BONUS QUESTIONS BONUS 1 (S PTS): Refer to the circuit in Problem P1. Use a single current division equation to find the

[answerhappygod]

Bonus Questions Bonus 1 S Pts Refer To The Circuit In Problem P1 Use A Single Current Division Equation To Find The 1 (38.2 KiB) Viewed 51 times

Bonus Questions Bonus 1 S Pts Refer To The Circuit In Problem P1 Use A Single Current Division Equation To Find The 2 (45.31 KiB) Viewed 51 times

Bonus Questions Bonus 1 S Pts Refer To The Circuit In Problem P1 Use A Single Current Division Equation To Find The 3 (45.31 KiB) Viewed 51 times

BONUS QUESTIONS BONUS 1 (S PTS): Refer to the circuit in Problem P1. Use a single current division equation to find the transfer function G(p) so that i(t) = G(p) (1) 1/Cp i(t) = -iz(t) + R, + R2 бр +1 (R. + R,)Cp +1'46) BONUS 2 (10 PTS): Write the complete solution y().e> of the following DFQ: leave any arbitrary constants unspecified. 1. (p+2)y(t) = 10u(t) Homogeneous component char. Rootss=-2. Then y(t) = Ke Particular component Y = Kz Complete solution: y(t) = (Kx + Ke-2).(t) 2. (P+3 - j4)(p + 3 + 14)y(t) = 10u (1) Homogeneous component: char. Roots s = -3 14: Then y(t) = Kie-*cos(4 + ) Particular componenty, Ry Complete solution: y(t) = (Kx + Kxe-cas (4 + 0)),() BONUS 3 (SPTS): Use Laplace Transforms to find the complete solution y(6),t> 0 of the following DFQ FA 1 1 (p+2)(p+ 3)y(t) = 100,(),y(0*) = 7: py(0*) = 0 The DFQ is: (2 + 5p + 6)y(t) = 10,(0) Apply LT: s*Y($) - 7s - 0 +SsY(s) - 5(7) + 6(s) = 75* +355 + 10 5/3 16 32/3 y(t) 5+2 5+3 - (+16+2 -* *).( - -6) Y(s) - 5(8 + 2)(8 + 3) s 32 como sanat stu.

P1 50 PTS In the given circuit valid fort 20*. 1,(t) = 2u_(t); R = 40; R = 20; C = 0.5F. The capacitor is initially charged 50 v.(0)=3V a. Convert the circuit to the s-domain. Clearly label all circuit elements. ke 카 를 IG th OR ie 10 in alte WIPE 2 b. Use Nodal OR Mesh analysis to find i(t).t> 0 Clearly label all variables and indicate which method you want graded. Nodal on LEFT circuit: the single node equation is *+- (s+) () =*+- (s+) V() = ***V() = So: v(t) = (12-9%),(e) Then (t) =* = (2-). A Nodal on RIGHT circuit: the single node equation is So: v(t) = 12 - 9e Then 2 3 9/6 (4+ 68 +2 5 **(s+v6) -(s+) (3) - ****V(5)-23- = (12 - 9e"),(6) (8)="* = (2-4), () (A) Mesh on LEFT circuit: Label, the right loop, so that I =*- The single right-loop equation is (4+2+3).-(4+2) 1- = -16) = -1)= 4,60 50:6() = 2 -4(0)-(2-1),(0) (A) Mesh on RIGHT circuit: Label, the MIDDLE loop, so that I = -1, The single MIDDLE-loop equation is =) 96 (0) s+ $0: e(t) = 2-1() = (2-*)(0 (A) 2 23 65 +2 9 52 24-3-0-564-76 =24*4,09