Bode Plot and Exact Phase Margin of a Lightly-Damped Second-Order System: 20-Bio-A2 Process Dynamics and Control, May 2018, Q1

Problem 1: Bode Plot and Exact Phase Margin of a Lightly-Damped Second-Order System (20%)

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. $G(s)=\dfrac{10}{s^2+s+9}$, a second-order transfer function with unity (proportional, $K_c=1$) feedback assumed for the phase-margin calculation.

Find. (1) the Bode magnitude and phase plots, with the exact values at $\omega=0$ and $\omega\to\infty$; (2) the exact phase margin PM.

Magnitude Bode plot. Flat near $0.92\,\text{dB}$ at low $\omega$, a resonant peak of $10.58\,\text{dB}$ near $\omega\approx2.92\ \text{rad/s}$ (underdamped, $\zeta<0.707$), then rolling off at $-40\,\text{dB/decade}$ beyond $\omega_n=3\ \text{rad/s}$.

Phase Bode plot. Starts at $0^{\circ}$, crosses $-90^{\circ}$ at $\omega=\omega_n=3\ \text{rad/s}$, and approaches $-180^{\circ}$ as $\omega\to\infty$.

Approach. Write $G(s)$ in standard second-order form $K/(\tau^2s^2+2\zeta\tau s+1)$ to read off $\zeta$ and $\omega_n$, evaluate the extreme-frequency magnitude/phase directly, locate the underdamped resonant peak, then solve the exact gain-crossover frequency algebraically and evaluate the phase there to get PM.

Standard form. $G(s)=\dfrac{10}{s^2+s+9}=\dfrac{10/9}{\tfrac19 s^2+\tfrac19 s+1}$, so $$\boxed{K=\tfrac{10}{9}=1.111,\quad \omega_n=3\ \text{rad/s},\quad \zeta=\tfrac{1}{6}=0.1667.}$$ Since $\zeta<0.707$ the system is underdamped and its Bode magnitude shows a resonant peak.
Extreme-frequency values. At $\omega=0$: $|G|=10/9=1.111$ ($0.92\ \text{dB}$), phase $=0^{\circ}$. As $\omega\to\infty$: $|G|\to10/\omega^2\to0$ (rolling off at $-40\ \text{dB/decade}$), phase $\to-180^{\circ}$. At $\omega=\omega_n=3$: $G(j3)=10/(j3)=-j3.333$, phase exactly $-90^{\circ}$.
Resonant peak. $\omega_r=\omega_n\sqrt{1-2\zeta^2}=2.915\ \text{rad/s}$; peak ratio $M_r=\dfrac{1}{2\zeta\sqrt{1-\zeta^2}}=3.043$, so $$\boxed{|G(j\omega_r)|=K\,M_r=3.381\ (10.58\ \text{dB}).}$$
Exact gain crossover. Setting $|G(j\omega)|=1$: $100=(9-\omega^2)^2+\omega^2$. With $x=\omega^2$, $x^2-17x-19=0\Rightarrow x=\dfrac{17+\sqrt{365}}{2}=18.05$, so $$\boxed{\omega_{gc}=\sqrt{18.05}=4.249\ \text{rad/s}.}$$
Exact phase margin. At $\omega_{gc}$, $9-\omega_{gc}^2=-9.05$, so the denominator vector $(-9.05+j4.249)$ has angle $154.86^{\circ}$; hence $\angle G(j\omega_{gc})=0-154.86^{\circ}=-154.86^{\circ}$, and $$\boxed{\text{PM}=180^{\circ}+(-154.86^{\circ})=25.14^{\circ}.}$$

Result	Value
Standard form	$K=1.111$, $\omega_n=3$, $\zeta=0.1667$
DC / high-freq magnitude	$0.92\ \text{dB}$ at $\omega=0$; $\to-\infty$ (roll-off $-40\ \text{dB/dec}$) as $\omega\to\infty$
Resonant peak	$10.58\ \text{dB}$ at $\omega_r=2.915\ \text{rad/s}$
Gain-crossover frequency	$\omega_{gc}=4.249\ \text{rad/s}$
Phase margin	$\text{PM}=25.14^{\circ}$

Concept & Theory

Topic. Bode-plot construction and exact phase-margin computation for an underdamped second-order transfer function.

Key relations. Standard form $K/(\tau^2s^2+2\zeta\tau s+1)$; resonant peak $M_r=1/(2\zeta\sqrt{1-\zeta^2})$ for $\zeta<0.707$; $\text{PM}=180^{\circ}+\angle G(j\omega_{gc})$ where $|G(j\omega_{gc})|=1$.

Why this works. A pair of complex poles with light damping produces a magnitude peak near $\omega_n$ before the $-40\,\text{dB/decade}$ roll-off, and the exact crossover can be found in closed form here because $|G(j\omega)|=1$ reduces to a quadratic in $\omega^2$.

Common pitfall. Reading the resonant peak frequency as $\omega_n$ instead of $\omega_r=\omega_n\sqrt{1-2\zeta^2}$, or evaluating PM at $\omega_n$ instead of the true gain-crossover frequency.

Source. Seborg et al., Process Dynamics and Control, Ch. 13–14 (frequency response, Bode diagrams, resonance, phase margin).

Question 1 of 8: Bode Plot and Exact Phase Margin of a Lightly-Damped Second-Order System

Problem 1: Bode Plot and Exact Phase Margin of a Lightly-Damped Second-Order System (20%)