Dead-Time Process — Stability and Gain Margin: 23-Chem-A6 Process Dynamics and Control, May 2013, Q6

Problem 6: Dead-Time Process — Stability and Gain Margin (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. $G_p(s)=\dfrac{e^{-\theta s}}{s+1}$ with dead time $\theta=\dfrac{3\pi}{4}$; proportional gain $K_c$. Open loop $L(j\omega)=\dfrac{K_c\,e^{-j\theta\omega}}{j\omega+1}$.

Find. (a) stability at $K_c=1$ and the critical (phase-crossover) frequency $\omega_c$; (b) the gain margin at $K_c=1$.

Problem 6: polar plot of $L=e^{-\frac{3\pi}{4}s}/(s+1)$ at $K_c=1$. The locus crosses the negative-real ($-180^\circ$) axis at $\omega_c=1$ with magnitude $0.707$, inside the unit circle — stable. The gain margin is $1/0.707=\sqrt2\approx1.41$.

Approach. Use the Bode (frequency-response) stability criterion exactly, with no dead-time approximation: find the phase-crossover frequency $\omega_c$ where $\angle L=-180^{\circ}$, evaluate the amplitude ratio there, and compare with unity; the gain margin is the reciprocal of that amplitude ratio.

Phase and the crossover frequency. $\angle L=-\theta\omega-\tan^{-1}\omega$. Setting $\angle L=-\pi$: $$\theta\omega_c+\tan^{-1}\omega_c=\pi\ \Rightarrow\ \tfrac{3\pi}{4}\omega_c+\tan^{-1}\omega_c=\pi.$$ Try $\omega_c=1$: $\tfrac{3\pi}{4}(1)+\tan^{-1}(1)=\tfrac{3\pi}{4}+\tfrac{\pi}{4}=\pi.$ Exactly satisfied, so $$\boxed{\omega_c=1\ \mathrm{rad/time}}.$$
Amplitude ratio at $\omega_c$ (with $K_c=1$). The dead time has unit magnitude, so $\mathrm{AR}=\dfrac{K_c}{\sqrt{\omega_c^2+1}}=\dfrac{1}{\sqrt{1^2+1}}=\dfrac{1}{\sqrt2}=0.707.$
(a) Stability verdict. By the Bode criterion the closed loop is stable iff $\mathrm{AR}<1$ at the phase crossover. Here $\mathrm{AR}=0.707<1$, so for $K_c=1$ the system is stable, with critical frequency $\omega_c=1$.
(b) Gain margin. $$\mathrm{GM}=\dfrac{1}{\mathrm{AR}\big|_{\omega_c}}=\dfrac{1}{0.707}=\boxed{\sqrt2\approx1.41}.$$ The gain may be increased by a factor of $\sqrt2$ before the amplitude ratio reaches unity at $\omega_c$; equivalently the ultimate gain is $K_{cu}=\sqrt2\approx1.41$ (in decibels, $\mathrm{GM}=20\log_{10}\sqrt2\approx3.0\ \mathrm{dB}$).

Quantity	Value
Critical (phase-crossover) frequency	$\omega_c=1\ \mathrm{rad/time}$ (exact)
Amplitude ratio at $\omega_c$ ($K_c=1$)	$1/\sqrt2=0.707$
Stability at $K_c=1$	stable ($\mathrm{AR}<1$)
Gain margin	$\sqrt2\approx1.41$ ($\approx3.0\ \mathrm{dB}$)
Ultimate gain $K_{cu}$	$\sqrt2\approx1.41$

Concept & Theory

Topic. Exact frequency-response (Bode) stability analysis of a first-order-plus-dead-time (FOPDT) loop; phase crossover and gain margin.

Key relations. $\mathrm{AR}=|L(j\omega)|=\dfrac{K_c}{\sqrt{\omega^2+1}}$; $\angle L=-\theta\omega-\tan^{-1}\omega$; phase crossover at $\angle L=-180^{\circ}$; $\mathrm{GM}=1/\mathrm{AR}(\omega_c)$; a pure dead time contributes phase $-\theta\omega$ but unit magnitude.

Why this works. Because the dead time only adds phase (never magnitude), the exact criterion needs no Padé approximation: find where the total phase hits $-180^{\circ}$ and check whether the loop still amplifies there. The neat result $\omega_c=1$ arises because $\theta=3\pi/4$ was chosen so the lag phase ($\pi/4$ at $\omega=1$) plus the dead-time phase ($3\pi/4$) sum to exactly $\pi$.

Common pitfall. Approximating $e^{-\theta s}$ by a Padé polynomial (the question forbids it and it is unnecessary), or confusing the gain margin (a ratio $\ge1$) with the phase margin.

Source. Seborg et al., Ch. 14 (Bode stability, gain/phase margins, dead time); Stephanopoulos, Ch. 18 (frequency response of dead-time systems).

Question 6 of 8: Dead-Time Process — Stability and Gain Margin

Problem 6: Dead-Time Process — Stability and Gain Margin (20%)