Continuous and Discrete Time Signals and Systems (Mandal & Asif) Solutions - Chap06

(1)

Problem 6.1 (a)

∫

∞ ∞ − ∞ ∞ − −∞ − ∞ + − − − − ∞ ∞ − − ₌ ₊ ₋ ₌ ₊ = II t s I t s st t st t st_dt _e _u _t _e _dt _e _u _t _e _dt _e _dt _e _dt e t x s X 0 ) 4 ( 0 ) 5 ( 4 5 ₍ ₎ ₍ ₎ ) ( ) ( . Integral I reduces to

[

]

( 5) ( 5) 1 0 0

1 _{0 1} 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)

∫

∞ ∞ − ∞ + − ∞ − − ∞ − − ∞ − − − − ∞ ∞ − − ₌ ₌ ₊ ₌ ₊ = II t s I t s st t st t st t st_dt _e _e _dt _e _e _dt _e _e _dt _e _dt _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 − − −∞ −∞ − = = = − = − > ⇒ < − − −

∫

,

[

]

( 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 ∞ ∞ − + − + − = = = − = + > ⇒ > − − + + +

∫

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 − = + = − = = − < < + − − ∩ .

(2)

(c)

(

)

0 2 1 2 10 10 2

( )

st

cos(10 ) ( )

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 j s s − − − − + − + = + = + (d)

∫

∞ ∞ − ∞ + − ∞ − − − − ∞ ∞ − − ₌ ₌₌ ₊ = II t s I t s st t st_dt _e _t _e _dt _e _t _dt _e _t _dt e t x s X 0 ) 3 ( 0 ) 3 (

3 _cos(₅ ₎ _cos(₅ ₎ _cos(₅₎

) ( ) ( . 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 − − − −∞ −∞   = = _ − + _ − + − − = − + − + = − > ⇒ < − + − +

∫

[

]

[

]

Re{( 3)} 0 :Re{} 3 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 s t s t s t

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 + − = + = − = − < < + + − + ∩ .

(3)

(e) 7 ( 7) 0 ( ) ( ) st tcos(9 ) ( ) st s tcos(9 ) 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 ∞ − − − −   = _− − + _ − > − + = + − − − + − + − = − + ROC R: Re{ } 7s > (e) 0 7 ( 7) ( ) ( ) st tcos(9 ) ( ) st s tcos(9 ) X s x t e dt e t u t e dt e t dt ∞ ∞ − − − − −∞ −∞ −∞ =

∫

=

∫

− =

∫

[

]

0 ( 7) ( 7) 2 2 2 2 2 2 1

( 7) cos(9 ) 9 sin(9 ) provided Re{( 7)} 0

( 7) 9 1 ( 7) 0 (0 0) ( 7) 9 ( 7) ( 7) 9 s t s t s e t e t s s s s s s − − − − −∞   = _− − + _ − < − + = − − + − + − + − − = − + ROC R: Re{ } 7s < (f) 1 1 2 0 0 0 ( )_s ( ) 2 (1 ) 2 0.5 1 X s x t dt t dt t t ∞ = −∞   =

_∫

=

_∫

− = _ − _ = 0 1 0 1 0 ( ) ( ) st (1 ) st (1 ) st s I II X s x t e dt t e dt t e dt ∞ − − − ≠ −∞ − =

∫

=

∫

+ +

∫

− where

(

)

(

)

2 2 2 2 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 (1 ) ( 1) 1 1 ( 1) 1 ( 1) 1 st st st st st s s s s s s s s _s _s _s I t e dt e dt te dt e e st e e s se s e s e s − − − − − − − − − − −     = + = + = _ _ + _ − − _     = − + _− − − _= _ − − − − _= − −

∫

and

(

)

(

)

2 2 2 2 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 (1 ) ( 1) 1 ( 1) 1 ( 1) 1 1 st st st st st s s s s s s s s s s s II t e dt e dt te dt e e st e e s s se e s e s − − − − − − − − − − −     = − = − = _ _ − _ − − _     = − − _ − − + _= _ − + + − =_ + −

∫

(

)

2 1 1 0 ( ) : Entire s-plane 2 0 s s s s X s ROC e e− s =  =  ₊ ₋ _≠  ▌

(4)

Problem 6.2 (a) ∞ − − − − − − ∞ −         − − − + − − − + − − − = =

∫

0 6 5 4 2 3 3 2 4 5 0 5 ) ( 120 ) ( 120 ) ( 60 ) ( 20 ) ( 5 ) ( ) ( s e s e t s e t s e t s e t s e t dt e t s X st st st st st st st .

Applying the limits, we obtain 6 6 ! 5 120 ) ( s s s

X = = with ROC: Re{s} > 0.

(b) 1

(

6 6

)

2 0 0 0 ( ) sin(6 ) ( ) st sin(6 ) st j t j t st j X s t u t e dt t e dt e e e dt − ∞ ∞ ∞ − − − − =

_∫

=

_∫

=

_∫

− ( 6) ( 6) 1 1 2 2 0 0 0.5 ( 6) 0.5 ( 6) 0 6 ₀ 6 ₀ 0.5 0.5 6 6 1 1 6 s Re {s j6}>0 0.5 s j t s j t j j j s j t j s j t s j s j j j s j s j s j e dt e dt e e j j ω ∞ ∞ − − − + ∞ ∞ − − − + − + − + + = −     = _ _ − _ _ ≠ ± = − + ± = −

∫

2 2 12 6 36 6 36 0.5 ROC: Re{s}>0 j s j s s j − − + +   = ×   =

c)

∞ − ∞ − ∞ − ∞ −             + + − + = + = =

∫

0 2 2 0 0 0 2 12 ) 12 sin( 12 ) 12 cos( 2 1 2 1 ) 12 cos( 2 1 2 1 ) 6 ( cos ) ( st st st e st s t t s s dt t e dt e dt e t s X which reduces to ) 144 ( ) 72 ( 12 2 1 2 1 ) ( ₂ 2 2 2 ₊ + = + × + = s s s s s s s

X with ROC: Re{s} > 0.

