# How can I simplify this Convolution through a limit change?

1. Feb 12, 2013

### ElijahRockers

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

Compute the convolution y[n] = x[n] * h[n]

if discrete signals:
$x[n] = \alpha^nu[n]$
$h[n] = \beta^nu[n]$

Where $\alpha \neq \beta$.

2. Relevant equations

$y[n] = \sum_{k=-\infty}^{\infty} h[n-k]x[k]$

3. The attempt at a solution

I plugged the two equations into the above convolution formula and got

$y[n] = \sum_{k=-\infty}^{\infty} \beta^{h-k}u[n-k]\alpha^ku[k]$

Simplifying, I get

$y[n] = \beta^n \sum_{k=-\infty}^{\infty} (\frac{\alpha}{\beta})^ku[n-k]u[k]$

And that's as far as I understand. The solution is supposed to be:

$y[n] = \beta^n \sum_{k=0}^n(\frac{\alpha}{\beta})^k , n\geq0$
$y[n] = (\frac{\beta^{n+1} - \alpha^{n+1}}{\beta - \alpha})u[n] , \alpha \neq \beta$

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.

2. Feb 12, 2013

### Ray Vickson

Are the equations $x[n] = \alpha^nu[n], \; y[n] = \beta^nu[n]$ supposed to hold for all integer n from -∞ to +∞, or are they supposed to hold only for integer n ≥ 0? (The answer makes a BIG difference.)

3. Feb 12, 2013

### rude man

Why? The u(n) says n > = 0 only, does it not?
.

4. Feb 12, 2013

### rude man

Z transforms, anyone? Hint: anu[n] → 1/(1 - az-1).

5. Feb 12, 2013

### jbunniii

It seems to me that you're almost there. You just need to take into account the values of $k$ for which $u[k]$ and $u[n-k]$ are nonzero, and change the limits on the sum accordingly. Assuming the definition
$$u[k] = \begin{cases} 1 & \textrm{ if }k \geq 0 \\ 0 & \textrm{ otherwise} \\ \end{cases}$$
then for what values of $k$ is $u[n-k]$ nonzero? And then what about the product $u[k]u[n-k]$?

6. Feb 12, 2013

### ElijahRockers

u[n-k] = 1 when k is ≤ n
u[k] is 1 when k ≥ 0

ok, so k only matters from 0≤k≤n, so i see where the limits are.... and 1 times 1 is 1....

ok. its clicking.

but why is the restriction on n≥0?

Also, I used a formula for the sum when A/B ≠ 1 to get

$\beta^n(\frac{1-(\frac{\alpha}{\beta})^{n+1}}{1-\frac{\alpha}{\beta}})$

But... is it just a matter of simplifying this to get the expression given as the solution?

7. Feb 12, 2013

### jbunniii

This is because if $n < 0$, there are no values of $k$ which satisfy $0 \leq k \leq n$, or putting it another way, $u(k)u(n-k) = 0$ for all $k$. Therefore $y[n] = 0$ if $n < 0$. This is why there is a $u[n]$ in the expression for the solution.
Yes. Try multiplying through by $\beta^n$, and clearing denominators.

8. Feb 12, 2013

### ElijahRockers

Ok, after grinding my gears for a bit, I get why n>0 for the first expression, duh, the sum upper limit cant be less than the lower limit,

but where does the u[n] come from in the second expression?

9. Feb 12, 2013

### jbunniii

By the way, as you noted, your expression is invalid if $\alpha = \beta$. But there is a valid solution if $\alpha = \beta$. For completeness, you might consider finding a separate formula for that special case.

10. Feb 12, 2013

### ElijahRockers

That's part B), already figured that one out,

y[n] = (n+1)αn...

but the solution has u[n] also, just like in part A).... not sure where either of those comes from.

11. Feb 12, 2013

### jbunniii

See the remark I made above regarding why the expression only holds for $n \geq 0$:

12. Feb 12, 2013

### ElijahRockers

Ok, that makes some kind of sense... it's kind of hard to visualize what's actually happening but I see where you are coming from.