Continuous and Discrete Time Signals and Systems (Mandal & Asif) Solutions - Chap05

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Problem 5.1

(a) The CTFT for x1(t) is given by

[ ]

. 2 sinc ) 2 / ( ) 2 / ( sin ) 2 / ( sin 2 2 ) cos( 1 2 ) cos( 2 2 ) ( 1 ) ( 1 ) ( ) ( ) ( 1 ) ( 1 ) ( ) ( 1 ) ( ) ( 1 1 1 ) ( ) ( 2 2 2 2 2 2 2 2 2 1 2 1 2 1 2 1 0 2 1 0 2 1 0 0 1 1       π ωτ τ = ωτ ωτ τ = τ ω ωτ × = τ ω ωτ − × = τ ω ωτ − τ ω =         ω − × − ω − − ω − × +         ω − × + ω − × − ω − =         ω − × − − ω − × − +         ω − × − ω − × + = − + + = = ω τ ωτ − τ ωτ τ τ τ ω − τ ω − τ τ − ω − τ ω − τ τ − τ ω − τ ω − τ ∞ ∞ − ω −

_∫

∫

j j j e j e j j j e j e j e j e dt e dt e dt e t x X j j t j t j t t j t j t t j t t j t t j

(b) The CTFT for x2(t) is given by

[

]

( ) ( ) ( ) ( ) ( ) 2 3 4 5 4 4 ( ) 2 2 0 4 3 2 ( ( )) (( ( )) ( ( )) (( ( )) ( ( )) ₀

( )

4

12

24

24 0 0 0 0 0

0 0 0 0

a j t a j t a j t a j t a j t j t at j t a j t e e e e e a j a j a j a j a j

X

x t e

dt

t e u t e

dt

t e

dt

t

ω

t

ω

t

ω

t

ω ω ω ω ω ω ω ω ω ω

ω

− + − + − + − + − + ∞ ∞ ∞ − − − − + −∞ −∞ ∞ − + − + − + − + − +

=





=

_

+

_

= + + + + −

+ + +

∫

5 5 1 24 ( ( )) ( )

24 .

a jω a jω − + +



₊



₌





(c) The CTFT for x3(t) is given by

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

( )

cos(

) ( )

0

a j j t a j j t j t j t j t at j t a j t a j j t a j j t e e a j j a j j

X

x t e

dt

e

t u t e

dt

e

dt

e

dt

e

dt

ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω

ω

− + − − + + ∞ ∞ ∞ − − − − − + −∞ −∞ ∞ ∞ _∞ _∞ − + − − + + − + − − + +





=

_

+

_









=

+

=

_

_

+

_

_

=

−

∫

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

0

( ( )) 2 ( ) ( ) ( )

.

a j a j j a j j a j j a j j a j ω ω ω ω ω ω ω ω ω ω ω + − + − − + + + − + + ₊ ₊





+



−



=



+



=













(2)

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

( ) ( ) exp( ) exp[ ] exp exp

exp exp exp 2 2 exp

t j t t j t j j j t t j t j t j j X x t e dt e dt dt dt dt ωσ ωσ ωσ ωσ ω ω σ σ σ σ ωσ ωσ ωσ ω σ σ σ σ ω πσ πσ ∞ ∞ ∞ ∞ + + + − − −∞ −∞ −∞ −∞ ∞ + −∞     = = − = − = _− _ _ _       = _ _⋅ _− _ = _ _ = −

∫

 2 . ▌ Problem 5.2 (a) By definition,

[ ]

[

]

[

]

[

]

[

]

[

]

). 2 / ( sinc 3 6 6 ) 2 / sin( 2 1 3 3 ) ( 2 / 2 / ) 2 / sin( / 21 2 / ) 2 / sin( 2 / 2 / 3 0 2 / 2 / 2 / 3 3 0 ) ( 1 ω π = × = = ωπ − − = − − = − − = = = ω ωπ − ωπωπ π ωπ − ω ωπ ωπ − ωπ − ω π ωπ ωπ − ωπ − ω ωπ − ω π ω − ω −

∫

−ω j j j j j j j j j j j j e t j e e e j e e e e e dt e X jt (b) By definition,

[

]

[

]

0.5 1.5 0.5 1.5 2 ( ) _0.5 ( ) _0.5 0.5 0.5 0.5 0.5 1.5 0.5 0.5 1 0.5 1 sin 0.5 0.5

( )

0.5

0.5 2 sin(0.5

)

2 sin(0.5

)

j t j t T T T T j t j t e e j _T j _T T T j T j T j T j T j j j T j j T T

X

e

dt

e

dt

e

j

T

e

j

T

ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω

ω

− − − − − ₋ − − − − − −









=

+

=

_

_

+

_

_









= −

_

−

_

−

_

−

_

= −

−

=

×

∫

(

)

(

)

(0.5 ) 0.5 sin(0.5 ) 0.5 0.5 T 0.5 T

2 0.5 sinc

sinc

.

T j T T T T j T

e

T

Te

ω ω ω ω ω ω ω ω π π − −





+



×











=

+

(c) By definition,

( )

(

1

)

. 0 1 1 ) ( 2 2 2 2 2 2 1 1 1 1 1 ) ( 1 1 ) ( 1 ) ( 1 0 ) ( 1 ) ( 0 3 T j T j T j T j T j T j j e T T j e T j e Tt T t j Tt e e dt e X T j t j t j ω − ω ω ω ω ω − ω ω − ω − ω − ω − ω − ω − − + = + + − =     ₋ ₋ −     ₋ ₋ =     ₋ ₋ ₋ = − = ω ω − ω − ω −

∫

For ω = 0,

( )

₂ ₂ 0 2 2 0 3( ) 1 1 0 T T T Tt T T Tt dt X ω =

∫

− =_− − _ = + = .

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(d) By definition,

( )

[

1 cos( )

]

sinc ( ). 0 0 1 1 1 1 ) ( 5 . 0 2 ) 5 . 0 ( ) 5 . 0 ( sin ) 5 . 0 /( 1 4 ) 5 . 0 ( sin 2 2 2 ) ( 1 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 1 ) ( 1 0 ) ( 1 ) ( 0 ) ( 1 ) ( 0 0 4 2 2 2 2 2 2 2 2 2 2 2 2 πω ω ω ω ω × ω ω − ω − ω − ω − ω − ω − ω − ω − − ω − ω − ω − − ω − = × = = ω − =     ₋ ₋ ₋ ₊ ₋ +     ₋ ₋ ₊ =     ₋ ₋ ₋ +     ₊ ₋ = − + + = ω ω − ω ω − ω − ω − ω −

∫

T T T T T T T j T j j e T j e T j T j T j e T j e Tt T j e T j e Tt T t j Tt T t j Tt T T dt e dt e X T j T j t j t j t j t j (e) By definition,

( )

5 0 0 0 ( ) 1 0.5sin 0.5 sin T T T j t j t j t t t T T A B X _ω π e−ωdt e−ωdt π e−ωdt = =   =

∫

_ − _ =

∫

−

∫

We consider different cases for the above integral. Case I: ( ω = 0)

( )

[

]

5 0 0 0 2 / 0.5 1 (0) ( ) 1 0.5sin 0.5 sin

cos( ) cos( ) cos(0) (1 )

