归一化频率 和实际频率的 怎么对应
本帖最后由 wdhd 于 2016-9-12 13:41 编辑求助做EMD和HHT的
function =instfreq(x,t,L,trace);
%INSTFREQ Instantaneous frequency estimation.
% =INSTFREQ(X,T,L,TRACE) computes the instantaneous
% frequency of the analytic signal X at time instant(s) T, using the
% trapezoidal integration rule.
% The result FNORMHAT lies between 0.0 and 0.5.
%
% X : Analytic signal to be analyzed.
% T : Time instants (default : 2:length(X)-1).
% L : If L=1, computes the (normalized) instantaneous frequency
% of the signal X defined as angle(X(T+1)*conj(X(T-1)) ;
% if L>1, computes a Maximum Likelihood estimation of the
% instantaneous frequency of the deterministic part of the signal
% blurried in a white gaussian noise.
% L must be an integer (default : 1).
% TRACE : if nonzero, the progression of the algorithm is shown
% (default : 0).
% FNORMHAT : Output (normalized) instantaneous frequency.
% T : Time instants.
%
% Examples :
% x=fmsin(70,0.05,0.35,25); =instfreq(x); plot(t,instf)
% N=64; SNR=10.0; L=4; t=L+1:N-L; x=fmsin(N,0.05,0.35,40);
% sig=sigmerge(x,hilbert(randn(N,1)),SNR);
% plotifl(t,); grid;
% title ('theoretical and estimated instantaneous frequencies');
%
% See alsoKAYTTH, SGRPDLAY.
% F. Auger, March 1994, July 1995.
% Copyright (c) 1996 by CNRS (France).
%
% ------------------- CONFIDENTIAL PROGRAM --------------------
% This program can not be used without the authorization of its
% author(s). For any comment or bug report, please send e-mail to
% f.auger@ieee.org
if (nargin == 0),
error('At least one parameter required');
end;
= size(x);
if (xcol~=1),
error('X must have only one column');
end
if (nargin == 1),
t=2:xrow-1; L=1; trace=0.0;
elseif (nargin == 2),
L = 1; trace=0.0;
elseif (nargin == 3),
trace=0.0;
end;
if L<1,
error('L must be >=1');
end
= size(t);
if (trow~=1),
error('T must have only one row');
end;
if (L==1),
if any(t==1)|any(t==xrow),
error('T can not be equal to 1 neither to the last element of X');
else
fnormhat=0.5*(angle(-x(t+1).*conj(x(t-1)))+pi)/(2*pi);
end;
else
H=kaytth(L);
if any(t<=L)|any(t+L>xrow),
error('The relation L<T<=length(X)-L must be satisfied');
else
for icol=1:tcol,
if trace, disprog(icol,tcol,10); end;
ti = t(icol); tau = 0:@L;
R = x(ti+tau).*conj(x(ti-tau));
M4 = R(2:@L+1).*conj(R(1:@L));
diff=2e-6;
tetapred = H * (unwrap(angle(-M4))+pi);
while tetapred<0.0 , tetapred=tetapred+(2*pi); end;
while tetapred>2*pi, tetapred=tetapred-(2*pi); end;
iter = 1;
while (diff > 1e-6)&(iter<50),
M4bis=M4 .* exp(-j*2.0*tetapred);
teta = H * (unwrap(angle(M4bis))+2.0*tetapred);
while teta<0.0 , teta=(2*pi)+teta; end;
while teta>2*pi, teta=teta-(2*pi); end;
diff=abs(teta-tetapred);
tetapred=teta; iter=iter+1;
end;
fnormhat(icol,1)=teta/(2*pi);
end;
end;
end;
这是求瞬时频率的程序，请问怎么修改能去掉频率限制在between 0.0 and 0.5之间？
谢谢大家 ！很着急啊 查看下帖子，恢复显示实际频率。
http://forum.vibunion.com/forum/thread-61930-1-16.html
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