Laplace Transform
4.5 Laplace Transform of Piecewise Functions
A. Step function
In this section, we explore how to apply the Laplace Transform to piecewise continuous functions. In the next section, we will address solving initial value problems that involve second-order differential equations with constant coefficients where the forcing function [asciimath]f(t)[/asciimath] is a continuous piecewise function.
Jump discontinuities often occur in physical situations like switching mechanisms or abrupt changes in forces acting on the system. To handle such discontinuities in the Laplace domain, we utilize the unit step function to transform piecewise functions into a form amenable to Laplace transforms and subsequently find piecewise continuous inverses of Laplace transforms for the solution.
The unit step function (Heaviside function) [asciimath]u(t)[/asciimath] is defined as
[asciimath]u(t) = {(0,t lt 0),(1,t ge 0):}[/asciimath]
It steps (or jumps) from 0 to 1 at [asciimath]t=0[/asciimath]. By shifting the input argument [asciimath]t[/asciimath], we can move the step to different locations.
[asciimath]u(t-a) = {(0,t-a lt 0),(1,t-a ge 0):}[/asciimath] [asciimath]->[/asciimath] [asciimath]u_a(t)=u(t-a) = {(0,t lt a ),(1,t ge a ):}[/asciimath]
The step function can also be transformed, e.g., shifted, stretched, or compressed. For example, by multiplying [asciimath]u(t)[/asciimath] by some constant [asciimath]Mgt1[/asciimath], we can stretch it vertically.
[asciimath]Mu(t-a) = {(0,t lt a ),(M,t ge a ):}[/asciimath]
Or by combined shifting and reflecting [asciimath]u(t)[/asciimath], we can opposite the way the function switches on and off.
[asciimath]1-u(t-a) = {(1,t lt a ),(0,t ge a ):}[/asciimath]
The step function enables us to represent any piecewise continuous function conveniently. For instance, consider the function
[asciimath]f(t) = {(f_0(t), \ 0 le t lt a),(f_1(t), t ge a):}[/asciimath]
[asciimath]= f_0(t){(1, \ 0 le t lt a),(0, t ge a):}[/asciimath] [asciimath]+f_1(t){(0, \ 0 le t lt a),(1, t ge a):}[/asciimath]
[asciimath]=f_0(t)(1-u(t-a))+f_1(t)u(t-a)[/asciimath]
It can be rewritten as
[asciimath]f(t)= f_0(t) + u(t-a)(f_1(t)-f_0(t))[/asciimath] (4.5.1)
We can extend Equation 4.5.1 to more general continuous piecewise functions.
[asciimath]f(t) = {(f_0(t), \ 0 le t lt a),(f_1(t), a le t lt b), (f_2(t), t ge b) :}[/asciimath]
[asciimath]f(t)= f_0(t) + u(t-a)(f_1(t)-f_0(t))[/asciimath] [asciimath]+u(t-b)(f_2(t)-f_1(t))[/asciimath] (4.5.2)
B. Laplace Transform of Piecewise Functions
The Laplace Transform of the step-modulated function is key in solving differential equations with piecewise forcing functions.
Theorem: Laplace Transform of a Step-Modulated Function. Let [asciimath]g(t)[/asciimath] be defined on [asciimath][0,oo)[/asciimath], suppose [asciimath]a ge 0[/asciimath], and assume [asciimath]\mathcal{L}{g(t+a)}[/asciimath] exists for [asciimath]sgts_0[/asciimath]. Then
[asciimath]\mathcal{L}{u(t-a)g(t)}=e^(-as)\mathcal{L}{g(t+a)}[/asciimath] (4.5.3)
This theorem enables the transformation of step-modulated functions into the Laplace domain, which can then be manipulated algebraically.
Find the Laplace transform of [asciimath]u(t-1)3t^2[/asciimath].
