Unformatted text preview: Classical PID Control
P: Proportional, I: Integral, D: Derivative. proportional to the error signal e. That is, u = Ke.
r
 Proportional Feedback Control: the actuating signal u is
e K u G y Example 1 Speed control revisisted: The closedloop transfer function is AK KGs T s = 1 + KGs = s2 + + s + AK + 1 1 2 1 2 which is a prototype except a constant factor. The closedloop poles are functions of K : A Gs = s + 1 s + 1 : 1 2 , 1 + 2 1 + 22 , 4 1 21 + AK p1;2 = 212 8 , 1,1; if K = 0: = : ,1 ,2 ; r 1 Location of poles for K 0: As K increases from 0, the distance from the the pole to imaginary axis increases, and thus ts decreases. However at 0 1 2 2 1 B 1 + 2 C = 1 , 2 , 1A 4A K = [email protected] 4 1 2 1 2 the two poles meet at + = 1 2: 212 Further increase of K value does not reduce ts, but increase the overshot from 0 by 1. On the otherhand, the increase of overshot reduces tr . The following plot shows the pole location for K 0:
1 + 2 2 1 2 2
1 1 1 For this 2nd order system, increase of K does not lead to instability. But for high order systems, it is very likely to result in unstable feedback syste if high gain control is used.
2 Consider proportional control for plant 1 Gs = ss + 1s + 2 : Then the poles for feedback system with gain K are the roots of ss + 1s + 2 + K = s3 + 3s2 + 2s + K = 0: At K = 6, then the poles are roots of as functions of K and plot them on complex plane, then we have:
K= p that has imaginary roots: s1;2 = 2j . If we compute roots s3 + 3s2 + 2s + 6 = s + 3s2 + 2 = 0 K= K= Tradeo s: High gain is desired to obtain good sensitivity, and disturbance rejection, but may lead to instability. Thus proportional control is inadequate. 3 Integral Feedback Control: The actuating signal is integral
of the error signal: with TI the integral, or reset time.
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 K Z t e d ; or Ds = K ; ut = T t TI s I
0 w e
K TIs u G y Noted that the error signal is accumulated in actuating signal. Proof: The output due to disturbance input W s = w=s is G Gs Yws = 1 + DsGs W s = s + KGss=T w: I Hence by Final Value Theorem, Fact: For stable feedback systems, integral control rejects the step disturbance completely as t ! 1. yw 1 = slim0 sYw s = slim0 sTI = 0: ! ! K
= 0: For the D.C. motor, the closedloop poles are roots of
3 2 1 2s + 1 + 2s + s + AK=TI 4 ProportionalIntegral Feedback Control:
For integral control of the D.C. motor, the closedloop poles are function of K=TI which are plotted next: 2 1 1 1 Hence the closedloop system has worse settling time, and may even be unstable. The reason lies in the fact that the actuating signal requires accumulation of the error. This motivates PIcontrol: 0 1 0 1 1 1 Zt Ds = K @1 + A ; or ut = K @et + t0 e d A : The closedloop poles are now roots of
1 2s 3+ 1 + 2 s TI s TI 2 + 1 + AK s + AK=T I = 0: Thus two coe cients can be altered independently by choosing K and TI which may result in better transient response. 5 Derivative Feedback Control:
ut = KTD e_ t; e = r , y; TD : derivative time:
The control action is proportional to the rate of change of the error signal predictive control. For D.C. Motor example, the characteristic polynomial is
2 1 2s + 1 + 2 + AKTD s + 1 s2 + 1 + 2 + AKTD s + 1 1 2 1 2 = 0; or = 0: Thus the gain K changes = !n: the time constant of the feedback system. The larger the gain, the faster the system responses to the input. This is often true for higher order systems as well. Derivative control is not practical. First the derivative operation ampli es the noise. So often Ds = KTDss ; 1+ 1; is employed. Second, the actuating signal depends only on et, but not et. Third if et = constant, no control action _ is taken. So PD control is more preferred. 6 PID Feedback Control:
TI s Three parameters are involved: K , TI , and TD . The control
action is
0 ut = K @et + 0 Ds = K @1 + 1 1 + TD sA : where e = r , y. For D.C. motor example, PID control implies 0 1 Zt de 1 : vat = K @et + 0 e d + TD A The characteristic polynomial is de 1 ; A 0 e d + TD dt TI
1
Zt TI dt 3 2 1 2TI s + TI 1 + 2 + AKTD s + TI 1 + AK s + AK = 0: The closedloop transfer functions are AK 1 + I s Try s = T s3 + T + + AKT Ts2 + T 1 + AK s + AK ; 1 2 I I 1 2 D I TI Bs Twy s = T s3 + T + + AKT s2 + T 1 + AK s + AK : 1 2 I I 1 2 D I PID o ers more exibility, but the design is not easy. 7 ZieglerNichols Tuning
The plant is assumed to be of the form: Y s = He,t s Gs = U s 1 + s with td: time delay, and : time constant.
d Step response: Hence the step response is: He,t s = 0 H , H 1 e,t s: Y s = Gs s = s1 + s @ s 1 + s A
1
d d yt = H
H t,t 1 , e, d ! 1t , td: R= H td t We identify the model using L = td; R = H ; where H; can be easily determined from step response plot. 8 Quater decay ratio method: This method aims to achieve dominant transient decay 25 in the next cycle.
