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IRWIN 9e 13_24

# IRWIN 9e 13_24 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 13.24 Given the i'niiuwing ['metinm Fir). i'iml fir]. [a] Fm : “1:32:13 [b] HS} 2 ii: i g]: SOLUTION: (a) m): 3+3 :AwLB +3.. s(s+7,)L 5 5+2 {94).}; 5:0 3: QUJ) A:3/ k,’ S:—— i: -Z C: -| 2 CC ) A 3:; H:A(3)L+B(3) +C H: 3 x9+\$3~l TI 6 B: *3 4, H3) 7— 3/;, + “3/9 Jr 4/7. 3 (3+2) (8%)" = 3 —3 *1t_i+e'2““)q t. (/9 £16 2 () Chapter 13: The Laplace Transform Problem 13.24 2 Irwin, Basic Engineering Circuit Analysis, 9/E (b) F65) = 3+6 HS): R+ B + C 3+6 : li’rjiﬁ. + ,9, W 3 3+1 (\$4.91 SH 3 MHz)" + B(s)(s+2) + c (s) Let\$=o LeJc 5‘ "i ..|+6 : AH“? @0091“) +C(—i) Problem 13.24 Chapter 13: The Laplace Transform Irwin, Basic Engineering Circuit Analysis, 9/E 3 8: 340+W 2. 1:65): 31+ ~3zL 4r -1 b 5 5+1 (SW1 pm: (3/ - 4 8‘” ~21 é“) am '7. Chapter 13: The Laplace Transform r Problem 13.24 ...
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