74.1 LaPlace Transforms Concept Overview
LAPLACE TRANSFORMS | CONCEPT OVERVIEW The TOPIC of LAPLACE TRANSFORMS can be referenced on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: We have seen as we have progressed through DIFFERENTIAL EQUATIONS, that these equations can become get quite complex and harry, putting us in a position to execute quickly, or straight up pass over the problem altogether. Come exam day, passing over the problem doesn’t need to be the option…we can hack the solutions to these complex DIFFERENTIAL EQUATIONS using LAPLACE TRANSFORMS. Although the term itself may incite some uncomfortable feelings, LAPLACE TRANSFORMS can essentially break any DIFFERENTIAL EQUATION down in to a simple ALGEBRAIC expression for us to solve. For the more simple DIFFERENTIAL EQUATIONS, such as FIRST ORDER LINEAR, HOMOGENEOUS, and most HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATIONS, solving using the methods that we laid out in previous reviews will suffice, employing the LAPLACE TRANSFORM, although possible, may be more work than is needed.
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It’s when the DRIVING FUNCTIONS of the HIGHER ORDER NONHOMOGENOUS DIFFERENTIAL EQUATIONS become more and more complicated, that the power and time savings capacity of deploying the LAPLACE TRANSFORM will really be witnessed. In previous reviews, the DRIVING FUNCTIONS we were dealing with were continuous over an infinitely defined interval. However, these DRIVING FUNCTIONS may not be continuous, but rather, CONTINIOUS PIECEWISE, over a number of different subintervals. A CONTINUOUS PIECEWISE function can be broken up in to a number of subintervals on which each PIECE of that function is continuous across it’s particular SUBINTERVAL, but does not maintain that continuity across that SUBINTERVALS border. Illustrating a CONTINUOUS PIECEWISE function, we have:
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As you can see, at the border of each SUBINTERVAL, the ENDPOINT and STARTING POINT of that function in the next subinterval won’t necessarily match, and it’s distribution across that SUBINTERVAL will be unique from the others. This is what makes it a PIECEWISE function, and what makes it impossible to solve using methods we have learned up to this point. This is where LAPLACE TRANFORMS will come in to play. The algebra of deriving LAPLACE TRANSFORMS can be very gritty and convoluted and we won’t spend much time doing that. Fortunately, we don’t need to, we are provided a TABLE of the most common LAPLACE TRANFORMS, and the ones that we should expect to need and use on the exam, on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Solving a HIGHER ORDER NONHOMOGENEOUS DIFFERNTIAL EQUATION that has a DRIVING FUNCTION that is CONTINIOUS PIECEWISE is not the only scenario in which using LAPLACE TRANFORMS can be power. In some cases, our typical continuous DRIVING FUNCTION may be so complex that using LAPLACE TRANFORMS will be much more simple than the typical processes. A LAPLACE TRANSFORM will be presented using the NOMENCLATURE: ℒ 𝑓(𝑡) = 𝐹 𝑠
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This shows the through the TRANFORMATION, we produce a new function of the variable “s”, where all the “t” variables in the original function will drop out through the INTEGRATION process. This INTEGRATION, or more properly stated, the LAPLACE TRANFORM, is generally written as: !
𝑓 𝑡 𝑒 !!" 𝑑𝑡
𝐹 𝑠 = !!
This GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The UNILATERAL LAPLACE TRANSFORM PAIR represents a powerful tool for the transient and frequency response of linear time invariant systems, and is expressed as:
𝑓 𝑡 =
1 2𝜋𝑗
!!!∞
𝐹 𝑠 𝑒 !" 𝑑𝑡 !!!!∞
Where: • 𝑠 = 𝜎 + 𝑗𝜔 The usefulness of LAPLACE TRANSFORMS spans mathematics and can be used in a number of other subject areas including the analysis of circuits.
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The TABLE of COMMON LAPLACE TRANSFORMS can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This table provides us the most common LAPLACE TRANSFORMS that we will need and encounter on the day of our exam. This table is huge, although possible, we don’t want to go through the process of manually deriving each of these TRANFORMATIONS in real time. Some useful LAPLACE TRANSFORMS are represented as: 𝑓(𝑡)
𝐹(𝑠)
𝛿 𝑡 , 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 𝑎𝑡 𝑡 = 0
1
𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0
1/𝑠
𝑡 𝑢 𝑡 , 𝑅𝑎𝑚𝑝 𝑎𝑡 𝑡 = 0
1/𝑠 !
