IRWIN 9e 13_46

# IRWIN 9e 13_46 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin. Basic Engineering Circuit Analysis, 9/E 1 ____________.___________..—__ 1346 Determine the initial and final values iil'ﬂt} ii' H3} i5 given by the expres'etimts }_ _ at: + 2) {a} m _ st; +1) {b} HS] = 2(5‘ + 23 + I5} [5' +1“; + 2n: + 3} _ _ 2:? m “53' _ [5' +11iigi2 + 2:: + 2) SOLUTION: (a) PCS): 2 5+1 SCS+0 F(Q) : le SFCS) 3—900 FLO): ﬁi'm 3[2(s+2)] s->oo 5 (5-H) Fm) : 2 F-(oo) : um 8H5) 590 Mm): ti‘m S 1(£+2) 3—50 S(S+I) F(Oo) 3 hm 2(2) 5—)0 I F‘CoO) : H Chapter 13: The Laplace Transform Problem 13.46 " 2 Irwin, Basic Engineering Circuit Analysis. 9/E ()3) HS): 11 slag-is) @ﬂJszxm) FCO) 7— ,Qi‘m SF“) s~>0° Ho) 2 arm 9 .] s~>6° 3+0 (2+2)(H3) F[O): 2 Woo) I ﬁrm SFG) 3—30 140°) 2 Mm §[1(51+25+6) 5'50 (gm (S+1)CS+3) : O (C) PCS): .321...“ (S+i)(\$2+25+2) Pb) I ,Qnm SFG) 5—300 Ho) 2 ﬁrm 5 [ 231 ] Sac» 6—H)( \$1425 + 2) Fan): 2 Problem 13.46 Chapter 13: The Laplace Transform irwin, Basic Engineering Circuit Analysis, 9/E 3 PH»): Ilium SFCS) 3‘50 F030): Ql‘m g ’251 360 (Hi) (6+2 5+ 2) F(Oo) 3 O Chapter 13: The Laplace Transform Problem 13.46 ...
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IRWIN 9e 13_46 - Irwin Basic Engineering Circuit Analysis...

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