# impulse response of a causal LTI system

This is a difference equation to a causal LTI system:

$$y[n] = ay[n - 1] + x[n] - a^Nx[n - N]$$

Where N is a positive integer. I need to determine the impulse response of the system, so I have the equation:

$$h[n] = ah[n-1] + \delta[n] - a^N\delta[n - N]$$

h[n] is simple to find for the case where n < 0, n = 0, or n = 1. However, the response could be different for the case n = 2, depending on the value of N (in this case, N = 1 or N = 2):

$$h = ah + 0 - a^1\delta[2-1] = a^2$$
$$h = ah + 0 - a^2\delta[2-2] = 0$$

So which one is it? Intuition tells me the top one as the impulse response will be decaying, but we only know that N is positive.

• Is $\sigma$ an impulse? I ask because I've seen that being used for a step. (and a discretre pulse is typically denoted $\delta[n]$) Sep 17, 2020 at 8:28
• Well, of course the impulse response depends on $N$. It's a parameter, after all. Like $a$, which it also depends on. You can start looking at the impulse response for a large value of $N$ and then see what happens when you decrease it. Sep 17, 2020 at 8:45
• @MarcusMüller missed that - I edited my post. It is an impulse. Sep 17, 2020 at 10:28

Just write down the values of the output signal $$y[n]$$ for $$x[n]=\delta[n]$$ for values of $$n$$ from $$0$$ to $$N-1$$ (you don't need any specific value for $$N$$, just use $$n=0,1,2,\ldots$$ and then you'll see what happens at $$n=N-1$$). Then figure out what happens at $$n=N$$, and what consequences this has on the output values for $$n>N$$. You may be surprised to find out that the given system has a finite length impulse response, even though it is recursive.