I have a plant whose transfer function has all the poles in the left half of the s-plane (obtained the poles using Wolfram Alpha). However, its bode plot gives negative phase and gain margins; I obtained the bode plot from both MATLAB and Wolfram Alpha and they agree with each other. I am not able to get around this fact. How is this possible? My characteristic polynomial is Characteristic polynomial (Denominator of TF)

Obtained bode plot

  • \$\begingroup\$ Matlab's site has some information on when you have multiple 0dB crossings (which is the case here): mathworks.com/help/control/ug/… \$\endgroup\$
    – esilk
    Commented Nov 8, 2019 at 16:18
  • \$\begingroup\$ @jDAQ But this is the plot just for the open loop system (plant itself), not the closed loop. Then how do we relate this to closed loop being unstable? And could you please explain why the reason you gave (involving gain and phase) holds? \$\endgroup\$
    – ModCon
    Commented Nov 8, 2019 at 18:01
  • \$\begingroup\$ I retract my previous comments, I reread cds.caltech.edu/~murray/books/AM08/pdf/… and the stability margin is just a way to evaluate for how much more gain or phase delay the system will remain (if it was at all) stable. And the author show some examples that that can also be a very poor metric to it, since it does not account for both a change in gain and phase. \$\endgroup\$
    – jDAQ
    Commented Nov 9, 2019 at 5:24

1 Answer 1


The only thing the Barkhausen criterion (which is what gain and phase margin analysis is based on) says is that in order to be oscillating a system's loop gain must be exactly 1 + 0j. In control systems we generally assume a subtraction in there someplace and turn that into the open-loop gain with a sign change must be exactly -1.

You've just discovered that the Barkhausen criterion, by itself, cannot predict stability -- it can only predict stable oscillation.

The Nyquist stability criterion is the more general test that -- if you know the number of unstable zeros in the system -- tells you whether the system is stable. I'm going to leave it to you to do the searching (a good introductory book on classical controls should have it, as does the Internet). Basically, you plot the values of the open-loop transfer function for all frequencies, and count the number of times that -1 is encircled, then compare that to the number of unstable zeros.

Personally, I prefer to start with the system in a known-stable state (found by looking at it and saying "garsh! it ain't movin'!", or by calculating the transfer function for one tuning, etc.), and then looking for gain and phase margin changes from there.

  • \$\begingroup\$ I am confused because till now I thought that a negative stability margin means instability. If I understand what you said correctly, does it mean that, since this in no way implies anything about the number of encirclements of -1+0j it does not tell anything about stability? Also, isn't Nyquist criterion just for evaluating closed loop systems (it might be uncontrolled but closed loop nonetheless)? How do I evaluate the stability of just my open loop system to say a step input? \$\endgroup\$
    – ModCon
    Commented Nov 8, 2019 at 18:09
  • \$\begingroup\$ You can use the Nyquist stability criterion on a bare system -- it's just that then, instead of counting encirclements of -1, you count encirclements of 0. You still need to know how many unstable zeros there are, though. A more direct way if you have a system model is to get the characteristic polynomial and assess it for unstable poles. An even more direct way if you have the actual system is to give it a step input and see if it goes wonky. \$\endgroup\$
    – TimWescott
    Commented Nov 8, 2019 at 18:49
  • \$\begingroup\$ Got it. Now that I know for a fact that the phase and gain margins themselves don't predict stability of a system, does it mean that a system with positive margins can be unstable? \$\endgroup\$
    – ModCon
    Commented Nov 8, 2019 at 19:07
  • \$\begingroup\$ Yes. Most normal systems work the way you think -- but that still leaves the occasional oddball system that is the exception. \$\endgroup\$
    – TimWescott
    Commented Nov 8, 2019 at 19:12

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