Compute convolution y[n]=x[n]*h[n]: x[n]={2,0,1,-1,3}; h[n]={1,2,0,1}

64 次查看（过去 30 天）

Li Hui Chew 2021-6-24

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/864275-compute-convolution-y-n-x-n-h-n-x-n-2-0-1-1-3-h-n-1-2-0-1

编辑： Paul 2023-12-3

My approach:

x=[2 0 1 -1 3];
h=[1 2 0 1];
% therefore 
y=conv(x,h)

y =

2 4 1 3 1 7 -1 3

4 个评论
显示 2更早的评论隐藏 2更早的评论

Walter Roberson 2023-12-2

在 MATLAB Online 中打开

@James

It would depend on whether that u[n-1] is the unit step function or not.

If it is then x[n] would be 0 for n < 0, and 1 for n >= 0 -- an infinite stream of 1's. And h[n] would be 0 for n <= 1, and 0.9^n for n > 1 -- an infinite stream of non-negative numbers. You cannot express that as a finite convolution sequence.

Let us see what it would turn out like for continuous functions:

sympref('heavisideatorigin', 1)

ans =

syms n integer

x(n) = heaviside(n)

x(n) =

h(n) = (sym(9)/sym(10))^n * heaviside(n-1)

h(n) =

syms t tau

C(t) = int(x(tau) * h(t-tau), tau, 0, t)

C(t) =

[C(-1), C(0), C(1)]

ans =

assume(sign(t-1) == 1)

simplify(C)

ans(t) =

Paul 2023-12-2

编辑：Paul 2023-12-3

The follow-up problem posed by James can be solved for a closed form solution using symsum and the shift property of the convolution sum. Seem like a HW problem so won't show it now.

请先登录，再进行评论。

请先登录，再回答此问题。