Solution manual fundamentals of electric circuits 3rd edition chapter18

Find the Fourier transform of the waveform shown in Fig... Obtain the Fourier transforms of the signals shown in Fig... Determine the Fourier transforms of the signals in Fig... Obtain t

f = δ + − δ + − δ − + δ −

2 j j j 2

e ) ( F

j ω ω = ω− ω − − ω + − ω = 2 cos 2 ω − 2 cos ω

F(ω) =

ω

ω

− ω

j

] cos 2

[cos 2

1 t 0 , ) t (

or ω = ∫1 − ω

0 t

j dt e t ) (

F

But ∫ = ( ax − 1 ) + c

a

e dx e

ax ax

=

0 2

j

) 1 t j ( j

e )

(

2 + ω − ω

1 ) t ( ' f , 2 t 2 , 2

1 ) t ( = − < < = − < <

t j t

) j ( 2

e dt e t 2

1 ) ( F

12 F(ω) = j2( 2 ω cos 2 ω − sin 2 ω )

ω

Chapter 18, Problem 4

Find the Fourier transform of the waveform shown in Fig 18.29

Figure 18.29

For Prob 18.4

Chapter 18, Solution 4

4 sin 4 cos 4

e j 2 e

2 4 e j 2 e 2 ) ( G ) j (

) 1 t ( 2 ) 1 t ( 2 ) t ( 4 ) 1 t ( 2 ) 1 t ( 2 g

jj

jj

2

+ ω ω

− ω

−

= ω ω

− δ′

−

− δ

− δ + + δ′

+ + δ

−ω

ω

) 1 sin (cos

4 ) (

ω

= ω

t

0

2 g’

–2 –1

–2 –1

2δ’(t+1)

–2δ(t–1)

1 –2δ(t+1)

–2δ’(t–1)

− ω

= ω

−

−

= ω ω

δ′

−

− δ

− + δ

) t ( 2 ) 1 t ( ) 1 t ( ) t ( h

jj

h”(t)

0 1

–δ(t–1) δ(t+1)

t

1

–2δ(t) –1

h’(t)

0 1

Find the Fourier transforms of both functions in Fig 18.31 on the following page

10δ(t)

0 1 2 -5

Take the Fourier transform of each term

(jω)2G(jω) = 10jω – 5jωe–jω – 5e–jω + 5e–j2ω which leads to

G(jω) = (–10jω + 5jωe–jω + 5e–jω – 5e–j2ω )/ω2

Chapter 18, Problem 7

Find the Fourier transforms of the signals in Fig 18.32

2 ( )

Obtain the Fourier transforms of the signals shown in Fig 18.33

j2

21tj1

0tj

21

tj1

0

tj

) 1 t j ( e

2 e

j

4 e

j

2

dt e ) t 2 4 ( dt e 2

)

(

F

− ω

− ω

−

− ω

−

+ ω

−

=

− +

=

ω

ω

−ω

−ω

−

ω

−ω

−ω

ω

− ω

− ω

+ ω

+ ω

=

j

4 j

2 e

j

2 2 ) ( F

(b) g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]

ω

ω

− ω

ω

=

ω ) 4 sin 2 2 sin (

G

Chapter 18, Problem 9

Determine the Fourier transforms of the signals in Fig 18.34

=

ω ) 2 sin 2 4 sin (

Y

(b) Z ( ) ( 2 t ) e dt 2 e ( j t 1 ) 2 2 e ( 1 j )

2

j2

101

tjt

ω

− ω

=

− ω

− ω

Obtain the Fourier transforms of the signals shown in Fig 18.35

0 t , e e

t

t )

ω

−

− ω

1 0

t j t t

j t t

e ) t ( y ) ( Y

0

t j 1 ( 0 1

t j 1 (

) j 1 (

e j

1

e

ω +

−

+ ω

ω

− ω + ω

−

ω + ω

− ω +

j 1

sin j cos j

1

sin j cos e 1

+

−

Chapter 18, Problem 11

Find the Fourier transform of the “sine-wave pulse” shown in Fig 18.36

t j t j t j t

j 2

1 dt e t sin )

e ( j 2 1

−

+ π

− ω

−

0

t ) ( j 2 0 t ) ( j

) ( j

e e

) ( j

1 j

2 1

= ⎜⎜ ⎝ ⎛ π − − ω + − π + ω ⎟⎟ ⎠ ⎞

ω

− ω

− j 2 1 e j 2

e 1 2 1

( π + π − ω)

