Question

matlab利用递归求解差分方程

function y = recur(a,b,n,x,x0,y0);
%
% y = recur(a,b,n,x,x0,y0)
% solves for y[n] from:
% y[n] + a1*y[n-1] + a2*y[n-2]... + an*y[n-N]
% = b0*x[n] + b1*x[n-1] + ... + bm*x[n-M]
%
% a, b, n, x, x0 and y0 are vectors
% a = [a1 a2 ... aN]
% b = [b0 b1 ... bM]
% n contains the time values for which the solution will be computed
% y0 contains the initial conditions for y, in order,
% i.e., y0 = [y[n0-N], y[n0-N+1], ...,y[n0-1]]
% where n0 represents the first element of n
% x0 contains the initial conditions on x, in order
% i.e., x0 = [x[n0-M],...,x[n0-1]]
% the output, y, has length(n)
%

N = length(a);
M = length(b)-1;
y = [y0 zeros(1,length(n))];
x = [x0 x]
a1 = a(length(a):-1:1) % reverses the elements in a
b1 = b(length(b):-1:1)
for i=N+1:N+length(n),
y(i) = -a1*y(i-N:i-1)' + b1*x(i-N:i-N+M)';
end
y = y(N+1:N+length(n))

求各位大神详细解释此段程序具体如何运行，看到循环那有些看不明白

Answer 1

首先，这个不是matlab利用递归求解差分方程，而是递推；差分方程其实就是递推关系式。
然后这个循环：
for i=N+1:N+length(n),
y(i) = -a1*y(i-N:i-1)' + b1*x(i-N:i-N+M)';
end
其实是因为：
y[n] + a1*y[n-1] + a2*y[n-2]... + an*y[n-N] = b0*x[n] + b1*x[n-1] + ... + bm*x[n-M]
所以：
y[n] = -（a1*y[n-1] + a2*y[n-2]... + an*y[n-N] ）+ b0*x[n] + b1*x[n-1] + ... + bm*x[n-M]

具体来说，就是：
我们已知了y1、y2、y3。。。yN，然后通过循环依次求得yN+1、yN+2等等。。。