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 ´ ´ 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 ) ) 1 0 0 11 0 1 provided Re{( 5)} 0 : Re{ } 5 ) (5 ) 5 st e Ie d t s R O C R s ss s −+ === = + > > + + , while integral II reduces to 0 0 (4 ) ) 1 10 p r o v i d e dR e { ( 4 ) }0 :R e {}4 ) ) 4 e II e dt s ROC R s s −∞ −∞ = > < −− . The Laplace transform is therefore given by 12 9 () : : (5 R e {} 4 ) 54 ( 5 ) ( 4 ) X s I II with ROC R R R or R s s s =+ = = = −< < +− + . (b) + + = + = = = ´ ´ 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 ) ) 1 R e { ( 3 ) : R e { }3 ) ) 3 e I e dt provided s ROC R s s −∞ −∞ = > < , while integral II reduces to ) ) 1 0 0 01 R e { ( 3 ) : R e { } 3 ) ) 3 e II e dt provided s ROC R s s = + > > + + The Laplace transform is therefore given by 2 6 : : (3 R e {} 3 ) 33 9 X s I II with ROC R R R or R s s = = < .

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2 Chapter 6 (c) () 0 2 2 10 10 1 2 co s ( 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 00 2(1 0 ) 2( 1 0 ) 11 22 0 0.5 (1 0 ) 2 2 0 ) 0 0.5 0 ) 2 2 0 ) 0.5 0.5 0 ) ( 10) 2( 10) 2 ( 10) 2( 10) 2 s 10 20 sj t jj j sj j te d t d t es j t s j t j t s j t j −+ −∞ −∞ −∞ + −∞  + +  + + + ± 3 33 0 ) ( 10) ( 2 0 Re {s j10}<0 ROC: Re{s}<0 j j + + −± = 33 2 2 3 3 23 0) ( 10) 6 10 2000 60 2000 ( 10) ( 10) ( 100) ( 100) ROC: Re{s}<0 j s s s j + + (d) + + == = = ± ± ±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 ) ) ) 1 1 cos(5 ) (3 ) cos(5 ) 5 sin( ) ) 5 1( 3 ) 0) (0 0) Re{(3 )} 0 :Re{ } 3 ( 3)5 )5 st Ie t d t s e t e b t s s sp r o v i d e d s R O C R s ss −∞ −∞ + + + = > ⇒< 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 12 : : (3 R e {} 3 ) )5(3 Xs I I I w i t hROCR R Ro rR s +− =+ = = −< < ++ .
Solutions 3 (e) 7( 7 ) 0 ( ) ( ) cos(9 ) ( ) cos(9 ) st t st s t Xs xte d t e tute d t e td t ∞∞ −− −∞ −∞ == = ∫∫ [] (7 ) ) 22 0 1 ( 7) cos(9 ) 9 sin(9 ) provided Re{( 7)} 0 )9 1 (0 0) ( ( 7) 0) ) st se te t s s s s s s  =− + >  −+ =+ + = : Re{ } 7 ROC R s > (e) 0 7 ) ( ) ( ) cos(9 ) ( ) cos(9 ) st t st s t t e tu te d t e t −∞ −∞ −∞ = 0 ) ) 1 ( 7) cos(9 ) 9 sin(9 ) provided Re{( 7)} 0 1 )0 ( 00 ) ) t s s s s s s −∞ + < + + = Re{ } 7 ROC R s < (f) 1 1 2 0 0 0 () 2( 1 ) 2 0 .5 1 s xtd t t t t = −∞ = = 01 0 10 ( ( st st st s I II x t ed t t t t t −−− −∞ + + ±²²³ ² ²´ ±²²³ ²²´ where 2 2 0 11 1 1 1 (1 ) ( 1) 1 1 ) 1 ) 1 st st st st st s s ss s s s s s I t e dt e dt te dt e e st ee s s e s e s e s = + = + + =

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## This note was uploaded on 01/19/2012 for the course EE 421 taught by Professor Zhicheng during the Spring '11 term at Ohio State.

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

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