-1
$\begingroup$

I have a function $y$ defined as $$ y(t) = \left\{\begin{array} ~t~ \mbox{ where} |t| \le 3\\ 0~ \mbox{ otherwise} \end{array} \right . $$ With a system defined as

$$G(t) = ty(t), $$

is it BIBO stable? I know that if my function is defined at all points along the time axis then it will not be stable. But if $t=4, y(t)=0$, would this be considered bounded and thus BIBO stable?

Thank you in advance.

$\endgroup$
  • $\begingroup$ What is $G(t)$? The statement "if my function is defined at all points along the time axis then it will not be stable" is wrong. $\endgroup$ – Matt L. Jan 24 '18 at 11:15
  • $\begingroup$ 1. Take definition of BIBO stable from textbook. 2. put in terms you've got. 3. Profit. What exactly do you need help with? $\endgroup$ – Marcus Müller Jan 24 '18 at 12:21
1
$\begingroup$

I'll assume that you are referring to an LTI system with impulse response $G(t)$. The system is stable if the impulse response is absolutely integrable:

$$\int_{-\infty}^{\infty}|G(t)| \ \mathrm{d}t<\infty$$

Because $G(t)$ has a closed form, we can actually compute that integral:

$$\int_{-\infty}^{\infty}|G(t)| \ \mathrm{d}t=\int_{-3}^3|t^2| \ \mathrm{d}t=18 <\infty$$

So the system is BIBO stable.

$\endgroup$

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.