📄 abf.m
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%求出ntfz;stfz得幅频特性曲线、说明于fb中stfz是全通得,而ntfz是大衰减得
clc;
clear;
w=0:0.01:pi;
z1=exp(-i.*w);
a=[0.5 0.5 0.2 1/6];
A1=a(1);
A2=-3*a(1)+a(1)*a(2);
A3=3*a(1)-2*a(2)*a(1)+a(1)*a(2)*a(3);
A4=-a(1)+a(1)*a(2)-a(1)*a(2)*a(3)+a(1)*a(2)*a(3)*a(4);
x1=(1-z1).^4;
ntfz=x1./(1+(A1-4)*z1+(A2+6)*(z1.^2)+(A3-4)*(z1.^3)+(A4+1)*(z1.^4));
stfz=(A1*z1+A2*z1.^2+A3*z1.^3+A4*z1.^4)./(1+(A1-4)*z1+(A2+6)*(z1.^2)+(A3-4)*(z1.^3)+(A4+1)*(z1.^4));
plot(w,20*log10(abs(ntfz)),w,20*log10(abs(stfz)));
%subplot(2,1,1);
%plot(w,abs(ntfz),'.');
%subplot(2,1,2);
%plot(w,20*log10(abs(ntfz)));
grid on;
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