Lista 3 - Solução homegêna, Particular,

8/8/2019 Lista 3 - Soluo homegna, Particular, ...

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3 Lista de Exerccios Soluo Homognea, Particular, Completa e Diagrama de Blocos

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Disciplina: IM 144 Prof. Dr. Janito Vaqueiro Ferreira

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QUESTO 1.a:

a) 2; (0) 0; (0) 2y y y y = = =gg g

Resposta:

Soluo Homognea (sh):

0y y =gg

2 ( ) ( ) 0y t y t = (3.1)

Como:

( )t

y t e

= (3.2)

Substituindo a equao (3.2) na equao (3.1):

( )

2

2

0

1 0

t t

t

e e

e

=

=(3.3)

2

1 0 =Q 1,2 1 = (3.4)

Soluo Particular (sp):

( ) y t C = (3.5)

2y y =gg

(3.6)

Resolvendo a derivada da equao (3.6):

0 ( ) 2y t = 2C = (3.7)

Soluo Completa (sc):

sc sh sp= + (3.8)

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5/23


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( ) 2t ty t Ae Be= + (3.9)

Derivando a equao (3.9):

( ) t ty t Ae Be= g

(3.10)

Extraindo os coeficientes da equao (3.9) e da equao (3.10), temos:

( ) 2y t A B= + (3.11)

( ) y t A B= g

(3.12)

Substituindo as condies iniciais, onde (0) 0; (0) 2y y= =g

nas equaes (3.11) e (3.12):

0 2A B= + (3.13)

2 A B= (3.14)

Resolvendo o sistema linear de duas equaes, obtemos:

2A = e 0B =

Substituindo na equao (3.9), temos como resposta:

( ) 2 2ty t e= (3.15)

QUESTO 1.b:

b) sin( ); (0) 3y y t y+ = =g

Resposta:


0y y+ =g

( ) ( ) 0y t y t + = (3.16)

Como:

( )ty t e= (3.17)

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( )

0

1 0

t t

t

e e

e

+ =

+ =(3.18)

1 0+ =Q 1 = (3.19)


( ) ( ) cos( )y t Bsin t C t = + (3.20)

sin( )y y t + =g

(3.21)


cos( ) sin( ) ( ) cos( ) sin( )B t C t Bsin t C t t + + =

( ) ( )cos( ) sin( ) sin( )B C t B C t t + + =

0B C+ = e 1B C = (3.22)

Resolvendo o sistema linear de duas equaes, obtemos:

1

2

B = e1

2

C= (3.23)


sc sh sp= + (3.24)

( ) ( ) cos( )ty t Ae Bsin t C t = + + (3.25)

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7/23


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Substituindo as variveis da equao (3.23) na equao (3.25):

1 1( ) ( ) cos( )

2 2

t y t Ae sin t t

= + (3.26)

Aplicando as condies iniciais, onde (0) 3y = na equao (3.26):

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

y Ae sin= +

13

2A= e

7

2A = (3.27)


7 1 1( ) ( ) cos( )

2 2 2

ty t e sin t t = + (3.28)

QUESTO 1.c:

c) 2 3 5; (0) 2; (0) 4y y y y y+ + = = =gg g g

Resposta:


2 3 0y y y+ + =gg g

2 ( ) 2 ( ) 3 ( ) 0y t y t y t + + = (3.29)

Como:

( )ty t e

= (3.30)


2 2 3 0t t te e e + + = (3.31)

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( )2 2 3 0te + + =

2 2 3 0 + + =Q 1,2 1 2j = (3.32)


( ) y t C = (3.33)

2 3 5y y y+ + =gg g

(3.34)


3 5C= 5

3C= (3.35)


sc sh sp= + (3.36)

( ) ( )1 2 1 2( )

j t j t y t Ae Be C

+ = + +

( ) ( )1 2 1 2 5( )

3

j t j t y t Ae Be

+ = + + (3.37)


( )( )

( )( )1 2 1 2

( ) 1 2 1 2

j t j t

y t j Ae j Be

+

= + +

g

(3.38)


5( )

3y t A B= + + (3.39)

( ) ( )( ) 1 2 1 2y t j A j B= + + g

(3.40)

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nas equaes (3.39) e (3.40):

5

23

A B= + + 1

3A B+ = (3.41)

( ) ( )4 1 2 1 2j A j B= + +

( ) ( )1

4 1 2 1 23

j B j B = + +

1 14 2 2 2

3 3B j Bj B Bj= + +

1 14 2 2 2

3 3j Bj+ =

13 12 2 2

3 3j Bj =

13 1 266 2 2

j Bj j

+ =

1 13

6 6 2B

j=

(3.42)

