Question

{ProBlow #4} For this Given Gneust Fan ilt). 60 k h 30h 8 I to

Answer 1

initial conditions : for tto C cap is open ) : -vc61+2=0 VC 6 )= PV 60k * м Кл i Yi2 + 60kn Vco) 1.Uc6 )=2V1 SV L (OR) (OR modal analysis at at node vi lit iz + 13 = 0 V-S 60K + V + V -2 30k =0 60K - is = Vzok V - 2 = 3-2 30k All v-8 + v + 9V-4 co 60 4V-12 = 0 v= 1214 = 3 volls. ; = IMA 30 -V + i3 (306) + VC 6) = v= (i3 ) 30 K VCO )= = 3- M 60K) = 3-1 Por t o the circuil.:. 30 kr GORR with ucco-) = 2V wh 60 kr QUF SV (+) for

Jn s-Domain : GOKN v iz 30kn white V GORN jy is nodal at node vi itiz+ i3 =0 is = 165) | is) | 1 t s V- Slov V-215 con 60K 30K+ 1 905 v Bordt ek + sökt Veus] = *) (2017qus) J = lok) * $(30K+ plus) r. (30K+ 1|205) + 60K 1 s 30K+ 1/245 ) +120K (COK) (30k4 ||2us) rooms +1 pus ) +120K s[e (content') +60k]

240m 6+4 + 120ms 5 rooms +1+ Goms] 360ms +4 5 ( 180m 5 +1) Erein 360m 5+ 4 GOK is) - - using - Bom StToomstu partial fraction 75) = 202 - + . goms +1 A = sis) S=0 - (non 12075 +12) Is=0

A = 4 = 66.67 el B= (1.90ms +1 ) [s? | s2-1/100m = (180ms A11 (360ms +12 GOK $ (120996+1) |-- - 360m cicom) + sok t 'lizom) 3 + = -0.5 1106. = tei 7:51106 -0.5*106 =-gle

= 15! r£9.99 alu 120msti ; G6. 671 16.67 1 a st 4120m

ce (t) a 1 I e un 16 는 sta inverse Trang form Taking

ist) = 66.67 .414) + - 16.67 u ē 1200 U CH) (66. 670 + 16.67 - tl 120m Ju(t) ē ar UA.