LMD局域均值分解的matlab程序及示例
说明：研究LMD局域均值分解有3个月左右，能找到的相关文章也基本上看了一遍，觉得是个很好的方法，号称是EMD经验模态分解的改进版。但是网络上一直没有找到该算法的matlab程序，只见文章说的天花乱坠。后来自己写了一个，但是使用有一个问题没有解决，就是分解的时候怎么去除骑行波的问题。先把自己写的程序贡献出来，让大家分析下，一起讨论下，到底LMD的程序怎样写才能如文献中说的那样达到目的。欢迎大家热烈讨论！
程序仍存在不收敛的问题，拿出来分享只是希望高手指点一二，写的不好欢迎拍砖！
想要M文件的，给出下载地址：
文件夹包含，找出纯调频函数，计算瞬时幅值，瞬时频率的函数等
%%%%%%%%%%%主程序%%%%%%%%%%
%lmd1原始版
%通过emd.m的三次样条+镜像延拓求出上下包络及均值
%局域均值函数=（上包络+下包络）/2
%局域包络函数=|上包络-下包络|/2
%相关文章见《一种基于EMD的振动信号时频分析新方法研究》-胡劲松，杨世锡
function[PF,A,SI]=lmd1(m)
%最后一个PF是残余分量
%A是瞬时赋值
%SI是纯调频函数，求它的瞬时频率就是需要的频率
wucha1=0.001;
n_l=nengliang(m);
&&& k=k+1;
[pf,a,si]=zhaochun1(a,h,wucha1);
c_pos=pos(c);
n_c=nengliang(c);
n_pf=nengliang(pf);
&&& %停止调节
%1.emd用的是三次样条求包络，要求至少3个极值点，所以这里c的极值点个数也应该至少为3
%2.如果上一个PF的极值点数比下一个PF的极值点数少，说明结果也不正确（这个也可以作为停止条件考虑进去）
%上面一句是否可以等价于当前PF的极值点个数一定要大于等于残量（c）的极值点个数（目前是用这个作为停止条件的一个参考写入程序）
%3.当前PF分量的能量应该大于残量c的能量(这个有待商榷)
%4.残余能量不能大于信号能量
length(c_pos)&3 ||
n_c&n_l/10& ||k==3%||
n_pf&n_c%||n_c&n_l
PF(k+1,:)=c;
%%%%%%%%%%%%%%%%%%nengliang函数%%%%%%%%%%
function p=nengliang(y)
% my=mean(y);
% p=(y-my).*(y-my);
% p=sum(p);
p=sum(abs(y).^2);
%%%%%%%%%%%%%找纯函数%%%%%%%%%%%%%%%%%
function [pf,a,si]=zhaochun1(a,h,wucha1)
chun_num=0;
chun_num=chun_num+1;
t=1:length(h);
[envmin,envmax,envmoy,indmin,indmax,indzer] =
envelope(t,h,'spline');
mi=(envmax+envmin)./2;
ai=abs(envmax-envmin)./2;
si=(h-mi)./
ai_funmax=max(ai);
ai_funmin=min(ai);
(ai_funmax&=1+wucha1&&ai_funmin&=1-wucha1)
function [envmin, envmax,envmoy,indmin,indmax,indzer] =
envelope(t,x,INTERP)
%computes envelopes and mean with various interpolations
NBSYM = 2;&&% 边界延拓点数
DEF_INTERP = 'spline';
if nargin & 2
&t = 1:length(x);
&INTERP = DEF_INTERP;
if nargin == 2
&if ischar(x)
&&INTERP =
&&t = 1:length(x);
if ~ischar(INTERP)
&error('interp parameter must be ''linear'''',
''cubic'' or ''spline''')
if ~any(strcmpi(INTERP,{'linear','cubic','spline'}))
&error('interp parameter must be ''linear'''',
''cubic'' or ''spline''')
if min([size(x),size(t)]) & 1
&error('x and t must be vectors')
s = size(x);
if s(1) & 1
s = size(t);
if s(1) & 1
if length(t) ~= length(x)
&error('x and t must have the same length')
lx = length(x);
[indmin,indmax,indzer] = extr(x,t);
%boundary conditions for interpolation
[tmin,tmax,xmin,xmax] =
boundary_conditions(indmin,indmax,t,x,NBSYM);
% definition of envelopes from interpolation
envmax = interp1(tmax,xmax,t,INTERP);&
envmin = interp1(tmin,xmin,t,INTERP);
if nargout & 2
&&& envmoy =
(envmax + envmin)/2;
function [tmin,tmax,xmin,xmax] =
boundary_conditions(indmin,indmax,t,x,nbsym)