T T t t T T T t T T T T X x t dt dt dt dt T T T T π π π π π π π π −∞ ∞ ∞ −∞   = = _ − _ = − = + _ _ = + − = − = −

∫

Case II: ( ω ≠ 0, ω ≠ π/T):

[

]

0 0 1 1 ₁ 1 ₁ ₀ T T j t j t j T j T j j j A=

∫

e−ωdt=₋ _ω_e−ω _ =₋_ω_e−ω − =_ _ω_ −e−ω _ ω≠

( )

{

}

( )

2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0.5 0 0, ₀ 0.5 0.5

0.5 sin cos for 0,

sin cos 1 j t T T e t t T T T T T j t T t t T T T T at t T j T j T T T T T T T B j e j e e ω π π π π π ω ω π π π π ω ω ω π π π π ω π ω ω ω ω − − − − = = − − − −   = _ − − _ ≠ ±    _ _   = _ + _  _ _        = _− − _= _ + _ Case III: ( ω = π/T):

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( )

2 2 2 ( ) ( ) 0.5 0.5 2 2 0 0 0 0.5 2 0 0.5 2 0 0.5 sin 1 , 1 t t T T T T t T j t T t T T T T j j j t j t j t j t t T j j T j j T j j T T B e dt e e e dt e e dt e dt

As e is periodic with period T e e dt π π π π π π π ω ω ω ω π π π ω ω − − − − + − − − ± ±     = = _ − _ = _ − _  _ _ − =  _ _  =  ₋ _ ₋ _ _{= −}  _ _ 

∫

[ ]

2 0 0.5 0.5 2 0 2 0 j t T T T _T j j dt t π   =     = ± = ±

∫

Combining, the above results, the CTFT can be expressed as

2 2 2 1 0.5 5 2 0.5 1 1 (1 ) 0 ( ) 1 1 1 j T T j T j T T j T T j j T X e e e π ω π ω π ω π ω ω ω ω ω − ω − − − − =   = _ − _ = ±  − −  +      ∓ 2 2 2 1 4 0.5 2 1 (1 ) 0 1 1 T j T j T T j T T j j T otherwise T e e π π ω π ω π ω π ω ω ω − − −      − = = ± = ±  − −  +      ∓ otherwise      ▌ Problem P5.3

From magnitude and phase spectra shown in Fig. P5.3, the individual CTFT’s can be expressed as follows Fig. P5.3(b):    _× ₋ _≤_ω_≤ = ω ω otherwise 0 1 ) ( 0.5 1 e W W X j Fig. P5.3(c): 0.5 0.5 2

1

0 ( )

1

0

0 otherwise

j j

e

W

X

e

W

ω ω

ω

−

 ×

− ≤ ≤



=

_

×

≤ ≤





Fig. P5.3(d): /3 / 3 3

1

0 ( )

1

0

0 otherwise

j j

e

W

X

e

W

π π

ω

−

 ×

− ≤ ≤



=

_

×

≤ ≤





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[

]

[

]

0.5 (0.5 ) 1 (0.5 ) (0.5 ) (0.5 ) 1 1 1 ( ) ( ) 2 2 2 2 sin (0.5 ) 1 1 1 2 (0.5 ) 2 (0.5 ) 2 (0.5 ) sin (0.5 ) sinc ( 0.5) (0.5 ) W W j t j j t j t W W W j t j t W j t W W W x t X e d e e d e d j t W e e e j t j t j t t W W W t t W ω ω ω ω ω π

ω

π

∞ + −∞ − − + + − + − = = = +    −  = _ _ = _ _= + + +     + = × = _ + +

∫

. 

Using the CTFT synthesis Eq. (5.9), the function

x t

₂

( )

is calculated as follows. 0 ( 0.5 ) (0.5 ) 2 0 0 ( 0.5 ) (0.5 ) ( 0.5 ) (0.5 ) 0 0. 2 1 1 1 ( ) ( ) 2 2 2 1 1 1 1 1 2 ( 0.5 ) 2 (0.5 ) 2 ( 0.5 ) (0.5 ) 1 1 2 ( 0.25) W j t j t j t W W j t j t j t W j t W W j x t X e d e d e d e e e e j t j t j t j t e j t ω ω ω ω ω

ω

π

∞ − + + −∞ − − + + − − + + − = = +      − −  = _ _ + _ _ = _ + _ − + + − + +       = + −

∫

(

)

5 2 0.5 2 2 sin( ) cos( ) ( 0.25) 1 1 2 sin( ) cos( ) . 2 ( 0.25) W j W jt tW tW j t e jt tW tW j

π

t  −   ₋      = _ + − _ −

Clearly, at (t = ±0.5), x2(t) is undefined in the above expression. Computing directly, we obtain

At t = 0.5: 1 0.5 0.5 1 1 1 2 2 2 2 ₀ 2 0 0 (0.5) 1 W W W j j j j jW j j x = _π

_∫

e ωe ωdω= _π

_∫

e dω ω= _π _eω_ = _π_e − _. At t = −0.5: 0 0 0 0.5 0.5 1 1 1 1 2( 0.5) 2 2 2 2 1 j j j j jW j _W j W W x _π e− ωe− ωd

ω

_π e−ωd

ω

₋ _π e−ω ₋ _π e − −     − =

∫

=

∫

= _ _ = _ − _.

Using the CTFT synthesis Eq. (5.9), the function

x t

₃

( )

is calculated as follows.

[

]

0 0 / 3 / 3 3 0 / 3 / 3 / 3 / 3 0 1 1 1 ( ) ( ) 2 2 2 1 1 1 1 1 2 2 2 1 sin( / 3) sin( / 3) 2 sin( / 3) 2 sin( / 3) 2 j t j j t j j t W W W j t j t jWt jWt j j j j W x t X e d e e d e e d e e e e e e e e jt jt jt jt Wt j Wt j j t ω π ω π ω ω ω π π π π

ω

π

∞ − −∞ − − − − − − = = +      − −  = _ _ + _ _ = _ + _       + − = + − =

∫

t

π

     

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0 1 3 4 4 2 0 (0) (1 3) (1 3) 1 3 1 3 W W W W x _π j dω j dω _π j j _π −   _ _ = _ − + + _= _ − + + _= 

∫

 .

Although the functions

x t

₁

( )

,

x t

₂

( )

, and

x t

₃

( )

have the same magnitude spectra, their phase spectra are different. As a result, the time domain representations of these functions are different.

