資源簡介
對信號進行局部均值分解,得到一系列具有物理意義的pf分量
代碼片段和文件信息
function?[envmin?envmaxenvmoyindminindmaxindzer]?=?envelope(txINTERP)
%computes?envelopes?and?mean?with?various?interpolations
NBSYM?=?2;?%?邊界延拓點數(shù)
DEF_INTERP?=?‘spline‘;
if?nargin?2
x?=?t;
t?=?1:length(x);
INTERP?=?DEF_INTERP;
end
if?nargin?==?2
if?ischar(x)
INTERP?=?x;
x?=?t;
t?=?1:length(x);
end
end
if?~ischar(INTERP)
error(‘interp?parameter?must?be?‘‘linear‘‘‘‘?‘‘cubic‘‘?or?‘‘spline‘‘‘)
end
if?~any(strcmpi(INTERP{‘linear‘‘cubic‘‘spline‘}))
error(‘interp?parameter?must?be?‘‘linear‘‘‘‘?‘‘cubic‘‘?or?‘‘spline‘‘‘)
end
if?min([size(x)size(t)])?>?1
error(‘x?and?t?must?be?vectors‘)
end
s?=?size(x);
if?s(1)?>?1
x?=?x‘;
end
s?=?size(t);
if?s(1)?>?1
t?=?t‘;
end
if?length(t)?~=?length(x)
error(‘x?and?t?must?have?the?same?length‘)
end
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件?????????343??2013-01-11?10:35??Unti
?????文件????????3957??2013-01-11?10:32??envelope.m
?????文件????????1705??2013-01-11?10:33??extr.m
?????文件??????????58??2010-12-15?23:23??li
?????文件?????????843??2013-01-11?10:29??lmd1.m
?????文件????????1916??2013-01-14?17:17??move.m
?????文件?????????100??2013-01-11?10:29??nengliang.m
?????文件?????????189??2013-01-11?10:34??pos.m
?????文件?????????991??2013-01-11?10:35??position.m
?????文件?????????971??2010-12-16?10:56??smove.m
?????文件?????????386??2013-01-11?10:30??zhaochun1.m
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