Question

Problem 4 (50 pts): Consider the electrical circuit represented in figure, with R = 102, C = 1 F and L=1 H. Assume V (t) as input and i(t) as output E ER V.(t) a) (5 pts) Determine the differential equation of the system b) (5 pts) Find the transfer function of the system c) (20 pts) Write the symbolic expressions of magnitude and phase of the transfer function d) (20 pts) Determine the expression of steady-state response of the current i(t) correspondent to Vi(t) = 2 cos(100t) [V]. (the angular rate is expressed in rad/s)

Answer 1

(4) (a) ilt) i te Toh * Voc ] Vilt) ☆ ♡ Volt) for inductor, V=Latest for capacitor, 0 = CVC or del = c dve or Lift) = dve Now, apply KVL in the loop. Vilt) = V2 +Vc + Volt) On dt or, Vilt) = L dilt) t Vet Ri(t) dt Differentiate the obore equation wirit. t. dvice) - Led ) + dixe - at + R dilt) at Substitute dve at

LOL + i(t) + R dilt) out2 dit dvilt) dt Substitute, L= IH, Celf and R=1012 dt at dilt) + 10 dilt) + ilt) = dvilu) Which is required differential equation. ) Draw the impedance circuit. luse Laploce transforms) I(s) SL 75C t o mm Vi (s) © Vols) I(s) f Use ohon's love ( I = VR). I(s) = Vi (s) SL + C + R = se Vils) s²LC + SRC + 1

Now, Vols) = R ils) or, Vols) = R1 sc Vils) 1 I s²LC + SRC+1) or Vols) / SRC (5) (SLC +SRC+1) Put the values of Lc and R. Vols) - 105 Vi (s) s? + 10s +1) Which is the required transfer O function. (6) Put s=ju in the transfer function. - judo : Glju) (jue)2 + jlow tt j som -w² + 120w +2 j low (2-4²) + jsou

Magnitude, lG(ju) = 1 (200)? v (1-62)² + (20w/? (since, 1(a+jb)} = 7a2+62) low | Gljuy) = ✓ w 4+98m²+1 Rearrange transfer function as below for phase. . Glju) = 0 + jsou (1-42) +j tom So, phase, L Glju) = tan! ( 10u) – 20w e 1 LG(ju) = so'rtant (20w) - Therefore, equations ① and gives the expression of magnitude and phose.

(d) ilt) jul tot Vilt) ) S UC Total impedance of the circuit, 2 = ( jul - j_ +R) Criven, Vilt) = 2 Cos (100+) Compare Vilt) with Vs. Coscut. 1. So, w=200 2 = j Loo + 10 = Lo + j(100-100) = to + j ( 10000=4) = 10 + j 99.99 Convert it to poler form using, N20 = Va2+62 _tom! ( ) for atib. Z = 1660)2+(99.9972 L ten' (1989)

Z So, Z = 100.489 / 84.289 Now, consider Vils) as base. So, Vilt) = Vs Lo = Vs Coscut where, Vs=2 v and w= 100 rad's So, current is given by, ilt) – Vs Los - 226 A 100.489 <84.289 = 0.019903 _-84.289*A = 19.903 1-84.289' ma So, ilt) legs vilt) by 84.2857 Therefore, the required expression of current is given by, ilt) = 19.903 L-84.289 = 19.903 Cosleut -84.289)