Topic: Stability of a System

## Question

The input x(t) and the output y(t) of a system are related by the equation

$y(t)=\frac{ {\color{red} t }}{1+x^2(t)}.$

OOPS, I actually meant to put a "t" on top of the fraction (now in red). -pm

This system is stable. I'm actually not sure how to show this, does the following logic work?

$\lim_{x(t) \to 0}\frac{1}{1+x^2(t)} = 1$ and $\frac{1}{1+x^2(t)} < 1$ for all x(t), thus the system is stable.

I'm not sure that the justification works here...

--Cmcmican 17:44, 24 January 2011 (UTC)

Unfortunately no. Here is how you should go about answering such questions. If you think it is stable,
then assume that x(t) is bounded (i.e., |x(t)|<m ) and then try to show that y(t) is also bounded (|y(t)<M ).
If you think it is not stable, then try to think of a bounded signal x(t) for which y(t) would not be bounded.

Hint for this case: Look at the constant signal x(t)=1. -pm

Now that it has a t on top, it's not bounded.

If you consider the constant signal x(t)=1, then $y(t) = \frac{{t }}{1+1^2} = \frac{{t }}{2}$, which is not bounded.

--Cmcmican 19:26, 24 January 2011 (UTC)

Good! And what if there was no t on top? -pm

If there is not a t on top (i.e it is back to being a '1'), then the signal is bounded*.

Considering the case where $|x(t)| \le \infty$ then $0<\frac{{1}}{1+x^2(t)}\le1$.

$\therefore y(t)$ is bounded by $M = \pm 1$

*Addendum: This only works for $x(t) \in \Re$ as there are imaginary values that cause it to be unstable.

--Darichar 14:05, 26 January 2011 (UTC)

TA's comment: BIBO stability requires that the response doesn't diverge for any bounded input signal, including complex signals. Therefore, we just say that this system is unstable.

--Ahmadi 22:00, 27 January 2011 (UTC)

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