# PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada

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1 PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1

2 Outline PD compensation. PI compensation. PID compensation. 2

3 PD Control L= loop gain s cl = desired closed-loop pole location. Find a controller C s.t. CL s cl = ±180 or its odd multiple (angle condition). Controller Angle for s cl θ c = ±180 L s cl Zero location a = ω d tan θ c + ζω n 3

4 Remarks Graphical procedures are no longer needed. CAD procedure to obtain the design parameters for specified ζ& ω n (s cl known).» evalfr(l,scl) % Evaluate L at scl» polyval( num,scl) % Evaluate num at scl The complex value and its angle can also be evaluated using any hand calculator. 4

5 Procedure:MATLAB or Calculator 1. Calculate π L s cl»theta = pi angle(evalfr(l, scl) ) 2. Calculate zero location using a = ω d + ζω tan θ n c 3. Calculate new loop gain (with zero)» Lc = tf( [1, a],1)*l 4. Calculate gain (magnitude condition)» K = 1/abs( evalfr( Lc, scl)) 5. Check time response of PD-compensated system. Modify the design to meet the desired specifications if necessary (MATLAB). 5

6 Procedure 5.2: Given e & ζ 1. Obtain error constant K e from e and determine a system parameter K f that remains free after K e is fixed for the system with PD control. 2- Rewrite the closed-loop characteristic equation of the PD controlled system as 1 + K f G f s = 0 6

7 Procedure 5.2 (Cont.) 3. Obtain K f corresponding to the desired closedloop pole location. As in Procedure 5.1, K f can be obtained by clicking on the MATLAB root locus plot or applying the magnitude condition using MATLAB or a calculator. 4. Calculate the free parameter from the gain K f. 5. Check the time response of the PD compensated system and modify the design to meet the desired specifications if necessary. 7

8 Example 5.5 Use a CAD package to design a PD controller for the type I system 1 G s = s s + 4 to meet the following specifications (a) ζ = 0.7 & ω n = 10 rad/s. (b) ζ = 0.7 & e = 4% due to a unit ramp. 8

9 (a) ζ = 0.7 & ω n = 10 rad/s MATLAB Calculate pole location & corresponding gain >> scl = 10*exp( j*( pi-acos(0.7) ) ) scl = I >> g=zpk([ ],[0,-4],1); >> theta=pi ( angle( evalfr( g, scl) ) ) theta =

10 (a) Design (cont.)» a = 10 * sqrt(1-0.7^2)/ tan(theta) + 7 a = » k =1/abs(evalfr(tf( [1, a],1)*g,scl)) % k = Gain at scl: K = 10 10

11 RL of Uncompensated System 8 Root Locus Imaginary Axis Real Axis 11

12 RL of PD-compensated System 8 Root Locus 0.7 Imaginary Axis System: gcomp Gain: 10 Pole: i Damping: Overshoot (%): 4.56 Frequency (rad/sec): Real Axis 12

13 (b) ζ = 0.7, e % = 4% for unit ramp K v = Lim s 0 = Ka sc s G s = Lim s 0 sk 4 = 100 e % = 25 Ka = 100, a = 100/K s + a s s + 4 Closed-loop characteristic equation with PD 1 + K s+a s s+4 = 0 13

14 (b) Design (cont.) 1 + K s+a = 0 s s+4 Assume that K varies with K a fixed, then s s Ks + Ka = 0, Ka = 100 s 1 + K s 2 + 4s = 0 RL= circle centered at the origin 14

15 RL of PD-compensated System K a fixed 15

16 (b) Design Cont. Desired location: intersection of root locus with ζ = 0.7 radial line. 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 = Time responses of two designs are identical. 16

17 PI Control Integral control to improve e (increases type by one) worse transient response or instability. Add proportional control controller has a pole and a zero. Transfer function of proportional-plusintegral (PI) controller C s = K p + K i s = K s + a p, a = K i s K p 17

18 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 is the same as 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). 18

19 Procedure Design a proportional controller for the system to meet the transient response specifications, i.e. place the dominant closed-loop poles at s cl = ζω n ± jω d. 2. Add a PI controller with the zero location a such that φ = 3 5 (small angle) a = ζ+ ω n 1 ζ 2 tan φ or a = ζω n Tune the gain of the system to moved the closed-loop pole closer to s cl. 19

20 Comments Use PI control only if P-control meets the transient response but not the steady-state error specifications. Otherwise, use another control. Use pole-zero diagram to prove zero location formula. 20

21 Pole-zero Diagram of PI Controller s cl tan 180 θ p = ω d ζω n j d z s cl a p s cl a n tan 180 θ z = ω d ζω n a 21

22 Proof From Figure tan 180 θ p = ω d ζω n = tan 180 θ z = ω d ζω n a = x = a/ω n 1 ζ2 ζ 1 ζ2 ζ x Controller angle at s cl φ = θ z θ p = 180 θ p 180 θ z 22

23 Proof (Cont.) Trig. Identity tan A B = tan φ = 1 ζ 2 ζ x 1 ζ2 ζ 1 1 ζ2 ζ x Solve for x = a ω n = ζ+ Multiplying by ω n gives a 1 ζ 2 ζ 1 tan A tan B 1+tan A tan B = 1 ζ 2 tan φ x 1 ζ2 1 ζx 23