(d) 3 ( 3) 0 ( ) ( ) st tcos(9 ) ( ) st s tcos(9 ) X s x t e dt e t u t e dt e t dt ∞ ∞ ∞ − − − − + −∞ −∞ =

∫

=

∫

=

∫

[

]

( 3) ( 3) 2 2 0 2 2 2 2 1

( 3) cos(9 ) 9 sin(9 ) provided Re{( 3} 0

( 3) 9 1 _{(0 0) ( (} _{3) 0)} ( 3) 9 3 ( 3) 9 s t s t s e t e t s s s s s s ∞ − + − +   = _− + + _ + < + + = + − − + + + + + = + + : ROC R Re{ }s < −3

(e)

II j s I j s st_dt _t _e _dt _t _e _dt e t t s X

∫

∞ + − ∞ − − ∞ − ₌ ₊ = 0 ) 10 ( 2 0 ) 10 ( 2 0 2 2 1 2 1 ) 10 cos( ) ( . Integral I reduces to

(5)

3 0 3 ) 10 ( 2 ) 10 ( ) 10 ( 2 ) 10 ( 2 ) 10 ( 2 ) 10 ( 2 ) 10 ( _s _j _s _j e j s e t j s e t I t j s t j s t j s − =         − − + − − − − = ∞ − − − − − −

with ROC: Re{s} > 0. Integral II reduces to 3 0 3 ) 10 ( 2 ) 10 ( ) 10 ( 2 ) 10 ( 2 ) 10 ( 2 ) 10 ( 2 ) 10 ( _s _j _s _j e j s e t j s e t II t j s t j s t j s + =         + − + + − + − = ∞ + − + − + −

with ROC: Re{s} > 0. The Laplace transform is, therefore, given by

3 3 ₍ ₁₀₎ 1 ) 10 ( 1 ) ( j s j s s X + + −

= with ROC: Re{s} > 0.

(f)

1 2 1 1 2 0 0 0 0 ( )_s ( ) (1 ) 0.5 X s x t dt t dt t t − ∞ = =

∫

=

∫

− = −  =

(

)

(

)

2 2 2 2 1 1 1 0 0 0 0 0 1 1 1 1 1 1 0 0 1 1 ( ) ( ) (1 ) ( 1) 1 ( 1) 1 ( 1) 1 1 st st st st s st st s s s s s s s s s s s X s x t e dt t e dt e dt te dt e e st e e s s se e s e s − ∞ − − − − ≠ − − − − − − − − = = − = −       = _ _ − _ − − _ = − − _ − − + _   = _ − + + − =_ + −

∫

Therefore,

(

)

2 1 2 1 0 ( ) : Entire s-plane 1 0 s s s X s ROC e− s s =  =  + − ≠  ▌ Problem 6.3

(a) Using partial fraction expansion and associating the ROC to individual terms, yields 2 2 2 ( 1) 2 1 1 ( 1)( 2)( 3) ( 2)( 3) 2 3 ( 1)( 5 6) ROC:Re{ } 2 ROC:Re{ } 3

( )

s s s s A B s s s s s s s s s s s s

X s

+ + + + + + + + + + + + + + >− >−

=

+

where 1 1 3 ₂

1,

2 ₃

2

s s s _s s _s

A

+

B

+ + ₌₋ + ₌₋

=

 

_{ }

= −

=

 

_{ }

=

. Calculating the inverse transform of X(s), yields

(

)

2 3 3 2

( )

t

( ) 2

t

( )

2

t t

( ).

x t

_{= −}

e u t

−

₊

e u t

−

₌

e

−

₋

e

−

u t

(b) Using partial fraction expansion and associating the ROC to individual terms, yields 2 22 1 ₍ ₂₎₍1 ₃₎ ₂ ₃ ( 1)( 5 6) ROC:Re{ } 2 ROC:Re{ } 3

( )

s s s A B s s s s s s s s s

X s

+ + + + + + + + + + <− <−

=

+

where constants A, and B were computed in part (a) as A = −1, and B = 2.

In other words, 1 2 2 3 ROC:Re{ } 2 ROC:Re{ } 3

( )

_s _s s s

X s

− + + <− <−

=

+

(6)

Using the transform pair s a a s t u e at L <− + → ← −

− − with ROC: Re{ }

) ( 1 ) ( , the inverse transform of X(s) yields

(

)

2 3 2 3

( )

t

( )

t

( )

t t

( ).

x t

₌

e u t

−

_{− −}

e u t

−

_{− =}

e

−

₋

e

−

u t

₋

Note that the same rational fraction for X(s) yields different time domain representations if the associated ROC is different.

(c) ₍ ₁₎₍2 23 ₅4 ₆₎ ₍ ₁₎₍2 3₂₎₍4 ₃₎ ₁ ₂ ₃

ROC:Re{ } 1 ROC:Re{ } 2 ROC:Re{ } 3

( )

s s s s A B C s s s s s s s s s s s s

X s

+ − + − + + + + + + + + + >− >− >−

=

+

where 2 3 4 2 3 4 2 3 4 (ss 2)(ss 3)_s ₁

3,

( 1)(ss ss 3)_s ₂

6,

(ss 1)(ss 3) _s ₂

2 A

+ −

B

+ −

C

+ − + + ₌₋ + + ₌₋ + + ₌₋

=

= −

=

= −

.

( )

X s

can be expressed as: 3 6 2

1 2 3

( )

_s _s _s s s s

X s

− − + + + >− >− >−

=

+

.

The inverse transform of X(s) yields

(

2 3

)

( )

3

t

6

t

2

t

( )

x t

= −

e

−

+

e

−

e

−

u t

(d) _{( 1)(}2 23 ₅4 ₆₎ _{( 1)(}2 3₂₎₍4 ₃₎ ₁ ₂ ₃

( )

s s s s A B C s s s s s s s s s s s s

X s

+ − + − + + + + + + + + + <− <− <−

=

+

where constants A, B, and C were computed in part (a) as A = -3, B = 6, and C =-2.

( )

X s

can be expressed as: 3 6 2

1 2 3

( )

_s _s _s s s s

X s

− − + + + <− <− <−

=

+

.

Using the transform pair s a

a s t u e at L <− + → ← −

− − with ROC: Re{ }

) ( 1 ) ( , the inverse transform of X(s) yields

(

2 3

)

( ) 3 t 6 t 2 t ( ) x t ₌ e− ₋ e− ₊ e− u t_{− .}

(e) Using partial fraction expansion and associating the ROC to individual terms, yields

} 4 1 Re{ } Re{ : ROC ) 17 2 ( 1 } Re{ : ROC ) 1 ( 0 } Re{ : ROC ) 17 2 )( 1 ( 1 2 2 2 ) ( j s s s D Cs s sB s s A s s s s s s X ± − > + + + − > + > + + + + ₌ ₊ ₊ = where . 8 1 ) 1 ( and 17 1 1 ) 17 2 ( 1 1 ) 17 2 )( 1 ( 1 0 ) 17 2 )( 1 ( 1 0 ) 17 2 )( 1 ( 1 2 2 2 2 2 2 2 2 − =     =     ₊ = =     =     = − = + ++ − = + + + + = + + + + = + + + + s s s s s s s s s s s s s s s s s s s s s s s B s A

