一个IIR带通滤波器，希望得到大家的帮助

[zesairu]

要做一个可容纳100～700Hz声音频率的IIR带通滤波器仿真，现有程序如下，希望大家帮学生改成可容纳100～700Hz声音频率的，并且加一些注释，不胜感激！
clear;

fs=20;fpa=2;fpb=4;fsa=1.5;fsb=4.5;

Ap=0.0877;As=16.9897;

wpa=2*pi*fpa/fs;wpb=2*pi*fpb/fs;wsa=2*pi*fsa/fs;wsb=2*pi*fsb/fs;

c=sin(wpa+wpb)/(sin(wpa)+sin(wpb));

omegap=abs((c-cos(wpb))/sin(wpb));

omegasa=(c-cos(wsa))/sin(wsa);omegasb=(c-cos(wsb))/sin(wsb);

omegas=min(abs(omegasa),abs(omegasb));

ep=sqrt(10^(Ap/10)-1);es=sqrt(10^(As/10)-1);

N=ceil(log(es/ep)/log(omegas/omegap));

omega0=omegap/ep^(1/N);

K=floor(N/2);

for i=1:K

theta(i)=pi*(N-1+2*i)/(2*N);

end

for i=1:K

G(i)=omega0^2/(1-2*omega0*cos(theta(i))+omega0^2);

end

for i=1:K

a1(i)=4*c*(omega0*cos(theta(i))-1)/(1-2*omega0*cos(theta(i))+omega0^2);

end

for i=1:K

a2(i)=2*(2*c^2+1-omega0^2)/(1-2*omega0*cos(theta(i))+omega0^2);

end

for i=1:K

a3(i)=-(4*c*(omega0*cos(theta(i))+1))/(1-2*omega0*cos(theta(i))+omega0^2);

end

for i=1:K

a4(i)=(1+2*omega0*cos(theta(i))+omega0^2)/(1-2*omega0*cos(theta(i))+omega0^2);

end

if K<(N/2)

G0=omega0/(1+omega0);a0(1)=-2*c/(1+omega0);a0(2)=(1-omega0)/(1+omega0);

end

w=(0+eps):pi/300:pi;

Hw2=1./(1+((c-cos(w))./(omega0*sin(w))).^(2*N));

plot(w/pi,Hw2);

grid;