pitch_test.m

滤波器的设计以及零极点对应频率响应分析

l=size(data);
%san=data(:,6:14);
%st=12*log10(san./65)/log10(2);
%deta=san-data(:,6:14);
zeo=sum((data>0),2);
t=1;
for i=1:l(1)
   if zeo(i) == 15
       san(i,:)=data(i,6:15);
     t=t+1;    
   end
   
end
mea=mean(san,2);


%stm=12*log10(kk./65)/log10(2);
kk=zeros(size(san));
for i=1000:2000
    
    plot(san(i,:));
    hold on
 % kk(i,:)=san(i,:)-mea(i);
end
%zz=sum(san)/t;
%plot(zz,'*'); hold on