- #1
ElijahRockers
Gold Member
- 270
- 10
Homework Statement
Compute the convolution y[n] = x[n] * h[n]
if discrete signals:
[itex] x[n] = \alpha^nu[n][/itex]
[itex] h[n] = \beta^nu[n][/itex]
Where [itex]\alpha \neq \beta[/itex].
Homework Equations
[itex]y[n] = \sum_{k=-\infty}^{\infty} h[n-k]x[k][/itex]
The Attempt at a Solution
I plugged the two equations into the above convolution formula and got
[itex]y[n] = \sum_{k=-\infty}^{\infty} \beta^{h-k}u[n-k]\alpha^ku[k][/itex]
Simplifying, I get
[itex]y[n] = \beta^n \sum_{k=-\infty}^{\infty} (\frac{\alpha}{\beta})^ku[n-k]u[k][/itex]
And that's as far as I understand. The solution is supposed to be:
[itex]y[n] = \beta^n \sum_{k=0}^n(\frac{\alpha}{\beta})^k , n\geq0[/itex]
[itex]y[n] = (\frac{\beta^{n+1} - \alpha^{n+1}}{\beta - \alpha})u[n] , \alpha \neq \beta[/itex]
But I'm not sure how to get there. I see that the change in limits somehow facilitates this, but I just can't wrap my head around how to arrive at that, especially the part where α≠β.
I see that for each n, k goes through all possible values and sums each result, but I don't see why the lower limit for the solution is 0, or the upper limit is n.
There's obviously something fundamental about convolutions that I'm not getting.