How to convolve h[n] = δ(n) + δ(n-1) with itself

[FrogPad]

The following is all in discrete time, n is an integer

We are given that:
$$h_2(n) = \delta ( n ) + \delta ( n-1 )$$

I want to find the convolution of h2[n]*h2[n].

I don't really understand how to solve this properly.

So,
$$y(n) = \Sigma_{k=-\infty}^{k=\infty} (\delta(n)+\delta(n-1)) \times (\delta(n-k)+\delta(n-k-1))$$

So the $$(\delta(n)+\delta(n-1))$$ pulls out because it is constant.

So,

$$y(n) = (\delta(n)+\delta(n-1)) \Sigma_{k=-\infty}^{k=\infty} \delta(n-k)+\delta(n-k-1)$$

How do I even solve this?
The book gets
h_2(n)*h_2(n) = $$\delta(n) + \2\delta(n-1) + \delta(n-2)$$

I don't understand how they get this.

 

[D H]

$$y(n) = \Sigma_{k=-\infty}^{k=\infty} (\delta(n)+\delta(n-1)) \times (\delta(n-k)+\delta(n-k-1))$$

So the $$(\delta(n)+\delta(n-1))$$ pulls out because it is constant.

You are not calculating the convolution correctly.

$$(h_2\ast h_2)(n) = \sum_{k=-\infty}^{\infty}\cdot h_2(k) h_2(n-k)$$

 

[FrogPad]

omg...

haha


I'm going to to take a walk. That was a ridiculous mistake.

thanks man :)