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note3 - Zero-degree Loci for Negative K We consider sm +...

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Unformatted text preview: Zero-degree Loci for Negative K We consider sm + b1sm,1 +    + bm : KGs = ,jK j sn + a sn,1 +    + a 1 n By convention, we have 1 + KGs = 0; K = 0 ! ,1: The angle of Gs is 0o + 360o  l Change: 180o + 360o is replaced by 0o + 360o  l. On real axis: the number of poles and zeros on RHS of a testing point so = even number 0 is counted even. Centroid is the same but the angle of asymptotes are l = l  360o=n , m; l = 0; 1; :::; n , m , 1: This is rotated by 180o=n , m from 180o loci. j !-axis crossing: locus crosses for K 0. 1 The departure and arriving angles: 1X X =  i , i , l  360o ; dep q 1X X arr =  i , i + l  360o q where l = 0; :::; q , 1 also rotated by 180o=n , m from 180o loci. Example: Find root locus of + KGs = K s s2 1 for both negative and positive K . -1 Solid line for positive K , and dashed line for negative K 2 Example: Find zero-degree root locus for K+ K + KGs = s2s +s9s1 3 = K s4 + s s3 1 27 : + 12 + Step 1, 2: Mark poles zeros and raw locus on real axis. Step 3: Find centroid of asymptotes: P pi , P zi = 0 + 0 , 9 , 3 + 1 = ,3:667; = n,m 4,1 and angles for asymptotes: 8 0o 360o l = l  n , m = 120o : ,120o l = 0; l = 1; l = 2: Step 4: Departure angle: not necessary as they are clear from locus on real axis. The same is true for arriving angle. Step 5: j!-axis crossing: aj! + Kbj! = !4 , 12j!3 , 27!2 + K j! + 1 = 0  K! , 12!3 = 0; !4 , 27!2 + K = 0: The solutions are ! = 0 and K = 12!2 that are not crossing as K 0 for zero-degree root locus. 3 Step 6: Breaking points, or multiple roots: 0 = b0sas , a0sbs = s4 + 12s3 + 27s2 ,4s3 + 36s2 + 54ss + 1 = ,s3s3 + 28s2 + 63s + 54 = 0: The possible roots are 0; ,6:545; ,1:395  j 0:8980. The only possible breaking point is ,6:545. Step 7: Complete sketch. -9 -3 4 -1 Selecting Gain from Loci: K = 0 ! 1 K = , G1s  K = jG1sj :   Hence if so is a desired closed-loop pole, then K= Q jso , pij Q jso , zij L L :::L = 12 n S1S2:::Sm where Li: length of vector from so to pi and Si: from so to zi. Example: Determine K such that = 0:5 where 1 Gs = s s + 42 + 16 ; p1 = 0; p2;3 = ,4  4j: First sketch root locus: Since n , m = 3 , 0 = 3, there are three braches: one on real axis, and two approaching asymptotes: 8 60o; l = 0; 8 = , ; l = 180o; l = 1; 3 : ,60o; l = 2: We need also determine departure angle and j!-axis crossing: dep = , 1 , 3 + 180 j! j! + 42 + 16 = 0 o = !3 , 32! = 0; ,135o , 90o + 180o = ,45o;  K = 8!2 p that gives !c =  32 = 5:66, and Kc = 8  32 = 256. 5 Since the desired closed-loop poles are stable, K Kc = 256. = 0:5 implies that the angle between the vector to the desired pole and j!-axis is = sin,10:5 = 30o: To achieve the desired pole location, we need K = jG1s j = jsojjso , p2jjso , p3j: o Since jsoj  4, jso , p2j  2:1, and jso , p3j  7:7, we obtain desired K as K = 65. ζ=0.5 | s o -p 2 | 4j |so -p 1 | |so -p 3| −4 6 Dynamic Compensators Often static gain may not achieve the desired performance, or stability. In this case, dynamic compensators need be employed. 2 Example: Gs = 1=s2 + !o . Root locus shows that the closed-loop is unstable no matter what K we choose. ωo j −ωo j 7 However, if we replace K by PD controller Ds = K s + z; z 0: Then we have stable closed-loop system for any K 0. Since such Ds is not realizable, we may use lead compensator s Ds = K + + z ; p z: 1 s=p -p -z Root locus with lead compensator. 8 Lead Compensator: We consider compensators of the form KDs where Ds = s + z : s+p If p z 0, then we call it lead compensator. If z p then we call it lag compensator. 0, The argument to the desired pole location generated by lead compensator is given by m by + m + 180 = , o ,  = 180o. φm φ ψ 9 Design of Lead Compensator: The purpose is to improve transient response, through assigning the dominant poles at the desired location, while all other poles far away from j!-axis. Step 1: Sketch root locus for KDsGs. Step 2: Translate speci cations into desired pole location. Step 3: Compute the argument to be compensated for by using the formula: m= X , X i + 180o: i i i Make sure that the rest of the poles are at least 3  5 times away from j!