laplace_transform_step_fun
TRANSCRIPT
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logo1
Transforms and New Formulas A Model The Initial Value Problem Double Check
Laplace Transforms of Step Functions
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
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7/29/2019 Laplace_transform_step_fun
2/91
logo1
Transforms and New Formulas A Model The Initial Value Problem Double Check
Everything Remains As It Was
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
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7/29/2019 Laplace_transform_step_fun
3/91
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 differentialequations with Laplace transforms stays the same.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
-
7/29/2019 Laplace_transform_step_fun
4/91
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 differentialequations with Laplace transforms stays the same.
Time Domain (t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
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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 differentialequations with Laplace transforms stays the same.
Time Domain (t)
Original
DE & IVP
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
-
7/29/2019 Laplace_transform_step_fun
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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 differentialequations with Laplace transforms stays the same.
Time Domain (t)
Original
DE & IVP EL
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
T f d N F l A M d l Th I iti l V l P bl D bl Ch k
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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 differentialequations with Laplace transforms stays the same.
Time Domain (t)
Original
DE & IVP
Algebraic equation for
the Laplace transformEL
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
T f d N F l A M d l Th I iti l V l P bl D bl Ch k
-
7/29/2019 Laplace_transform_step_fun
8/91
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 differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
Original
DE & IVP
Algebraic equation for
the Laplace transformEL
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
9/91
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 differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
Original
DE & IVP
Algebraic equation for
the Laplace transformEL
Algebraic solution,
partial fractions
c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
10/91
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 differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
Original
DE & IVP
Algebraic equation for
the Laplace transform
Laplace transform
of the solution
EL
Algebraic solution,
partial fractions
c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
11/91
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 differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
Original
DE & IVP
Algebraic equation for
the Laplace transform
Laplace transform
of the solution
E
'
L
L1
Algebraic solution,
partial fractions
c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
12/91
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 differentialequations with Laplace transforms stays the same.
Time Domain (t) Transform domain (s)
Original
DE & IVP
Algebraic equation for
the Laplace transform
Laplace transform
of the solutionSolution
E
'
L
L1
Algebraic solution,
partial fractions
c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
The Unit Step Function
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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logo1
The Unit Step Function
1. U(t) =1; for t 0,0; for t< 0,
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
The Unit Step Function
1. U(t) =1; for t 0,0; for t< 0,
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
The Unit Step Function
1. U(t) =1; for t 0,0; for t< 0,
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
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 Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
18/91
logo1
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 Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
19/91
logo1
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 Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
20/91
logo1
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 Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
21/91
logo1
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(ta)f(t)U(tb) is equal to f on
[a,b) and equal to zero outside [a,b).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
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(ta)f(t)U(tb) is equal to f on
[a,b) and equal to zero outside [a,b).5. Lf(ta)U(ta)(s) = e
asF(s).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
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(ta)f(t)U(tb) is equal to f on
[a,b) and equal to zero outside [a,b).5. Lf(ta)U(ta)(s) = e
asF(s).
6. Keep the exponential separate when working in thetransform domain.
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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logo1
An Application Problem
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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logo1
An Application Problem(Dimensions fictitious.)
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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logo1
An Application Problem(Dimensions fictitious.)
In an RC circuit with resistance R = 1 and capacitance
C=1
3F initially, the charge of the capacitor is 2C. At time
t= 2 seconds, a sine shaped external voltage is activated. Attime t= 5 seconds, the external voltage is turned off.
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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7/29/2019 Laplace_transform_step_fun
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logo1
An Application Problem(Dimensions fictitious.)
In an RC circuit with resistance R = 1 and capacitance
C=1
3F initially, the charge of the capacitor is 2C. At time
t= 2 seconds, a sine shaped external voltage is activated. Attime t= 5 seconds, the external voltage is turned off. Find thecharge of the capacitor as a function of time.
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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d
d
ERI
'
E(t) = Rq +1
Cq
q
C
f
ff
f
ff
f
ff
f
ff
E(t)
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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logo1
Underlying Equations
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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logo1
Underlying Equations Ry
+
1
Cy = E(t),
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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7/29/2019 Laplace_transform_step_fun
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logo1
Underlying Equations Ry
+
1
Cy = E(t), y
+ 3y = E(t)
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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7/29/2019 Laplace_transform_step_fun
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logo1
Underlying Equations Ry +
1
Cy = E(t), y
+ 3y = E(t)
y(0) = 2
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
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7/29/2019 Laplace_transform_step_fun
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logo1
Underlying Equations Ry +
1
Cy = E(t), y
+ 3y = E(t)
y(0) = 2 E(t) is sin(t), activated at t= 2 and deactivated at t= 5.