(7)

) 1 ( ) ( ) 17 2 ( ) 17 2 )( 1 ( 1 2 2 2 ₊ ₌_A_s₊ _s ₊ _s₊ ₊_Bs _s ₊ _s₊ ₊ _Cs₊_D _s _s₊ s

and compare the coefficients of s3_{and s}2_{. We get}

D C B A C B A + + + ==3 + 2+ 1 0

which has a solution C = 9/136 and D = 137/136. The Laplace transform may be expressed as

1 } Re{ : ROC ) 4 ) 1 (( 136 4 32 1 } Re{ : ROC ) 4 ) 1 (( 136 ) 1 ( 9 1 } Re{ : ROC ) 1 ( 8 1 0 } Re{ : ROC 171 2 2 2 2 ) ( − > + + × − > + + + − > + > + + − = s s s s s s s s s s X

Taking the inverse transform of X(s) yields

(

)

9 1 1 4 17 8 136 17 9 1 1 4 17 8 136 17

( )

cos(4 ) ( )

sin(4 ) ( )

cos(4 )

sin(4 ) ( ).

t t t t t t

x t

u t

e u t

e

t u t

e

t u t

e

t

e

t u t

− − − − − −

=

−

+

=

−

+

(f) Using partial fraction expansion and associating the ROC to individual terms yields

4 } Re{ ) 4 ( 3 } Re{ ) 3 ( 2 } Re{ ) 2 ( 2 } Re{ ) 2 ( ) 4 )( 3 ( ) 2 ( 1 2 2 ) ( − > + − > + − > + − > + + + + + ₌ ₊ ₊ ₊ = s sD s sC s s B s sA s s s s s X where

[

]

[

]

[

] [

]

[

] [

]

. 4 3 ) 4 ( and 2 ) 3 ( 2 1 ) 2 ( 4 ) 3 ( ) 2 ( 1 4 ) 4 )( 3 ( ) 2 ( 1 3 ) 4 ( ) 2 ( 1 1 ) 4 )( 3 ( ) 2 ( 1 2 ) 4 )( 3 ( 1 1 2 ) 4 )( 3 ( ) 2 ( 1 2 2 2 2 2 = = + = − = = + = − = = + = − = + + + − = + + + + − = + + + − = + + + + − = + + + − = + + + + s s s s s s s s s s s s s s s s s s s s s s s s s s s s D s C s B To evaluate A, expand X(s) as ) 3 ( ) 2 ( ) 4 ( ) 2 ( ) 4 )( 3 ( ) 4 )( 3 )( 2 ( 1= + + + + + + + + 2 + + + 2 + + A s s s B s s C s s D s s s

and compare the coefficients of s3_{. We get}

D C

A+ +

= 0

which has a solution A = 5/4.

Taking the inverse transform of X(s) yields

(

)

2 2 3 3 5 1 3 4 2 4 2 2 3 3 5 1 3 4 2 4

( )

( ) 2

( )

2 ( ).

t t t t t t t t

x t

e u t

te u t

e u t

e

te

e

u t

− − − − − − − −

=

−

+

=

−

+

(g) Using partial fraction expansion and associating the ROC to individual terms, gives

0 } :Re{ ROC ) 16 ( ) ( 1 } :Re{ ROC ) 1 ( 1 } :Re{ ROC ) 1 ( 1 } :Re{ ROC ) 1 ( ) 16 ( ) 1 ( 1 2 2 3 2 2 3 2 ) ( < + + − < + − < + − < + + + + − ₌ ₊ ₊ ₊ = s s E Ds s s C s s B s sA s s s s s X where

(8)

[

] [ ]

. 17 4 ) 1 ( 2 ) 16 ( 1 2 1 3 ) 16 ( ) 1 ( 1 2 2 2 2 3 2 = = + = − = + + − − = + + + − s s s s s s s s s _s C

To evaluate A, B, and C expand X(s) as

3 2 2 2 2 2 ₋₂_s₊₁₌_A₍_s₊₁₎ ₍_s ₊₁₆₎₊_B₍_s₊₁₎₍_s ₊₁₆₎₊_C₍_s ₊₁₆₎₊₍_Ds₊_E₎₍_s₊₁₎ s

and compare the coefficients of s4_{, s}3_{, s}2_{, and s. We get}

) of nts (coefficie 3 16 32 2 ) of nts (coefficie 3 3 17 1 ) of nts (coefficie 3 2 0 ) of nts (coefficie 0 2 3 4 s E D B A s E D C B A s E D B A s D A + + + = − = + + + + + + + = + = , or, E D B A E D B A B D E A D A 3 16 32 2 3 3 172 3 0 0 17 13 + + + = − + + + = + + + = + = .

which has a solution of A = 0.0206, B = −0.2076, D = −0.0206, and E = 0.2282. Taking the inverse transform of X(s) yields

2 ( ) 0.0206 t ( ) 0.2076 t ( ) 0.1176 t ( ) 0.0206cos(4 ) ( ) 0.057 sin(4 ) ( ) x t _{= −} e u t− _{− +} te u t− _{− −} t e u t− _{− +} t u t_{− −} t u t₋ 2 0.0206_e−t 0.2076_te−t 0.1176_{t e}−t 0.0206 cos(4 ) 0.057 sin(4 )_t _{t u t}( ).   = −_ + − + − _ − ▌ Problem 6.4

The Laplace transform of the combined signal x1(t) +2x2(t) is given by

2 2 2 2 2 2 1

( ) 2 ( )

2 ₅ ₆ ₅ ₆ ₅ ₆ ( 2)( 3) L s s s s s s s s s s s

x t

+ + + + + + + + + +

+

←→

+

=

.

The ROC of L x t

{

₁( ) 2 ( )+ x t₂

}

includes the region (R1 ∩ R2), or, Re{s} > −2. However, simplifying the expression of L x t

{

₁( ) 2 ( )+ x t₂

}

, we obtain 1 1

( ) 2 ( )

2 3 L s

x t

+

x t

←→

₊ .

Since the pole at s = −2 cancels out, the overall ROC is greater than the intersection of the two individual

ROC’s and is given by R: Re{s} > −3. ▌

Problem 6.5

Using partial fraction expansion, the Laplace transform is given by

) 5 4 ( 125 . 0 375 . 0 ) 5 4 ( 125 . 0 375 . 0 ) 1 (.025 0 ) 1 (.025 0 2 2 ) ( + ++ + − + − − + + + + − = _s s_s s s s s s s X .