% computes the boundary conditions for interpolation (mainly mirror
&lx = length(x);
&% 判断极值点个数
&if (length(indmin) + length(indmax)
&&error('not enough
% 插值的边界条件
&if indmax(1) & indmin(1)%
第一个极值点是极大值
&if x(1) & x(indmin(1))%
以第一个极大值为对称中心
fliplr(indmax(2:min(end,nbsym+1)));
fliplr(indmin(1:min(end,nbsym)));
indmax(1);
如果第一个采样值小于第一个极小值，则将认为该值是一个极小值，以该点为对称中心
fliplr(indmax(1:min(end,nbsym)));
[fliplr(indmin(1:min(end,nbsym-1))),1];
&&if x(1) &
x(indmax(1))% 以第一个极小值为对称中心
fliplr(indmax(1:min(end,nbsym)));
fliplr(indmin(2:min(end,nbsym+1)));
indmin(1);
如果第一个采样值大于第一个极大值，则将认为该值是一个极大值，以该点为对称中心
[fliplr(indmax(1:min(end,nbsym-1))),1];
fliplr(indmin(1:min(end,nbsym)));
序列末尾情况与序列开头类似
&if indmax(end) & indmin(end)
&&if x(end) &
x(indmax(end))
fliplr(indmax(max(end-nbsym+1,1):end));
fliplr(indmin(max(end-nbsym,1):end-1));
indmin(end);
[lx,fliplr(indmax(max(end-nbsym+2,1):end))];
fliplr(indmin(max(end-nbsym+1,1):end));
&&if x(end) &
x(indmin(end))
fliplr(indmax(max(end-nbsym,1):end-1));
fliplr(indmin(max(end-nbsym+1,1):end));
indmax(end);
fliplr(indmax(max(end-nbsym+1,1):end));
[lx,fliplr(indmin(max(end-nbsym+2,1):end))];
将序列根据对称中心，镜像到两边
&tlmin = 2*t(lsym)-t(lmin);
&tlmax = 2*t(lsym)-t(lmax);
&trmin = 2*t(rsym)-t(rmin);
&trmax = 2*t(rsym)-t(rmax);
&% in case symmetrized parts do not extend enough%
如果对称的部分没有足够的极值点
&if tlmin(1) & t(1) | tlmax(1)
& t(1)% 对折后的序列没有超出原序列的范围
&&if lsym == indmax(1)
fliplr(indmax(1:min(end,nbsym)));
fliplr(indmin(1:min(end,nbsym)));
&&if lsym == 1%
这种情况不应该出现，程序直接中止
&&&error('bug')
&&lsym = 1;&%
直接关于第一采样点取镜像
2*t(lsym)-t(lmin);
2*t(lsym)-t(lmax);
序列末尾情况与序列开头类似
&if trmin(end) & t(lx) | trmax(end)
&&if rsym == indmax(end)
fliplr(indmax(max(end-nbsym+1,1):end));
fliplr(indmin(max(end-nbsym+1,1):end));
&if rsym == lx
&&error('bug')
2*t(rsym)-t(rmin);
2*t(rsym)-t(rmax);
% 延拓点上的取值
&xlmax =x(lmax);
&xlmin =x(lmin);
&xrmax =x(rmax);
&xrmin =x(rmin);
% 完成延拓
&tmin = [tlmin t(indmin) trmin];
&tmax = [tlmax t(indmax) trmax];
&xmin = [xlmin x(indmin) xrmin];
&xmax = [xlmax x(indmax) xrmax];
%---------------------------------------------------------------------------------------------------
% 极值点和过零点位置提取
function [indmin, indmax, indzer] = extr(x,t);
%extracts the indices corresponding to extrema
if(nargin==1)
& t=1:length(x);
m = length(x);
if nargout & 2
&x1=x(1:m-1);
&x2=x(2:m);
&indzer = find(x1.*x2&0);
&if any(x == 0)
&& iz = find( x==0 );
&& indz = [];
&& if any(diff(iz)==1)
zer = x == 0;
dz = diff([0 zer 0]);
debz = find(dz == 1);
finz = find(dz == -1)-1;
indz = round((debz+finz)/2);
&& indzer = sort([indzer