𝑒 !!"
1/(𝑠 + 𝛼)
𝑡𝑒 !!"
1/ 𝑠 + 𝛼
!
𝑒 !!" sin 𝛽𝑡
𝛽/ 𝑠 + 𝛼
𝑒 !!" cos 𝛽𝑡
(𝑠 + 𝛼)/ 𝑠 + 𝛼
!
+ 𝛽! !
+ 𝛽!
!!! !
𝑠 𝐹 𝑠 −
𝑑! 𝑓 𝑡 𝑑𝑡 !
𝑠
!!!!!
!!!
!
𝑑 ! 𝑓(0) 𝑑𝑡 !
(1/𝑠)𝐹(𝑠)
𝑓 𝜏 𝑑𝜏 ! !
𝑥 𝑡 − 𝜏 ℎ(𝜏) 𝑑𝜏
𝐻 𝑠 𝑋(𝑠)
!
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𝑓 𝑡 − 𝜏 𝑢(𝑡 − 𝜏)
𝑒 !!" 𝐹(𝑠)
limit 𝑓(𝑡)
limit 𝑠𝐹(𝑠)
limit 𝑓(𝑡)
limit 𝑠𝐹(𝑠)
!→! !→!
!→!
!→!
The LAPLACE TRANSFORM is a LINEAR OPERATOR, so when given a function with multiple terms, we can break it down term by term to determine what the complete transform will look like.
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LAPLACE TRANSFORMS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The Laplace Transform of the given function is best represented as: 𝑓 𝑡 = 𝑒 !! + 𝑒 !! + 5𝑡 !
A.
! !!!
B.
!
C.
!
+
! !!!
!
+
! !!
!
+ !!! + ! ! ! !
!
+ ! − !! ! !
!
!"
D. !!! + !!! + ! !
SOLUTION: The GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Throughout our studies, we have learned a number of techniques to deploy when solving various forms of DIFFERENTIAL EQUATIONS. These methods are solid, very useful, but at times, fall short depending on the structure or complexity of the DRIVING FUNCTION.
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Time is always of the essence when it comes to performing on the FE Exam, so quickly realizing the most efficient route to take when working DIFFERENTIAL EQUATIONS, is extremely crucial. In most cases, our typical methods of solving DIFFERENTIAL EQUATIONS will suffice. However, there are two instances where the LAPLACE TRANFORM will either significantly cut down on our effort, or flat make it possible to find a solution. These two scenarios are: 1. When the DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is so complex that a starting PARTICULAR FORMULA can not be found in the table that is provided to us on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. 2. When the DRIVING FUNCTION is a PIECEWISE FUNCTION of a defined INTERVAL. In this problem, we are given the function: 𝑓 𝑡 = 𝑒 !! + 𝑒 !! + 5𝑡 ! We are not presented this function as a DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, but rather, just simply as a function.
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We will carry out the process of determining the LAPLACE TRANSFORM, but let’s not lose sight of what we are doing. Carrying out the process will be equivalent to defining the PARTICULAR SOLUTION of HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION if that’s what we are given. Although the term itself may incite some uncomfortable feelings, LAPLACE TRANSFORMS can essentially break any DIFFERENTIAL EQUATION down in to a simple ALGEBRAIC expression for us to solve. The algebra of deriving LAPLACE TRANSFORMS can be very gritty and convoluted, we could spend time on doing each derivation, but it wouldn’t be good use for our time. Fortunately, we are provided a TABLE of the most common LAPLACE TRANFORMS, and the ones that we should expect to need and use on the exam, on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A LAPLACE TRANSFORM can be presented using the NOMENCLATURE: ℒ 𝑓(𝑡) = 𝐹 𝑠 Or in our case: ℒ 𝑒 !! + 𝑒 !! + 5𝑡 ! = 𝐹 𝑠 The LAPLACE TRANSFORM will produce a new function of the variable “s”, where all the “t” variables in the original function will drop out through the INTEGRATION process.
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This INTEGRATION, or more properly stated, the LAPLACE TRANFORM, is generally written as: !
𝑓 𝑡 𝑒 !!" 𝑑𝑡
𝐹 𝑠 = !!
And in our case: !
(𝑒 !! + 𝑒 !! + 5𝑡 ! )𝑒 !!" 𝑑𝑡
𝐹 𝑠 = !!
This GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The LAPLACE TRANSFORM is a LINEAR OPERATOR, so when given a function with multiple terms, as we are here, we can break it down term by term to determine what the complete transform will look like. Therefore, we can rewrite our LAPLACE TRANSFORM as: ℒ 𝑒 !! + ℒ 𝑒 !! + ℒ 5𝑡 ! = 𝐹 𝑠
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Or in the INTEGRATION form: !
!
(𝑒 !! )𝑒 !!" 𝑑𝑡 +
𝐹 𝑠 = !!
!
(𝑒 !! )𝑒 !!" 𝑑𝑡 + !!
(5𝑡 ! )𝑒 !!" 𝑑𝑡 !!
From this point, we can just go down the route of assessing each of these INTEGRALS, which we are capable of doing, and determine what the solution would be for this particular function. However, the algebra can get a bit convoluted, and very very time consuming…so we will rely on the TABLES…at least to the extent which we can. The TABLE of COMMON LAPLACE TRANSFORMS can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This table provides us the most common LAPLACE TRANSFORMS that we will need and encounter on the day of our exam. Some useful LAPLACE TRANSFORMS are represented as:
𝑓(𝑡)
𝐹(𝑠)
𝛿 𝑡 , 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 𝑎𝑡 𝑡 = 0
1
𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0
1/𝑠
𝑡 𝑢 𝑡 , 𝑅𝑎𝑚𝑝 𝑎𝑡 𝑡 = 0
1/𝑠 !
𝑒 !!"
1/(𝑠 + 𝛼)
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𝑡𝑒 !!"
!
1/ 𝑠 + 𝛼
𝑒 !!" sin 𝛽𝑡
𝛽/ 𝑠 + 𝛼
𝑒 !!" cos 𝛽𝑡
(𝑠 + 𝛼)/ 𝑠 + 𝛼
!
+ 𝛽! !
+ 𝛽!
!!! !
𝑠 𝐹 𝑠 −
𝑑! 𝑓 𝑡 𝑑𝑡 !
𝑠
!!!!!
!!!
!
𝑑 ! 𝑓(0) 𝑑𝑡 !
(1/𝑠)𝐹(𝑠)
𝑓 𝜏 𝑑𝜏 ! !
𝑥 𝑡 − 𝜏 ℎ(𝜏) 𝑑𝜏
𝐻 𝑠 𝑋(𝑠)
!
𝑓 𝑡 − 𝜏 𝑢(𝑡 − 𝜏)
𝑒 !!" 𝐹(𝑠)
limit 𝑓(𝑡)
limit 𝑠𝐹(𝑠)
limit 𝑓(𝑡)
limit 𝑠𝐹(𝑠)
!→!
!→!
!→!
!→!
Revisiting our function written in LAPLACE TRANSFORM terms, we have: ℒ 𝑒 !! + ℒ 𝑒 !! + ℒ 5𝑡 ! = 𝐹 𝑠 We will cycle term by term creating each TRANSFORM and combining them along the way. Let’s start with: ℒ 𝑒 !!
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Referencing the table, we match this term up with the row: 𝑓(𝑡)
𝐹(𝑠)
𝑒 !!"
1/(𝑠 + 𝛼)
This allows us to conclude that: ℒ 𝑒 !! = 1/(𝑠 − 4) Cycling to our next term, we have: ℒ 𝑒 !! Referencing the table, we match this term up with the row: 𝑓(𝑡)
𝐹(𝑠)
𝑒 !!"
1/(𝑠 + 𝛼)
This allows us to conclude that: ℒ 𝑒 !! = 1/(𝑠 − 2) Lastly, we have: ℒ 5𝑡 !
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Referencing the table, we can’t directly relate this term to one that is given, however, this is a VERY COMMON transform that should be included…and certainly one we should memorize. Generally speaking: ℒ 𝑡 ! = 𝑛!/𝑠 !!! This allows us to conclude: 5ℒ 𝑡 ! = 5(2!/𝑠 !!! ) Or:
5ℒ 𝑡 ! = 5
2! 𝑠!
