While trying to determine the unit step and impulse response for a discrete LTI system I had several problems:

$$ \begin{aligned} (Hx)[n]&=(Hx)[n-1]+\frac{1}{N}(x[n]-x[n-N]) \\x[n]&=(Hx)[n]=0\;\forall\;n\lt0 \end{aligned} $$

Starting with the Impulse response:


Can I argue that because of the condition $x[n]=(Hx)[n]=0\;\forall\;n\lt0$ the impulse response only exists where $n\gt 1\;\&\;n\gt N?$ But what about the other cases?

Concerning the Step response I am quite lost and am not even sure if this is how it is calculated:


Help as to how to proceed and solutions are greatly appreciated


1 Answer 1


This is a homework type problem, so I'll give you a few hints to help you solve it yourself (and learn something while doing so).

From what is given, we can assume that the given difference equation describes a causal discrete-time system. Let $h[n]$ denote the impulse response. It must satisfy

$$h[n]=h[n-1]+\frac{1}{N}\big(\delta[n]-\delta[n-N]\big),\qquad n\ge 0\tag{1}$$

with $h[-1]=0$. Note that the second term on the right-hand side of $(1)$ only contributes to $h[n]$ for $n=0$ and $n=N$. Now start at $n=0$ and continue up to $n=N$ to see what happens. It will turn out that the impulse response has a finite length, i.e., the given system is an FIR filter.

The step response is the convolution of the impulse response with the unit step sequence, which simplifies to


So the step response $a[n]$ is just the cumulative sum of the impulse response $h[n]$.

  • $\begingroup$ The recursive part of the response would be carried over to the next response? Therefore, starting from 0 for the 1st I would have 1/N up until n=N? $\endgroup$
    – dsisko
    Jan 4, 2020 at 16:20
  • $\begingroup$ @dsisko: That's it. At $n=N$ that value gets subtracted so $h[n]$ equals zero for all $n\ge N$. $\endgroup$
    – Matt L.
    Jan 4, 2020 at 16:52
  • $\begingroup$ @Matt L : In this particular case, is the step response just a moving average in the time domain ? $\endgroup$
    – mark leeds
    Jan 6, 2020 at 2:24
  • $\begingroup$ @markleeds: The filter itself is a moving average. The step response increases linearly up to its final value which is just the sum of all impulse response coefficients. $\endgroup$
    – Matt L.
    Jan 6, 2020 at 6:28

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