Question

Res) flant Gain / 425t 2 Ca) wlen PID-Contrel Gainsks) Klsts)(it Spetel the Yoot - locus (Plot the keoKeoe) : abnt oPen-loop Plart's respause YS)-GCS)us) /(mit step Tu) x ax3:time t) Plot the g raph (usng muene (aplace fam) Y RTS pesfonse YIE) ConPare on a bant Closes-loop system's resparse ys= T19)r)/(gan k = 14) Plot the Yaxis: pespaye E aPh Cushy inere larhce torm) L4:CGiven Gis)SH Plot the root 10cus on whan p-lantnGain is es)=k Pectively aud Po-Cantrol Gain is ks)=K (S+2) Res) flaunt Cain s42512 La) When PID-Contrel Gain s Ks) Klsts)(it) Gpoteh the Yoot - acus (Plot the KeoKeco) : abnt ofen-10p Plant's respanse (s) -GCS)us)/mit step Tp) x aT3:Ene r) Plot the graph (usng invene (aplace fam) Y ATS resfonse Ye) Cu Pare on abnt Closel -loop system's resparse 1S=TIS)rs)/(Jank=14) Plot the yaris pesae yePh luhg mere larhce torm) L): Caiven Gis) Plot tha rot lous whan P-lntnGain is Kcs)-k Jrapeiey aud PD-antrol Gain is ks)=K (st2)

Answer 1

first page of solution is uploaded at last.uploaded pages of order is 2,3,4,5,6,7,8,9,1. The page number is written at the top right corner of the each page.please adjust.if you have any doubt comment to me.thank you.

(-1+15) C1t1t0-2) (2 26H S-1t) t an tan +tan -90 14.036-51.34-9 +4s 2.304 lab7 ( 80 82.3oq 97 696 97.696 PLA ko - 0.2 420 - 5 262-304 trom 6realc poink It open loop TF z0 Cs+25+2s) ditf. co.r ts on both k Sides and sV45.25 + equal to z e0 d e 5P

as2) Cs4525t1)- s34241s) C2s+52) dind 2 0-8s+ 4S +26t lo 4st) 84+ AS.6 S+ ss+ 4s -2s-5. 2s3 453-10-45-4sY-164 sニo S9 10.45 +1-9s+4st2 0 S-9.20 Brealc in point Valid 1 b 4is)E GES) uts) 1 SCt25+2) 7+5Tt A yes) S++ (s+-i AlSEOt 2j C-1-) Ei-5) C--jーり Sニー1-j 2iC+) (-11) C-1+11 +51) 2(-111) 1 -1 + Gti45リ 25C1+5) 2i C-11) St1- (7-1)st1-1) (11) Cst1+1)7 t 2j 『+リ CJ-1) Cs+」t5) (st1-J)

ftst) - +5Csti) 1) t aj (P-C22) 25cst1) +2 ytu aj (1-1 Ct25t 2) yesr S41t) + 1 2 -2 (s+2St2 S+2 20 S+1 I 2 S 2 St1) + 1 Mpply invesse aplace Trans form uct) sint) ut) CoS t 1 Cost-e Sint e u ct) 2 yu- t(1-e CostStot)) utt) yt) o yt o312 t 5- t1o Ct) O.499

14. KY42S42 k. 6cs) Ces) TIS) - KGCS) RCS) 14 425t Ces) TOS): S+2S+16 RCS) 14 4ESE TLS) RCS) RIS) sCs2st16) SC s+25+16) s(S+1+j3.87) (St1-38) C Sc125t16) S St-3.87) (Stltis.87) 14 A 16 14 -S383)(イー3.8aガー234) s -1-5384 14 1-53.81) -j7.74) 14 『ネそ4 - 29.9538 14 SE-1+3.8 t = F1ti3 ) (74) 14 - ft79-29.9538 -4 29.9538+J子 44

14 yes) SCチ74)-299538 S1+53.89 14 S+1-f3.89 了チ44+29.9538 +2 9.95 6) + S+1) +J 3 82)(474-299536) 8 S 7 2St1 RC2.9538) Cst)+ 2(349 X744 14 $39.322 91+5てt S+1 14X2X29TS38 32 323 91 5 t1+ yes)= SY425+16 S8 s'TE) Apply fouerse loplaee Tans fom Cos JIst e sin uct) S1 N ye - Sin t

tzo -1 o.12S 七こ1 → yu)2 O.367 0.9971-0.0675 - 0.51S 9]= te5 44t)= - 0. 006 0.943-0 085 = 0.869 tz 1o t) 0-7801-0.1(6LS 0-8749 -0.0000953 4Cs) GCS) uus) 0-499 O.416 -312 o-3 t 10 uts) TCS) Rts) .87 49 ,0.869 Kb.515 1 -0.125 -0-S

GUS Ststi) i + kcs)E k. Open Loop T.F: Sts+1) No.0 poler z, No-o zeros ニ2. No-of tinite No.of RLB poles poles zevos L No.ot Acymptodes N P-212 (29 t1) x tRo ao1,2 P-21 P-2 2 0. O81 3 x180 240 ; 9と1 Real part ot pole Sumot Real part of zers Sum of Centoid o = P 2 O 24 -1 2 2 20 RLB 2

If kLs) kCst 2) KCst2) Open oop TF = SiStI) No.0 poles = 2 No. of zeTOS RetaB (N) P-2) = , Asymptodes. = 180 9o No of にこD GEGEISIO -3 414 -2 -1 70.585 o -1+ 2 o-1-C-2) Ceptoid o 1 Break pointsfrm open Loop TF 0 (5 t- di ff oY to s (25+1) CS4 2) - Cs+ s)C1t0) an d equate to o dic 2 S+ 4St C Bocak acoay point o. 585 Break in point -3.414

plant Gain a tet 4 assume Negative feedback GeD Gain (s+s)Ct K Cs+5) Cs+0-2) Gain Open loop TF kCs+5)(s+o.2) て+Sてよ KCS+5) CSt o.2) SC SY+25t+ 2) NNo of poles= 3 No.0f zesos 2 Noof finite No.0f Root Loccy Branche pole if poles 7 z eros a No of ASymp tode N P-21 3-2) 1 29t1) 180 angle of asymp todes P 2 9E0.1 P-21-1 o, 081- Angle of deperture = cd 180 L6H ve comple pole, . S2 5t 2 +1 rO i 180 + 26け S-1t