IRWIN 9e 13_38

IRWIN 9e 13_38 - lnrvin, Basic Engineering Circuit...

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Unformatted text preview: lnrvin, Basic Engineering Circuit Analysis. 9/E 1 13.38 Fiml fir) if Fir} is given by thc MUM-"mg functions: EIIF + lje" [a] H3}: [r+2][r+4j [b F _ ID($+2)€_2’ 1' m‘fi:+1j(s+4j 1 _ 3-9" m H”— [r+4)[5+3] SOLUTION: , _, VS (a) Us) —- 28+; (3+2)(s+w) 61a): 2(5’H2 I (H1) (5+H) (5+2) (3H) 9‘2 SW 2am : Amy) + MHZ) w23: ~6 'Bz2 Leis: *2 Chapter 13: The Laplace Transform Problem 13.38 Irwin, Basic Engineering Circuit Analysis. 9/E 2A :—2 =—l 61(3) =1 “I +4.3.“ 3+2. 51w 9%): ~e’2t + 36% Fa): [— e‘m‘W “we”? LLCt-l) PCs) : [0 3+1 6'25” (9+1) (UH) 6a) : /O(§+2) (“08%) fits): L'fifl, Sf! 3+9 Io($+2) : A, Jr "L (HUGH!) 3+; 3+4 loam : M5794) r 80+!) Lei §:’Lf onwrz) = B («q-H) Problem 13.38 Chapter 13: The Laplace Transform Irwin, Basic Engineering Circuit Analysis. 9/E 3 3A 2 (0 A:- (0/: 67(5) 3 L9 22. 3 + 3 5+; <+q gee) : 49... 8* + 2.9 a?“ 3 3 at) : cm) 3 3 (C) F65) = ,_§_C;j.______. (5+9) (3+2) 63(3) : 5 (S‘HUCS‘H?) Chapter 13: The Laplace Transform Problem 13.38 4 Irwin, Basic Engineering Circuit Analysis, 9/E A+P S : (qu) (5+8) 81W 5+5 5: Mm?) 7‘ 8(S+Li) Let‘s: ~X ~33 B(—¥+H) "7'8: ~56 8:7. Let S: “1 *LI: Maw”) HA: *9 A: vi 6,55): L + 1 5+9 *5 90c)?— -e'%+2€'5uC -234) Far) : few") + 2e Judi") Problem 1338 Chapter 13: The Laplace Transform ...
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This note was uploaded on 02/22/2011 for the course EEE 202 taught by Professor Cihan during the Spring '06 term at ASU.

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IRWIN 9e 13_38 - lnrvin, Basic Engineering Circuit...

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