Phase Margin, Gain Margin and Maximum Stable Gain for a Dead-Time Process: 20-Bio-A2 Process Dynamics and Control, May 2018, Q3

Problem 3: Phase Margin, Gain Margin and Maximum Stable Gain for a Dead-Time 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. Open-loop $L(s)=K\dfrac{e^{-0.5s}}{s(s+1)}$, dead time $\theta=0.5\ \text{s}$, integrator plus first-order lag.

Find. (1) PM and GM at $K=1$; (2) the exact maximum $K$ for closed-loop stability.

Approach. Write the exact magnitude and phase of $L(j\omega)$ (the dead time contributes phase only, never magnitude); solve the gain-crossover condition $|L|=1$ in closed form; solve the transcendental phase-crossover condition numerically to full precision; then GM $=1/|L(j\omega_{pc})|$ and, since $K$ only scales magnitude, $K_{\max}=$ GM.

Frequency response. $$|L(j\omega)|=\frac{K}{\omega\sqrt{\omega^2+1}},\qquad \angle L(j\omega)=-0.5\omega\,(\text{rad})-90^{\circ}-\arctan\omega.$$
(1) Gain crossover at $K=1$. $|L(j\omega_{gc})|=1\Rightarrow\omega^2(\omega^2+1)=1$. With $x=\omega^2$: $x^2+x-1=0\Rightarrow x=\tfrac{\sqrt5-1}{2}$, so $$\boxed{\omega_{gc}=\sqrt{\tfrac{\sqrt5-1}{2}}=0.786\ \text{rad/s}.}$$
(1) Phase margin. At $\omega_{gc}=0.786$: $-0.5(0.786)\cdot\tfrac{180}{\pi}=-22.52^{\circ}$, $-\arctan(0.786)=-38.16^{\circ}$; total phase $=-22.52-90-38.16=-150.68^{\circ}$, so $$\boxed{\text{PM}=180^{\circ}-150.68^{\circ}=29.31^{\circ}.}$$
Phase crossover (exact, solved numerically). Setting $\angle L=-180^{\circ}$ gives the transcendental equation $0.5\omega+\arctan\omega=\pi/2$. Newton–Raphson converges to full precision: $$\boxed{\omega_{pc}=1.3065\ \text{rad/s}.}$$
(1) Gain margin at $K=1$. $|L(j\omega_{pc})|=\dfrac{1}{1.3065\sqrt{1.3065^2+1}}=0.4652$, so $$\boxed{\text{GM}=\frac{1}{0.4652}=2.150\ (6.65\ \text{dB}).}$$
(2) Maximum stable gain. The controller gain $K$ scales $|L|$ uniformly without changing its phase, so the phase-crossover frequency is unchanged for any $K$; instability begins exactly when $K|L_1(j\omega_{pc})|=1$, i.e. $$\boxed{K_{\max}=\text{GM}=2.150.}$$

Result	Value
Gain-crossover frequency ($K=1$)	$\omega_{gc}=0.786\ \text{rad/s}$
Phase margin ($K=1$)	$\text{PM}=29.31^{\circ}$
Phase-crossover frequency	$\omega_{pc}=1.3065\ \text{rad/s}$
Gain margin ($K=1$)	$\text{GM}=2.150\ (6.65\ \text{dB})$
Maximum stable gain	$K_{\max}=2.150$

Concept & Theory

Topic. Frequency-response stability margins (PM, GM) for a dead-time-plus-integrator process.

Key relations. Dead-time phase contribution $-\theta\omega$ (radians); $\text{PM}=180^{\circ}+\angle L(j\omega_{gc})$; $\text{GM}=1/|L(j\omega_{pc})|$; for pure-gain control, $K_{\max}=\text{GM}$ evaluated at $K=1$.

Why this works. A proportional controller multiplies the open-loop magnitude at every frequency by the same constant $K$ without altering phase, so the phase-crossover frequency is $K$-independent and the stability limit is exactly the reciprocal of the unity-gain phase-crossover magnitude.

Common pitfall. Forgetting that dead time contributes only phase (never attenuates magnitude), or reporting GM/PM in the wrong units (ratio vs. dB, radians vs. degrees).

Source. Seborg et al., Process Dynamics and Control, Ch. 14 (Bode stability criterion, gain and phase margins, dead-time processes); Stephanopoulos, Ch. 16 (Nyquist/Bode criteria for dead-time systems).

Question 3 of 8: Phase Margin, Gain Margin and Maximum Stable Gain for a Dead-Time Process

Problem 3: Phase Margin, Gain Margin and Maximum Stable Gain for a Dead-Time Process (20%)