Sure! Let's work through each part step-by-step. --- # 1. **Nature of the Transfer Functions** A standard second-order system has the form: \[ T(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \] where: - \(\omega_n\) = natural frequency - \(\zeta\) = damping ratio The nature depends on \(\zeta\): - **Overdamped:** \(\zeta > 1\) - **Critically damped:** \(\zeta = 1\) - **Underdamped:** \( < \zeta < 1\) - **Undamped:** \(\zeta = \) Let's analyze each transfer function: --- ### (a) \(T(s) = \frac{1}{s^2 + 3s + 7}\) Compare to standard form: - \(2\zeta\omega_n = 3\) - \(\omega_n^2 = 7\) → \(\omega_n = \sqrt{7} \approx 2.65\) So, \[ 2\zeta\omega_n = 3 \implies \zeta = \frac{3}{2\omega_n} = \frac{3}{2 \times 2.65} \approx .57 \] **Conclusion**: \(\zeta \approx .57\) (\( < \zeta < 1\)), so it is **underdamped**. --- ### (b) \(T(s) = \frac{13}{s^2 + 4s + 37}\) - \(2\zeta\omega_n = 4\) - \(\omega_n^2 = 37 \implies \omega_n = \sqrt{37} \approx 6.08\) So, \[ \zeta = \frac{4}{2 \times 6.08} \approx .33 \] **Conclusion:** \(\zeta \approx .33\) (\( < \zeta < 1\)), so it is **underdamped**. --- ### (c) \(T(s) = \frac{1}{s^2 + 4s + 13}\) - \(2\zeta\omega_n = 4\) - \(\omega_n^2 = 13 \implies \omega_n = \sqrt{13} \approx 3.61\) So, \[ \zeta = \frac{4}{2 \times 3.61} \approx .55 \] **Conclusion:** \(\zeta \approx .55\) (\( < \zeta < 1\)), so it is **underdamped**. --- # 2. **Step Response Analysis for \(T(s) = \frac{13}{s^2 + 4s + 37}\)** Let’s analyze for a unit step input: ## **a) Steady-State (Final) Value** For a unit step input (\(R(s) = \frac{1}{s}\)), use the Final Value Theorem: \[ y_{ss} = \lim_{t \to \infty} y(t) = \lim_{s \to } sY(s) \] \[ Y(s) = T(s) \cdot R(s) = \frac{13}{s^2 + 4s + 37} \cdot \frac{1}{s} \] \[ y_{ss} = \lim_{s \to } s \cdot \frac{13}{s(s^2 + 4s + 37)} = \frac{13}{37} \] \[ \approx .351 \] --- ## **b) 10-90% Rise Time** Rise time (\(t_r\)) for underdamped system is approximately: \[ t_r \approx \frac{\pi - \theta}{\omega_d} \] where - \(\omega_d = \omega_n \sqrt{1-\zeta^2}\) - \(\theta = \arccos(\zeta)\) Given: - \(\omega_n = 6.08\) - \(\zeta = .33\) Calculate: - \(\omega_d = 6.08 \sqrt{1-.33^2} = 6.08 \times .944 = 5.74\) - \(\theta = \arccos(.33) \approx 1.23\) radians \[ t_r \approx \frac{\pi - 1.23}{5.74} = \frac{1.91}{5.74} \approx .33 \text{ s} \] --- ## **c) 2% Settling Time** Settling time (\(t_s\)): \[ t_s \approx \frac{4}{\zeta \omega_n} = \frac{4}{.33 \times 6.08} = \frac{4}{2.01} \approx 2. \text{ s} \] --- ## **d) Maximum % Overshoot** \[ \%OS = e^{-\frac{\zeta \pi}{\sqrt{1-\zeta^2}}} \times 100\% \] \[ \sqrt{1-.33^2} \approx .944 \] \[ -\frac{.33 \pi}{.944} \approx -1.097 \] \[ \%OS = e^{-1.097} \times 100 \approx 33.4\% \] --- ## **Summary Table** | Quantity | Formula | Value | |--------------|---------|---------| | Steady-State | \(\frac{13}{37}\) | .4 | | Rise Time | \(\frac{\pi - \arccos(\zeta)}{\omega_d}\) | .3 s | | Settling Time| \(\frac{4}{\zeta\omega_n}\) | 2. s | | % Overshoot | \(e^{-\frac{\zeta \pi}{\sqrt{1-\zeta^2}}} \times 100\%\) | 33.4% | --- ## **Sketch** - The step response will rise quickly, overshoot by ~33%, and settle to .4 after about 2 seconds. - Mark .4 on the y-axis, .3 s for rise time, 2. s for settling time, and the overshoot peak at ~.53. --- **Let me know if you want a hand-drawn sketch or further clarifications!**