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Continuous and Discrete Time Signals and Systems (Mandal &amp; Asif) solutions - chap06

# Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap06

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Chapter 6: Laplace Transform Problem 6.1 (a) + + = + = = ±² ±³ ´ ±² ±³ ´ II t s I t s st t st t st dt e dt e dt e t u e dt e t u e dt e t x s X 0 ) 4 ( 0 ) 5 ( 4 5 ) ( ) ( ) ( ) ( . Integral I reduces to [ ] ( 5) ( 5) 1 0 0 1 1 0 1 provided Re{( 5)} 0 : Re{ } 5 ( 5) ( 5) 5 s t s t e I e dt s ROC R s s s s + + = = = = + > > − + + + , while integral II reduces to [ ] 0 0 (4 ) (4 ) 1 1 1 1 0 provided Re{(4 )} 0 : Re{ } 4 (4 ) (4 ) 4 s t s t e II e dt s ROC R s s s s −∞ −∞ = = = = > < . The Laplace transform is therefore given by 1 2 1 1 9 ( ) : :( 5 Re{ } 4) 5 4 ( 5)( 4) X s I II with ROC R R R or R s s s s s = + = = = < < + + . (b) + + = + = = = ±² ±³ ´ ±² ±³ ´ II t s I t s st t st t st t st dt e dt e dt e e dt e e dt e e dt e t x s X 0 ) 3 ( 0 ) 3 ( 0 3 0 3 3 ) ( ) ( . Integral I reduces to [ ] 0 0 (3 ) (3 ) 1 1 1 1 0 Re{(3 )} 0 : Re{ } 3 (3 ) (3 ) 3 s t s t e I e dt provided s ROC R s s s s −∞ −∞ = = = = > < , while integral II reduces to [ ] ( 3) ( 3) 1 0 0 1 1 0 1 Re{( 3)} 0 : Re{ } 3 ( 3) ( 3) 3 s t s t e II e dt provided s ROC R s s s s + + = = = = + > > − + + + The Laplace transform is therefore given by 1 2 2 1 1 6 ( ) : :( 3 Re{ } 3) 3 3 9 X s I II with ROC R R R or R s s s s = + = = = < < + .

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2 Chapter 6 (c) ( ) 0 2 2 10 10 1 2 ( ) ( ) cos(10 ) ( ) st st j t j t st j X s x t e dt t t u t e dt t e e e dt −∞ −∞ −∞ = = = ( ) ( ) [ ] 3 3 3 0 0 2 ( 10) 2 ( 10) 1 1 2 2 0 0.5 ( 10) 2 2 ( 10) 0 0.5 ( 10) 2 2 ( 10) 0.5 0.5 ( 10) ( 10) 2( 10) 2 ( 10) 2( 10) 2 s 10 2 0 s j t s j t j j j s j t s j j s j t s j j j s j t e dt t e dt e s j t s j t e s j t s j t j + −∞ −∞ −∞ + + −∞ = = + + + + + + ≠ ± = [ ] 3 3 3 ( 10) 1 1 ( 10) ( 10) ( 1 2 0 Re {s j10}<0 ROC: Re{s}<0 s j s j s j s j j j + + + ± = = 3 3 2 2 3 3 2 3 2 3 0) ( 10) 6 10 2000 60 2000 ( 10) ( 10) ( 100) ( 100) ROC: Re{s}<0 s j s j j s s j s j s s j + + + + = = (d) + + == = = ± ± ± ² ± ± ± ³ ´ ± ± ± ² ± ± ± ³ ´ II t s I t s st t st dt t e dt t e dt e t e dt e t x s X 0 ) 3 ( 0 ) 3 ( 3 ) 5 cos( ) 5 cos( ) 5 cos( ) ( ) ( . Integral I reduces to [ ] 0 0 (3 ) (3 ) (3 ) 2 2 1 2 2 2 2 1 cos(5 ) (3 ) cos(5 ) 5 sin( ) (3 ) 5 1 ( 3) (3 0) (0 0) Re{(3 )} 0 : Re{ } 3 (3 ) 5 ( 3) 5 s t s t s t I e t dt s e t e bt s s s provided s ROC R s s s −∞ −∞ = = + + = + + = > < + + while integral II reduces to [ ] [ ] 3 } Re{ : 0 )} 3 Re{( 5 ) 3 ( ) 3 ( ) 0 ) 3 ( ( ) 0 0 ( 5 ) 3 ( 1 ) 5 sin( 5 ) 5 cos( ) 3 ( 5 ) 3 ( 1 ) 5 cos( 1 2 2 2 2 0 ) 3 ( ) 3 ( 2 2 0 ) 3 ( > > + + + + = + + + + + = + + + + = = + + + s R ROC s provided s s s s t e t e s s dt t e II t s t s t s The Laplace transform is therefore given by 1 2 2 2 2 2 3 3 ( ) : :( 3 Re{ } 3) ( 3) 5 ( 3) 5 s s X s I II with ROC R R R or R s s s + = + = = < < + + + .
Solutions 3 (e) 7 ( 7) 0 ( ) ( ) cos(9 ) ( ) cos(9 ) st t st s t X s x t e dt e t u t e dt e t dt −∞ −∞ = = = [ ] ( 7) ( 7) 2 2 0 2 2 2 2 1 ( 7) cos(9 ) 9 sin(9 ) provided Re{( 7)} 0 ( 7) 9 1 (0 0) ( ( 7) 0) ( 7) 9 ( 7) ( 7) 9 s t s t s e t e t s s s s s s = + >

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