22-Mec-A3 System Analysis and Control · December 2016

Question 2 of 6: Steady-State Level of a Proportional Liquid-Level Loop

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Steady-State Level of a Proportional Liquid-Level Loop (25 marks)

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 level control system with

Find. The steady-state controlled level $h_{css}(t)$ — the value the tank level settles to long after the step is applied.

Approach. Reduce the loop to a single closed-loop transfer function, confirm that the closed loop is stable so that the Final Value Theorem may legitimately be applied, then take the limit of $s\,H_{c}(s)$ as $s\to 0$.

1. Write the open-loop (forward-path) transfer function. The controller and the tank are in cascade, so their transfer functions multiply:$$G(s)=K_{p}\cdot\frac{R}{RCs+1}=\frac{K_{p}R}{RCs+1}$$
2. Close the loop. For unity negative feedback the closed-loop transfer function is $T=G/(1+G)$:$$T(s)=\frac{H_{c}(s)}{H_{r}(s)}=\frac{\dfrac{K_{p}R}{RCs+1}}{1+\dfrac{K_{p}R}{RCs+1}}=\frac{K_{p}R}{RCs+1+K_{p}R}$$Dividing through by $(1+K_{p}R)$ puts this in the standard first-order form $T(s)=K_{cl}/(\tau_{cl}s+1)$ with$$K_{cl}=\frac{K_{p}R}{1+K_{p}R},\qquad \tau_{cl}=\frac{RC}{1+K_{p}R}$$
3. Confirm the Final Value Theorem may be used. The single closed-loop pole is $s=-(1+K_{p}R)/(RC)$, which is strictly negative for any physically meaningful $K_{p},R,C>0$. The closed loop is therefore stable and the limit below is legitimate.
4. Apply the Final Value Theorem to the step response. With $H_{r}(s)=H_{R}/s$,$$h_{css}=\lim_{s\to0}s\,H_{c}(s)=\lim_{s\to0}s\cdot\frac{K_{p}R}{RCs+1+K_{p}R}\cdot\frac{H_{R}}{s}=\frac{K_{p}R\,H_{R}}{1+K_{p}R}$$Because this is a constant, the steady-state response as a function of time is that constant held indefinitely:$$\boxed{\;h_{css}(t)=\frac{K_{p}R}{1+K_{p}R}\,H_{R}\;=\;\frac{K_{p}R\,H_{R}}{1+K_{p}R}\,u(t)\;}$$
5. Quantify the offset the proportional controller leaves behind. The steady-state error is the difference between the demand and what is achieved:$$e_{ss}=H_{R}-h_{css}=H_{R}\left(1-\frac{K_{p}R}{1+K_{p}R}\right)=\frac{H_{R}}{1+K_{p}R}$$This is the standard type-0 result $e_{ss}=H_{R}/(1+K_{p}^{\text{pos}})$ with position error constant $K_{p}^{\text{pos}}=\lim_{s\to0}G(s)=K_{p}R$. The level never reaches the set point; raising $K_{p}$ shrinks the offset but cannot remove it.
6. Illustrate with representative numbers. Taking $K_{p}=4$, $R=2$ and $C=3$ in consistent units, $K_{p}R=8$, so $h_{css}=\tfrac{8}{9}H_{R}=0.889\,H_{R}$, an offset of 11.1 %, reached with closed-loop time constant $\tau_{cl}=RC/(1+K_{p}R)=6/9=0.667$ — nine times faster than the open-loop tank ($RC=6$). Proportional feedback has bought speed and accuracy together, but has not eliminated the offset.