For the special case W = π, the three functions are plotted in Fig. S5.3. Since x2(t) is a complex function,

its magnitude is plotted in Fig. S5.3. The Matlab code is also included below. ▌

-5

-4

-3

-2

-1

0

1

2

3

4

5 -1

-0.5

0

0.5

11 Problem 5.3

t

x

1

(t)

-5

-4

-3

-2

-1

0

1

2

3

4

5

0

0.25

0.5

0.75

11 t

ab

s(

x

2

(t

))

-5

-4

-3

-2

-1

0

1

2

3

4

5 -1

-0.5

0

0.5

11 t

x

3

(t

)

Fig. S5.3. Plots of functions in Problem P5.3. % MATLAB code to plot the functions in Problem 5.3

del = 0.01; t = -5:del:5; W = pi ; x1 = (W/pi)*sinc((W/pi)*(t+0.5)) ; x2 = 1./(j*2*pi*(t.^2-0.25)).*(1+exp(j*0.5*W)*(2*j*t.*sin(t*W)-cos(t*W))); x2(t==0.5) = 1./(j*2*pi)*(exp(j*W)-1); x2(t==-0.5) = 1./(j*2*pi)*(exp(j*W)-1); x3 = (sin(W*t+pi/3)-sin(pi/3))./(pi*t); x3(t==0) = W/(2*pi) ;

subplot(3,1,1), plot(t, x1), grid on title('Problem 5.3');

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xlabel('t') % Label of X-axis ylabel('x_1(t)') % Label of Y-axis %

subplot(3,1,2), plot(t, abs(x2)), grid on xlabel('t') % Label of X-axis ylabel('abs(x_2(t))') % Label of Y-axis subplot(3,1,3), plot(t, x3), grid

xlabel('t') % Label of X-axis ylabel('x_3(t)') % Label of Y-axis

Problem 5.4

(a) The partial fraction expansion is given by

) 3 ( 2 ) 2 ( 1 ) 3 )( 2 ( ) 1 ( ) ( 1 ω = ₊ _ω+ ω₊ _ω ≡ ₊− _ω + ₊ _ω j j j j j X

Calculating the inverse CTFT, we obtain

) ( 2 ) ( ) ( 2 3 1 t e u t e u t x =− −t + −t . (b) The partial fraction expansion is given by

) 3 ( 5 . 0 ) 2 ( 1 ) 1 ( 5 . 0 ) 3 )( 2 )( 1 ( 1 ) ( 2 ₊ _ω + ₊ _ω − + ω + ≡ ω + ω + ω + = ω j j j j j j X

) ( 5 . 0 ) ( ) ( 5 . 0 ) ( 2 3 2 t e u t e u t e u t x = −t − −t + −t . (c) The partial fraction expansion is given by

) 3 ( 5 . 0 ) 2 ( 1 ) 2 ( 0 ) 1 ( 5 . 0 ) 3 ( ) 2 )( 1 ( 1 ) ( ₂ ₂ 3 + −₊ _ω ω + − + ω + + ω + ≡ ω + ω + ω + = ω j j j j j j j X

) ( 5 . 0 ) ( ) ( 5 . 0 ) ( 2 3 3 t e u t te u t e u t x = −t − − t + −t .

(d) The partial fraction expansion is given by

) ) ( 2 2 ( 1 ) 1 ( 1 ) ) ( 2 2 )( 1 ( 1 ) ( ₂ ₂ 4 ω + ω + ω + − ω + ≡ ω + ω + ω + = ω j j j j j j j X or, ₄ ₂ ) 1 ( 1 1 ) 1 ( 1 ) ( ω + + ω + − ω + = ω j j j X

) ( cos ) ( ) ( 4 t e u t e tu t x = −t − −t . (e) The partial fraction expansion is given by

2 2 2 2 2 2 2 2 5 ) ) ( 2 2 ( ) ) ( 4 ( 25 . 0 ) ) ( 2 2 ( 50 . 1 ) 1 ( 1 ) ) ( 2 2 ( ) 1 ( 1 ) ( ω + ω + ω + ω + ω + ω + − ω + ≡ ω + ω + ω + = ω j j j j j j j j j j X

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or, ₂ ₂ 2 2 2 5 ) ) 1 ( 1 ( ) ) ( 4 ( 25 . 0 ) 1 ( 1 50 . 1 ) 1 ( 1 ) ( ω + + ω + ω + ω + + − ω + = ω j j j j j X

        ω + + ω + ω ℑ + − = − − −1 ₂ ₂2 5 ) ) 1 ( 1 ( ) ) ( 4 ( 25 . 0 ) ( sin 50 . 1 ) ( ) ( j j j t u t e t u te t x t t . ▌ Problem 5.5

Consider an arbitrary function φ(t), and assume that

dt e t p

∫

j t ∞ ∞ − ω = ) ( .

Now, consider the integral

{ }

( )

-Φ( ) ( ) Changing the order of integration

, ( ) ( ) ( ) ( ) Φ( ) Φ( )( ) j t T j T j t j T t j T d d t p t T dt t e d dt e t e dt d e d e d ω ω ω ω ω φ ω ω ω ω ω

φ

ω

φ

ω

ω ω

ω

∞ ∞ ∞ ∞ ∞ ∞ − − − −∞ −∞ −∞ ∞ −∞ −∞ =ℑ −∞ ′ ∞ ′ ′ =− =−     − = _ _ = _ _ = −     ′ ′ = −

∫

Φ( ) Φ( ) ...(1) j T j T e d e d ω ω

ω ω

ω

∞ ′ −∞ ∞ −∞ ′ ′ = =

∫

Note that the right hand side of Eq. (1) is the inverse CTFT of Φ(ω) computed at t = T, i.e., φ(T). Hence, ) ( 2 ) ( ) ( ) (t p t−T dt= Φ ωej Tdω= πφT φ

∫

∞ ∞ − ω ∞ ∞ − .

The above equation is valid for any arbitrary φ(t) if and only if p(t) = 2πδ(t) as can be seen from the following property of the impulse response

) ( 2 ) ( ) ( 2π

∫

φt δ t−T dt= πφT ∞ ∞ − . In other words, ) ( 2 t dt ej t = πδ

∫

∞ ∞ − ω _.

Interchanging the variables, t and ω, we obtain the required identity ) ( 2πδω = ω

∫

∞ ∞ − ω _d ej t . ▌

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Alternate Proof:

Note that the above result can be proved directly from the CTFT pair ) ( 2 1← →CTFT πδ ω . By definition,

∫

∞ ∞ − ω − × = ω πδ( ) 1 e j tdt 2 Since δ(−ω) = δ(ω),

∫

∞ ∞ − ω = ω − πδ = ω πδ( ) 2 ( ) ej tdt 2 . ▌ Problem 5.6

Using Eq. (5.40), the CTFT for a real-valued even function x(t) can be expressed as

∫

∞ ∞ ∞ − ω − ₌ _ω = ω 0 ) cos( ) ( 2 ) ( ) ( x t e dt x t t dt X j t .

Since there is no complex value in the above equation, X(ω) is real valued, i.e., Im{X(ω)} = 0.

Also, )( ) 2 ( )cos( ) 2 ( )cos( ) (

0 0 ω = ω = ω − = ω −

∫

∞ ∞ X dt t t x dt t t x X .

Therefore, X(ω) is also an even function with respect to ω. Since, X(ω) is real valued, Re{X(ω)} =

Re{X(−ω)}. ▌

Problem 5.7

Using Eq. (5.40), the CTFT for a real-valued odd function x(t) can be expressed as

∫

∞ ∞ ∞ − ω − ₌₋ _ω = ω 0 ) sin( ) ( 2 ) ( ) ( x t e dt j x t t dt X j t .

Since x(t) is real, the product x(t)sin(ωt) is also real and so is the integral. Therefore, X(ω) is pure imaginary, i.e., Re{X(ω)} = 0.

Also, ( ) 2 ( )sin( ) 2 ( )sin( ) ( )

0 0 ω − = ω − = ω − = ω −

∫

∞ ∞ X dt t t x dt t t x X .

