Laplace Transform
4.1 Definitions
A. Introduction
In this section, we delve into an integral operator known as the Laplace Transform. This powerful tool is employed to convert initial value problems described by differential equations in one domain (e.g., t domain) into algebraic equations in another domain (s domain). Doing so facilitates a more efficient solution process, particularly for linear differential equations with constant coefficients and discontinuous or impulsive forcing terms. For instance, consider an initial value problem in the time domain
t-Domain: [asciimath]y'(t)+5y(t)=f(t),[/asciimath] [asciimath]y(0)=10[/asciimath]
Applying the Laplace Transform, the differential equation is transmuted into an algebraic equation in the s domain:
s-Domain: [asciimath]sY(s)-10+5Y(s)[/asciimath] [asciimath]=F(s)[/asciimath]
This algebraic representation in the s domain is often simpler to solve, and the solution can then be transformed back to the original t domain.
B. Definition
Let [asciimath]f(t)[/asciimath] be a function defined on [asciimath][0,oo)[/asciimath], and let [asciimath]s[/asciimath] be a real number. The Laplace transform of [asciimath]f[/asciimath] is the function [asciimath]F[/asciimath] defined by the integral
[asciimath]F(s)=int_0^ooe^(-st)f(t)dt[/asciimath] (4.1.1)
The Laplace transform of [asciimath]f[/asciimath] is denoted by both [asciimath]F[/asciimath] and [asciimath]\mathcal{L}{f}[/asciimath]. The functions can also be expressed as a transform pair [asciimath]f(t)harrF(s)[/asciimath].
The improper integral in the definition 4.1.1 is more precisely defined as
[asciimath]int_0^ooe^(-st)f(t)dt[/asciimath] [asciimath]:=lim_(T->oo)int_0^Te^(-st)f(t)dt[/asciimath]
The integral converges, meaning it results in a finite number when this limit exists and is finite.
Find the Laplace transform of the constant function [asciimath]f(t)=2[/asciimath].
Show/Hide Solution
Substituting [asciimath]f(t)=1[/asciimath] into integral 4.1.1 of the definition of the Laplace transform, we obtain
[asciimath]F(s)=int_0^ooe^(-st)(2)dt[/asciimath] [asciimath]=lim_(T->oo)int_0^Te^(-st)(2)dt[/asciimath]
[asciimath]=lim_(T->oo)[(-2e^(-st))/s]_0^T[/asciimath] [asciimath]=lim_(T->oo)[2/s-e^(-sT)/s][/asciimath] [asciimath]= {(2/s if s>0),(oo if s<=0):}[/asciimath]
Note that the integral diverges for [asciimath]s<=0[/asciimath] , so the domain of [asciimath]F(s)[/asciimath] is [asciimath]s>0[/asciimath] . Since [asciimath]e^(-sT)->0[/asciimath] when [asciimath]T->oo[/asciimath] for a fixed [asciimath]s[/asciimath] , we then get
[asciimath]F(s)=2/s[/asciimath] for [asciimath]s>0[/asciimath] or [asciimath]2harr2/s[/asciimath] as a transform pair
In general, the Laplace transform of the constant function [asciimath]f(t)=C[/asciimath] is [asciimath]\mathcal{L}{C}=C/s[/asciimath].
Find the Laplace transform of function[asciimath]f(t)=e^(at)[/asciimath].
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Substituting [asciimath]f(t)=e^(at)[/asciimath] into integral 4.1.1 of the definition of the Laplace transform, we obtain
[asciimath]F(s)=int_0^ooe^(-st)(e^(at))dt[/asciimath] [asciimath]=lim_(T->oo)int_0^Te^(-(s-a)t)dt[/asciimath]
[asciimath]=lim_(T->oo)[(-e^((s-a)t) )/(s-a)]_0^T[/asciimath] [asciimath]=lim_(T->oo)[1/(s-a)-e^((s-a)T)/(s-a)][/asciimath] [asciimath]= {(1/(s-a) if s>0),(oo if s<=a):}[/asciimath]
Note that the integral diverges for [asciimath]s<=a[/asciimath], so the domain of [asciimath]F(s)[/asciimath] is [asciimath]s>a[/asciimath]. Therefore,
[asciimath]F(s)=1/(s-a)[/asciimath] for [asciimath]s>a[/asciimath] or [asciimath]e^(at)harr1/(s-a)[/asciimath] as a transform pair
In practice, while the definition of the Laplace Transform involves an integral, it is rarely computed directly via integration due to the complexity and time-consuming nature of the process. Instead, we typically use precomputed tables of Laplace Transforms. These tables list common functions and their corresponding transforms, allowing for quick and accurate application of the Laplace Transform to solve differential equations and analyze systems. Table 4.1.1 includes the Laplace Transform of some common functions. A more comprehensive table can be found in Section 4.8.
