Gain margin of a third-order open-loop transfer function
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A unity-feedback system has $$G(s) = \frac{2}{s\,(s+1)(s+2)}.$$
The gain margin, in decibels, is ________.
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Step-by-step derivation
- Gain margin is measured at the phase crossover frequency $\omega_{pc}$, where the phase reaches $-180^\circ$. Write the phase: $$\angle G(j\omega) = -90^\circ - \arctan(\omega) - \arctan\!\left(\frac{\omega}{2}\right).$$
- Set it to $-180^\circ$: $$\arctan(\omega) + \arctan\!\left(\frac{\omega}{2}\right) = 90^\circ.$$
- 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: $$1 - \omega \cdot \frac{\omega}{2} = 0 \;\Longrightarrow\; \omega^2 = 2 \;\Longrightarrow\; \omega_{pc} = \sqrt{2}\ \text{rad/s}.$$
- Evaluate the magnitude at that frequency: $$|G(j\omega)| = \frac{2}{\omega\sqrt{1+\omega^2}\,\sqrt{4+\omega^2}}.$$
- 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.)
- Gain margin is the reciprocal of that magnitude: $$GM = \frac{1}{|G(j\omega_{pc})|} = 3.$$
- Convert to decibels: $$GM_{dB} = 20\log_{10}(3) = 20(0.4771) = \mathbf{9.54\ dB}.$$
- 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. ✓
The idea behind this question
The gain margin says how much the loop gain could be multiplied before the closed loop becomes unstable. Find the phase-crossover frequency, where the open-loop phase is $-180^\circ$, and measure how far below 1 the magnitude is there. A magnitude of $1/3$ at that frequency means the gain could be tripled: $20\log_{10}3 = 9.54$ dB.
Try a variation
What is the gain margin of $G(s) = \dfrac{4}{s(s+1)(s+2)}$, in dB?
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Answer: $3.52$ dB
The phase crossover is still $\omega = \sqrt{2}$, where $|G| = 4/6$; $20\log_{10}(6/4) = 3.52$ dB.
Other mistakes to avoid
- Giving the answer as a plain ratio when decibels are asked for, or the reverse.
- Evaluating the phase without the $-90^\circ$ from the pole at the origin.