22-Mec-A3 System Analysis and Control · Undated paper

Question 3 of 6: Steady-state response of a proportional liquid-level loop

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

Notes on this paper

Paper format. 16-Mec-A3 System Analysis and Control, National Exam, May 2019 sitting. Three hours, closed book, Casio or Sharp approved calculator only. Six questions; any four constitute a complete paper and all questions are of equal value (25 marks each). A table of Laplace transforms is supplied with the paper. Every one of the six questions is worked below.

Reference texts. Ogata, Modern Control Engineering, 5th ed. (Chs. 5–8); Nise, Control Systems Engineering, 8th ed. (Chs. 4, 6, 8, 10); Dorf & Bishop, Modern Control Systems, 13th ed.; Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed.

Question 3: Steady-state response 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 unity-feedback loop in which the error $E(s)=H_r(s)-H_c(s)$ drives a proportional controller of gain $K_p$, which in turn drives the tank whose transfer function is the standard first-order lag $R/(RCs+1)$, with $R$ the outlet valve resistance and $C$ the tank capacitance. The reference is a step of height $H_R$. The paper supplies no numerical values, so the answer is required symbolically.

Find. The steady-state level $h_{css}(t)$ reached after the step, and with it the closed-loop time constant and the residual offset.

[Figure not reproduced: Question 3 block diagram: proportional control of a first-order liquid-level process, redrawn from the exam paper. See the official exam paper.]

Approach. Reduce the single loop to a closed-loop transfer function, confirm that the closed loop is stable, and then apply the final-value theorem to the step response.

Form the closed-loop transfer function. The forward path is $G(s)=K_pR/(RCs+1)$ and the feedback is unity, so $$T(s)=\frac{H_c(s)}{H_r(s)}=\frac{G}{1+G}=\frac{\dfrac{K_pR}{RCs+1}}{1+\dfrac{K_pR}{RCs+1}}=\frac{K_pR}{RCs+1+K_pR}$$
Put it in standard first-order form. Dividing through by $(1+K_pR)$ exposes the closed-loop gain and time constant: $$T(s)=\frac{\dfrac{K_pR}{1+K_pR}}{\dfrac{RC}{1+K_pR}\,s+1}\qquad\Longrightarrow\qquad K_{cl}=\frac{K_pR}{1+K_pR},\quad \tau_{cl}=\frac{RC}{1+K_pR}$$ The single closed-loop pole sits at $s=-(1+K_pR)/RC$, which is in the left half-plane for every $K_p>0$, so the loop is unconditionally stable and the final-value theorem may be applied.
Apply the final-value theorem. With $H_r(s)=H_R/s$, $$h_{css}=\lim_{s\to0}s\,T(s)\,\frac{H_R}{s}=T(0)\,H_R=\frac{K_pR}{1+K_pR}H_R$$ so that, as a function of time, $$\boxed{h_{css}(t)=\frac{K_pR}{1+K_pR}\,H_R\,u(t)}$$ a constant level held for all $t$ large compared with $\tau_{cl}$.
Quantify the offset. The steady-state error left by the proportional controller is the difference between reference and level: $$e_{ss}=H_R-h_{css}=\frac{H_R}{1+K_pR}=\frac{H_R}{1+K_v}$$ where $K_p^{\text{pos}}=\lim_{s\to0}G(s)=K_pR$ is the position error constant. The complete transient, for reference, is $h_c(t)=\frac{K_pR}{1+K_pR}H_R\left(1-e^{-t/\tau_{cl}}\right)$.
Interpret the result. Feedback divides the open-loop time constant $RC$ and the offset by the same factor $(1+K_pR)$. Raising $K_p$ therefore buys speed and accuracy together, but never eliminates the offset: a proportional controller produces actuating flow only in proportion to error, so a zero error would command zero inflow and could not hold the tank against its own outflow. Removing the offset entirely requires integral action.

Check: numerical illustration. The paper gives no numbers, so the boxed symbolic result is the answer. As a check of the algebra, take $K_p=2$, $R=0.5$, $C=4$ and $H_R=1.5$ m: then $K_pR=1$, the closed-loop gain is $0.5$, $\tau_{cl}=RC/2=1.0$ s, the steady level is $0.75$ m and the offset is $0.75$ m — and level plus offset reproduces the $1.5$ m reference exactly.

Quantity	Result
Closed-loop transfer function	$T(s)=K_pR/(RCs+1+K_pR)$
Steady-state response	$h_{css}(t)=\dfrac{K_pR}{1+K_pR}H_R\,u(t)$
Closed-loop time constant	$\tau_{cl}=RC/(1+K_pR)$
Steady-state offset	$e_{ss}=H_R/(1+K_pR)$
Position error constant	$K_{pos}=K_pR$
Stability	Stable for every $K_p>0$

Topic. Proportional control of a first-order process: offset and the final-value theorem.

Key relations. $T=G/(1+G)$; $\tau_{cl}=\tau_{ol}/(1+K_{loop})$; $e_{ss}=R/(1+K_{pos})$ with $K_{pos}=\lim_{s\to0}G(s)$; final-value theorem $\lim_{t\to\infty}f(t)=\lim_{s\to0}sF(s)$, valid only when $sF(s)$ has all its poles in the open left half-plane.

Why this works. A type-0 loop has finite low-frequency gain, so the error signal must remain non-zero to sustain the controller output that holds the level. The factor $(1+K_pR)$ is the loop gain at zero frequency; it appears identically in the pole location and in the error because both come from the same characteristic polynomial $RCs+1+K_pR$. This is the cleanest illustration of the general rule that feedback trades loop gain for both bandwidth and accuracy at once.

Common pitfall. Applying the final-value theorem without first confirming closed-loop stability. Here the loop is stable for all positive gain so the check is trivial, but on a higher-order plant the theorem returns a finite, entirely fictitious answer for a diverging response. A second error is quoting the open-loop time constant $RC$ for the closed-loop settling time.

Source. Ogata, Modern Control Engineering, 5th ed., §§4–3 (liquid-level systems) and 5–10; Nise, Control Systems Engineering, 8th ed., Ch. 7.

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