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Laplace Transforms of Step Functions Bernd Schr oder logo1 Bernd - - PowerPoint PPT Presentation

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Transforms and New Formulas A Model The Initial Value Problem Double Check Laplace Transforms of Step Functions Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Laplace Transforms of

slide-1
SLIDE 1

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Laplace Transforms of Step Functions

Bernd Schr¨

  • der

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-2
SLIDE 2

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-3
SLIDE 3

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-4
SLIDE 4

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-5
SLIDE 5

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t)

Original DE & IVP

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-6
SLIDE 6

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t)

Original DE & IVP ✲ L

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-7
SLIDE 7

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t)

Original DE & IVP Algebraic equation for the Laplace transform ✲ L

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-8
SLIDE 8

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform ✲ L

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-9
SLIDE 9

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform ✲ L Algebraic solution, partial fractions ❄

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-10
SLIDE 10

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform Laplace transform

  • f the solution

✲ L Algebraic solution, partial fractions ❄

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-11
SLIDE 11

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform Laplace transform

  • f the solution

✲ ✛ L L −1 Algebraic solution, partial fractions ❄

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-12
SLIDE 12

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Everything Remains As It Was

No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s)

Original DE & IVP Algebraic equation for the Laplace transform Laplace transform

  • f the solution

Solution ✲ ✛ L L −1 Algebraic solution, partial fractions ❄

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-13
SLIDE 13

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

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SLIDE 14

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-15
SLIDE 15

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-16
SLIDE 16

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-17
SLIDE 17

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

  • 2. The unit step function can be shifted and then used to

model the switching on and off of another function.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-18
SLIDE 18

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

  • 2. The unit step function can be shifted and then used to

model the switching on and off of another function.

  • 3. The function U (t −a)−U (t −b) is equal to 1 on [a,b)

and equal to zero outside [a,b).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-19
SLIDE 19

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

  • 2. The unit step function can be shifted and then used to

model the switching on and off of another function.

  • 3. The function U (t −a)−U (t −b) is equal to 1 on [a,b)

and equal to zero outside [a,b).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-20
SLIDE 20

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

  • 2. The unit step function can be shifted and then used to

model the switching on and off of another function.

  • 3. The function U (t −a)−U (t −b) is equal to 1 on [a,b)

and equal to zero outside [a,b).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-21
SLIDE 21

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

  • 2. The unit step function can be shifted and then used to

model the switching on and off of another function.

  • 3. The function U (t −a)−U (t −b) is equal to 1 on [a,b)

and equal to zero outside [a,b).

  • 4. The function f(t)U (t −a)−f(t)U (t −b) is equal to f on

[a,b) and equal to zero outside [a,b).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-22
SLIDE 22

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

  • 2. The unit step function can be shifted and then used to

model the switching on and off of another function.

  • 3. The function U (t −a)−U (t −b) is equal to 1 on [a,b)

and equal to zero outside [a,b).

  • 4. The function f(t)U (t −a)−f(t)U (t −b) is equal to f on

[a,b) and equal to zero outside [a,b).

  • 5. L
  • f(t −a)U (t −a)
  • (s) = e−asF(s).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-23
SLIDE 23

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

The Unit Step Function

  • 1. U (t) =
  • 1;

for t ≥ 0, 0; for t < 0,

  • 2. The unit step function can be shifted and then used to

model the switching on and off of another function.

  • 3. The function U (t −a)−U (t −b) is equal to 1 on [a,b)

and equal to zero outside [a,b).

  • 4. The function f(t)U (t −a)−f(t)U (t −b) is equal to f on

[a,b) and equal to zero outside [a,b).

