Ive got a question about identifying the overdamped, underdamped, and critically damped waveforms of an RLC circuit.

From my (very basic understanding), underdamped decays while oscillating. Critically damped decays the fastest without oscillating, and the overdamped decays without oscillating (but critically damped decreases faster)

In the question below, I feel like the the graph the key shows as overdamped is the critically damped response since it decays the fastest. Which is correct? enter image description here

  • 4
    \$\begingroup\$ I feel like that question is ill-posed. I'm pretty sure 1 and 3 are both underdamped. \$\endgroup\$
    – Hearth
    May 2, 2017 at 1:38
  • 1
    \$\begingroup\$ I agree with Felthry because critical and overdamped should both have no over shoot. You have two underdamped. \$\endgroup\$
    – owg60
    May 2, 2017 at 1:42
  • \$\begingroup\$ Agree with these two, the correct answer is not in the list there. \$\endgroup\$ May 2, 2017 at 6:16
  • \$\begingroup\$ The answers are ill-posed too. :-) There is only one answer with "underdamped" as third option, and that is definitely the easier one to get. So all other answers should be ruled out by exclusion. They could have just asked the student "can yo tell which of the three waveform is underdamped?" \$\endgroup\$ May 2, 2017 at 10:09
  • \$\begingroup\$ The critically damped system DOES possibly under/overshoot! Because the solution is something like ae^-t+bte^-t. Note that the second term does have a t as a multiplier. This is due to the fact, that when the root term vanishes, you have a double zero, which has two fundamental solutions, e^-t, t*e^-t. So the answer of your paper is right. Sometimes, however critically damped solutions do not over/undershoot, depending on the parameters. Search for "critical under over damping", the link is too long for this. Sorry, all above commenters, but you're wrong. \$\endgroup\$
    – jjstcool
    May 2, 2017 at 16:52

1 Answer 1


Only graph 1 can be critically damped since there is an undershoot, but no subsequent overshoot.

Assuming the capacitor has an initial condition, then the voltage across the three components in parallel is:

\$v=-\frac{1}{C}\large \int \small I_C \: dt \: =L\large\frac{dI_L}{dt}\small \:=R\left(I_C-I_L\right) \$

Solving the equations simultaneously for \$\small I_L\$ gives:

\$\large \frac{d^2I_L}{dt^2}\small +\large\frac{1}{RC}\frac{dI_L}{dt}\small +\large\frac{1}{LC}\small I_L\$=0

Let \$\small LC=1\$, i.e. natural frequency = 1 rad/sec, for simplicity, but without losing generality.

Then, for critically damped (equal roots) the solution is of the form:

\$\small I_L=(A+Bt)e^{-t}\$, where \$\small A\$ and \$\small B\$ are constants.

The voltage across the inductor (and R and C) is \$v=\small L\large \frac{dI_L}{dt}\$, giving:

\$v=\small L(B-A-Bt)e^{-t}\$

Which has the characteristic shape of Graph 1

  • \$\begingroup\$ Thumbs up for the nice TeXing - and the only correct answer.... \$\endgroup\$
    – jjstcool
    May 2, 2017 at 16:55

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.