Question

Question 1 An electronic circuit is shown in Figure 1. In order to simulate the circuit, a transfer function is needed. a) Is this an open loop or closed loop process? Justify your answer. (3 marks) b) Find the transfer function V.(s) for the electrical network shown in Figure 1. V:(s) (10 marks) c) What is the order of the process? (2 marks) d) Determine the roots of the transfer function and comment on the stability of the process. (5 marks) 192 0.5F 292 T V, (t) 522 0.2F V (t) 0.5H Figure 1

Answer 1

1) iJp loop 1.2 A 0.5F unt B 22 um C tilp loop 0.8F 20.5H Pilt F E D Fig1 @ the circut shown in olp loop. Clearly have 3 looks ЛАВЕРА OVILA) i ABCDEFA Volt) riil BCDEB ole loop And all the 3 looks Que conneeled to each other. | No breakage of link in belween the looks. - So clearly It is an {"It is an Closed Loop System .* You can think other ways As some of the output loop current ialt is some how connected to the input loop o aho. So the encuit has some fee back mechanisum and feedback eust in close loop Dyplem only. So, el is a close loop Syplem. So clearly @ To connects on indu chor mlo Laplace equivalent s-domcein L LS A Capacitor is convaled inlo Laplace equivalent as C s-domain 1 CS. where Li induchince of Coil C: Cabacilance of coil Scarent, with Renishonce in S-domain Remains same. -2 anos

@ Using above concept draw the s-domain (Laplace equivalent) Crcuit in dig 2 - Capacitances & V(s) s-domain i 22 А 11 w C th B 0.5F 0.55 WWW 52 5) Vils) alu bla Vols) - s-domain colo 1024 0.25 F E - 55 ulos = 0 + Inductance Fig2 S-domain Equivalent 0.5H 5-domein 0.5 5 = of Grcuit. -> Now we can use all inductance and capacitor Value simply as a impedance in S-domein Analyse). -s dot voltage of Node B is v/ Applying Nodal Analysis a node B VIS) VIS) - Vils) Vis) 5+ On Simplifying S [v/s) – Vils)] & Vis) S.VIS) (5+2) Now adding, we gel |${+5/S+10) [vis) – Vils)] + 8V(5) (S12)605 +5) + Sv/s)[Sta)/S+10) Scanned with (S+2)(S+10) (QS+5) =0 + 2+5 wolen 1 + 2 S + (5+10) 125+5) = 0 C CamScanner

لا Taking Denemindor to other side and separahng vis) and Vils) term in numerator we get V(1) [s fes45)(4b) +8 (5+2)2845) + s{S-2/3+1)] +15)75425 15)$#*]=0 Taking vils) term other side and Multipleng all vis) tremo V(9[25°+2552 +505 + 48° +185 +20 + s +125°+8057 Vilsbl2515) (S+10) Adding same Power terms of : L.HS. co 1 VIS) [ 3s8+ 4152 + 885 +20] = Vi(s). S (25+5)(S110) S(25+5)/S410) :: V/S) Vils). (389+418 + 88S+20 Fig 3, will show the helahon blw V(s) and ole Vols). co Ri= 22 mm VIS) C 1 voltage dinsion R2=5 Vols) Applying Rule Ra vals) = Gana where Ra= Jur s) D R2 figa cafer : R=22 Pueling Ri, Pa and Vils) foom eq" D 5ls 5 Vols) = (245) X VIS) - c Scanned with Cam Scanner (25+5)*V/s)

Pulting value 5 Vols) X = Vols) of V(s) Vilo (28+5)(8+10) (25+5) (354+4182+885+2D) Cancelling out common term 68845 ) and taking vils) do lus 55 (S+10) And Vils) (33+ 418+885+20 Transfer funchen of given elechical Network. © Order Order of the Syplem! Order of the syslem in the heghest power of dexiyatine in a différendel Equahan or highest power rof Š' in the denominator of tranfes Junchon. s in the denomindor of Vols=2. Vils) Here Highest Power & Order of System = 2 And @ If we solve for Zeros le rool of Numerator of Tranfer Junchen. 5 S(S+10) = 0 s=ok first zero Le = 0 S+10) S = -10 They are known as zeroes of Transfer fun chon cs Scanned js = 0,-10 Ang Zeroes of Transfer Junchen CamScanner

a g Poles of a transfer funchon. When we find mools of the denominator of transfer Junchon le 35?+ 4152 + 88S + 20 = 0 To find the root of eqn we the calculator and find -0.26 the transfer junchm. Poles of S - - 11.07 Am u - 2. 34 the ove - A suplem is called a stable syslem it all oth Poles Real lies in of jw Axis of S-klane or of all the Poles. All the poles of given Transfer funchon lees in re of jw Axis So This Sylem is stable Am Paul is-ve Clearly -> -re C$ Scanned with CamScanner