  • 5. L
  • f(t −a)U (t −a)
  • (s) = e−asF(s).
  • 6. Keep the exponential separate when working in the

transform domain.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

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SLIDE 24

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

An Application Problem

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-25
SLIDE 25

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

An Application Problem

(Dimensions fictitious.)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

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SLIDE 26

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

An Application Problem

(Dimensions fictitious.) In an RC circuit with resistance R = 1Ω and capacitance C = 1 3F initially, the charge of the capacitor is 2C. At time t = 2π seconds, a sine shaped external voltage is activated. At time t = 5π seconds, the external voltage is turned off.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

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SLIDE 27

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

An Application Problem

(Dimensions fictitious.) In an RC circuit with resistance R = 1Ω and capacitance C = 1 3F initially, the charge of the capacitor is 2C. At time t = 2π seconds, a sine shaped external voltage is activated. At time t = 5π seconds, the external voltage is turned off. Find the charge of the capacitor as a function of time.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

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SLIDE 28

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

❞ ❞

∼

✲

RI

✛

E(t) = Rq′ + 1 Cq q C

✂ ✂ ✂❇ ❇ ❇✂ ✂ ✂ ✂ ✂ ✂ ✂ ✂ ✂ ❇ ❇ ❇ ❇ ❇ ❇ ❇ ❇ ❇

E(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-29
SLIDE 29

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Underlying Equations

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

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SLIDE 30

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Underlying Equations

◮ Ry′ + 1

Cy = E(t),

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-31
SLIDE 31

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Underlying Equations

◮ Ry′ + 1

Cy = E(t), y′ +3y = E(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-32
SLIDE 32

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Underlying Equations

◮ Ry′ + 1

Cy = E(t), y′ +3y = E(t)

◮ y(0) = 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-33
SLIDE 33

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Underlying Equations

◮ Ry′ + 1

Cy = E(t), y′ +3y = E(t)

◮ y(0) = 2 ◮ E(t) is sin(t), activated at t = 2π and deactivated at t = 5π.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-34
SLIDE 34

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Underlying Equations

◮ Ry′ + 1

Cy = E(t), y′ +3y = E(t)

◮ y(0) = 2 ◮ E(t) is sin(t), activated at t = 2π and deactivated at t = 5π. ◮ E(t) = sin(t)U (t −2π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-35
SLIDE 35

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Underlying Equations

◮ Ry′ + 1

Cy = E(t), y′ +3y = E(t)

◮ y(0) = 2 ◮ E(t) is sin(t), activated at t = 2π and deactivated at t = 5π. ◮ E(t) = sin(t)U (t −2π)−sin(t)U (t −5π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-36
SLIDE 36

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-37
SLIDE 37

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Adjusting the right side.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-38
SLIDE 38

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Adjusting the right side. sin(t)U (t −2π)−sin(t)U (t −5π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-39
SLIDE 39

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Adjusting the right side. sin(t)U (t −2π)−sin(t)U (t −5π) = sin(t −2π +2π)U (t −2π)−sin(t −5π +5π)U (t −5π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-40
SLIDE 40

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Adjusting the right side. sin(t)U (t −2π)−sin(t)U (t −5π) = sin

  • (t −2π)+2π
  • U (t −2π)−sin(t −5π +5π)U (t −5π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-41
SLIDE 41

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Adjusting the right side. sin(t)U (t −2π)−sin(t)U (t −5π) = sin

  • (t −2π)+2π
  • U (t −2π)−sin
  • (t −5π)+5π
  • U (t −5π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-42
SLIDE 42

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Adjusting the right side. sin(t)U (t −2π)−sin(t)U (t −5π) = sin

  • (t −2π)+2π
  • U (t −2π)−sin
  • (t −5π)+5π
  • U (t −5π)

= sin(t −2π)U (t −2π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-43
SLIDE 43

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Adjusting the right side. sin(t)U (t −2π)−sin(t)U (t −5π) = sin

  • (t −2π)+2π
  • U (t −2π)−sin
  • (t −5π)+5π
  • U (t −5π)

= sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-44
SLIDE 44

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-45
SLIDE 45

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y′ +3y = sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π), y(0) = 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-46
SLIDE 46

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y′ +3y = sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π), y(0) = 2 sY −2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-47
SLIDE 47

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y′ +3y = sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π), y(0) = 2 sY −2+3Y

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-48
SLIDE 48

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y′ +3y = sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π), y(0) = 2 sY −2+3Y = e−2πs 1 s2 +1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-49
SLIDE 49

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y′ +3y = sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π), y(0) = 2 sY −2+3Y = e−2πs 1 s2 +1 +e−5πs 1 s2 +1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-50
SLIDE 50

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y′ +3y = sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π), y(0) = 2 sY −2+3Y = e−2πs 1 s2 +1 +e−5πs 1 s2 +1 (s+3)Y −2 = e−2πs 1 s2 +1 +e−5πs 1 s2 +1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-51
SLIDE 51

