I'm just getting my head around RLC dynamics.

After some analysis and math i get,

\$\frac{d^{2}v_{\text{c}}}{dt^{\text{2}}} + \frac{R}{L}\frac{dv_{\text{c}}}{dt} + \frac{1}{LC}v_{c} = \frac{1}{LC}v_{\text{S}}\$

and the characteristic equation:

\$s^{2} + 2\alpha s + \omega_{0}^{2}\$


\$ v_c = V_S + A_1e^{s_1t} + A_2e^{s_2t}\$

3 cases:

case 1: \$ \alpha = \omega_{0}\$ --- critically damped, then,

\$v_c = V_S + A_1e^{-\alpha t} + A_2te^{-\alpha t}\$

I understand that no sinusoids here. But I note texts drawing some overshoot before the response continues to the final value, with no clear explanation. I tried to crunch my head with little maths to explain the overshoot, but failed. Any one to explain the overshoot?


i.e. isn't the figure below the case? enter image description here

or in the case of source-free RLC.

enter image description here

  • 2
    \$\begingroup\$ "Critically damped" = no overshoot so maybe you are not reading the drawing correctly or the drawing is wrong. Is your question, in reality, not about deriving the formula but the drawing (that is confusing you)? It appears to be the only question you have so maybe you could focus on this bit? \$\endgroup\$ – Andy aka Mar 13 '14 at 12:03
  • \$\begingroup\$ Critically damped is not the same thing as minimum settling time, which is actually achieved with slight under-damping. Your texts might be confusing these two issues. \$\endgroup\$ – Dave Tweed Mar 13 '14 at 12:49

The usual blanket statement of no overshoot for a critically damped system makes an implicit assumption about the initial conditions. A critically damped system will overshoot at most once.

To intuitively see this, imagine a mass spring damper system where the mass is released at t=0 and finds an equilibrium condition. That response will have no overshoot (say a door is release and it closes automatically).

On the other hand, if the mass is moving rapidly at t=0 (say the door is given a mighty push just before t=0) it may well overshoot before reaching a final equilibrium.

For a detailed mathematical treatment, minus hand waving, see, for example, this.

If it's a step response from equilibrium there should be NO overshoot, and any overshoot is an indication of underdamping.

As Dave Tweed mentions, if some overshoot can be tolerated, then a faster rise time can be achieved by slightly underdamping. For example, for a tolerable overshoot of 5%, \$\zeta \$ should be about 0.69. The 10% to 90% rise time is then about 35% faster than the critically damped case. Perhaps they are illustrating such a design decision, which is pretty common.

There seems to be quite a few bad examples of formulas out there, so be careful. Here's one example from here, which is totally incorrect.

enter image description here

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  • 1
    \$\begingroup\$ For electronic systems in particular, the "no overshoot" statement applies specifically to the step response of the system, which does indeed imply that the system is at equilibrium immediately before the step. \$\endgroup\$ – Dave Tweed Mar 13 '14 at 13:27
  • \$\begingroup\$ it is a step response. so in conclusion, no overshoot for critically damped circuit? \$\endgroup\$ – r m Mar 14 '14 at 10:55
  • \$\begingroup\$ Agreed. There should be no overshoot \$\endgroup\$ – Spehro Pefhany Mar 14 '14 at 12:39

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