Transformada de Laplace

INSTITUTO UNIVERSITARIO POLITÉCNICO
“SANTIAGO MARIÑO”
EXTENSIÓN MATURÍN
ESCUELA DE INGENIERÍA ELÉCTRICA
Y ELECTRÓNICA
TRANSFORMADA DE
LAPLACE
Profesora: Realizado por:
Mariangela Pollonais. Jesús Figueras.
Materia:
Teoría de Control Semestre: VII
Sección: V
Maturín, enero del 2017.

1
TRANSFORMADA DE LAPLACE.
1) 𝒇( 𝒕) = 𝟐 𝐬𝐢𝐧 𝒕 + 𝟑 𝐜𝐨𝐬 𝟐𝒕
𝑳[sin 𝐾𝑡] =
𝐾
𝑆2 + 𝐾2
𝑳[cos 𝐾𝑡] =
𝑆
𝑆2 + 𝐾2
𝑳[2sin 𝑡 + 3cos2𝑡] = 2𝑳[sin 𝑡] + 3𝑳[cos2𝑡] = 2(
1
𝑠2 + 1
) + 3(
𝑆
𝑠2 + 4
)
=
2
𝑠2 + 1
+
3𝑆
𝑠2 + 4
4) 𝒑( 𝒕) = ( 𝒕 + 𝒂) 𝟑
= 𝑡3
+ 3𝑡2
𝑎 + 3𝑡𝑎2
+ 𝑎3
𝑳[ 𝑡 𝑛] =
𝑛!
𝑆 𝑛+1
𝑳[ 𝑡3
+ 3𝑡2
𝑎 + 3𝑡𝑎2
+ 𝑎3] = 𝑳[ 𝑡3] + 3𝑎𝑳[ 𝑡2
𝑎] + 3𝑎2
𝑳[ 𝑡] + 𝑎3
𝑳[1]
=
6
𝑆4
+
(3𝑎)2
𝑆3
+
(3𝑎2)1
𝑆2
+
( 𝑎3)1
𝑆
=
6
𝑆4
+
6𝑎
𝑆3
+
3𝑎2
𝑆2
+
𝑎3
𝑆
5) 𝒒( 𝒕) = 𝐬𝐢𝐧 𝟐
𝒂𝒕
sin2
𝑥 =
1 − cos2𝑥
2
sin2
𝑎𝑡 =
1 − cos2𝑎𝑡
2
=
1
2
−
cos2𝑎𝑡
2
𝑳[1] =
1
𝑆

2
𝑆
𝑆2 + 𝐾2
𝑳[sin2
𝑎𝑡] = 𝑳 [
1
2
−
cos2𝑎𝑡
2
] =
1
2
𝑳[1] −
1
2
𝑳[cos2𝑎𝑡] =
1
2
(
1
𝑆
) −
1
2
(
𝑆
𝑆2 + 4𝑎2
)
=
1
2𝑆
−
𝑆
2𝑆2 + 8𝑎2
=
2𝑆2
+ 8𝑎2
− 2𝑆2
4𝑆3 + 16𝑎2 𝑆
=
8𝑎2
4( 𝑆3 + 4𝑎2 𝑆)
=
2𝑎2
𝑆3 + 4𝑎2 𝑆
9) 𝒒( 𝒕) = 𝟒 𝐜𝐨𝐬 𝟐
𝟑𝒕
cos2
𝑥 =
1 + cos2𝑥
2
4 cos2
3𝑡 = 4(
1 + cos6𝑡
2
) = 2 + 2 cos6𝑡
𝑳[1] =
1
𝑆
𝑆
𝑆2 + 𝐾2
𝑳[4cos2
3𝑡] = 𝑳[2 + 2cos6𝑡] = 2𝑳[1] + 2𝑳[cos6𝑡] = 2(
1
𝑆
) + 2 (
𝑆
𝑆2 + 36
)
=
2
𝑆
+
2𝑆
𝑆2 + 36
=
2𝑆2
+ 72 + 2𝑆2
𝑆3 + 36𝑆
=
4𝑆2
+ 72
𝑆3 + 36𝑆
10) 𝒓( 𝒕) = 𝐜𝐨𝐬 𝟑
𝒕
cos2
𝑥 =
1 + cos2𝑥
2
cos 𝐴 × cos 𝐵 =
cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵)
2

