Question 6 of 6: Asymptotic Bode magnitude plot, phase margin and gain margin

Question 6: Asymptotic Bode magnitude plot, phase margin and gain margin

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. Loop gain function $G(s)=20/[(s+5)(0.1s+1)(0.025s+1)]$, assumed to be closed with unity negative feedback. Note that the first factor is not in time-constant form.

Find. (a) the asymptotic (straight-line) Bode magnitude plot; (b) the phase margin and gain margin of the loop.

Approach. Put every factor into time-constant form so the d.c. gain and the corner frequencies can be read directly, build the asymptotic magnitude by accumulating $-20$ dB/decade at each corner, then find the gain-crossover and phase-crossover frequencies and evaluate the two margins there.

1. Convert to time-constant (Bode) form. The factor $(s+5)$ must be written $5(0.2s+1)$: $$G(s)=\frac{20}{5(0.2s+1)(0.1s+1)(0.025s+1)} =\frac{4}{(0.2s+1)(0.1s+1)(0.025s+1)}$$ The d.c. gain is therefore $K=4$, i.e. $20\log_{10}4 = 12.04\ \text{dB}$, not 26 dB as the un-normalised numerator might suggest.
2. List the corner frequencies. Each factor $(\tau s+1)$ corners at $\omega=1/\tau$: $$\omega_1=\frac{1}{0.2}=5,\qquad \omega_2=\frac{1}{0.1}=10,\qquad \omega_3=\frac{1}{0.025}=40\ \ \text{rad/s}$$ All three are simple poles, so each adds $-20$ dB/decade.
3. Build the asymptotic magnitude. Starting flat at 12.04 dB (there is no integrator, so the low-frequency asymptote is horizontal), the slope steepens at each corner:

The straight-line plot crosses 0 dB on the $-40$ dB/decade segment: solving $6.02-40\log_{10}(\omega/10)=0$ gives $\omega\approx 14.1\ \text{rad/s}$. The exact curve lies below the asymptotes near the corners (each simple pole costs 3 dB at its own corner), so the true crossover is somewhat lower, as the next step shows.

4. Find the exact gain-crossover frequency. Setting $$|G(j\omega)|=\frac{4}{\sqrt{1+(0.2\omega)^2}\sqrt{1+(0.1\omega)^2}\sqrt{1+(0.025\omega)^2}}=1$$ and solving numerically gives $$\omega_{gc}=11.54\ \text{rad/s}$$ which is about 18 % below the straight-line estimate of 14.1 rad/s — the expected penalty for reading a crossover off asymptotes alone when two corners are close together.
5. Evaluate the phase margin. The phase of three simple poles is $$\angle G(j\omega)=-\left[\tan^{-1}(0.2\omega)+\tan^{-1}(0.1\omega)+\tan^{-1}(0.025\omega)\right]$$ At $\omega_{gc}=11.54\ \text{rad/s}$ the three terms are $66.6^\circ$, $49.1^\circ$ and $16.1^\circ$, totalling $131.8^\circ$, so $$\text{PM}=180^\circ+\angle G(j\omega_{gc})=180^\circ-131.8^\circ \quad\Longrightarrow\quad \boxed{\;\text{PM}=48.2^\circ\;}$$
6. Find the phase-crossover frequency. Solving $\angle G(j\omega)=-180^\circ$, i.e. $\tan^{-1}(0.2\omega)+\tan^{-1}(0.1\omega)+\tan^{-1}(0.025\omega)=180^\circ$, gives $$\omega_{pc}=25.50\ \text{rad/s}$$ which lies on the $-40$ dB/decade segment, well above gain crossover.
7. Evaluate the gain margin. The magnitude there is $$|G(j\omega_{pc})|=0.2370 \quad\Longrightarrow\quad \text{GM}=\frac{1}{0.2370}=4.22 \quad\Longrightarrow\quad \boxed{\;\text{GM}=12.5\ \text{dB}\;}$$ Both margins are positive and the open loop has no right-half-plane poles, so the closed loop is stable. As a check, multiplying the loop gain by exactly 4.22 (raising the numerator from 20 to 84.4) places the closed-loop poles precisely on the imaginary axis, confirming the gain margin independently.
8. Assess the design. A phase margin near $48^\circ$ and a gain margin above 12 dB is a comfortable, well-damped design by the usual industrial guideline (PM 45–60°, GM 6–12 dB). The closed-loop response would show a modest overshoot — the rule of thumb $\zeta\approx\text{PM}/100$ gives $\zeta\approx 0.48$, close to the $\zeta=0.5$ targeted in Question 4 — and a bandwidth of roughly $\omega_{gc}$, about 12 rad/s.