Question 6 of 7: Additional poles/zeros and step-response specifications

Question 6 — Additional poles/zeros and step-response specifications [12 + 8]

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 — Fourth-order system and its dominant model [12]

Given. Zero $-12$; poles $-20,-16,-1\pm j5$; DC gain $0.95$.

Find. $G(s)$ polynomial form; reduced $G_m(s)$; step-response estimates.

1. (a) Transfer function. $-1\pm j5$ give the factor $s^2+2s+26$, so $G(s)=\dfrac{K(s+12)}{(s+20)(s+16)(s^2+2s+26)}$. The DC gain fixes $K$: $\dfrac{K(12)}{(20)(16)(26)}=0.95\Rightarrow K=658.7$. Expanding, $$G(s)=\frac{658.7\,s+7904}{s^4+38s^3+418s^2+1576s+8320}.$$
2. (b) Why a second-order model. The real poles $-16,-20$ have real parts $16$–$20\times$ larger than the dominant pair's ($-1$); their modes decay $\ge16\times$ faster and are nearly cancelled by the neighbouring zero at $-12$, so they contribute negligibly to the transient. Keeping only $-1\pm j5$ and preserving the DC gain, $$G_m(s)=K_{dc}\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}=\frac{0.95(26)}{s^2+2s+26}=\frac{24.7}{s^2+2s+26},$$ with $\boxed{K_{dc}=0.95},\ \omega_n=\sqrt{26}=\boxed{5.10\ \text{rad/s}},\ \zeta=\dfrac{2}{2\omega_n}=\boxed{0.196}.$
3. (c) Step-response estimates. $e_{ss}=(1-K_{dc})\times100=\boxed{5\%}$; $\text{PO}=e^{-\zeta\pi/\sqrt{1-\zeta^2}}\times100=\boxed{53.4\%}$; $T_{s(\pm2\%)}=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{1.0}=\boxed{4.0\ \text{s}}$ (here $\zeta\omega_n=1$ exactly).

Part B — Model identified from the step response [8]

Given (read from Figure Q6.1). Steady-state value $y_{ss}=0.85$; first peak $\approx1.16$ at $t_p\approx1.2$ s; first trough $\approx0.75$ at $t\approx2.2$ s.

Find. $K_{dc},\zeta,\omega_n$ and $G_m(s)$.

[Figure not reproduced: Figure Q6.1 (redrawn) — measured step response, showing $y_{ss}=0.85$, the first peak ($1.16$ at $1.2$ s) and the $\pm2\%$ settling band. See the official exam paper.]

1. DC gain. $K_{dc}=y_{ss}=\boxed{0.85}$ (unit-step input).
2. Damping from overshoot. $\text{PO}=\dfrac{1.16-0.85}{0.85}=0.365$, so $\zeta=\dfrac{-\ln0.365}{\sqrt{\pi^2+\ln^2 0.365}}=\boxed{0.306}.$
3. Natural frequency from peak time. $t_p=\dfrac{\pi}{\omega_n\sqrt{1-\zeta^2}}=1.2\ \text{s}\Rightarrow \omega_n=\dfrac{\pi}{1.2\sqrt{1-0.306^2}}=\boxed{2.75\ \text{rad/s}}.$ (The peak-to-trough spacing of $1.0$ s gives a consistent damped period $T_d\approx2.0$ s.)
4. Model. $$G_m(s)=\frac{K_{dc}\,\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}=\frac{0.85(7.56)}{s^2+1.68s+7.56}=\boxed{\frac{6.43}{s^2+1.68s+7.56}}.$$