1 / 12

A Study on Motor Speed Control

This study at Boğaziçi University delves into motor speed control using system equations. Transfer functions and numerical values are examined for open-loop and PID control systems to analyze disturbance rejection and steady-state responses, with a focus on PID controller design. Study link: http://www.engin.umich.edu/group/ctm/index.html

maughan
Download Presentation

A Study on Motor Speed Control

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. A Study on Motor Speed Control ME 335 Boğaziçi University

  2. System Equations KVL and Newton’s 2’nd law gives : Let : Output : y = Input : u = va Disturbance w = Tl Laplace transforming, we obtain : ME 335 Boğaziçi University

  3. Transfer Functions Where: ME 335 Boğaziçi University

  4. System Parameters For this example, the following values for the physical parameters will be assumed. • moment of inertia of the rotor (J) = 0.0001 kg.m2/s2* damping ratio of the mechanical system (b) = 0.0 Nms* electromotive force constant (K=Ke=Kt) = 0.01 Nm/Amp* electric resistance (R) = 0.1 ohm * electric inductance (L) = 0.0005 H* input (V): Source Voltage* output (): angular velocity of shaft * disturbance (Tl) : Load torque • * The rotor and shaft are assumed to be rigid Reference : http://www.engin.umich.edu/group/ctm/index.html ME 335 Boğaziçi University

  5. Numerical Values J=0.0001; b=0.0; Ke=0.01; Kt=0.01; R=0.1; L=0.0005; A=Kt/(b*R+Kt*Ke);A=100 B=1/(b*R+Kt*Ke);B=10000 den=[J*L/(b*R+Kt*Ke) (J*R+b*L)/(b*R+Kt*Ke) 1]; r = roots(den); tau1= -1/r(1); tau2=-1/r(2);tau1 = 0.0947 s.: mech. time constant tau2 = 0.0053 s.: electrical time constant. ME 335 Boğaziçi University

  6. Open Loop Step Response step(A,den,0:.05:1.5) title('Step Response for the Open Loop System') ME 335 Boğaziçi University

  7. Steady State Model yss = 100 u + 10000 w where y is in rad/s, u is in Volts and w is in N/m. For example, u = va = 1.5 Volts gives ss = 150 rad/s = 1432 rpm. If w = Tl = -0.002 N m., then ss = 150 - 20 = 130 rad/s = 1241 rpm. Result : Without control, the speed decreases (or increases) proportional to w. Controller must increase (or decrease) u, to compensate for the effect of disturbance. ME 335 Boğaziçi University

  8. 100 4.2 Disturbance : Gain Load Torque Tl Ref. 1.5 signal (Volts) Steady state Voltage input x' = Ax+Bu PID y = Cx+Du Saturation Motor PID Controller DC MOTOR speed Voltage input 0.028 TACHOMETER Feedback Control Simulink block diagram Input saturation : Input voltage va can be between 0 and 2.5 volts Measurement : Tachometer gain : 0.028 Volts/rad/s Reference signal : 150 rad/s * 0.028 Volts/rad/s = 4.2 Volts; is compared with tachometer output. ME 335 Boğaziçi University

  9. Proportional Control yref is increased from 150 to 160 rad/s. There is ss. error, ss. error decreases as KP is increased ME 335 Boğaziçi University

  10. Disturbance Rejection A step torque disturbance of Tl = -0.002 Nm. is applied Compared to no control (130 rad/s), there is improvement, still there is ss. error. ME 335 Boğaziçi University

  11. PID Controller Control signal is a linear combination of error, integral of error and time rate of change of error. ME 335 Boğaziçi University

  12. PID Control of DC Motor No steady state error, good response. ME 335 Boğaziçi University

More Related