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note1 - Classical Feedback Control In classical control,...

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Unformatted text preview: Classical Feedback Control In classical control, feedback is the key concept. In fact the concept of feedback has been applied to many other branches of science. Question: Why feedback? Plant uncertainty: parameter variations, nonlinearities etc. Environmental uncertainty: disturbances, noises. Any mathematical model is an approximation of reality. Open-loop Control: dt rt - C s ut -  ? y tP 1 +  - Y s = 1 : xed gain Therefore Tdy s = Ds = tracking error in presence of disturbance. Consequently large gain has to be employed in C s. = the system is sensitive to perturbation s. 1 Closed-loop Control: rt - Cs 6 - dt ut -  ? y tP 1 +  - S Tdy s = 1 + C sP 11 + s = 1 + T osss s o s  C s Try s = 1 + sPs1 + + s = T1o+1+s C s P s1 To s s 1 C sP s . If both where Sos = and Tos = 1 + C sP s 1 + C sP s So and To jSoj!j; jToj!j are small = the e ect of both uncertainties is small. Design Trade-o : Because So + To 1, it is impossible to make both So and To small. = There is a trade-o in feedback control system design. 2 Question: How to Trade-o ? Spectral analysis of command input rt: Generically, Rj! has a low pass characteristic rt is unit step, or slowly varying signals. = jSoj!j should be small in operating low frequency range. Spectral analysis of plant uncertainty j!: Recall if ut = cos!t, the steady-state response of the plant is yst = jPtj!j cos !t + 6 Ptj! Hence the smaller the gain jPtj!j is, the more measurement error it results in. Consequently, jj!j has a high pass characteristic. = jToj!j should be small in high frequency range. 3 Gain of the loop transfer function Ls = C sP s: = Sos = L s ; Tos = 1 + Ls : 1 + Ls 1 Therefore small jSoj!j small jToj!j  large jLj!j; ! 2 LFB ;  small jLj!j; ! 2 HFB ; Typical desirable magnitude response of loop gain jLj!j: 4 Concluding Remark: Classical design is most e ective for those stable plant models of low order. Since last decade, there has been a renewed interest in classic method applied to modern control theory such as optimal control of multivariable systems. 5 x5 The Root-locus Design Method Review: Two methods in classical control system design: 1. Dominate mode analysis; 2. Frequency domain method. Root-locus belongs to dominate mode method. Stability: A system is represented by bs = sm + b1sm,1 + :::bm P s = as sn + a sn,1 + :::a 1 n Let pi's be roots of as = 0 = as = n Y s , pi. i=1 The system is said to be stable if Re pi 0 8i. Example 1: P1s = ss2 + 51 : unstable imaginary poles. + , Example 2: P2s = s +s1s2+ 4 : stable. 6 Performance: Consider a typical 2nd order system: 2 !n P s = s2 + 2 ! s + !2 ; !n 0 : natural frequency; n n with input rt = ut: unit step. 2 !n = Y s = 2 2 s s + 2 !ns + !n . 0 1 underdamped: e, ! t sin "! r1 , 2t ,  ; t  0; yt = 1 + p1 , 2 n p where = tan,1 ,2 , 1. n p 1=e= ,2,1 = 1=e= tan ; maximum overshot = 0:8 + 0:25 rise time  !n " ; r  1 3 settle time = , log 0:05 1 , 2  !n !n 5; 4 4:6 or 2; or 1 1 !n !n Therefore, larger !n = faster response; larger = larger overshot. 7 ω n θ Figure 1: Pole location for second order plant. ξω n θ Figure 2: Desirable region for pole location. 8 Connection of pole location with performance Further away from imaginary axis, the faster of the response; Larger of the angle, the larger of the overshot. = desirable poles location is the clipped sector. Conclusion: Not only stability, but also performance is determined by pole location. If the plant does not satis y the stability and performance requirement, we need employ feedback to relocate the poles. Feedback Systems: - h 6 - P s K - h ? Figure 3. Feedback Control System Clearly the poles of the closed-loop system are the roots of Kbs + as = 0: Hence the closed-loop poles are functions of parameter K . 