Therefore, X(ω) is also an odd function with respect to ω. Since, X(ω) is imaginary-valued, Im{X(ω)} =

−Im{X(−ω)}. ▌ Problem 5.8 (a) Since

( )

₍ ₎ ₂ ₍5 ₅₎ 5 2 5 1 −ω = ₊j₋_ω₋ = ₋j _ω₊ X is not equal to

( )

₂ ₍5 ₅₎ 1∗ ω = ₋_j_ω₋ X ,

(10)

X1(ω) does not satisfy the Hermitian property. Its inverse CTFT x1(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x1(t) from the Hermitian property.

(b) Since

( )

cos

(

2

)

cos(2 ) 1₂sin(2 )

23 6

2 −ω = − ω+π = ω + ω

X

is not equal to

( )

cos

(

2

)

cos(2 ) ₂1sin(2 )

23 6

2∗ ω = ω+ π = ω − ω

X ,

(c) Since X₃

( )

−ω =5sin₍[₋4_ω(−₋ω_π−₎π)]=5sin₍_ω[4₊(ω_π+₎π)]=5₍sin_ω₊[ ]4_πω₎ is not equal to ∗

( )

ω =5sin₍[_ω4₋(ω_π₎−π)]=5₍sin_ω₋[ ]4_πω₎

3

X ,

(d) Since X₄

( ) (

−ω = 3+2j

) (

δ −ω−10

) (

+ 1−2j

) (

δ−ω+10

) (

= 3+2j

) (

δω+10

) (

+ 1−2j

) (

δω−10

)

is not equal to X₄∗

( ) (

ω = 3+2j

) (

δω−10

) (

+ 1−2j

) (

δω+10

)

,

X4(ω) does not satisfy the Hermitian property. Its inverse CTFT x4(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x4(t) from the Hermitian property. (e) Since X5

( )

−ω =₍₁₋_j_ω₎₍₃₋1_j_ω₎2

₍

₅₊_ω2

₎

is equal to

( )

₍₁ ₎₍₃ ₎2

₍

₅ 2

₎

1 5∗ ω = ₋_j_ω ₋_j_ω ₊_ω X ,

X4(ω) satisfies the Hermitian property. Its inverse CTFT x4(t) is real valued.

Since X4(ω) is complex (neither pure real-valued or pure imaginary), x4(t) is neither even nor odd

with respect to t. ▌

Problem 5.9

(a) Applying the linearity property,

( )

{

5 3cos(10) 7 2 sin(3) ()

}

5

{}

1 3

{

cos(10)

}

7

{

2 sin(3 ) ()

}

1 t e t u t t e t u t

X ω =ℑ + − − t = ℑ + ℑ − ℑ − t .

By selecting the appropriate CTFT pairs from Table 5.2, we get

( )

{}

₂ ₂ 1 3 ) 2 ( 21 ) 10 ( 3 ) 10 ( 3 1 ) ( 10 + ω + − − ω πδ + − ω πδ + ℑ ω δ = ω j X .

(b) Entry (8) of Table 5.2 provides the CTFT pair

ω   → ← _j t₎ CTFT 2 sgn( .

Using the duality property, 2 _←_{ →}CTFT_ 2_πsgn(₋_ω)

(11)

or, 1 _←_{ →}CTFT_ ₋ sgn(_ω)

πt j .

(c) Entry (7) of Table 5.2 provides the CTFT pair

ω + − _←_{ →}_ j t e 4 CTFT ₄8 . Using the time shifting property, − − ← → ₊ _ω − 5ω

48 CTFT 5 4 j j t e e .

Using the frequency differentiation property,

{

− ω ₊ _ω

}

ω − − _←_{ →}_ j j d d t e j e t2 4 5 CTFT 2 5 ₄8 2 2 ) ( or, ₍₄ 1 ₎3 5 41 5 CTFT 5 4 2 ₂₀₀ ₁₆ ω + ω − ω + ω − − − _←_{ →}_ ₊ j j j j t e e e t .

(d) Entry (17) of Table 5.2 provides the CTFT pair

( )

ω_π ππ ← → = CTFT ₆ 3 ) 3 sin( _rect 3 ) 3 ( sinc 3 t _tt

and 5sinc(5t)=5sin(₅_π5_tπt)← →CTFT rect

( )

₁₀ω_π

Using the multiplication property

( )

[

ω_π

]

π ω π π π π π π _× _←_{ →}_ _∗ ×

π2 sin(3 ) sin(5 ) CTFT ₂2 _rect ₆ _rect ₁₀

t t t

t

or, sin(3π)_t2sin(5π)← →CTFT ₂π

[

rect

( )

₆ω_π ∗rect

( )

₁₀ω_π

]

t t , or, 2

( )

sin(3 )sin(5 ) CTFT 5 2 6 10

5

t t

rect

t π π π ω ω π π





←→

_

∗

_

,

where * is the convolution operation.

(e) Entry (17) of Table 5.2 provides the CTFT pair

( )

ω_π ππ ← → =3sin(₃3 ) CTFT rect ₆ ) 3 ( sinc 3 t _tt

and 4sinc(4t)=4sin(₄_π4_tπt)← →CTFT rect

( )

₈ω_π . Using the time differentiation property,

( )

ω_π π π

←

 →



ω

8 CTFT ) 4 sin( 1

_{( j}

₎

_rect

t t dtd .

Using the convolution property

( )

[

ω_π

]

π ω π π π π ππ ∗ ← → × ω ×

π2 sin(3 ) 1 sin(4 ) CTFT ₂2 _rect ₆ _rect ₈

j t t dtd t t

or, sin(3π)∗_dtd sin(_t4πt)← →CTFT ₂π

[

rect

( )

₆ω_π × jωrect

( )

₈ω_π

]

t t

,

(12)

Problem 5.10

Using the linearity property,

{

}

{

}

{

}

{

}

2 2 2 2 2 6 6 4 13 13 13 2 6 6 4 13 13 13 6 1 6 6 4 4 3 13 2 13 2 2 139 13 2 2 139 6 1 2 13 2 9 9 26 ( ) cos(3 ) sin(3 ) ( ) ( ) cos(3 ) ( ) sin(3 ) ( ) ( 3) ( 3) ( 3) ( 3) 6 t t j j j j j j X e t t u t e u t t u t t u t ω π π π π ω ω ω ω π ω ω ω

ω

δ ω

− − − + − − + ₋ ₋   = ℑ _ − + _ = ℑ − ℑ + ℑ   = − _ − + + _− + _ − + + _+   = _ − + _−

[

]

[

]

[

]

2 2 2 (9 ) (2 )(2 ) 6 13 (2 )(9 ) 26 6 13 (2 )(9 ) ( 3) 6 ( 3) 4 ( 3) 4 ( 3) (6 4) ( 3) (6 4) ( 3) (3 2) ( 3) (3 2) ( 3) j j j j j j j j j j ω ω ω π ω ω π ω ω

δ ω

− + − + + − + − − + + + − − +   = _ _− + − + − + = − + − + − +

which is the required result. ▌

Problem 5.11

From the definition of CTFT,

{

( )

}

∫

∞ ∞ − ω − = x at e dt at x F ₍ ₎ j t _.

