.Instituto Universitario Politécnico“Santiago Mariño”Extensión Maturín

Esc. Ingeniería Eléctrica y ElectrónicaTransformada de la Place

EJERCICIOS: TRANSDORMADA DE LA PLACE

Autor:Oscar Ariza, C.I:26.117.819

Asesora:Mariangela Pollonais

Maturín, Enero del 2017

A. Determine la transformada de Laplace de las siguientes funciones.

3. H (t) = e−2 t . sen5 t

K(s−a )2+k 2

= 5( s+2 ) 2+52

= 5( s+2 ) 2+25

5. Q (t) = Sen2.at

Sen2 x=1−cos2x2

L {1−cos2at2 } = L 12 {1s }−L 12 {cos2at } = 1s− S2¿¿

6. 3e−t+Sen6 t

3 L {e−t }+L {Sen6 t }= 3s+1

+ 6S2+36

1s−a

kS2+k2

8. H (t) = -3cos2t + 5sen4t

L {−3cos2t+5 sen 4 t }=−3 L {−3cos2 t }+5 L {sen 4 t }

¿−3.( SS2+2¿2 )+5.( 4

S2+4¿2 )¿− 3 sS2+4

+ 20S2+16

12. G (t )=e4 t (t−cost )e4 t (t−cost )=t e4 t−e4 t cost

L {t e4 t }−L {e4 t cost }= 1( s−4 ) 2

−(S−4)

(S−4 )2+1

B. En los siguientes ejercicios calcule la transformada inversa de Laplace de la función s dada

2. G (s )= 1S (S+1)

1S (S+1)

=¿ L−1 { 1s (s+1)}

L−1 { 1s (s+1 ) }= 1

s (s+1 )= As

+ B(s+1 )

∗s (s+1 )

1=A ( s+1 )+BSParaS=−1=¿1=−B=¿ B=−1ParaS=0=¿ A=1

L−1 {1s }−L−1{ 1s+1 }=1−e−t

3. H (s )= 2 s

(S¿¿2+1)2tsenkt= 2Ks(S¿¿2+k2)2¿

¿ Senkt+ktcoskt=2k S2

¿¿¿

L−1 ¿

6. R (s )= 3S2

(S2+1)2= 3.2. S2

2(S2+12)2

L−1{32∗2S2S2+1¿

2¿}=32L−1¿

7. R (s )= 2S4 {1s+ 3S2+ 4S6 }= 2

S5+ 6S6

+ 8S10

2 L−1{ 1S5 }+6 L−1{ 1S6 }+8 L−1 { 1S10 }=2 L−1 {4 !S5 }+6 L−1 {5 !S6 }+8L−1{ 9 !S10 }

2+t 4+6 t 5+8 t9

15. H (s )= S2−2 s+3

s (S2−3 s+2)= S2−2 s+3s (s−1)(s−2)

S2−2 s+3s (s−1)(s−2)

= As+ B

(S−1)+ C(S−2)

∗s (s−1)(s−2)

S2−2 s+3=A (s−1 ) (s−2 )+BS (s−2 )+Cs ( s−1 )

s=03=2 A=¿ 32

s=1

1−2+3=−B=¿ B=−2

s=2

4−4+3=2C=¿C=32

32L

−1 {1S }−2L−1 { 1(S−1 ) }+32 L

−1{ 1(S−2 ) }=32−2et+ 32 e2 t

C. Resuelva las siguientes ecuaciones diferenciales28) y '−2 y=1−t y (0 )=1

L [1 ]= 1S

L [ tn ]= n !Sn+1

L [eat ]= 1S−a

L [ y ' (t ) ]=S y ( S)− y (0 )=S y (S )−1

−2 L [ y ( t ) ]=−2 y (S )

L [1 ]= 1S

L [−t ]=−1S2

S y (S )−1−2 y ( S )=1S− 1S2

y (S )(S−2)=S2−SS3

+1

y ( S )=S2−S+S3

S3(S−2)=S(S

2+S−1)S3(S−2)

= S2+S−1S2(S−2)

S2+S−1S2(S−2)

