22-Mec-A3 System Analysis and Control · Undated paper

Question 5 of 6: Bode diagrams of two open-loop transfer functions

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5: Bode diagrams of two open-loop transfer functions (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two open-loop transfer functions: (a) a type-1 loop with an integrator and a lightly damped quadratic factor; (b) a type-2 loop with a double integrator and one real pole.

Find. Asymptotic and corrected magnitude plots and the phase plots, together with the gain- and phase-crossover frequencies and the resulting stability margins.

Approach. Normalise each transfer function to time-constant (Bode) form so that the low-frequency asymptote can be read directly, mark the corner frequencies and the slope changes they cause, apply the corrections at the corners, then locate the crossovers and cross-check each gain margin against the Routh gain limit of the same loop.

1. Normalise (a). The quadratic must be written with a unity constant term before the Bode gain can be read: $$G H=\frac{150}{s\left(s^2+2s+9\right)}=\frac{150/9}{s\left[\left(\dfrac{s}{3}\right)^2+\dfrac{2}{9}s+1\right]}$$ so the Bode gain is $K=150/9=16.67$, and the quadratic has $\omega_n=3$ rad/s with $2\zeta/\omega_n=2/9$, giving $\zeta=1/3$.
2. Draw the asymptotes for (a). Below the corner the magnitude is that of the integrator alone, $20\log_{10}(16.67/\omega)$: a $-20$ dB/decade line passing through $24.44$ dB at $\omega=1$ rad/s and reaching $14.89$ dB at the corner $\omega_n=3$ rad/s. Above the corner the quadratic adds $-40$ dB/decade, so the asymptote breaks to $-60$ dB/decade.
3. Correct at the break point. Because $\zeta=1/3<1/\sqrt2$ the quadratic resonates. The peak of the correction is $$M_r=\frac{1}{2\zeta\sqrt{1-\zeta^2}}=1.591=4.03\text{ dB}\quad\text{at}\quad \omega_r=\omega_n\sqrt{1-2\zeta^2}=2.646\text{ rad/s}$$ so the actual curve rises about $4$ dB above the asymptote just before the corner and joins it again about a decade higher.
4. Locate the phase crossover of (a) exactly. The integrator contributes a constant $-90^\circ$, so the phase reaches $-180^\circ$ precisely when the quadratic contributes $-90^\circ$, which happens at its own natural frequency: $$\omega_{pc}=\omega_n=3\text{ rad/s}$$ No numerical search is needed. There the magnitude is $$\left|GH(j3)\right|=\frac{150}{3\left|9-9+j6\right|}=\frac{150}{18}=8.333\quad\Longrightarrow\quad \boxed{GM=\frac{1}{8.333}=0.12=-18.4\text{ dB}}$$
5. Cross-check the gain margin against Routh. Closing the loop gives $s^3+2s^2+9s+K=0$, whose Routh limit is $K_{max}=2\times9=18$. The gain margin as a ratio must equal $K_{max}/K=18/150=0.12$ — it does, which confirms that the margin has not been inverted.
6. Find the gain crossover and phase margin of (a). Solving $\left|GH(j\omega)\right|=1$, i.e. $\omega\sqrt{\left(9-\omega^2\right)^2+4\omega^2}=150$, gives $\omega_{gc}=5.71$ rad/s. The phase there is $$\angle GH=-90^\circ-\tan^{-1}\!\frac{2\omega}{9-\omega^2}=-244.2^\circ\quad\Longrightarrow\quad \boxed{PM=180^\circ-244.2^\circ=-64.2^\circ}$$ Both margins are negative, and the gain crossover lies above the phase crossover: the closed loop is unstable, with two right-half-plane roots. That is a legitimate answer, not an arithmetic error — the gain of 150 is more than eight times the largest gain this plant can tolerate.

Part (b) has no quadratic factor, so the asymptotic construction is exact except at the single real corner, but its phase behaviour is more severe.

7. Normalise (b). Factoring $10$ out of the real pole, $$GH=\frac{50}{s^2(s+10)}=\frac{5}{s^2\left(0.1s+1\right)}$$ so the Bode gain is $K=5$ ($13.98$ dB) and the only corner is at $\omega=10$ rad/s. Forgetting to normalise here would place the low-frequency asymptote $20$ dB too high.
8. Draw the asymptotes for (b). The double integrator gives $-40$ dB/decade from the lowest frequency, passing through $20\log_{10}5=13.98$ dB at $\omega=1$ rad/s and reaching $-26.02$ dB at the corner. Beyond $\omega=10$ rad/s the real pole steepens the slope to $-60$ dB/decade. The only correction needed is the standard $-3$ dB at the corner itself.
9. Examine the phase of (b). The double integrator contributes a fixed $-180^\circ$ and the real pole adds a further lag, so $$\angle GH=-180^\circ-\tan^{-1}\frac{\omega}{10}$$ The phase therefore starts at $-180^\circ$ and only falls further; it never crosses $-180^\circ$ from above, so there is no conventional phase-crossover frequency and no positive gain margin exists at any gain.
10. Find the gain crossover of (b). Setting $50=\omega^2\sqrt{\omega^2+100}$ gives $\omega_{gc}=2.21$ rad/s, where the phase is $-180^\circ-12.5^\circ=-192.5^\circ$: $$\boxed{PM=180^\circ-192.5^\circ=-12.5^\circ,\qquad \text{no finite }GM}$$
11. Confirm with the characteristic equation. Closing the loop on (b) gives $s^3+10s^2+0\cdot s+50=0$. The $s$ term is missing entirely, so the necessary coefficient condition of the Routh criterion fails outright: the loop has two right-half-plane roots for every positive gain, and no simple gain adjustment can stabilise it. A double integrator with only one lag always needs phase-lead compensation.