Now the final step is to gather all of our LAPLACE TRANFORMS and combine them together in to a single solution, giving us:
𝐹 𝑠 =
1 1 2! + + 5 !!! 𝑠−4 𝑠−2 𝑠
Simplifying we get,
𝐹 𝑠 =
1 1 10 + + ! 𝑠−4 𝑠−2 𝑠 𝟏
𝟏
𝟏𝟎
The correct answer choice is D. 𝒔!𝟒 + 𝒔!𝟐 + 𝒔𝟑
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CONCEPT INTRO: We have seen as we have progressed through DIFFERENTIAL EQUATIONS, that these equations can become get quite complex and harry, putting us in a position to execute quickly, or straight up pass over the problem altogether. Come exam day, passing over the problem doesn’t need to be the option…we can hack the solutions to these complex DIFFERENTIAL EQUATIONS using LAPLACE TRANSFORMS. Although the term itself may incite some uncomfortable feelings, LAPLACE TRANSFORMS can essentially break any DIFFERENTIAL EQUATION down in to a simple ALGEBRAIC expression for us to solve. For the more simple DIFFERENTIAL EQUATIONS, such as FIRST ORDER LINEAR, HOMOGENEOUS, and most HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATIONS, solving using the methods that we laid out in previous reviews will suffice, employing the LAPLACE TRANSFORM, although possible, may be more work than is needed.
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It’s when the DRIVING FUNCTIONS of the HIGHER ORDER NONHOMOGENOUS DIFFERENTIAL EQUATIONS become more and more complicated, that the power and time savings capacity of deploying the LAPLACE TRANSFORM will really be witnessed. In previous reviews, the DRIVING FUNCTIONS we were dealing with were continuous over an infinitely defined interval. However, these DRIVING FUNCTIONS may not be continuous, but rather, CONTINIOUS PIECEWISE, over a number of different subintervals. A CONTINUOUS PIECEWISE function can be broken up in to a number of subintervals on which each PIECE of that function is continuous across it’s particular SUBINTERVAL, but does not maintain that continuity across that SUBINTERVALS border. Illustrating a CONTINUOUS PIECEWISE function, we have:
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As you can see, at the border of each SUBINTERVAL, the ENDPOINT and STARTING POINT of that function in the next subinterval won’t necessarily match, and it’s distribution across that SUBINTERVAL will be unique from the others. This is what makes it a PIECEWISE function, and what makes it impossible to solve using methods we have learned up to this point. This is where LAPLACE TRANFORMS will come in to play. The algebra of deriving LAPLACE TRANSFORMS can be very gritty and convoluted and we won’t spend much time doing that. Fortunately, we don’t need to, we are provided a TABLE of the most common LAPLACE TRANFORMS, and the ones that we should expect to need and use on the exam, on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Solving a HIGHER ORDER NONHOMOGENEOUS DIFFERNTIAL EQUATION that has a DRIVING FUNCTION that is CONTINIOUS PIECEWISE is not the only scenario in which using LAPLACE TRANFORMS can be power. In some cases, our typical continuous DRIVING FUNCTION may be so complex that using LAPLACE TRANFORMS will be much more simple than the typical processes. A LAPLACE TRANSFORM will be presented using the NOMENCLATURE: ℒ 𝑓(𝑡) = 𝐹 𝑠
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This shows the through the TRANFORMATION, we produce a new function of the variable “s”, where all the “t” variables in the original function will drop out through the INTEGRATION process. This INTEGRATION, or more properly stated, the LAPLACE TRANFORM, is generally written as: !
𝑓 𝑡 𝑒 !!" 𝑑𝑡
𝐹 𝑠 = !!
This GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The UNILATERAL LAPLACE TRANSFORM PAIR represents a powerful tool for the transient and frequency response of linear time invariant systems, and is expressed as:
𝑓 𝑡 =
1 2𝜋𝑗
!!!∞
𝐹 𝑠 𝑒 !" 𝑑𝑡 !!!!∞
Where: • 𝑠 = 𝜎 + 𝑗𝜔 The usefulness of LAPLACE TRANSFORMS spans mathematics and can be used in a number of other subject areas including the analysis of circuits.
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The TABLE of COMMON LAPLACE TRANSFORMS can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This table provides us the most common LAPLACE TRANSFORMS that we will need and encounter on the day of our exam. This table is huge, although possible, we don’t want to go through the process of manually deriving each of these TRANFORMATIONS in real time. Some useful LAPLACE TRANSFORMS are represented as: 𝑓(𝑡)
𝐹(𝑠)
𝛿 𝑡 , 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 𝑎𝑡 𝑡 = 0
1
𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0
1/𝑠
𝑡 𝑢 𝑡 , 𝑅𝑎𝑚𝑝 𝑎𝑡 𝑡 = 0
1/𝑠 !
𝑒 !!"
1/(𝑠 + 𝛼)
𝑡𝑒 !!"
1/ 𝑠 + 𝛼
!
𝑒 !!" sin 𝛽𝑡
𝛽/ 𝑠 + 𝛼
𝑒 !!" cos 𝛽𝑡
(𝑠 + 𝛼)/ 𝑠 + 𝛼
!
