可以详细讲讲每行都干了点什么吗？% Evaluating an LCCDE algorithmically% y[n] + y[n-1] - 6y[n-2] = x[n]% with y[-2] = -1, y[-1] = 1, x[n] = 8.u[n]% Calculate for n = 0..4, but include -2 and -1nn = -2:4;% So setup initial y values; zero is index 3zz = 3;y(zz+ -2) = -1;y(zz+ -1) = 1;% Now evaluate the LCCDE step by stepx = zeros(1,7);x(zz+ [0:4]) = 8;for n = 0:4; y(zz+ n) = x(zz+ n) - y(zz+ n-1) + 6*y(zz+ n-2); endy% y =% -1 1 1 13 1 85 -71
% Evaluating an LCCDE algorithmically% y[n] + y[n-1] - 6y[n-2] = x[n]--可以看出，这是一个差分方程，对于x向量的初值(x[1])来说，n=1时，y向量出现了y[-2]，但是MATLAB里面是没有负索引的，首地址的索引为1，所以要在后面用3-2来实现这个1.这也就是zz=3的原因。--当然写成y[n]的形式是可以的，你老师这样做想必是为了使程序更有意义一些。% with y[-2] = -1, y[-1] = 1, x[n] = 8.u[n]% Calculate for n = 0..4, but include -2 and -1nn = -2:4;% So setup initial y values; zero is index 3zz = 3;y(zz+ -2) = -1;y(zz+ -1) = 1;--给y向量赋初值。% Now evaluate the LCCDE step by stepx = zeros(1,7);x(zz+ [0:4]) = 8;--给x向量赋初值。for n = 0:4; y(zz+ n) = x(zz+ n) - y(zz+ n-1) + 6*y(zz+ n-2); end--根据差分方程计算y向量的后几个值。y% y =% -1 1 1 13 1 85 -71
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