Question

The objective of the controller design in Fig. 1 is to find the controller Gc(s) such that I.)The closed-loop system is stable ii.)The output of the system above (y(t)) can track the reference input r(t) = At (A>0 is any real number). Use the Nyquist plot and Nyquist criterion to show that: a.)The portional controller Gc(s)=K can achieve asymptotic tracking of the ramp input r(t)=At but cannot meet the stability requirement for 0<K<+inf. b.) Can the PI controller Gc(s)=K(1+10/s) be used to meet the requirements I.) - ii.)? Please use the Nyquist plot and Nyquist criterion, together with the steady-state analysis, to prove your conclusion. c.) Can the PI controller Gc(s)=K(1+10/s) be used to track the input signal r(t)=At^2 for any A>0? Prove or disprove your answer.

plant [ problem 1 CPI compensation ) RS) +9+40 stopy Controller Fig. I The objective of the controller design in Fig. 1 is to find the controller Gels) such that i) The closed - loop system is stable in The Output of the system above (ylt) can track the reference input ret) = At (4>0 is any real number) Use the Nyquiat plot and Nyquist criterion to show that? a) The portional controller Gels)=k can achieve asymptotic tracking of the ramp input rut) = At but can not meet the stability requirement for ock<tos. 6) Car the Pl controller Gels) = K (1+ 18) be used to meet the requirements ;) ;;)7 please use the Nyquist plot & Nyquist Criterion, together with the steady-state analysis to prove your conclusion, Can the PI Controller Ge(s) = KCH ) be used to track the input signal rit)=At for any A>o ? prove or disprove your answer.

Answer 1

1)

S+10 S2 Controller plant let GeCS)= kp wat kpisto s? fps + 10kp KPS+ 82 = $? + Kopisto) RIS) Itke (st10 1 stable for all closed loop The above Transfer function is KPTO @ Tetat e. in no 7 Now yes) kuollep (670) SV + 10 KAKS Res) (+14(5+10) sh+ks +10k ) 15 405) = (+051641 Ak[s+to] AlfS710) = o Carnestory s41 ks joks' The controller accs) = k on meet the stability initeral for all K70 but not for ked because 461) Ris k(410) & lostlokp bet k=-=>(E becomes as s_s-10 /5H) (stio) S=-1,10 poles is on the Right side of the Root locy So it is constable for all ico Scanned with CamScanner

2)

Alyqurile yu) kisho) we draw Nyquist plot for Oltr ie open loop Transfer function of al system put Sujuvuelo 1 (100+10) - Magnet Klipco' tio?.M Rujug juje s noguerit .M (H = - 1800 + tant (w/10) = f - phase ' -180 -90 WO mooo tre are ve an's -90

3)

O QC3) = K (H+.10/3) = *( + 10) x (stre) - K(SH10)2 yes) K (840)2 53 - K(s+10)2 ROD SP+ x(st 2012 5P+k[s+206 +100] = K (stid? s3+ks?+2068 +100* in order to be stable bc² rad => 20k'>100k 2017100 wyQuest'; oltr of Ce System is K75 outf yos) -s S) K (5+0)2 1 (M1 = Magnituda Ris) W =0)-270 MER = K (83+102) wito a-stant (20/10) 000 MO o to 1-2 tots tant (w/10 308 I-240 1011 - 90 Stable sts Lyscanned with CamScanner

4)

© g-tolt): At? | Ges) = ( 14 iols) Res) - 42 $ 3 wikit 11) - Gels). (6+10) K(Sto/S+10) ROS) it Gers) (Strol HKISHO)? S3 yeo) - Krstio) 2 Kist 10) 2 Res) gat k (St 102 834 ks'+20KS + 100k 419) - Oxk (5+1012 Sº [03+1st+2018+100k ? 2 (s + 10)2 S9+ kB & trok s t + 100ks By RH Criteria Ti 0 0 Look To K 20k The above RH contero unstable. characterste equation is not satisfying the So for the provided input the system is becoming CS Scanned with CamScanner

SUMMARY: here we have to find out characteristic equation of closed loop trnsfer function and cheack for stability before continuing furthur analysis we must ensure weather the system is stable or not.We have to apply laplase trasform to convert time domain input into s domain.

trick:while drawing nyquist plot you can follow these three simple steps

i)draw polar plot

ii)draw its mirror image

iii)join the mirror image end direction with polar plot starting direction

HOPE THIS HELPS YOU!!

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