Old spec 4.2 worked example, open and closed loop control of a DC motor (ENGR5001)

Old spec 4.2 worked example, open and closed loop control of a DC motorTutorialOld spec ENGR2022:212 min

ENGR202 (old spec) Control 4.2 Example DC Motor, worked example and exercise (slides 5-18)

A DC motor has input voltage $U$ and shaft speed $X$ (measured as a voltage by a tachometer), modelled as $\tau\dot x + x = Ku$ , i.e. $X = H(s)U$ with $H(s) = K/(\tau s + 1)$ . In an open-loop test with no load disturbance, a constant input $U = 0.5$ V gives a steady motor speed $X = 20$ V, which is the desired speed $D = 20$ .

Formulas you may need

$x(t \to \infty) = K \times u_0$ ; open-loop proportional control $U = K_pD$ (learn this)
Closed loop with $U = K_p(D - X)$ : $X = \dfrac{HK_p}{1 + HK_p}D$ (learn this)

(a)
What is the steady state gain $K$ of the motor?
K =
(b)
An open-loop proportional controller $U = K_pD$ is used. What control gain $K_p$ gives the desired speed at steady state? Why is this sensitive to load disturbances?
Kp (open loop) =
(c)
The loop is now closed: $U = K_p(D - X)$ . Derive the closed-loop relationship between $D$ and $X$ by algebra, and find the closed-loop steady state gain with $K_p = 0.025$ and with $K_p = 2.5$ .
CL gain, Kp = 0.025 =
CL gain, Kp = 2.5 =