I need to solve a bunch of these questions and would appreciate a solid direction.
I have an impulse response $h[n]=\left(\frac{1}{5}\right)^nu[n]$. Firstly we are to find a whole number A which satisfies $h[n]-A h[n-1]=\delta [n]$.
After it is found we need to find the impulse response of the inverse system to the original one.
I dont know how to find the original system from the data given.
In regards to the first part, I know that the discrete unit impulse is 1 when n=0 and 0 otherwise and I gathered that A should be such that it satisfies this structure but have failed to find it.
Thank you for your attention.
Well regarding the first part, just solve for $A$:
$A = (h[n]-\delta[n]) / h[n-1]$
Ok, in general case you would not know that the right hand side is independent of $n$. But since the question is already tells us, that such an $A$ exists, it will be found by just solving. When you plug everything in, $n$ cancels out and you obtain $A=1/5$.
For the second part, on how to find the inverse, just recall that the inverse to $h[n]$ is defined by
$h[n] * g[n] = \delta[n]$
So the task is to find $g[n]$. We notice that this equation looks already very similar to the equation in your question (i.e. $h[n] - A h[n-1] = \delta[n]$) so we should think how to convert it to the form.
The answer is easily found by noticing that $h[n] = h[n] * \delta[n]$ and $h[n-1] = h[n] * \delta[n-1]$. So by plugging this in and rearranging, we get
$h[n] * (\delta[n] - A \delta[n-1]) = \delta[n]$
which yields the inverse
$g[n] = \delta[n] - A \delta[n-1]$
Maybe a word of interpretation: The system in your original question resembles a first order lowpass, because it just has a exponentially decaying impulse response. The inverse system is just a first order highpass, which is just given by the system $g[n]$. Since a first order highpass amplifies the detail of a signal, while ignoring the average value (i.e. the opposite of a lowpass), it makes sense that the $g[n]$ outputs the difference of consecutive input samples with some weighting.