9 De nition: Root locus is a set of roots of Kbs + as = 0 on the complex plane as K varies from 0 to 1. Root-locus is a tool to determine closed-loop poles. Example 5.1: KGs = ssK 1 . + The roots of the closed-loop system is: as + Kbs = ss + 1 + K = s2 + s + K = 0: K = 0: the roots are as = 0, or equivalently, the roots are poles of the plant Gs: s = 0; ,1. K = 1, the roots are bs = 0, or equivalently, the roots are the zeros of the plant Gs: at 1. Hence the root locus begin from s = 0; 1, and end at in nity. 10 K = constant: the roots are obtained as p p 1 , 4K 1 1 ; or s = ,  j 4K , 1 : s=,  2 2 2 K = 1=4: the two roots are both in ,1=2. 0 2 K 1=4: the two roots are both real with p p 1 1 , 4K 1 r= + ; r = , 1 , 4K : 1 K with 22 2 2 2 1=4, the roots are complex with xed real part: ,1=2 p p 1 1 s1 = 2 + j 4K , 1 ; s2 = 2 , j 4K , 1 : 2 2 j -1 Figure 4: Root locus for Example 5.1. 11 Example 5.2: Find root-locus for 1 Gs = ss + K  with respect to parameter K . Analysis: Closed-loop poles are the roots of 1 + Gs = 0 = s2 + Ks + 1 = 0 = s2 + 1 + Ks = 0 We can then set as = s2 + 1 and bs = s. Root-locus: K = 0 : roots are s = j ; K = 1 : roots are s = 0; 1: Thus the root locus begin from j and end at 0 and 1. 12 For K 2 0; 1, the roots are p2 p K  K , 4 ; or s = , K  j 4 , K 2 : s=,2 2 2 2 = K 2: complex roots satisfying 1 12 0 p 0 K A + B 4 , K 2 C2 @, @ A 2 For K  2: real roots with s1 = , K + 2 p K2 , 4 2 2 = 1: p K , K2 , 4: ; s2 = , 2 2 j -1 -j Figure 4: Root locus for Example 5.2. 13 x5.2 Guideline for Root Locus K  0. Root locus is de ned as the set of s such that Qm i s , zi  = 0: 1 + KGs = 1 + K Qn=0 i=0s , pi As convention, poles pi's are marked by cross  and zeros zi's are marked by circle on the complex plane. Computer usage: rlocusnum,den; Question: Why we study how to plot root locus now days? Phase Condition for s = so to lie on root locus: A test point so is on the root locus  m n X6 X so , zi , 6 so , pi = 180o + `  360o: i=1 i=1 The root locus of Gs is the set of points in the s-plane where the phase of Gs is 180o. 14 + Example Gs = s + 22s+ 41s + 5s . Poles and zeros are: z1 = ,1; p1 = 0; p2 = ,2 + j 2; p3 = ,2 , j 2; p4 = ,5: Thus 1 + Gso = 0 implies that 6 G = 6 so +1 , 6 so , 6 so +22 +4 , 6 so +5 = 180o + `  180o: φ2 2j φ 4 -5 φ1 ψ -4 -2 -1 φ3 -2j For so = ,1 + 2j in the above gure, we have that 1 = 90o; 1 = 116:6o; 2 = 0o; 3 = 76o; 4 = 26:6o: Therefore, we obtain 6 G = 1 , 1 , 2 , 3 , 4 = 90o = 90o , 116:6o , 0o , 76o , 26:6o = ,129:2o = so is not on the root locus. 15 Question: Suppose we are given a set of poles pi's denoted by  on s-plane and a set of zeros zi's denoted by on the s-plane, which part of the real axis lies on the root locus? Fact 1: The test point must be on the left of an odd number of real axis poles plus zeros. Because so lies on real axis, the angles generated by complex poles and zeros cancelled each other. Same reason, the angles generated by poles and zeros on the left of the test point are zero. Each pole on the right of the test point generates ,180o, and each zero on the right of the test point generates 180o. 16 + + Example: Gs = ss + 1s +s3s2s5 4+ 1:52 + 4 . +s 2j −φ j −5 −4 −3 −2 −1 φ -2j 17 Question: Where do the in nity roots go as K ! 1 for the case n , m 1? Note that if n = m, root locus lie on nite region of s-plane; if n , m = 1: previous result can be used to nd the in nity root. Fact 2: These roots go to in nity along asymptotes centered at and leaving at l where n , m = number P asymptotes; of Pn n i pi , i zi = ,a1 + b1 ; = n,m n,m o + l , 1  360o ; l = 1; 2; :::; n , m: l = 180 n , m These are the in nity roots as K ! 1: sm + b1sm,1 + bm = 0: lim 1 + K n K !1 s + a1sn,1 + ::: + an 18 We are interested in those s at in nity and thus 0 1 K B 1 + b1s,1 +    + bms,m C A 1 + KGs = 1 + n,m @ s 1 + a1s,1 +    + ans,n  1 + nKm ; n , m 1: , s Take test point so = Rej where Rn,m = K ! 