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  • 語言: Matlab
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資源簡介

FAM的matlab的實現 可以對循環普進行估計 來計算循環普 和其他方法做比較

資源截圖

代碼片段和文件信息

function?[alpha0alpha_profile]=fam(FsdfX)

%Sx=fam(FsdfdalphaNpLPNX);

%?computes?spectral?auto?correlation?(SCD)?of?Signal?X

%?X?input
%?Fs?sampling?frequency
%?df?Desired?frequency?resolution
%?dalpha?Desired?cyclic?frequency?resolution

%?SX?output
%?alpha?Cyclic?Frequency
%?fo?Spectrum?Frequency

%-------------ver?description-------------%
%?input?SSB?modulation?also?added.
%-----------------------------------------%



%?clear?all;clc;?close?all;
%?load?lp_512Hz_fs8192Hz.dat;
%?Fs=8192;
%?df=256;
%?dalpha=64;
Np=pow2(nextpow2(Fs/df));
L=Np/4;?%?This?should?be?chosen?to?be?less?than?or?equal?to?Np/4
N=length(X);
N=pow2(nextpow2(N));
P=pow2(nextpow2(N/L));
%N=P*L;
%?
%?%?Insert?input?signal?here%
%?%%----------------------------------------------------%%
%?fm=512;
%?fc=2048;
%?A_dc=.5;
%?
%?%??sig=sin(2*pi*(fm/Fs).*(1:N))‘;
%???c=cos(2*pi*(fc/Fs).*(1:N))‘;
%?%??sig_phaseshift=cos(2*pi*(fm/Fs).*(1:N))‘;
%?%??c_phaseshift=sin(2*pi*(fc/Fs).*(1:N))‘;
%?
%??x=randn(N1);
%??sig=filter(lp_512Hz_fs8192Hz1x);
%?%AM_sig=(A_dc+sig).*c;?%%%%%%%%%%ordinary?AM?Modulation
%?AM_sig=sig.*c;%%%%%%%%%DSB-SC
%?%AM_sig=sig.*c+sig_phaseshift.*c_phaseshift;?%%%%%%%%%SSB
%?%%----------------------------------------------------%%
AM_sig=X;
if?length(AM_sig)????AM_sig(N)=0;
elseif?length(AM_sig)>N
????AM_sig=AM_sig(1:N);
end
NN=(P-1)*L+Np;
xx=AM_sig;
xx(NN)=0;
xx=xx(:);
X=zeros(NpP);

for?k=0:P-1
????X(:k+1)=xx(k*L+1:k*L+Np);
end

%?windowing?----------------------------------%
w=hamming(Np);
%XW=diag(w)*X;
XW=X;

%-----------------First?FFT--------------------%
XF1=fft(XW);
XF1=fftshift(XF1);
XF1=[XF1(:P/2+1:P)?XF1(:1:P/2)];

%-----------------Down?conversion--------------%
E=zeros(NpP);
i=sqrt(-1);
for?k=-Np/2:Np/2-1
????for?m=0:P-1
????????E(k+Np/2+1m+1)=exp(-i*2*pi*k*m*L/Np);
????end
end
XD=XF1.*E;
XD=conj(XD‘);
clear?(‘XF1‘‘E‘‘X‘);

%----------------Multiplication----------------------------%

XM?=?zeros(PNp^2);
for?k?=?1:Np
????for?c?=?1:Np
????????XM(:(k-1)*Np+c)?=?(XD(:k).*conj(XD(:c)));
????end
end
%---------------------------------------------------------%

%-------------------Second?FFT----------------------------%
XF2?=?fft(XM);
XF2?=?fftshift(XF2);
XF2?=?[XF2(:Np^2/2+1:Np^2)?XF2(:1:Np^2/2)];
XF2?=?XF2?(P/4:3*P/4:);
MM?=?abs(XF2);

alpha0?=?-Fs:Fs/N:Fs;
f0?=?-Fs/2:Fs/Np:Fs/2;
Sx?=?zeros(Np+1?2*N+1);

for?k1?=?1:P/2+1
????for?k2?=?1:Np^2
????????if?rem(k2Np)?==?0
????????????c?=?Np/2?-?1;
????????else
????????????c?=?rem(k2Np)?-?Np/2?-?1;
????????end
????????k?=?ceil(k2/Np)-Np/2-1;
????????p?=?k1-P/4-1;
????????alpha?=?(k-c)/Np+(p-1)/N;
????????f?=?(k+c)/2/Np;
????????if?alpha<-1?|?alpha>1
????????????k2?=?k2+1;
????????elseif?f<-.5?|?f>.5
????????????k2?=?k2+1;
????????else
????????????kk?=?1+Np*(f?+?.5);
????????????ll?=?1+N*(alpha?+?1);
????????????Sx(round(kk)?round(ll))?=?MM(k1k2);
????????e

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????4134??2009-06-03?22:08??fam.m

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