See figure below for the problem statement.

My idea was to trace backwards from the ω→+∞ at the origin until I reach a vector from the origin to a point on the curve with magnitude one. This ω will correspond to the largest gain cross over frequency and provide me with the phase margin.

I could then link this value of the phase margin back to zeta through the following equation,

\(\text{P.M.} = \gamma = tan^{-1}\left( \frac{2\zeta}{\sqrt{-2\zeta^{2}+\sqrt{1+4\zeta^{4}}}}\right)\)

The first point I could find that would give me a vector of magnitude one resides at (0.6, -0.8) yielding a phase margin of 126.87°. Unfortunately this PM yields a negative value for zeta.

Any idea where I went wrong or an easier way to solve the problem?

 

I think you messed up on the conversion from complex values to p.m. You got 127 deg, but I get -53 deg which is a p.m. of +53 deg which is 180-127. The p.m. is the phase that must increase to get to 180 deg. Understand that due to the inherent 180 due to the presence of neg feedback its actually 360 total phase till instability. You are correct to use the formula to get zeta from p.m. Just be careful about the phase number used. Try 53 deg. Regards, Kamran Kazem.

 

I think you messed up on the conversion from complex values to p.m. You got 127 deg, but I get -53 deg which is a p.m. of +53 deg which is 180-127. The p.m. is the phase that must increase to get to 180 deg. Understand that due to the inherent 180 due to the presence of neg feedback its actually 360 total phase till instability. You are correct to use the formula to get zeta from p.m. Just be careful about the phase number used. Try 53 deg. Regards, Kamran Kazem.

I'm sure you're correct, but I'm still not seeing it.

The purple angle there was what I found to be ~127°, which makes sense because its definitely more than 90 degrees out from the 180° angle.

This angle can also be thought of as -53°. (or 180° + 127° = 307°)

PM is the difference between this angle and -180°

PM = -53° - (-180°) ≈ 127°

Am I still making another mistake?

 

Can anyone clarify this for me?

 

I agree with jegues.

The phase margin is ~127°.

The function relating phase margin to damping factor shown in post #1 can't be used to find the damping factor in this instance, since the range of the phase margin is limited to 0-90° by the given equation structure.

Assuming the Nyquist curve has an exact solution at point (-0.6,-j0.8), I calculated the damping factor ζ as √(0.4) or 0.6324555. This would give a peak percentage overshoot of ~7.7% for a step input.

Attached is a Nyquist plot for the normalized second order function below with the aforementioned damping factor value.

\(\text{G(s)=\frac{1}{s^2+2\zeta s+1}}\)

This closely resembles the plot shown in post #1.

In this instance the PM would be 126.87°.