We consider two different cases (a > 0) and (a < 0)

Case 1: Assume a > 0. Substitute r = at in the above expression. The upper and lower limits of integration stay the same and dr = a dt. The final result is

( )

{

}

( )

j

( )

r _a

( )

_a a a dr r j X dr e r x e r x at x F a a ω ∞ ∞ − − ∞ ∞ − −

∫

= = = ₍ ₎ ω 1 ₍ ₎ ω 1 _.

Case 2: Assume a < 0. Substitute r = at in the above expression. The upper limit of integration is

r→ −∞and the lower limit of integration is r→ ∞, and dr = a dt. The final result is

( )

{

}

( )

j

( )

r _a

( )

_a a r j a a dr r j X dr e r x dr e r x e r x at x F a a a ω ∞ ∞ − − −∞ ∞ − ∞ ∞ − − ₌ ₌₋ ₌₋ =

∫

₍ ₎ ω 1

∫

₍ ₎ ω 1

∫

₍ ₎ ω 1 _.

Combining the two cases, yields F

{

x

( )

at

}

= _a1 X

( )

ω_a _. _▌

Problem 5.12

Comparing with Fig. 5.9(a), we observe that

( )

₂

1

) (_t _x t

h = .

Using the scaling property, H(ω)=2X₁

( )

2ω

or, H(ω)=_ω22

[

2ωsin(4ω)+cos(2ω)−1

]

,

(13)

Problem 5.13

Using the definition of CTFT, we obtain

{

0

( )

}

=

∫

0 () =

∫

() ( 0) = (ω−ω₀) ∞ ∞ − ω − ω − ∞ ∞ − ω − ω ω _x_t _e _x _t _e _dt _x _t _e _dt _X e F j t j t j t j t . ▌ Problem 5.14

Using the convolution property,

( )

t ∗u t ← → X ω

[

πδ ω + _j_ω

]

x ₍ ₎ CTFT ₍ ₎₂ ₍ ₎ 1 _, or, ( ) ( ) CTFT 2 (0) ( ) X( ) j x

τ

u t

τ τ

d

π

X

δ ω

_ωω ∞ −∞ − ←→ +

∫

, or, ( ) ( ( )) CTFT 2 (0) ( ) X( ) j x

τ

u

τ

t d

τ

π

X

δ ω

_ωω ∞ −∞ − − ←→ +

∫

, or, ( ) CTFT 2 (0) ( ) ( ) t X j x

τ τ

d

π

X

δ ω

_ωω −∞ ←→ +

∫

. ▌ Problem 5.15

(a) Using the time scaling property, _x

( )

₂_t _←_{ →}CTFT__ ₂1 _X

( )

_ω₂ _.

Using the frequency shifting property,

( )

₂5

2 1 CTFT 5 ₂ ω+ − _x _t _←_{ →}_ _X e j t _.

Substituting the value of X(ω), we obtain

{

5

}

1 35 2 11 12 1 12

1 5 3

(2 )

0 elsewhere

11

5

1

0 elsewhere.

j t

e

x t

ω ω ω

ω

+ − + −

 −

+ ≤



ℑ

_{= }



− ≤ ≤ −





=

_

− ≤ ≤ −





(b) Using the frequency differentiation property,

( )

2 2 CTFT 2 ) ( ← → _ω d X d t x jt , or,

( )

2 2 CTFT 2 ω −   → ← d X d t x t . The CTFT of t2_{x(t) is given by}

{ }

2 ( ) 2

[ ]

( )

₃

[

rect

( )

₃

]

[

( 3) ( 3)

] [

( 3) ( 3)

]

2 + ω δ − − ω δ = − ω δ − + ω δ − = − = ∆ − = ω ω ω ω dd d d t x t F .

(14)

(c) Express

(

)

dx_dt dt dx dt dx _t t+5 = +5 .

Using the time differentiation property, the CTFT of dx

dt is given by

)

(

CTFT

_

_ω

 →

←

j

X

dt dx _.

Applying the frequency differentiation property to the above CTFT pair, gives

[

]

_ω ω

ω

=

−

ω

−

ω



 →

←

dX_d dd dt dx

_j

_X

t

CTFT

(

)

(

)

. The CTFT of

(

5 )

dx dt

t

+

is given by

(

)

{

+5

}

=− (ω)−ω +5 ω (ω) ℑ t X dX_d_ω j X dt dx _.

Substituting the value of X(ω), we obtain

(

)

{

}

( ) (

)

     ≤ ω ≤ − + − + ω ≤ ω ≤ − − − ω = + ℑ ω ω ω ω elsewhere. 0 0 3 1 1 5 3 0 1 1 5 5 2₃ 3 3 2 3 j j t dx_dt

(d) Using the time multiplication property,

( ) ( )

t ⋅x t ← → _π

[

X

( )

ω ∗X

( )

ω

]

x CTFT ₂1 _,

which implies that

{

₍ ₎ ₍ ₎

}

₂1

[

( ) ( )

ω₃ ω₃

]

π ∆ ∗∆ = ⋅ tx t x F .

(e) Using the time convolution property,

( ) ( )

t x t ← → X

( ) ( )

ω ⋅X ω x * CTFT , which reduces to

{

}

    ₊ ₋ _ω _≤ =       ≤ ω     − = ∗ ω ω ω elsewhere. 0 3 1 elsewhere 0 3 1 ) ( ) ( 3 2 9 2 3 2 t x t x F

(f) Using the time multiplication property,

( )

t ⋅cosω0t← →CTFT ₂1_πX

( )

ω ∗πδ

(

ω−ω0

)

+ ₂1_πX

( )

ω ∗πδ

(

ω+ω0

)

x , or,

( )

(

)

₂1

(

₀

)

0 2 1 CTFT 0 cosω ← → ω−ω + ω+ω ⋅ t X X t x

Case I: For ω0 = 3/2, we obtain

( )

(

)

(

₂3

)

2 1 2 3 2 1 CTFT ) 2 / 3 cos( ← → ω− + ω+ ⋅ t X X t x .

(15)

{

}

       ≤ ω ≤ + ≤ ω ≤ − − ≤ ω ≤ − + = _ω₋ + ω elsewhere. 0 1 ) 2 / 3 cos( ) ( 2 9 2 3 63/2 2 1 2 3 2 3 2 3 2 9 63/2 2 1 t t x F

Case II: For ω0 = 3, we obtain

( )

cos3

(

3

)

₂1

(

3

)

2 1 CTFT_ _ω₋ ₊ _ω₊  → ← ⋅ t X X t x .

Since there is no overlap between the two shifted replicas,

{

}

       ≤ − ω − ≤ + ω − = ω− + ω elsewhere. 0 3 3 1 3 3 1 3 cos ) ( ₃3 3 3 2 1 t t x F or,

{

}

       < ω ≤ − < ω ≤ − − = ω− + ω elsewhere. 0 6 0 0 6 3 cos ) ( ₂1 ₆3 6 3 2 1 t t x F

Case III: For ω0 = 6, we obtain

( )

cos6

(

6

)

₂1

(

6

)

2 1 CTFT_ _ω₋ ₊ _ω₊  → ← ⋅ t X X t x .

Since there is no overlap between the two shifted replicas,

{

}

       ≤ − ω − ≤ + ω − = ω− + ω elsewhere. 0 3 6 1 3 6 1 3 cos ) ( ₃6 3 6 2 1 t t x F or,

{

}

       < ω ≤ − − < ω ≤ − − = ω− + ω elsewhere. 0 9 3 3 9 3 cos ) ( 1₂ ₆6 6 6 2 1 t t x F ▌ Problem 5.16

(a) From Table 5.2, inverseCTFT 5 2 ( )

25 e u t t j − ω + ←→ .

