convolution product calculater impulse response

2022 2 17
1 个回答
回答已采纳
更新时间：2022 2 19
2 次查看（30 天）

0 个投票

can anyone tell me where is the problem
LDKI
x(n)=[1,3,-2,5,7]
h(n)= [1,5,8,-5,13]
x=[1,3,-2,5,7]
h= [1,5,8,-5,13]
dx=length(x)
dh=length(h)
dy=dx+dh-1
for i=1:dy
y1(i)=0;
for j=1:dx
y1(i)=y1(i)+x(j)*h(i-j+1)
end
end
check1=conv(x1,h1)

请先登录，再回答此问题。

 采纳的回答

David Goodmanson
David Goodmanson 2022-2-19
编辑：David Goodmanson 2022-2-19

0 个投票

Hi amjad,
You have to keep the indices within the bounds set by dx and dh. For the j loop, you can use the line
for j=max(1,i-dh+1):min(i,dx)
Also, in case you run the script more than once with different x and h, it pays to initialize y1 with
y1 = zeros(1,dy);

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by