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Dr. Hatem Elaydi Digital Control, EELE 4360 Dec. 16, 2014

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Presentation on theme: "Dr. Hatem Elaydi Digital Control, EELE 4360 Dec. 16, 2014"— Presentation transcript:

1 Dr. Hatem Elaydi Digital Control, EELE 4360 Dec. 16, 2014
Islamic University of gaza electrical engineering department Control design Dr. Hatem Elaydi Digital Control, EELE 4360 Dec. 16, 2014

2 Lead Or Phase-lead Compensator Using Root Locus
 C(s) can be designed using the root locus. A lead compensator in root locus form where the magnitude of z0 is less than the magnitude of p0

3  Root Locus Based Controller Design Using MATAB
Design a digital controller such that the closed loop system has zero steady state error to step input with a reasonable dynamic performance. Velocity error constant of the system should at least be 5.

4 Root locus of the uncompensated system

5 Pole zero map of the uncompensated system

6 One of the design criteria is that the closed loop system should have a zero steady state error for unit step input. Thus a PI controller is required  Let us take Ki = 1. With Ki = 1 , the characteristic equation becomes

7 Root locus of the system with PI controller
The zeros of the system are 1 and and the poles of the system are ±  i and respectively.

8 It is clear from the figure that the system is stable for a very small range of Kp.

9 The figure shows that the stable range of Kp is 0. 239 < Kp < 6
The figure shows that the stable range of Kp is < Kp < The best achievable overshoot is 45.5% , for Kp = 1 , which is very high for any practical system. To satisfy velocity error constant, Ki ≥ 1. If we assume 15% overshoot (corresponding to and 2 sec settling time (corresponding to the desired dominant poles can be calculated as, Thus the closed loop poles in z-plane

10 Pole zero map including poles of the PID controller

11 Let us denote the angle contribution starting from the zero to the right most pole as θ1, θ2 , θ3 , θ4and θ5 respectively. The angles can be calculated as θ1 = 9.5° , θ2 = 20° , θ3 = 99.9° , θ4 =115.7° and θ5 = 129.4°.  Net angle contribution is A = 9.5° - 20° ° ° ° = ° . Angle deficiency is ° + 180° = °  Thus the two zeros of PID controller must provide an angle of 175.5°. Let us place the two zeros at the same location, zpid .  Since the required angle by individual zero is 87.75° , we can easily say that the zeros must lie on the left of the desired closed loop pole. 

12 The controller is then written as
This figure shows that the desired closed loop pole corresponds to K = 4.33 .

13


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