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A system has the following open loop bode plot: -

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Which one of the plots below describe the closed loop step response for the entire system?

enter image description here

My attempt

My initial thought was to look at the static gain from the open loop bode plot, which is $-4\text{dB} = 0.63$. I interpret this as this will be the final value of the step response when it is settled. The only graph that sort of matches this description is graph 3.

However, it turns out that this is not the correct answer. And therefore my reasoning about the static gain can not be correct either. So my question is, how can I identify the closed loop step response from the open loop bode plot?

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  • $\begingroup$ That's a bit unusual to consider wc when the phase is -120 degrees. Or did they just chose it this way because of the 0 dB response? That would also be unusual. $\endgroup$
    – a concerned citizen
    Commented Mar 14, 2021 at 16:03
  • $\begingroup$ @aconcernedcitizen Isn't $\omega_c$ defined to be the frequency where the gain is 0dB? Also called the crossover frequency? $\endgroup$
    – Carl
    Commented Mar 15, 2021 at 17:56
  • $\begingroup$ If that were true, a strictly decreasing monotonic passband filter (e.g. Butterworth) will always have wc=0, while an equiripple (e.g. Chebyshev) will be problematic. The -3 dB point is also only valid for a Butterworth-like response, and even that can differ (e.g. Bessel). This is why choosing the phase over the magnitude can be less prone to error by choosing the point where it reaches the half-point (-90 deg). It's not written in stone, though. $\endgroup$
    – a concerned citizen
    Commented Mar 15, 2021 at 18:09
  • $\begingroup$ @aconcernedcitizen Are you sure you aren't talking about the cut-off frequency when you talk about -3dB point and -90 deg point? $\omega_c$ is the crossover frequency and that is the frequency where the gain is 0dB and not the same as cut-off frequency. Or perhaps it's just me who is confused. $\endgroup$
    – Carl
    Commented Mar 15, 2021 at 19:06
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    $\begingroup$ Oh damn! You're right. I've seen wc used for both cross-over and corner/cut-off frequency, so that didn't help, either. $\endgroup$
    – a concerned citizen
    Commented Mar 15, 2021 at 19:20

2 Answers 2

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The final value of the step response is the DC gain of the closed-loop transfer function, which is generally different from the open-loop DC gain.

Assuming unity gain feedback, the feed-forward transfer function $G(s)$ equals the open-loop transfer function, and the closed-loop transfer function is given by

$$C(s)=\frac{G(s)}{1+G(s)}\tag{1}$$

The final value of the closed-loop step response equals $C(0)$. With the given value of $G(0)$, which I would interpret as $-3$dB, the final value of the closed-loop step response should be around $\sqrt{2}-1\approx 0.4$. The step response in Figure $1$ seems like a good match.

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Open loop gain at DC is -3dB or .707 and 0 degrees. We don’t know the forward gain but assuming it is the open loop gain, the closed loop gain would be $.707/(1+.707)= .4148$, matching the first plot.

(With 60 degrees of phase margin however I would have expected a response closer to the third plot, but as explained in the comments this is due to my familiarity of 2nd order loops with high DC gain; here the gain at the natural frequency of the loop, which is at the rate of the ringing, is much higher leading to its higher response in the closed loop system)

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  • $\begingroup$ OK, now our answers seem to agree ... :) +1 $\endgroup$
    – Matt L.
    Commented Mar 14, 2021 at 14:46
  • $\begingroup$ @MattL. Still I am surprised by the amount of ringing with 60 degrees of phase margin $\endgroup$
    – Dan Boschen
    Commented Mar 14, 2021 at 14:46
  • $\begingroup$ (+1 to you for cleaner answer) $\endgroup$
    – Dan Boschen
    Commented Mar 14, 2021 at 14:56
  • $\begingroup$ I guess the excessive ringing is explained by the gain difference at DC vs the natural frequency; I am used to loops with high DC gain $\endgroup$
    – Dan Boschen
    Commented Mar 14, 2021 at 15:00
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    $\begingroup$ Yes, it must be that 5dB peak ... $\endgroup$
    – Matt L.
    Commented Mar 14, 2021 at 15:01

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