Control Systems 2 marks Frequency response Numerical answer

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 ________.

Show the step-by-step derivation
Answer 9.54 dB (accept 9.4 to 9.7) dB (accept 9.4 to 9.7)

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. ✓
The trap this question is built around Computing the magnitude at the gain crossover instead, or forgetting that the pole at the origin contributes a fixed $-90^\circ$ that never changes with frequency. Without that term the crossover equation has no solution.

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?

Show the answer

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.

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