What does it mean if you've been told to design a compensator to ensure that the closed loop phase margin is at least 120 degrees? Isn't a compensator typically designed using the open loop transfer function? Using the matlab command allmargin() for both closed loop and open loop transfer functions of the same system yields different results!


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Phase margin is indeed read from an open-loop transfer function, but the rationale behind it is to prevent the closed-loop from going unstable. As you might recall: $$T(s) = \frac{GC}{1+GC}$$ Looking at this function from a mathematical viewpoint, this function explodes when $GC = -1$. In bode-plot terms, $GC = -1$ if $|GC| = 0dB$ and $\angle GC = -180 \deg$. As most dynamical systems cross this $0dB$ point somewhere, one should prevent the phase is anywhere close to $-180\deg$. Phase margin describes the distance to $-180\deg$ when the magnitude crosses the $0dB$ line.

As you see, the importance of it is to save the closed-loop transfer function from exploding, but determining it can be done directly from the open-loop transfer function. Because of this, the compensator is also designed in open-loop, but only works properly if implemented in a closed-loop system.

  • $\begingroup$ Let's say you start with an open loop transfer function. You need to improve say, the settling time and overshoot, so you find a new gain using the root locus method. You also need to improve the phase margin. When designing a lead compensator, do you use the original open loop transfer function, or do you use the one multiplied by the gain you just designed? My lead compensator is churning out weird solutions because i used the latter open loop transfer funtion. $\endgroup$
    – misajaja
    Nov 27, 2020 at 12:15
  • $\begingroup$ if you apply a gain on the open loop tf as you explained, the point where it crosses $0dB$ shifts. Therefore you need to recompute you lead compensator with the gain included. However, it could be that because of this gain, the phase margin is already 120, which means the lead compensator is not needed. $\endgroup$
    – Petrus1904
    Nov 27, 2020 at 15:27

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