(a) For ROC R: Re{s} < −2, the ROC’s associated with individual terms are given by

2 } Re{ ) 5 4 ( 125 . 0 375 . 0 2 } Re{ ) 5 4 ( 125 . 0 375 . 0 1 } Re{ ) 1 (.025 0 1 } Re{ ) 1 (.025 0 2 2 ) ( − < + ++ < + − + − < − − < + + + + − = s s s s s s s s s s s s s X .

Taking the inverse Laplace transform, the time domain representation is obtained as ). ( sin 125 . 0 ) ( cos 375 . 0 ) ( sin 125 . 0 ) ( cos 375 . 0 ) ( 025 . 0 ) ( 025 . 0 ) ( 2 2 2 2 t tu e t tu e t tu e t tu e t u e t u e t x t t t t t t − − − − − − − + − − − = − − −

(9)

(b) For ROC R: −2 < Re{s} < −1, the ROC’s associated with individual terms are given by 2 } Re{ ) 5 4 ( 125 . 0 375 . 0 2 } Re{ ) 5 4 ( 125 . 0 375 . 0 1 } Re{ ) 1 (.025 0 1 } Re{ ) 1 (.025 0 2 2 ) ( − > + ++ < + − + − < − − < + + + + − = s s s s s s s s s s s s s X .

Note that the ROC associated with the last term is changed. Taking the inverse Laplace transform, the time domain representation is obtained as

). ( sin 125 . 0 ) ( cos 375 . 0 ) ( sin 125 . 0 ) ( cos 375 . 0 ) ( 025 . 0 ) ( 025 . 0 ) ( 2 2 2 2 t tu e t tu e t tu e t tu e t u e t u e t x t t t t t t − − − + + − − − + − − − =

(c) For ROC R: −1 < Re{s} < 1, the ROC’s associated with individual terms are given by

2 } Re{ ) 5 4 ( 125 . 0 375 . 0 2 } Re{ ) 5 4 ( 125 . 0 375 . 0 1 } Re{ ) 1 (.025 0 1 } Re{ ) 1 (.025 0 2 2 ) ( − > + ++ < + − + − < − − > + + + + − = s s s s s s s s s s s s s X .

Taking the inverse Laplace transform, the time domain representation is obtained as ). ( sin 125 . 0 ) ( cos 375 . 0 ) ( sin 125 . 0 ) ( cos 375 . 0 ) ( 025 . 0 ) ( 025 . 0 ) ( 2 2 2 2 t tu e t tu e t tu e t tu e t u e t u e t x t t t t t t − − − + + − − − + − − − =

(d) For ROC R: 1 < Re{s} < 2, the ROC’s associated with individual terms are given by

2 } Re{ ) 5 4 ( 125 . 0 375 . 0 2 } Re{ ) 5 4 ( 125 . 0 375 . 0 1 } Re{ ) 1 (.025 0 1 } Re{ ) 1 (.025 0 2 2 ) ( − > + ++ < + − + − > − − > + + + + − = s s s s s s s s s s s s s X .

Taking the inverse Laplace transform, the time domain representation is obtained as ). ( sin 125 . 0 ) ( cos 375 . 0 ) ( sin 125 . 0 ) ( cos 375 . 0 ) ( 025 . 0 ) ( 025 . 0 ) ( 2 2 2 2 t tu e t tu e t tu e t tu e t u e t u e t x t t t t t t − − − + + − − − + + − =

(e) For ROC R: Re{s} > 2, the ROC’s associated with individual terms are given by

2 } Re{ ) 5 4 ( 125 . 0 375 . 0 2 } Re{ ) 5 4 ( 125 . 0 375 . 0 1 } Re{ ) 1 (.025 0 1 } Re{ ) 1 (.025 0 2 2 ) ( − > + ++ > + − + − > − − > + + + + − = s s s s s s s s s s s s s X .

Calculating the inverse Laplace transform, the time domain representation is obtained as

2 2 2 2

( )

0.025 ( ) 0.025

( ) 0.375

cos ( ) 0.125

sin ( )

0.375 cos ( ) 0.125

sin ( )

t t t t t t

x t

e u t

e

tu t

e

tu t

e

tu t

e

tu t

− − −

= −

+

−

+

2 2 2 2

0.025_e−t 0.025_et 0.375_e tcos_t 0.125_e tsin_t 0.375_e−tcos_t 0.125_e−tsin_{t u t}( ).

  = −_ + − + + + _ ▌ Problem 6.6 Assume that R s X t x()←→L ( ) with ROC: . By definition

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) ( ) ( ) ( ) ( 0 ( 0) ₀ 0 x t e x t e dt x t e dt X s s est ←→L

∫

st st =

∫

s s t = − ∞ ∞ − − − ∞ ∞ − − _.

with ROC: R + Re{s0} because the new transform is a shifted version of X(s). For any s in the ROC R of

x(t), the values of s + Re{s0} are in the ROC of exp[s0t] x(t). ▌

Problem 6.7

Unilateral Laplace Transform: By definition, the unilateral Laplace transform is given by

∫

∞ − − = 0 ) ( ) (s x t e dt X st .

Dividing both sides with s and integrating the right hand side by parts, we get

[ ]

∫ ∫

∫

∞ − ∞ − ∞ ₋ − − − − − −         α α − α α = = 0 0 0 0 0 ) ( ) ( ) ( ) ( dt e d x s e d x dt s e t x s s X t _st I Term st t st .

The time integration property for unilateral Laplace transform follows directly from the above relationship by noting that Term I is zero at both the upper and lower limits.

Bilateral Laplace Transform: The proof for the bilateral Laplace transform is similar except that Term I

is nonzero. ▌

Problem 6.8

To prove the initial value theorem, we expand x(t) using the Taylor series expansion about t = 0+_as follows. 2 (1) (2) ( ) 2! ! ( ) (0 ) (0 ) _t (0 ) _tn _n (0 ) n x t ₌x + ₊tx + ₊ x + ₊ ₊ x + ₊

where x(n)_{denotes the n’th order derivative of x(t).}

Because, x(t) is a causal function, it can be expressed as follows. 2 (1) (2) ( ) 2! ! ( ) ( ) ( ) (0 ) ( ) (0 ) ( ) (0 )_t ( ) _n(0 )_tn ( ) n x t ₌x t u t ₌x + u t ₊x + tu t ₊x + u t ₊ ₊x + u t ₊ _,

Taking the Laplace transform of both sides of the above equation, we get

2 3 1 (1) (2) ( ) 1 1 1 1 ( ) (0 ) (0 ) (0 ) (0 ) n n s s s s X s x x x x + + + + + = + + + + + ,

Multiplying both sides of the above equation with s and applying the limit, t → ∞, gives

2 2 ( ) (1) (2) ( ) (1) (2) ( ) 0, assuming (0 ) for 1,2,3,... 1 1 1 1 1 1 lim ( ) lim (0 ) (0 ) (0 ) (0 ) lim (0 ) lim (0 ) (0 ) (0 ) lim n n r n s s n s s x r s s s s s s s sX s x x x x x x x x + + + + + →∞ →∞ + + + + →∞ →∞ = <∞ =   = _ + + + + + _   = + _ + + + + _ = x(0 )+ →∞ .