ω

− π

2

) (

2 1

Chapter 18, Problem 12

Find the Fourier transform of the following signals

j

ω ω

Find the Fourier transform of the following signals:

(a) We know that F [cos at ] = π [ δ ( ω − a ) + δ ( ω + a )]

Using the time shifting property,

) a ( e ) a ( e

)]

a ( ) a ( [ e

)]

a 3 / t ( a

[cos − π = π −jωπ/ a δ ω − + δ ω + = π −jπ/3δ ω − + π jπ/3δ ω + F

(b) sin π ( t + 1 ) = sin π t cos π + cos π t sin π = − sin π t

g(t) = -u(t+1) sin (t+1)

Let x(t) = u(t)sin t, then 2 2

1

1 1 ) j (

1 )

( X

ω

−

= + ω

= ω Using the time shifting property,

1

e e

1

1 )

(

G

2

jj

)]

b ( Y ) b (

1 t j (

e j

e dt e ) t 1

j2

402

tj4

0

tjt

ω

− ω

− ω

=

− ω

− ω

−

− ω

Find the Fourier transforms of these functions:

(a) f(t) = e−t cos(3t + π )u(t)

(b) g(t) = sin π t[u(t + 1) – u(t – 1)]

(c) h(t) = e− 2tcos π tu(t – 1)

(d) p(t) = e− 2tsin 4tu(–t)

(e) q(t) = 4 sgn(t – 2) + 3 δ (t) – 2u(t – 2)

Chapter 18, Solution 14

(a) ) cos( 3 t + π ) = cos 3 t cos π − sin 3 t sin π = cos 3 t ( − 1 ) − sin 3 t ( 0 ) = − cos( 3 t

) t ( u t 3 cos e ) t ( = − − tF(ω) = ( )

j 1

2 + ω +

ω +

−

(b)

[ u ( t 1 ) u ( t 1 ) ]

t cos )

− ω π

−

= ω ω

ω π

Alternatively, we compare this with Prob 17.7

π

= ω

=

2 2

e ) ( F G

j 22 sin2

π

− ω

ω π

ω π

(c) t cos π ( t − 1 ) = cos π t cos π + sin π t sin π = cos π t ( − 1 ) + sin π t ( 0 ) = − cos π Let x ( t ) = e− 2 t − 1 )cos π ( t − 1 ) u ( t − 1 ) = − e2h ( t )

and y ( t ) = e− 2 tcos( π t ) u ( t )

2 2) j 2 (

j 2 )

( Y

π + ω +

ω +

= ω

) 1 t ( x ) t (

ω

−ω

=

ω ) X ( ) e j(

Y

jj 2

e j 2 ) ( X

π + ω +

ω +

=

) ( H e ) (

X ω = − 2 ω

) ( X e ) (

H ω = − − 2 ω

2 jj 2

e j 2

π + ω +

ω +

(d) Let x ( t ) = e− 2 tsin( − 4 t ) u ( − t ) = y ( − t )

) t ( x ) t (

p = − where ) y ( t ) = e2 tsin 4 t u ( t

4 j

2

j 2 )

( Y

+ ω +

ω +

= ω

j 2 )

( Y ) (

+ ω

−

ω

−

= ω

−

= ω

= ω

−

=

ω ) X ( ) (

p

2 j

2 +

− ω

− ω

j

1 ) ( 2 3 e

j

8 ) (

− + ω

= ω

Q(ω) = ej 2 3 2 ( ) e j 2

j

6 ω + − πδ ω − ωω

Chapter 18, Problem 15

Find the Fourier transforms of the following functions:

) ( ) 0 ( F j

) ( G

ω δ π + ω

ω

=

) ( ) 1 ( 2 j

e

2 j

ω δ

− πδ + ω

= ω

1 ) ( F

2

j 3

−

Chapter 18, Problem 16

* Determine the Fourier transforms of these functions:

ω

−

→

ω π

→

−

2 t

4

ω π

− 4

(b) e−at 2 2

a

a 2 ω +

e 4

Chapter 18, Problem 17

Find the Fourier transforms of:

(a) cos 2tu(t) (b) sin 10tu(t)

Chapter 18, Solution 17

(a) Since H(ω) = F ( 0 ) [ F ( 0) ( F 0) ]

2

1 ) t ( t

where F(ω) = F [ ] ( ) ( ) , 2

j

1 t

ω + ω πδ

− ω πδ + + ω + + ω πδ

=

ω

2 j

1 2

2 j

1 2

2

1 H

− ω + + ω

−

− ω δ + + ω δ

π

=

2 2

2 2

2

j 2 2

2

4

j 2 2

ω

−

− ω δ + + ω δ π

(b) G(ω) = F [ 0 ] [ F ( 0) ( F 0) ]

2

j ) t ( t

where F(ω) = F [ ] ( ) ( )

ω + ω πδ

=

j

1 t

u

( ) ω = ⎢ ⎣ ⎡ πδ ( ω + ) ( + ω + ) − πδ ( ω − ) ( − j ω − 10 ) ⎥ ⎦ ⎤

1 10

10 j

1 10

2

j G

j 2

j 10 10

10 2

j

2 − ω

−

− ω δ

− + ω δ π

Chapter 18, Problem 18

Given that F( ω ) = F[f(t)], prove the following results, using the definition of Fourier

Find the Fourier transform of

f(t) = cos 2 π t[u(t) – u(t – 1)]

− π

0

t 2 j t 2 j t

j

( ) ω = ∫1[ − ( ω + π ) + − ( ω − π ) ]

0

t 2 j t 2

e 2

1 F

1 0

t 2 j t

2

2 j

1 e

2 j

1 2

−

+ π

+ ω

− +

π + ω

−

−

= − ω+ π − ω− π

2 j

1 e

2 j

1 e

+ π + ω

1 j

1 e 2

1 F

Chapter 18, Problem 20

(a) Show that a periodic signal with exponential Fourier series

t jn

n c

(b) T = 2 π 1

T

2

o = π = ω

jn

2

1 dt e t f T

1

( e 1 )

n 2

j e

jn

1 2

≠

= π

−, n even0

0 n , odd n , n j

=

0 n

2

1 dt 1 2

1 c

jnt

e n

j 2

1 )

− δω

odd

n 0 n n

n n

j 2

1

Chapter 18, Problem 21

Show that

∫−∞∞

2sin

1 dt ) t (

ω π

1 a 2 dt ) 1 ( dt ) t ( f

22

aa22

or

a a 4

a 4 d a

a sin

e e ) t ( t

sin ) t

o

o o

− ω

− ω

e ) t ( j 2

= [ F ( o) ( F o) ]

j 2

1 ω − ω − ω + ω

Chapter 18, Problem 23

If the Fourier transform of f(t) is

F( ω ) =

) 5 )(

/ j 5 3 / j 2

10 3

/ j 15 2 / j 2

10 2

1

+ ω + + ω

5 2

j 5 2 j 2

5

− ω +

− ω +

+ + ω + + ω +

(d) F [ ] ( ) ( ) ( + ω )( + ω )

ω

= ω ω

=

j 5 j 2

10 j F

j t ' f

t

dt t

( ) ω + π ( ) ( ) δ ω

ω

0 F j

F

10 x j

5 j 2 j

ω + ω + ω

=

= j ω ( 2 + j ω )( 5 + j ω ) + πδ ( ) ω

10

Chapter 18, Solution 24

(a) X ( ) ( ) ω = F ω + F [3]

ω + ω

(b) y ( ) ( t = f t − 2 ) ( ) ω = e− ω F ( ) ω =

Y j 2 je j 2 ( e j − 1 )

ω

ω

− ω

−

(c) If h(t) = f '(t)