d = diff(x);
n = length(d);
d1 = d(1:n-1);
d2 = d(2:n);
indmin = find(d1.*d2&0 &
indmax = find(d1.*d2&0 &
% when two or more consecutive points have the same value we
consider only one extremum in the middle of the constant area
% 当连续多个采样值相同时，把最中间的一个值作为极值点，处理方式与连0类似
if any(d==0)
& imax = [];
& imin = [];
& bad = (d==0);
& dd = diff([0 bad 0]);
& debs = find(dd == 1);
& fins = find(dd == -1);
& if debs(1) == 1
length(debs) & 1
debs = debs(2:end);
fins = fins(2:end);
debs = [];
fins = [];
& if length(debs) & 0
&&& if fins(end)
if length(debs) & 1
debs = debs(1:(end-1));
fins = fins(1:(end-1));
debs = [];
fins = [];
& lc = length(debs);
& if lc & 0
&&& for k =
if d(debs(k)-1) & 0
if d(fins(k)) & 0
imax = [imax round((fins(k)+debs(k))/2)];
if d(fins(k)) & 0
imin = [imin round((fins(k)+debs(k))/2)];
& if length(imax) & 0
&&& indmax =
sort([indmax imax]);
& if length(imin) & 0
&&& indmin =
sort([indmin imin]);
%%%%%%%%%%%%%%%%%%pos%%%%%%%%%%%%
function poss=pos(y)
%功能：找序列极值点位置坐标
%y:输入序列
%pos:极值点的序列位置坐标
[indmin,indmax]=position(y);
minmax=cat(2,indmin,indmax);
poss=sort(minmax);
%%%%%%%%%%%%%%position%%%%%%%
%返还极大值和极小值位置下标
function [indmin,indmax]=position(y)
m = length(y);
d = diff(y);
n = length(d);
d1 = d(1:n-1);
d2 = d(2:n);
indmin = find(d1.*d2&0 &
indmax = find(d1.*d2&0 &
if any(d==0)
& imax = [];
& imin = [];
& bad = (d==0);
& dd = diff([0 bad 0]);
& debs = find(dd == 1);
& fins = find(dd == -1);
& if debs(1) == 1
length(debs) & 1
debs = debs(2:end);
fins = fins(2:end);
debs = [];
fins = [];
& if length(debs) & 0
&&& if fins(end)
if length(debs) & 1
debs = debs(1:(end-1));
fins = fins(1:(end-1));
debs = [];
fins = [];
& lc = length(debs);
& if lc & 0
&&& for k =
if d(debs(k)-1) & 0
if d(fins(k)) & 0
imax = [imax round((fins(k)+debs(k))/2)];
if d(fins(k)) & 0
imin = [imin round((fins(k)+debs(k))/2)];
& if length(imax) & 0
&&& indmax =
sort([indmax imax]);
& if length(imin) & 0
&&& indmin =
sort([indmin imin]);
%说明每个程序要单独保存成m文件，放在同一个文件夹下调用
使用实例：
(2+cos(90*t).*cos(500*t+1800.*t.*t));
t=0:1/fs:0.341;
subplot(5,1,1);plot(t,y);xlabel('原始信号');
[pf,a,si]=lmd1(y);
subplot(5,1,2);plot(t,pf(1,:));xlabel('PF1');
subplot(5,1,3);plot(t,pf(2,:));xlabel('PF2');
subplot(5,1,4);plot(t,pf(3,:));xlabel('PF3');
subplot(5,1,5);plot(t,pf(4,:));xlabel('残量信号');
从上图可以看出，成功将第一个分量和第二个分量分离出来，但是残余分量存在很大的问题，这是因为分解过程中的骑行波没有去处导致的，至于怎样完善LMD局域均值分解算法，目前个人没有时间研究了，希望得到指点。
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Which Windows OS versions does it run on?
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