I have a continuous system described as:


and I'm trying to understand why the system is stable. I know that a system is unstable when you provide an input and as a result the output is not bounded. So, when I looked at the system I set $t=\infty$ and assumed that the output goes to infinity because I'd have

$$ y(\infty)=2x(\infty)+0.5x(\infty-2)+0.25x(\infty+2)$$

making it unbounded. Looking at this equation, isn't the output going to infinity resulting in an unstable system? The solutions state it's stable and I can't see why. Thanks

  • 1
    $\begingroup$ It ain't what you don't know that will kill you; it is what you know that just ain't so. BIBO stability is not defined in the way you assert it is. If you assume tha $x(t)$ is bounded for all choices of $t$ (and so, just so's you don't misunderstand, so are $x(t-2)$ and $x(t+2)$ bounded), meainng that there is a finite number $M$ such that $|x(t)| \leq M$ (ditto $|x(t-2)|$ and $|x(t+2)|$ $\leq M$) for all $t$, and you can prove that there is a finite number $M^\prime$ such that $|y(t)|\leq M^\prime$ for all $t$, then the system is BIBO stable. Try your system for $M^\prime = 2.75M$. $\endgroup$ – Dilip Sarwate Jul 11 at 20:42
  • $\begingroup$ Bound too tight youtube.com/watch?v=stlKHh_f0-0 $\endgroup$ – Laurent Duval Jul 11 at 20:50
  • 1
    $\begingroup$ As someone as said, BIBO stable doesn't have to do with $t \rightarrow \infty$, you can read this article en.wikipedia.org/wiki/BIBO_stability or any text to get that part. Hopefully using the right definition of BIBO stable you can head down the correct path! $\endgroup$ – Engineer Jul 11 at 20:51

Triangle inequality is your friend: if $y = a_1x_1+ a_2x_2+ a_3x_3 $, then $$|y| \le |a_1||x_1|+ |a_2||x_2|+ |a_3||x_3|$$ Hence if $x_{.}$ is bounded, the sum is bounded as well.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.