Bode Plot and Gain-Margin Design for a FOPDT Process: 23-Chem-A6 Process Dynamics and Control, December 2015, Q8

Problem 8: Bode Plot and Gain-Margin Design for a FOPDT Process (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. A first-order-plus-dead-time process: time constant $\tau=0.5$ s (corner $\omega=1/\tau=2$ rad/s), dead time $\theta=0.1$ s, unity process gain; $L(j\omega)=k_c\,e^{-0.1j\omega}/(0.5j\omega+1)$.

Find. (a) the Bode asymptotes and extremes; (b) $k_c$ for $\text{GM}=1.7$.

Approach. Read the magnitude asymptotes and phase limits directly from the FOPDT form, then locate the phase-crossover frequency ($\angle L=-180^\circ$) and invert the gain-margin definition to solve for $k_c$.

(a) Bode shape. The normalised magnitude $|G_p|=1/\sqrt{1+0.25\omega^2}$ is flat at $0$ dB (value $k_c$) below the corner $\omega=2$ rad/s, then rolls off at $-20$ dB/decade. The phase $\phi=-0.1\omega-\tan^{-1}(0.5\omega)$ starts at $0^\circ$; the first-order lag contributes $-45^\circ$ at $\omega=2$ and tends to $-90^\circ$, while the dead-time term $-0.1\omega$ keeps subtracting phase, so $\phi\to-\infty$ — the signature of a transport lag.

Problem 8(a): Bode plot of $G_p=e^{-0.1s}/(0.5s+1)$. Magnitude (blue) with $0$ dB and $-20$ dB/dec asymptotes (dashed) and corner $\omega=2$; the phase crosses $-180^\circ$ at $\omega_{co}=16.9$ rad/s and continues toward $-\infty$ because of the dead time.

(b) Phase crossover. Solve $\angle L=-0.1\omega-\tan^{-1}(0.5\omega)=-\pi$, i.e. $0.1\omega+\tan^{-1}(0.5\omega)=\pi$. Because $\tan^{-1}$ saturates near $\pi/2$, most of the phase comes from the delay, giving $\omega_{co}=16.9$ rad/s.
Amplitude ratio there. $\dfrac{|L|}{k_c}=\dfrac{1}{\sqrt{1+0.25\cdot16.9^2}}=0.1176$.
Solve for the gain. The gain margin is $\text{GM}=\dfrac{1}{|L(j\omega_{co})|}=\dfrac{1}{k_c\cdot0.1176}$. Setting $\text{GM}=1.7$: $$\boxed{k_c=\frac{1}{1.7\times0.1176}=5.0}.$$ (The ultimate gain, $\text{GM}=1$, would be $k_{cu}=1/0.1176=8.5$; a gain margin of $1.7$ backs the design off to $k_c=5.0$.)

Quantity	Value
Corner frequency	$\omega=1/\tau=2$ rad/s
High-frequency slope	$-20$ dB/decade
Phase-crossover frequency	$\omega_{co}=16.9$ rad/s
$\|L\|/k_c$ at $\omega_{co}$	$0.1176$
$k_c$ for $\text{GM}=1.7$	$\mathbf{5.0}$ (ultimate $k_{cu}=8.5$)

Concept & Theory

Topic. Bode plot of a FOPDT system; gain margin; proportional-gain design from a target margin.

Key relations. $|G_p|=1/\sqrt{1+(\tau\omega)^2}$, corner $\omega=1/\tau$; $\phi=-\theta\omega-\tan^{-1}(\tau\omega)$; phase crossover $\angle L=-180^\circ$; $\text{GM}=1/|L(j\omega_{co})|$; $k_c=k_{cu}/\text{GM}$.

Why this works. Since $|L|$ scales linearly with $k_c$ but the phase does not depend on $k_c$, the phase-crossover frequency is fixed by the process alone; the gain margin is then simply the reciprocal of the loop gain there, so a target GM sets $k_c$ directly. The dead time both lowers $\omega_{co}$ and forces the unbounded phase roll-off.

Common pitfall. Solving the phase equation in degrees instead of radians for the $0.1\omega$ term; confusing gain margin (a ratio $>1$) with its dB value; or ignoring the dead-time phase and badly under-estimating $\omega_{co}$.

Source. Seborg et al., Ch. 14 (Bode plots, gain and phase margins, FOPDT frequency response); Coughanowr & LeBlanc, Ch. 16–17 (frequency response, stability margins).

Question 8 of 8: Bode Plot and Gain-Margin Design for a FOPDT Process

Problem 8: Bode Plot and Gain-Margin Design for a FOPDT Process (20%)