3
𝑆
𝑆2 + 𝐾2
cos3
𝑡 = cos 𝑡 × cos2
𝑡 = cos 𝑡 × (
1 + cos2𝑡
2
) =
1
2
cos 𝑡 +
1
2
cos 𝑡 × cos2𝑡
=
1
2
cos 𝑡 +
1
2
[
cos(3𝑡) + cos(−𝑡)
2
] =
1
2
cos 𝑡 +
1
4
cos3𝑡 +
1
4
cos(−𝑡)
1
2
𝑳[cos𝑡] +
1
4
𝑳[cos3𝑡] +
1
4
𝑳[cos(−𝑡)] =
1
2
(
𝑆
𝑆2 + 1
) +
1
4
(
𝑆
𝑆2 + 9
) +
1
4
(
𝑆
𝑆2 + 1
)
=
1
4
(
𝑆
𝑆2 + 9
) +
3
4
(
𝑆
𝑆2 + 1
) =
1
4
(
𝑆
𝑆2 + 9
+
𝑆
𝑆2 + 1
)
=
1
4
[
𝑆( 𝑆2
+ 1) + 3𝑆(𝑆2
+ 9)
(𝑆2 + 9)(𝑆2 + 1)
] =
1
4
[
𝑆3
+ 𝑆 + 3𝑆3
+ 27𝑆
(𝑆2 + 9)(𝑆2 + 1)
]
=
1
4
[
4𝑆3
+ 28𝑆
(𝑆2 + 9)(𝑆2 + 1)
] =
𝑆3
+ 7𝑆
(𝑆2 + 9)(𝑆2 + 1)
TRANSFORMADA DE LAPLACE INVERSA.
4) 𝒑( 𝒔) =
𝟏
𝑺( 𝑺+𝟐) 𝟐
𝑳−𝟏
[
1
𝑆( 𝑆 + 2)2
]
1
𝑆( 𝑆 + 2)2
=
𝐴
𝑆
+
𝐵
𝑆 + 2
+
𝐶
( 𝑆 + 2)2
1 = 𝐴( 𝑆 + 2)2
+ 𝐵𝑆( 𝑆 + 2) + 𝐶𝑆
1 = 𝐴( 𝑆2
+ 4𝑆 + 4) + 𝐵𝑆2
+ 2𝐵𝑆 + 𝐶𝑆

4
{
𝐴 + 𝐵 = 0 → 𝐵 = −𝐴 = −
1
4
4𝐴 + 2𝐵 + 𝐶 = 0 → 𝐶 = −4 (
1
4
) − 2(
−1
4
) = −1 +
1
2
= −
1
2
4𝐴 = 1 → 𝐴 =
1
4
1
4
𝑳−𝟏
[
1
𝑆
] −
1
4
𝑳−𝟏
[
1
𝑆 + 2
] −
1
2
𝑳−𝟏
[
1
( 𝑆 + 2)2
] =
1
4
−
𝑒−2𝑡
4
−
𝑡𝑒−2𝑡
2
5) 𝒑( 𝒔) =
𝟏
𝑺 𝟐 −𝟒𝑺+𝟓
1
𝑆2 − 4𝑆 + 5
=
1
( 𝑆2 − 4𝑆 + 4) + 1
=
1
( 𝑆 − 2)2 + 1
𝑳[ 𝑒 𝑎𝑡
sin 𝑘𝑡] =
𝐾
( 𝑆 − 𝑎)2 + 𝑘2
𝑳−𝟏
[
1
( 𝑆 − 2)2 + 1
] = 𝑒2𝑡
sin 𝑡
12) 𝒇( 𝒔) =
𝒆−𝟒𝑺
𝑺(𝑺 𝟐+𝟏𝟔)
𝑒−4𝑆
𝑆(𝑆2 + 16)
=
𝑒−4𝑆
𝑆(𝑆2 + 42)
=
16
16
(
𝑒−4𝑆
𝑆(𝑆2 + 42)
) =
42
𝑒−4𝑆
16𝑆(𝑆2 + 42)
𝑳[1 − cos 𝑘𝑡 ] =
𝐾2
𝑆(𝑆2 + 𝑘2)
1
16
𝑳−𝟏
[
42
𝑆(𝑆2 + 42)
× 𝑒−4𝑆
] =
1
16
[1 − cos4( 𝑡 − 4)] 𝒖( 𝑡 − 4)
= [
1
16
−
1
16
cos4( 𝑡 − 4)] 𝒖(𝑡 − 4)

