HW_10_Solution

# HW_10_Solution - EE 3/5 Hw#10 Sela-Lam 4.309 Wow F(s)=;—,md I Yo “‘5 ‘ 5 3 3 2 y(t =(3e'2‘ — 3e-3‘ — 21e"3‘)u(t 4.4-2 Before

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Unformatted text preview: EE 3/5 Hw #10 Sela-Lam; 4.309 Wow) F(s)=,;—,md I Yo)- “‘5 ‘+5 3 3 2 y(t) = (3e'2‘ — 3e-3‘ — 21e"3‘)u(t) 4.4-2 Before the switch is opened, the inductor current is 5A, that is y(0) = 5. Figure S4.4—2b shows the transformed circuit for t 2 0 with initial condition generator. The current Y(s) is given by Fig. 84.4-2 4.4-4 At t = 0, the steady-state values ofcurrents 311 and y; is 511(0) = 2. y3(0.) = 1. g I Figure 54.4-4 shows the transformed circuit for t Z 0 with initial condition generators. The loop equations are 6‘ (3+ 2)Y1(s) - Y2(s) = 2+ ; —K(:) + (s + 2mm =1 Cramer’s rule yields “(‘>=m':‘m‘s+s Y2“): s(:2:1iis++63) - ' 3% + 5% M) = (4 — gs" - %e"')u(t) M) = (2 — %e" + ; ’"MU S 2 s :1 Fig. 84.4-4 Hw 41:10 Solcchfows Fig. 54.4.7 4.4—7 Figure 84.4—7 shows the transformed circuit with parallel form of initial condition generators. The admittance W(s) seen by the source is ' - 2 The voltage across terminals ab is _ 1(3) _ é+3 _ 3.1+1 Vab(5) -— m — ’_—.3:4:113 ' :2 +45 +13 Also - —\$ .5] ‘V 1 3a+1 ‘gé’e +- 'aee 0(3) - 5 3°“) “ 2(,2 +4: +13) ~ 54. 1.0 3 5+ 2113 and um) =Wt) I. 726'21605 (3t +. 5/) u (H ...
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## This note was uploaded on 09/18/2008 for the course EE 313 taught by Professor Cardwell during the Spring '07 term at University of Texas at Austin.

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HW_10_Solution - EE 3/5 Hw#10 Sela-Lam 4.309 Wow F(s)=;—,md I Yo “‘5 ‘ 5 3 3 2 y(t =(3e'2‘ — 3e-3‘ — 21e"3‘)u(t 4.4-2 Before

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