In proving the theorem, we assumed x(t)u(t) = x(t) which is valid only for causal signals, and therefore, the initial value theorem holds true for the unilateral Laplace transform and not for the bilateral Laplace

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transform. In addition, x(t) should not contain an impulse function or any other discontinuity at t=0 so

that _x( )r _{(0 )}+ _{< ∞}_for_r₌_{1, 2,3,...}_. _▌

Problem 6.9

From the time differentiation property, we know

R x s sX dt dx_←_→L ₍ ₎₋ ₍₀−₎ _with_ROC_: or, −

( )

− ∞ − =

∫

₋ ( ) 0 0 x s sX dt e st dt dx _.

Applying the limit, s → 0, on both sides of the equation, we get or, 0 0 0 lim dx st lim ( ) (0 ) dt s ₋ e dt s sX s x ∞ − − → →     = −   _ _   

∫



which simplifies to lim lim

[

( ) (0 )

]

0 0 0 − → − → ∞ − =    

∫

₋ e dt sX s x s st s dt dx or, 0 0 0 lim ( ) (0 ) lim st 1 dx dtdt _s sX s x _s e − ∞ − − →    →  = _ − _ _ = _

∫

∵ .

Applying the limits to

0 ₀ ( ) lim ( ) (0 ) s x t − sX s x ∞ − → = − , we get 0 ( ) lim ( ) s x sX s → ∞ = ,

which proves the final value theorem. ▌

Problem 6.10

(a) From Table 6.1,

2 2 0 ) ( ) cos( ₀_t _u _t L s_s + ω → ← ω .

Using the s-domain differentiation property (see Table 6.2), we get 2 2 2 0 2 2 0 2 2 0 ( ) ) 2 ( ) ( 0 ) ( ) cos( s_s s _ss s dsd L t u t t ω ←→ _ω ₊ = ω+_ω ₊− − or, 2 02 2 2 2 0 0 ₍ ₎

cos(

) ( )

L s s

t

ω

t u t

←→

₊−_ωω . (b) From Table 6.1, 2 2 0 0 ) ( ) sin( ₀ s L t u t ←→_ωω₊ ω .

Using the s-domain differentiation property (see Table 6.2), we get 2 2 2 0 0 2 2 0 0 ) ( ) 2 ( 0 ) ( ) sin( s s s dsd L t u t t ω ←→ _ωω₊ =− _ωω₊ −

(12)

or, 0 2 2 2 0 2 0 ₍ ₎

sin(

) ( )

L s s

t

ω

t u t

←→

₊ω_ω .

(c) Using the results from part (b) and using the linearity property, we get

(

)

_   ₋ → ← − ₊ 2₊−2 2 2 2 2 2 3 3 ₍ ₎ ) ( ) ( 2 1 2 1 _sin( ₎ _cos( ₎ ₍ ₎ s a a s s a a a L a at at at u t a which reduces

(

)

₍

₂ ₂

₎

2 3 1 2 1 _sin( ₎ _cos( ₎ ₍ ₎ a s L a at −at at u t ←→ ₊ . ▌ Problem 6.11 (a) ( ) cos(10 ) ( ) ( ) 5 ( ). 2 1 5 2 1 1 t t x t e x t e x t f ₌ ₌ j t ₊ −j t

Using the s-shifting property, we obtain

[

shiftedby 5

] [

shiftedby 5

]

: ROC ) 5 ( ) 5 ( ) ( ₂1 2 1 1 s = X s− + X s+ R s= R s= F _x ∩ _x . (b) )_f₂(_t)=_e−5t_x(4_t−3 _.

Using the time-shifting property,

x s

L _e _X _s _R

t

x( −3)←→ −3 ( ) ROC: . Using the scaling property,

x s s L _e _X _R t x(4 3) ₄1 4 (₄) ROC:2 3 − → ← − .

Using the s-shifting property, we obtain

. 5 by shifted 2 : ROC ) ( ) ( ) 3 4 ( ) ( ( 5) ₄5 4 1 2 5 2 4 3 − = = → ← − =_e− _x _t _F _s _e− + _X + _R _s t f t L s s _x

(c) Using the time-shifting property,

x s

L _e _X _s _R

t

x( ₋4)_←_→ −4 ( ) ROC: _. Using the time differentiation property,

[ ]

L x dtd x(t) sX(s) x(0 ) ROC:R − − → ← .

We now use the s-plane differentiation property

If L _dsd

[

]

_x

x

L _X _s _R _tx_t _X _s _R

t

x()←→ ( ) ROC: then − ()←→ ( ) ROC: . Using the s-plane differentiation property,

[ ]

L _dsd

[

]

x dtd x t sX s R t) ( ) ( ) ROC: ( 4 4 4 _←_→ − .

Using the time shifting property,

[

]

L s _dsd

[

]

x dtd x t e sX s R t 4) ( 4) ( ) ROC: ( 4 4 4 4 ₋ _←_→ − − .

(13)

(d)

[

]

2 2

4

( )

( ) 2

( ) 4 ( ) 4

f t

=

x t

+

=

x t

+

x t

+

Using the s-convolution property, we obtain

{ }

2

_{( )}

1 _{( )* ( ) ROC :}

2

L x

x t

X s

R

π

←→

.

{ }

( ) L ( ) ROC : x x t ←→X s R

1

L

ROC : Re{ } 0

s

←→

>

Using the linearity property, we obtain

{

}

4

1

4 ( )

( )* ( ) 4 ( )

ROC :

Re( ) 0 .

2

x

F s

X s

R

s

π

=

+



_

∩

>



_

(e) Using the s-shifting property, we obtain

{

}

0 0 0 ( ) ( ) ROC : Re( ) s t L x e− x t ←→X s s+ R − s .