[

t

]

j t j e u t e 5 2 CTFT inverse ) 5 ( 2+ 5ω− ←→5 − ( ) × implying that x₁(t)=5e(−2+j5)tu(t).

(16)

Using the duality property,

[

δ − +δ +

]

← → π

(

− ω

)

= π

( )

ω

π (t 2) (t 2) CTFT 2 cos 2 2 cos 2 .

[

δ( −2)+δ( +2)

]

− 12 ← →CTFT 2πcos

(

2(ω+₁₂π)

)

π t t e jπt , implying that

[

]

_  _δ ₋ ₊_δ ₊ = + δ + − δ = t t e−jπt t e−jπt t e−jπt t x ( ) ( 2) ( 2) 12 ( 2) 12 ( 2) 12 2 1 2 1 2 or,     ₋ _δ ₋ ₊ ₊ _δ ₊ =    _δ ₋ ₊_δ ₊ = − π π 3π 2 6 6 ( 2) ( 3 ) ( 2) ( 3 ) ( 2) ) 2 ( ) ( ₄1 2 1 2 t t e j t ej j t j t ej x .

(c) From Table 5.2,

( )

_←_{ →}_

( )

ω_π ₌ _πωπω _ππ ₌ sin(_ω2ω) ) 2 / 4 ( ) 2 / 4 sin( 2 4 CTFT 4 4sinc 4 2 rect t _.

Using the time scaling property,

( )

_ωω ωω ⋅ ← → ⋅ = ) 4 sin( 2 ) 4 sin( CTFT 2 4 2 2 2 rect t _.

( )

CTFT sin(₍4( ₎ )) 8 2 rect t _ejπt← → _ω₋ω_π−π _, implying that

( )

t _ej t t x = ₂5 ₈ π 3( ) rect .

(d) Using the linearity property, we obtain

{

}

{

}

{

}

1 4 1 1 (3 2 ) 10 (1 2 ) 10 2 2 ( ) (3 2 ) ( 10) (1 2 ) ( 10) (3 2 ) ( 10) (1 2 ) ( 10) . j j t j j t x t j j j j e e π π

δ ω

− − − + − − = ℑ + − + − + = + ℑ − + − ℑ + = +

Expanding the exponential terms using the Euler’s formula, we obtain ) 10 sin 10 (cos ) 10 sin 10 (cos ) ( (3₂2 ) (1₂2 ) 4 t t j t t j t x = +_πj + + −_πj −

or, x (t) 2cos10t (2 j)sin10t

4 =_π − _π− .

(e) Taking the partial fraction expansion

( )

₍₁ ₎₍₃ 1 ₎ ₍₅ ₎ ₍₁ ₎ ₍₃ ₎ ₍₃ ₎ ₍₅ ₎ 5 2 2 2 ₊_ω2 + ω ω + ω + ω + ω + ω + ω + ≡ + + + = ω jD E j C j B j A j j X where A=0.0625, B=0.25, C=0.125, D=−0.3125, and E=0.6876. Calculating the inverse CTFT transform yields

( )

() 3 () 3 () cos( 5 ) ()

(

/ 5

)

sin( 5 ) ()

5 t Ae u t Be u t Cte u t D t u t E t u t

(17)

Problem 5.17

(i) The functions are plotted in Fig. S5.17. The MATLAB code used to generate the plots is given below. % MATLAB code to plot the functions in Problem 5.17

t = -10:0.01:10 ;

t4 = 0:0.001:10000 ; % for plotting x4(t) t = t + eps;

t4 = t4 + eps; %for plotting x4(t) % x1 = exp(-2*abs(t)); % a=2 x2 = exp(-2*t).*(cos(5*t)).*(t>=0); % a=2, w=5 x3 = (t.^4).*exp(-2*t).*(t>=0); % a=2 x4 = sin(log(t4)); x5 = 1./t ; x6 = cos(pi./(2*t)) ; x7 = exp(-(t.^2)./(2*3*3)) ; % sigma=3 %

subplot(4,2,1), plot(t, x1), grid

xlabel('t') % Label of X-axis ylabel('x1(t)') % Label of Y-axis axis([-5 5 0 1.3])

%

xlabel('t') % Label of X-axis ylabel('x2(t)') % Label of Y-axis % axis([-5 5 -0.5 1.3])

%

xlabel('t') % Label of X-axis ylabel('x3(t)') % Label of Y-axis % axis([-4 4 0 0.4])

%

subplot(4,2,5), plot(t4, x4), grid

xlabel('t') % Label of X-axis ylabel('x4(t)') % Label of Y-axis % axis([0.001 10000 -1.3 1.3])

%

xlabel('t') % Label of X-axis ylabel('x5(t)') % Label of Y-axis axis([-1 1 -100 100])

%

xlabel('t') % Label of X-axis ylabel('x6(t)') % Label of Y-axis % axis([-5 5 -1.3 1.3])

%

xlabel('t') % Label of X-axis ylabel('x7(t)') % Label of Y-axis % axis([-5 5 0 1.3])

(18)

-5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.5 1 t x1 (t ) -5 -4 -3 -2 -1 0 1 2 3 4 5 -0.5 0 0.5 1 t x2 (t ) -4 -3 -2 -1 0 1 2 3 4 0 0.1 0.2 0.3 0.4 t x3 (t ) 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 -1 -0.5 0 0.5 1 t x4 (t ) -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -100 -50 0 50 100 t x5 (t ) -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -0.5 0 0.5 1 t x6 (t ) -5 -4 -3 -2 -1 0 1 2 3 4 5 0 0.5 1 t x7 (t )

Fig. S5.17: Time-domain Waveforms for Problem 5.17

(ii) (a)

[

] [

]

0 0 1 1 1 1 2 ( ) ₀ 0 1( ) a t at at at at 1 0 0 1 . a a a a a x t dt e dt e dt e dt e e ∞ ∞ ∞ _∞ − − − − −∞ −∞ −∞ −∞ = = + = + = − − − = < ∞

∫

Since the condition in Eq. (5.59) is satisfied, the CTFT for x1(t) exists.

(b) 1 0 0 1 ( 0) 1 ( 0) 0 2 2 2 0 0 0 2( ) atcos( ) ( ) at j t j t a j t a j t I II x t dt e ω t u t dt e eω e ω dt e ω dt e ω dt ∞ ∞ ∞ ∞ ∞ − − − − + − − −∞ −∞   = = _ + _ ≤ +

∫

Integral I is given by

[

( )

] [

₀ 21 ( 1 )

] [ ]

21 ( 1 ) 2 1 0 ) ( 2 1 0 0 0 ) 0 ( 0 0 ω − ω − − ∞ ω − − ∞ ω − − ₌ ₌ ₋ ₌ = − −ω

∫

ea j a j a j t j a a j t dt e I ,

(19)

[

( )

]

₀ 12

[

( 1 )

] [

21 ( 1 )

]

2 1 0 ) ( 2 1 0 0 0 ) 0 ( 0 0 ω + ω + − ∞ ω + − ∞ ω + − ₌ ₌ ₋ ₌ =

∫

a j t e−_aa+j_jω t _a _j _a _j dt e II . Therefore, . ) ( ₂ 0 2 0 0 1 ) ( 1 2 1 ) ( 1 2 1 4 = + = −ω + +ω = ₊_ω <∞ ∞ ∞ −

∫

a j a j a II I dt t x

Since the condition in Eq. (5.59) is satisfied, the CTFT for x2(t) exists.

