Question 1 of 6: Bode plot, gain margin and phase margin

Question 1: Bode plot, gain margin and phase margin (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. A unity-feedback loop whose open-loop transfer function is $G(s) = 200K/[s(s+2)(s+5)]$, evaluated at $K = 1$, so $G(s) = 200/[s(s+2)(s+5)]$. The loop is type 1 (one pole at the origin) with real open-loop poles at $s = 0,\ -2,\ -5$ and no finite zeros.

Find. The Bode magnitude and phase plots of $G(j\omega)$, the gain-crossover and phase-crossover frequencies, and from them the gain margin (GM) and phase margin (PM), together with the resulting stability verdict.

Approach. Put $G(j\omega)$ into Bode (time-constant) form so the asymptotes can be drawn by inspection, locate the phase-crossover frequency analytically from the arctangent sum, read the gain margin there, then solve $|G(j\omega)| = 1$ for the gain-crossover frequency and evaluate the phase margin.

1. Normalise the transfer function to Bode form. Every binomial must be written as $(1 + s/p)$, otherwise the low-frequency asymptote is misplaced — a classic slip that shifts the whole magnitude curve. Factoring the pole constants out, $$G(s)=\frac{200}{s\,(s+2)(s+5)}=\frac{200}{s\cdot 2\left(1+\frac{s}{2}\right)\cdot 5\left(1+\frac{s}{5}\right)} =\frac{20}{s\left(1+\frac{s}{2}\right)\left(1+\frac{s}{5}\right)}$$ so the Bode gain constant is $K_B = 20$ (not 200), and because the loop is type 1 this same constant is the velocity error constant $K_v = 20\ \text{s}^{-1}$.
2. Sketch the asymptotic magnitude. The low-frequency asymptote is the integrator line $20/\omega$, i.e. a $-20$ dB/decade slope passing through $20\log_{10} 20 = 26.02$ dB at $\omega = 1$ rad/s and crossing 0 dB at $\omega = 20$ rad/s if it were extended. The slope steepens to $-40$ dB/decade at the first corner $\omega = 2$ rad/s and to $-60$ dB/decade beyond the second corner $\omega = 5$ rad/s. The plotted curve above is the exact magnitude, which lies about 3 dB below each corner.
3. Write the exact magnitude and phase. With $s = j\omega$, $$|G(j\omega)|=\frac{200}{\omega\sqrt{\omega^{2}+4}\;\sqrt{\omega^{2}+25}},\qquad \angle G(j\omega)=-90^\circ-\tan^{-1}\frac{\omega}{2}-\tan^{-1}\frac{\omega}{5}$$ Note that the phase must be accumulated as a sum of arctangents; taking a principal-value argument of the complex number wraps past $-180^\circ$ and destroys the crossover search.
4. Locate the phase-crossover frequency. Setting $\angle G = -180^\circ$ requires $\tan^{-1}(\omega/2)+\tan^{-1}(\omega/5)=90^\circ$, i.e. the two arctangents are complementary, which happens when $(\omega/2)(\omega/5) = 1$. Hence $\omega^{2}=2\times 5=10$ and $$\boxed{\;\omega_{pc}=\sqrt{10}=3.162\ \text{rad/s}\;}$$ This closed form — $\omega_{pc}=\sqrt{p_1 p_2}$ for a type-1 plant with two real poles — is worth memorising for the exam.
5. Read the gain margin. Substituting $\omega_{pc}$ into the magnitude, and using $\omega_{pc}^{2}+4=14$ and $\omega_{pc}^{2}+25=35$, $$|G(j\omega_{pc})|=\frac{200}{\sqrt{10}\sqrt{14}\sqrt{35}}=\frac{200}{\sqrt{4900}}=\frac{200}{70}=2.857$$ so the magnitude at phase crossover is $20\log_{10}2.857 = +9.12$ dB and $$\boxed{\;\text{GM}=\frac{1}{2.857}=0.35\quad\Longleftrightarrow\quad -9.12\ \text{dB}\;}$$ A negative gain margin in decibels means the loop gain must be reduced by a factor of 2.857 to reach the stability boundary — the system is already unstable.
6. Locate the gain-crossover frequency. Setting $|G(j\omega)| = 1$ gives the algebraic condition $\omega\sqrt{\omega^{2}+4}\sqrt{\omega^{2}+25}=200$, or equivalently the cubic-in-$\omega^{2}$ relation $\omega^{6}+29\omega^{4}+100\omega^{2}-40000=0$. Solving numerically (bisection between 1 and 20 rad/s), $$\boxed{\;\omega_{gc}=5.105\ \text{rad/s}\;}$$ which lies above $\omega_{pc}$ — the signature of an unstable loop.
7. Evaluate the phase margin. At $\omega_{gc}=5.105$ rad/s, $$\angle G(j\omega_{gc})=-90^\circ-\tan^{-1}(2.5525)-\tan^{-1}(1.0210)=-90^\circ-68.60^\circ-45.60^\circ=-204.20^\circ$$ and therefore $$\boxed{\;\text{PM}=180^\circ+\angle G(j\omega_{gc})=-24.2^\circ\;}$$
8. Cross-check with Routh–Hurwitz. The closed-loop characteristic equation is $s(s+2)(s+5)+200K = s^{3}+7s^{2}+10s+200K = 0$, whose Routh condition $7\times 10 > 200K$ gives $K < 0.35$. At $K = 1$ the loop is indeed unstable, and the maximum permissible gain $K_{\max}=0.35$ is exactly the gain-margin ratio computed in Step 5 — an independent confirmation that the Bode reading is right.

Both margins are negative and the gain crossover sits above the phase crossover, so with $K = 1$ the closed loop has two poles in the right half-plane and the response grows without bound. To obtain a serviceable design one would reduce $K$ to about $0.35/2.5 \approx 0.14$, which buys roughly 8 dB of gain margin — though the phase margin is then still only $25.7^\circ$, because backing the gain off also drags the gain crossover down to $\omega_{gc}=1.899$ rad/s, where the plant phase has already reached $-154.3^\circ$. A phase margin near $45^\circ$ needs a further cut to $K \approx 0.073$ ($\omega_{gc}=1.217$ rad/s, GM $=13.6$ dB): on this plant the two margins cannot be traded independently, and buying the usual $45^\circ$ costs about 5.6 dB more gain margin than the 2.5-fold gain cut alone delivers.