ω ω

= ω ω

3 x 10 2

3 F 2

3 x 4 ) ( G , t 3

5 f 10 t 3

2 f 4 t g

5 3

j 6 1 e

2 3

⋅

= j 4 ( e j 3 / 2 1 ) j 10 ( e j 3 / 5− 1 )

ω +

− ω

ω

− ω

−

1 (

1 +

⎯→

⎯

−

− +

1 ( u 2 ) 1 (

− + ω

=

ω

Chapter 18, Problem 27

Find the inverse Fourier transforms of the following functions:

(a) F(ω ) =

) 10 (

100 +

2 (

10

+ +

ω

j j

j

(c) H(ω ) =

1300 40

1 (

) ( +

ω

ω δ

j j

Chapter 18, Solution 27

+ +

= +

10 s

B s

A 10 s s

100 s

F

10 10

100 B , 10 10

10 j

10 F

+ ω

− ω

= ω f(t) = 5 sgn ( ) t − 10 e−10 tu ( ) t

−

3 s

B s 2

A s 3 s 2

s 10 s

G

6 5

30 B

, 4 5

20

( )

3 j

6 2 j

4 G

+ ω

− + ω

−

=

= ω g(t) = 4 e2 tu ( ) − t − 6 e−3 tu ( ) t

(c) ( )

60 1300

40 j j

60

+ + ω

= +

ω + ω

= ω

= + ω ω +

ω ω

δ π

=

2

1 2

1 1 j j 2

d e 2

1 t y

t j

π

4 1

Chapter 18, Problem 28

Find the inverse Fourier transforms of:

(a)

) 2

)(

5

(

) (

ω ω

(

) 2

)(

2

(

) 1 (

20

ω ω

5

ω

ω πδ π

= ω ω

π

) j 2 )(

j 5 (

e ) ( 2

1 d e ) ( F 2

1 )

t

(

t j t

5

(

1 2

1

0.05

− ω

+

−

− π

= ω +

ω ω

+ ω δ π

=

) 1 2 j )(

2 j (

e 2

10 d e ) 1 j ( j

) 2 ( 10 2

1 )

t

(

t 2 j t

j

=

− π

2 j 1

e 2

ω

+ + π

= ω ω + ω +

− ω δ π

=

) j 3 )(

j 2 (

e 2

20 d

) 5 3 )(

j 2 (

e ) 1 ( 20 2

1 )

t

(

jt t

j

= + π

=

) j 5 5

− ) ejt1 (

) j 5 ( j

5 )

j 5 (

) ( 5 ) (

ω + ω

+ ω +

ω πδ

= ω

+

ω πδ π

5

1 2

5 d e j 5

) ( 5 2

1 ) t (

1

1 B , 1 A , 5 s

B s

A ) s 5 ( s

5 ) s (

+ +

= +

=

5 j

1 j

1 ) (

F2

+ ω

− ω

= ω

t 5 t

5

2

1 e

) t sgn(

2

1 ) t (

= +

= f ( t ) f ( t ) )

t ( 1 2 u ( ) − e−5 t

= 4 cos 3 t 2

(

t

t 2 sin 8 2

1 ) t (

π

= g(t) =

t

t 2 sin 4

π (c) Since

cos(at) πδ ( ω + a ) + πδ ( ω − a )

Using the reversal property,

) 2 t ( ) 2 t ( 2

cos

2 π ω ↔ πδ + + πδ −

or F -1[ 6 cos 2 ω ] = 3 δ t + 2 ) + 3 δ t − 2 )

= ω ω

= ω

⎯→

⎯

=

j a

1 ) ( X , j

2 ) ( Y )

t sgn(

a 2 2 j

) j a ( 2 ) ( X

) ( Y )

(

ω +

= ω

ω +

= ω

= ω ω

+

=

ω

j 2

1 ) ( Y , j 1

1 )

(

X

) t ( u e ) t ( ) t ( h j

2

1 1 j 2

j 1 )

(

ω +

−

= ω +

ω +

=

ω

(c) In this case, by definition, h ( t ) = y ( t ) = e−atsin bt u ( t )

