Proportional Temperature Control — Type Number, Gain and Steady-State Error: 22-Mec-A3 System Analysis and Control, May 2015, Q3

Question 3: Proportional Temperature Control — Type Number, Gain and Steady-State Error (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. Unity-feedback loop of Fig. 1 with plant $G(s)=\dfrac{1}{(s+1)(s+5)}$ and proportional controller $G_c(s)=K_c$; error signal $E = R - C$; specification $\zeta = 0.5$; test input a unit step.

Find. (a) the type number of the loop and its gain; (b) the value of $K_c$ that places the closed-loop poles at $\zeta = 0.5$, and the resulting steady-state error to a unit step.

Figure 1 — unity-feedback temperature loop with proportional controller $G_c=K_c$ and plant $G(s)=1/[(s+1)(s+5)]$.

Approach. Count the free integrators in the loop gain to get the type number, normalise the loop gain to time-constant form to get the Bode gain (which for a type-0 loop is the position error constant), then match the closed-loop characteristic polynomial to the standard second-order form to fix $K_c$, and finish with $e_{ss}=1/(1+K_p)$.

Form the loop gain and count integrators. $$L(s)=G_c(s)G(s)=\frac{K_c}{(s+1)(s+5)}$$ There is no factor of $s$ in the denominator, so the loop contains no free integrator and the system is type 0. A type-0 loop tracks a step with finite error and cannot follow a ramp at all.
Read the gain in the normalised (Bode) form. The question asks for "the gain" of the loop, which is the constant left when every binomial is written as $(1+s/p)$: $$L(s)=\frac{K_c}{5\,(s+1)\left(\tfrac{s}{5}+1\right)} =\frac{K_c/5}{(s+1)(0.2s+1)}\quad\Rightarrow\quad \boxed{K = \frac{K_c}{5}}$$ This is not the same as the numerator coefficient of the pole-zero form (which is $K_c$). For a type-0 loop this normalised constant is exactly the position error constant, $K_p=\lim_{s\to 0}L(s)=K_c/5$ — which is why the examiner asks for it here and uses it again in part (b).
Write the closed-loop characteristic equation. With unity feedback, $$1+L(s)=0 \;\Longrightarrow\; (s+1)(s+5)+K_c=0 \;\Longrightarrow\; s^2+6s+(5+K_c)=0$$ Comparing with the standard form $s^2+2\zeta\omega_n s+\omega_n^2=0$ identifies $2\zeta\omega_n = 6$ and $\omega_n^2 = 5+K_c$. Note that the damping term is fixed by the plant alone; proportional gain can only raise $\omega_n$, which is what lowers $\zeta$.
Solve for the gain that gives $\zeta = 0.5$. From $2(0.5)\omega_n = 6$ we get $\omega_n = 6\ \text{rad/s}$, hence $$5+K_c=\omega_n^2=36 \quad\Longrightarrow\quad \boxed{K_c = 31}$$ The closed-loop poles are then $s = -3 \pm j3\sqrt3 = -3 \pm j5.196$, which indeed lie on the $60^\circ$ rays from the negative real axis ($\cos^{-1}0.5 = 60^\circ$). The resulting transient carries $16.3\ \%$ overshoot and settles to within 2 % in about $4/3 = 1.33\ \text{s}$.
Compute the steady-state error. For a type-0 loop driven by a unit step, $$K_p=\lim_{s\to 0}L(s)=\frac{K_c}{5}=\frac{31}{5}=6.2,\qquad e_{ss}=\frac{1}{1+K_p}=\frac{1}{7.2}=\boxed{0.1389}$$ i.e. about 13.9 % of the commanded temperature step is never removed. The same figure follows directly from the final value of the output, $c(\infty)=K_c/(5+K_c)=31/36=0.8611$, so $e_{ss}=1-0.8611=0.1389$ — an independent confirmation that costs one line.

Check: the figure labels the summing junction $+/-$ with the feedback path taken directly from $C$, so unity feedback ($H=1$) is assumed throughout. If a temperature sensor of gain $H \ne 1$ were present, both $K_p$ and the achievable $\zeta$ would scale by $H$, and the error would have to be defined relative to the sensed variable.

Quantity	Symbol	Value
System type number	—	Type 0
Loop gain (normalised / Bode form)	$K$	$K_c/5$
Controller gain for $\zeta = 0.5$	$K_c$	31
Undamped natural frequency	$\omega_n$	6.00 rad/s
Closed-loop poles	$s_{1,2}$	$-3 \pm j5.196$
Position error constant	$K_p$	6.20
Steady-state error, unit step	$e_{ss}$	0.1389 (13.9 %)

Topic. System type, static error constants, and second-order pole placement with proportional control.

Key relations. Type number $=$ number of free integrators in $L(s)$; $K_p=\lim_{s\to0}L(s)$, $K_v=\lim_{s\to0}sL(s)$; for a unit step $e_{ss}=1/(1+K_p)$; standard form $s^2+2\zeta\omega_n s+\omega_n^2$ with $\zeta = \cos\theta$ where $\theta$ is the pole angle from the negative real axis.

Why this works. The final-value theorem converts the steady-state question into a limit at $s \to 0$, where only the free integrators and the DC gain survive; everything about the transient becomes irrelevant. The pole-placement half of the problem is equally structural: with a fixed second-order plant and a single scalar gain, the sum of the closed-loop poles is pinned by the plant while their product moves with $K_c$, so damping and bandwidth are not independently adjustable. That coupling is exactly the limitation that motivates PI, lead and lag compensation later in the syllabus.

Common pitfall. Quoting the pole-zero-form coefficient $K_c$ when the question asks for "the gain" of a Bode-normalised loop; here that would give 31 instead of 6.2 and would then produce the wrong $e_{ss}$. A second trap is trying to improve both $\zeta$ and $e_{ss}$ with the same knob — raising $K_c$ shrinks the error but degrades the damping.

Source. Ogata, Modern Control Engineering, 5th ed., §5.3 and §5.8; Dorf and Bishop, Modern Control Systems, 13th ed., §5.6.

Question 3 of 6: Proportional Temperature Control — Type Number, Gain and Steady-State Error

Question 3: Proportional Temperature Control — Type Number, Gain and Steady-State Error (25 marks)