5
13) 𝒇( 𝒔) =
𝒆−𝑺
( 𝑺−𝟓) 𝟑
𝑳[ 𝑡 𝑛
𝑒 𝑎𝑡] =
𝑛!
( 𝑠 − 𝑎) 𝑛+1
𝑳−𝟏
[
𝑒−𝑆
( 𝑆 − 5)3
] =
1
2
𝑳−𝟏
[
2
( 𝑆 − 5)3
× 𝑒−𝑆
] =
( 𝑡 − 1)2
𝑒5(𝑡−1)
2
× 𝒖(𝑡 − 1)
14) 𝒈( 𝒔) =
𝑺𝒆−𝟏𝟎𝑺
( 𝑺 𝟐
−𝟒)
𝟐
𝑆𝑒−10𝑆
( 𝑆2 − 4)2
=
4
4
[
𝑆𝑒−10𝑆
( 𝑆2 − 4)2
] =
1
4
[
2 × 2 × 𝑆
( 𝑆2 − 22)2
× 𝑒−10𝑆
]
𝑳[ 𝑡 sin 𝑘𝑡 ] =
2𝑘𝑆
( 𝑆2 + 𝐾2)2
1
4
𝑳−𝟏
[
2 × 2 × 𝑆
( 𝑆2 − 22)2
× 𝑒−10𝑆
] =
1
4
( 𝑡 − 10)sin 2( 𝑡 − 10) 𝒖(𝑡 − 10)
ECUACIONES DIFERENCIALES.
𝑳[𝐹′′′(𝑡)] = 𝑆3
𝐹( 𝑆) − 𝑆2
𝐹(0) − 𝑆𝐹′
(0) − 𝐹′′
(0)
𝑳[𝐹′′(𝑡)] = 𝑆2
𝐹( 𝑆) − 𝑆𝐹(0) − 𝐹′(0)
𝑳[𝐹′(𝑡)] = 𝑆𝐹( 𝑆) − 𝐹(0)
𝑳[𝐹(𝑡)] = 𝐹( 𝑆)

6
26) 𝒚′′′
+ 𝟒𝒚′′
+ 𝟓𝒚′
+ 𝟐𝒚 = 𝟏𝟎 𝐜𝐨𝐬 𝒕 𝒚( 𝟎) = 𝒚′( 𝟎) = 𝟎; 𝒚′′( 𝟎) = 𝟑
𝑳[𝑦′′′(𝑡)] = 𝑆3
𝑦( 𝑆) − 𝑆2
𝑦(0) − 𝑆𝑦′
(0) − 𝑦′′
(0) = 𝑆3
𝑦( 𝑆) − 3
4𝑳[𝑦′′(𝑡)] = 4(𝑆2
𝑦( 𝑆) − 𝑆𝑦(0) − 𝑦′(0)) = 4𝑆2
𝑦( 𝑆)
5𝑳[𝑦′(𝑡)] = 5(𝑆𝑦( 𝑆) − 𝑦(0)) = 5𝑆𝑦( 𝑆)
2𝑳[𝑦(𝑡)] = 2𝑦( 𝑆)
10𝑳[cos𝑡] =
10𝑆
𝑆2 + 1
𝑆3
𝑦( 𝑆) − 3 + 4𝑆2
𝑦( 𝑆) + 5𝑆𝑦( 𝑆) + 2𝑦( 𝑆) =
10𝑆
𝑆2 + 1
𝑦( 𝑆)( 𝑆3
+ 4𝑆2
+ 5𝑆 + 2) =
10𝑆
𝑆2 + 1
+ 3
Factorizar 𝑆3
+ 4𝑆2
+ 5𝑆 + 2 aplicando Ruffini:
1 4 5 2 𝑆3
+ 4𝑆2
+ 5𝑆 + 2 = ( 𝑆 + 2)( 𝑆 + 1)2
-2 -2 -4 -2
1 2 1 0
-1 -1 -1
1 1 0
-1 -1
1 0
𝑦( 𝑆) =
10𝑆 + 3𝑆2
+ 3
( 𝑆2 + 1)( 𝑆 + 2)( 𝑆 + 1)2
10𝑆 + 3𝑆2
+ 3
( 𝑆2 + 1)( 𝑆 + 2)( 𝑆 + 1)2
=
𝐴𝑆 + 𝐵
( 𝑆2 + 1)
+
𝐶
( 𝑆 + 2)
+
𝐷
( 𝑆 + 1)
+
𝐸
( 𝑆 + 1)2