Show/Hide Solution
To apply Equation 4.5.3, we take [asciimath]g(t)=3t^2[/asciimath] and [asciimath]a=1[/asciimath]. Therefore, we have
[asciimath]g(t+1)=3(t+1)^2[/asciimath] [asciimath]=3t^2+6t+3[/asciimath]
From the table then, we find [asciimath]\mathcal{L}{g(t+1)}[/asciimath].
[asciimath]\mathcal{L}{g(t+1)}[/asciimath] [asciimath]=\mathcal{L}{3t^2+6t+3}[/asciimath]
[asciimath]=3\mathcal{L}{t^2}[/asciimath] [asciimath]+6\mathcal{L}{t}[/asciimath] [asciimath]+\mathcal{L}{3}[/asciimath]
[asciimath]=6/s^3+6/s^2+3/s[/asciimath]
Thus by Equation 4.5.3, we have
[asciimath]\mathcal{L}{u(t-1)3t^2}=e^(-s)(6/s^3+6/s^2+3/s )[/asciimath]
Find the Laplace transform of
[asciimath]f(t) = {(2t-1, \ \ 0 le t lt 2),(4t, t ge 2):}[/asciimath]
Show/Hide Solution
We first write [asciimath]f(t)[/asciimath] in terms of the step function using Equation 4.5.1 with [asciimath]a=2[/asciimath], [asciimath]f_0(t)=2t-1[/asciimath], and [asciimath]f_1(t)=4t[/asciimath].
[asciimath]f(t)= 2t-1+ u(t-2)(4t-2t+1)[/asciimath]
[asciimath]=2t-1+u(t-2)(2t+1)[/asciimath]
Taking the Laplace transform, we have
[asciimath]\mathcal{L}{f(t)}=\mathcal{L}{2t-1}[/asciimath] [asciimath]+\mathcal{L}{u(t-2)(2t+1) }[/asciimath]
To apply Equation 4.5.3 to the second term, we take [asciimath]g(t)=2t+1[/asciimath] and [asciimath]a=2[/asciimath] .
[asciimath]g(t+2)=2(t+2)+1[/asciimath] [asciimath]=2t+5[/asciimath]
We have then
[asciimath]\mathcal{L}{f(t)}=\mathcal{L}{2t-1}[/asciimath] [asciimath]+e^(-2s)\mathcal{L}{g(t+2)}[/asciimath]
[asciimath]=\mathcal{L}{2t}-\mathcal{L}{1}[/asciimath] [asciimath]+e^(-2s) \mathcal{L}{2t+5 }[/asciimath]
[asciimath]=2/s^2-1/s+e^(-2s)(2/s^2+5/s)[/asciimath]
Try an Example
C. Inverse Laplace Transform of Piecewise Functions
The previous theorem also allows us to determine the inverse Laplace Transform of functions that arise from piecewise functions. However, it will be more convenient to shift the argument of [asciimath]g(t)[/asciimath] and replace [asciimath]g(t)[/asciimath] with [asciimath]g(t-a)[/asciimath].
Translation in [asciimath]t[/asciimath] Theorem. If [asciimath]a ge 0[/asciimath] and [asciimath]L(g)[/asciimath] exists for [asciimath]sgts_0[/asciimath], then
[asciimath]\mathcal{L}{u(t-a)g(t-a)}=e^(-as)\mathcal{L}{g(t)}[/asciimath]
Given [asciimath]G(s)=\mathcal{L}{g(t)}[/asciimath], it is equivalent to
[asciimath]u(t-a)g(t-a)[/asciimath] [asciimath]harr[/asciimath] [asciimath]e^(-as) G(s)[/asciimath] (4.5.4)
Find the inverse Laplace transform of the given function and find distinct formulas for [asciimath]h(t)[/asciimath] on appropriate intervals.
[asciimath]H(s)=2/s-s/(s^2+1)+e^(-pi/2 s)((s-1)/(s^2+1))[/asciimath]
Show/Hide Solution
Since [asciimath]H(s)[/asciimath] has [asciimath]e^(-as)[/asciimath] as a factor, we use Equation 4.5.4 to determine the inverse.
