0
\$\begingroup\$

I'm currently studying a course of systems and signals. we talked about IAR ODE systems and their properties, and I have a question about something I don't quite get - how can an Initially at rest ODE system be dynamic (time dependent) and time invariant (doesn't change in time) at the same time? can anyone please give me an example?

thank you for your time and attention!

\$\endgroup\$
2
  • \$\begingroup\$ I assume ODE = ordinary diffferetial equation, LTI = linear time-invariant... could you please explain IAR? (please by updating the question instead of answering in comments?) \$\endgroup\$
    – Huisman
    Commented Oct 27, 2019 at 21:14
  • \$\begingroup\$ Is it the software tool the iar tag refers to? \$\endgroup\$
    – Huisman
    Commented Oct 27, 2019 at 21:23

1 Answer 1

Reset to default
1
\$\begingroup\$

Your definition of time invariant is incorrect, because a time invariant system can change in time.

Time invariant means that if a time-delay is applied on the input \$x(t+\delta )\$, it directly equates to a time-delay of the output \$y(t+\delta )\$.

Example:
When I charge a capacitor with a constant voltage source and constant resistance in series, the voltage is time dependent.
It is "an Initially at rest ODE system" and "dynamic (time dependent)". After e.g. 2 seconds, the voltage is x V.

But (with the same initial conditions) it doesn't not matter when I start charging this capacitor. The system is invariant to the (start)point in time when it is "running".

If I charge the capacitor tomorrow, after the same 2 seconds, the voltage will be same x V.

This system is also LTI.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ Time invariant means that the system does not change with time. The system parameters are constant. The response(s) may be, and usually are, functions of time - that means the systems is dynamic. \$\endgroup\$
    – Chu
    Commented Oct 27, 2019 at 21:54
  • \$\begingroup\$ Very nice addition, maybe even more clear than my answer \$\endgroup\$
    – Huisman
    Commented Oct 27, 2019 at 21:57
  • \$\begingroup\$ thank you very much for your answers! I understand it much better now. \$\endgroup\$
    – E. Ginzburg
    Commented Oct 29, 2019 at 7:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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