Old spec 4.6 example, derivative action on a second order plant (ENGR5001)

Old spec 4.6 example, derivative action on a second order plantTutorialOld spec ENGR2022:210 min

ENGR202 (old spec) Control 4.6 Derivative and Velocity, derivative action example (slides 5-13)

Derivative action $u(t) = k_d\dfrac{d}{dt}\big(d(t) - x(t)\big)$ is applied, with unity negative feedback, to the plant $X = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}U$

Formulas you may need

Derivative action $u(t) = k_d\dfrac{de}{dt}$ , i.e. $U = k_ds(D - X)$ (learn this)
Negative feedback rule; Final Value Theorem (learn this)

(a)
Find the closed-loop Transfer Function and characteristic equation.
(b)
Show how the control gain changes the damping ratio. For a plant with $\zeta = 0.1$ and $\omega_n = 2$ rad/s, what $k_d$ gives a closed-loop damping ratio of 0.7?
kd =s
(c)
Find the steady state output for a unit step in the set point, and explain the practical problem with a step in the set point. What is used instead?
steady state output =