On real axis: the number of poles and zeros on RHS of a testing point s o = even number (0 is counted even).

Transcription

1 Zero-degree Loci for Negative K We consider KG(s) = K sm + b 1 s m b m. s n + a 1 s n a n By convention, we have 1+KG(s) =0, K =0. The angle of G(s) is0 o o l Change: 180 o o is replaced by 0 o o l. On real axis: the number of poles and zeros on RHS of a testing point s o = even number (0 is counted even). Centroid is the same but the angle of asymptotes are θ l = l 360 o /(n m), l =0, 1,..., n m 1. This is rotated by 180 o /(n m) from 180 o loci. jω-axis crossing: locus crosses for K<0. 1

2 The departure and arriving angles: φ dep = 1 q ( Ψ i φ i l 360 o ), Ψ arr = 1 q ( φ i Ψ i + l 360 o ) where l =0,..., q 1 (also rotated by 180 o /(n m) from 180 o loci). Example: Find root locus of (s +1) KG(s) =K s 2 for both negative and positive K. -1 Solid line for positive K, and dashed line for negative K 2

3 Example: Find zero-degree root locus for KG(s) = K(s +1) s 2 (s +9)(s +3) = K K(s +1) s 4 +12s Step 1, 2: Mark poles/zeros and raw locus on real axis. Step 3: Find centroid of asymptotes: α = p i z i n m and angles for asymptotes: θ l = l = o n m = 0 o l =0, 120 o l =1, 120 o l =2. = 3.667, Step 4: Departure angle: not necessary as they are clear from locus on real axis. The same is true for arriving angle. Step 5: jω-axis crossing: a(jω)+kb(jω)=ω 4 12jω 3 27ω 2 + K(jω +1)=0 Kω 12ω 3 =0, ω 4 27ω 2 + K =0. The solutions are ω = 0 and K =12ω 2 that are not crossing as K<0for zero-degree root locus. 3

4 Step 6: Breaking points, or multiple roots: 0 = b (s)a(s) a (s)b(s) =(s 4 +12s 3 +27s 2 ) (4s 3 +36s 2 +54s)(s +1) = s(3s 3 +28s 2 +63s + 54) = 0. The possible roots are 0, 6.545, 1.395±j The only possible breaking point is Step 7: Complete sketch

5 Selecting Gain from Loci: (K =0 ) K = 1 G(s) K = 1 G(s). Hence if s o is a desired closed-loop pole, then K = s o p i s o z i = L 1L 2...L n S 1 S 2...S m where L i : length of vector from s o to p i and S i :froms o to z i. Example: Determine K such that ζ =0.5 where G(s) = 1 s[(s +4) ], p 1 =0,p 2,3 = 4 ± 4j. First sketch root locus: Since n m =3 0 = 3, there are three braches: one on real axis, and two approaching asymptotes: α = 8 3, θ l = 60 o, l =0, 180 o, l =1, 60 o, l =2. We need also determine departure angle and jω-axis crossing: φ dep = φ 1 φ o = 135 o 90 o o = 45 o, jω[(jω +4) ] = 0 ω 3 32ω =0, K =8ω 2 that gives ω c = ± 32 = ±5.66, and K c =8 32 =

6 Since the desired closed-loop poles are stable, K<K c = 256. ζ = 0.5 implies that the angle between the vector to the desired pole and jω-axis is θ =sin 1 (0.5) = 30 o. To achieve the desired pole location, we need K = 1 G(s o ) = s o s o p 2 s o p 3. Since s o 4, s o p 2 2.1, and s o p 3 7.7, we obtain desired K as K = 65. ζ=0.5 -p s o 2 4j s -p o 3 4 s -p o 1 6

7 Dynamic Compensators Often static gain may not achieve the desired performance, or stability. In this case, dynamic compensators need be employed. Example: G(s) = 1/(s 2 + ωo). 2 Root locus shows that the closed-loop is unstable no matter what K we choose. ω o j ω o j 7

8 However, if we replace K by PD controller D(s) =K(s + z), z > 0. Then we have stable closed-loop system for any K>0. Since such D(s) is not realizable, we may use lead compensator D(s) = K(s + z), p >> z. 1+s/p -p -z Root locus with lead compensator. 8

