資源簡介
ceemd分解,還是很不錯的,先將數據精心ceemd分解,得到imf分量,然后通過相關系數帥選分量,在求出他們的樣本熵的特征,完美運行,你值得擁有,可以的話,給一個好評,謝謝。
代碼片段和文件信息
%?Y:?Inputted?data;
%?Nstd:?ratio?of?the?standard?deviation?of?the?added?noise?and?that?of?Y;
%?NE:?Ensemble?member?being?used
%?TNM:?total?number?of?modes?(not?including?the?trend)
%
function?allmode=ceemd(YNstdNETNM)
%?find?data?length
xsize=length(Y);
dd=1:1:xsize;
%?Nornaliz?data
Ystd=std(Y);
Y=Y/Ystd;
%?Initialize?saved?data
TNM2=TNM+2;
for?kk=1:1:TNM2
????for?ii=1:1:xsize
????????allmode(iikk)=0.0;
????end
end
for?iii=1:1:NE
%?adding?noise
????for?i=1:xsize
????????temp=randn(11)*Nstd;
????????X1(i)=Y(i)+temp;
????????X2(i)=Y(i)-temp;
????end
????%?sifting?X1
????xorigin?=?X1;
????xend?=?xorigin;
%?save?the?initial?data?into?the?first?column
????for?jj=1:1:xsize
????????mode(jj1)?=?xorigin(jj);
????end
????nmode?=?1;
????while?nmode?<=?TNM
?????????xstart?=?xend;
????????iter?=?1;
????????while?iter<=5
?????????????[spmax?spmin?flag]=extrema(xstart);
?????????????upper=?spline(spmax(:1)spmax(:2)dd);
?????????????lower=?
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件??????59820??2018-12-27?22:47??C-樣本熵\20151124_08_15Bin粗卡閥.xlsx
?????文件???????2603??2014-04-01?21:11??C-樣本熵\ceemd.m
?????文件???????1053??2019-01-14?13:22??C-樣本熵\CEEMD分解-樣本熵特征.m
?????文件???????2180??2011-10-21?09:02??C-樣本熵\extrema.m
?????文件???????1521??2017-07-16?10:40??C-樣本熵\SampEn.m
?????目錄??????????0??2019-01-14?14:16??C-樣本熵
-----------?---------??----------?-----??----
????????????????67177????????????????????6
- 上一篇:基于粒子群算法的PID控制器優化
- 下一篇:自制-A4標準圖框(帶明細表).dwg
評論
共有 條評論