Consider an LTI (linear and time invariant) system that is BIBO (bounded input bounded output) stable and is such that $x[n] = 0$ for all $n < 0$ (note: this is sometimes referred to as a relaxed system). Show that

  1. Suppose that $x[n] = x_0[n] + au[n]$ for some constant a where $x_0[n]$ is a bounded signal with $$\lim_{n\rightarrow\infty} x_0[n] = 0.$$ Then $x[n]$ is bounded and tends to a constant (namely, $a$). Show that the output $y[n]$ will also tend to a constant.

  2. Suppose that $x[n]$ has finite energy. Show that $y[n]$ will also have finite energy.

  • For 1, I'm kinda stuck at the step proving $$\sum_{k=-\infty}^{\infty}|x_0[n]h[n-k]| \rightarrow C $$ where $C$ is a constant. I can prove that it's below ($\le$) a constant, but how do I prove it is actually approach to it when $n \rightarrow \infty$?

  • For 2, can we do it without using the Fourier transform (continuous version here)?

  • 1
    $\begingroup$ Looks like you left out the definition of y[n]. $\endgroup$ Apr 15, 2021 at 3:49
  • $\begingroup$ It's the output of the LTI system $\endgroup$
    – DrustZ
    Apr 15, 2021 at 19:44
  • $\begingroup$ Of course. My mistake. $\endgroup$ Apr 15, 2021 at 23:32
  • $\begingroup$ @DrustZ For [1] take a look here on "Discrete-time sufficient condition" section, this is the way to prove it. For [2], you don't have to use fourier transform, but it's much easier since with fourier you multiply the input and the impulse response instead of convolving them. $\endgroup$
    – Nitzan
    Apr 17, 2021 at 16:42
  • $\begingroup$ Thanks @Nitzan, I got the idea of using the sufficient condition to prove that $y[n]$ is less or equal than a constant for [1], but I wonder that's equivalent to $y[n]$ will approach a constant? $\endgroup$
    – DrustZ
    Apr 18, 2021 at 0:34

1 Answer 1


So I'm answering my question here; turned out that I have to solve (2) then solve (1).

  1. To prove $y[n]$ has finite energy, we write

$$\sum_{n=-\infty}^{\infty}y^2[n] = \sum_{n=-\infty}^{\infty}(\sum_{k=-\infty}^{\infty}h[k]x[n-k])^2 = \sum_{n=-\infty}^{\infty}(\sum_{k=-\infty}^{\infty}h[k]x[n-k])(\sum_{l=-\infty}^{\infty}h[l]x[n-l])$$

$$ = \sum_{k=-\infty}^{\infty}\sum_{l=-\infty}^{\infty}h[k]h[l]\sum_{n=-\infty}^{\infty}x[n-k]x[n-l]$$

Now we know that $x[n]$ has finite energy, so

$$ \sum_{n=-\infty}^{\infty}x[n-k]x[n-l] \leq \sum_{n=-\infty}^{\infty}x^2[n] = E$$

(a generalization of $a^2 + b^2 \geq 2ab$)


$$ \sum_{n=-\infty}^{\infty}y^2[n] \leq E\sum_{k=-\infty}^{\infty}|h[k]|\sum_{l=-\infty}^{\infty}|h[l]| $$

For a BIBO system, $$\sum_{l=-\infty}^{\infty}|h[l]| \leq M \leq \infty$$

So we approve $$\sum_{n=-\infty}^{\infty}y^2[n] \leq M^2E \leq \infty$$

  1. To prove $x[n]$ tends to a constant:

$$y[n] = a\sum_{k=-\infty}^{\infty}h[k]u[n-k] + \sum_{k=-\infty}^{\infty}h[k]x_0[n-k] = a\sum_{k=-\infty}^{\infty}h[k]+y_0[n]$$

Using the conclusion of 1, we know that $$\sum_{n=-\infty}^{\infty}x^2[n] < \infty \rightarrow \sum_{n=-\infty}^{\infty}y^2[n] < \infty$$

which implies that $$\lim_{n \rightarrow \infty} |y_0[n]| = 0.$$


$$ \lim_{n \rightarrow \infty} y[n] = a \times \lim_{n \rightarrow \infty} \sum_{k=0}^{\infty}h[k] = ac$$

where $c$ is a constant (because of BIBO).


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