Using the time integration property,

{

} {

}

0 0 0 0 ( ) ₁ 0

( ) ( ) ROC : Re( ) Re{ } 0 .

t X s s s L s x s s e α xα αd e αxα αd R s s − + − − −∞ −∞ ←→ + _ − > _

∫

∩ ▌ Problem 6.12

To determine the ROC, we use Property 2 that states:

For a right sided (causal) function, the ROC takes the form Re{s} > σ0 and consists of the right side of the complex s-plane

(a) Poles lie at s = −6.8541 and s = −0.1459, and hence the ROC is given by R: s > −0.1459. Since the ROC contains both s = 0 and s = ∞, the initial and final value theorems can be applied.

Initial value: lim ₀ ( ) lim ( ) lim lim 1.

2 1 7 2 2 1 1 1 7 = →∞ = ∞ → = ∞ → = → + + + + s s s s s sX s t x t s s s

Final value: lim ( ) lim ₀ ( ) lim ₀ 0.

1 7 2 2 = → = → = ∞ → s + s+ s s s sX s t x t Note that x(t)=1.0217e−6.8541tu(t)−0.0217e−0.1459tu(t), and therefore the initial and final value theorems compute the correct answer.

(b) Poles lie at s = −5.7016 and s = 0.7016, and hence the ROC is given by R: s > 0.7016. Since the ROC contains s = ∞, the initial value theorem can be applied. The final value theorem may give an incorrect answer.

Initial value: lim ₀ ( ) lim ( ) lim lim 1.

2 4 5 2 2 1 1 4 5 = →∞ = ∞ → = ∞ → = → + + + − s s s s s sX s t x t s s s

4 5 2 2 = → = → = ∞ → s + s− s s s sX s t x t

(14)

Note that x(t)=0.8904e−5.7016tu(t)+0.1096e0.1096tu(t),

and therefore the value of x(∞) obtained from the final value theorem is incorrect. The initial value theorem computes the correct answer.

(c) Poles lie at s = −5 and s = 5, and hence the ROC is given by R: s > 5. Since the ROC contains s = ∞, the initial value theorem can be applied. The final value theorem may give an incorrect answer.

Initial value: lim ₀ ( ) lim ( ) lim lim .

2 25 2 9 2 2 1 1 25 ) 9 ( ₌_∞ ∞ → = ∞ → = ∞ → = → −       + − + s s s s s s s s s sX s t x t

25 ) 9 ( 2 2 = → = → = ∞ → − + s s s s s sX s t x t Note that x(t)=δ(t)+3.4e5tu(t)−3.4e−5tu(t),

and therefore the value of x(∞) obtained from the final value theorem is incorrect. The initial value theorem computes the correct answer of x(0) = ∞.

(d) Poles lie at s = −1.5 + j1.323 and s = −1.5 − j1.323, and hence the ROC is given by R: s > −1.5. Since the ROC contains both s = 0 and s = ∞, the initial and final value theorems can be applied.

Initial value: lim ₀ ( ) lim ( ) lim

(

)

lim

(

2

)

.

2 2 2 / 4 / 3 1 / 1 / 2 1 4 3 1 2 ₌_∞ ∞ → = ∞ → = ∞ → = → + + + + + + + + s s s s s s s s s s s s s sX s t x t

Final value: lim ( ) lim ₀ ( ) lim ₀

(

2 ₃2 ₄1

)

0. 2 = → = → = ∞ → + + + + s s s s s s s sX s t x t

Note that x(t) (t) e t cos( 2.75t)u(t) e 1.5t sin( 2.75t)u(t

75 . 2 3 5 . 1 − − ₋ − δ = ,

and therefore the initial and final value theorems compute the correct answer.

(e) Poles lie at s = 0, −1, s = −2, and s = −3, and hence the ROC is given by R: s > 0. Since the ROC contains s = ∞, the initial value theorem can be applied. The final value theorem may give an incorrect answer. Initial value: 2 5 2 ( 4) (1 4 / ) 5 ( 1)( 2)( 3) _{(1 1/ )(1 2/ )(} ₃₎ 0

lim ( ) lim ( ) lim

lim

s

0.

s s s s s s _e _s _{s s} t

x t

s

sX s

s

e

s + + − + + + + + + →

=

→∞

=

→∞

=

→∞

=

Final value: lim ( ) lim ₀ ( ) lim ₀ ₃2.

) 3 )( 2 )( 1 ( ) 4 ( 5 2 = → = → = ∞ → + + + + − s s s s s e s s sX s t x t Note that ( ) ( 4) ( 5) ( 5) 4 2( 5) ( 5) 13₃ 3( 5) ( 5) 2 5 3 2_δ ₋ ₋ ₋ ₊ ₋ ₋ ₋ = t e− − u t e− − u t e− − u t t x t t t ,

and therefore the initial value theorem computes the correct answer. The value of x(∞) obtained

from the final value theorem is incorrect. ▌

Problem 6.13

(a) Calculating the Laplace transform of both sides, we obtain 2 0 0 0

( )

(0 )

3 ( )

(0 )

2 ( ) 1

s Y s

s y

−

y

−

sY s

y

−

Y s

= = =









−

+

−

+

=

























or, 1(_s2 ₊3_s₊2)_Y(_s)₌ _or 2 1 _{( 1)(}1 ₂₎ 1₁ 1₂ ( 3 2)

( )

_s _s _s _s _s _s

Y s

=

_{+ +}

=

₊ ₊

=

₊

−

₊ .

(15)

Calculating the inverse Laplace transform, we obtain

(

)

2 2

( )

t

( )

t

( )

t t

( ).

y t

=

e u t

−

e u t

−

=

e

−

e

−

u t

(b) Calculating the Laplace transform of both sides, we obtain

2 1 0 0 0 ( ) (0 ) (0 ) 4 ( ) (0 ) 4 ( ) _s s Y s s y − y − sY s y − Y s = = =     − − + − + =             or, ₍s2₊₄s₊₄₎Y₍s₎₌_s1_or 2 2 1 2 ( 2) ( 2)

( )

A B C s s s s s

Y s

=

₊

= +

₊

+

₊ ,

where the partial fraction coefficients are calculated as 2 1 1 ( 2) ₀ 2

1

1 , and

4

s

2

s _s s

A

₊

C

₌₋ =

=

 

_{ }

= −

Expanding Y(s), and comparing the numerator of both sides, we get 2 2 1 ( 2) ( 2) ( ) (4 2 ) 4 A s Bs s Cs A B s A B C A = + + + + = + + + + +

Comparing the coefficients of s2_{in both sides, we get (A + B) = 0 or B = −1/4.} In other words,

Y s

( )

=

_{s s}₍ 1₊₂₎2

=

1/ 4_s

−

_s1/ 4₊₂

−

₍_s1/ 2₊₂₎2

Calculating the inverse Laplace transform of Y(s) yields

2 2 2 1 1 4 4

( )

1

t

2

t

( )

1 (2

1)

t

( )

y t

₌

_

₋

e

−

₋

te

−

_

u t

₌

_

₋

t

₊

e

−

_

u t









.

