Discrete convolution evalutation.

1. Jan 23, 2013

perplexabot

Hello all. I have a hw question but since I have no idea how to go about solving it I have started with an exercise problem from the book (with the solution and vague steps provided). Here is my attempt of the exersize question.

1. The problem statement, all variables and given/known data

Compute the convolution y[n] = x[n] * h[n]
where:
x[n] = u[n - 4] (-1/2)^n
h[n] = u[2 - n] 4^n

2. Relevant equations

y[n] = summation of x[k]h[n - k] from k = -inf to k = inf

3. The attempt at a solution
Sketching the graphs, I was able to realize that there are two intervals. n <= 6 and n> 6. I continue on and imagine "sweeping" or shifting the h[n] function over x[n] for the n <= 6 region:
summation of x[n]h[n - k] from k = -inf to k = inf
=> summation of ( (-1/2)^k ) * ( 4^(n-k) ) from k = -inf to k = 6 (where "*" means multiply)
=> solving the summations using a calculator yields ( (.5^n)/3 ) - ( (2^n)(4/3) )

I could continue on to solve for n > 6 region but there is no point since my convolution is wrong. Can anyone lead me to the right path? I will be working on it in the mean time. Thank you.

Note: I have attached an image of the solution for convenience. I do not understand it.

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Last edited: Jan 23, 2013
2. Jan 23, 2013

haruspex

What's the sequence u?

3. Jan 23, 2013

perplexabot

Sorry, I should have mentioned that u[n] is a unit step function.

4. Jan 23, 2013

perplexabot

I don't understand why the second summation upper limits go to 3 and (n - 1) for both equations. I would have thought 4 and n. Anyone?

Last edited: Jan 23, 2013
5. Jan 23, 2013

haruspex

Ok. u[0] = 0.5?

6. Jan 23, 2013

perplexabot

no, u[0] = 1.

7. Jan 23, 2013

haruspex

In that case I get the given answer for n <= 6.
Where we diverge is in the summation range for k. I have k > 3.