Basic question of convolution

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??

Answer 1

[Matt J]

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.

[JjjO]

Thank you. So you mean I don't have to change my code for y[n] = x[n-1] * h[n+1]?

[Matt J]

That is what I mean.

[JjjO]

Thank you :)

Answer 2

[AB WAHEED LONE]

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])