Maximum Stabilising Gain for a Triple-Lag Loop, With and Without Sensor Delay: 20-Bio-A2 Process Dynamics and Control, May 2017, Q5

Problem 5: Maximum Stabilising Gain for a Triple-Lag Loop, With and Without Sensor Delay (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. $G_p(s)=\dfrac{1}{(s+1)^3}$, proportional controller gain $k_c$; case (a) unity sensor $H=1$; case (b) sensor with pure delay $H=e^{-0.7s}$.

Find. The maximum $k_c$ that keeps the closed loop stable, for each sensor case.

Approach. Case (a): the characteristic polynomial is a plain cubic, so apply Routh–Hurwitz directly. Case (b): the loop phase $-3\tan^{-1}\omega-0.7\omega$ is transcendental, so find the phase-crossover frequency ($-180^\circ$) by Newton–Raphson (3 iterations, as instructed) and read the ultimate gain off the amplitude ratio there.

(a) Characteristic equation. $1+k_cG_p=0\Rightarrow(s+1)^3+k_c=0\Rightarrow s^3+3s^2+3s+(1+k_c)=0$.
(a) Routh array. Row $s^3$: $1,\ 3$. Row $s^2$: $3,\ 1+k_c$. Row $s^1$: $\dfrac{3(3)-1(1+k_c)}{3}=\dfrac{8-k_c}{3}$. Row $s^0$: $1+k_c$. Stability needs every first-column entry positive: $3>0$ (always), $\dfrac{8-k_c}{3}>0\Rightarrow k_c<8$, and $1+k_c>0\Rightarrow k_c>-1$. Hence $$\boxed{0\le k_c<8\quad\text{(ultimate gain }K_u=8\text{)}.}$$ At $k_c=8$ the row-$s^1$ entry vanishes and $s^3+3s^2+3s+9=(s^2+3)(s+3)=0$ has poles $s=\pm j\sqrt3,\,-3$ — sustained oscillation, confirming the boundary.
(b) Phase-crossover condition. With $H=e^{-0.7s}$, the loop phase is $\angle L(j\omega)=-3\tan^{-1}\omega-0.7\omega$. Setting this to $-180^\circ=-\pi$ gives the transcendental equation $$3\tan^{-1}\omega+0.7\omega=\pi.$$
Newton–Raphson (first 3 iterations, $\omega_0=1.5$). With $f(\omega)=3\tan^{-1}\omega+0.7\omega-\pi$, $f'(\omega)=\dfrac{3}{1+\omega^2}+0.7$: $$\omega_1=1.5-\frac{f(1.5)}{f'(1.5)}=0.972,\qquad \omega_2=0.972-\frac{f(0.972)}{f'(0.972)}=1.038,\qquad \omega_3=1.038-\frac{f(1.038)}{f'(1.038)}=1.0393.$$ The iteration has essentially converged by the third step: $\boxed{\omega_c\approx1.039\ \text{rad/s}}$ (continuing further changes $\omega_c$ by less than $0.001$).
(b) Ultimate gain. At $\omega_c$, $|H(j\omega_c)|=1$ (a pure delay adds no attenuation), so $|G_p(j\omega_c)|=(1+\omega_c^2)^{-3/2}$ and the ultimate gain is its reciprocal: $$\boxed{K_u=(1+\omega_c^2)^{3/2}=(1+1.039^2)^{1.5}\approx3.00.}$$

Case	Maximum $k_c$
(a) $H=1$	$K_u=8$ (Routh boundary, oscillation at $\omega=\sqrt3$ rad/s)
(b) $H=e^{-0.7s}$	$K_u\approx3.00$ (phase crossover $\omega_c\approx1.039$ rad/s)

Concept & Theory

Topic. Ultimate gain of a triple-lag loop; Routh–Hurwitz for a cubic; sensor dead time lowers the achievable gain.

Key relations. Routh array first-column sign test; phase-crossover $\angle L(j\omega_c)=-180^\circ$; ultimate gain $K_u=1/|L(j\omega_c)|_{k_c=1}$.

Why this works. Every extra unit of phase lag (here, the sensor delay adds $-0.7\omega$ radians on top of the three lags' $-3\tan^{-1}\omega$) reaches $-180^\circ$ at a lower frequency, and the amplitude ratio of a low-pass process is larger at lower frequency — so the delay both moves the crossover down and shrinks the gain margin available there, cutting the ultimate gain from $8$ to about $3$.

Common pitfall. Forgetting that the delay contributes phase only (magnitude $1$) so it never appears in the amplitude-ratio calculation, or truncating Newton–Raphson before the mandated iteration count and reporting an under-converged $\omega_c$.

Source. Seborg et al., Process Dynamics and Control, Ch. 10 (Routh stability, ultimate gain); Ch. 14 (frequency-response stability with dead time).

Question 5 of 8: Maximum Stabilising Gain for a Triple-Lag Loop, With and Without Sensor Delay

Problem 5: Maximum Stabilising Gain for a Triple-Lag Loop, With and Without Sensor Delay (20% total)