0
$\begingroup$

I have a continuous system described as:

$$y(t)=2x(t)+0.5x(t-2)+0.25x(t+2)$$

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

Improve this question
$\endgroup$
3
  • 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 $M^\prime = 2.75M$. $\endgroup$ Commented Jul 11, 2020 at 20:42
  • $\begingroup$ Bound too tight youtube.com/watch?v=stlKHh_f0-0 $\endgroup$ Commented Jul 11, 2020 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! $\endgroup$ Commented Jul 11, 2020 at 20:51

1 Answer 1

Reset to default
1
$\begingroup$

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
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.