Understanding Convolution in Discrete Time: Solving the Homework Problem

FrogPad
Messages
801
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
FrogPad said:
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)
 
omg...

haha


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

thanks man :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top