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2 Pages

### IRWIN 9e 14_69

Course: EE 2120, Fall 2008
School: LSU
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LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is d.v f (t ) K 12 r 2 4r 4 0 r 2 2r 2 0 2 48 r 1 j1 2 v n (t ) K 2 e t cos t K 3 e t sin tv(t ) K 1 K 2 e t cos t K 3 e t sin tChapter 14: Application of the Laplace Transf
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is c.4 I ( s) 2 sI ( s ) 2 I ( s ) s 4 V s ( s ) 2 s 2 I ( s ) s V0 ( s ) 2 I ( s ) V (s) I (s) 0 2 4 V ( s) Vs ( s ) 2 s 2 0 s 2 V0 ( s ) s 2 Vs ( s ) s s 2 Vs ( s ) Chapter
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is a. s4 I s (s) I 0 ( s) s4 3 s I 0 (s) s( s 4) 2 I s (s) s 4s 3Chapter 14: Application of the Laplace Transform To Circuit AnalysisProblem 14.FE-3
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b. KVL:4s 4 s 2 I (s) s s 1 2I (s) 4s 2 ( s 2 1)( s 2 2 s 4)Chapter 14: Application of the Laplace Transform To Circuit AnalysisProblem 14.FE-4
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is d. KVL: Vs ( s ) 4 I ( s) 2 sI ( s )Vs ( s ) (2 s 4) I ( s ) V0 ( s) 2 sI ( s) V (s) I (s) 0 2sV s ( s ) ( 2 s 4) V0 ( s ) 2sV0 ( s ) s Vs ( s ) s 2Chapter 14: Applicatio
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
bn = 0
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
(s)(s)-(s) =(s) =(s) =(s)(s)(s)(s)o (s)(s) =(s)(s)-(s)(s)(s)(s)(s)(s)(s)(s)(s)(s)-
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b. By observation, the waveform has odd symmetry. Therefore, an = 0 for all n due to odd symmetry.Chapter 15: Fourier Analysis TechniquesProblem 15.FE-1