Letting [asciimath]H_0(s)=2/s-s/(s^2+1)[/asciimath] and [asciimath]H_1(s)=(s-1)/(s^2+1)[/asciimath], we obtain
[asciimath]h_0(t)=\mathcal{L}^-1{2/s-s/(s^2+1) }[/asciimath] [asciimath]=2-cos(t)[/asciimath]
[asciimath]h_1(t)=\mathcal{L}^-1{s/(s^2+1) -1/(s^2+1) }[/asciimath] [asciimath]=cos(t)-sin(t)[/asciimath]
Using Equation 4.5.4 with [asciimath]a=pi/2[/asciimath] and linearity of [asciimath]\mathcal{L}^-1[/asciimath], we have
[asciimath]h(t)=[/asciimath] [asciimath]\mathcal{L}^-1{H_0(s)}[/asciimath] [asciimath]+ \mathcal{L}^-1{e^(-pi/2s)H_1(s)}[/asciimath]
[asciimath]=h_0(t)[/asciimath] [asciimath]+ u(t-pi/2)(h_1(t-pi/2))[/asciimath]
[asciimath]=2-cos(t)[/asciimath] [asciimath]+u(t-pi/2)(cos(t-pi/2)-sin(t-pi/2))[/asciimath]
We simplify it using trigonometric identities: [asciimath]cos(t-pi/2)=sin(t)[/asciimath] and [asciimath]sin(t-pi/2)=-cos(t)[/asciimath]. Applying these identities yields
[asciimath]h(t)=2-cos(t)[/asciimath] [asciimath]+u(t-pi/2)(sin(t)+cos(t))[/asciimath]
From Equation 4.5.1, we recognize that
- The expression without a unit function, [asciimath]2-cos(t)[/asciimath], corresponds to [asciimath]f_0(t)[/asciimath], the function active before the step.
- The expression multiplied by the unit function, [asciimath]sin(t)+cos(t)[/asciimath], represents the change in the function at the step, thus corresponding to [asciimath]f_1-f_0[/asciimath].
Given [asciimath]f_0(t)=2-cos(t)[/asciimath] , we can solve for [asciimath]f_1(t)[/asciimath].
[asciimath]f_1-f_0=sin(t)+cos(t)[/asciimath]
[asciimath]f_1-(2-cos(t) )=sin(t)+cos(t)[/asciimath]
[asciimath]f_1(t)=sin(t)+2[/asciimath]
We can now express [asciimath]h(t)[/asciimath] as a piecewise function.
[asciimath]h(t) = {(2-cos(t), \ \ 0 le t lt pi/2),(sin(t)+2, t ge pi/2):}[/asciimath]
Try an Example
Section 4.5 Exercises
- Find the Laplace transform, [asciimath]F(s)[/asciimath] of [asciimath]f(t)[/asciimath].
[asciimath]f(t) = {(0,"if",t lt 3),(2 (t-3),"if",3 le t lt 7),(8,"if",t gt 7):}.[/asciimath]
Show/Hide Answer
[asciimath]F(s)=(2 e^(-3s))/s^2-(2 e^(-7s))/s^2[/asciimath]
- Take the inverse Laplace transform to determine [asciimath]y(t)[/asciimath]. Enter [asciimath]u_a(t)[/asciimath] for [asciimath]u(t-a)[/asciimath] if the unit function is a part of the inverse.
[asciimath]Y(s)=e^(-2s)/(s^2+4s+8)[/asciimath]
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[asciimath]y(t)=1/2sin(2(t-2))e^(-2(t-2))u_2(t)[/asciimath]
- Apply the Laplace transform to the differential equation, and solve for [asciimath]Y(s)[/asciimath].
[asciimath]y'' + 9 y = 4 (t-2) u_2(t) - 4 (t-3) u_3(t), \quad y(0) = y'(0) = 0[/asciimath]
Show/Hide Answer
[asciimath]Y(s)=(4 e^(-2s)-4 e^(-3s))/(s^2(s^2+9))[/asciimath]