24 Controller Angle at s cl = φ < 3 vs. ζ x = a ω n = ζ 10 x 1 ζ2 1 φ = tan 1 ζx ζ 1 ζ2 = tan 1 10 ζ 2 < 3 24

25 Example 5.6 Design a controller for the position control system to perfectly track a ramp input with a dominant pair with ζ = 0.7 & ω n = 4 rad/s. G s = 1 s s

26 Solution To use the PD procedure for PI design, consider G s 1 G i s = = s s 2 s + 10 Procedure 5.1 with modified transfer function. Unstable for all gains (see root locus plot). Controller must provide an angle 249 at the desired closed-loop pole location. Zero at Cursor at desired pole location on compensated system RL gives a gain of about

27 RL of System With Integrator 6 Root Locus 5 4 Imaginary Axis Real Axis 27

28 RL of PI-compensated System 28

29 Analytical Design Closed-loop characteristic polynomial s s 2 + Ks + Ka = s + α s 2 + 2ζω n s + ω n 2 = s 3 + α + 2ζω n s 2 + 2ζω n α + ω n 2 s + αω n 2 α = 10 2ζω n = = 4.4 K = ω n α + 2ζω n = = a = αω n = K = Values obtained earlier, approximately. 29

30 MATLAB» scl = 4*exp(j*(pi acos( 0.7)) ) scl = i» theta = pi + angle( polyval( [ 1, 10, 0, 0], scl ) ) theta = » a = 4*sqrt(1-.7^2)/tan(theta)+ 4*.7 a = » k = abs(polyval( [ 1, 10, 0,0], scl )/ polyval([1,a], scl) ) k =

31 Design I Results Closed-loop transfer function for Design I s G cl s = s s s s = s s s + 16 Zero close to the closed-loop poles excessive PO (see step response together with the response for a later design: Design II). 31

32 Step Response of PI-compensated System Design I (dotted), Design II (solid). 32

33 Design II: Procedure 5.3 For ζ = 0.7, K & ω n = rad/s. zero location ω a = n ζ + 1 ζ 2 tan φ = Closed-loop transfer function 50 s G cl s = s s s s = s s s Gain slightly reduced to improve response / tan

34 Compare Designs I & II Design II dominant poles: ζ = 0.7, ω n = rad/s Time response for Designs I and II Percentage overshoot for Design II (almost pole-zero cancellation) << Percentage overshoot for Design I (II better) 34

35 PID (proportional-plus-integralplus-derivative) Control Improves both transient & steady-state response. K p, K i, K d = proportional, integral, derivative gain, resp. Controller zeros: real or complex conjugate. C s = K p + K i s + K s 2 + 2ζω n s + ω n ds = K d s 2ζω n = K p, ω 2 K n = K i d K d 2 35

36 PID Design Procedures 1. Cancel real or complex conjugate LHP poles. 2. Cancel pole closest to (not on) the imaginary axis then add a PD controller by applying Procedures 5.1 or 5.2 to the reduced transfer function with an added pole at the origin. 3. Follow Procedure 5.3 with the proportional control design step modified to PD design. The PD design meets the transient response specifications and PI control is then added to improve the steady-state response. 36

37 Example 5.7 Design a PID controller for the transfer function to obtain zero e due to ramp, ζ = 0.7 and ω n = 4 rad/s. 1 G s = s s + 1 s

38 Design I Cancel pole at 1 with a zero & add an integrator 1 G i s = s 2 s + 10 Same as transfer function of Example 5.6 Add PD control of Example 5.6 PID controller C s = 50 s + 1 s s 38

39 Design II Design PD control to meet transient response specs. Select ω n = 5 rad/s (anticipate effect of adding PI). Design PD controller using MATLAB (see book: pdcon)» [k,a,scl] = pdcon(0.7, 5, tf(1, [1,11,10,0])) k = a = scl = i 39

40 Design II (Cont.) Obtain PI zero (approx. same as ζω n /10)» b =5 /(0.7 + sqrt(1-.49)/tan(3*pi/ 180) ) % b = Transfer function for Design II (reduce gain to 40 to correct for PI) s s C s = 40 s 40

41 Conclusions Compare step responses for Designs I and II: Design I is superior because the zeros in Design II result in excessive overshoot. Plant transfer function favors pole cancellation in this case because the remaining real axis pole is far in the LHP. If the remaining pole is at 3, say, the second design procedure would give better results. MORAL: No easy solutions in design, only recipes to experiment with until satisfactory results are obtained. 41

42 Time response for Design I(dotted) and Design II (solid) 42

43 Example 5.8 Design a PID controller for the transfer function to obtain zero e due to step, ζ = 0.7 and ω n 4 rad/s G s = 1 (s+10)(s 2 +2s+10) 43

44 Solution Complex conjugate pair slows down the time response. Third pole is far in the LHP. Cancel the complex conjugate poles with zeros and add an integrator G s = 1 s(s+10) 44

45 Solution (cont.) Stable for all gains. Closed-loop characteristic polynomial s s + K = s 2 + 2ζω n s + ω n 2 Equate coefficients with ζ = 0.7 ω n = 10 2ζ = = PID controller (meets design specs) G s = s2 + 2s + 10 s 45

46 Step Response of PIDcompensated System In practice, pole-zero cancellation may not occur but near cancellation is sufficient to obtain a satisfactory time response 46

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