= AS+BS2

+ CS−2

S2+S−1=(AS+B ) (S−2 )+C S2

S2+S−1=A S2+BS−2 AS−2B+C S2

{ A+C=1→C=1+ 14=54

−2 A+B=1→A=1−12

−2=−14

−2B=−1→B=12

−14L−1[ 1S ]+ 12 L−1 [ 1S2 ]+ 54 L−1[ 1

S−2 ]−14

+ 24t+ 54e2 t=1

4(2 t+5 e2 t−1)

29) y ' '−4 y'+4 y=1 y (0 )=1 , y ' (0)=4

L [1 ]= 1S

L [eat ]= 1S−a

L [ tneat ]= n!(s−a )n+1

L [ y ' '( t )]=S2 y (S )−S y ( 0)− y'(0 )=S

2 y (S )−S−4

−4 L [ y '( t )]=−4 (S y (S )− y ( 0) )=−4 S y ( S )+4

4 L [ y ( t ) ]=4 y (S )

L [1 ]= 1S

S2 y (S )−S−4−4S y (S )+4+4 y (S )=1S

S2 y (S )−4S y (S )+4 y (S )=1S+S

y (S ) (S2−4 S+4 )=1+S

2

S

y (S )=1+S2

S (S−2 )2

S2+1S (S−2 )2

= AS

+ BS−2

+ B(S−2 )2

S2+1=A (S−2 )2+BS (S−2 )+CS

S2+1=A S2−4 AS+4 A+BS2−2BS+CS

{ A+B=1→B=1−14= 34

−4 A−2B+C=0→C=1+ 64=104

=52

4 A=1→A=14

14 L

−1[ 1S ]+ 34 L−1[ 1(S−2) ]+ 52 L−1 [ 1

(S−2 )2 ]14+ 34e2 t+ 5

2t e2 t=1

4(10 t e2 t+3e2 t+1)

30) y ' '+9 y=t y (0)= y ' (0)=0

L [ y ' '( t )]=S2 y (S )−S y ( 0)− y'(0 )=S

2 y (S )

9 L [ y(t )]=9 y ( S )

L [t ]= 1S2

S2F (S)+9 F(S )=1S2

F(S )=1

S2(S2+32)= 127 [ 27

S2(S2+32) ]L [kt−sin kt ]= K3

S2(S2+K 2)

127L−1[ 27

S2(S2+32) ]=3 t−sin 3 t27

32) y ' '−6 y '+8 y=e t y (0 )=3 , y ' (0 )=9

L [eat ]= 1S−a

L [ y ' '(t )]=S2 y (S )−S y ( 0)− y'(0 )=S

2 y (S )−3 S−9

−6 L [ y ' (t ) ]=−6 (S y (S )− y ( 0) )=−6S y (S )+18

8 L [ y (t ) ]=8 y ( S )

L [et ]= 1S−1

S2 y (S )−3 S−9−6 S y ( S )+18+8 y (S )=1S−1

S2 y (S )−6 S y (S)+8 y ( S )=1S−1

+(3S−9)

y (S ) (S2−6 S+8 )= 1

S−1+(3S−9)

y (S )=1+(S−1)(3S−9)

(S−1 ) (S−2 ) (S−4 )= 3S2−9S−3 S+10

(S−1 ) (S−2 ) (S−4 )= 3S2−12S+10

(S−1 ) (S−2 ) (S−4 )

3S2−12S+10(S−1 ) (S−2 ) (S−4 )

= A(S−1 )

+ B(S−2 )

+ B(S−4 )

3S2−12S+10=A (S−2 ) (S−4 )+B (S−1 ) (S−4 )+C (S−1 ) (S−2 )

Para S=2

12−24+10=B (1 ) (−2 )→B=1

Para S=4

48−48+10=C (3 ) (2 )→C=106

=53

Para S=1

3−12+10=A (−1 ) (−3 )→A=13

13L−1[ 1

(S−1 ) ]+L−1[ 1(S−2) ]+ 53 L−1[ 1

(S−4) ]13e t+e2 t+ 5

3e4 t=1

3(5e4 t+3e2 t+e t)