-axis than the dominant poles. Step 3: Compute z and p. Step 4: Compute required gain via gain condition: K = , Ds 1Gs  0: o o In general, each lead compensator can have at most 90o compensation. If the required m 90o, more than one lead compensator need be employed. 10 1 Example: Gs = performance requirement ss + 1 with p and !n 7rad=sec. Note !d = 1 , 2!n = 6:0622. 0:5 The root locus is sketched rst which has three branches. = 0:5 implies gives the dashed ray with = 30o, and !n = 7 gives a dashed circle of radius 7. The intersection is the desired pole location. Compute the argument m m as = 1 + 2 + 180o = 30o + 20:92o  50:92o: where 1 = 120o and 2 = 90o + tan,12:5=6:022. φ ψ 11 φ2 φ1 The desired poles are so = ,3:5  6:06j by = 30o. Since = 0 is the lower bound, 51o. A general rule is that = o m+3  5o = 55o; = 3o  5o = 4o: p The distance from so to the real axis is 72 , 3:52 = 6:0622. See the gure below. Thus z , 3:5 = 6:0622  ctan55o = 4:2448; p , 3:5 = 6:0622  ctan4o = 86:6932: that gives z = 7:7448, and p = 90:1932. Compute K as s K = ss +s1z + p = 618:0427: + s=,3:5+j 6:06 7 55 o 4o -p -3.5 -z 12 Veri cation of the Design: After the design, we need verify if the design satis es the performance requirement. Find the poles of the closed-loop system: cs = ss + 1s + p + K s + z = s3 + p + 1s2 + p + K s + Kz = s + 83:4s + 3:9 + 6:5j s + 3:9 , 6:5j : The dominant poles are not exactly ,3:5  7j , but the corresponding damping ratio and natural frequency are p2 3:9 = = 0:5148; !n = 3:9 + 6:52 = 7:5763 !n that satisfy the performance requirement. Since the third pole is more than 20 times away from the j!-axis than the dominant poles, the design is satisfactory. The design can also be veri ed using step response of the closed-loop system: K z Tcs = ss + 1s +sp++K s + z ;  with MATLAB by the fact 0:5  Mp 16:3, and !n 7  ts 0:86 sec. 13 Lag Compensator: In time domain response, steady-state response is also important. The purpose of lag compensator is to improve steady-state errors. Indeed with lag compensator Ds, we have that lim Ds = s!0 z 1: p The above implies that 1 1 Ep = 1 +1K ; Ev = K ; Ea = K p v a Kp = slim0 GsKz=p; Kp = slim sGsKz=p; ! !0 2GsKz=p; Ka = slim0 s ! and thus tracking error is reduced by lag compensator. 14 1 Lead and lag compensators are used to improve the time response of the system. Transient response: the dominant poles are in the desired location, with rest of poles at least 3  5 times away from the j!-axis as the dominant ones. Lead Compensator is an approximation of PD controller p 0 s + z 1 = 1 + s=z: @ A plim z s + p !1 A consequence is that it is likely increasing the overshot. We note that PD controller is not realizable. Even it can be realized, we would not use because it ampli es the noise or disturbance. A compromise is to use lead compensator with large p. Steady-state error: the gain at ! = 0 is enlarged so that the steady-state error is reduced. Recall that the steady-state errors are: 1 1 Ep = 1 +1K ; Ev = K ; Ea = K p v a Kp = slim Gs; Kp = slim sGs; Ka = slim s2Gs: !0 !0 !0 15 1 Example: Gs = ss + 1 with performance requirement  0:5, and !n  2rad=sec, and steady-state error for unit ramp input ess  3:33  Kv  30. The root locus is skipped that has three branches. = 0:5 implies gives the dashed ray with = 30o, and !n = 2 gives a dashed circle of radius 2. The intersection is so = ,1  1:7321j that is the desired pole location. Compute the argument m m as = 1 + 2 + 180o = 120o + 90o + 180o  30o: where 1 = 120o and 2 = 90o. ζ=0.5 φ ψ φ2 φ1 ωn =2 16 The desired poles are so = ,1  1:7321j by = 30o. Since = 0 is the lower bound, 30o. To ensure the dominance of so, we set = o m+5 = 35o; = m p = 5o: The distance from so to the real axis is 3 = 1:7321. Thus p z , 1 = 3ctan35o = 2:4736; p p , 1 = 3ctan5o = 19:7974: that gives z = 3:4736, and p = 20:7974. Compute K as s K = ss +s1z + p + s=,1+1:7321j = 22:7974: For steady-state performance, we have that  Kv = slim0 s ssK s +sz+ p = Kz = 3:8077: ! + 1 p But