(c) Calculating the Laplace transform of both sides, we obtain 2 2 1 ( 3) 1 1 1

( )

(0 )

6 ( )

(0 )

8 ( )

_s

s Y s

s y

−

y

−

sY s

y

−

Y s

₊ = = =









−

+

−

+

=

























or, ( 6 8) ( ) ₍ 1₃₎2 ( 1 6) 2 ₊ ₊ ₌ ₊ ₊ ₊ + s s Y s s s or

Y s

( )

=

₍_s₊₂₎₍_s₊1_{3) (}2 _s₊₄₎

+

₍_s₊₂₎₍s+7_s₊₄₎. Calculating the partial fraction expansion of the two terms separately, we obtain

2 1/ 2 0 2 1/ 2 1 1 2 3 4 ( 2)( 3) ( 4) ( 3) 7 5/ 2 3/ 2 ( 2)( 4) 2 4 and

s s s s s s s s s s s s + + + + + + + + + + + +

=

+

−

=

−

Expanding Y(s) as 2 2 1/ 2 1 1/ 2 5/ 2 3/ 2 3 1 2 2 ₍ ₃₎ 4 2 4 2 ₍ ₃₎ 4

( )

_s _s _s _s _s _s s s

Y s

=

₊

−

₊

−

₊

+

₊

−

₊

=

₊

−

₊

−

₊ .

Calculating the inverse Laplace transform of Y(s) yields

(

2 3 4

)

( )

3

t t

2

t

( )

(16)

(d) Calculating the Laplace transform of both sides, we obtain 2 3 2 2 1 1 0 0 1 0 1 ( ) (0 ) (0 ) (0 ) 8 ( ) (0 ) (0 ) 19 ( ) (0 ) 12 ( ) _s s Y s s y − s y − y − s Y s s y − y − sY s y − Y s = = = = = =       − − − + − − + − + =                   or, ( 3 8 2 19 12) ( ) 1 ( 2 8 19) 2 + + + = + + + s s Y s s s s s or 2 4 3 2 3 5 1 2 4 2₍ 3 ₈ 12 _{19 12)} 3 ₈28 19_{19 12} 2₍ 8₁₎₍ 19₃₎₍ 1₄₎ 2 ₁ ₃ ₄ ( ) s s s s s k k k k k s s s s s s s s s s s s s s s s Y s + + + + + + + + + + + + + + + + + = + = = + + + + where

(

)

(

)

(

)

(

)

4 3 2 4 3 2 2 2 2 8 19 1 1 ( 1)( 3)( 4) 0 4 3 2 8 19 1 1 ( 1)( 3)( 4) ( 1) ( 3) ( 4) 0 3 2 19 1 1 1 12 144 144 144 0 8 19 1 ( 1)( 3)( 4) 0 8 19 12 19 s s s s s s s s s s s s s s s s s s d k ds d d s s s s s s ds ds d s s s ds + + + + + + = + + + + + + + + + = =   = _ _   =_ + + + − + + + _     =_ × − × + + + _ = − × = −   4 ₈3 ₁₉2 ₁ ₁ 2 (ss 1)(ss 3)(ss 4)_s ₀ 12

k

+ + + + + + ₌

=

4 3 2 28 19 1 1 8 19 1 13 3 ( 3)( 4) ₁ 1 2 3 6 s s s s s s _s

k

+ + + − + + × × + + ₌₋

=

4 3 2 28 19 1 81 216 171 1 37 4 _{( 1)(} ₄₎ 9 ( 2) 1 18 3 s s s s s s _s

k

+ + + − + + × − × − + + ₌₋

=

4 3 2 28 19 1 256 512 304 1 49 5 ( 1)( 3) ₄ 16 ( 3) ( 1) 48 s s s s s s _s

k

+ + + − + + × − × − + + ₌₋

=

In other words, Y s( )= −19 /144_s +1/12_s2 +13/ 6_s₊₁ −37/18_s₊₃ + 49 / 48_s₊₄ . Therefore,

(

19 1 13 37 3 49 4

)

144 12 6 18 48 ( ) t t t ( ) y t _{= −} ₊ t₊ e− ₋ e− ₊ e− u t _. An Alternative (Equivalent) Solution of (d)

Calculating the Laplace transform of both sides, we get

2 3 2 2 1 1 0 0 1 0 1 ( ) (0 ) (0 ) (0 ) 8 ( ) (0 ) (0 ) 19 ( ) (0 ) 12 ( ) _s s Y s s y − s y − y − s Y s s y − y − sY s y − Y s = = = = = =       − − − + − − + − + =             _ _     or, ( 3 8 2 19 12) ( ) 1 ( 2 8 19) 2 + + + = + + + s s Y s s s s s or 2 2 2₍ 3 ₈ 21 _{19 12)} 3 ₈ 28 19_{19 12} 2_{( 1)(}1 ₃₎₍ ₄₎ _{( 1)(} 8 19₃₎₍ ₄₎

( )

s s s s s s s s s s s s s s s s s s

Y s

+ + + + + + + + + + + + + + + +

=

+

=

+

Taking the partial fraction expansion of the two terms separately

2 3 2 2 2 3 2 0.0208 0.0556 0.1667 0.1319 0.0833 1 ( 4) ( 3) ( 1) ( 8 19 12) 8 19 1 2 2 ( 4) ( 3) ( 1) ( 8 19 12)

and

s s s s s s s s s s s s s s s s s + + + + + + + + + + + + + +

=

−

+

−

+

=

−

+

(17)

Expanding Y(s), we get 2 2 0.0208 0.0556 0.1667 0.1319 0.0833 1 2 2 ( 4) ( 3) ( 1) ( 4) ( 3) ( 1) 1.0208 2.0556 2.1667 0.1319 0.0833 ( 4) ( 3) ( 1)

( )

_s _s _s _s _s _s _s _s s s s s _s

Y s

₊ ₊ ₊ ₊ ₊ ₊ + + +

=

−

+

−

+

−

+

=

−

+

−

+

.

