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):

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

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.

  • $\begingroup$ Is $\sigma$ an impulse? I ask because I've seen that being used for a step. (and a discretre pulse is typically denoted $\delta[n]$) $\endgroup$ Sep 17, 2020 at 8:28
  • $\begingroup$ 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. $\endgroup$
    – Florian
    Sep 17, 2020 at 8:45
  • $\begingroup$ @MarcusMüller missed that - I edited my post. It is an impulse. $\endgroup$
    – dnclem
    Sep 17, 2020 at 10:28

1 Answer 1



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.


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