Question on convolution

Here is another question my teacher set for me but I can't solve:

**broken link removed**

Of course I could just z transform h(n) and x(n) but my teacher wants me to solve the question without doing so.

So far what I've got is,I changed the range of the summation from k=0 to infinity,I also know that for the h(k) part its u(k) since h(n)=u(n),and we can strike u(k) off as it will always be 1 since k does not go below 0.But after that point I'm kinda stuck,not too sure about the x(n-k) part.

Thanks alot!

Hi,

If you know what x(n) is you should know what x(n-k) is right?

For a different example, if you know that f(x) is x+1 and you want to know what f(m+x) is, you would end up with (m+x)+1.

I'm still abit unsure.But I'm guessing if thats the case then x(n-k) is 2(0.5)^(n-k) * u(n-k)? But if thats the case how do I continue from there?
I'm guessing you sub in values of K and it should be somehow equal to the geometric progression?

Hi again,

It looks like they want you to relate the 2*(0.5)^n part to the solution to the progression which is on the right.
It also looks like the sum parts will be limited to certain values depending on n, because u(n-k) is not always 1, which if it were, would make the series divergent, so u(n-k) sometimes going to 0 must make the convergence possible. So in other words it looks like the sum will only have to go between certain values of k rather than to infinity.

It also might make more sense if you do a couple little examples making n a particular value, like 1, 2, 3, 5 and 10 or something like that. You should see a pattern emerge.

You might also do a search for "Discrete Convolution".

Hmm,I think I get what you're trying to say.I'll give the question another go