9 Lead Compensator: We consider compensators of the form KD(s) where D(s) = s + z s + p. If p>z>0, then we call it lead compensator. If z>p>0, then we call it lag compensator. The argument (to the desired pole location) generated by lead compensator is given by φ m =Ψ φ by φ + φ m + (180 o Ψ) = 180 o. φ m φ ψ 9

10 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 jω-axis. Step 1: Sketch root locus for KD(s)G(s). Step 2: Translate specifications into desired pole location. Step 3: Compute the argument to be compensated for by using the formula: φ m = i φ i i Ψ i o. Make sure that the rest of the poles are at least 3 5times away from jω-axis than the dominant poles. Step 3: Compute z and p. Step 4: Compute required gain via gain condition: 1 K = D(s o )G(s o ) > 0. In general, each lead compensator can have at most 90 o compensation. If the required φ m > 90 o, more than one lead compensator need be employed. 10

11 1 Example: G(s) = with performance requirement ζ>0.5 s(s +1) and ω n > 7rad/sec. Note ω d = 1 ζ 2 ω n = The root locus is sketched first which has three branches. ζ = 0.5 implies gives the dashed ray with θ = 30 o, and ω n = 7 gives a dashed circle of radius 7. The intersection is the desired pole location. Compute the argument φ m as φ m = φ 1 + φ o =30 o o o. where φ 1 = 120 o and φ 2 =90 o +tan 1 (2.5/6.022). φ ψ φ 2 φ 1 11

12 The desired poles are s o = 3.5 ± 6.06j by θ =30 o. Since φ = 0 is the lower bound, Ψ > 51 o. A general rule is that Ψ=φ m +3 o 5 o =55 o, φ =3 o 5 o =4 o. The distance from s o to the real axis is = See the figure below. Thus z 3.5 = ctan(55 o )=4.2448, p 3.5 = ctan(4 o )= that gives z =7.7448, and p = Compute K as K = s(s +1)(s + p) s + z s= 3.5+j6.06 = p o 4 55 o -z

13 Verification of the Design: After the design, we need verify if the design satisfies the performance requirement. Find the poles of the closed-loop system: λ c (s) = s(s +1)(s + p)+k(s + z) = s 3 +(p +1)s 2 +(p + K)s + Kz = (s +83.4)(s j)(s j). The dominant poles are not exactly 3.5 ± 7j, but the corresponding damping ratio and natural frequency are ζ = 3.9 =0.5148, ω n = 3.9 ω = n that satisfy the performance requirement. Since the third pole is more than 20 times away from the jω-axis than the dominant poles, the design is satisfactory. The design can also be verified using step response of the closed-loop system: T c (s) = K(s + z) s(s +1)(s + p)+k(s + z), with MATLAB by the fact ζ>0.5 M p < 16.3%, and ω n > 7 t s < 0.86 sec. 13

14 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 D(s), we have that The above implies that lim s 0 D(s) =z p > 1. 1 E p =, E v = 1, E a = 1 1+K p K v K a K p = lim G(s)Kz/p, K p = lim sg(s)kz/p, s 0 s 0 K a = lim s 2 G(s)Kz/p, (1) s 0 and thus tracking error is reduced by lag compensator. 14

15 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 jω-axis as the dominant ones. Lead Compensator is an approximation of PD controller p lim s + z =1+s/z. p z s + p 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 amplifies the noise or disturbance. A compromise is to use lead compensator with large p. Steady-state error: the gain at ω = 0 is enlarged so that the steady-state error is reduced. Recall that the steady-state errors are: 1 Ep =, E v = 1, E a = 1 1+K p K v K a K p = lim G(s), K p =limsg(s), K a = lim s 2 G(s). s 0 s 0 s 0 15

16 1 Example: G(s) = with performance requirement ζ s(s +1) 0.5, and ω n 2rad/sec, and steady-state error for unit ramp input ess 3.33% K v 30. The root locus is skipped that has three branches. ζ = 0.5 implies gives the dashed ray with θ = 30 o, and ω n = 2 gives a dashed circle of radius 2. The intersection is s o = 1 ± j that is the desired pole location. Compute the argument φ m as φ m = φ 1 + φ o = 120 o +90 o o 30 o. where φ 1 = 120 o and φ 2 =90 o. ζ=0.5 φ ψ φ φ 2 1 ωn =2 16

