Question

PD & PID controller design Consider a unity feedback system with open loop transfer function, G(s) = 20/s(s+2)(8+4). Design a PD controller so that the closed loop has a damping ratio of 0.8 and natural frequency of oscillation as 2 rad/sec. b) 100 Consider a unity feedback system with open loop transfer function, aus. Design a PID controller, so that the phase margin of (S-1) (s + 2) (s+10) the system is 45° at a frequency of 4 rad/scc and the steady state error for unit ramp input is 0.1. G(s)

Answer 1

a) consider a unity feed back system with open loop tyans fer function, Ges) = 20/s (s+2) (s+4), Design la po controller so that the closed loop has a damping ratio of 0.8 and natural frequency of oscillation as 2 rad sec. sol: To find dominant pole, sd Dominant pole, Sda SWn jandi-SZ Given that, Ŝ = 0.8 and celo = 2 rad /sec i Sj =-0.8 x 2 + j2x VI-0182 = -16 + Jl.2 Let D= Island ß = 25d. By considering dominant pole at (-106+j1.2) we get S8 = -1.6 tj 1.2 = 2 % 1439 . De 2 and B = 1439 (note : use polar to rectangular conversion ) To find magnitude and phase of Gcs) at sa Sy. Let Ad = 1G (S)] and 4d = LG (9) Given that, Gcs) 2018 ° (s+2) {5+4). G(Sd) = 20 sd (Sd +2) (sd +4) F1.6+j112) (-1.6 +31.2 +2) (-116+j1.241) 20

20 20 F1.6+j1.2) (0.4+j1.2)(2.4 +312) 22143°x1.26 27142168226° - _ 20 2x 1.26 X 2.68 ce < (-1430-71-26*) = 2.96 L-240 and 4d = -240 i Ad = 2.96 Determine the trans fer function of PD controller Derivative constant, kde sin od DAJ sinf sin (-240) 242.96 * sin(143) L0.243 - 0.243 proportional constant, Kp Sim Kp = - sin (B+dd) -sin (143 - 240) Ãa sin ß = 0.557 . 2.96 sin (143) Transfer function of pip controller G,CS) = Kp + Kg8 = (0-557 +0.2438) = 0.243( .557 +57 = 0.243 ( 21292 +5) verify the design open loop tronsfer function of compensated system Go (s) = Gr (9) G(s) 0.243 (5+2.292) > s(s+2) (5+4) • 4186 (5+2.292) s(s+2) (8+4) 20

Gc (sa) = 0.243 (98 +21292) - 0.243 (-1.6+31.2 +2:292) -0.243 (0.692+ 31.2) = 0.243(1.385 260°) = 0.337 260° . Go (Sa) = Go (sa) G (SJ) = 0.337 260°x 2.962- 240° - 12-180° --|| since Go (Sd) = -1, the root locus will pass through Sd and the design is accepted. Transfer function of PD controller Gc (S) - (0.557 + 0.2438)= 0.243 (s+ 2.292) open loop transfer function of compensated system GO (S) - 486 (s+ 2.292) s(s+2) (s+4)

consider a unity feedback system with open loop transfer function. Gcs) = 100 - Design a PID (s+l) (S+2) (S+10)' controller. so that the phase margin of the system is 45° at a frequency of Arad /sec and the steady state error for unit ramp input is ooli 301 to find magnitude and phase of G (jw)) Given that, GCS) - - (Sti) (s+2) (S+10) · (hts)x9x (1+ $_)x10x{1+) 100 100 (ts) (1+0:58)(1+0:13) (1+5)(1+0.55) (1+0.15) put s = jw in Gos) . Q (jw) = (i+j) (1+ jo.5w) (1+30.16) + q• Llan' +0:25 a* LỆan o 56, I+ovo tai qua la ciu)) = Tites Vitossa Vitoon LG (j) = -lan'w - tan! 0.5W – tant 0.16.

of compensated The gain crossover frequency system, W, = 4 yad /sec. let AE l (jw) at = , 0. = LG (jw) at 6= w A = - - 0.5 Tota2 Vita? x 51+0.25x 42 x V 170.01x42 0. = -tan 4 - tan! (0.5x4) - tani! (0.1x4) = -161° To find You and o Let Yo = phase margin of uncompensated system 74 - Desired phase margin of compensated system. o = phase of Ge (jw) at C = , NOW, Ya = 180° + $i = 180°- 161° = 19 0 = Ye -Y = 45°-199= 26° . To find transfer function of PID - controller. Given that, steady state extoy, ess = 0.1 for unit ramp input. .: velocity error constant;' Kva da ti =10 The - velocity error constant of compensated system is given by. Kva lt SG (S) GCS)

Here Ge (S) = kp + Kys + ki = kys? + KpS + Ky S and G15) = (1+5) (1+0.55) (1+0.15) . Ky = its (kys? + KpS +K, ) Sto (1+5)(1+0.55 (1+0.15) = 10 . 5k, = 10 (or) Ki= 102 Derivative constant, ka: sino C, A 4X0.5 = 0.344 Proportional constant, Kp = coso , CoS(26= 18 A 0.5 Transfer function, S + 8 +0.344 otvolley f QCS) = (xp trys + ) = (1.8+0.3445 +}) of PID controlley = 0.3445² +1.85+2 0.344 (s2+ 18 5+2|| 344 0.344) lon = 0.344 (s?+ 5.235+5.81) - s To find open bop transfer function of compensated system. The PID. controlley is connected in cascade with Gics) as shown in figure

formuy 6945 23.5+58) He 0.344 (3+5.235 +5.81) (1+5) (+0.55) (1+0.15) Block diagram of system with Cascade PID controller open loop transfer of compensated function la " } Go(s) = G, (S) X G(S) system Joo = 0.344 (32+5.235 +5.81) - X - s (1+5)(1+0.55) (1 +0:18) To verify 1172 (s? + 5.235 +5.81) 5C1+5) (1 +0.55) (1+0.18) the design put s= jw in Go (S), 1.72 (-62 + 35.236 +5.81) ja (1+jW) (1+ 30.5W) (1+jo.16) i Glo (jw) = 1.72,\(5.81-60?)?+ (5.236) Ltant _5.236 : 5.81-02 2909114602 tant cot 1+(0.562 ltarlo.sw / 1760.16032 Land 0.16. Let Ao = 1 Go(jw) and do = L Go (jw) i do = 172/65.81 - Co2) +(5.23612 w/1t co2 lit(0.562 ) + (0.160.12

do = tan! 5.23W - 90°- tan-'w - tan' 0.5W - tanlo.16, . 5.81-62 for wc 15.81 = 180° + tan! 5.236 - 90°- taniw - tan' 0.50 - tanto.6, - - 5.81-02 for > √5.81 At W = W,, Ao = Aoi = 172x1(5.81-43)2 + 15:23x4j2 4x +1442 x 11+(0.5x432x]17 (1189) 2 At C = C,, $o - Po = 180° + tan! 5.2384 -90°- tanil 4 - tant - 5.81-4² - (0.5x4) - tan-! (01x4) = -135° phase margin of compensated system, Yo = 180+ do, = 180°-135º = 45 The phase margin of the compensated system, meets the given specification. H Hence the design is acceptable. 1) Transfer function of PID controller Ge(s) = (1.8+ 0.3448 +2) - 0.344 (5²+5.235 +5.81) 2) open loop transfer function). 172 (s+5.235+5.81) of compensated system sci+s) (1+0.55) (+018)