1, then phase condition yields KG = ,n , m l = 180o + l  360o: 180o + l  360o = l = n , m ; l = 0; 1; 2; :::; n , m , 1: Claim: There are only n , m independent angles: 6 360o 180o l = 0 + l ; 0 = n , m ;  = n , m ; n,X,1 m e l=0 l n,X,1 jl m = ej 0 e l=0 j n,m j 0 1 , e =e 1 , ej  = 0: To determine , we need the fact that for Qs = sn + q1sn,1 +    + qn = s , r1s , r2    s , rn thereholds q1 = ,r1 + r2 +    + rn: 19 Now we note that lim as + Kbs K !1 nYm , m Y s , , Rej  s , zi i=1 l=1 sn + a1sn,1 +    + an: = Klim !1 = l It follows that a1 = = = n X , i=1 pi n,X,1 j m e l=0 ,n , m , R ,n , m + b1: This is due to the fact that n,X,1 j m e l=0 l m j 0 n,X,1 ejl =e l=0 l 1,e = ej 0 Therefore we have veri ed Fact 2: j n,m 1 , ej  b1 , a1 = Pm pi , Pn=1 zi : i=1 i = n,m n,m 20 m , iX zi =1 = 0: Question: Where does locus across the j! axis? Routh array can be employed to determine the corresponding K and those roots on the j! axis can be computed exactly. Example: Consider 1 + s s + K2 + 16 = 0 4  s3 + 8s2 + 32s + K = 0: We have following Routh array: s3 : 1 32 s2 : 8 K s1 : 832,K 0 8 s0 : K In order to have roots on j! axis, it is necessary that we have an entire zero row = K = 8  32 = 256: Since for K = 256 the entire row corresponding s1 is zero, we must have roots s2 + 32 = 0 = 21 p s = j 32: Question: What are the departure angles? As K = 0, root locus leaves from poles. If a pole is a complex number, we need determine the departure angle. Answer: By phase condition, X X = i , dep i , 180 i6=dep o , l  360o: Taking the same example as last page, we have that 1 = 90 3 = ,135 o; o: By the phase condition, we have ,90o , 2 , 135o = 180o = 2 = ,90o + 135o + 180o = ,45o: φ 2 4j φ3 -4 φ1 -4j 22 A similar formula is the arriving angle to zero as K ! 1: arr = Example: 1= = X i, X i6=arr i + 180o + l  360o: s + 12 + 4 s + 1 + 2j s + 1 , 2j  ss2 + 5s + 6 = ss + 2s + 3 . ,90o , ; 2 = ,90o + ; 3 = ,45o; = ,90o: arr = 180o + 1 + 2 + 3 , 1 = 45o 2j ψ 1 -3 -2 -1 φ φ 3 φ1 2 θ θ ψarr 23 -2j If the complex pole or zero is not simple and has multiplicity q, then the formulas are modi ed as follows: P i , P i , 180o , l  360o ; dep = for l = 0; 1; :::; q , 1, and arr = P i, q P i + 180o + l  360o q where l = 0; 1; :::; q , 1. 24 ; Breaking point on real axis: If we consider s on real axis, then we have a function K = f s := , G1s ; s is real  that is smooth for those s 6= zi. If a break point so takes place as two roots merge together and then break away, the function K = f s de ned above has a maximum or minimum value K = Km at that breaking point so. We thus have the following condition for the breaking point df d 0, 1 1 = 0: =@A ds s=s ds G s=s The corresponding Km can be computed from K = ,1=Gso. o o Example: as = s2 + s; bs = 1. The function f s is df K = f s = ,s2 , s; = ds = ,2s , 1 = 0: We conclude that there is a breaking point at so = ,1=2 and Km = ,,1=22 , ,1=2 = 1=4. 25 Breaking point on complex plane: The key fact is that beaking point corresponds to multiple roots at s = so for some K = Km. That is as + Kmbs = s , solps; l 1; for some polynomial ps. The above is equivalent to aso + Kmbso = 0  Km = ,aso=bso: 2 Hence taking derivative gives a0s + Kmb0s = s , sol,1ps ~ 3 for some polynomial ps. ~ It follows from 3, and 2 that a0s o + Km b0s a0sobso , asob0so = 0 o = 0 = bso that is the same as d ds 0 1 B, as C @ A bs 26 s=s = 0: o Departure angle at a breaking point: To compute departure angle, we use a concept of continuation locus. We consider K = Km + K1 where Km = f so = , G1s  o with so a breaking point. Then as + Kbs = as + Kmbs + K1bs = 0: We set a1s = as + Kmbs and b1s = bs and compute the departure angle for root locus of a1s + K1b1s = 0: 4 Since at K1 = 0, the roots begin at roots of a1s = 0, including the breaking point so, the departure angle at the breaking point are the same as the departure angle for root locus of 4. For our example in the previous page, we have that Km = 1=4 = a1s = as + Kmbs = s2 + s + 1=4; and b1s = 1. It follos that 2 dep = ,180o , l  360o = dep = 90o: Skecth of root locus: Page 260 of the text. 27 Example 1: Gs = s + 1=s2 = z1 = ,1; p1 = p2 = 0 and n , m = 1. Step 1: Mark the poles and zeros on the s-plane. Step 2: Draw locus on the real axis: left of s = ,1. Step 3: Draw asymptotes: n , m = 1: included in Step 2. Step 4: Compute departure and arriving angles: ,2 1 = 180o + l  360o = 1 = ,90o; 2 = 90o: Step 5: Compute points where locus croesses j! axis: s2 : 1 s1 : K s0 : K Clearly for K K 0 0, there is no roots on j!-axis. Step 6: Determine possible breaking point: s2 K = f s = ,1=G = , s + 1 df = , 2ss + 1 , s2 = 0 = s = 0; ,2: = ds s + 12 Hence so = ,2 is the breaking point and Ko = ,4=,1 = 4. Step 7: Complete the sketch. 28 ...
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