Bode Diagram and Gain-Margin Design for an FOPDT Loop: 20-Bio-A2 Process Dynamics and Control, May 2017, Q8

Problem 8: Bode Diagram and Gain-Margin Design for an FOPDT Loop (20% total)

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. First-order-plus-dead-time (FOPDT) plant $G_p=\dfrac{e^{-0.1s}}{0.5s+1}$ under proportional control; target gain margin $\mathrm{GM}=1.7$.

Find. (a) the qualitative open-loop Bode diagram (corner frequency, slopes, extreme values); (b) $k_c$ for $\mathrm{GM}=1.7$.

Problem 8(a): asymptotic amplitude-ratio and phase sketch for $k_cG_p$ (shown at $k_c=1$, i.e. $0\ \text{dB}$ low-frequency asymptote). The gain is flat below the corner $\omega_c=1/0.5=2\ \text{rad/s}$ and falls at $-20\ \text{dB/decade}$ above it; the phase runs from $0^\circ$ toward $-\infty$ (a first-order lag contributes up to $-90^\circ$, and the dead time adds unbounded phase $-0.1\omega\ \text{rad}$), crossing $-180^\circ$ at $\omega_{180}\approx16.9\ \text{rad/s}$.

Approach. Read the amplitude ratio and phase of $G_p(j\omega)$ from its two factors (a first-order lag and a pure delay), identify the corner frequency and asymptotic slopes for the qualitative sketch, then locate the phase-crossover frequency by solving $\angle G_p(j\omega_{180})=-180^\circ$ and set $k_c$ so the amplitude ratio there equals $1/\mathrm{GM}$.

(a) Amplitude ratio and phase. With $s=j\omega$, $$\mathrm{AR}=|k_cG_p|=\frac{k_c}{\sqrt{1+(0.5\omega)^2}},\qquad \angle G_p=-0.1\,\omega-\tan^{-1}(0.5\,\omega)\ \text{(}\omega\text{ in rad/s, first term in radians).}$$ The delay contributes unit magnitude but phase that grows without bound; the lag alone contributes the $-20\ \text{dB/decade}$ roll-off and a phase that saturates at $-90^\circ$.
(a) Corner and asymptotes. Corner frequency $\omega_c=1/0.5=2\ \text{rad/s}$. For $\omega\ll2$: gain flat at $k_c$ ($0\ \text{dB}$ if $k_c=1$), phase $\to0^\circ$. For $\omega\gg2$: gain falls at $\boxed{-20\ \text{dB/decade}}$, phase tends to $-\infty$ (dead-time dominated). At the corner the lag alone gives $-3\ \text{dB}$, $-45^\circ$, plus a small extra $-0.1(2)=-0.2\ \text{rad}=-11.5^\circ$ from the delay. See the figure.
(b) Phase-crossover frequency. Solve $0.1\,\omega_{180}+\tan^{-1}(0.5\,\omega_{180})=\pi$ (Newton–Raphson): $$\boxed{\omega_{180}\approx16.89\ \text{rad/s}}\qquad\big(0.1(16.89)+\tan^{-1}(8.44)=1.689+1.452=3.14\approx\pi\big).$$
Amplitude ratio at $\omega_{180}$. $\mathrm{AR}_{k_c=1}(\omega_{180})=\dfrac{1}{\sqrt{1+(0.5\times16.89)^2}}=\dfrac{1}{\sqrt{1+71.4}}=0.1176$, so the ultimate gain ($\mathrm{GM}=1$) is $k_{cu}=1/0.1176=8.50$.
(b) Gain for $\mathrm{GM}=1.7$. By definition $\mathrm{GM}=1/[k_c\,\mathrm{AR}(\omega_{180})]$, so $$k_c=\frac{1}{\mathrm{GM}\cdot\mathrm{AR}(\omega_{180})}=\frac{1}{1.7\times0.1176}=\boxed{k_c\approx5.00}\ \ (\text{equivalently }k_{cu}/\mathrm{GM}=8.50/1.7).$$

Result	Value
Corner frequency	$\omega_c=1/0.5=2$ rad/s
High-frequency slope	$-20$ dB/decade
Phase-crossover frequency	$\omega_{180}\approx16.89$ rad/s
AR at $\omega_{180}$ ($k_c=1$)	$0.1176$
Ultimate gain (GM$=1$)	$k_{cu}\approx8.50$
Gain for GM$=1.7$	$k_c\approx5.00$

The derivation and every number above were reproduced independently here and match that already-verified solution.

Concept & Theory

Topic. Frequency response of a first-order-plus-dead-time process; Bode stability; gain margin.

Key relations. For $G_p=Ke^{-\theta s}/(\tau s+1)$: $\mathrm{AR}=K/\sqrt{1+(\tau\omega)^2}$, $\angle G_p=-\theta\omega-\tan^{-1}(\tau\omega)$; $\mathrm{GM}=1/[k_c\,\mathrm{AR}(\omega_{180})]$ at the phase crossover.

Why this works. Proportional gain shifts the log-magnitude curve up or down by $20\log_{10}k_c$ without touching the phase, so the phase-crossover frequency is fixed by the plant alone; the gain margin is then simply the headroom in amplitude ratio at that fixed frequency, and scaling $k_c$ trades directly against it.

Common pitfall. Forgetting to convert the delay's phase contribution to radians before combining with the arctangent term, or expecting a lone first-order lag (without the delay) to ever reach $-180^\circ$ — it asymptotes to only $-90^\circ$, so the dead time is what makes a finite gain margin/ultimate gain possible here.

Source. Seborg et al., Process Dynamics and Control, Ch. 13–14 (frequency response, Bode stability, gain/phase margins).

Question 8 of 8: Bode Diagram and Gain-Margin Design for an FOPDT Loop

Problem 8: Bode Diagram and Gain-Margin Design for an FOPDT Loop (20% total)