(c) 2 3 4 5 4 4 4 3 2 ( ) ( ) ( ) ( ) ( ) ₀ 0

3( )

_at

( )

_at _eat

4

_eat

12

_eat

24

_eat

24

_eat a a a a a

x t dt

t e u t dt

t e dt

t

−

t

−

t

−

t

− − ∞ ∞ ∞ _∞ − − − − − − − −∞ −∞





=

_

+

_

∫

[

]

5 5 1 24 ( )

0 0 0 0 0



0 0 0 0 24

₋_a



_a

.

= + + + + − + + + +

_

_

= < ∞

Since the condition in Eq. (5.59) is satisfied, the CTFT for x3(t) exists.

(d) The function

4( ) sin(ln( )) ( )

x t = t u t

is plotted in Fig. S5.17. Note that the horizontal axis uses a logarithmic scale. It is observed that the function oscillates like a sine wave (although not with a constant period). Therefore, the function has an infinite number of maximas and minimas. In addition,

0 4( ) sin(ln( )) x t dt t dt ∞ ∞ −∞ = →∞

∫

Therefore, the CTFT for x4(t) does not exist.

(e) 1 1

[

]

0 0 5( ) _t 2 _t 2 ln( ) x t dt dt dt t ∞ ∞ ∞ ∞ −∞ −∞ = = = →∞

∫

Since the condition in Eq. (5.59) is not satisfied, the CTFT for x5(t) does not exist.

(f) _{6( )} _cos

( )

/ 2 _. t x t dt π dt ∞ ∞ −∞ −∞ = → ∞

∫

Clearly the area enclosed by the cosine term would be infinite. Since the condition in Eq. (5.59) is not satisfied, the CTFT for x6(t) does not exist. Also, it can be checked that x6(t) has an infinite number of maximas and minimas, which is a second violation of the existence of the CTFT.

(g) 2 2 2 2 2 2 7( ) exp( t ) exp( t ) 2 x t dt _σ dt _σ dt π σ ∞ ∞ ∞ −∞ −∞ −∞ = − = − = < ∞

∫

.

In evaluating the above result, we used the fact that the area enclosed by a bell curve is 1. Mathematically, this implies that

[ ]

exp

[ ]

1 exp 2 2 2 2 2 ) ( 2 1 2 2 1 ₌

∫

₌

∫

∞ ∞ σ − σ π ∞ ∞ πσ σ dt dt t m t _,

(20)

Problem 5.18

(a) From the solution of Problem P4.11(a), the CTFS coefficients Dn of the rectangular pulse train are obtained as       ≠ = = π odd , 0 , even 0 0 3 2 3 n n n n D jn n

with fundamental frequency ω0 = 1 radians/s. Therefore, the CTFT is given by

∑

∞ −∞ = ∞ −∞ = − ω δ − ω πδ = ω − ω δ π = ω n n n n n j n D X odd 0) 3 ( ) 6 ( ) ( 2 ) ( 1 .

(b) From the solution of Problem P4.11(b), the CTFS coefficients Dn of the rectangular pulse train are obtained as     ≠ π − = = πsin(0.5 ) 0 0 5 . 0 4 3 n n n D n n

with fundamental frequency ω0 = π/T radians/s. Therefore, the CTFT is given by

∑

∞ ≠−∞ = π ∞ −∞ = − ω δ π − ω πδ = ω − ω δ π = ω 0 1 0) 1.5 ( ) sin(0.5 ) ( ) ( 2 ) ( 2 n n T n n n n n n D X .

(c) From the solution of Problem P4.11(c), the CTFS coefficients Dn of the rectangular pulse train are obtained as     ≠ = = π, 0. 0 , 21 2 1 n n D n j n

with fundamental frequency ω0 = 2π/T radians/s. Therefore, the CTFT is given by

∑

∞ ≠−∞ = π ∞ −∞ = − ω δ − ω πδ = ω − ω δ π = ω 0 2 1 0) ( ) ( ) ( 2 ) ( 3 n n T n n n n n j D X .

(d) From the solution of Problem P4.11(d), the CTFS coefficients Dn of the rectangular pulse train are obtained as        ≠ ≠ = = π odd , 0. 0 , even , 0 0 , 2 ) ( 2 2 1 n n n n n D n n

(21)

∑

∞ ≠−∞ = π π ∞ −∞ = − ω δ + ω πδ = ω − ω δ π = ω n n n T n n n n n D X odd0 1 4 0) ( ) ( ) ( 2 ) ( 4 2 .

(e) From the solution of Problem P4.11(e), the CTFS coefficients Dn of the rectangular pulse train are obtained as

(

)

(

)

2 1 1 2 1 8 1 2 ( 1) 1 1 0.3408 0 1 0 0.1933 1 0 1 n n jn n n j j n D n even n odd π π π π − =  − =  − = ±  =_ = ≠ =   _{± ≠ =}  ∓ ∓ 2 0.1592 1 0.3183 1 0 1 n n j n n even n odd − −   = ±   _{≠ =}   ± ≠ = 

with fundamental frequency ω0 = π/T radians/s. Therefore, the CTFT is given by

∑

∞ ≠−∞ = π ∞ ≠−∞ = π − π π ∞ −∞ = − ω δ − − ω δ + − ω πδ − + ω πδ + ω πδ = = ω − ω δ π = ω n n n T n n n n n T n n T T n n j j j n D X odd0,1 1 even 0,1 1 1 0 ) ( 2 ) ( ) ( 3866 . 0 ) ( 3866 . 0 ) ( 6816 . 0 6816 . 0 ) ( 2 ) ( 5 2 . ▌ Problem 5.19

(a) From the solution of Problem P5.2(a), the CTFT of the aperiodic signal is given by

    ≠ ω − = ω π = ω π = ω ₋ _ωπ ω ωπ − . 0 ) 1 ( 0 3 ) 2 / ( sinc 3 ) ( /2 ₃ 1 j j j e e X

The signal shown in Fig. P4.6(a) is a periodic signal with a fundamental period T0 = 2π with one period matching the function shown in Fig. P5.2(a). The fundamental frequency ω0 = 1 and the exponential CTFS coefficients are given by

    ≠ − = =     ≠ − = π = ω = ₋ _π π π ω − ω π ω = ω (1 ) 0 0 0 ) 1 ( 0 3 ) ( 23 2 3 3 21 1 1 0 0 0 0 e n n n e n X D _jn n j jn jn n T n which simplifies to       ≠ = = = _π . 0 , even 0 odd 0 3 2 3 n n n n D_n _jn

(b) From the solution of Problem P5.2(b), the CTFT of the aperiodic signal is given by

( )

₍

₎

    ≠ ω + = ω = + = ω ₋_ω ω ω πω ω − πω ₁ ₂ ₀_. 0 5 . 1 sinc sinc 5 . 0 ) ( 0.5 T 0.5 T _sin( _/₂₎ 2 T _j _T T j e T Te T X

