1. ## limits of summation

This problem is in the context of discrete systems and signals.

My goal is to find an equation for y(n) without the summation.

$y(n)=\sum_k k(n-k)u(k-4)u(k+2-n)$

where $u(k)$ is the unit step function

So I need to eliminate the unit step functions from the equation by changing the limits of summation:

if $k-4\geq 0$ then $u(k-4)=1$. And so it's eliminated from my summation above.
if $k+2-n\geq 0$ then $u(k+2-n)=1$. And it's also eliminated.

So my limits of summation are $k\geq 4$ and $k\geq n-2$. But the limits don't define a finite range! Can anyone help?

2. Originally Posted by jut
This problem is in the context of discrete systems and signals.

My goal is to find an equation for y(n) without the summation.

$y(n)=\sum_k k(n-k)u(k-4)u(k+2-n)$

where $u(k)$ is the unit step function

So I need to eliminate the unit step functions from the equation by changing the limits of summation:

if $k-4\geq 0$ then $u(k-4)=1$. And so it's eliminated from my summation above.
if $k+2-n\geq 0$ then $u(k+2-n)=1$. And it's also eliminated.

So my limits of summation are $k\geq 4$ and $k\geq n-2$. But the limits don't define a finite range! Can anyone help?
Would you please explicitly write the limits of the sum, i.e., is it $\sum_{k=0}^{\infty}\cdot$, $\sum_{k=1}^{\infty}\cdot$ or $\sum_{k=0}^{n-1}\cdot$?
And what you mean by the step function, i.e., $u(n)=\begin{cases}1,n\geq0\\0,n<0\end{cases}$?

3. Sorry, let me restate the original problem and be more explicit.

Find the convolution of x(n) and h(n).
$x(n) = (-1/2)^n u(n-4)$
$h(n) = 4^n u(2-n)$

where $u(n)=\begin{cases}1,n\geq0\\0,n<0\end{cases}$

where the convolution is $y(n)=x(n)\ast y(n)=\sum_{k=-\infty}^{\infty}x(k)h(n-k)$
note the $\ast$ symbol is not multiply, it is the convolution operator. Also y(n) is causal, which means y(n) is only defined for $n\geq 0$, where n=0,1,2,3....

Now,
$x(k) = (-1/2)^k u(k-4)$
$h(k) = 4^{n-k} u(2-n+k)$

Plugging in,
$x(n)\ast y(n)=\sum_{k=-\infty}^{\infty} (-1/2)^k u(k-4) 4^{n-k} u(2-n+k)$

Simplifying,
$x(n)\ast y(n)=4^n\sum_{k=-\infty}^{\infty}(-1/2)^k 4^{-k} u(k-4) u(2-n+k)$

The argument passed into the unit step must be $\geq 1$ or else the value of the unit step function is 0. So if I need to change my limits of summation so that the two unit step functions resolve to 1. That is,

if $k-4\geq 0$ then $u(k-4)=1$.
if $k+2-n\geq 0$ then $u(k+2-n)=1$.

So the limits on k would be,
if $k\geq 4$
if $k\geq n-2$

I am bamboozled by the two lower limits on k. How can I do a summation with two lower limits? Would the summation diverge?

4. Originally Posted by jut
Sorry, let me restate the original problem and be more explicit.

Find the convolution of x(n) and h(n).
$x(n) = (-1/2)^n u(n-4)$
$h(n) = 4^n u(2-n)$

where $u(n)=\begin{cases}1,n\geq0\\0,n<0\end{cases}$

where the convolution is $y(n)=x(n)\ast y(n)=\sum_{k=-\infty}^{\infty}x(k)h(n-k)$
note the $\ast$ symbol is not multiply, it is the convolution operator. Also y(n) is causal, which means y(n) is only defined for $n\geq 0$, where n=0,1,2,3....

Now,
$x(k) = (-1/2)^k u(k-4)$
$h(k) = 4^{n-k} u(2-n+k)$

Plugging in,
$x(n)\ast y(n)=\sum_{k=-\infty}^{\infty} (-1/2)^k u(k-4) 4^{n-k} u(2-n+k)$

Simplifying,
$x(n)\ast y(n)=4^n\sum_{k=-\infty}^{\infty}(-1/2)^k 4^{-k} u(k-4) u(2-n+k)$

The argument passed into the unit step must be $\geq 1$ or else the value of the unit step function is 0. So if I need to change my limits of summation so that the two unit step functions resolve to 1. That is,

if $k-4\geq 0$ then $u(k-4)=1$.
if $k+2-n\geq 0$ then $u(k+2-n)=1$.

So the limits on k would be,
if $k\geq 4$
if $k\geq n-2$

I am bamboozled by the two lower limits on k. How can I do a summation with two lower limits? Would the summation diverge?
I hope I am not doing wrong, my answer is convergent.
Consider the following:
$(x\ast y)(n)=\sum_{k=-\infty}^{\infty}\bigg(-\frac{1}{2}\bigg)^{k}u(k-4)4^{n-k}u(2-n+k)$....for $n\in\mathbb{Z}$.
Clearly, if $k<4$, then $u(k-4)=0$ and if $k, then $u(2-n+k)=0$.
These yields for $n\in\mathbb{Z}$ that
$(x\ast y)(n)=\sum_{k=4}^{\infty}\bigg(-\frac{1}{2}\bigg)^{k}4^{n-k}u(2-n+k)$
..............' $=\sum_{k=\max\{4,n-2\}}^{\infty}\bigg(-\frac{1}{2}\bigg)^{k}4^{n-k}$
..............' $=4^{n}\sum_{k=\max\{4,n-2\}}^{\infty}\bigg(-\frac{1}{2^{3}}\bigg)^{k}$ (a geometric series)
..............' $=\frac{8}{9}4^{n}\bigg(-\frac{1}{2^{3}}\bigg)^{\max\{4,n-2\}}.$
Clearly, the convolution converges and as $n\to\infty$, $(x\ast y)(n)\to{\color{red}{0}}$.

5. Nice solution. Thank you sir!

I was also able to solve it with a bunch more work and sloppy notation... but our solutions agree. I wrote y(n) in piecewise notation.

Originally Posted by bkarpuz
Clearly, the convolution converges but as $n\to\infty$, $(x\ast y)(n)\to\infty$.
Are you sure? For large values of n, the numerator has a $4^n$ and the denominator has a $8^n$ ... which would make it converge to 0.

6. Originally Posted by jut
Nice solution. Thank you sir!
Are you sure? For large values of n, the numerator has a $4^n$ and the denominator has a $8^n$ ... which would make it converge to 0.
Corrected now. :]