convolution of two unit step functions.

Question

justin stephens 2018-2-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/383554-convolution-of-two-unit-step-functions

回答： justin stephens 2018-2-20

alright folks, the issue i am having is that i am trying to use convolution on two step functions but for one i have an odd interval that i cannot figure out how to program in matlab. Here is the set up: x[n]= 1 ; 0<=n<=9 otherwise 0, h[n]= 1; 0<=n<=N where N is <= 9 otherwise 0. my problem in i do not know how to express this extra boundary in matlab. here is the code i have written thus far.

 code
end
nx=-1:9;
N<=9;
nh=0:N;
x=usD(n);
h=usD(n);
y=conv(x,h);
ny=(nx(1) + nh(1) + (0:length(nx) + length(nh) - 2));
plot (ny,y);

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Shounak Shastri 2018-2-20

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/383554-convolution-of-two-unit-step-functions#answer_306022

在 MATLAB Online 中打开

1. The first thing you need to know is that in Matlab, the array index starts from 1 and not from 0. So if you write x(0)=1, you would get an error.

So you need to modify your limits to x[n]= 1 ; 1<=n<=10 and N<=10.

2. The statement

N<=9;

will give you a true or false answer as it is a logical statement. Also, the answer would be of no use as it is not being stored or used as a flag in any way.

3. As for the boundary, you can simply add a if statement inside a for loop.

 for ii=1:length(h)
     if ii>=1 && ii<=N % considering you have added 1 as described in 1
         h(ii)=1;
     else
         h(ii)=0;
     end
 end