Do the rising waveforms for overdamped circuits of different \$\zeta >= 1\$ actually have a different "curviness"? Or does it just look that way because they're all shown on the same time scale?

enter image description here

You know, how you can get any square wave to look like any one of these by adjusting the horizontal scale on the scope. I'm wondering if they actually have more rounded shapes as the damping factor increases or whether they're all exponentials with a different time constant or something and you could overlay one on top of the other if you had a "fine" mode on your scope horizontal controls.

The reason I'm asking is I had the preconception that a waveform for a damping factor of 1.0 was going to be better (like straighter until you got to the top) than one for say 2.0, even if they both exhibited the same rise time. So I was thinking I would have to get my pole placement just right to get about 75° phase shift at exactly the crossover frequency, even if I could have say 85° at crossover without careful tuning.

  • \$\begingroup\$ In what application would "straighter until you got to the top" be better? \$\endgroup\$ Commented Sep 10, 2016 at 14:24
  • \$\begingroup\$ @BruceAbbott: I'm designing a pulse-driven electronic load for applying a load step to power supplies for load transient testing. So I'm thinking, in general, the closer to a pure square wave the better (like sharper corner). But I need to keep the frequency response at about 330 kHz (1us rise). So I was wondering, is 330kHz crossover at 73° PM qualitatively better than 330kHz crossover at 88° PM (because 88° is easier to get reliably in this circuit). \$\endgroup\$
    – scanny
    Commented Sep 10, 2016 at 18:32
  • \$\begingroup\$ "I need to keep the frequency response at about 330 kHz (1us rise)" - how much overshoot (if any) is acceptable? If the rise time is 1us (with 330KHz bandwidth) then it meets your needs, so why do you care how it gets there? Aesthetics? \$\endgroup\$ Commented Sep 10, 2016 at 19:44
  • \$\begingroup\$ @BruceAbbott: Zero overshoot. The idea is to make the step load (current) waveform absolutely clean so you don't distort the response of the supply under test, making it harder to interpret. I suppose it doesn't matter how "curvy" it is, as long as it gets there in time. But I'm starting to wonder now whether the rise time doesn't degrade with greater PM, even at the same (0 dB) crossover frequency. I'm figuring I'll need to place a pole right before crossover so I can trim the phase margin to right about 75°. I'm just putting together a quick simulation now. \$\endgroup\$
    – scanny
    Commented Sep 10, 2016 at 19:58

2 Answers 2


I was thinking I would have to get my pole placement just right to get about 75° phase shift at exactly the crossover frequency, even if I could have say 85° at crossover

For a simple 2nd order low pass filter here's what the phase shift is (bottom left): -

enter image description here

Unless you are envisaging a more complex filter, the phase response (\$\omega_{natural}\$ normalized to 1) is always going to be -90 degrees. And, of course for this example \$\omega_{natural}\$ = \$\dfrac{1}{\sqrt{LC}}\$

The amplitude response i.e. how peaky the spectrum is (top left picture) is always equal to the Q of the circuit.

So as damping increases (Q decreases) the maths becomes less that of a tuned resonant (and ringing) formula to that of an RC (exponential) circuit. In other words (in this example), L starts to be swamped by the dominance of R.

The upshot of this is that at damping ratios of 1 and 2 (but with a higher natural resonant frequency for the higher damping ratio), the two responses will "overlay" and look quite similar BUT, they won't be the same. For damping ratios of 10 and 20 (with the appropriate increase of resonant frequency for the damping ratio of 20), the responses will look extremely similar because of the dominance of R (over L) in both these cases.

It becomes just an RC circuit and therefore, providing the product of R and C are equal, the response will be progressively more similar as damping increases.


If you solve de DEq. analytically

  • with \$\zeta<1\$ the solution will be an exponential function with complex exponent, i.e. exponentially decaying sin/cos (infinitely many zero-crossings).
  • with \$\zeta>1\$ the solution will be just an exponential function with a real (negative) exponent (i.e. 0 or 1 zero-crossing).

So the two cases \$\zeta<1\$ and \$\zeta>1\$ have different "curvyness" (the \$\zeta>1\$ solution will cross the time-axis never or at most once).

Looking only at different \$\zeta>1\$ cases:
The solutions have the form \$A_1e^{-s_1 t} + A_2e^{-s_2 t}\$.

I.e. there is more variability than just a time scaling factor, e.g.
\$A_1 > A_2\$ and \$s_1 > s_2\$ will look qualitatively different to
\$A_1 > A_2\$ and \$s_1 < s_2\$
(as you said in your comment: when the faster component dominates or not)

or another example: when \$s_1\$ becomes close to \$s_2\$ (none of the two components dominate (much)) the solution will be close to the degenerate case of critically dampening which is a different waveform that differs not just by a time scale.

  • \$\begingroup\$ My question was about the \$\zeta >= 1\$ case only. "for overdamped circuits of different ζ>=1". I understand, of course, that over and underdamped ones can't be overlayed on each other :) \$\endgroup\$
    – scanny
    Commented Sep 10, 2016 at 18:33
  • \$\begingroup\$ @scanny: ok, I see. I misread the question. I will edit my answer. \$\endgroup\$
    – Curd
    Commented Sep 10, 2016 at 20:49
  • \$\begingroup\$ I like the diff-eq perspective, I think that promises some deeper insights. As I'm starting to understand it, the overdamped response is the sum of two real exponentials, where perhaps the faster one dominates at \$\zeta=1\$ or something; but haven't been able to completely make sense of it yet :) \$\endgroup\$
    – scanny
    Commented Sep 10, 2016 at 20:52

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