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Unformatted text preview: EE 3/5 Hw #10 SelaLam;
4.309
Wow) F(s)=,;—,md
I Yo) “‘5 ‘+5 3 3 2 y(t) = (3e'2‘ — 3e3‘ — 21e"3‘)u(t) 4.42 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. 84.42 4.44 At t = 0, the steadystate values ofcurrents 311 and y; is 511(0) = 2. y3(0.) = 1. g I
Figure 54.44 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. 84.44 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.
 Spring '07
 Cardwell

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