2020 - Q.7(a): Solve the differential equation $$\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 3x = 1$$ with initial conditions $$x(0) = 1$$ and $$x'(0) = -1$$ using Laplace transform. 1. Take Laplace transform of both sides: $$s^2X(s) - sx(0) - x'(0) + 2(sX(s) - x(0)) + 3X(s) = \frac{1}{s}$$ 2. Substitute initial conditions: $$s^2X(s) - s(1) - (-1) + 2(sX(s) - 1) + 3X(s) = \frac{1}{s}$$ 3. Simplify: $$s^2X(s) - s + 1 + 2sX(s) - 2 + 3X(s) = \frac{1}{s}$$ 4. Group terms: $$(s^2 + 2s + 3)X(s) = \frac{1}{s} + s - 1$$ 5. So, $$X(s) = \frac{\frac{1}{s} + s - 1}{s^2 + 2s + 3}$$ 6. Complete the square in denominator: $$s^2 + 2s + 3 = (s+1)^2 + 2$$ 7. Rewrite numerator: $$\frac{1}{s} + s - 1 = \frac{1}{s} + (s - 1)$$ 8. Use partial fraction or inverse Laplace transform techniques: Split: $$X(s) = \frac{1}{s((s+1)^2 + 2)} + \frac{s-1}{(s+1)^2 + 2}$$ 9. Inverse Laplace transform of each term: - For $$\frac{s-1}{(s+1)^2 + 2}$$, write as $$\frac{(s+1)-2}{(s+1)^2 + 2} = \frac{s+1}{(s+1)^2 + 2} - \frac{2}{(s+1)^2 + 2}$$ - Using known transforms: $$\mathcal{L}^{-1}\left\{\frac{s+a}{(s+a)^2 + b^2}\right\} = e^{-at} \cos(bt)$$ $$\mathcal{L}^{-1}\left\{\frac{b}{(s+a)^2 + b^2}\right\} = e^{-at} \sin(bt)$$ So, $$\mathcal{L}^{-1}\left\{\frac{s+1}{(s+1)^2 + 2}\right\} = e^{-t} \cos(\sqrt{2}t)$$ $$\mathcal{L}^{-1}\left\{\frac{2}{(s+1)^2 + 2}\right\} = \sqrt{2} e^{-t} \sin(\sqrt{2}t)$$ Therefore, $$\mathcal{L}^{-1}\left\{\frac{s-1}{(s+1)^2 + 2}\right\} = e^{-t} \cos(\sqrt{2}t) - \sqrt{2} e^{-t} \sin(\sqrt{2}t)$$ 10. For $$\frac{1}{s((s+1)^2 + 2)}$$, use convolution theorem or partial fraction expansion. Let $$F(s) = \frac{1}{s}, G(s) = \frac{1}{(s+1)^2 + 2}$$ Inverse Laplace: $$f(t) = 1, g(t) = \frac{1}{\sqrt{2}} e^{-t} \sin(\sqrt{2}t)$$ Convolution: $$h(t) = \int_0^t f(\tau) g(t-\tau) d\tau = \int_0^t \frac{1}{\sqrt{2}} e^{-(t-\tau)} \sin(\sqrt{2}(t-\tau)) d\tau$$ Change variable: $$= \frac{1}{\sqrt{2}} \int_0^t e^{-u} \sin(\sqrt{2} u) du$$ This integral evaluates to: $$\frac{1}{\sqrt{2}} \left[ \frac{1 - e^{-t}(\cos(\sqrt{2} t) + \frac{1}{\sqrt{2}} \sin(\sqrt{2} t))}{1 + 2} \right] = \frac{1}{3\sqrt{2}} \left(1 - e^{-t} \cos(\sqrt{2} t) - \frac{1}{\sqrt{2}} e^{-t} \sin(\sqrt{2} t) \right)$$ 11. Combine all parts: $$x(t) = e^{-t} \cos(\sqrt{2} t) - \sqrt{2} e^{-t} \sin(\sqrt{2} t) + \frac{1}{3\sqrt{2}} \left(1 - e^{-t} \cos(\sqrt{2} t) - \frac{1}{\sqrt{2}} e^{-t} \sin(\sqrt{2} t) \right)$$ Simplify: $$x(t) = \frac{1}{3\sqrt{2}} + e^{-t} \left( \cos(\sqrt{2} t) \left(1 - \frac{1}{3\sqrt{2}} \right) - e^{-t} \sin(\sqrt{2} t) \left( \sqrt{2} + \frac{1}{3 \times 2} \right) \right)$$ Final boxed answer: $$\boxed{x(t) = \frac{1}{3\sqrt{2}} + e^{-t} \left( \left(1 - \frac{1}{3\sqrt{2}}\right) \cos(\sqrt{2} t) - \left( \sqrt{2} + \frac{1}{6} \right) \sin(\sqrt{2} t) \right)}$$ --- 2020 - Q.7(b)(i): For $$T(s) = \frac{16}{s^2 + 3s + 16}$$ find damping ratio $$\zeta$$, natural frequency $$\omega_n$$, settling time $$T_s$$, peak time $$T_p$$, rise time $$T_r$$, and percent overshoot %OS. 1. Standard form denominator: $$s^2 + 2\zeta \omega_n s + \omega_n^2 = s^2 + 3s + 16$$ 2. Equate coefficients: $$2\zeta \omega_n = 3$$ $$\omega_n^2 = 16 \Rightarrow \omega_n = 4$$ 3. Calculate $$\zeta$$: $$2\zeta \times 4 = 3 \Rightarrow \zeta = \frac{3}{8} = 0.375$$ 4. Settling time (2% criterion): $$T_s = \frac{4}{\zeta \omega_n} = \frac{4}{0.375 \times 4} = \frac{4}{1.5} = 2.6667$$ seconds 5. Peak time: $$T_p = \frac{\pi}{\omega_n \sqrt{1 - \zeta^2}} = \frac{\pi}{4 \sqrt{1 - 0.375^2}} = \frac{\pi}{4 \times 0.927} = 0.847$$ seconds 6. Rise time (approximate for underdamped): $$T_r \approx \frac{\pi - \theta}{\omega_d}$$ where $$\theta = \arccos(\zeta) = \arccos(0.375) = 1.186$$ radians $$\omega_d = \omega_n \sqrt{1 - \zeta^2} = 4 \times 0.927 = 3.708$$ So, $$T_r = \frac{\pi - 1.186}{3.708} = \frac{1.955}{3.708} = 0.527$$ seconds 7. Percent overshoot: $$\%OS = e^{-\frac{\zeta \pi}{\sqrt{1 - \zeta^2}}} \times 100 = e^{-\frac{0.375 \pi}{0.927}} \times 100 = e^{-1.27} \times 100 = 28.1\%$$ Final boxed answers: $$\boxed{\zeta = 0.375, \quad \omega_n = 4, \quad T_s = 2.67\,s, \quad T_p = 0.85\,s, \quad T_r = 0.53\,s, \quad \%OS = 28.1\%}$$ --- 2020 - Q.7(b)(ii): For $$T(s) = \frac{0.04}{s^2 + 0.02 s + 0.04}$$ find $$\zeta$$, $$\omega_n$$, $$T_s$$, $$T_p$$, $$T_r$$, and %OS. 1. Compare denominator: $$s^2 + 2\zeta \omega_n s + \omega_n^2 = s^2 + 0.02 s + 0.04$$ 2. Equate: $$2\zeta \omega_n = 0.02$$ $$\omega_n^2 = 0.04 \Rightarrow \omega_n = 0.2$$ 3. Calculate $$\zeta$$: $$2 \zeta \times 0.2 = 0.02 \Rightarrow \zeta = \frac{0.02}{0.4} = 0.05$$ 4. Settling time: $$T_s = \frac{4}{\zeta \omega_n} = \frac{4}{0.05 \times 0.2} = \frac{4}{0.01} = 400$$ seconds 5. Peak time: $$T_p = \frac{\pi}{\omega_n \sqrt{1 - \zeta^2}} = \frac{\pi}{0.2 \times \sqrt{1 - 0.0025}} \approx \frac{\pi}{0.2} = 15.71$$ seconds 6. Rise time: $$\theta = \arccos(0.05) = 1.52$$ radians $$\omega_d = 0.2 \times \sqrt{1 - 0.0025} = 0.1997$$ $$T_r = \frac{\pi - 1.52}{0.1997} = \frac{1.62}{0.1997} = 8.11$$ seconds 7. Percent overshoot: $$\%OS = e^{-\frac{\zeta \pi}{\sqrt{1 - \zeta^2}}} \times 100 = e^{-\frac{0.05 \pi}{0.9987}} \times 100 = e^{-0.157} \times 100 = 85.5\%$$ Final boxed answers: $$\boxed{\zeta = 0.05, \quad \omega_n = 0.2, \quad T_s = 400\,s, \quad T_p = 15.7\,s, \quad T_r = 8.11\,s, \quad \%OS = 85.5\%}$$