the desired performance is Kv  1=ess = 1=:033 = 30. Hence lag compensator is required: 30 Dlags = s + z ; z = 3:8077 = 7:8788: s+p p Set z=p = 7:8788. A general rule is to choose z ten time smaller or equal to the distance of dominant pole to the j!axis This gives z = 1=10 = 0:1; p = z=8 = 0:0127: 17 Thus the dynamic compensator is s +3 KDleadsDlags = 22:7974 ss+ 20::4736 s ++:0:1 7974 0 0127 Veri cation: The closed-loop poles are the roots of ss +1s +20:7974s +0:0127+22:7974s +3:4736s +0:1 that are ,19:7915; ,1:0578  1:7337j , 0:097 which satisfy the transient requirement by p !n = 1:73372 + 1:05782 = 2:0309 2; = 1:0578=2:0309 = 0:5209 0:5: The steady-state error is 20:7974  0:0127 ess = = 0:0328 22:7974  3:4736  0:1 18 0:0333: 8:7s + 0:5 Example: Gs = 2 s + 2s + 26 with performance requirement P:O:  16:3; ts  1:333s; and steady-state error for unit step input ess  0:05. Design Procedure: We consider dynamic compensator of the form Ds = DleadsDlags 8:7sK 0:5 : + It follows that the loop transfer function is 0 10 1 0 s + zd 1 B s + zg C B K A DsGs = @ s + p A @ s + p A @ s2 + 2s + 26 C : d g Hence it is type 0 that has nite ess-value for step input. Our objective is to design lead lag compensator s+z s+z Dleads = s + pd ; Dlags = s + pg ; d g and the gain K such that specs. are met. 19 We rst design the lead compensator. Percentage overshot requirement gives j = 0:5:  r 2 logP:O:j 2  + logP:O: P:O:=0:163 Settling time requirement gives !n  4:6 = 6:1333: ts Thus the dominant poles are at r , !n  j!n 1 , 2 = ,3:0667  j 5:3116: We choose so = ,4  5j to have some design margin = p2 2 !n = 4 + 5 = 6:4031 6:1333; = 4=!n = 4=6:4031 = 0:6247 0:5: We now compute angles generated by p1 = ,1 + 5j , and p1 = ,1 , 5j to the dominant pole at so = ,4 + 5j : 1 = 180 o; 2 = 90 o + arctan3=10 = 106:7o: Hence the required phase compensation is see the gure next page o o m = 1 + 2 + 180 = 106:7 : Thus two lead compensator is required. 20 We take two identical lead compensators: 0 s + zd 12 Dleads = @ s + p A d Each needs compensate for 53:4o. We take = 58o to ensure the dominance of so. from the gure below we obtain zd = 4 + 5= tan58o = 7:1243; pd = 4 + 5= tan4:6o = 66:1442: The required gain is 0 s + pd 12 s2 + 2s + 26 K = @s + z A = 3502: d s=,4+5j φ1 5j ψ φ -p -z φ2 21 Design of lag compensator: the position constant is now 1 0 s + zd 12 0 K Kp = slim @ s + p A @ s2 + 2s + 26 A = 1:5626 19: !0 d Therefore we need lag compensator s+z z 19 Dlags = s + pg ; pg = 1:5626 = 12:16: g g Set zg = 0:4 yields pg = 0:4=12:16 = 0:0329. The dynamic compensator is 0 402:5287 1 0 s + 7:1242 12 0 s + 0:4 1 Ds = @ s + 0:5 A @ s + 66:1442 A @ s + 0:0329 A : Veri cation: Closed-loop poles are ,63:07  55:84j , ,3:97  5:10j , and ,0:25 which satisfy the requirement. MATLAB is used to simulate the closed-loop step response: 0.8 1 0.9 0.7 0.8 0.6 0.7 0.5 0.6 0.4 0.5 0.4 0.3 0.3 0.2 0.2 0.1 0 0 0.1 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 0 0 2 22 2 4 6 8 10 12 14 16 18 20 Analysis: With lead compensator only, the step response satis es transient requirement left gure. With both lead and lag compensator, ess-error is satis ed, but the ts spec. is not met right gure. In some design case, lag compensator may give such problem. A common way to remedy it is design of pre- lter F s such that ts is improved. r(t) F(s) + D(s) - G(s) y(t) Note that the closed-loop pole at ,0:25, and the zero at ,0:4 are the main source of causing large ts. Thus we choose F s = 11+ s=00::25 : + s= 4 The step response is plotted in next page 23 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Both steady-state error and settling time now satisfy the design requirement. But the percentage overshot exceeds the speci cation. Why? Because the zeros of lead compensator are also zeros of the closed-loop system that cause the increase of the overshot. Hence we need re-adjust the dominant pole position using the knowledge learned from Section x3.5, and do a second design. However we skip in the class lecture, and leave you to work out in the project design assignment. 24 ...
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