Question

3. The differential equation describing the current in a circuit is given by: dai(t) d+2° +2-06 + 5i(t) = v(t) dt a) Find i(t) fort > 0 using Laplace transforms if v(t) = 10u(t) and given that initial conditions are i(0+) = 4 and dico*) = -2. b) write an expression for i(t) if: 3 3 sin (2nnt - tan-1(4nt)) v(t) = nv1 + 16n2t2

part B asap

Answer 1

The given offj erectial equation is rLt) +2d110 + 5llt) - VH) dt at adace on both on sides we get applying Laplace on bo 5?10 f2.105 + 5.900) = y). > Im ? s25+5y = YO > I = VCD ${+25+5 Now het to - s²+25 +5 = I10 = V06. HC putting s= jw in +(6) we get +(jv) = ; (w) + 2 (jo) +5 W²+2jwth ti(w) = (5-2) +20W. Now finding the magnitude of "Heju] Cand o angle of "+1116)

→ Hilju)) = . V 6-w)+(21052 75+w+-10w1 462 25-6w²twy swt bu+25 I 2w => [#cja) = - tant en tant - © 5-un em. Now we know that for any Sinusoidal Input for any system the output is also sinusoidal. but itd Maginitude is multiplied by Hlyw) at sits Sinusai dal frequency and added [+1(jw) to its phase at its Sinusoidal frequency

In the given question. VH) - + Ź Cén (27nt whom (40 x)) ny 16 n'm from the find that the w is Sinusoidal ob 14 (52an) = - 1 tejzan) = equation of V ) we can for the de Value (3/2) ol, And the cofor input is (w= ann)- V (2xn't - 6 (27957 +25 1 V1634n4-247?n?+27 and Lucjan) - - Homile % +16) = -tamil ) Hence the magnitude is multiplied with 14(joan)) and L416jean) tip added to the phase.

AVH)= 35 sinf 2īt-tan'(411) -tan(unā / 6-4any) at an in-1 H1672211 - Is th4 2420 725 - here for the dc Value (3/2) - Wzo 6 14cjo) tos and L+(jo) = -fami! (0) = 0. 8. The de Value is mulApplied with 1/5 és V H) - czy to Hence the overall Vt Value is Ž 3a sin (ant-tan! (Anx) - tan " ( Anh 15-4222²)) (n-1 H167202) [1167424_247²1725 de 1 >llo M18 Vit