Question 1 of 8: Basic Definitions and Concepts of Control

Question 1: Basic Definitions and Concepts of Control (20 marks, compulsory)

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 printed step-response trace (parts 1–2), an explicit second-order transfer function (parts 3–4), a Type 0 block diagram with a disturbance input (part 5), a printed step-response time function to be matched against four candidate transfer functions (part 6), two multiple-choice conceptual items on Proportional Control (parts 7, 9–10), and a closed-loop transfer function to be converted back to an equivalent open-loop form (part 8).

Find. Every quantity requested in the ten sub-parts above.

Approach. Read transient specs directly off the graph (parts 1–2) or from the standard second-order form $G_m(s)=K_{dc}\omega_n^2/(s^2+2\zeta\omega_n s+\omega_n^2)$ (parts 3–4); apply the final-value theorem with superposition of the reference and disturbance inputs (part 5); match pole structure by inspection of the time-domain terms (part 6); apply the standard sensitivity/loop-gain arguments (parts 7, 9–10); and invert $G_{closed}=G/(1+G)$ to recover the open-loop $G(s)$ before reading off its Type and error constants (part 8).

1. Part 1) — read the graph. Peak value $\approx1.24$ at $t_p\approx0.35$ s; the response settles on a final value $y_{ss}\approx0.90$ (the tight band of horizontal reference lines on the plot). $PO=\dfrac{1.24-0.90}{0.90}\times100\%\approx\boxed{38\%}$. Steady-state error to the unit reference: $e_{ss(step)}\%=(1-y_{ss})\times100\%=\boxed{10\%}$. Settling time is read where the oscillation enters the $\pm5\%$ band around $0.90$, which the trace reaches by about $t\approx1.1$ s, i.e. $T_{settle(5\%)}\approx\boxed{1.1\ \text{s}}$.
2. Part 2) — damping ratio, natural frequency, DC gain from the same trace. Inverting the overshoot formula $\zeta=\dfrac{-\ln(PO/100)}{\sqrt{\pi^2+\ln^2(PO/100)}}$ at $PO=38\%$ gives $\zeta\approx0.30$. The peak time $t_p=\pi/\omega_d$ with $t_p\approx0.35$ s gives $\omega_d\approx8.98$ rad/s, so $\omega_n=\omega_d/\sqrt{1-\zeta^2}\approx\boxed{9.4\ \text{rad/s}}$ (with $\zeta\approx0.30$). Since the reference is a unit step and the response settles at $0.90$, $K_{dc}=y_{ss}/r=\boxed{0.90\ \text{V/V}}$ — consistent with part 1's $e_{ss}=10\%=1-K_{dc}$.
3. Part 3) — read $\zeta,\omega_n,K_{dc}$ directly off $G(s)=\dfrac{22.5}{s^2+s+25}$. Matching to $K_{dc}\omega_n^2/(s^2+2\zeta\omega_n s+\omega_n^2)$: $\omega_n^2=25\Rightarrow\omega_n=\boxed{5\ \text{rad/s}}$; $2\zeta\omega_n=1\Rightarrow\zeta=\boxed{0.10}$; $K_{dc}=22.5/\omega_n^2=22.5/25=\boxed{0.90\ \text{V/V}}$.
4. Part 4) — step-response specs for the model of Part 3. $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=100\,e^{-0.1\pi/\sqrt{0.99}}\approx\boxed{72.9\%}$; $e_{ss(step)}\%=(1-K_{dc})\times100\%=\boxed{10\%}$; $T_{settle(5\%)}=3/(\zeta\omega_n)=3/(0.1\times5)=\boxed{6.0\ \text{s}}$. This system has the same DC gain and steady-state error as Parts 1–2's graph but a much lighter damping ratio, so it overshoots far more severely and settles six times slower — the two sub-parts are deliberately independent illustrations of the same formulas, not the same plant.
5. Part 5) — steady-state output of the Type 0 diagram by superposition. With $G_1(s)=3/(s+2)$ and $G_2(s)=3/(s^2+s+1)$, the reference-to-output and disturbance-to-output closed-loop gains at DC are $G_1(0)=1.5$, $G_2(0)=3$, loop gain $G_1(0)G_2(0)=4.5$: $$\left.\frac{Y}{R}\right|_{s=0}=\frac{4.5}{1+4.5}=\frac{9}{11},\qquad \left.\frac{Y}{D_{dist}}\right|_{s=0}=\frac{G_2(0)}{1+4.5}=\frac{6}{11}.$$ By superposition, $y_{ss}=\dfrac{9}{11}(2)+\dfrac{6}{11}(1)=\dfrac{24}{11}=\boxed{2.18\ \text{V}}$.
6. Part 6) — match the step response to its transfer function. A ramp term ($0.0833t$) together with a constant term ($-0.139$) in a unit-STEP response comes from a DOUBLE pole at the origin of $Y(s)=G(s)/s$, which needs a SIMPLE pole at the origin in $G(s)$ itself (Type 1). The paired $t\,e^{-3t}$/$e^{-3t}$ and $t\,e^{-2t}$/$e^{-2t}$ terms come from DOUBLE poles at $s=-3$ and $s=-2$ in $G(s)$. Only option (iv), $G(s)=\dfrac{3}{s(s+2)^2(s+3)^2}$, has this exact pole structure (a single pole at the origin plus double poles at $-2$ and $-3$).
7. Part 7) — ideal tracking. For unity-feedback Proportional Control the tracking error is $e_{ss}=R/(1+K_pG(0))$; making the error vanish in the limit needs the open-loop DC gain $K_pG(0)\to\infty$. $\boxed{\text{(c) as close to infinity as possible}}$
8. Part 8) — System Type and error constants from $G_{closed}(s)$. Inverting $G(s)=G_{closed}/(1-G_{closed})$ (the coefficient “$1.2s^2$” is read as $12s^2$, the reading that makes the quartic factor cleanly as $(s+3)(s+1)^3$ — a genuine Hurwitz-stable closed loop) gives $$G(s)=\frac{5s^2+10s+3}{s^2(s^2+6s+7)}.$$ The $s^2$ factor in the denominator makes this a $\boxed{\text{Type }N=2}$ system: both the position constant $K_{pos}$ and velocity constant $K_v$ are infinite (so $e_{ss(step)}=\boxed{0}$ and $e_{ss(ramp)}=\boxed{0}$), while the acceleration constant is finite, $K_a=\lim_{s\to0}s^2G(s)=3/7\approx\boxed{0.429\ \text{s}^{-2}}$, giving $e_{ss(parab)}=1/K_a=\boxed{7/3\approx2.33\ (233\%)}$.
9. Part 9) — ideal disturbance rejection, open-loop-gain framing. The disturbance-to-output closed-loop gain is $Y/D=G_2/(1+K_pG_1G_2)$ (Part 5's algebra with $K_p$ inserted); driving the OPEN-loop gain $K_pG_1G_2\to\infty$ drives this ratio to zero, exactly the same sensitivity argument as Part 7. $\boxed{\text{(c) as close to infinity as possible}}$
10. Part 10) — ideal disturbance rejection, closed-loop-gain framing. This sub-part asks about the disturbance-to-output CLOSED-loop transfer function itself (not the plant's open-loop gain): for the disturbance to have no effect on the output, that closed-loop gain must itself be zero, not one or infinite. $\boxed{\text{(a) as close to zero as possible}}$