Feedback Loop — Closed-Loop Transfer Function and Sensor-Gain Stability Range: 23-Chem-A6 Process Dynamics and Control, May 2018, Q2

Problem 2: Feedback Loop — Closed-Loop Transfer Function and Sensor-Gain Stability Range (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 single-loop feedback system with a proportional controller and a first-order sensor:

Element	Transfer function
Controller	$G_c=K_c=1$
Process	$G_1(s)=\dfrac{2}{(5s+1)(3s+1)}$
Sensor (measurement)	$G_2(s)=\dfrac{-k_2}{10s+1}$

Find. (1) $C(s)/R(s)$ in terms of $K_c,G_1,G_2$; (2) the values of $k_2$ giving a stable closed loop.

Problem 2: standard single-loop feedback. The sensor $G_2$ sits in the feedback path; its measurement is subtracted from the set-point $R$, and the error drives $K_c$ then the process $G_1$.

Approach. Reduce the loop to $C/R=K_cG_1/(1+K_cG_1G_2)$; the denominator set to zero gives the characteristic polynomial, and a third-order Routh test bounds $k_2$.

(1) Closed-loop transfer function. With forward path $K_cG_1$ and feedback $G_2$, block-diagram algebra gives $$\boxed{\frac{C(s)}{R(s)}=\frac{K_cG_1(s)}{1+K_cG_1(s)G_2(s)}.}$$
(2) Characteristic equation. Set $1+K_cG_1G_2=0$ with $K_c=1$. Here $G_1G_2=\dfrac{2}{(5s+1)(3s+1)}\cdot\dfrac{-k_2}{10s+1}=\dfrac{-2k_2}{(5s+1)(3s+1)(10s+1)}$, so $$(5s+1)(3s+1)(10s+1)-2k_2=0.$$
Expand the polynomial. $(5s+1)(3s+1)=15s^2+8s+1$; multiplying by $(10s+1)$: $$150s^3+95s^2+18s+(1-2k_2)=0.$$
Routh test (third order). For $a_0s^3+a_1s^2+a_2s+a_3$ with $a_0=150,\,a_1=95,\,a_2=18,\,a_3=1-2k_2$, stability requires every coefficient positive and $a_1a_2>a_0a_3$: $$1-2k_2>0\ \Rightarrow\ k_2<0.5,\qquad 95(18)>150(1-2k_2)\ \Rightarrow\ 1710>150-300k_2.$$
Solve the inequalities. The second gives $300k_2>150-1710=-1560$, i.e. $k_2>-5.2$. Combining, $$\boxed{-5.2<k_2<0.5.}$$ At $k_2=0.5$ the constant term vanishes (a root through the origin); at $k_2=-5.2$ the $s^1$ Routh entry vanishes (an imaginary-axis crossing, sustained oscillation).

Quantity	Value
(1) Closed-loop TF	$C/R=K_cG_1/(1+K_cG_1G_2)$
Characteristic polynomial	$150s^3+95s^2+18s+(1-2k_2)$
(2) Stable range of $k_2$	$-5.2<k_2<0.5$

Concept & Theory

Topic. Single-loop closed-loop transfer function; Routh–Hurwitz stability as a function of a loop gain (here embedded in the sensor).

Key relations. $C/R=G_cG_1/(1+G_cG_1G_2)$; third-order Routh condition: all $a_i>0$ and $a_1a_2>a_0a_3$.

Why this works. Stability depends only on the roots of $1+G_cG_1G_2$, so the sign and size of the sensor gain $k_2$ act exactly like a loop gain. The two Routh conditions capture the two ways a cubic loses stability: a real root crossing the origin (constant term $\to0$) and a complex pair crossing the imaginary axis ($s^1$ entry $\to0$).

Common pitfall. Quoting only the upper bound $k_2<0.5$; a sufficiently negative sensor gain also destabilises the loop by sending a complex pair into the right half-plane. Also, do not drop $K_c=1$ from the denominator by assuming unity feedback — the sensor $G_2$ is the feedback element.

Source. Stephanopoulos, Ch. 14–15 (closed-loop TF; Routh test); Seborg et al., Ch. 11.

Question 2 of 8: Feedback Loop — Closed-Loop Transfer Function and Sensor-Gain Stability Range

Problem 2: Feedback Loop — Closed-Loop Transfer Function and Sensor-Gain Stability Range (20%)