Substituindo a equao (3.42) na equao (3.41), temos:

13 1 1

6 36 2A j + =

13 1 1

6 36 2A

j= +

1 13

6 6 2A

j= + (3.43)

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( ) ( )1 2 1 21 13 1 13 5( )

6 6 36 2 6 2

j t j t y t e e

j j

+ = + + +

2 21 13 1 13 5( )6 6 36 2 6 2

t j t t j t y t e e e ej j

= + + +

2 21 13 13 5( ) 1 16 32 2

t j t j t y t e e ej j

= + + +

Como:2

2

cos

cos( 2 ) ( 2 )

cos( 2 ) ( 2 ) cos( 2 ) ( 2 )

j t

j t

jae a jsena

e t jsen t

e t jsen t t jsen t

= +

= +

= + =

( ) ( )1 13 13 5

( ) 1 cos( 2 ) ( 2 ) 1 cos( 2 ) ( 2 )6 32 2

t y t e t jsen t t jsen t

j j

= + + + +

13 13cos( 2 ) s n( 2 ) cos( 2 ) s n( 2 ) cos( 2 ) ( 2 )

2 21 5( )

13 136 3cos( 2 ) ( 2 )

2 2

t

t j e t t j e t t jsen t j j

y t e

t jsen t j j

+ + + + = + +

1 26 5

( ) 2 cos( 2 ) s n( 2 )6 32

t y t e t e t

= + +

1 26 5( ) cos( 2 ) s n( 2 )

3 36 2

t y t e t e t

= + +

1 13 5( ) cos( 2 ) s n( 2 )

3 33 2

t t y t e t e e t

= + + (3.44)

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Disciplina: IM 144 Prof. Dr. Janito Vaqueiro Ferreira3 Lista de Exerccios Soluo Homognea, Particular, Completa e Diagrama de Blocos

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11/23

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QUESTO 1.d:

d) 4 4 5 ; (0) 1; (0) 6y y y t y y + = = =gg g g

Resposta:


4 4 0y y y + =gg g

2 ( ) 4 ( ) 4 ( ) 0y t y t y t + = (3.45)

Como: ( )ty t e= (3.46)


2 4 4 0t t te e e + = (3.47)

( )2 4 4 0te + =

2 4 4 0 + =Q 1,2 2 = (3.48)


( )y t C Dt = + (3.49)

4 4 5y y y t + =gg g

(3.50)


4 4 4 5 D C Dt t + + = (3.51)

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Assim:

4 5Dt t = 5

4

D = (3.52)

4 4 0C D + = 5

4C D= = (3.53)


sc sh sp= + (3.54)

2 2 5 5( )4 4

t ty t Ae Bte t = + + + (3.55)


2 2 2 5( ) 2 24

t t ty t Ae Bte Be= + + +g

(3.56)


5 5( )

4 4y t A Bt t = + + + (3.57)

5( ) 2 2

4 y t A Bt B= + + +

g

(3.58)


nas equaes (3.57) e (3.58):

5

14

A= + 1

4A = (3.59)

56 2

4A B= + +

192

4A B+ = (3.60)

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Substituindo a equao (3.60) na equao (3.59), temos:

21

4

B = (3.61)


2 21 21 5 5( )4 4 4 4

t ty t e te t = + + + (3.62)

QUESTO 1.e:

e) 25 500 ; (0) 20ty y e y

+ = =g

Resposta:


5 0y y+ =g

( ) 5 ( ) 0y t y t + = (3.63

Como:

( )t

y t e

= (3.64)


( )

5 0

5 0

t t

t

e e

e

+ =

+ =(3.65)

5 0+ =Q 5 = (3.66)


2( ) ty t Be= (3.67)

25 500 ty y e+ =g

(3.68)

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2 2 2

2 2

2 5 500

3 500

3 500

t t t

t t

Be Be e

Be e

B

+ =

=

=

500

3B = (3.69)


sc sh sp= + (3.70)

5 2( ) t ty t Ae Be = + (3.71)

Aplicando as condies iniciais, onde (0) 20y = na equao (3.71):

0 0500(0) 3y Ae e= +

50020

3A= +

440

3A = (3.72)

Substituindo na equao (3.37) o valor da equao (3.35) e (3.38), temos como resposta:

5 2440 500( )3 3

t ty t e e = + (3.73)

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QUESTO 1.f:

d)2

10 24 50 cos(3 ); (0) 4; (0) 1t

y y y e t y y

+ + = = =

gg g g

Resposta:


10 24 0y y y+ + =gg g

2 ( ) 10 ( ) 24 ( ) 0y t y t y t + + = (3.74)

Como:

( )

t

y t e

= (3.75)


( )

2

2

10 24 0

10 24 0

t t t

t

e e e

e

+ + =

+ + =(3.76)

2 10 24 0 + + =Q 1 6 = e 2 4 = (3.77)


2 2( ) cos(3 ) sin(3 )t ty t Ce t De t = + (3.78)

210 24 50 cos(3 )ty y y e t + + =

gg g(3.79)

Resolvendo a derivada primeira da equao (3.78):

2 2 2 2( ) 2 cos(3 ) 3 (3 ) 2 sin(3 ) 3 cos(3 )t t t t y t Ce t Ce sin t De t De t = +g

( ) ( )2 2( ) 2 3 cos(3 ) 3 2 sin(3 )t ty t C D e t C D e t = + +g

(3.80)

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Resolvendo a derivada segunda da equao (3.78):

( ) ( )

( ) ( )

2 2

2 2

( ) 2 2 3 cos(3 ) 3 2 3 sin(3 )

2 3 2 sin(3 ) 3 3 2 cos(3 )

t t

t t

y t C D e t C D e t

C D e t C D e t

= + +

+ + +

gg

( ) ( )

( ) ( )

2 2

2 2

( ) 4 63 cos(3 ) 6 9 sin(3 )

6 4 sin(3 ) 9 6 cos(3 )

t t

t t

y t C D e t C D e t

C D e t C D e t

= +

+ + +

gg

( ) ( )2 2( ) 4 6 9 6 cos(3 ) 6 9 6 4 sin(3 )t ty t C D C D e t C D C D e t = + + +gg

( ) ( )2 2( ) 5 12 cos(3 ) 12 5 sin(3 )t ty t C D e t C D e t = + gg

(3.81)

Aplicando as condies iniciais, onde (0) 4; (0) 1y y= =g

na equao (3.83):

210 24 50 cos(3 )ty y y e t + + =gg g

( ) ( ) ( ) ( )2 2 2 2

2

5 12 cos(3 ) 12 5 sin(3 ) 10 2 3 cos(3 ) 3 2 sin(3 )

24 ( ) 50 cos(3 )

t t t t

t

C D e t C D e t C D e t C D e t

y t e t

+ + + + + =

( ) ( )2 2 25 12 20 30 cos(3 ) 12 5 30 20 sin(3 ) 24 ( ) 50 cos(3 )t t tC D C D e t C D C D e t y t e t + + + =

( ) ( )2 2 225 18 cos(3 ) 18 25 sin(3 ) 24 ( ) 50 cos(3 )t t tC D e t C D e t y t e t + + + =

Como:

2 2( ) cos(3 ) sin(3 )t ty t Ce t De t = + (3.82)

( ) ( )2 2 2 2 225 18 cos(3 ) 18 25 sin(3 ) 24 cos(3 ) 24 sin(3 ) 50 cos(3 )t t t t t C D e t C D e t Ce t De t e t + + + + =

Assim:

( )

( )

2 2 2

2 2

25 18 cos(3 ) 24 cos(3 ) 50 cos(3 )

18 25 sin(3 ) 24 sin(3 ) 0

t t t

t t

C D e t Ce t e t

C D e t De t

+ + =

+ =

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( )

( )

2 2

2

25 18 24 cos(3 ) 50 cos(3 )

18 25 24 sin(3 ) 0

t t

t

C D C e t e t

C D D e t

+ + =

+ =

18 50

18 0

C D

C D

+ =

=(3.83)

Resolvendo o sistema linear, referente equao (3.83):

2

13C = e

36

13D = (3.84)


sc sh sp= + (3.85)

4 6 2 22 36( ) cos(3 ) sin(3 )13 13

t t t t y t Ae Be e t e t = + + (3.86)

4 6 2 2 2 24 6 72 108( ) 4 6 cos(3 ) sin(3 ) sin(3 ) cos(3 )13 13 13 13

t t t t t t y t Ae Be e t e t e t e t = + + +g

(3.87)

Aplicando as condies iniciais, onde (0) 4; (0) 1y y= =g

na equao (3.83):

0 0 0 02 36(0) cos(0) sin(0)13 13

y Ae Be e e= + +

24

13A B= +

54

13A B+ = (3.88)

0 0 0 0 0 04 6 72 108(0) 4 6 cos(0) sin(0) sin(0) cos(0)13 13 13 13

y Ae Be e e e e= + + +g

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4 1081 4 6

13 13A B= + +

1124 6 1

13A B+ = +

994 6

13A B+ = (3.89)