(22)

The signal shown in Fig. P4.6(b) is a periodic signal with a fundamental period T0 = 2T with one period matching the function shown in Fig. P5.2(b). The fundamental frequency ω0 = π/T and the exponential CTFS coefficients are given by

(

)

_ ≠ − = =     ≠ ω + = ω = ω = ₋ _π π π ω − ω ω ω = ω ₍₁ ₎ ₀ 0 0 2 1 0 5 . 1 ) ( 2 ) 2 / sin( 4 3 ) 2 / sin( 21 2 1 0 0 0 0 0 e n n e T X D _jn n n T jn n T n T n T n which simplifies to        ≠ + = + = − = = = π π . 0 , even 0 3 4 1 4 0 21 21 4 3 n n k n k n n D n n n

(c) From the solution of Problem P5.2(c), the CTFT of the aperiodic signal is given by

(

)

    ≠ ω − + = ω = ω ₋ _ω ω ω 1 0. 0 5 . 0 ) ( 2 1 1 3 _j _T T j e T X

The signal shown in Fig. P4.6(c) is a periodic signal with a fundamental period T0 = T with one period matching the function shown in Fig. P5.2(c). The fundamental frequency ω0 = 2π/T and the exponential CTFS coefficients are given by

(

)

_

(

)

 ≠ − − = =     ≠ ω − − = ω = ω = ₋ _π π π ω − ω ω ω = ω ₁ ₀ 0 5 . 0 0 1 0 5 . 0 ) ( ₂ 4 1 21 1 1 1 3 1 2 2 0 2 0 2 0 0 0 e n n e T X D _j _n n n j T jn T n jn T n T n which simplifies to     ≠ = = π 0. 0 5 . 0 21 n n D n j n

(d) From the solution of Problem P5.2(d), the CTFT of the aperiodic signal is given by

    ≠ ω = ω = = ω πω πω _sinc ₍ ₎ ₀_. 0 ) ( sinc ) ( 2 0.5 ₂ ₀_.₅ 3 T _T T T T X

The signal shown in Fig. P4.6(d) is a periodic signal with a fundamental period T0 = 2T with one period matching the function shown in Fig. P5.2(d). The fundamental frequency ω0 = π/T and the exponential CTFS coefficients are given by

   ≠ = =     ≠ ω = ω = ω = π ω ω = ω 0.5sinc (0.5 ) 0 0 5 . 0 . 0 ) ( sinc 0 ) ( ₂1 ₂ ₀_.₅ ₂ 3 1 0 0 0 n n n T T X D _T _n _T n T n which simplifies to       ≠ ≠ = = π odd , 0. 0 , even 0 0 5 . 0 2 ) ( 2 _n _n n n n D n n

(23)

(e) From the solution of Problem P5.2(e), the CTFT of the aperiodic signal is obtained as 2 2 2 1 5 4 0.5 2 1 (1 ) 0 ( ) 1 1 T j T j T T j T T j j T T X e e otherwise π π ω π ω π ω π ω ω ω ω − − − − = = ± = ±  − −  +      ∓     

The signal shown in Fig. P4.6(e) is a periodic signal with a fundamental period T0 = 2T and whose one period is identical to the function shown in Fig. P5.2(e). Therefore, the fundamental frequency ω0 = π/T and the exponential CTFS coefficients are given by

0 0 0 0 2 0 0 1 1 0 2 0 1 0.5 1 2 2 0.5 ( ) 1 1 ( ) ( ) (1 ) 0 1 1 1 n T T j T T T j jn T T n j jn D X n X n T n e n e π ω ω π π ω ω ω ω ω − − = = − =   = × ± _ − _ = ±  − −   ∓ _∓ 0 2 2 1 0.5 1 2 2 1 (1 ) 0 1 1 1 jn T T j T T j j jn T T e otherwise T n e n e ω π π π π − −     _ _ +     − =   = × ± _ − _ = ± − ∓ _∓ 2 2 2 0.5 1 1 2 2 0.5 2 1 1 0 jn T jn n T j j e otherwise n π π π π π π π − −     _ _ _ _ − +       − = = ± ∓

(

)

(

)

2 1 4 ( 1) 1 1 2 1 8 21 1 1 1 ( 1) 1 ( 1) 1 0 n n n j n n otherwise n j π π π π −   _{= ±}   _ _{− −} _₊ _ _{+ −} _       − = − = ∓ 2 1 2 ( 1) 1 1 0 1 n jn n n even n odd π π −   = ±   ≠ =   _{± ≠ =}  ▌ Problem 5.20

(a) Calculating the CTFT of both sides and applying the time differentiation property, yields

( ) ( ) ( ) ( )

3 2

( ) ( )

( )

6

11

6 j

ω

Y

ω

+

j

ω

Y

ω

+

j

ω

Y

ω

+

Y

ω

=

X

ω

, or,

(

( )

j

ω

3+6

( )

j

ω

2+11

( )

j

ω

+6

)

Y

( )

ω

=X

( )

ω

, or,

( )

( ) ( )

3

( )

2

( )

1

6

11

6 Y

H

X

j

ω

=

+

.

The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as ) 3 ( 5 . 0 ) 2 ( 1 ) 1 ( 5 . 0 ) 3 )( 2 )( 1 ( 1 ) ( ω + + ω + − + ω + ≡ ω + ω + ω + = ω j j j j j j H

(24)

Calculating the inverse CTFT, we obtain ) ( 5 . 0 ) ( ) ( 5 . 0 ) (t e u t e 2u t e 3u t h = −t − − t + − t .

(b) Calculating the CTFT of both sides and applying the time differential property, yields

( ) ( ) ( ) ( )

jω 2Y ω +3 jωY ω +2Y

( )

ω =X

( )

ω , or,

(

( )

jω 2 +3

( )

jω +2

)

Y

( )

ω =X

( )

ω , or,

( )

( ) ( )

3

( )

2 1 2 ₊ _ω ₊ ω = ω ω = ω j j X Y H .

The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as

( )

(2 ) 1 ) 1 ( 1 2 3 1 ) ( ₂ ω + − ω + ≡ + ω + ω = ω j j j j H

) ( ) ( ) (t e u t e 2 u t h = −t − −t .

(c) Calculating the CTFT of both sides and applying the time differentiation property, yields

( ) ( ) ( ) ( ) ( )

jω 2Y ω +2 jωY ω +Y ω =X

( )

ω , or,

(

( )

jω 2 +1

( )

jω +1

)

Y

( )

ω =X

( )

ω , or,

( )

( ) ( )

2

( )

1 1 2 ₊ _ω ₊ ω = ω ω = ω j j X Y H .

The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as

(

₁

)

2 1 ) ( ω + = ω j H

) ( )

(t te u t h = −t .

(d) Calculating the CTFT of both sides and applying the time differentiation property, yields

( ) ( ) ( ) ( )

jω 2Y ω +6 jωY ω +8Y

( ) ( ) ( )

ω = jω X ω +4X

( )

ω , or,

(

( )

jω 2 +6

( )

jω +8

)

Y

( ) ( )

ω =

(

jω +4

) ( )

X ω , or,

( )

_{( )}

( )

_ω ₊

( )

_ω ₊ = + ω + ω = ω ω = ω j j j j X Y H 2 1 8 6 4 2 .

The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which is given by