overlapsav.m

基于双线性变换法的IIR滤波器范例 该滤波器为低通原型 可根据需要自行转换成高通 带通 以及重叠累加和重叠保留的例子

function [yn] = overlapsav(xn,hn,M)
N = length(hn);
N1 = length(xn);
p = (N1+N-1)/M
x = zeros(1,M+N-1)
y = zeros(p+1,M+N-1)
% split input xn. N-1 zeros are added in the first new input segment and in each
%head of segment there also will have N-1 overlap
for k = [0:1:p-2]
    if k == 0
        x(N:N+M-1) = xn(M*k+1:M*(k+1));
        y(k+1,:) = x;
    else
        x(1:M+N-1) = xn(M*k-N+2:M*(k+1));
        y(k+1,:) = x;
    end
end
%if the last few elements of xn are in the overlap part a new segment still will 
%be created to calculate the last N-1 elements
if N1>M*(k+1)
    l = 1;
    for i = [M*(k+1)-N+2:N1]
        y(k+2,l) = xn(i)
        l = l+1;
    end
    l = 1;
    for i = [M*(k+2)-N+2:N1]
        y(k+3,l) = xn(i)
        l = l+1;
    end
end

h = zeros(1,N+M-1);
for i = [1:N]
    h(i) = hn(i)
end
b=0
for j = [1:p+1]
    y(j,:) = circonv(y(j,:),h)
    b=b+1;
end
yn = y;