作业帮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
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![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](/uploads/image/z/3935801-65-1.jpg?t=matlab%E5%88%A9%E7%94%A8%E9%80%92%E5%BD%92%E6%B1%82%E8%A7%A3%E5%B7%AE%E5%88%86%E6%96%B9%E7%A8%8Bfunction+y+%3D+recur%28a%2Cb%2Cn%2Cx%2Cx0%2Cy0%29%3B%25%25+y+%3D+recur%28a%2Cb%2Cn%2Cx%2Cx0%2Cy0%29%25+solves+for+y%5Bn%5D+from%3A%25+y%5Bn%5D+%2B+a1%2Ay%5Bn-1%5D+%2B+a2%2Ay%5Bn-2%5D...%2B+an%2Ay%5Bn-N%5D%25+%3D+b0%2Ax%5Bn%5D+%2B+b1%2Ax%5Bn-1%5D+%2B+...%2B+bm%2Ax%5Bn-M%5D+%25%25+a%2Cb%2Cn%2Cx%2Cx0+and+y0+are+vectors%25+a+%3D+%5Ba1+a2)
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
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))
看到循环那有些看不明白
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
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.