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Underlying Equations Ry +
1
Cy = E(t), y
+ 3y = E(t)
y(0) = 2 E(t) is sin(t), activated at t= 2 and deactivated at t= 5. E(t) = sin(t)U(t2)
Bernd Schroder Louisiana Tech University, College of Engineering and ScienceLaplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
35/91
logo1
Underlying Equations Ry +
1
Cy = E(t), y
+ 3y = E(t)
y(0) = 2 E(t) is sin(t), activated at t= 2 and deactivated at t= 5. E(t) = sin(t)U(t2) sin(t)U(t5)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3y = sin(t)U
(t2
) sin(t)U
(t5
),y(0) = 2.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3y = sin(t)U
(t2
) sin(t)U
(t5
),y(0) = 2.Adjusting the right side.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3y = sin(t)U
(t2
) sin(t)U
(t5
),y(0) = 2.Adjusting the right side.
sin(t)U(t
2)
sin(t)U(t
5)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3y = sin(t)U
(t2
) sin(t)U
(t5
),y(0) = 2.Adjusting the right side.
sin(t)U(t
2)
sin(t)U(t
5)
= sin(t2+ 2)U(t2) sin(t5+ 5)U(t5)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3y = sin(t)U
(t2
) sin(t)U
(t5
),y(0) = 2.Adjusting the right side.
sin(t)U(t
2)
sin(t)U(t
5)
= sin
(t2) + 2U(t2) sin(t5+ 5)U(t5)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3y = sin(t)U
(t2
) sin(t)U
(t5
),y(0) = 2.Adjusting the right side.
sin(t)U(t
2)
sin(t)U(t
5)
= sin
(t2) + 2U(t2) sin(t5) + 5U(t5)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3y = sin(t)U
(t2
) sin(t)U
(t5
),y(0) = 2.Adjusting the right side.
sin(t)U(t
2)
sin(t)U(t
5)
= sin
(t2) + 2U(t2) sin(t5) + 5U(t5)= sin(t2)U(t2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3y = sin(t)U
(t2
) sin(t)U
(t5
),y(0) = 2.Adjusting the right side.
sin(t)U(t
2)
sin(t)U(t
5)
= sin
(t2) + 2U(t2) sin(t5) + 5U(t5)= sin(t2)U(t2) + sin(t5)U(t5)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y + 3
y= sin(
t)U
(t2
) sin(t)U
(t5
),y(0) = 2.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
45/91
logo1
Solve the Initial Value Problem
y +
3y=
sin(t)U
(t
2)
sin(t)U
(t
5)
,
y(0) = 2.y + 3y = sin(t2)U(t2) + sin(t5)U(t5),
y(0) = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y +
3y=
sin(t)U
(t
2)
sin(t)U
(t
5)
,
y(0) = 2.y + 3y = sin(t2)U(t2) + sin(t5)U(t5),
y(0) = 2
sY2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
47/91
logo1
Solve the Initial Value Problem
y +
3y=
sin(t)U
(t
2)
sin(t)U
(t
5)
,
y(0) = 2.y + 3y = sin(t2)U(t2) + sin(t5)U(t5),
y(0) = 2
sY2 + 3Y
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
48/91
logo1
Solve the Initial Value Problem
y+ 3y = sin(t)U(t
2)
sin(t)U(t
5),
y(0) = 2.y + 3y = sin(t2)U(t2) + sin(t5)U(t5),
y(0) = 2
sY2 + 3Y = e2s 1s2 + 1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
49/91
logo1
Solve the Initial Value Problem
y+ 3y = sin(t)U(t
2)
sin(t)U(t
5),
y(0) = 2.y + 3y = sin(t2)U(t2) + sin(t5)U(t5),
y(0) = 2
sY2 + 3Y = e2s 1s2 + 1
+ e5s 1s2 + 1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
-
7/29/2019 Laplace_transform_step_fun
50/91
logo1
Solve the Initial Value Problem
y+ 3y = sin(t)U(t
2)
sin(t)U(t
5),
y(0) = 2.y + 3y = sin(t2)U(t2) + sin(t5)U(t5),
y(0) = 2
sY2 + 3Y = e2s 1s2 + 1
+ e5s 1s2 + 1
(s+ 3)Y2 = e2s 1s2 + 1
+ e5s1
s2 + 1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
S l h I i i l V l P bl
-
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logo1
Solve the Initial Value Problem
y+ 3y = sin(t)U(t
2)
sin(t)U(t
5),
y(0) = 2.y + 3y = sin(t2)U(t2) + sin(t5)U(t5),
y(0) = 2
sY2 + 3Y = e2s 1s2 + 1
+ e5s 1s2 + 1
(s+ 3)Y2 = e2s 1s2 + 1
+ e5s1
s2 + 1
(s+ 3)Y = e2
s
1
s2 + 1 + e5
s
1
s2 + 1 + 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
S l h I i i l V l P bl
-
7/29/2019 Laplace_transform_step_fun
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logo1
Solve the Initial Value Problem
y+ 3y = sin(t)U(t
2)
sin(t)U(t
5),
y(0) = 2.y + 3y = sin(t2)U(t2) + sin(t5)U(t5),
y(0) = 2
sY2 + 3Y = e2s 1s2 + 1
+ e5s 1s2 + 1
(s+ 3)Y2 = e2s 1s2 + 1
+ e5s1
s2 + 1
(s+ 3)Y = e2
s
1
s2 + 1 + e5
s
1
s2 + 1 + 2
Y = e2s1
(s2+1) (s+3)+ e5s
1
(s2+1) (s+3)+
2
s+3
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
P ti l F ti D iti
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logo1
Partial Fraction Decomposition.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
P ti l F ti D iti
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 :
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 :
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 : 1 = 3B+1
10
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 : 1 = 3B+1
10, B =
3
10
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition.