7
10𝑆 + 3𝑆2
+ 3
= ( 𝐴𝑆 + 𝐵)( 𝑆 + 2)( 𝑆 + 1)2
+ 𝐶( 𝑆2
+ 1)( 𝑆 + 1)2
+ 𝐷( 𝑆2
+ 1)( 𝑆 + 2)( 𝑆 + 1) + 𝐸( 𝑆2
+ 1)( 𝑆 + 2)
10𝑆 + 3𝑆2
+ 3
= 𝐴𝑆4
+ 2𝐴𝑆3
+ 𝐴𝑆2
+ 𝐵𝑆3
+ 2𝐵𝑆2
+ 𝐵𝑆 + 2𝐴𝑆3
+ 4𝐴𝑆2
+ 2𝐴𝑆
+ 2𝐵𝑆2
+ 4𝐵𝑆 + 2𝐵 + 𝐶𝑆4
+ 2𝐶𝑆3
+ 𝐶𝑆2
+ 𝐶𝑆2
+ 2𝐶𝑆 + 𝐶 + 𝐷𝑆4
+ 3𝐷𝑆3
+ 2𝐷𝑆2
+ 𝐷𝑆2
+ 3𝐷𝑆 + 2𝐷 + 𝐸𝑆3
+ 2𝐸𝑆2
+ 𝐸𝑆 + 2𝐸
10𝑆 + 3𝑆2
+ 3
= 𝑆4( 𝐴 + 𝐶 + 𝐷) + 𝑆3(4𝐴 + 𝐵 + 2𝐶 + 3𝐷 + 𝐸)
+ 𝑆2(5𝐴+ 4𝐵 + 2𝐶 + 3𝐷 + 2𝐸) + 𝑆(2𝐴 + 5𝐵 + 2𝐶 + 3𝐷 + 𝐸)
+ (2𝐵 + 𝐶 + 2𝐷 + 2𝐸)
{
𝐴 + 𝐶 + 𝐷 = 0
4𝐴 + 𝐵 + 2𝐶 + 3𝐷 + 𝐸 = 0
5𝐴 + 4𝐵 + 2𝐶 + 3𝐷 + 2𝐸 = 3
2𝐴 + 5𝐵 + 2𝐶 + 3𝐷 + 𝐸 = 10
2𝐵 + 𝐶 + 2𝐷 + 2𝐸 = 3
Aplicando Gauss para resolver el sistema de ecuaciones 5x5:
(
1 0 1 1 0
4 1 2 3 1
5 4 2 3 2
2 5 2 3 1
0 2 1 2 2
|
|
0
0
3
10
3 )(
𝐹1
𝐹2
𝐹3
𝐹4
𝐹5
|
|
𝐹2 − 4𝐹1 → 𝐹2
𝐹3 − 5𝐹1 → 𝐹3
𝐹4 − 2𝐹1 → 𝐹4
)
(
1 0 1 1 0
0 1 −2 −1 1
0 4 −3 −2 2
0 5 0 1 1
0 2 1 2 2
|
|
0
0
3
10
3 )(
𝐹1
𝐹2
𝐹3
𝐹4
𝐹5
|
|
𝐹3 − 4𝐹2 → 𝐹3
𝐹4 − 5𝐹2 → 𝐹4
𝐹5 − 2𝐹2 → 𝐹5
)
(
1 0 1 1 0
0 1 −2 −1 1
0 0 5 2 −2
0 0 10 6 −4
0 0 5 4 0
|
|
0
0
3
10
3 )(
𝐹1
𝐹2
𝐹3
𝐹4
𝐹5
|
| 𝐹4 − 2𝐹3 → 𝐹4
𝐹5 − 𝐹3 → 𝐹5
)