Table 4.1.1: Brief Table of Laplace Transform
| [asciimath]f(t)[/asciimath] | [asciimath]F(s)=[/asciimath] [asciimath]\mathcal{L}{f}[/asciimath] | Domain of [asciimath]F(s)[/asciimath] |
| [asciimath]C[/asciimath] | [asciimath]C/s[/asciimath] | [asciimath]s>0[/asciimath] |
| [asciimath]t[/asciimath] | [asciimath]1/s^2[/asciimath] | [asciimath]s>0[/asciimath] |
| [asciimath]t^n,[/asciimath] [asciimath]n=1,2, ...[/asciimath] | [asciimath](n!)/s^(n+1)[/asciimath] | [asciimath]s>0[/asciimath] |
| [asciimath]e^(at)[/asciimath] | [asciimath]1/(s-a)[/asciimath] | [asciimath]s>a[/asciimath] |
| [asciimath]t^n\e^(at),[/asciimath] [asciimath]n=1,2, ...[/asciimath] | [asciimath](n!)/(s-a)^(n+1)[/asciimath] | [asciimath]s>a[/asciimath] |
| [asciimath]sin(bt)[/asciimath] | [asciimath]b/(s^2+b^2)[/asciimath] | [asciimath]s>0[/asciimath] |
| [asciimath]cos(bt)[/asciimath] | [asciimath]s/(s^2+b^2)[/asciimath] | [asciimath]s>0[/asciimath] |
| [asciimath]e^(at)sin(bt)[/asciimath] | [asciimath]b/((s-a)^2+b^2)[/asciimath] | [asciimath]s>a[/asciimath] |
| [asciimath]e^(at)cos(bt)[/asciimath] | [asciimath](s-a)/((s-a)^2+b^2)[/asciimath] | [asciimath]s>a[/asciimath] |
| [asciimath]sinh(bt)[/asciimath] | [asciimath]b/(s^2-b^2)[/asciimath] | [asciimath]s>b[/asciimath] |
| [asciimath]cosh(bt)[/asciimath] | [asciimath]s/(s^2-b^2)[/asciimath] | [asciimath]s>b[/asciimath] |
Use the table of Laplace Transform to determine the Laplace Transform of the following function:
a) [asciimath]f(t)=sin(2t)[/asciimath]
b) [asciimath]g(t)=cos(5t)[/asciimath]
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a) From the table
[asciimath]\mathcal{L}{sin(bt)}[/asciimath] [asciimath]=b/(s^2+b^2)[/asciimath] for [asciimath]s>0[/asciimath]
Recognizing that [asciimath]b=2[/asciimath], the transformation is
[asciimath]\mathcal{L}{sin(2t)}[/asciimath] [asciimath]=2/(s^2+2^2)[/asciimath] for [asciimath]s> 0[/asciimath]
b) From the table
[asciimath]\mathcal{L}{cos(bt)}[/asciimath] [asciimath]=s/(s^2+b^2)[/asciimath] for [asciimath]s>0[/asciimath]
Recognizing that [asciimath]b=5[/asciimath], the transformation is
[asciimath]\mathcal{L}{cos(5t)}[/asciimath] [asciimath]=s/(s^2+5^2)[/asciimath] for [asciimath]s> 0[/asciimath]
Try an Example
Section 4.1 Exercises
- Find the Laplace transform, [asciimath]F(s)[/asciimath], of the function [asciimath]f(t) = e^(4t) , quad t gt 0[/asciimath].
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[asciimath]F(s)=1/(s-4)[/asciimath]
- Find the Laplace Transform of, [asciimath]F(s)[/asciimath], of the function [asciimath]f(t) = cos(4t) , quad t gt 0[/asciimath].
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[asciimath]F(s)=s/(s^2+4^2)[/asciimath]
- Find the Laplace transform of the function [asciimath]f(t)=6 cosh(2 t)[/asciimath] [asciimath], \ t gt 0[/asciimath] .
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[asciimath]F(s)=(6s)/(s^2-2^2)[/asciimath]