= ω ω

+

= ω

j a

1 ) ( H , ) j a (

1 )

(

Y

2

) t ( u e ) t ( x j

a

1 ) ( H

) ( Y ) (

ω +

= ω

ω

= ω

(b) By definition, x ( t ) = y ( t ) = u ( t + 1 ) − u ( t − 1 )

(c )

ω

= ω ω

+

= ω

j

2 ) ( H , ) j a (

1 )

( Y

) t ( u e 2

a ) t ( 2

1 ) t ( x )

j a ( 2

a 2

1 ) j a ( 2

j )

−

= ω +

ω

= ω

ω

=

ω

Chapter 18, Problem 32

* Determine the functions corresponding to the following Fourier transforms:

(a) F1( ω ) =

1 +

2 1

) (

e j

+ ω

ω

−

e−( )t − 1u ( t − 1 ) and F ( ) − ω f(-t)

( )

1 j

e F

2

2 +

ω

−πe 2

If F2( ) ω = 2 e−ω, then

f2(t) = ( t 1 )

2

2 + π (d) By partial fractions

4 1 1

j 4 1 1

j 4 1 1

j 4 1 1

j 1 j

1

− ω

−

− ω

+ + ω

+ + ω

=

− ω + ω

= ω

Hence ( ) ( te e te e ) u ( ) t

4

1 t

ω +

ω δ π

= ω ω

π

=

2 j 1

e 2

1 d e F 2

1 t

* An asterisk indicates a challenging problem

ω π

= ω Applying duality property,

t

t sin j 2 t X 2

1 t f

− π

−

=

− π

= f(t) = 2 2

t

t sin j 2 π

−

ω

− ω

− ω ω

=

− ω

j

e j

e e

e

j 2 j

2

1 1 t sgn 2

1 t

1 1 t u t

= u ( t − 1 ) ( − u t − 2 )

t 2

1 –2 –1

20

0

g ‘(t)

t 2

e e

3 / j 2

1 3

1 ) (

ω +

= ω

+ + ω +

=

− ω + + ω

=

ω

) 5 ( j 2

1 )

5 ( j 2

1 2

1 )]

5 ( F ) 5 ( F [

ω

= ω ω

= ω

j 2

j ) ( F j ) ( Z

) j 2 (

1 )

( F ) ( F ) ( H

ω +

= ω ω

= ω

(e)

2

1 )

j 2 (

) j 0 ( j ) ( F d

d j ) (

ω +

= ω +

−

= ω ω

= ω

If the input signal to the circuit is vs(t) = t

e−4 u(t) V find the output signal Assume all

initial conditions are zero

=

ω

j 2

2 j j

2

By current division,

( ) ( ) ( )

ω + + ω

ω

= ω +

ω +

ω +

ω

= ω

ω

=

ω

4 j 8 2 j

2 j j

2

2 j 4

j 2

2 j I

I H

s o

H(ω) =

ω +

ω

3 j 4 j

j

1 dt e ) t 1 ( ) ( V

+ ω ω +

= ω

10 10

x j

10

) ( V

)

(

v && = δ − δ − + δ −

2 j j

2V ( ω ) = 1 − 2 e−ω + eωω

= ω +

= ω

j

2 j 1 j

1 2 Z

( )

ω

⋅ ω

−

−

= ω

ω

2 j 1

j 1 e e 2 Z

V

2 j j

ω + ω

= 0 5 0 5 e j 2 e j

j 5 0 j 1

But

5 0 s

B s

A ) 5 0 s ( s

1

+ +

=

ω +

−

− +

ω

=

j 5 0

2 e

e 5 0 5 0 j

2 I

i(t) = sgn( t 2 ) sgn( t 1 ) e u ( ) e u t 2 ) 2 e u t 1 )

