Question

Q2) If the machine lab have the parameters listed in Table 1, i) Develop the transfer function m9 and draw the corresponding block diagram. ii) Compute the steady state speed corresponding to a step change of Vt = 134 v Raa Rob Ricc 3.82 2.2 3.21 Table 1: Machine Parameters Te:electrical time const. 0.049 s Motor const. Im: mechanical time const. 1.027 s B: Viscous friction coefficient 0.758 *103 N.m.s Km0.97 Field Circuit Fig. 1: Schematic diagram of the separately excited DC motor Va(s) 1.(S) 2.() Rate B(1+ ) E.(S) Fig. 2: Block diagram of the separately excited DC motor

Answer 1

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Given. Ra Moom E field Circuit sla -Te. X - Wm. Schematic diagram of the separately. I encited DC motor. i) Transfer function wmls) S = 1415) Ra La By Krichoff's voltage , law. î s a VE SaRa + La. dia + Eb = Vti mm 1. dt In Separately excited De motor, flux is constant. Eb = po nj 0x3 = constant. wm > Eb = km wm. ☺. Substitute. equ ® is equo → V = km wom + li Ra + la. dia — Torque = I= Kg P la = km fa ho Komula.., ☺ Apply Laplace Transform Azny L{ Aimlay > focs) - Kom 9aly)]

orm Apply Laplace transform to equ☺ » Aluey = { known the lakay +2{La dialy => 418) - Km wm(s) + Ra Pa(s) + 6/a-Dals). > 14(3) = km wmls) + Dal 3) Ra+sha] > Vt (s) = km um(s) + Ra Dals () + Rs 2,15)(-+ 5 cm Te = electrical Time Constant = Lalka > 415) – Knwn(s) + Ba Tals) [1tsve] - © In te W10 MA - dt The dynamic equation for the mechanical system Te = J. derem + B.wm+ Th. — Here, J = Moment of inertia of rotor in Kigame. B = Damping ratio of the mechanical system in Nas wn= Angular speed. Bum sepresents rotational loss torque of System. Apply Laplace transform to equa a Hrej. 443. domnis + 4B.com) +{ vay. ar olt

LEOLU → 915) - J. S urls) + B wmls) +7215) » 7(s) – T(3) = m(s) (8+Js). Won(s) = T(S)-I(S) (B+Js) = 715) - TL(S) B(1+5.7/6) Mechanical Time Constant = Im= 7/8. um 13) = 7(5) – U15) B(1+57) Substitute Equ ☺ in above equation » lum (s). Km Dals) - TU15) Bl 1457) Equ ② can be written as => 1 Ta(s) - 14(5) – Kmwem(s) Ralitste) V (5) - input to the motor. wmls) = Output to the motor.

• By using equ , eau , equ ® and equ6 , draw the block diagram. - Equ ② » Eb (s) = Im wenis) 5 km k wonis) Equ 3 = T(8) = km Dals) = Dals) y KmH otel Equ * > Homls) = km Jals) - TL() B(1+57m) 31(s) - P(s) The obstant and , Els y Talms B (itsTm) Bals Equm > Dals)V715) - Kmuenls) Rall+852) vislt@ les = VHS) - Ebls) le bls) Rall+stre) . Combine all the blocks. ?(s) Ebls) Vtis) + Wnis) 1 km RA Ebis) km k

amis -? 115)=0. VH15) Tls=0 100 a patien) 4, 5, Eb(3) kim RSG G2 / Cng ]_119) 3219-1 G1419). 1 G = G, GzG3. km Here G, =- - Rall+ste) Gz = Km ; G3 = . - B( 145 m) ara Бю) . Ga Ra B(1+SFe) (148Tm) VE (3) Vzis) wmls). God "Brall+578)11496 I komt If RIS G Y15) R(s) G els) 1AGH i : 4(s) G tot It GA

km km Here G=0 H -Km RaBl1+57e) (1457m) . wmls) G ( Rablluste) (175Tmn)) V415) I+GH 11+ (km)[km) RaB (17ste)(1+SFm) » wmls) Vxls) Rabli+ste)41455m) | Pabll+59e) (145Tom) +km Rab [Lustre)(lts For wmls). Km 1415) Rabl 1787e) (148Tm) + km 1 ani . This is the required Transfer function. ii) steady state speed =? when 4102= 1344 04/) = 1347 Steady state response :- 1. It is that part of response. which is fixed when time "t" tends to infinite. Here the accuracy of the system is analyzed Cloo) = \t clt) = 1 S.C(s) 1. to S >

« (یا مهما -F 1415) Dorina Rab[1457e) (1485m) +km. linge Substitute 4+15) = 134. » wmls) . Kam RaBC1+57e) (1455m) + Kom? inloo) = It s. m(s) = It unlt) sto to = s km It sto Rabli+8te) (14 STm) +km) ! Kr) 134 RaB (1+0) +1+0+ km² ) wnlo) a (134) Km RaB + km² Substitute given values. B- viscous friction coefficient = 0.758*10* N-M-S. kma Motor Constast - 0.97 Raa = 308.2 Rbb=2.22 Rec=3.22 Kaa Ra = Raaf Rbh t R = 3.8+2.2 +3.2=9.22

Wn100) = 134 (0.97) (7-2)(0.7587109) +(0197) 129.98 0.0069736 +0.9409 129.98 0-9478736 = 137.127 wmloo) = 137.127 rad/sec This is the steady state. speed. Correspanding to given isput.