I am having trouble deriving what this project needs.
Problem for PBL
1. Based on “door opening control system” as shown by Figure 1, do the following:
a. Derive it mathematical model and block diagram (input is (Va) Rated motor terminal voltage
and output is door position 8L).
b. Analyze it stability
c. Implement P, PI, PD and PID controller as a controller of the system. Especially for PID
controller, gain tuning must done manual and automatic. Analyze each type controller
responses and compare with system response without controller.
d. By using PID controller you should ensure maximum overshoot (OS%) ≤ 10%, minimize Rise
time (Tr), Settling time (Tc), Peak Time (Tp) and reduce steady state error (ecc).
DC Motor data
1. (Va) Rated motor terminal voltage is 220 V
2. (Jm) Moment of inertia of the rotor is 0.222 kg.m2
3. (JL) Moment of inertia of Load is 0.022 kg.m2
4. (Bm) Motor viscous friction constant is 0.003 N.m.s/rad
5. (kb) Electromotive force constant is 1.25 V/rad/sec
6. (ka) Motor torque constant is 1.28 N.m/Amp
7. (Ra) Electric resistance is 11.2 Ohm
8. (La) Electric inductance is 0.121 H
9. (N) Rated motor speed is 1500 rpm.
Gear Data Gear 1
n1=50 , r1=10cm Gear 2
n2=750 , r1=150cm
Noted: no friction on load (between door rail, pully and gear). Door opening control system
The file I uploaded is the picture of the door opening control
If you are interested in the position only, the gears could be easily replaced with a gain block in Simulink, followed by backlash.
Hi Derrick Jeremiah Dave,
Newton's 2nd law: m·a = Σ Fᵢ
Neglecting all frictions and backlash effect on the gears, and assuming the roller shutter door as a point mass, and x as the vertical shutter opening as measured from the ground, then the simplified mechanical model should look as follows:
m·x" = F – m·g,
where m is the mass of the shutter, F is the lifting force produced by the motor, and g is the gravitational acceleration. You can complete the full model by adding the actuator dynamics (motor). The weight of the roller shutter (m·g) is the disturbance that pull the shutter downwards.
If the controller is unavailable or malfunction, F = 0, then
m·x" = – m·g
x" = – g.
Assuming g = 10 m/s2, and solving the 2nd-order ODE, if the roller shutter is partially opened at a certain height, x(0) = h, then the shutter will drop according to x(t) = h – 5·t2 until x = 0 (shutter hits the ground). Hence, if h = 9/5 = 1.8 m above ground, then the shutter will hit the ground in t = √(3/5) = 0.6 sec (by solving the quadratic equation).
Hi Prof. Michael Short,
Thanks for clearing that up. I should have known the rack and pinion implies the mechanism of a sliding door. Hope Derrick Jeremiah Dave could catch my mistake. The question asks about the system response without the controller. In the practical setting, if the controller is unavailable, the sliding door without a restoring mechanism (auto-shut) would stay still. But in the friction-less environment, there is no equilibrium position.
Hi Michael Short and Yew-Chung Chak , thank you for assisting me in this question and the explanations. Your answers really solved my complication for this question.
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