+ 𝛽! !
+ 𝛽!
!!! !
𝑠 𝐹 𝑠 −
𝑑! 𝑓 𝑡 𝑑𝑡 !
𝑠
!!!!!
!!!
!
𝑑 ! 𝑓(0) 𝑑𝑡 !
(1/𝑠)𝐹(𝑠)
𝑓 𝜏 𝑑𝜏 ! !
𝑥 𝑡 − 𝜏 ℎ(𝜏) 𝑑𝜏
𝐻 𝑠 𝑋(𝑠)
!
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𝑓 𝑡 − 𝜏 𝑢(𝑡 − 𝜏)
𝑒 !!" 𝐹(𝑠)
limit 𝑓(𝑡)
limit 𝑠𝐹(𝑠)
limit 𝑓(𝑡)
limit 𝑠𝐹(𝑠)
!→! !→!
!→!
!→!
The LAPLACE TRANSFORM is a LINEAR OPERATOR, so when given a function with multiple terms, we can break it down term by term to determine what the complete transform will look like.
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LAPLACE TRANSFORMS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The Laplace Transform of the given function is best represented as: 𝑓 𝑡 = 𝑒 !! + 𝑒 !! + 5𝑡 !
A.
! !!!
B.
!
C.
!
+
! !!!
!
+
! !!
!
+ !!! + ! ! ! !
!
+ ! − !! ! !
!
!"
D. !!! + !!! + ! !
SOLUTION: The GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Throughout our studies, we have learned a number of techniques to deploy when solving various forms of DIFFERENTIAL EQUATIONS. These methods are solid, very useful, but at times, fall short depending on the structure or complexity of the DRIVING FUNCTION.
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by Prepineer | Prepineer.com
Time is always of the essence when it comes to performing on the FE Exam, so quickly realizing the most efficient route to take when working DIFFERENTIAL EQUATIONS, is extremely crucial. In most cases, our typical methods of solving DIFFERENTIAL EQUATIONS will suffice. However, there are two instances where the LAPLACE TRANFORM will either significantly cut down on our effort, or flat make it possible to find a solution. These two scenarios are: 1. When the DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION is so complex that a starting PARTICULAR FORMULA can not be found in the table that is provided to us on page 31 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. 2. When the DRIVING FUNCTION is a PIECEWISE FUNCTION of a defined INTERVAL. In this problem, we are given the function: 𝑓 𝑡 = 𝑒 !! + 𝑒 !! + 5𝑡 ! We are not presented this function as a DRIVING FUNCTION of a HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION, but rather, just simply as a function.
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We will carry out the process of determining the LAPLACE TRANSFORM, but let’s not lose sight of what we are doing. Carrying out the process will be equivalent to defining the PARTICULAR SOLUTION of HIGHER ORDER NONHOMOGENEOUS DIFFERENTIAL EQUATION if that’s what we are given. Although the term itself may incite some uncomfortable feelings, LAPLACE TRANSFORMS can essentially break any DIFFERENTIAL EQUATION down in to a simple ALGEBRAIC expression for us to solve. The algebra of deriving LAPLACE TRANSFORMS can be very gritty and convoluted, we could spend time on doing each derivation, but it wouldn’t be good use for our time. Fortunately, we are provided a TABLE of the most common LAPLACE TRANFORMS, and the ones that we should expect to need and use on the exam, on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A LAPLACE TRANSFORM can be presented using the NOMENCLATURE: ℒ 𝑓(𝑡) = 𝐹 𝑠 Or in our case: ℒ 𝑒 !! + 𝑒 !! + 5𝑡 ! = 𝐹 𝑠 The LAPLACE TRANSFORM will produce a new function of the variable “s”, where all the “t” variables in the original function will drop out through the INTEGRATION process.
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This INTEGRATION, or more properly stated, the LAPLACE TRANFORM, is generally written as: !
𝑓 𝑡 𝑒 !!" 𝑑𝑡
𝐹 𝑠 = !!
And in our case: !
(𝑒 !! + 𝑒 !! + 5𝑡 ! )𝑒 !!" 𝑑𝑡
𝐹 𝑠 = !!
This GENERAL FORMULA for the LAPLACE TRANSFORM can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The LAPLACE TRANSFORM is a LINEAR OPERATOR, so when given a function with multiple terms, as we are here, we can break it down term by term to determine what the complete transform will look like. Therefore, we can rewrite our LAPLACE TRANSFORM as: ℒ 𝑒 !! + ℒ 𝑒 !! + ℒ 5𝑡 ! = 𝐹 𝑠
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Or in the INTEGRATION form: !