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 : 1 = 3B+1
10, B =
3
10
s = 1 :
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition.
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 : 1 = 3B+1
10, B =
3
10
s = 1 : 1 = A+ 3104 + 1
10 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition.
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logo1
Partial Fraction Decomposition.
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 : 1 = 3B+1
10, B =
3
10
s = 1 : 1 = A+ 3104 + 1
10 2, A =
1
10
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition.
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p
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 : 1 = 3B+1
10, B =
3
10
s = 1 : 1 = A+ 3104 + 1
10 2, A =
1
101
(s2 + 1) (s+ 3)=
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition.
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p
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 : 1 = 3B+1
10, B =
3
10
s = 1 : 1 = A+ 3104 + 1
10 2, A =
1
101
(s2 + 1) (s+ 3)=
1
10
s+ 3s2 + 1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Partial Fraction Decomposition.
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p
1(s2 + 1) (s+ 3)
= As+Bs2 + 1
+ Cs+ 3
1 = (As+B)(s+ 3) +Cs2 + 1
s = 3 : 1 = 10C, C=1
10
s = 0 : 1 = 3B+1
10, B =
3
10
s = 1 : 1 = A+ 3104 + 1
10 2, A =
1
101
(s2 + 1) (s+ 3)=
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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g p
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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g p
Y = e2s
110s+ 3s2 + 1
+ 110
1s+ 3
+e5s
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
+
2
s+ 3
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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Y = e2s
110s+ 3s2 + 1
+ 110
1s+ 3
+e5s
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
+
2
s+ 3
= 110
e2s ss2 + 1
+ 3 1s2 + 1
+ 1s+ 3
+
1
10e5s
ss2 + 1
+ 31
s2 + 1+
1
s+ 3
+ 2
1
s+ 3
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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Y = e2s
110s+ 3s2 + 1
+ 110
1s+ 3
+e5s
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
+
2
s+ 3
= 110
e2s ss2 + 1
+ 3 1s2 + 1
+ 1s+ 3
+
1
10e5s
ss2 + 1
+ 31
s2 + 1+
1
s+ 3
+ 2
1
s+ 3
y = 110
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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Y = e2s
110s+ 3s2 + 1
+ 110
1s+ 3
+e5s
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
+
2
s+ 3
= 110
e2s s
s2 + 1+ 3 1
s2 + 1+ 1
s+ 3
+
1
10e5s
ss2 + 1
+ 31
s2 + 1+
1
s+ 3
+ 2
1
s+ 3
y = 110U(t2)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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Y = e2s
110s+ 3s2 + 1
+ 110
1s+ 3
+e5s
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
+
2
s+ 3
= 110
e2s s
s2 + 1+ 3 1
s2 + 1+ 1
s+ 3
+
1
10e5s
ss2 + 1
+ 31
s2 + 1+
1
s+ 3
+ 2
1
s+ 3
y = 110U(t2)
cos(t) + 3sin(t) + e3t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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Y = e2s
110s+ 3s2 + 1
+ 110
1s+ 3
+e5s
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
+
2
s+ 3
= 110
e2s s
s2 + 1+ 3 1
s2 + 1+ 1
s+ 3
+
1
10e5s
ss2 + 1
+ 31
s2 + 1+
1
s+ 3
+ 2
1
s+ 3
y = 110U(t2)
cos(t) + 3sin(t) + e3t
tt2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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Y = e2s
110s+ 3s2 + 1
+ 110
1s+ 3
+e5s
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
+
2
s+ 3
= 110
e2s s
s2 + 1+ 3 1
s2 + 1+ 1
s+ 3
+
1
10e5s
ss2 + 1
+ 31
s2 + 1+
1
s+ 3
+ 2
1
s+ 3
y = 110U(t2)
cos(t) + 3sin(t) + e3t
tt2
+1
10U(t5)
cos(t) + 3sin(t) + e3t
tt5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Inverting the Laplace transform.