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y′ +3y = sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π), y(0) = 2 sY −2+3Y = e−2πs 1 s2 +1 +e−5πs 1 s2 +1 (s+3)Y −2 = e−2πs 1 s2 +1 +e−5πs 1 s2 +1 (s+3)Y = e−2πs 1 s2 +1 +e−5πs 1 s2 +1 +2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-52
SLIDE 52

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y′ +3y = sin(t −2π)U (t −2π)+sin(t −5π)U (t −5π), y(0) = 2 sY −2+3Y = e−2πs 1 s2 +1 +e−5πs 1 s2 +1 (s+3)Y −2 = e−2πs 1 s2 +1 +e−5πs 1 s2 +1 (s+3)Y = e−2πs 1 s2 +1 +e−5πs 1 s2 +1 +2 Y = e−2πs 1 (s2+1)(s+3) +e−5πs 1 (s2+1)(s+3) + 2 s+3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-53
SLIDE 53

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-54
SLIDE 54

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-55
SLIDE 55

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-56
SLIDE 56

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-57
SLIDE 57

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-58
SLIDE 58

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-59
SLIDE 59

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-60
SLIDE 60

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-61
SLIDE 61

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 :

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-62
SLIDE 62

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 : 1 = 3B+ 1 10

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-63
SLIDE 63

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 : 1 = 3B+ 1 10, B = 3 10

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-64
SLIDE 64

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 : 1 = 3B+ 1 10, B = 3 10 s = 1 :

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-65
SLIDE 65

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 : 1 = 3B+ 1 10, B = 3 10 s = 1 : 1 =

  • A+ 3

10

  • 4+ 1

10 ·2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-66
SLIDE 66

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 : 1 = 3B+ 1 10, B = 3 10 s = 1 : 1 =

  • A+ 3

10

  • 4+ 1

10 ·2, A = − 1 10

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-67
SLIDE 67

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 : 1 = 3B+ 1 10, B = 3 10 s = 1 : 1 =

  • A+ 3

10

  • 4+ 1

10 ·2, A = − 1 10 1 (s2 +1)(s+3) =

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-68
SLIDE 68

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 : 1 = 3B+ 1 10, B = 3 10 s = 1 : 1 =

  • A+ 3

10

  • 4+ 1

10 ·2, A = − 1 10 1 (s2 +1)(s+3) = 1 10 −s+3 s2 +1

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-69
SLIDE 69

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Partial Fraction Decomposition.

1 (s2 +1)(s+3) = As+B s2 +1 + C s+3 1 = (As+B)(s+3)+C

  • s2 +1
  • s = −3 :

1 = 10C, C = 1 10 s = 0 : 1 = 3B+ 1 10, B = 3 10 s = 1 : 1 =

  • A+ 3

10

  • 4+ 1

10 ·2, A = − 1 10 1 (s2 +1)(s+3) = 1 10 −s+3 s2 +1 + 1 10 1 s+3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-70
SLIDE 70

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-71
SLIDE 71

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Y = e−2πs 1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +e−5πs

1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +

2 s+3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-72
SLIDE 72

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Y = e−2πs 1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +e−5πs

1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +

2 s+3 = 1 10e−2πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • + 1

10e−5πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • +2

1 s+3

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-73
SLIDE 73

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Y = e−2πs 1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +e−5πs

1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +

2 s+3 = 1 10e−2πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • + 1

10e−5πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • +2

1 s+3 y = 1 10

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-74
SLIDE 74

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Y = e−2πs 1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +e−5πs

1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +

2 s+3 = 1 10e−2πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • + 1

10e−5πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • +2

1 s+3 y = 1 10U (t −2π)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-75
SLIDE 75

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Y = e−2πs 1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +e−5πs

1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +

2 s+3 = 1 10e−2πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • + 1

10e−5πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • +2

1 s+3 y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-76
SLIDE 76

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Y = e−2πs 1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +e−5πs

1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +

2 s+3 = 1 10e−2πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • + 1

10e−5πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • +2

1 s+3 y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3t

t→t−2π

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-77
SLIDE 77

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Y = e−2πs 1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +e−5πs

1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +

2 s+3 = 1 10e−2πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • + 1

10e−5πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • +2

1 s+3 y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3t

t→t−2π

+ 1 10U (t −5π)

  • −cos(t)+3sin(t)+e−3t

t→t−5π

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-78
SLIDE 78

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Inverting the Laplace transform.

