问题描述
|
我有24个Y值,相应的24个Y值是通过实验测量的,
而t的值是:
t=[1,2,3........24]
我想使用傅立叶分析找到Y和t之间的关系作为方程式,
我尝试过的是:
我编写了以下MATLAB代码:
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,1.5);grid % plot of original data Y
figure;bar(f,M);grid % plot of harmonic components as bar shape
条形图的结果是:
现在,我想找到这些代表数据的谐波分量的方程式。之后,我想用从拟合函数中找到的数据绘制原始数据Y,两条曲线应彼此靠近。
我应该使用cos或sin还是-sin或-cos?
换句话说,将这些谐波表示为函数的规则是什么:is2ѭ?解决方法
使用离散正弦变换对数据和Mathematica进行处理的示例。希望您可以推断到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]]
系数为:
{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}
更详细的视图:
编辑
但是,由于偶数函数似乎更好,所以我还进行了类型3的离散傅立叶余弦变换,其效果更好:
在这种情况下,系数为:
{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}
并通过以下公式获得系数和函数的图:
coef = FourierDCT[fg,3]/Sqrt[n];(*Fourier transform*)
f[x_]:= Sum[coef[[r]]*Cos[Pi (r - 1/2) x],n - 1}]
您将不得不尝试一下...,取决于MATLAB给您的回报。它可以是正弦和余弦,也可以是复杂的指数。
我知道的大多数FFT算法通常都要求数据点的数量为2的整数次幂。您的数据集最接近的一个是32,因此您应该用零填充它。,谢谢你的帮助。
我找到了我想要的解决方案,但由于某种原因,一切都偏移了1
这是代码:
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,1.5),grid
xlabel(\'frequency (Hz)\');ylabel(\'amplitude (MW)\')
subplot(212),phi,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,1.5);grid on
hold on;
plot(t,Pmodel,\'r\',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,1.5);grid on
hold on;
plot(t+1,\'model\');xlabel(\'time (hours )\');ylabel(\'Power (MW)\')
为什么会发生这种变化的问题,我该如何解决?,问题出在
2号线
\“ t = [1:ts:24]; \”
它应该是\“ t = 0:ts:23; \”