8
(
1 0 1 1 0
0 1 −2 −1 1
0 0 5 2 −2
0 0 0 2 0
0 0 0 2 2
|
|
0
0
3
4
0)(
𝐹1
𝐹2
𝐹3
𝐹4
𝐹5
|
| 𝐹5 − 𝐹4 → 𝐹5
)
(
𝐴 𝐵 𝐶 𝐷 𝐸
1 0 1 1 0
0 1 −2 −1 1
0 0 5 2 −2
0 0 0 2 0
0 0 0 0 2
|
|
0
0
3
4
−4)
{
𝐴 + 𝐶 + 𝐷 = 0
𝐵 − 2𝐶 − 𝐷 + 𝐸 = 0
5𝐶 + 2𝐷 − 2𝐸 = 3
2𝐷 = 4
2𝐸 = −4
𝐸 = −
4
2
= −2
𝐷 =
4
2
= 2
5𝐶 + 2(2) − 2(−2) = 3 → 𝐶 =
3 − 8
5
= −1
𝐴 = −𝐶 − 𝐷 = −(−1)− (2) = −1
𝐵 = 2𝐶 + 𝐷 − 𝐸 = 2(−1) + 2 − (−2) = 2
−𝑳−𝟏
[
𝑆
( 𝑆2 + 1)
] + 2𝑳−𝟏
[
1
( 𝑆2 + 1)
] − 𝑳−𝟏
[
1
( 𝑆 + 2)
] + 2𝑳−𝟏
[
1
( 𝑆 + 1)
]
− 2𝑳−𝟏
[
1
( 𝑆 + 1)2
] = −cos 𝑡 + 2 sin 𝑡 − 𝑒−2𝑡
+ 2𝑒−𝑡
− 2𝑡𝑒−𝑡

9
28) 𝒚′
− 𝟐𝒚 = 𝟏 − 𝒕 𝒚( 𝟎) = 𝟏
𝑳[1] =
1
𝑆
𝑳[ 𝑡 𝑛] =
𝑛!
𝑆 𝑛+1
𝑳[ 𝑒 𝑎𝑡] =
1
𝑆 − 𝑎
𝑳 [𝑦′
( 𝑡)] = 𝑆𝑦( 𝑆) − 𝑦(0) = 𝑆𝑦( 𝑆) − 1
−2𝑳[𝑦( 𝑡)] = −2𝑦( 𝑆)
𝑳[1] =
1
𝑆
𝑳[−𝑡] = −
1
𝑆2
𝑆𝑦( 𝑆) − 1 − 2𝑦( 𝑆) =
1
𝑆
−
1
𝑆2
𝑦( 𝑆)(𝑆 − 2) =
𝑆2
− 𝑆
𝑆3
+ 1
𝑦( 𝑆) =
𝑆2
− 𝑆 + 𝑆3
𝑆3(𝑆 − 2)
=
𝑆(𝑆2
+ 𝑆 − 1)
𝑆3(𝑆 − 2)
=
𝑆2
+ 𝑆 − 1
𝑆2(𝑆 − 2)
𝑆2
+ 𝑆 − 1
𝑆2(𝑆 − 2)
=
𝐴𝑆 + 𝐵
𝑆2
+
𝐶
𝑆 − 2
𝑆2
+ 𝑆 − 1 = ( 𝐴𝑆 + 𝐵)( 𝑆 − 2) + 𝐶𝑆2
𝑆2
+ 𝑆 − 1 = 𝐴𝑆2
+ 𝐵𝑆 − 2𝐴𝑆 − 2𝐵 + 𝐶𝑆2

