- #1

ElijahRockers

Gold Member

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## 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.