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I have the following open-loop frequency response data of a system:

Frequency (rad/s) Gain (dB) Phase (degrees)
0.0628 9.3080 -137.4711
0.1257 8.2370 -126.3069
0.2513 7.8477 -151.0235
0.6283 9.4192 -149.3126
1.2566 3.1785 -172.3820
2.5133 -4.4930 -180.8220
5.0265 -17.5587 -200.7328
10.0531 -32.5263 -224.8463

Additional info:

  • All poles of the open loop system are in the left half-plane.
  • In low frequencies, the system is strictly proper.
  • Phase of G(jω) tends to 0+ as ω -> 0.

Edited following answer given: -

I need to close the loop with ONLY a proportional gain K. (negative feedback)

enter image description here

How can I find the interval of values of K where the closed loop system remains stable?

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    \$\begingroup\$ Well, you know that your compensator - providing it is an active filter - will invert the signal and adds another 180° phase shift. You know that the total phase shift of the loop phase (your system and the compensator) cannot approach 360° or you will be unstable. For instance, if your K is -7.8477 dB, then at 0.2513 rad/s, the loop gain is 1 (0 dB) and the total phase will be -151-180 = 331° and the phase margin (PM) is 29°, not much but it is stable. You can now look at all other combinations. Remember, the point at which the PM is considered is only for a loop gain magnitude of 1 or 0 dB. \$\endgroup\$ Nov 20, 2022 at 17:08
  • \$\begingroup\$ @VerbalKint k doesn't introduce any phase shift (proportional gain) so how can it introduce -180? The op doesn't say that it feedback gets inverted anywhere from what I can tell. \$\endgroup\$
    – Andy aka
    Nov 20, 2022 at 17:22
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    \$\begingroup\$ Hi Andy, in lack of block diagram, I assumed that K was inverting or at least there was a neg. sign in the return path as it is usually the case. But it can be a different scenario of course. \$\endgroup\$ Nov 20, 2022 at 17:26
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    \$\begingroup\$ Yes, indeed it is a negative feedback scenario. I've updated the question to make it clear. \$\endgroup\$
    – Gustavo L.
    Nov 20, 2022 at 17:34
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    \$\begingroup\$ Ah, ok, then I stand by my comment : ) So the sum of your system lag and -180° of the neg. side should stay away from -360° by the amount of phase margin you want when observed at a 0-dB loop gain magnitude. A system is unstable with 0° phase margin in theory but you usually cannot accept less than 30° of PM otherwise the transient response is unacceptably ringing. So check the PM at the various 0-dB points obtained with different values of K. \$\endgroup\$ Nov 20, 2022 at 17:48

3 Answers 3

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Plot the bode diagram (phase and magnitude) for your system's open loop frequency response data.

The amount of gain (k) which can be added to the loop to make the closed loop system marginally stable (on the point of instability) is equal to the gain margin where the gain margin is equal to the number of dBs required to be added to the magnitude plot to make the magnitude equal to 0 dB (unity) at the -180 degrees frequency (-360 degrees frequency if the inversion at the summing junction is included in the loop).

A cursory glance at your table of data informs me that you can raise the magnitude plot by about 4.493 dB to make the gain about 0dB at -180 degrees and so k could have a value as large as 4.493 dB before the closed loop system becomes technically unstable.

Although technically, instability occurs when the gain is increased above unity at the frequency where the phase is -180 degrees (-360 degrees for the whole loop), for low gain and phase margins there would be considerable peaking at the top end of the closed loop frequency response and considerable overshoot in the closed loop step response and so in a real system we must ensure that the gain and phase margins are sufficiently large.

If you were to derive the open loop transfer function of your system by fitting asymtotes to your open loop plots (this is called system identification) and then derive the closed loop transfer function equal to k.TFol/(1+k.TFol) then you could use the Root Locus technique and by solving the characteristic equation for various value of k, you can find the value of k which forces the closed loop poles to be positioned on the imaginary axis of the s-plane (marginally stable closed loop system).

The value of k which, using the Root Locus technique, puts the closed loop poles on the imaginary axis should be the same value for k which is equal to the gain margin obtained from the open loop bode plots.

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This answer was given before the op said that the feedback (called "proportional") is in fact negative proportional. A late amendment by the op says this: -

Sorry, that was my fault. In this case, it is negative feedback. I've updated the question.


There is nothing in your question that suggests the feedback signal is inverted for instance...

I need to close the loop with ONLY a proportional gain K

The feedback (k) being "proportional" will introduce no inherent phase shift so, looking at your phase angles, it looks like the main forward amplifying control block is basically inverting (-180°) and, I see nothing that would convince me that it will oscillate providing k is attenuating. However, if k is amplifying, the phase margin could start to move towards 0° (or 360°) above the highest listed frequency in your table but, there is nothing in the actual table that might confirm this.

In conclusion, it's impossible to say whether the system-with-feedback will become unstable when k is greater than unity.

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  • \$\begingroup\$ Sorry, that was my fault. In this case, it is negative feedback. I've updated the question. \$\endgroup\$
    – Gustavo L.
    Nov 20, 2022 at 17:33
  • \$\begingroup\$ @Delaunay your diagram is still contradictory to your words; you said "close the loop with ONLY a proportional gain K" and that means "k" is in the feedback path. Your late added diagram shows unity gain negative feedback. \$\endgroup\$
    – Andy aka
    Nov 20, 2022 at 18:10
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How can I find the interval of values of K where the closed loop system remains stable?

First use a system ID tool to fit the open loop data to a model of the transfer function. You could actually do this in Excel or anything with a fitting function by picking different constants and try and do it by hand. Try first or second order transfer function first. Once you have the open loop transfer function then you can solve for k

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