10
{
𝐴 + 𝐶 = 1 → 𝐶 = 1 +
1
4
=
5
4
−2𝐴 + 𝐵 = 1 → 𝐴 =
1 −
1
2
−2
= −
1
4
−2𝐵 = −1 → 𝐵 =
1
2
−
1
4
𝑳−𝟏
[
1
𝑆
] +
1
2
𝑳−𝟏
[
1
𝑆2
] +
5
4
𝑳−𝟏
[
1
𝑆 − 2
]
−
1
4
+
2
4
𝑡 +
5
4
𝑒2𝑡
=
1
4
(2𝑡 + 5𝑒2𝑡
− 1)
29) 𝒚′′
− 𝟒𝒚′
+ 𝟒𝒚 = 𝟏 𝒚( 𝟎) = 𝟏, 𝒚′( 𝟎) = 𝟒
𝑳[1] =
1
𝑆
1
𝑆 − 𝑎
𝑳[ 𝑡 𝑛
𝑒 𝑎𝑡] =
𝑛!
( 𝑠 − 𝑎) 𝑛+1
𝑳[𝑦′′(𝑡)] = 𝑆2
𝑦( 𝑆) − 𝑆𝑦(0) − 𝑦′
(0) = 𝑆2
𝑦( 𝑆) − 𝑆 − 4
−4𝑳 [𝑦′
( 𝑡)] = −4(𝑆𝑦( 𝑆) − 𝑦(0)) = −4𝑆𝑦( 𝑆) + 4
4𝑳[𝑦( 𝑡)] = 4𝑦( 𝑆)
𝑳[1] =
1
𝑆
𝑆2
𝑦( 𝑆) − 𝑆 − 4 − 4𝑆𝑦( 𝑆) + 4 + 4𝑦( 𝑆) =
1
𝑆

11
𝑆2
𝑦( 𝑆) − 4𝑆𝑦( 𝑆) + 4𝑦( 𝑆) =
1
𝑆
+ 𝑆
𝑦( 𝑆)( 𝑆2
− 4𝑆 + 4) =
1 + 𝑆2
𝑆
𝑦( 𝑆) =
1 + 𝑆2
𝑆( 𝑆 − 2)2
𝑆2
+ 1
𝑆( 𝑆 − 2)2
=
𝐴
𝑆
+
𝐵
𝑆 − 2
+
𝐵
( 𝑆 − 2)2
𝑆2
+ 1 = 𝐴( 𝑆 − 2)2
+ 𝐵𝑆( 𝑆 − 2) + 𝐶𝑆
𝑆2
+ 1 = 𝐴𝑆2
− 4𝐴𝑆 + 4𝐴 + 𝐵𝑆2
− 2𝐵𝑆 + 𝐶𝑆
{
𝐴 + 𝐵 = 1 → 𝐵 = 1 −
1
4
=
3
4
−4𝐴 − 2𝐵 + 𝐶 = 0 → 𝐶 = 1 +
6
4
=
10
4
=
5
2
4𝐴 = 1 → 𝐴 =
1
4
1
4
𝑳−𝟏
[
1
𝑆
] +
3
4
𝑳−𝟏
[
1
(𝑆 − 2)
] +
5
2
𝑳−𝟏
[
1
( 𝑆 − 2)2
]
1
4
+
3
4
𝑒2𝑡
+
5
2
𝑡𝑒2𝑡
=
1
4
(10𝑡𝑒2𝑡
+ 3𝑒2𝑡
+ 1)
30) 𝒚′′
+ 𝟗𝒚 = 𝒕 𝒚( 𝟎) = 𝒚′( 𝟎) = 0
𝑳[𝑦′′(𝑡)] = 𝑆2
𝑦( 𝑆) − 𝑆𝑦(0) − 𝑦′
(0) = 𝑆2
𝑦( 𝑆)
9𝑳[𝑦(𝑡)] = 9𝑦( 𝑆)
𝑳[ 𝑡] =
1
𝑆2

