Why Is n = 3 Not Considered in Convolution?

[Larrytsai]

Homework Statement

δ = dirac delta
x[n] = δ[n] + 2δ[n-1] - δ[n-3]
h[n] = 2δ[n+1] + 2δ[n-1]

y[n] = x[n]*h[n]

Homework Equations

y[n] = x[n]h[n] = \sumh[k]x[n-k]

The Attempt at a Solution

I have graphed the x[-k] and h[n], the solution saids

y[n] = h[-1]*x[n+1] + h[1]*x[n-1]
= 2x[n+1] + 2x[n-1]

I do not understand why, it seems like the solution forgot about the value at n = 3 for x[n].

can anyone help explain to me what happened in the solution and why n = 3 is not considered in the convolution.

 

[ckutlu]

It is hard to visualize convolution problems. Did you solve it by hand?

 

[Jaynte]

y[n] is longer than what's shown in the answer but most of them are zero. that's why n=3 is not in the answer because that prduct is zero. you can solve the product of sum with -1 \leq n \leq 4 and see what you get

 

[Ecthelion]

Would you mind posting your solution? Graphing convolution problems - particular with a system consisting of delays - generally shows the solution relatively simply. Try showing the x, h, and then as they overlap. You'll see, as Jaynte pointed out, that the product at n=3 is zero, therefore negating the need to write it out.