17 The desired poles are s o = 1 ± j by θ =30 o.since φ = 0 is the lower bound, Ψ > 30 o. To ensure the dominance of s o, we set Ψ=φ m +5 o =35 o, φ = φ m =5 o. The distance from s o to the real axis is 3= Thus z 1 = 3ctan(35 o )=2.4736, p 1 = 3ctan(5 o )= that gives z =3.4736, and p = Compute K as K = s(s +1)(s + p) s + z = s= j For steady-state performance, we have that K(s + z) K v = lim s s 0 s(s +1)(s + p) = Kz p = But the desired performance is K v 1/ess = 1/.033 = 30. Hence lag compensator is required: D lag (s) = s + z s + p, z p = = Set z/p = A general rule is to choose z ten time smaller or equal to the distance of dominant pole to the jωaxis/ This gives z =1/10 = 0.1, p = z/8 =

18 Thus the dynamic compensator is KD lead (s)d lag (s) = s s +0.1 s s Verification: The closed-loop poles are the roots of s(s+1)(s )(s ) (s )(s+0.1) that are , ± j which satisfy the transient requirement by ω n = = > 2, ζ = / = > 0.5. The steady-state error is ess = = <

19 8.7(s +0.5) Example: G(s) = with performance requirement s 2 +2s +26 P.O. 16.3%, t s 1.333s, and steady-state error for unit step input ess Design Procedure: We consider dynamic compensator of the form K D(s) =D lead (s)d lag (s) 8.7(s +0.5). It follows that the loop transfer function is D(s)G(s) = s + z d s + z g K s + p d s + p g (s 2. +2s + 26) Hence it is type 0 that has finite ess-value for step input. Our objective is to design lead/lag compensator D lead (s) = s + z d s + p d, D lag (s) = s + z g s + p g, and the gain K such that specs. are met. 19

20 We first design the lead compensator. Percentage overshot requirement gives ζ log(p.o.) π2 +log(p.o.) 2 Settling time requirement gives ω n 4.6 = ζt s Thus the dominant poles are at =0.5. P.O.=0.163 ζω n ± jω n 1 ζ 2 = ± j We choose s o = 4 ± 5j to have some design margin = ω n = = > , ζ = 4/ω n =4/ = > 0.5. We now compute angles generated by p 1 = 1 +5j, and p 1 = 1 5j to the dominant pole at s o = 4+5j: φ 1 = 180 o, φ 2 =90 o + arctan(3/10) = o. Hence the required phase compensation is (see the figure next page) φ m = φ 1 + φ o = o. Thus two lead compensator is required. 20

21 We take two identical lead compensators: D lead (s) = s + z d s + p d Each needs compensate for 53.4 o.wetakeψ=58 o to ensure the dominance of s o. from the figure below we obtain z d = 4+5/ tan(58 o )=7.1243, p d = 4+5/ tan(4.6 o )= The required gain is K = s + p d s + z d 2 2 (s 2 +2s + 26) = s= 4+5j φ 1 5j -p φ -z ψ φ 2 21

22 Design of lag compensator: the position constant is now K p = lim s + z d 2 K s 0 s + p d s 2 = << s +26 Therefore we need lag compensator D lag (s) = s + z g s + p g, z g = 19 p g = Set z g =0.4 yields p g =0.4/12.16 = The dynamic compensator is D(s) = s +0.5 s s s +0.4 s Verification: Closed-loop poles are ± 55.84j, 3.97 ± 5.10j, and 0.25 which satisfy the requirement. MATLAB is used to simulate the closed-loop step response:

23 Analysis: With lead compensator only, the step response satisfies transient requirement (left figure). With both lead and lag compensator, ess-error is satisfied, but the t s spec. is not met (right figure). In some design case, lag compensator may give such problem. A common way to remedy it is design of pre-filter F (s) such that t s 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 t s. Thus we choose F (s) = 1+s/ s/0.4. The step response is plotted in next page 23

24 Both steady-state error and settling time now satisfy the design requirement. But the percentage overshot exceeds the specification. 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 3.5, and do a second design. However we skip in the class lecture, and leave you to work out in the project design assignment. 24