Question 8 of 8: Analytical step response and derivation of the 2nd-order specs

Question 8 — Analytical step response and derivation of the 2nd-order specs [10 + 10]

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 — Analytical step response [10]

Approach. Partial-fraction $Y(s)=G(s)/s$ (with a repeated pole at $-2$) and invert term-by-term using the Laplace table.

1. Partial fractions. $$\frac{2s+5}{s(s+2)^2(s+4)}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{(s+2)^2}+\frac{D}{s+4}.$$
2. Residues. $A=\dfrac{5}{16}$; $C=\left.\dfrac{2s+5}{s(s+4)}\right|_{-2}=-\dfrac14$; $D=\left.\dfrac{2s+5}{s(s+2)^2}\right|_{-4}=\dfrac{3}{16}$; $B=\dfrac{d}{ds}\!\left[\dfrac{2s+5}{s(s+4)}\right]_{-2}=-\dfrac12$.
3. Invert. Using $\tfrac1s\to1$, $\tfrac{1}{s+a}\to e^{-at}$, $\tfrac{1}{(s+a)^2}\to t\,e^{-at}$: $$\boxed{y(t)=\tfrac{5}{16}-\tfrac12e^{-2t}-\tfrac14t\,e^{-2t}+\tfrac{3}{16}e^{-4t}.}$$
4. Checks. $y(0)=\tfrac{5}{16}-\tfrac12+\tfrac{3}{16}=0$ ✓ and $y(\infty)=\tfrac{5}{16}=G(0)$ ✓ (initial and final value both consistent).

Part B — Derivation of the overshoot and settling-time formulas [10]

This part is a derivation, so it is written as flowing steps rather than a numeric table. Start from the unit-step response of the standard system (Laplace-table entry): $$y(t)=1-\frac{1}{\sqrt{1-\zeta^2}}\,e^{-\zeta\omega_n t}\,\sin\!\big(\omega_d t+\cos^{-1}\zeta\big),\qquad \omega_d=\omega_n\sqrt{1-\zeta^2}.$$

1. Peak time. Setting $\dot y=0$, the derivative reduces to $\dfrac{\omega_n}{\sqrt{1-\zeta^2}}e^{-\zeta\omega_n t}\sin(\omega_d t)=0$, whose first positive root is $t_p=\dfrac{\pi}{\omega_d}$.
2. Overshoot. Evaluate $y(t_p)$. At $\omega_d t_p=\pi$, $\sin(\pi+\cos^{-1}\zeta)=-\sin(\cos^{-1}\zeta)=-\sqrt{1-\zeta^2}$, so $$y(t_p)=1-\frac{e^{-\zeta\omega_n\pi/\omega_d}}{\sqrt{1-\zeta^2}}\big(-\sqrt{1-\zeta^2}\big)=1+e^{-\zeta\pi/\sqrt{1-\zeta^2}}.$$ Since the final value is $1$, the fractional overshoot is $y(t_p)-1$, hence $$\boxed{PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}\ \%.}$$
3. Settling time. The response is bounded by the exponential envelope $1\pm\dfrac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}}$. Entering the $\pm2\%$ band requires $\dfrac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}}\le0.02$. Taking the dominant exponential ($\sqrt{1-\zeta^2}\approx1$, $\ln(0.02)\approx-3.9\approx-4$): $$e^{-\zeta\omega_n T_s}=0.02\ \Rightarrow\ \boxed{T_{settle(\pm2\%)}=\frac{4}{\zeta\omega_n}.}$$