归一化频率 和实际频率的 怎么对应

hziee_hz

　　求助做EMD和HHT的

function [fnormhat,t]=instfreq(x,t,L,trace);
%INSTFREQ Instantaneous frequency estimation.
%        [FNORMHAT,T]=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); [instf,t]=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,[instfreq(sig,t,L),instfreq(x,t)]); grid;
%         title ('theoretical and estimated instantaneous frequencies');
%
%        See also  KAYTTH, 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;
[xrow,xcol] = 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
[trow,tcol] = 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

songzy41

查看下帖子，恢复显示实际频率。
http://forum.vibunion.com/forum/thread-61930-1-16.html