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Does less settling time means that the response dies out quickly? If so the settling time for a just correct Resistance such that the circuit is slightly underdamped is least, then it's most damped(considering the definition of damped losing energy gradually, what it intuitively means). Then shouldn't it be called overdamped instead? Or maybe I'm confused with the relation between damping and settling time?

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    \$\begingroup\$ Under/overdamped and settling time are not correlated to each other since you could have an underdamped system that rises/falls very quickly and CROSSES the steady state value very quickly, but then overshoots/undershoots and rings for nearly forever before settling at the steady state value. Or you can also have system so overdamped that it doesn't ring at all, but it takes nearly forever to reach it's steady state value. \$\endgroup\$ – DKNguyen Mar 7 '19 at 19:54
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Settling time means the time it takes to get to and stay within some distance from the final value. It does not mean how long it takes for the system to settle out completely, because (for linear systems, at least) the response always gets ever closer, but never actually reaches the final value. (In practice things settle out to being indistinguishable to the final value -- but that's not what theory says).

An underdamped system rings -- it has a (hopefully!) decaying sinusoidal component to its response. As with settling, this component rings forever in theory.

There's a trade-off between settling time, overshoot, and damping. An underdamped system (usually) overshoots, but a mild amount of underdamping will often make the system settle faster, usually as long as the overshoot is less than the settling time limit. A severely underdamped system will have a long settling time, because it'll be ringing like mad.

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  • \$\begingroup\$ But then why an underdamped rlc is called so? \$\endgroup\$ – Adarsh Kumar Mar 7 '19 at 20:19
  • \$\begingroup\$ It's the presence of undershoot/overshoot and ringing that defines whether a system is underdamped or overdamped, not settling time. \$\endgroup\$ – DKNguyen Mar 7 '19 at 20:24
  • \$\begingroup\$ Whats the case of shooting with critically and over damped circuits? \$\endgroup\$ – Adarsh Kumar Mar 7 '19 at 20:38
  • \$\begingroup\$ "Whats the case of shooting with critically and over damped circuits?" Complicated, because a critically or over-damped system can still have zeros that make it overshoot. This should all be covered in a decent course on dynamic systems theory (usually just called "systems" or "signals and systems" in an EE program). \$\endgroup\$ – TimWescott Mar 7 '19 at 20:58

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