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Y = e2s
110s+ 3s2 + 1
+ 110
1s+ 3
+e5s
1
10
s+ 3s2 + 1
+1
10
1
s+ 3
+
2
s+ 3
= 110
e2s s
s2 + 1+ 3 1
s2 + 1+ 1
s+ 3
+
1
10e5s
ss2 + 1
+ 31
s2 + 1+
1
s+ 3
+ 2
1
s+ 3
y = 110U(t2)cos(t) + 3sin(t) + e3t tt2
+1
10U(t5)
cos(t) + 3sin(t) + e3t
tt5
+ 2e3t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Solve the Initial Value Problem
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y + 3y = sin(t)U(t
2)
sin(t)U(t
5),
y(0) = 2.
y =1
10U(t2)cos(t) + 3sin(t) + e
3t
tt2+
1
10U(t5)
cos(t) + 3sin(t) + e3t
tt5
+ 2e3t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Solve the Initial Value Problem
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y + 3y = sin(t)U(t
2)
sin(t)U(t
5),
y(0) = 2.
y =1
10U(t2)cos(t) + 3sin(t) + e
3t
tt2+
1
10U(t5)
cos(t) + 3sin(t) + e3t
tt5
+ 2e3t
=1
10U(t2)
cos(t) + 3sin(t) + e3(t2)
+ 1
10U(t5)
cos(t)3sin(t) + e3(t5)
+ 2e3t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
y =1
10U(t2)
cos(t) + 3sin(t) + e3(t2)
1 3(t 5 ) 3t
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+1
10U(t5)cos(t)3sin(t) + e
3(t5)
+ 2e3t
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Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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y = 110U(t2)cos(t) + 3sin(t) + e3(t2)
+1
10U(t5)
cos(t)3sin(t) + e3(t5)
+ 2e3t
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions
Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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y = 110U(t2)cos(t) + 3sin(t) + e3(t2)
+1
10U(t5)
cos(t)3sin(t) + e3(t5)
+ 2e3t
Initial value: By inspection. The function y = e3t solves the differential equationy + 3y = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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y = 110U(t2)cos(t) + 3sin(t) + e3(t2)
+1
10U(t5)
cos(t)3sin(t) + e3(t5)
+ 2e3t
Initial value: By inspection. The function y = e3t solves the differential equationy + 3y = 0.
So all exponential terms in the solution are o.k., provided
that the rest, which is1
10
U(t2)U(t5)
cos(t) + 3sin(t)
produces a sine function that only exists on [2,5).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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Laplace Transforms of Step Functions Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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1
10 cos(t) + 3sin(t) + 3 110 cos(t) + 3sin(t)
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Laplace Transforms of Step Functions Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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1
10 cos(t) + 3sin(t) + 3 110 cos(t) + 3sin(t)
=1
10
sin(t) + 3cos(t)
+ 3
1
10
cos(t) + 3sin(t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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1
10 cos(t) + 3sin(t) + 3 110 cos(t) + 3sin(t)
=1
10
sin(t) + 3cos(t)
+ 3
1
10
cos(t) + 3sin(t)
=
1
10 (33) cos(t) +1
10 (1 + 9) sin(t)
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Laplace Transforms of Step Functions Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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1
10 cos(t) + 3sin(t) + 3 110 cos(t) + 3sin(t)
=1
10
sin(t) + 3cos(t)
+ 3
1
10
cos(t) + 3sin(t)
=
1
10 (33) cos(t) +1
10 (1 + 9) sin(t)= sin(t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions Transforms and New Formulas A Model The Initial Value Problem Double Check
Checking the Solution.
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1
10 cos(t) + 3sin(t) + 3 110 cos(t) + 3sin(t)
=1
10
sin(t) + 3cos(t)
+ 3
1
10
cos(t) + 3sin(t)
=
1
10 (33) cos(t) +1
10 (1 + 9) sin(t)= sin(t)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laplace Transforms of Step Functions