2

1 ) sgn(

ω + ω

= + ω

− ω

=

− ω + ω + ω +

−

j 2

9 j 4 j

2

j 4 j V 4 2 j

0 2 j

V 2 V j 2

j 2

1 V

22

V(ω) =

) j 2 4 )(

j 2 (

) 2 j 5 4 ( j 2

ω

− ω +

ω + ω

2

+ V

0.5s 1/s

ω + j

2

−

Chapter 18, Problem 42

Obtain the current io(t) in the circuit of Fig 18.44

(a) Let i(t) = sgn(t) A

(b) Let i(t) = 4[u(t) – u(t – 1)] A

Figure 18.44

For Prob 18.42

Chapter 18, Solution 42

By current division, ⋅ ( ) ω

ω +

j 2

2

Io(a) For i(t) = 5 sgn (t),

( )

) j 2 ( j

20 j

10 j 2

2 I

j

10 I

o = + ω ⋅ ω = ω + ω

ω

= ω

2 s

B s

A ) 2 s ( s

20

+ +

= +

=

( )

ω +

− ω

= ω

j 2

10 j

− ω

= ω + ω

−

j 2

1 j

1 4 ) j 2 ( j

e 1 8

I

ω +

+ ω

− ω +

− ω

j 2

e 4 j

e 4 j 2

4 j

⎯→

⎯

= ω

= ω

= ω

⎯→

j 1

5 I

e 5 i , j

50 10

x 20 j

1 C

j

1

mF

ω

= +

+

= ω

• ω +

) 25 1 s )(

1 s (

250 j

50 I j

50 40

= +

+ +

=

25 1 s

1 1 s

1 1000 25

1 s

B 1

s

A

Vo

V ) t ( u ) e e ( 1000 )

t (

vo = −1t − −1.25t

We transform the voltage source to a current source as shown in Fig (a) and then

combine the two parallel 2Ω resistors, as shown in Fig (b)

, 1

2

2 = Ω

2

V j 1

1

ω +

=

) j 1 ( 2

V j I j

=

( ) t 10 ( t 2 ) 10

10 j 2

ω +

− ω +

= ω +

−

j 1

5 j

1

5 j

1

e 1 5 V

) 2 t ( u e 5 ) t ( u e 5 ) t (

1 (

Chapter 18, Problem 46

Determine the Fourier transform of io(t) in the circuit of Fig 18.48

Figure 18.48

For Prob 18.46

Chapter 18, Solution 46

F 4

1

ω

−

= ω

4 j 4

1 j 1

2H jω2 )

t (

) t ( u

e− t

ω + j 1 1 The circuit in the frequency domain is shown below:

At node Vo, KCL gives

ω

= ω

− − +

− ω

+

2 j

V 4 j

V 3 2

V j

1

1

o o o

ω

−

= ω

− ω +

− ω

V 2 j V j 3 j V 2 j

1

2

ω

− ω +

ω + ω +

=

2 j j 2

3 j j 1

ω +

ω

− ω +

= ω

=

ω

2 j j 2 2 j

j 1

3 3 j 2 2

j

V I

2

o o

Io(ω) =

) 2 8 ( 6

4

3 j 2

3 2

2 2

ω

− ω + ω

−

ω

− ω +

Io( ω)

2 Ω

j2 ω

–j4/ ω+

−

+

−

1 I

ω +

=

ω + ω +

= ω +

ω

= ω

=

j 1

8 j 2

2 I

j

2 1 j

2 I

+

) 2 s )(

1 s

5 j C j 1

As an integrator,

4 0 10 x 20 x 10 x 20

−

j

V RC

−

=

j 2 j

2 4

0 1

−

=

=

j 2 j

2 125

0 mA 20

V

o

( ) ω πδ

− ω +

+ ω

−

j 2

125 0 j

125 0

π

− +

−

2

125 0 t u e 125 0 ) t sgn(

125 0 )

io(t) = 0 625 − 0 25 u ( ) + 0 125 e−2 tu ( ) mA

= 2

I 3

I j 3 j 1 2 2

ω +

ω + ω +

=

2 2

+

2 s 4 s

V 2 s

2 j

2 + ω + ω

− ω δ + + ω δ π + ω

( )

∫−∞∞

ω ω ω

π

2

1 )

t

(

o o

+ ω + ω

ω

− ω δ +

ω + +

ω + ω

ω + ω δ +

ω

=

2 4 j j

d 1 e

2 j 2 1 2

4 j j

d 1 e

2 j 2 1

2

t j 2

t j

2 4 j 1

e 2 j 2 1

2 4 j 1

e 2 j 2

+ +

−

+ + +

1 e 17

4 j 1 j 2 2

1 )

t

(

v

jt jt

o

−

− +

Chapter 18, Problem 50

Determine vo(t) in the transformer circuit of Fig 18.52

Figure 18.52

For Prob 18.50.