Resolvendo o sistema linear composto pelas equaes (3.88) e (3.89), temos:

225

26A = e

117

26D = (3.90)

Substituindo na equao (3.86) os valores da equao (3.90), temos como resposta:

4 6 2 2225 117 2 36( ) cos(3 ) sin(3 )26 26 13 13

t t t t y t e e e t e t = + (3.91)

QUESTO 2: Construa o diagrama de blocos para o sistema descrito pela equao

diferencial

4 11 15 2 5 y y y y u u+ + + = +

ggg gg g g

Resposta:

Isolando o termo de numerador da equao diferencial e adicionando o nmero de

integradores da ordem do sistema, temos:

2 5 4 11 15y u u y y y= + ggg g gg g

(3.92)

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18

yggg

ygg

yg

y


19/23

QUESTO 3.a: Determine a equao diferencial relacionada a cada um dos diagramas

de blocos:

Resposta:

Dado nomes a todas as entradas e sadas do sistema, escreve-se a equao de cada

componente:

9 32 60y u y y y= ggg gg g

9 32 60u y y y y= + + +ggg gg g

(3.93)

7 2y u u= +

g

(3.94)

Assim, a FT igual a:

7 2

9 32 60

N y u uFT

D u y y y y

+= = =

+ + +

g

ggg gg g (3.95)

Obtem-se ento a equao diferencial:

9 32 60 7 2y y y y u u+ + + = +ggg gg g g

(3.96)

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y

ggg

y

gg

y

g

y


20/23

QUESTO 3.b: Determine a equao diferencial relacionada a cada um dos diagramas

de blocos:

Resposta:


componente:

3 x u x= g

3u x x= +g

3u

px

= + (3.99)

4 12z x z z = gg g

4 12x z z z = + +gg g

2 4 12

xp p

z= + + (3.97)

3 2y z z = +g 3 2

yp

z= + (3.98)

Assim, a FT igual a:

( )2

1 13 2

3 4 12

N yFT p

D u p p p

= = = +

+ + +

( )23 2

3 4 12

N y pFT

D u p p p

+= = =

+ + +

3 2

3 2

7 24 36

y p

u p p p

+=

+ + +(3.99)

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20

xg

x zgg

zg

z

u yx z1

3p + 21

4 12p p+ +2 3p +


21/23


7 24 60 3 2y y y y u u+ + + = +ggg gg g g

(3.100)

QUESTO 3.c: Determine a equao diferencial relacionada a cada um dos diagramas

de blocos:

Resposta:


componente:

3 x u x= g

(3.101)

4 12z u z z = gg g

(3.102)

5 3 2 y z z x= + +g

(3.103)

Aplicando Laplace nas equaes (3.101), (3.102) e (3.103):

Primeira equao: 3u x x= +g

( ) ( ) 3 ( )U s SX s X s= + ( ) 1

( ) 3

X s

U s s=

+(3.104)

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xg

x

zgg

zg

z


22/23

Segunda equao: 4 12u z z z = + +gg g

2( ) ( ) 4 ( ) 12 ( )U s s Z s sZ s Z s= + + 2( ) 1

( ) 4 12

Z s

U s s s

=

+ +

(3.105)

Terceira equao: 5 3 2 y z z x= + +g

( ) 5 ( ) 3 ( ) 2 ( )Y s Z s sZ s X s= + +

( )( ) 5 3 ( ) 2 ( )Y s s Z s X s= + +

Substituindo as equaes (3.104) e (3.105), temos a FT, onde:

( )2

1 2( ) 5 3 ( ) ( )

4 12 3Y s s U s U s

s s s= + +

+ + +

2

( ) 5 3 2

( ) 4 12 3

Y s s

U s s s s

+= +

+ + +

( ) ( ) ( )( ) ( )

2

2

5 3 3 2 4 12( )

( ) 4 12 3

s s s sY s

U s s s s

+ + + + +=

+ + +

2 2

3 2 2

( ) 5 15 3 9 2 8 24

( ) 3 4 12 12 36

Y s s s s s s

U s s s s s s

+ + + + + +=

+ + + + +

2

3 2

( ) 5 22 39

( ) 7 24 36

Y s s s

U s s s s

+ +=

+ + + (3.106)

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23/23

Logo, a transformada inversa de Laplace :

2

3 2

5 22 39

7 24 36

y u u

u y y y

+ +=

+ + +(3.106)


7 24 36 5 22 39y y y y u u+ + + = + +ggg gg g gg g

(3.107)

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Lista 3 - Solução homegêna, Particular,