Question

The transfer function of a position control system, with load angular position as an output and motor armature voltage, is given as G(s) : s(s + 10) For this system design the following controllers 1. Proportional controller to obtain { = 0.7 2. PD controller to obtain { = 0.7 and 2% steady-state error due to a ramp input. 3. PI controller to have a dominant pair of poles with { = 0.7 , wn = 4 rad/sec and zero steady-state error due to a ramp input. 4. PID controller to obtain { = 0.7 and wn = 4 rad/sec Compare the controllers based on their time -domain response. State your conclusions on basis of the results and graphs. Submit a report which will contain a. The mathematical work used to design the controllers in each case. b. Simulation plots showing the closed-loop response in each case. c. Simulation plots showing clearly the rise time, percent overshoot, settling time and steady state error for all states in each case. d. Well commented MATLAB code for each controller. e. Provide a copy of the code in the project report.

Answer 1

GUS) = sestiel. combensato y = Gic(S) = Rp. ter Characteristic oan will be ! 1+ GLS) Giccs) = 0 It 116) = 0 S2+ 40 S+Kp =0 combering this ean with SP+2Ewns tonizo com? - KP won = V KP pewn = 10 on veequired &= 0+1=5 / lip = 0432 =151:02) 2-7 let PD controller i Giccs) = xp Cl + Tas). Characteristic eah w i It Gl(s) Giccs) = 0 It Sisto X Kp (1+Tds) =0. → SP + 105 + XP + RP TOS =0. SP+S CIO+ Word) + kp=0 on comhau'ng waith strewnst while o cons VIP. a coon = 10+ kptd com ste VAPE Kpid E = Yo + KpTd Puipp = 0.7 Mequined . Lc. ess in quien as: ess a lim SRCS) S0 1+ G1) Gc four ramp Inbot R(S) =Śé.

S70 essa lim sx SP + Istvoj* kp CI+Tds ) lim 1 S+ lyp CI+ Tols) $ C'S+10). c ol (equined ) O + 어 Rei 10 용 1 .02X VP AP = 10 Op Ap. = 500 from ean O. Lo + lyp Td - =0.7. Pulp 40+500Td -0-7 2 X 22.36 Td=043 Thus Giccs) = 500 CI+ 0043 ) PI controllen is Gels) = KP (I + c) = Rp (Ite 'TIS. Gls) = sest 10), stimes since pole at s=-10 iu move the distant than e smallest pole at s=o. hence by dominant Pole rule Pole at s-10 Can be cremoved it will not affect wewit. GIS) -Ź R(s) = 5 , Howe ramo Stó H Gils) Ge(s) inbut. B) ess = I'm s. Als) - en st 4*** MO CHE

ess = pt WP lo+0) = Jouver. Doy characteristic eqn coill be. it Grecs) Gils). - 0 1+ HP (TiS)) x š = 0 Tise + KpTiS + xp = 0; ☺ SP + (xpls +K 20: on comparing with SP + REVO) 5 + 60 = 0 one = ke DEwon = Rp. Mequined E- 0.7, lon= 4. Rp = 28.744 kp.= 5.6 come sql = Hp - Ti = Sob =0-35 16 Thus Geld = MP CI+) =f56 (+0355) 7 PID._ E=0.7, WA4. By clominat Pole vole we can write. Gush- + PID controller be = kp (1+ Tasty so characteau'stic eqn us! 1+ X RP CIS+ TITI S + KP (TIS + TOT T'S ite. Gls) = 5. KP (It Tastitis) I= 0 TiSP + Ipti's + APTIT'SP +WP = 0 ise Tº + IPTC Ti) + HpTiS + xp = 0. SP + KPIPs Ti + kptati CI+ kpta) + Rp

Up set out in s tem leheteta) = 0 On comparing this eah with SP + pewns tuose = get 2x0.7X 4) st 42 = 0. = SP + 5088 +16 = 0. Кр = 506 14 WPTD Rp = 16 A CIA hp ro] soj Ti 2 Sob =1035 it Wp =10 10 _Jo - sob. 1+1oTo Td = lorset = 444 3644 su PID can be 10 = Hp C1+ Tast = 40 C1+00445+ desis)