How Do You Solve a Discrete Convolution Sum with Step Functions?

[Xyius]

Homework Statement

Find $x[n] \ast h[n]$ when $x[n] = 3 u[2-n]$ and $h[n] = 4\left( \frac{1}{2} \right)^{n+2}u[n+4]$ where $u[n-k]$ is the unit step function.

Homework Equations

None really

The Attempt at a Solution

So I know this is probably simple but I am confused.

So the convolution is written as,

$$\sum_{k=-\infty}^{\infty} 4\left( \frac{1}{2} \right)^{k+2}u[k+4](3)u[2-n+k]$$

This simplifies to,

$$3\sum_{k=-\infty}^{\infty} \left( \frac{1}{2} \right)^{k}u[k+4]u[2-n+k]$$

The first step function of $u[k+4]$ means that the terms in the sum corresponding to $k < -4$ are zero, therefore this becomes the bottom limit.

$$3\sum_{k=-4}^{\infty} \left( \frac{1}{2} \right)^{k}u[2-n+k]$$

This form is confusing to me. I don't see an obvious limit on k except for that its LOWER limit is $n-2$. What I tried to do is do some change of variables, such as $k'=2-n+k$, this changes the lower limit to -2-n and the upper limit to infinity. So I have...

$$3\sum_{k'=-2-n}^{\infty} \left( \frac{1}{2} \right)^{k'-2+n}u[k']$$

So this means that $k'$ must be larger than zero, which means $n < -2$. So in this regime, the step function is 1 and I get..

$$3\sum_{k'=0}^{\infty} \left( \frac{1}{2} \right)^{k'-2+n}=3\left( \frac{1}{2} \right)^{n-1}$$

Am I doing this correctly? I am just not sure. Thank you in advance for anyone that takes the time to read this!

 

[fresh_42]

Homework Statement

Find $x[n] \ast h[n]$ when $x[n] = 3 u[2-n]$ and $h[n] = 4\left( \frac{1}{2} \right)^{n+2}u[n+4]$ where $u[n-k]$ is the unit step function.

Homework Equations

None really

The Attempt at a Solution

So I know this is probably simple but I am confused.

So the convolution is written as,

$$\sum_{k=-\infty}^{\infty} 4\left( \frac{1}{2} \right)^{k+2}u[k+4](3)u[2-n+k]$$

This simplifies to,

$$3\sum_{k=-\infty}^{\infty} \left( \frac{1}{2} \right)^{k}u[k+4]u[2-n+k]$$

The first step function of $u[k+4]$ means that the terms in the sum corresponding to $k < -4$ are zero, therefore this becomes the bottom limit.

$$3\sum_{k=-4}^{\infty} \left( \frac{1}{2} \right)^{k}u[2-n+k]$$

This form is confusing to me. I don't see an obvious limit on k except for that its LOWER limit is $n-2$. What I tried to do is do some change of variables, such as $k'=2-n+k$, this changes the lower limit to -2-n and the upper limit to infinity. So I have...

$$3\sum_{k'=-2-n}^{\infty} \left( \frac{1}{2} \right)^{k'-2+n}u[k']$$

So this means that $k'$ must be larger than zero, which means $n < -2$. So in this regime, the step function is 1 and I get..

$$3\sum_{k'=0}^{\infty} \left( \frac{1}{2} \right)^{k'-2+n}=3\left( \frac{1}{2} \right)^{n-1}$$

Am I doing this correctly? I am just not sure. Thank you in advance for anyone that takes the time to read this!

I don't see how the substitution is any improvement. Since both step functions have to be simultaneously $1$ I get:
$$3\sum_{k=max\{-4, n-2\}}^{\infty} \left( \frac{1}{2} \right)^{k}$$
which leaves you two cases. Is $n$ a natural number or another integer?

 

[Xyius]

I don't see how the substitution is any improvement. Since both step functions have to be simultaneously $1$ I get:
$$3\sum_{k=max\{-4, n-2\}}^{\infty} \left( \frac{1}{2} \right)^{k}$$
which leaves you two cases. Is $n$ a natural number or another integer?

$n$ is an integer. And I believe must be smaller than -2 from what I showed here. So that would mean the max value in the lower limit should be -4? Then I could just expand out until k=0 and use a geometric sum.

 

[fresh_42]

$n$ is an integer. And I believe must be smaller than -2 from what I showed here. So that would mean the max value in the lower limit should be -4? Then I could just expand out until k=0 and use a geometric sum.

Yes. Except that I'm not certain about your assumption on $n$. But that will only give you a factor depending on $n$. If you can calculate it for $k = -4$ you should as well be able to calculate it for $k=n-2$ regardless what $n$ is.