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1、1,PD, PI, PID Compensation,M. Sami Fadali Professor of Electrical Engineering University of Nevada,2,Outline,PD compensation. PI compensation. PID compensation.,3,PD Control,L= loop gain = N / D. scl = desired closed-loop pole location. Find a controller C s.t. L C(scl) = 180 or its odd multiple (an
2、gle condition). Controller Angle for scl,Zero location,4,Remarks,Graphical procedures are no longer needed. CAD procedure to obtain the design parameters for specified & n. evalfr(l,scl) % Evaluate l at scl polyval( num,scl) % Evaluate num at scl The complex value and its angle can also be evaluated
3、 using any hand calculator.,5,Procedure:MATLAB or Calculator,Calculate L (scl) theta = pi angle(evalfr(l, scl) ) Calculate zero location using (5.14). Calculate new loop gain (with zero) lc = tf( 1, a,1)*l Calculate gain (magnitude condition) K = 1/abs( evalfr( lc, scl) 5. Check time response of PD-
4、compensated system. Modify the design to meet the desired specifications if necessary (MATLAB).,6,Procedure 5.2: Given e() & ,Obtain error constant Ke from e() and determine a system parameter Kf that remains free after Ke is fixed for the system with PD control. 2- Rewrite the closed-loop character
5、istic equation of the PD controlled system as,7,Procedure 5.2 (Cont.),Obtain Kf corresponding to the desired closed-loop pole location. As in Procedure 5.1, Kf can be obtained by clicking on the MATLAB root locus plot or applying the magnitude condition using MATLAB or a calculator. Calculate the fr
6、ee parameter from the gain Kf . Check the time response of the PD compensated system and modify the design to meet the desired specifications if necessary.,8,Example 5.5,Use a CAD package to design a PD controller for the type I system,to meet the following specifications (a) = 0.7 & n = 10 rad/s. (
7、b) = 0.7 & e()= 4% due to a unit ramp.,9,(a) =0.7 & n=10 rad/s,MATLAB: calculate pole location theta=pi( angle( evalfr( g, scl) ) ) theta = 1.1731 a = 10 * sqrt(1-0.72)/ tan(theta) + 7 a = 10.0000 k =1/abs(evalfr(tf( 1, a,1)*l,scl) % Gain at scl k = 10.0000,10,RL of Uncompensated System,11,RL of PD-
8、compensated System,12,(b) =0.7, e()=4% for unit ramp,Closed-loop characteristic equation with PD,Assume that K varies with K a fixed, then,RL= circle centered at the origin,13,RL of PD-compensated System,K a fixed,14,(b) Design Cont.,Desired location: intersection of root locus with = 0.7 radial lin
9、e. K = 10, a =10 (same values as in Example 5.3). MATLAB command to obtain the gain k = 1/abs( evalfr( tf(1,10, 1, 4, 0), scl) ) k = 10.0000 Time responses of two designs are identical.,15,PI Control,Integral control to improve e(), (increases type by one) worse transient response or instability. Ad
10、d proportional control controller has a pole and a zero. Transfer function of proportional-plus-integral (PI) controller,16,PI Remarks,Used in cascade compensation (integral term in the feedback path is equivalent to a differentiator in the forward path) PI design for a plant transfer function G(s)
11、= PD design of G(s)/ s. A better design is often possible by “almost canceling“ the controller zero and the controller pole (negligible effect on time response).,17,Procedure 5.3,Design a proportional controller for the system to meet the transient response specifications, i.e. place the dominant cl
12、osed-loop system poles at scl = z wn j wd . Add a PI controller with the zero location (- a) = small angle ( 3 5). Tune the gain of the system to move the closed-loop pole closer to scl.,18,Comments,Use PI control only if P-control meets the transient response but not the steady-state error specific
13、ations. Otherwise, use another control. Use pole-zero diagram to prove zero location formula.,19,Pole-zero Diagram of PI Controller,20,Proof,Controller angle at scl,From Figure,21,Proof (Cont.),Solve for x,x = ( a/ n) Multiplying by n gives a.,Trig. identity,22,Controller Angle at scl ( 3) vs. ,23,E
14、xample 5.6,Design a controller for the position control system to perfectly track a ramp input with a dominant pair with = 0.7 & wn =4 rad/s.,24,Solution,Procedure 5.1 with modified transfer function. Unstable for all gains (see root locus plot). Controller must provide an angle 249 at the desired c
15、losed-loop pole location. Zero at 1.732. Cursor at desired pole location on compensated system RL gives a gain of about 40.6.,25,RL of System With Integrator,26,RL of PI-compensated System,27,Analytical Design,Values obtained earlier, approximately.,Closed-loop characteristic polynomial,28,MATLAB, s
16、cl = 4*exp(j*(pi acos( 0.7) ) scl = -2.8000 + 2.8566i theta = pi + angle( polyval( 1, 10, 0, 0, scl ) ) theta = 1.9285 a = 4*sqrt(1-.72)/tan(theta)+ 4*.7 a = 1.7323 k = abs(polyval( 1, 10, 0,0, scl )/ polyval(1,a, scl) ) k = 40.6400,29,Design I Results,Closed-loop transfer function for Design I,Zero close to the closed-loop poles excessive PO (see step response toge
