Question

Rs) Plant Gain Gls)= st25t2 Closed -ap System Cpin (a): when the P-Cntrol Gain kls)=k changes the PI-trol S)-K(1+ SOIVE the chaVacteristic eruentin, Sketd the rat locus atont chouging C6) when the P-tl Cain kis)=e donges the ID-Cantrl Gain Kis) -Ks5)( Solve the Charmcteristic equation, Sketch the reot ous abnt chenying e:when the PCwntrol Gain 1S KIS)-K, Determine and Pole's psition damping Jatio o. 5 and Plot the S0 that the donain rts have Yoot latus e PD-Lontnl Gain is kis)-kst5), Detemine and Pole's fositian d)when Go that the donain rats have a damPi Yatis Pnd Plot the 2 Yoot ocus o-2 e): Nov we use the PGon: -k, PD Gain ps)-Kises), PI Gain: KiS)= k(It Compare the stedy-eror storterusng Frnal value theoren abont -20 ) usng le), Plot the P,PP.PI GAn on aph Res Plant Cain Gls) St2st2 Cosed lop system Cpin Ca): when the P-Contl Gain e)= K changes the PI-tre-(1+ SoIve the chaacteristic euntin, etl the Yort acns alost Cb whan the P-Ctl Cain is)k dnges the D-Crtrl Gain Kes) = KUsts)(1) Solve the Chamcteristic equutim, Sketd the t us ant chrging Pole's forition (:when the-Cantrol Gain 1S K Determine and that the Jorain rts hare damping Jertio 5 and Plot the Yoot locus (3-e (d)har the PD-Contl Gain is kis) = KISHS), Deremne and Pele's fesition damPing Yatio nd Plot the Go that the doain ts have Yoot ocus e) Nov we use the PGin: -, PD Gain s)-Kises), PI Gan: kis)=K(It Cour Pave the Stenly -eror stmte using Final value theoren a bent k-20 ung Cel, Plot the P,PP.PI Gain on gafh

Answer 1

(1 ccs) HLS)= KCsto.2) 6C5) HCs) sCs+ 25+2) Yf 2st2 Characteristic cquation t 6es) HCS) o KCS+0.2) SCO+2St 2) SC2+2) tR(Sto.2) SC+2542) S3t2s+ 2s KS+ 02k zo S+25 + sC2+k) t0.2k 0 ROot Locus plot RCS+0.2) GCS) HCS Sc420+2) No.of poles 3 ND.O zeos 1 No.of Root Couy Bran chea poles t poles 7e No.ot asymptodes IP-21= 13-11 2 NS (29 t1)180 Angle of aymptoder P-2 Ip-z e 0, 1, 2

90 3 x180 270 Sum of Real partof poles- Sum of Real part of Centroid o Nenos P-z. O. 2) -1.8 -2+0.2 0.9 -1-1-0- RLE- Root loccy BYanch kzo Jo RLB + -eek 1-o.9 - o-2 C k-o 276,3a RiB we have Corm ple poles So we need angle of depertare ad +ve Comple1 pole 26H KC-1tit0.2) C-o-8+j1)

t80- tan ニーtan -90 t 180-180 +455139 1 96-34 Now Dd 83. 6S Gain kes) CS+5) (14 ) KC545) 42541) GCS)HCS) RCS+5)Csto2) Ges) HCS)= SCSV+ 25+2) Now characteristic equation 1+ G) HCS)0 KCST5) CS+ O 2) SCs+2 S+ 2) Sう+2s+ 2S+ K( SY402S45S+リ=0 S3+ 8s+ 2St Kst 5.2 KSt K=0 s34 5(2+ K) +SC 2+5.2K)+ Kro

KCS45) CSto 2) Root locey plot. Ges)HCS) Sc25 2) No.of poles 3, No of zeot2 No of Root toey B1anc ber poles po ler> z enos No.of asym ptode lP-el13-2) = 1. e (29t1) XIeo a01, P-2)- p-2 0 Centoid s Need only when No 0t arymp toder2 here So Need centcid NE 249-62 RLB -5 RLB H97+ 931 P 4Hjt KC-141+5) C141+0.2) KC-4+1) (-0i8+) L6tH S2C-14J1) '() tar0) - 1S0tan -tan 69.6 24 9. 62

Brealk point thom 14 GLS HCS- k . -+ 2s25) SY+5.25 + diff co.to s on both side and cuate to zero, e S2)(s45-251) - (242)+Sa) de de ニ0 (34S+2)s5.2s+1)-42842) (2s+5.2) =o 3s94 15-6 35+ 4s3+ 20. 8s +4s + 2s+ 10.45+2 2s4-52 3-4 -10.4-4 S 10.45 50 4 +10.4S t 11.4 s+95S +20 Break in polint Sニ-9.20|→ Nalid KCS)K Transtes func-ton 1 て+Sてt SV+2542 TF Compare with A &E0.5 Giveo 25, ton s ton R& ton 2

い K+2 Kt2 2 K2 4 ヒ=2 Now Ges) HCS)= +25+2, No-of Zeros 20 No 0f Poles 2 2 HP>z loccey Branche = p2 No.of Root No ot asymptodes 20/0 2 NEIP-2) 2. -1-1 o -2 Centboid 2 o (29 +1リメ180 e p-2 2 3 x19 2 Fo 2t0 YRLB

GH 90 FIHfi+i C-1ti-iv 2j ab Po KCS+5) ECS+5) +20+2 +25+2+ヒい+5K KeS+5) 14 S4252 KCS+5) 3 TF = given) - 0.866 SY+ SC2+ヒ)ナ2+5k wn 2+5 conJ 2t5c 2&00 2+K 」.チ32 2t Squaring on both side 31 2+5k (21SE) = 4ナビ+4に 2-00 20N ヒY+4ヒ+4-6 -15にニo Ic-11C 2 0 Ko1179 -O.17, K 11.1 79 ECS+5 No o pole 2, Root locey Gts) HCSJ +2042 No.ot zen RLB 2, P>Z No.0f N (P-2) P 2-11-1 -I80

H97 +031 PO KC 4+ji) kC-1+1+5) j2 2GH 190 + tan 75.96 d104.03 104-03 PLB 9.2 9-5 2.55.96 Bocal point Yom 1+Gts HS) o -Ct25+2) differentiate both ides coith nespect at and equate S+5 to z eo (s+2) Cs+5)-(42512) (Ho) de ds (+5) est2)(s45)-C4 20+2) 254105 ナ25ナ10-s-25-2こ0 S-9.129, S-0.87 Lvalid Breal point 9.129

Steady state eor (ess)- S ECS) e ct E Let Consides Res uct) Cinit step in put KIS)EK, ヒニ20 ECS) Ces) GLS) 425+2 . RCsD Ets) 1 Res) 2) RCS) ess 14 S. S+25+ O+0+ 2 ess = ス+20 Ot0t 2+k 2 2 ess z KIS) CS 5) Res ECS)= 「+ ヒCS+S) G(s) S+2 St 2 ess- (s2+2)+ K(S+5) 2 2 2 102 2 +100 ess 2+ k15) 5 1

Ct+ Kis Ctar O. 2 -2) s8nt 4s RCs) ECS)= Csto.2)GCS) 1-+ S( S25+2) ess = SCSV+2 5+2) +k(s+0.2) 7 Css 20 C0.2) o+ KCO.2) ess o ŁU KIS+5) KCs 20 kes 20 C+5) tis) 20 140Mgo Mgn = 20 20 20 223.606 S+5SC ad PID 2039 -1 -10 w