Question

Chapter 14, Problem 14.02 (Circuit Solution) Find the input impedance Z(s) of the network in the figure below. 8 Q Z(s) 1 7F 1Ω375 8 H 1Q 6 Q Find Z(s) at s = 1 s-1.

Answer 1

→ The circuit can be redraw os, I now a che Z(s) Į & Tour 385 Lois a Reduce the circuit. Z is series combination of 6 and 8s. Z(s) - Z , is ll z Litts N 2 1 (6+85) combination of I and As. Z = 3 xi- Ito = (1+7) I+75 It has

z(s) this la Zz is ll combination of t and Zz. Za = 1 *(6+85) 1 +6+85 = 16+85 ) 2 1+75 7+85) z ais series combination of 8 . and Zz. | 24 = 8+ Z2 = 8+ 6+85 z(s) zotas LI1+75 7+85 62 + 72 S 7 +85 Zizual 25 ) So, Zls) =- ( +757 +85 Z, +24 62+725 1+7S 7+85 62+725 7+85+62+728 +4345 +504 52 62 +72 S - 5045² +5145+ 69 - 62+72 504 +514 + 69 - Thus, z/s= 15") = 62772763 = 0.1233 22

- 250 d 1/85 la ( Given circuit can be redraws ab, (use Laplece realisations) mmA Use Superposition theorem to find the It 1² A output voltage Consider the Voltage source first. 25 A ( Current source is I open circuited) Apply Kirchoff's low at La Vol) Y F T los node A. Y V 0 Suppose voltage at node A is va VA -45 So, VA -0 Lt alss I 2VA - 2 + VA + 165 VA =0

65 - ZVA + 16 SVA = 3/5 » VA (3+16s) = 2/5 - VA = 3(3+165) Now, by voltage division, Vso (s) = 2 VA = 3 15 16 +33 0 Now, consider the current source only (valt ge source is ofen circuited). The current passes rooth i § 485 through to resistor. 0 %A Vzo (s) = -(²) (1) 1 & ₂ (s) negative sign, since Current enters the are temind) So, Vo(s) - V -*

Now, by superposition theorems add ® ond . * to get the result. Vo(s) = Viols) + V2015) s(165+3) - Partial factorse the term S1165+3) · Suppose, S(115+3) = Â + Bar ie 1 = A(165+3) + BS substitute s o in above equation. So, I = BA to (A = 13) Now, substitute s=(-3/16) in the same. I = A (-3+3) – 38 B=(16)

Ś V 16 Hense, Vols) = 1 15 - 21 est3) - 85 Or, Vols) = $(2) - (13) Cista vo(s) = $(5) - $ 8 - Take inverse Lopbce transform. (use, at (2/2) = , 2t (sta) = e-at) volt) = _-4e*-2 = "- fe *t v Thus, volt=1) = -5/5 - 1/3 e 16 volt) =-1.943 v