# Closed loop system stability with Bode

Stability of the closed loop system (with -1 feedback) can be assessed via Bode plot gain/phase margin analysis of its open loop system.

How can one deduce stability of the closed loop system directly its Bode plot?

One approach would be to fit a transfer function to the Bode (Frequency Response) and examine the poles' location of the fitted transfer function. But I'm looking for a rather intuitive approach using directly the Bode (frequency Response) plot of the closed loop system.

• The underlying concept also addressed here, but I'm not sure how much light it will shed on the question Commented Feb 15, 2021 at 18:47
• The intuitive method you might be looking for is the "graphical method", which is described around at 4:10 in this video. It is still a variation of the open loop bode plot analysis, except it looks at the intersection between the forward gain and 1/(feedback gain). I find it very convenient for designing loops. Commented Feb 15, 2021 at 18:52
• Why don't you add an example of a bode plot that is close to instability? Commented Feb 15, 2021 at 18:56
• @Andyaka, Here is an example of an unstable 2nd order TF which was obtained from from a tfest to a set of frequency response data. Num: [0 32.9957 -205.1263] Den: [1 15.1460 -216.8863] Commented Feb 15, 2021 at 21:03
• I don’t see an example. Commented Feb 15, 2021 at 22:21

You can easily find the phase margin by simply solving for the open loop gain from the closed loop gain. If $$\F(s)\$$ is the closed loop transfer function, and $$\G(s)\$$ is the open loop, then:

$$F(s)=\frac{G(s)}{1+G(s)}$$ so $$G(s)=\frac{F(s)}{1-F(s)}$$

You can work out the closed loop gain that corresponds to the open loop unity gain with a particular phase margin.

and just look at the closed loop gain bode plot to see where your loop is. The one you posted seems to have a phase margin of about 70 degrees.

The bode plot is not the easiest way to do this, if you plotted the closed loop gain on a polar plot, then the table above would be a curve and you could just find the point of intersection.

• Ah of course! Then the stability criterion can also be converted with that formula. nice. Commented Feb 15, 2021 at 22:48
• But the problem is: You must know in advance the loop gain function G(s). Does this really answers your question and solves your problem? In general, from the closed-loop sysytem function you cannot find the loop gain unambiguously. Or what is realy your problem?
– LvW
Commented Feb 16, 2021 at 10:17
• Tesla23 method is useful but still based on reverting back to loop gain and phase/gain margin information. @LvW, my problem statement is: Given the frequency response of a system, like an actuator shown in the original post, can one deduce whether the system is stable or not? And by intuitively I mean without fitting a transfer function or calculating the open loop gain. Commented Feb 16, 2021 at 13:42
– LvW
Commented Feb 16, 2021 at 14:04
• I took the problem as not wanting to determine if the system was unstable, but how stable it was - the sort of information you get from the open loop plot. Instability is usually pretty clear from the closed loop plot. I make some assumptions about the loop, but you can clearly get a rough indication as to how stable the loop is from the amount of peaking. I'm not sure if there more exact ways ways, but the table I provided does suggest, for example, that if the peaking is < 2.3dB then the phase margin is > 45 deg. Commented Feb 16, 2021 at 21:27

I am not sure, if this answer is what you expect. Nevertheless:

• Whether a system with feedback is stable (or not) can best be checked using the loop gain plot (BODE diagram). Now the well-known stability criterion can be applied.

• Simulation programs are able to calculate magnitude and phase functions also for closed-loop systems which are unstable. Now - the question arises if we can derive from these BODE plots (for an unstable system) some information about stability properties. And the answer is Yes.

• For stable closed-loop systems the phase function has - in general - a falling characteristic (in particular, in the frequency region where the pole frequency of the dominant pole pair is located.) In contrast, for unstable sysytems this phase function will exhibit a rising characteristic (posive slope). In case of an oscillator (pole on the imag. axis) this function will have a vertical form (phase jump).

• More than that, the finite (positive) slope of this rising phase function is a kind of "measure" for the degree of instability - that means: This slope is also an indication how far the system is away from the stability limit - hence an indication for the amount of the negative phase margin. Of course, for stable systems, the negative slope of the phase function is a measure for the positive phase margin.

• It should be noted that the max. phase slope can conveniently be expressed with the peak of the corresponding group delay figure. Therefore, we have quasi-linear relationship between the phase margin and the inverse group delay peak.

• referring to the example in my original post, there is no "apparent" rising characteristic in the phase plots. Actually, in the first glance, it looks like a very "nice (stable)" system. That being said, given the 2nd order of the system, the phase change is only 90deg (not 180 deg like a nominal second order system). This might take us in a right direction toward answering the "problem statement". Commented Feb 17, 2021 at 15:25
• I must admit that I do not understand the main point of your comment. Nevertheless, regarding the last comment (from your side) to your original post, my answer is YES. As you can read in my answer, you can find the phase margin with pretty good accuracy from the closed-loop response (without knowing the TF and the p/z-distribution).
– LvW
Commented Feb 17, 2021 at 15:44