Y = e−2πs 1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +e−5πs

1 10 −s+3 s2 +1 + 1 10 1 s+3

  • +

2 s+3 = 1 10e−2πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • + 1

10e−5πs

  • −

s s2 +1 +3 1 s2 +1 + 1 s+3

  • +2

1 s+3 y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3t

t→t−2π

+ 1 10U (t −5π)

  • −cos(t)+3sin(t)+e−3t

t→t−5π +2e−3t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-79
SLIDE 79

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3t

t→t−2π

+ 1 10U (t −5π)

  • −cos(t)+3sin(t)+e−3t

t→t−5π +2e−3t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-80
SLIDE 80

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Solve the Initial Value Problem y′ +3y = sin(t)U (t −2π)−sin(t)U (t −5π), y(0) = 2.

y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3t

t→t−2π

+ 1 10U (t −5π)

  • −cos(t)+3sin(t)+e−3t

t→t−5π +2e−3t

= 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3(t−2π)

+ 1 10U (t −5π)

  • cos(t)−3sin(t)+e−3(t−5π)

+2e−3t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-81
SLIDE 81

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3(t−2π)

+ 1 10U (t −5π)

  • cos(t)−3sin(t)+e−3(t−5π)

+2e−3t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-82
SLIDE 82

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-83
SLIDE 83

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3(t−2π)

+ 1 10U (t −5π)

  • cos(t)−3sin(t)+e−3(t−5π)

+2e−3t

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-84
SLIDE 84

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3(t−2π)

+ 1 10U (t −5π)

  • cos(t)−3sin(t)+e−3(t−5π)

+2e−3t

◮ Initial value: “By inspection.” ◮ The function y = e−3t solves the differential equation

y′ +3y = 0.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-85
SLIDE 85

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

y = 1 10U (t −2π)

  • −cos(t)+3sin(t)+e−3(t−2π)

+ 1 10U (t −5π)

  • cos(t)−3sin(t)+e−3(t−5π)

+2e−3t

◮ Initial value: “By inspection.” ◮ The function y = e−3t solves the differential equation

y′ +3y = 0.

◮ So all exponential terms in the solution are o.k., provided

that the rest, which is 1 10

  • U (t −2π)−U (t −5π)
  • −cos(t)+3sin(t)
  • produces a sine function that only exists on [2π,5π).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-86
SLIDE 86

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-87
SLIDE 87

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

1 10

  • −cos(t)+3sin(t)

′ +3 1 10

  • −cos(t)+3sin(t)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-88
SLIDE 88

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

1 10

  • −cos(t)+3sin(t)

′ +3 1 10

  • −cos(t)+3sin(t)
  • =

1 10

  • sin(t)+3cos(t)
  • +3 1

10

  • −cos(t)+3sin(t)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-89
SLIDE 89

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

1 10

  • −cos(t)+3sin(t)

′ +3 1 10

  • −cos(t)+3sin(t)
  • =

1 10

  • sin(t)+3cos(t)
  • +3 1

10

  • −cos(t)+3sin(t)
  • =

1 10(3−3)cos(t)+ 1 10(1+9)sin(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-90
SLIDE 90

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

1 10

  • −cos(t)+3sin(t)

′ +3 1 10

  • −cos(t)+3sin(t)
  • =

1 10

  • sin(t)+3cos(t)
  • +3 1

10

  • −cos(t)+3sin(t)
  • =

1 10(3−3)cos(t)+ 1 10(1+9)sin(t) = sin(t)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions

slide-91
SLIDE 91

logo1 Transforms and New Formulas A Model The Initial Value Problem Double Check

Checking the Solution.

1 10

  • −cos(t)+3sin(t)

′ +3 1 10

  • −cos(t)+3sin(t)
  • =

1 10

  • sin(t)+3cos(t)
  • +3 1

10

  • −cos(t)+3sin(t)
  • =

1 10(3−3)cos(t)+ 1 10(1+9)sin(t) = sin(t) √

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Laplace Transforms of Step Functions