12
𝑆2
𝐹(𝑆) + 9𝐹(𝑆) =
1
𝑆2
𝐹(𝑆) =
1
𝑆2(𝑆2 + 32)
=
1
27
[
27
𝑆2(𝑆2 + 32)
]
𝑳[ 𝑘𝑡 − sin 𝑘𝑡] =
𝐾3
𝑆2(𝑆2 + 𝐾2)
1
27
𝑳−𝟏
[
27
𝑆2(𝑆2 + 32)
] =
3𝑡 − sin3𝑡
27
32) 𝒚′′
− 𝟔𝒚′
+ 𝟖𝒚 = 𝒆 𝒕
𝒚( 𝟎) = 𝟑, 𝒚′( 𝟎) = 𝟗
1
𝑆 − 𝑎
𝑳[𝑦′′(𝑡)] = 𝑆2
𝑦( 𝑆) − 𝑆𝑦(0) − 𝑦′
(0) = 𝑆2
𝑦( 𝑆) − 3𝑆 − 9
−6𝑳 [𝑦′
( 𝑡)] = −6(𝑆𝑦( 𝑆) − 𝑦(0)) = −6𝑆𝑦( 𝑆) + 18
8𝑳[𝑦( 𝑡)] = 8𝑦( 𝑆)
𝑳[ 𝑒 𝑡] =
1
𝑆 − 1
𝑆2
𝑦( 𝑆) − 3𝑆 − 9 − 6𝑆𝑦( 𝑆) + 18 + 8𝑦( 𝑆) =
1
𝑆 − 1
𝑆2
𝑦( 𝑆) − 6𝑆𝑦( 𝑆) + 8𝑦( 𝑆) =
1
𝑆 − 1
+ (3𝑆 − 9)
𝑦( 𝑆)( 𝑆2
− 6𝑆 + 8) =
1
𝑆 − 1
+ (3𝑆 − 9)
𝑦( 𝑆) =
1 + (𝑆 − 1)(3𝑆 − 9)
( 𝑆 − 1)( 𝑆 − 2)( 𝑆 − 4)
=
3𝑆2
− 9𝑆 − 3𝑆 + 10
( 𝑆 − 1)( 𝑆 − 2)( 𝑆 − 4)
=
3𝑆2
− 12𝑆 + 10
( 𝑆 − 1)( 𝑆 − 2)( 𝑆 − 4)

13
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 =
10
6
=
5
3
Para S=1
3 − 12 + 10 = 𝐴(−1)(−3) → 𝐴 =
1
3
1
3
𝑳−𝟏
[
1
( 𝑆 − 1)
] + 𝑳−𝟏
[
1
(𝑆 − 2)
] +
5
3
𝑳−𝟏
[
1
(𝑆 − 4)
]
1
3
𝑒 𝑡
+ 𝑒2𝑡
+
5
3
𝑒4𝑡
=
1
3
(5𝑒4𝑡
+ 3𝑒2𝑡
+ 𝑒 𝑡
)
DETERMINE CORRIENTE I APLICANDO TRANSFORMADA DE
LAPLACE.
𝐿 = 𝑆𝐿
𝐶 =
1
𝑆𝐶
𝑍1 = 𝑅2 +
1
𝑆𝐶
=
𝑅2 𝑆𝐶 + 1
𝑆𝐶

14
𝑍 𝑒𝑞 = 𝑍1||𝑅1 =
𝑅1 (
𝑅2 𝑆𝐶 + 1
𝑆𝐶
)
𝑅1 +
𝑅2 𝑆𝐶 + 1
𝑆𝐶
=
𝑅1 𝑅2 𝑆𝐶 + 𝑅1
𝑆𝐶
𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1
𝑆𝐶
=
𝑅1 𝑅2 𝑆𝐶 + 𝑅1
𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1
𝑉𝑅1
=
𝐸 (
𝑅1 𝑅2 𝑆𝐶 + 𝑅1
𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1
)
𝑆𝐿 + (
𝑅1 𝑅2 𝑆𝐶 + 𝑅1
𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1
)
=
𝐸𝑅1 𝑅2 𝑆𝐶 + 𝐸𝑅1
𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1
𝑆𝐿( 𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1) + 𝑅1 𝑅2 𝑆𝐶 + 𝑅1
𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1
=
𝑆𝐿( 𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1) + 𝑅1 𝑅2 𝑆𝐶 + 𝑅1
𝐼 𝑅1
=
𝑉𝑅1
𝑅1
=
[ 𝑆𝐿( 𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1) + 𝑅1 𝑅2 𝑆𝐶 + 𝑅1] 𝑅1
=
𝑅1( 𝐸𝑅2 𝑆𝐶 + 𝐸)
[ 𝑆𝐿( 𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1) + 𝑅1 𝑅2 𝑆𝐶 + 𝑅1] 𝑅1
=
𝐸𝑅2 𝑆𝐶 + 𝐸
𝑆𝐿( 𝑅1 𝑆𝐶 + 𝑅2 𝑆𝐶 + 1) + 𝑅1 𝑅2 𝑆𝐶 + 𝑅1

15
TABLA DE TRANSFORMADA DE LAPLACE UTILIZADA.

Transformada de Laplace

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Transformada de Laplace