= ω

−

ω +

=

j

I j 1 2 5

0 j

I j 1

1Substituting this into (1),

I j 1 2

ω

ω +

−

=

2

2 I 2

3 4 j 4 j

2 2

5 1 4 j 4

j 2 I

ω

− ω +

ω

=

( )2 2

o

j 5 1 4 j 4

j 2 I

V

ω + ω +

8 j 3 8

j 3

4 V

ω +

ω +

ω

=

2 2

2 2

3

8 j

3 4

3 16

3

8 j

3 4

j 3

4 4

3

8 sin e

657 5 ) ( u t 3

8 cos e

3

e 2 4

e 2 e

2 0

2( t ) dt 1 F ( ) d f

2

=

2 3

1 ) 3 / ( tan 3

1 d 9

1 1

0

1

π π

= ω

π

= ω ω + π

0 t , et 2

t 2

t40

t4

0 4t

4

e 4

e 2 dt e dt e

Chapter 18, Problem 54

Given the signal f(t) = 4e−t

u(t) what is the total energy in f(t)?

t 2

2( t ) dt 16 e dt 8 e

= ω ω + π

= ω ω

0 1 4

4 0

2

) ( tan e 25 d 1

1 e

25 d ) ( F 1

= 12.5e4 = 682.5 J

or W1Ω = ∫−∞∞ = ∫0∞ − =

t 2 4

2( t ) dt 25 e e dt

Chapter 18, Problem 56

The voltage across a 1- Ω resistor is v(t) = te− 2t

u(t) V (a) What is the total energy

absorbed by the resistor? (b) What fraction of this energy absorbed is in the frequency band –2 ≤ ω ≤ 2?

1 5

0 tan 5 0 4 4

+ ω

ω π

4 d ) 1 (

1 2

4 d ) ( 2

1

0

1 2

π

= ω π

= ω ω + π

= ω ω π

1

π

= π

= ω π

Determine the following:

(a) the carrier frequency,

(b) the lower sideband frequency,

(c) the upper sideband frequency

Chapter 18, Solution 58

ωm = 200π = 2πfm which leads to fm = 100 Hz (a) ωc = πx104 = 2πfc which leads to fc = 104/2 = 5 kHz

(b) lsb = fc – fm = 5,000 – 100 = 4,900 Hz

(c) usb = fc + fm = 5,000 + 100 = 5,100 Hz

Chapter 18, Problem 59

For the linear system in Fig 18.54, when the input voltage is vi(t) = 2δ (t) V, the output

is v0(t) = 10e− 2t – 6e− 4tV Find the output when the input is vi(t) = 4e−t

− ω +

= ω +

− ω +

= ω

ω

=

ω

j 4

3 j

2

5 2

j 4

6 j

2 10 )

( V

) ( V )

(

H

io

ω

= +

+

− + +

=

ω +

− ω +

= ω ω

= ω

j s , ) 4 s )(

1 s (

12 )

2 s )(

1 s ( 20

j 1

4 j 4

3 j

2

5 )

( V ) ( H )

(

Using partial fraction,

ω +

+ ω +

− ω +

= +

+ +

+ +

+ +

= ω

j 4

4 j

2

20 j

1

16 4 s

D 1 s

C 2 s

B 1 s

A )

(

Vo

Thus,

( 16 e 20 e 4 e ) u ( t ) V )

t (

vo = −t − −2t + −4t

Chapter 18, Problem 60

A band-limited signal has the following Fourier series representation:

is(t) = 10 + 8 cos(2 π t + 30º) + 5 cos(4 π t – 150º)mA

If the signal is applied to the circuit in Fig 18.55, find v(t)

Figure 18.55

For Prob 18.60.