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IRWIN 9e 13_16

# IRWIN 9e 13_16 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis. 9/E 1 13.16 Given the i'nJlam-‘ing t'LInciinna Fisj. i'iml fir). _ _ 5+1 (“1 }(5)'_ [3-+ 2}[3-+ a} _ _ 24 {b} }(5]'_ [;-+ 2}[3-+ 3} _ _ 4 {c} }{5j _'[s +-3}(3-+ 4) _ _ 1i]; “i “33' ‘W SOLUTION: (a) PCS) = 3+! (5+2) (5+6) SH :: A + @ (3+2) (5+6) 3 + 2. 3+6 FQ) : A ‘i’ #15)...” 8%? 3+6 3+1: A(S+6) +B(s+9.) Let 5:“6 _.5:: Q(:~6+?J Chapter 13: The Laplace Transform Problem 13.16 Irwin, Basic Engineering Circuit Analysis, 9/E m) : wt, + 5/9 S+2 316 ~t Wt): (4/1., 6*” + 5/9 6 )u' H) (b) PCS) 1 im... (5+2){S+ 3) F(3) 1'5 VA, ,. i} 312 S43 ’2“! : A + 8 (5+2) (5+3) 5‘” 5+3 9&1: A(&ﬁn T'8(5+2) Jﬁt §:r—3 29: B 0342.) 8 : ml” Lei S: “2— 29: A(—2f3) Problem 13.16 Chapter 13: The Laplace Transform Irwin. Basic Engineering Circuit Analysis, 9/E 3 a) Fﬁw': H (3+3){S+w) FR) 2: A -i' 8 :+3 3+L/ t4 : A + B (5+3)(s+%) (5+3) s+L4 Lét .5: -H H: 8(~H+9 g: ~Li Lat 52: ~3 L4: A(~3+q) Chapter 13: The Laplace Transform Problem 13.16 Inn/in, Basic Engineering Circuit Analysis, 9/E Azw FCS): ,1... + —M 8+3 5w! PG) WM“ -— (46% ) LL (t) (d) PCs) :- log ‘ (Sit) (H6) #5): A + B S‘H 5+6 WIQMSWW, = J,” + _5___ (Hours) 5+; 3+6 {05: A (3+6)‘i 3(51“) ‘ Lei s: -6 1006) =' 3('6+i) 8: ’7. Le: x: “I mm) : A(—I+6) Problem 13.16 Chapter 13: The Laplace Transform Irwin, Basic Engineering Circuit Analysis, 9/E 5 45 Chapter 13: The Laplace Transform Problem 13.16 ...
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