STRUCTURE IS:
% Pmodel=Ai*COS(i*w*t+phii)
% where, w=2*pi/24 and i is the harmonic order
% Here, up to the third harmonic is enough
% and using Parseval's Theorem, the model is:
% PP=12.6635+2*(1.9806*cos(w*tt+1.807)+0.86388*cos(2*w*tt+2.0769)+0.39683*cos(3*w*tt- 1.8132));
w=2*pi/24;
Pmodel=12.6635+2*(1.9806*cos(w*t+1.807)+0.86388*cos(2*w*t+2.0769)+0.39686*cos(3*w*t-1.8132));
figure
plot(t,P,'LineWidth',1.5);grid on
hold on;
plot(t,Pmodel,'r','LineWidth',1.5)
legend('original','model');xlabel('time (hours )');ylabel('Power (MW)')
% But here is a problem, the modeled signal is shifted
% by 1 comparing to the original one
% I redraw the two figures together by plotting Pmodeled vs t+1
% Actually, I don't know why it is shifted, but they are
% exactly identical with shifting by 1
figure
plot(t,P,'LineWidth',1.5);grid on
hold on;
plot(t+1,Pmodel,'r','LineWidth',1.5)
legend('original','model');xlabel('time (hours )');ylabel('Power (MW)')
Why has this shifting problem happened, and how can I solve it?
回答4:
The problem is with line 2 "t = [1:ts:24];" it should be "t= 0:ts:23;"
来源：https://stackoverflow.com/questions/5445499/using-fourier-analysis-to-fit-function-to-data