Mp .25 Mp 1 .25 1 With this performance requirement for the closedloop systems, = 0:21 is a good compromise between quick response, and percentage overshot. If proportional control is used: K = 1 ; TI = 1; TD = 0: If PI compensator is used: 0:9 K = RL ; TI = 0L3 : : If PID compensator is used: K = 1:2 ; TI = 2L; TD = 0:5L: RL RL 9 Ultimate gain method: First constant gain K is used in feedback system, and K is increased from 0. u
 K G(s) y As K increases, the feedback system eventually enters oscillation. The critical gain at which the output becomes sinusoidal is denoted as Ku, and the corresponding period is denoted as Pu. If proportional control is used: K = 0:5Ku; TI = 1; TD = 0:
If PI compensator is used: If PID compensator is used: K = 0:45Ku; TI = 112 Pu: : K = 0:6Ku; TI = 0:5Pu; TD = 1 Pu: 8 Homework: 4.11, 4.12, plus PID design for the D.C. motor in project 1. Due April 8, 1998. 10 Example heat exchanger: The step response is given by
1 100 t By inspection, we have that L = td R 1=90. 12, and the slope Quater decay ratio: For proportional compensator, 1 K = Rt = 90 = 7:5: 12 d For PI compensator, 0: 81 12 K = Rt9 = 12 = 6:75; TI = 0t:d3 = 0:3 = 40: d The step response for the feedback system using the above two compensators are plotted in a of Figure 4.23 page 194. If K := K=2 is used, then overshot is substantially reduced. See b of the same gure. 11 Ultimate gain: We rst determine parameter Ku and Pu using proportional control. When K = Ku = 15:3, the feedback system oscillates. The steadystate response is given below:
.009 40 80 120 160 t The period Pu 42. For proportional compensator, K = 0:5Ku = 7:65:
For PI compensator, The step response for the feedback system using the above two compensators are plotted in a of Figure 4.25 page 196. If K := K=2 is used, then overshot is reduced. See b of the same gure. K = 0:45Ku = 6:885; TI = 112 Pu = 35: : 12 Sensitivity Analysis
We consider PI control system:
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1 TI K v z 1 s 600 s + 60 y (a) r K z v 1 s 600 s + 60 y v 1 TI z (b) r y H(s)
v z (c)
13 What is H s?
We have relation in sdomain: 2 3 2 32 3 6 Y s 7 6 7 = 6 H11s H12s 7 6 Rs 7 6 76 7 4 5 4 54 Z s H21s H22s V s 5 2 3 2 3 6 H s 7 6 H s 7 = 6 11 7 Rs + 6 12 7 V s 4 5 4 H21s H22s 5 Hence if Rs = 0, then
2 3 2 3 6 Y s 7 6 H12s 7 6 7=6 7 V s 4 5 4 5 Z s H22s which is equivalent to If V s = 0, then Y s H12s = V s Z s R=0 Z ; H22s = V s s H21s R=0 : 2 3 2 3 6 Y s 7 6 H11s 7 6 7=6 7 Rs 4 5 4 5 which is equivalent to s H11s = Y s R V =0 Z ; H21s = Rs s V =0 : 14 Computation of H s:
Now for v = 0, we have
r z y K 600 s + 60 It follows that Y s = K s600 = 600 +60 H11s = Rs 1 + K 600 s + 60 +K K 600 s+60 K K s + 60 Z H21s = Rs = 1 + K 600 = s + 60 + 600K s s+60 Now for r = 0, we have
v
1 s 600 s + 60 y z K 15 It follows that Combinning together gives
2 6 H s = 6 4 Y s = 1 s600 = 600 +60 H12s = V s s 1 + K 600 ss + 60 + 600K s+60 Z K H22s = V s = ,KH12s = , ss + 600+ 600K s 60
600K K s + 60 600=s 1 s + 60 + 600K : ,600K=s
3 7 7 5 With = TI,1, we have
r v + v H(s) y + y z + z 16 Sensitivity Model:
Thus we obtain 2 3 2 32 3 6Y + Y 7 7 6 7 = 6 H11s H12s 7 6 R 6 76 7 4 5 4 54 Z+ Z H21s H22s V + V 5 2 32 3 2 3 6 H s H12s 7 6 R 7 + 6 H12 s 7 V 76 7 6 7 = 6 11 4 54 5 4 H21s H22s V H22s 5 Because 2 3 2 32 3 6 Y 7 6 H11s H12s 7 6 R 7 6 7=6 76 7 4 5 4 Z H21s H22s 5 4 V 5 we obtain 2 3 2 3 6 Y 7 6 H12s 7 6 7=6 7 V = H2s V 1 4 5 4 Z H22s 5 Note also that V + V = + Z + Z = Z + Z + Z + Z Z + Z + Z
BY V = Z , V = Z + Z: 2 17 Analysis:
Combinning 1 and 2 yields the block diagram:
y H2(s)
z z v H12(s) H22(s)
z y z v Therefore H Y = 1 , 12Z ; or H22 dY = lim Y = H12Z : !0 d 1 , H12
18 Overall System:
In connection with the original feedback system, we obtain:
r y
dy d H(s)
v H2(s)
z For our particular example, H12 ^ H s = 1 , Hss = ss + 60 + 600 K + 600K 600 22 T
I which measures the sensitivity of output with respect to the variation of the parameter = TI,1. For varying from 220 to 300, we have: 19 Integrator Antiwindup:
For PI control where the actuator has saturation, the actuating signal has the form
8 ua = uc jucj umax u : ju j umax jucj umax
c c The problem arises with the integrator which can not change its sign instantaneously.
