Discrete convolution evalutation.

[perplexabot]

Hello all. I have a homework 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.

Homework Statement

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

Homework Equations

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

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.

 

[haruspex]

What's the sequence u?

 

[perplexabot]

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

 

[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?

 

[haruspex]

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

Ok. u[0] = 0.5?

 

[perplexabot]

no, u[0] = 1.

 

[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.