# Finding an impulse response to an inverse system of that of which an impulse reponse is given

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.

• If $H(z)$ is a filter, $1/H(z)$ is its inverse filter. $H(z)$ is the z-transform of $h[n]$. Commented May 17, 2014 at 3:48
• 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) Commented May 17, 2014 at 5:13

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.

• A brilliant and well formed answer and interpretation. Thank you for your time and answer. Commented May 17, 2014 at 22:07
• 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 $u[n]$ Commented May 18, 2014 at 5:25
• @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! Commented May 18, 2014 at 15:31
• 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 $n=0$. Commented May 18, 2014 at 15:39
• Thank you for the clarification. You are very observant. Commented May 19, 2014 at 2:28