Let $x[n]$ be the input and $y[n]$ be the output of an LTI system. That is,

$y[n]=\displaystyle\sum_{k=-\infty}^{\infty} h[n-k]\space x[k]$

where $h[\cdot]$ is the impulse response function. We are interested in the value of:

$z[n] = \displaystyle\sum_{j=n}^{\infty} y[j]$

Are there methods for the analysis and computation of the value of $z[n]$ given some assumptions about $h[\cdot]$? Namely, it seems clear that the impulse response must decay sufficiently fast in order for the infinite sum to converge.


2 Answers 2


That depends a bit on what you know about the input and the impulse response. If both a are known, you can just calculate the output directly and be done with it. If you know nothing you can't predict anything.

This being said, we can make a few observations. In essence, $z[n]$ is just the integral over $y[n]$. To makes things a little easier to write, we assume that $h[n]$ is causal, i.e. $h[n] = 0, n < 0$ and we can rewrite the convolution as

$$y[n] = \sum_{k = 0} ^{\infty}h[k]\cdot x[n-k]$$

If both sums converge, we can swap integral and convolution and can write this

$$z[n] = \sum_{k = 0} ^{\infty}g[k] \cdot x[n-k]$$

where the impulse response $g[n]$ is the integral over $h[n]$, i.e.

$$g[n] = \sum_{k=0}^{n} h[k]$$

In other words, $g[n]$ is the step response of the system.

Convergence can indeed be an issue. The convolution will always converge if the impulse response is absolutely summable, i.e.

$$\sum_{k=0}^{\infty} |h[k]| < \infty$$

Since an integrator has infinite gain at DC, the step response will only converge of $h[n]$ is mean free. Even if $h[k]$ is NOT mean-free the sum for $z[n]$ might still converge if $x[n]$ is mean free.

  • $\begingroup$ Thank you for your help! My naive understanding of the step response is that step response functions settle at a stable value (say 1). If so, it seems that as $k$ becomes large, we are effectively summing past values of $x[n-k[$ since $g[k]\approx 1$. Is this understanding correct? If so, I am struggle to understand that result intuitively. $\endgroup$
    – singular
    Oct 21, 2023 at 16:35
  • $\begingroup$ I have a few doubts: the way you defined $g[n]$, it doesn't actually depend on $n$ (but it should), and the actual $g[n]$ is not equal to the step response (but it can be expressed by the step response). $\endgroup$
    – Matt L.
    Oct 21, 2023 at 19:59
  • $\begingroup$ Between the surprisingly unintuitive consequence of the result in the answer and the comment above, I've decided to hold off accepting this answer until there is some clarification of the result. Thank you all for the help! $\endgroup$
    – singular
    Oct 22, 2023 at 13:54
  • $\begingroup$ That was certainly a stupid typo on my part. The upper bound was supposed to be $n$ not $\infty$. I hope this is better now $\endgroup$
    – Hilmar
    Oct 23, 2023 at 0:30
  • $\begingroup$ If you assume that the system described by $h[n]$ is causal, the step response is also causal, so the relation between $z[n]$ and $x[n]$ as you describe it must also be causal. But that's a contradiction because the system in the question is clearly non-causal. I've written up an answer to clarify this. $\endgroup$
    – Matt L.
    Oct 23, 2023 at 10:00

Assuming that all the sums converge - which depends on the nature of the impulse response $h[n]$ - you can write the relation between $z[n]$ and $x[n]$ as a convolution, as correctly pointed out in Hilmar's answer. First, note that $z[n]$ is the convolution of $y[n]$ with $u[-n]$, where $u[n]$ is the unit step sequence:


And since $y[n]$ is the convolution of $x[n]$ and $h[n]$, $z[n]$ can be written as the convolution of $x[n]$ with a sequence $g[n]$ defined as the convolution of $h[n]$ with $u[-n]$:



Note that $g[n]$ is not the system's step response. However, it is related to the step response




where we assume that $a[\infty]$ exists. This is equivalent to assuming that the system's frequency response at DC is finite:


The sequence $g[n]$ relating the input $x[n]$ to the sequence $z[n]$ is generally non-causal, i.e., $g[n]$ does not vanish for $n<0$. If $h[n]$ is FIR then $g[n]$ will vanish for a certain index $n>n_0$, where $n_0$ depends on the support of $h[n]$. If $h[n]$ is causal and FIR with length $N$, then $g[n]=0$ for $n\ge N$. In that case, for $-\infty<n\le 0$ we have $g[n]=a[\infty]=a[N-1]$, i.e., $g[n]$ is constant for non-positive $n$.

  • $\begingroup$ I think for the third equation down you meant $k=n$ and not $k-\infty$? $\endgroup$
    – singular
    Oct 23, 2023 at 11:35
  • $\begingroup$ @singular: I actually meant $k=-\infty$. $\endgroup$
    – Matt L.
    Oct 23, 2023 at 12:21

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