Old spec 3.6 example, closed-loop control of vehicle speed (ENGR5001)

Old spec 3.6 example, closed-loop control of vehicle speedTutorialOld spec ENGR202Third6 min

ENGR202 (old spec) Control 3.6 Block Diagram Analysis, closed-loop control of vehicle speed (slides 5-7 and 16-19)

The speed $X$ of a vehicle is controlled by a proportional controller $U = K_p(D - X)$ , where $D$ is the desired speed and $U$ the engine force. The vehicle model is $X = \dfrac{0.8}{5s + 1}U$ (unity negative feedback of the speed).

Formulas you may need

Unity negative feedback: $X = \dfrac{C(s)G(s)}{1 + C(s)G(s)}D$ (learn this)

(a)
Use the rules of block diagram analysis to find the Closed-Loop Transfer Function $X/D$ .
(b)
For $K_p = 5$ , find the closed-loop steady state gain and time constant.
steady state gain =
closed-loop time constant =s