Old spec 3.2 worked example, mass-spring-damper step response by Laplace (ENGR5001)

Old spec 3.2 worked example, mass-spring-damper step response by LaplaceTutorialOld spec ENGR2022:215 min

The mass-spring-damper system is modelled by $M\frac{d^2x}{dt^2} + C\frac{dx}{dt} + Kx = u(t)$

Formulas you may need

$\mathcal{L}[\dot x] = sX(s) - x(0)$ , $\mathcal{L}[\ddot x] = s^2X(s) - sx(0) - \dot x(0)$ ; unit step $\to 1/s$ ; $e^{-\alpha t} \leftrightarrow \dfrac{1}{s + \alpha}$ (learn this)
Final Value Theorem $\lim_{t\to\infty} f(t) = \lim_{s\to 0} sF(s)$ (learn this)

(a)
Take Laplace transforms (keeping the initial conditions), then set the initial conditions to zero and find the Transfer Function $X(s)/U(s)$ .
(b)
For $M = 1$ , $C = 3$ and $K = 2$ , find the response $x(t)$ to a unit step input using partial fractions and inverse Laplace transforms. Sketch the response.
A (coefficient of 1/s) =
B (coefficient of 1/(s+1)) =
C (coefficient of 1/(s+2)) =
x(1 s) =
(c)
Find the steady state value of $x$ using the Final Value Theorem, and confirm it using the steady state gain of the Transfer Function.
x(infinity) =