Basic question of convolution
显示 更早的评论
Hi. I solved a problem using conv .
'y[n] = x[n] * h[n] when x[n] = h[n] = [1,3,2,4]'.
Like this:
x=[1,3,2,4];
h=[1,3,2,4];
y=conv(x, h)
stem(y,'r','LineWidth', 2);
But, when the equation is y[n] = x[n-1] * h[n+1], how can I make n-1 and n+1??
采纳的回答
But, when the equation is y[n] = x[n-1] * h[n+1], how can I make n-1 and n+1??
There is no need to in this case. because of the shift-invariance of convolution, the negative shift in x will be cancelled out by the positive shift in h.
更多回答（1 个）
AB WAHEED LONE
2021-4-17
编辑：AB WAHEED LONE
2021-4-17
1 个投票
Look Convolution for discrete case is defined as y[n]=Σ(x[k]*h[n-k])
so y[n-1]=Σ(x[k]*h[n-1-k]) and y[n+1]=Σ(x[k]*h[n+1-k])
类别
在 帮助中心 和 File Exchange 中查找有关 Numerical Integration and Differential Equations 的更多信息
Translated by 
