# 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]. – MackTuesday Apr 15 at 3:49
• It's the output of the LTI system – DrustZ Apr 15 at 19:44
• Of course. My mistake. – MackTuesday Apr 15 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. – Nitzan Apr 17 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? – DrustZ Apr 18 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).