Bode Plot and Gain Margin of a First-Order-Plus-Dead-Time Process: 23-Chem-A6 Process Dynamics and Control, December 2018, Q7

Problem 7: Bode Plot and Gain Margin of a First-Order-Plus-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. A FOPDT process under proportional control:

Quantity	Value
Static gain	$1$ ($0$ dB)
Time constant	$\tau=0.5$ s $\Rightarrow \omega_c=2$ rad/s
Dead time	$\theta=0.1$ s
Target gain margin	$\mathrm{GM}=1.7$

Find. (a) the asymptotic Bode magnitude and phase; (b) $K_c$ giving $\mathrm{GM}=1.7$.

Problem 7: the magnitude is flat at $0$ dB to the corner $\omega_c=2$ rad/s, then falls at $-20$ dB/dec. The phase is the first-order lag plus the linearly-growing delay term $-0.1\omega$ (rad), so it crosses $-180^\circ$ at a finite $\omega_{pc}\approx16.9$ rad/s — only the dead time makes a phase crossover possible.

Approach. Build the Bode asymptotes from the first-order lag and the delay, find the phase-crossover frequency $\omega_{pc}$ where $\angle G=-180^\circ$, evaluate $|G(\omega_{pc})|$, and set $K_c$ so that $\mathrm{GM}=1/(K_c|G(\omega_{pc})|)=1.7$.

(a) Magnitude asymptotes. The delay has unit magnitude, so $|G|=1/\sqrt{1+(0.5\omega)^2}$: flat at $0$ dB for $\omega\ll2$, breaking at the corner $\omega_c=1/\tau=2$ rad/s, then rolling off at $-20$ dB/decade. As $\omega\to\infty$, $|G|\to0$.
(a) Phase. $\angle G=-\arctan(0.5\omega)-0.1\omega$ (rad). The lag contributes $0^\circ\to-90^\circ$ (passing $-45^\circ$ at $\omega_c=2$); the delay adds $-0.1\omega$ rad $=-5.73\omega$ degrees, unbounded. Hence the phase decreases without limit — the delay is what allows a $-180^\circ$ crossover.
Phase crossover. Set $\angle G=-\pi$: $\arctan(0.5\omega)+0.1\omega=\pi$. Solving numerically gives $$\omega_{pc}=16.89\ \text{rad/s}.$$
Magnitude at crossover. $|G(\omega_{pc})|=\dfrac{1}{\sqrt{1+(0.5\cdot16.89)^2}}=\dfrac{1}{\sqrt{1+71.3}}=0.1176.$
(b) Gain for GM $=1.7$. With proportional control $L=K_cG$, the gain margin is $\mathrm{GM}=\dfrac{1}{K_c|G(\omega_{pc})|}$. Setting it to $1.7$: $$\boxed{K_c=\frac{1}{1.7\,|G(\omega_{pc})|}=\frac{1}{1.7(0.1176)}=5.0.}$$ At this gain the open-loop magnitude at the $-180^\circ$ frequency is $1/1.7\approx0.59$, i.e. $1.7\times$ below the stability limit.

Quantity	Value
Corner frequency	$\omega_c=2$ rad/s ($-20$ dB/dec above)
Phase-crossover freq.	$\omega_{pc}=16.89$ rad/s
$\|G(\omega_{pc})\|$	$0.1176$
(b) $K_c$ for GM $=1.7$	$K_c=5.0$

Concept & Theory

Topic. Frequency response of FOPDT; Bode asymptotes; gain margin from the phase-crossover frequency.

Key relations. $|G|=1/\sqrt{1+(\tau\omega)^2}$; $\angle G=-\arctan(\tau\omega)-\theta\omega$; $\mathrm{GM}=1/|L(j\omega_{pc})|$ where $\angle L(j\omega_{pc})=-180^\circ$.

Why this works. Only the dead time can drag the phase past $-180^\circ$ for a first-order plant (a lone first-order lag stops at $-90^\circ$), so the delay creates a finite $\omega_{pc}$ and hence a finite ultimate gain. The gain margin is the factor by which $K_c$ may rise before $|L|$ reaches $1$ at that frequency.

Common pitfall. Approximating the delay (the question forbids it) or mixing radians and degrees when combining the lag and delay phases. The corner frequency belongs to the magnitude plot; the crossover frequency (from the phase) is much higher here.

Source. Seborg et al., Ch. 13–14 (Bode plots, gain/phase margins); Stephanopoulos, Ch. 17.

Question 7 of 8: Bode Plot and Gain Margin of a First-Order-Plus-Dead-Time Process

Problem 7: Bode Plot and Gain Margin of a First-Order-Plus-Dead-Time Process (20%)