Gain margin of a third-order open-loop transfer function

Step-by-step derivation

1. Gain margin is measured at the phase crossover frequency $\omega_{pc}$, where the phase reaches $-180^\circ$. Write the phase: $$\angle G(j\omega) = -90^\circ - \arctan(\omega) - \arctan\!\left(\frac{\omega}{2}\right).$$
2. Set it to $-180^\circ$: $$\arctan(\omega) + \arctan\!\left(\frac{\omega}{2}\right) = 90^\circ.$$
3. Take the tangent of both sides using $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$. The right side is $\tan 90^\circ$, which is infinite, so the denominator must vanish: $$1 - \omega \cdot \frac{\omega}{2} = 0 \;\Longrightarrow\; \omega^2 = 2 \;\Longrightarrow\; \omega_{pc} = \sqrt{2}\ \text{rad/s}.$$
4. Evaluate the magnitude at that frequency: $$|G(j\omega)| = \frac{2}{\omega\sqrt{1+\omega^2}\,\sqrt{4+\omega^2}}.$$
5. Substitute $\omega = \sqrt{2}$: $$|G| = \frac{2}{\sqrt{2}\cdot\sqrt{3}\cdot\sqrt{6}} = \frac{2}{\sqrt{36}} = \frac{2}{6} = \frac{1}{3}.$$ (Note $\sqrt{2}\sqrt{3}\sqrt{6} = \sqrt{36} = 6$ exactly.)
6. Gain margin is the reciprocal of that magnitude: $$GM = \frac{1}{|G(j\omega_{pc})|} = 3.$$
7. Convert to decibels: $$GM_{dB} = 20\log_{10}(3) = 20(0.4771) = \mathbf{9.54\ dB}.$$
8. Interpretation. The loop gain can be raised by a factor of $3$ before the closed loop becomes marginally stable. At $K = 6$ instead of $2$, $|G|$ would be exactly $1$ at $\omega_{pc}$ - matching the Routh-Hurwitz answer for the same plant. ✓