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1 u sT I uc ua K G(s) y Ka One way to resolve this issue is to add a negative feedback path dashed line block which is turned on only when jucj exceeds the maximum value or uc is saturated. Such a nonlinear block is conceptually simple, but di cult to implement. 20 Implementation of Antiwindup
An easy way to implement the previous scheme is shown in the following gure:
r y K 1 u sT I uc G(s) Ka s Clearly, s = 0 whenever jucj umax, and s 6= 0 whenever jucj umax. If s 6= 0, then the PI controller is equivalent to K 1 u sT I uc K(sT I +1) sTI + Ka Ka s 21 Example: Consider a plant and a PI controller given by
respectively. The actuating signal is limited to 1:0. The simulation plots are given by the following picture. The left one is the step response, and the right one is the control input. The improvement with antiwindup is obvious. 1 Gs = s ; Ds = K + KI = 2 + 4 ; s s Figure 4.29, page 199 of text 22 A Di erent Design Method:
A design method of equal importance to PID is the pole zero cancellation method, or dominant poles method. Step 1: The plant model is rst approximated through dominant poles method direct truncation, or other model reduction method. Step 2: Design a dynamic compensator which cancel undesirable poles or zeros of the plant model, and place the poles in the desirable locations, speci ed by design requirements. Step 3: Verify the performance for the original feedback system. Why step 1? Without step 1, the resulting dynamic compensator will have high order, thereby increasing the complexity of the compensator. Moreover step 1 helps elimination of unnecessary pole zero cancellations which might be sensitive to the modeling error. Step 2 is a naive method for design, and thus step 3 is required make sure the design speci cation is met.
23 Example 1: Consider a plant model 25 104 Gs = s + 62 + 144 s + 202 + 100 Design a feedback PID compensator to achieve P.O. 10. : Answer: In le note4, we used direct truncation to obtain a
5 102 G2s = s + 62 + 144 : A much better approximation is: 5 102 G3s = s + 62 + 144 1 + s= 6 103 = s + 62 + 144 s + 12 where = 12. This serves as Step 1 of the design procedure. At the second step, we consider design using G3s. Suppose we use approximate PID control why?: Kps2 + as + b Ds = ss + p where p is su ciently large. second order approximation: 24 Design of Ds:
By pole zero cancellation, we choose s2 + as + b = s + 62 + 144 = s2 + 12s + 180: It follows that if p = 100 6 103K 6 103Kp Ls = DG3s = ss + ps + 12 ss + 12 :
The closedloop transfer function is: DG3s = 6 103K T s = 1 + DG s s2 + 12s + 6 103K 3 Compared with prototype model:
2 !n Tos = s2 + 2
! s + !2 n n we have: !n = 6 103K; = 6=!n = 6= 6 103K:
Since P.O. 10 is equivalent to r 2j log:1j 2 = 0:5912; + j log:1j p p we have = 6= 6 103K 0:5912 0 6 12 K @ 0:5912 A =6 103 = 0:0172:
25 p 3 Veri cation:
25 104Kp Ls = DGs = ss + ps + 202 + 100 where p = 100, and K 0:0172. With K = 0:015, the step response is give by:
1.4 1.2 We need use 1 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 The P.O. is 14:3 exceeding the speci cation. Through a few trials, we nd that with K = 0:0132, P.O. = 9:87.
1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 26 ...
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 Fall '07
 Chen
 KU

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