--- title: "Find a, b, c from time specifications" subject: "Control Systems" chapter: 3 key: "semester_3/control_systems:mech-abc-overshoot-tp-ess" source: https://www.bctengineer.com/q/semester_3/control_systems/mech-abc-overshoot-tp-ess site: BCT Engineering --- # Find a, b, c from time specifications Control Systems · Chapter 3: System Transfer Function and Responses - Asked in: 2080 Baisakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/mech-abc-overshoot-tp-ess - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3 ## Question For a mechanical system with closed loop transfer function θ(s)/T(s) = 1/(as² + bs + c), where θ(s) is the output of step input T(t) = 10 Nm. Determine values of a, b and c if maximum overshoot is 6%, peak time (tp) = 1 sec, and ess = 0.5. ## Short answer From steady-state output 0.5 for a 10 Nm step, c=20 Nm/rad. A 6% overshoot yields damping ratio 0.667; peak time 1 s gives natural frequency 4.217 rad/s. Then a=1.125 kg-m^2, b=6.33 Nm-s/rad. ## Answer Compare the system with the standard second order form. The final value fixes $c$; the overshoot fixes $\zeta$; the peak time then fixes $\omega_n$. **Interpretation:** "$e_{ss} = 0.5$" is taken as the steady state value of the output, $\theta(\infty) = 0.5$ rad, for the 10 Nm step (the usual reading of this problem, since $\theta$ and $T$ have different units). $$ \frac{\theta(s)}{T(s)} = \frac{1}{as^2+bs+c} = \frac{1/a}{s^2 + \frac{b}{a}s + \frac{c}{a}} \;\Rightarrow\; \omega_n^2 = \frac{c}{a},\quad 2\zeta\omega_n = \frac{b}{a} $$ #### Step 1: Find $c$ from the steady state $T(s) = 10/s$. By the final value theorem: $$ \theta(\infty) = \lim_{s\to 0} s\cdot\frac{10}{s}\cdot\frac{1}{as^2+bs+c} = \frac{10}{c} = 0.5 \;\Rightarrow\; c = 20\ \text{Nm/rad} $$ #### Step 2: Damping ratio from 6% overshoot $$ \zeta = \frac{-\ln M_p}{\sqrt{\pi^2 + (\ln M_p)^2}} = \frac{-\ln 0.06}{\sqrt{\pi^2 + (\ln 0.06)^2}} = \frac{2.8134}{\sqrt{9.8696 + 7.9153}} = 0.6671 $$ #### Step 3: Natural frequency from $t_p = 1$ s $$ t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} \;\Rightarrow\; \omega_n = \frac{\pi}{1\times\sqrt{1 - 0.6671^2}} = \frac{3.1416}{0.7449} = 4.217\ \text{rad/s} $$ #### Step 4: Find $a$ and $b$ $$ \begin{aligned} a &= \frac{c}{\omega_n^2} = \frac{20}{4.217^2} = \frac{20}{17.785} = 1.125\ \text{kg-m}^2 \\ b &= 2\zeta\omega_n a = 2(0.6671)(4.217)(1.125) = 6.328\ \text{Nm-s/rad} \end{aligned} $$ **Answer:** $a \approx 1.125$ kg-m², $b \approx 6.33$ Nm-s/rad, $c = 20$ Nm/rad. (If instead $e_{ss} = 10 - \theta(\infty) = 0.5$ is used, then $c = 10/9.5 = 1.053$, $a = 0.0592$, $b = 0.333$; $\zeta$ and $\omega_n$ are unchanged.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the answer, which step fixes the constant c in 1/(as^2+bs+c) for a 10 Nm step with theta(inf) = 0.5? A. Final value theorem: theta(inf) = 10/c = 0.5, so c = 20 B. Peak time equation, c = pi/tp C. Overshoot formula, c = Mp x 100 D. Settling time, c = 4/ts Answer: A. For a step of 10 Nm the steady state output is 10/c, so setting it equal to 0.5 gives c = 20 Nm/rad. (Key point: Final value gives c) 2. Which relation gives the damping ratio from the 6% overshoot? A. zeta = Mp/100 = 0.06 B. zeta = -ln(Mp)/sqrt(pi^2 + (ln Mp)^2), about 0.667 C. zeta = pi/tp = 3.14 D. zeta = c/(2a) = 10 Answer: B. Mp = exp(-pi zeta/sqrt(1-zeta^2)) is inverted to give zeta = 0.667. (Key point: Damping ratio from overshoot) 3. With zeta = 0.667 and tp = 1 s, how is wn found? A. wn = pi/tp = 3.14 rad/s B. wn = 4/(zeta tp) = 6 rad/s C. wn = pi/(tp sqrt(1 - zeta^2)) = 4.217 rad/s D. wn = sqrt(c) = 4.47 rad/s Answer: C. Peak time is pi/wd with wd = wn sqrt(1-zeta^2), so wn = 3.1416/0.7449 = 4.217. (Key point: Natural frequency from peak time) 4. After c = 20 and wn = 4.217, how are a and b obtained from comparing with the standard form? A. a = wn^2/c and b = 2 zeta wn B. a = c wn^2 and b = zeta wn C. a = 1 and b = 2 zeta wn D. a = c/wn^2 = 1.125 and b = 2 zeta wn a = 6.33 Answer: D. Since wn^2 = c/a and 2 zeta wn = b/a, a = c/wn^2 and b = 2 zeta wn a. 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