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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.

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  • $\begingroup$ If $H(z)$ is a filter, $1/H(z)$ is its inverse filter. $H(z)$ is the z-transform of $h[n]$. $\endgroup$
    – Batman
    Commented May 17, 2014 at 3:48
  • $\begingroup$ The second part is about the inverse system, not an inverse filter. Also, must be done without Laplase (dont know if it is called for here but in any case) $\endgroup$
    – SteelSoul
    Commented May 17, 2014 at 5:13

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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.

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  • $\begingroup$ A brilliant and well formed answer and interpretation. Thank you for your time and answer. $\endgroup$
    – SteelSoul
    Commented May 17, 2014 at 22:07
  • $\begingroup$ one problem though. For the solution of A, n doesnt cancel out plus A is supposed to be an integer (whole) number. When plugging in, the equation is $A=\frac{\left(\frac{1}{5}\right)^nu[n]-\delta [n]}{\left(\frac{1}{5}\right)^{n-1}u[n-1]}$.. Although $u[n]-\delta [n]=u[n-1]$ There is a coefficient to $u[n]$ $\endgroup$
    – SteelSoul
    Commented May 18, 2014 at 5:25
  • $\begingroup$ @SteelSoul: of course n cancels out! Check your math. For $n > 0$, we have $A = (1/5)^n / (1/5)^{n-1} = 1/5$. For $n=0$ we get $0/0$ which essentially means $A$ can be arbitrary. So $A=1/5$ is a solution. This is easily verified by plugging in: $\left(\frac{1}{5}\right)^n u[n] - \left(\frac{1}{5}\right) \left(\frac{1}{5}\right)^{n-1} u[n-1] = \left(\frac{1}{5}\right)^n (u[n] - u[n-1]) = \left(\frac{1}{5}\right)^n \delta[n] = \delta[n]$! Please check your math! $\endgroup$
    – Andreas H.
    Commented May 18, 2014 at 15:31
  • $\begingroup$ Maybe the point you did not realize is that $(1/5)^n \delta[n] = \delta[n]$. This is just because $\delta[n]$ is zero almost everywhere (so the equation is correct at $n \neq 0$). And at $n=0$ we have $(1/5)^0 = 1$, which makes it correct also for $n=0$. $\endgroup$
    – Andreas H.
    Commented May 18, 2014 at 15:39
  • $\begingroup$ Thank you for the clarification. You are very observant. $\endgroup$
    – SteelSoul
    Commented May 19, 2014 at 2:28

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