!
(𝑒 !! )𝑒 !!" 𝑑𝑡 +
𝐹 𝑠 = !!
!
(𝑒 !! )𝑒 !!" 𝑑𝑡 + !!
(5𝑡 ! )𝑒 !!" 𝑑𝑡 !!
From this point, we can just go down the route of assessing each of these INTEGRALS, which we are capable of doing, and determine what the solution would be for this particular function. However, the algebra can get a bit convoluted, and very very time consuming…so we will rely on the TABLES…at least to the extent which we can. The TABLE of COMMON LAPLACE TRANSFORMS can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. This table provides us the most common LAPLACE TRANSFORMS that we will need and encounter on the day of our exam. Some useful LAPLACE TRANSFORMS are represented as:
𝑓(𝑡)
𝐹(𝑠)
𝛿 𝑡 , 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 𝑎𝑡 𝑡 = 0
1
𝑢 𝑡 , 𝑆𝑡𝑒𝑝 𝑎𝑡 𝑡 = 0
1/𝑠
𝑡 𝑢 𝑡 , 𝑅𝑎𝑚𝑝 𝑎𝑡 𝑡 = 0
1/𝑠 !
𝑒 !!"
1/(𝑠 + 𝛼)
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𝑡𝑒 !!"
!
1/ 𝑠 + 𝛼
𝑒 !!" sin 𝛽𝑡
𝛽/ 𝑠 + 𝛼
𝑒 !!" cos 𝛽𝑡
(𝑠 + 𝛼)/ 𝑠 + 𝛼
!
+ 𝛽! !
+ 𝛽!
!!! !
𝑠 𝐹 𝑠 −
𝑑! 𝑓 𝑡 𝑑𝑡 !
𝑠
!!!!!
!!!
!
𝑑 ! 𝑓(0) 𝑑𝑡 !
(1/𝑠)𝐹(𝑠)
𝑓 𝜏 𝑑𝜏 ! !
𝑥 𝑡 − 𝜏 ℎ(𝜏) 𝑑𝜏
𝐻 𝑠 𝑋(𝑠)
!
𝑓 𝑡 − 𝜏 𝑢(𝑡 − 𝜏)
𝑒 !!" 𝐹(𝑠)
limit 𝑓(𝑡)
limit 𝑠𝐹(𝑠)
limit 𝑓(𝑡)
limit 𝑠𝐹(𝑠)
!→!
!→!
!→!
!→!
Revisiting our function written in LAPLACE TRANSFORM terms, we have: ℒ 𝑒 !! + ℒ 𝑒 !! + ℒ 5𝑡 ! = 𝐹 𝑠 We will cycle term by term creating each TRANSFORM and combining them along the way. Let’s start with: ℒ 𝑒 !!
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Referencing the table, we match this term up with the row: 𝑓(𝑡)
𝐹(𝑠)
𝑒 !!"
1/(𝑠 + 𝛼)
This allows us to conclude that: ℒ 𝑒 !! = 1/(𝑠 − 4) Cycling to our next term, we have: ℒ 𝑒 !! Referencing the table, we match this term up with the row: 𝑓(𝑡)
𝐹(𝑠)
𝑒 !!"
1/(𝑠 + 𝛼)
This allows us to conclude that: ℒ 𝑒 !! = 1/(𝑠 − 2) Lastly, we have: ℒ 5𝑡 !
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Referencing the table, we can’t directly relate this term to one that is given, however, this is a VERY COMMON transform that should be included…and certainly one we should memorize. Generally speaking: ℒ 𝑡 ! = 𝑛!/𝑠 !!! This allows us to conclude: 5ℒ 𝑡 ! = 5(2!/𝑠 !!! ) Or:
5ℒ 𝑡 ! = 5
2! 𝑠!
Now the final step is to gather all of our LAPLACE TRANFORMS and combine them together in to a single solution, giving us:
𝐹 𝑠 =
1 1 2! + + 5 !!! 𝑠−4 𝑠−2 𝑠
Simplifying we get,
𝐹 𝑠 =
1 1 10 + + ! 𝑠−4 𝑠−2 𝑠 𝟏
𝟏
𝟏𝟎
The correct answer choice is D. 𝒔!𝟒 + 𝒔!𝟐 + 𝒔𝟑
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