第六章作业答&aeli

第六章作业答&aeli

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(P257-260) 6 .1-6.9 6.1 1 e a>0 2 ) ( t u at t b e b>0 3 4 ) ( ) ( 2 t u e t u e t t + ) 3 ( t u 5 e 6 δ ) ( ) ( 5 3 t u e t u t t + ) ( 0 t t 7 ) ( ) ( δ t u t + 8 ) 2 ( ) 1 ( t u t u ROC Re{s}>a 1 a s t u e L at → 1 ) ( ROC Re{s}> a a s a s e dt e e dt e t u e s X t a s st at st at = = = = 1 ) ( ) ( 0 ) ( 0 2 2 2 2 b s b L t b → e ROC -b < Re{s}< b 2 2 0 ) ( 0 ) ( 0 0 2 ) ( b s b b s e b s e dt e e dt e e dt e e s X t b s t b s st bt st bt st t b = + = + = = + Re{s}< b Re{s}>-b ROC -b < Re{s}< b 3 ) 2 )( 1 ( 3 2 2 1 1 1 ) ( ) ( 2 + + + = + + + → + s s s s s t u e t u L t t e ROC Re{s}>-1 ) 2 )( 1 ( 3 2 2 1 ) ( ) ( ) ( 0 ) 2 ( 0 ) 1 ( 0 2 0 2 + + + = + + = + = + = + + s s s s e s e dt e e dt e e dt e t u e e s X t s t s st t st t st t t Re{s}>-1 Re{s}>-2 ROC Re{s}>-1 4 s e t s L 3 ) 3 ( → u ROC Re{s}>0 s e s e dt e dt e t u s X s st st st 3 3 3 ) 3 ( ) ( = = = = ROC Re{s}>0 u(t) u(t-3) 5 3 1 3 1 ) ( ) ( 1 ) ( 3 1 ) ( 3 3 = + → → + → s s t u e a s X a at x s t u L t L L t e ) 5 )( 3 ( 8 2 5 1 3 1 ) ( ) ( 5 3 = → + s s s s s t u e t u e L t t ROC Re{s}<3 ) 5 )( 3 ( 8 2 5 1 3 1 5 3 ) ( ) ( ) ( 0 ) 5 ( 0 ) 3 ( 0 5 0 3 5 3 = = = + = + = s s s s s s e s e dt e e dt e e dt e t u e e s X t s t s st t st t st t t
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Re{s}<3 Re{s}<5 ROC Re{s}<3 6 δ ROC R S s t L e t t 0 ) ( 0 → 0 0 ) ( δ ) ( 0 st t t st st e e dt e t t s X = = = = 7
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This note was uploaded on 08/22/2011 for the course EE 201 taught by Professor Yhm during the Spring '05 term at Zhejiang University.

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第六章作业答&aeli

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