I am currently trying to solve a convolution problem analytically
For the discrete-time input signal x(n) and the step response h(n)
$$ x(n) = \left. \begin{cases} 1, & \text{for } 0 \leq n \leq 9 \\ 0, & \text{otherwise} \end{cases} \right\} $$
$$ h(n) = \left. \begin{cases} 0.9^{n}, & \text{for } n \geq 0 \\ 0, & \text{otherwsie} \end{cases} \right\} $$
The convolution for y(n) is as follows: $$ y(n) = \sum_{k=-\infty}^{\infty} h(k) x(n-k) $$
I divided the problem into 3 cases
Case 1: $ n < 0 $. There would be no overlap between h(k) and x(n-k) so y(n) = 0
Case 2: $ 0 \leq n \leq 9 $. I resolved it using the geometric series identity. $$ y(n) = \sum_{k=0}^{n} 0.9^{k} = \frac{1-0.9^{n+1}}{1-0.9} $$ Case 3: $ n \geq 9 $. $$ y(n) = \sum_{k=n-9}^{n} 0.9^{k} $$
Now I am stuck in resolving this summation $ y(n) = \sum_{k=n-9}^{n} 0.9^{k} $. I know that this summation will always have 9 terms, but what identity can I use to get an analytical expression for this convolution?
I saw a solution once for it that resolves it as $ y(n) = 0.9^{n} \sum_{k=0}^{9} 0.9^{-k} $. However, I am not sure if this is correct and I don't know how it got to that shape.
Any thoughts? Thanks!
After a look at this post: Non-infinite geometric sum; does not start at 0 or 1
I found the answer!
The sum $$ y(n) = \sum_{k=n-9}^{n} 0.9^{k} $$
can be expressed as $$ y(n) = \sum_{k=n-9}^{n} 0.9^{k} = \sum_{k=0}^{n} 0.9^{k} - \sum_{k=0}^{n-9} 0.9^{k} $$
which then can be resolved as $$ y(n) = \frac{1 - 0.9^{n+1}}{1 - 0.9} - \frac{1-0.9^{n}}{1-0.9} = \frac{-0.9^{n+1}+0.9^{n}}{0.1} $$