Question 2 of 8: System type, error constants, dominant 2nd-order model (compulsory)

Question 2 — System type, error constants, dominant 2nd-order model (compulsory) [20]

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.

Part A — open loop, type and error constants

Given. the Q1 loop at $K=K_{op}=20$. Find. $G_{open}(s)$, the type, and $K_{pos},K_v,K_a$.

1. Equivalent unity-feedback open loop. Since the actuating error is $e_1=R-Y$, the equivalent forward transfer function is $G_{open}=G_{cl}/(1-G_{cl})$: $$G_{open}(s)=\frac{10K(s+0.5)}{s\,(s^3+13s^2+30s+10)}.$$ One free integrator ($s$ in the denominator) $\Rightarrow$ Type 1.
2. Error constants at $K_{op}=20$. $K_{pos}=\lim_{s\to0}G_{open}=\infty$ (Type 1); $K_v=\lim_{s\to0}sG_{open}=\dfrac{10K(0.5)}{10}=0.5K=\boxed{10}$; $K_a=\lim_{s\to0}s^2G_{open}=\boxed{0}$.
3. Consistency. $e_{ss(ramp)}=1/K_v=1/10=0.1$ V/V and $e_{ss(step)}=1/(1+K_{pos})=0$ — exactly the Q1 specs.

Part B — dominant-pole model and step specs

Given. factored $G_{cl}$: poles $-11.9,\,-0.505,\,-0.296\pm j4.068$ and a zero at $-0.5$. Find. $G_{cl}(0)$, the reduced model, and the step specs. Approach. the zero at $-0.5$ nearly cancels the slow real pole at $-0.505$ and the pole at $-11.9$ is far away, so the complex pair dominates.

1. DC gain. $G_{cl}(0)=\dfrac{200(0.5)}{(11.9)(0.505)(16.64)}=\dfrac{100}{100.0}=\boxed{1}.$ (A unity DC gain is expected from the zero step-error result.)
2. Cancel and drop. The near-cancelling factor $(s+0.5)/(s+0.505)\approx1$, and the far pole contributes its DC value $s+11.9\to11.9$. What remains is $\dfrac{200(s+0.5)}{(s+11.9)(s+0.505)}\Big|_{s=0}=16.64$ over the quadratic, so $K_{dc}\omega_n^2=16.64$.
3. Model parameters. From $s^2+0.5925s+16.64$: $\omega_n=\sqrt{16.64}=\boxed{4.08\ \text{rad/s}}$, $\zeta=\dfrac{0.5925}{2\omega_n}=\boxed{0.0726}$, and $K_{dc}=\dfrac{16.64}{\omega_n^2}=\boxed{1.0}$. Hence $$\boxed{G_m(s)=\frac{16.64}{s^2+0.5925\,s+16.64}.}$$
4. Overshoot. $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=100\,e^{-0.2288}=\boxed{79.6\%}$ — a very lightly damped response.
5. Rise / settle / period. With $\omega_d=\omega_n\sqrt{1-\zeta^2}=4.068$ rad/s and $\sigma=\zeta\omega_n=0.296$: the numerically evaluated 10–90% rise time is $T_{rise}=\boxed{0.265\ \text{s}}$; $T_{settle(\pm2\%)}=\dfrac{4}{\sigma}=\boxed{13.5\ \text{s}}$; $T_{period}=\dfrac{2\pi}{\omega_d}=\boxed{1.54\ \text{s}}$.
6. Step error. $K_{dc}=1\Rightarrow e_{ss(step)}=\boxed{0\%}$.