Taking the inverse Laplace transform of Y(s) gives

4 3 3 4

( ) 1.0208

( ) 2.0556

( ) 2.1667

( ) 0.1319 ( ) 0.0833 ( )

0.1319 0.0833

2.1667

2.0556

1.0208

( )

t t t t t t

y t

e u t

u t

tu t

t

e

u t

− − − − − −

=

−

+

−

+





= −

_

+

−

+

_

(e) 4 2 4

2

( )

( );

(0 )

(0 ) 0.

d y

y t

u t

y

dt

− − − −

+

=

Calculating the Laplace transform of both sides, we get

3 2 2 1 0 0

4 ( )

(0 )

2 ( )

(0 )

( )

_s

s Y s

s y

−

s y

−

sy

−

y

−

s Y s

sy

−

y

−

Y s

= =









−

+

−

+

=

























or, ₍s4 ₊₂s2 ₊₁₎Y₍s₎₌_s1_or 2 2 2 2 4 ) 1 ( ) 1 ( ) 1 2 ( 1 ) ( ₊+ + + + + = + + = s E Ds s C Bs s A s s s s Y . where

[

] [

]

1. 0 ) 1 2 ( 1 0 ) 1 2 ( 1 2 4 2 4 = = = = + + = + + s _s s s _s s s s A

Equating numerator of Y(s) in both sides, we get (Note: A=1)

2 2 2 4 3 2 1 ( 1) ( ) ( 1) ( ) (1 ) (2 ) ( ) 1 s Bs C s s Ds E s B s Cs B D s C E s = + + + + + + = + + + + + + + +

Comparing the coefficients of polynomials of different order we get

4 3 2 Coefficients of : 1 0 1 Coefficients of : 0 Coefficients of : 2 0 1 0 1 Coefficients of : 0 0 s B B s C s B D D D s C E E + = ⇒ = − = + + = ⇒ + = ⇒ = − + = ⇒ =

The partial fraction expansion of Y(s) is given by

2 2 2 1 1 ( 1)

( )

s s s s s

Y s

= −

₊

−

₊

Noting that (see Problem 6.10(b))

(

₂ 0₂

)

2 0 2 0 sin( ) ( ) L s s t t u t ω ω

ω

+

←→ , the inverse transform is obtained as

[

]

( ) 1 cos( ) 0.5 sin( ) ( )

(18)

Problem 6.14 (a) (i) _{( )}

{

_{4 ( )}

}

4 s X s =L u t = and

{

}

2 22 2 1 1 2 2 ( 2) ( ) ( ) t ( ) s s s s s s Y s L tu t e u t− + + + ₊ = + = + = .

The transfer function is given by 22 2

( ) 2 2 ( ) ( 2) 4 4 ( 2) ( ) Y s s s s s s X s s s s s H s + + + + + + = = × = .

(ii) The impulse response is given by

{

2

}

{

}

{

}

(

)

1 2 1 1 2 1 1 1 2 1 2 4 ( 2) 4 ( 2) 4 2 4

( )

s s

1

s

1 ( )

( ) 2

t

( )

s s s s s s

h t

L

− + +

L

− −

L

−

δ

t

u t

e u t

− + + +

=

+

=

+ −

=

+

−

(iii) In order to calculate the input-output relationship in the form of a differential equation, we represent the transfer function as follows.

2 ₂ _{( )} 4 ( 2) ( )

( )

s s Y s s s X s

H s

+ + +

=

or,

(4

s

2

+

8 ) ( ) (

s Y s

=

s

2

+ +

s

2) ( )

X s

Calculating the inverse Laplace transform and assuming zero initial conditions, the differential equation representing the system is given by

2 2

4 d y

8 dy

d x dx

2 ( )

x t

dt

+

dt

=

dt

+

dt

+

.

(b) (i) The Laplace transform of the input and output signals are given by

) 2 ( 1 4 ) 2 ( 1 and ( ) 3 ) ( = ₊ = − s _s₊ s Y s e s X .

Dividing Y(s) with X(s), the transfer function is given by

s s X s Y _e s H( )= (₍)₎ =3 −4 .

(ii) The impulse response is obtained by calculating the inverse Laplace transform. The impulse response is given by

h(t)=3δ(t−4).

(iii) In order to calculate the input-output relationship in the form of a differential equation, we represent the transfer function as

) ( ) ( 4 3 ) (s e s _XY s_s H = − = _.

Cross multiplying, we get Y(s)=3e−4sX(s).

Calculating the inverse Laplace transform, the input-output relationship of the system is given by ) 4 ( 3 ) (t = x t− y .

(c) (i) The Laplace transform of the input and output signals are given by

) 4 ( 1 2 1 _and ₍ ₎ ₃ ) (s = _s2 Y s =_s3 − _s₊ X .

Dividing Y(s) with X(s), the transfer function is given by ) 4 (3 2 ) ( ) ( 2 ) ( = = − _s₊s s s X s Y s H .

(19)

(ii) The impulse response is obtained by calculating the inverse Laplace transform. The impulse response is given by

[

( )

]

. 3 ) ( 2 ) ( 2 4 2 t u e t u t h t dt d − − =

) ( ) ( ) 4 ( 3 ) 4 ( 2 3 ) (s s_s _s s _XY s_s H = +₊− = .

Cross multiplying, we get s2Y(s)+4sY(s)=−3s3X(s)+2sX(s)+8X(s).

Calculating the inverse Laplace transform, the input-output relationship of the system is given by ) ( 8 2 3 4 3₃ 2 2 t x dt dx dt x d dt dy dt y d + + − = + .

(d) (i) The Laplace transform of the input and output signals are given by

3 1 1 1 2 1 _and ₍ ₎ ) (s =_s₊ Y s = _s₊ +_s₊ X .

Dividing Y(s) with X(s), the transfer function is given by

3 1 1 1 ) 3 ( ) 2 ( ) 1 ( ) 2 ( ) ( ) ( ₂ ) ( = = +₊ + s_s+₊ ≡ + _s₊ − _s₊ s s s X s Y s H .

(ii) The impulse response is obtained by calculating the inverse Laplace transform. The impulse response is given by ). ( ) ( ) ( 2 ) (t u t e u t e 3u t h = + −t − − t

) ( ) ( ) 3 )( 1 ( ) 3 1 )( 2 ( ) (s s _s s _s s _XY s_s H = + ₊ +₊+ + = .

Cross multiplying, we get2s2Y(s)+8sY(s)+8Y(s)=s2X(s)+4sX(s)+3X(s).

Calculating the inverse Laplace transform, the input-output relationship of the system is given by ) ( 3 4 ) ( 8 8 2 ₂ 2 2 2 t x dt dx dt x d t y dt dy dt y d + + = + + .

(e) Note that there is no overlap between the ROC’s of the two terms exp(t)u(−t) and exp(−3t)u(t), and

hence the Laplace transform for y(t) does not exist. ▌

Problem 6.15 (a) 22 2 ( )( ) 1 2 1 ( 1) ( ) s s j s j s s s H s + + − + + + = = Two zeros: at