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I'm studying system stability, and I know a closed loop system can be unstable due to the feedback. I also know that an open system can be BIBO unstable (e.g., a capacitor with a direct current input). What I'm not sure about is, can an open loop system be Lyapunov unstable?

Both Bode and Nyquist study instability via the open loop transfer function of a closed loop system, so they seem to require a closed loop; and in the books I have I didn't find anything about studying instability for open loop systems; I spoke with a couple of engineers, one of which said open loop systems can't be unstable (considering an impulsive input), while the other said they can; though, he wasn't really able to give an example of an open loop unstable system.

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Yes, open-loop systems can be unstable (examples below, there's innumerable ones).

"Open loop" can be a fuzzy term, and if there was a pitcher of beer on the line I could prove that any unstable system "has internal feedback, now gimme that beer".

The first real ground truth is that natural as-found systems can be stable or not.

The second real ground truth (and possibly what was misleading that first engineer) is that with open-loop control we cannot change the stability properties of a system -- you have what you have. In the case of some nonlinear systems you can avoid unstable operating points, but you can't make the system operate at such a point for any length of time, because it's unstable.

The third real ground truth is that with closed-loop control we can change the stability properties of a system: we can make a stable system unstable, or with active control we can make an unstable system stable (both of my examples below can be stabilized, BTW).

Examples

Find the nearest broom, place it vertically on the floor with the stick on the floor and the broomhead up. Let go.

Up until the moment that it smacks into the floor and stops moving, that's an unstable system, and it's operating in open loop. You can tell it's open-loop by writing the differential equations for its motion. If you ignore friction, air resistance (and the floor), and if you linearize the equations around the operating point, you'll see that it has two modes: \$e^{-at}\$ and \$e^{+at}\$ (i.e., it has poles at \$s = \pm a\$). That second mode is unstable.

Find the nearest NPN power transistor. Put it in a circuit with a healthy (for it) voltage on the collector, and bias it with a fixed voltage to flow a healthy (for it) current from collector to emitter.

In the absence of a truly massive heat sink, as it flows current, it'll heat up. As it heats up, it'll flow more current. That'll make it heat up more. While the precise dynamic equation is both complicated and nonlinear, if you simplify it and linearize it then the dominant behavior will once again be an unstable 1st-order response of \$e^{+at}\$.

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  • \$\begingroup\$ Isn't the NPN example a case of BIBO stability, though, since the bias voltage is kept applied? I was wondering, could a voltage-controlled electrical system be unstable also if the disturbance is impulsive? \$\endgroup\$
    – Mauro
    Nov 12, 2021 at 21:08
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    \$\begingroup\$ I count exponential growth until something breaks as being -- for practical purposes -- unstable. Note that I'm calling out that the base-emitter voltage be fixed, which is never done in practical circuits because it is thermally unstable and leads to thermal runaway. \$\endgroup\$
    – TimWescott
    Nov 12, 2021 at 21:22
  • \$\begingroup\$ I agree that would count as unstable, my doubt was that - as far as I know - there are two types of stability: BIBO, which considers a bounded input and checks if the output is bounded too; and impulsive, which consider an impulsive input and checks what happens to the output when the impulse is gone. Your example seems to me of the first type, and I was wondering if a voltage-controlled electrical circuit could be unstable also in the second type. Or I'm reading it wrong, and your example is of the second type? \$\endgroup\$
    – Mauro
    Nov 12, 2021 at 21:32
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    \$\begingroup\$ That's a good subject for another question (if it's not already out there). For linear systems, BIBO and impulsive stability are almost identical -- a system is just stable, or it's not. For nonlinear systems, things get more complicated (as, for instance, you could claim that the broom is stable because it's response is bounded by that floor). For practical systems, things are more complicated yet, because you probably stop caring about whether a system comes to rest after it's broken, so a strictly mathematically stable system can still be "unstable" for all practical purposes. \$\endgroup\$
    – TimWescott
    Nov 12, 2021 at 21:53
  • \$\begingroup\$ Thanks. I didn't find a question about that other subject, so I opened it: electronics.stackexchange.com/questions/594812/…. \$\endgroup\$
    – Mauro
    Nov 12, 2021 at 23:32

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