# Output of a stable LTI system (discrete)

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)?

• Looks like you left out the definition of y[n]. Apr 15, 2021 at 3:49
• It's the output of the LTI system Apr 15, 2021 at 19:44
• Of course. My mistake. Apr 15, 2021 at 23:32
• @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. Apr 17, 2021 at 16:42
• 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? Apr 18, 2021 at 0:34

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 $$a^2 + b^2 \geq 2ab$$)

Then

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

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

So

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

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