Question

Find i(t) for t> 0 in the op amp circuit of Fig. P4. (u(t) is unit step function. It is 1 for t> 0 and 0 for t<0.) 1/6F It 3 Ω 212 + 2 u(t) V 1/6 F 10 Ω Figure P4

Answer 1

/GF HC 3N 21 vo 0 Vo Vo li 2Ut) V 10.2 OF in the op- amp find ist) for f>0 circuit of above figure. [uct, is ant step function it is I footso and o fortco] ) solution: In the given op-amp circut at Point node 9 voltage is out put voltage. at Point 9 voltage indicated as Vo SO SO nou is vo. at point ③ voltage nothing but voltage at Point ① of-amp on-inverting (tues terminal voltage because no component between Point ① and Point , and They are short. Point voltage Point ③ voltage SO = Vo ideal of-amp so The giver of ramp is CS Scanned with CamScanner

OA Vo + OA 12 ideal of-amp for ideal op-amp infout impedance Zird so i= iz: 0 A voltage gain SO - L so VI-V₂ = 0 vd (VI-V2) so VI=Uz so in our circuit voltage at Point ① and equal. (ideal of acup) are voltage at Point ② SO voltage at point vo is say NOW voltage at Point 0 vi convert the circuit into Laplace domain NOW LIT supely voltage 2 Ut 2 alu ter LT calautor sct) con آیا Registor R R 2 CS Scanned with CamScanner

HE cola Hiz 14 Vos). 3 Vig) i 2 Vos) + it 165) colo 10 ula Laplace domain equivalent circuit" Now arely KCL at node ① I BJ & J G & 23 (S) where 1163) Vis)-2/ Iz(S) Vi (S) - VOG (615) 13(s) Vics) - Vo(s) 2 so VICS) – @ls) VI(S) - VO(S) + + VICS) - VOCs) -0 2 3 (61) vices [ } + { + 1] - Voss [++)- įs 3. CS Scanned with CamScanner

S 2 3 S VIES) [ + ] - Vo GSM + } ] Multicly both sides with "6" so VISI C 5+5) - Vo(s) [5+3) ų - Now arely kel at node ③ 14(8) + Is(s) + IGB) here Iuss) VOCs) - Vic) 2 158) Vos) - 0 (Gls) لا Ideal of-camp A 16 (3) 11 SO VOCs) - Vis) VO) to = 0 2 Cols) ViB) 2 Vols) { / 2 + 3 [ 34 ] ViB) vors) C 2 VC) VOCs) (3+) 3 so put vies from @ in o from Vo CSJ (3+5) (5+5) - VOKS) (5+3) = 5 3 CS Scanned with CamScanner

Vors) (5+3) (5+5) - 3 (5+3) ३)] יו 4 S 3 VO (S) (S+3) [S+5) - 3 a 4 3 voce) ( (5+3) ( 5+2)] slf 3 vos) 1) 12 S (5+2) (5+3) we know Vors) so out Put side current Ics) Vo (3) R 17 Vols 10 12 G SCS+2)(5+3) 10 103) 1.2 SCS+2) (5+3) Now apply partial fractions method. M Is ch + + c S +3 St2 5 CS Scanned with CamScanner

s 50) Is= 1.2 (S +2) CS+3) /s:0 1.2 (+2) Co +3) 0.2 BF (S+2) I (s) 15:02 1.2 1s=12 S (573) (22) (-2+3) 3 11 -0.6 С יו (5+3) I(S) /S=-3 12 S(5+2) 18=-3 1.2 (-3) (-3+2) C - 0.4 so 165) 0.2 0:4 0.2 S + - 0.6 (+2) (5+3) CS Scanned with CamScanner

Now apply Inverse Laplace trausform." ILIT ut Uft) DiLiT ㅗ Sta I eat act) SO ) 0.2 Act) - ooo e -0.6 elt ult) + 0.4 ē34 ust) est ict) [0.2 - 0.6 e 2 + + 0.4 € Jucts so icts for to is ict) -3t [0.2 -0.6 ēkt +0.4 e 11 est] uz CEL) 171 CS Scanned with CamScanner