Question 4 of 8: Second-order dominant model, system type, transient specs, effect of a zero

Question 4 — Second-order dominant model, system type, transient specs, effect of a zero [5 + 5 + 5 + 5]

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 pole closest to the imaginary axis dominates.

Find. $(\zeta,\omega_n,K_{dc})$; system type and errors; step specs; and the effect of moving the zero to $-2$.

Part 1 — Dominant second-order model [5]

1. Dominant pair. $-1\pm j4$ gives $\omega_n=\sqrt{1^2+4^2}=\sqrt{17}=\boxed{4.123\ \text{rad/s}}$ and $\zeta=\dfrac{1}{\omega_n}=\dfrac{1}{4.123}=\boxed{0.2425}$ (the quadratic factor is $s^2+2s+17$).
2. DC gain. $K_{dc}=G(0)=\dfrac{5000(20)}{(17)(15)(404)}=\dfrac{100000}{103020}=\boxed{0.9707}$, where $|{-1\pm j4}|^2=17$, $|{-20\pm j2}|^2=404$.
3. Model. Preserving the DC gain, $$G_m(s)=0.9707\cdot\frac{17}{s^2+2s+17}=\boxed{\frac{16.5}{s^2+2s+17}}.$$

Part 2 — System type and steady-state errors [5]

1. System Type. $G(s)$ has no pole at the origin, so it is $\boxed{\text{Type 0}}$.
2. Step error. $K_{pos}=G(0)=0.9707$, so $e_{ss(step)}=\dfrac{1}{1+K_{pos}}=\dfrac{1}{1.9707}=\boxed{50.7\%}$.
3. Ramp error. $K_v=\lim_{s\to0}sG(s)=0$ for a Type-0 system, so $e_{ss(ramp)}=\dfrac{1}{K_v}=\boxed{\infty}$ (the output cannot track a ramp).

Part 3 — Step-response specifications [5]

1. Percent overshoot. $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=100\,e^{-0.2425\pi/0.9701}=\boxed{45.6\%}$.
2. Settling time. The dominant real part is $\zeta\omega_n=1$, so $T_{settle}(\pm2\%)=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{1}=\boxed{4.0\ \text{s}}$.

Part 4 — Effect of moving the zero to $-2$ [5]

The original zero at $-20$ is remote ($20\gg5\zeta\omega_n=5$), so it barely alters the dominant response — that is why the pure second-order model is valid. Moving it to $-2$ places it close to the dominant poles (its magnitude $2$ is only twice the dominant real part $1$, well inside the “$\ge5\times$” negligibility rule), so it strongly reshapes the response:

• Overshoot rises sharply. A left-half-plane zero adds a derivative-like term to the response; near the poles it injects a large early transient. A numerical step simulation of the second-order model with the zero gives $PO\approx132\%$, up from $\approx46\%$ — roughly a threefold increase.
• Faster rise, similar settling. The zero speeds up the initial rise (shorter rise/peak time), but the decay envelope is still set by the unchanged pole real part ($e^{-t}$), so $T_{settle}$ stays near $4$ s (marginally longer because the larger overshoot must decay).
• DC gain and error worsen. The DC gain drops by the factor $\tfrac{2}{20}=\tfrac{1}{10}$ to $K_{dc}=0.0971$, so $K_{pos}=0.0971$ and $e_{ss(step)}=\dfrac{1}{1.0971}=91.2\%$ — far worse steady-state accuracy.