Question 2 of 8: Open-Loop Type, Error Constants and 2nd-Order Dominant-Pole Model

Question 2: Open-Loop Type, Error Constants and 2nd-Order Dominant-Pole Model (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. The same servo-positioning system as Question 1, with the PI gain fixed at $K_{op}=20$, and the exact factorized closed-loop transfer function printed above.

Find. Part A: $G_{open}(s)$, System Type, $K_{pos}$, $K_v$, $K_a$. Part B: $G_{cl}(0)$; the dominant 2nd-order model $K_{dc},\zeta,\omega_n,G_m(s)$; and the step-response specs PO, $T_{rise(10\%-90\%)}$, $T_{settle(\pm2\%)}$, $T_{period}$, $e_{ss(step)\%}$.

Approach. Break the outer loop to read $G_{open}(s)=C(s)\cdot P/(1+P)$ directly from Question 1's reduction, then take the standard $s\to0$ limits for the error constants. For Part B, evaluate the factored $G_{cl}(s)$ at $s=0$; then discard the fast pole $s=-11.9$ and note the near pole–zero pair $(s+0.5)/(s+0.505)$ nearly cancels, leaving the complex quadratic as the dominant 2nd-order model; read every step-response spec off the standard closed-form formulas for that model.

1. Part A.1) — open-loop transfer function and System Type. Breaking the outer feedback path (comparing $R$ and $Y$) but leaving the plant's own inner unity loop closed gives the forward path $$G_{open}(s)=C(s)\cdot\frac{P(s)}{1+P(s)}=\frac{K(s+0.5)}{s}\cdot\frac{10}{s^3+13s^2+30s+10} =\frac{10K(s+0.5)}{s(s^3+13s^2+30s+10)}.$$ The cubic factor is nonzero at $s=0$ (value $10$), so the only pole at the origin is the explicit $1/s$: $\boxed{\text{System Type }N=1}$.
2. Part A.2) — error constants at $K=K_{op}=20$. $K_{pos}=\lim_{s\to0}G_{open}(s)=\boxed{\infty}$ (Type 1, matching Question 1's zero-SSE-to-step spec). $K_v=\lim_{s\to0}sG_{open}(s)=\dfrac{10K\times0.5}{10}=0.5K$, so at $K_{op}=20$, $\boxed{K_v=10\ \text{s}^{-1}}$ — consistent with $e_{ss(ramp)}=1/K_v=1/10=0.1$, exactly Question 1's design target (this is how $K_{op}=20$ was solved for). $K_a=\lim_{s\to0}s^2G_{open}(s)=0$ since the numerator vanishes linearly in $s$ at the origin while only one pole sits there: $\boxed{K_a=0}$.
3. Part B.1) — DC gain. $$G_{cl}(0)=\frac{200(0.5)}{(11.9)(0.505)(16.64)}=\frac{100}{100.0}\approx\boxed{1.00\ \text{V/V}},$$ exactly matching the exact symbolic value from Question 1 ($G_{cl}(0)=5K/5K=1$) — consistent with the Type–1 loop having zero steady-state error to a step.
4. Part B.2) — 2nd-order dominant-pole model. The pole at $s=-11.9$ is far from the origin (fast, decays quickly) and the zero at $s=-0.5$ sits almost exactly on top of the pole at $s=-0.505$ (near cancellation) — both are dropped from the dynamics, replaced by their combined value AT $s=0$ so the reduced model keeps the full system's DC gain: $$K_{dc}=G_{cl}(0)\times\frac{\omega_n^2}{\omega_n^2}=1.00\ \ (\text{since evaluating the dropped factors at }s=0\text{ exactly reproduces }G_{cl}(0)).$$ Matching the surviving quadratic to $s^2+2\zeta\omega_ns+\omega_n^2=s^2+0.5925s+16.64$: $\omega_n=\sqrt{16.64}\approx\boxed{4.08\ \text{rad/s}}$, $\zeta=0.5925/(2\times4.08)\approx\boxed{0.0726}$ (lightly damped). So $$\boxed{G_m(s)=\frac{16.64}{s^2+0.5925s+16.64}}\qquad(K_{dc}=1.00).$$
5. Part B.3) — step-response specifications of the dominant model. $$PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=100\,e^{-0.0726\pi/\sqrt{1-0.0726^2}}\approx\boxed{79.6\%}$$ (a large overshoot, expected for such a lightly-damped $\zeta$). Using the exact underdamped step response $c(t)=1-\tfrac{e^{-\zeta\omega_nt}}{\sqrt{1-\zeta^2}}\sin(\omega_dt+\cos^{-1}\zeta)$ (evaluated numerically): $T_{rise(10\%-90\%)}\approx\boxed{4.82\ \text{s}}$; the response only re-enters and stays within the $\pm2\%$ band by $T_{settle(\pm2\%)}\approx\boxed{13.2\ \text{s}}$ (close to the estimate $4/(\zeta\omega_n)=13.5$ s); the ringing period is $T_{period}=2\pi/\omega_d=2\pi/(\omega_n\sqrt{1-\zeta^2})\approx\boxed{1.54\ \text{s}}$; and since $K_{dc}=1.00$ exactly, $\boxed{e_{ss(step)\%}=0\%}$.