Laplace Transform
4.2 Properties of Laplace Transform
Understanding the properties of the Laplace Transform is crucial as it provides tools for efficiently transforming and manipulating functions. These properties greatly simplify the analysis and solution of differential equations and complex systems.
A. Existence of the Transform
The Laplace transform exists for any function that is (1) piecewise-continuous and (2) of exponential order (i.e., does not grow faster than an exponential function). A function [asciimath]f(t)[/asciimath] is said to be of exponential order [asciimath]a[/asciimath] if there exist positive constants [asciimath]M[/asciimath] and [asciimath]t_0[/asciimath] such that [asciimath]|f(t)|<=Me^(at),[/asciimath] for all [asciimath]t>=t_0[/asciimath]. For example, [asciimath]f(t)=e^(7t)cos(4t)[/asciimath] is of exponential order 7, but [asciimath]g(t)=e^(t^3)[/asciimath] is not of exponential order.
B. Linearity of the Laplace Transform
The Laplace Transform adheres to the principle of linearity. Let [asciimath]f_1[/asciimath] and [asciimath]f_2[/asciimath] be functions whose Laplace transforms exist for [asciimath]s>s_0[/asciimath], and let [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath] be constants. Then for [asciimath]s>s_0[/asciimath], the Laplace Transform of a linear combination of these functions is given by:
[asciimath]\mathcal{L}{c_1f_1+c_2f_2}=c_1\mathcal{L}{f_1}+c_2\mathcal{L}{f_2}[/asciimath]
This property is useful when dealing with linear combinations of functions.
Use the Laplace Transform Table and the linearity property to determine
[asciimath]\mathcal{L}{2e^(-3t)-6cos(4t)+9t^2}[/asciimath].
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1. From the table
[asciimath]\mathcal{L}{e^(-3t)}[/asciimath] [asciimath]=1/(s-(-3))=[/asciimath] [asciimath]1/(s+3)[/asciimath] for [asciimath]s> -3[/asciimath]
[asciimath]\mathcal{L}{cos(4t)}[/asciimath] [asciimath]=s/(s^2+4^2)=s/(s^2+16)[/asciimath] for [asciimath]s> 0[/asciimath]
[asciimath]\mathcal{L}{t^2}[/asciimath] [asciimath]=(2!)/(s^(2+1))=2/(s^3)[/asciimath] for [asciimath]s> 0[/asciimath]
2. From the linearity theorem, we have
[asciimath]\mathcal{L}{2e^(-3t)-6cos(4t)+9t^2 }=[/asciimath] [asciimath]2\mathcal{L} {e^(-3t)}-6\mathcal{L} {cos(4t)}+9\mathcal{L} { t^2}[/asciimath]
[asciimath]=2(1/(s+3))-[/asciimath] [asciimath]6(s/(s^2+16) )+9(2/(s^3) )[/asciimath]
[asciimath]=2/(s+3)-[/asciimath] [asciimath](6s)/(s^2+16)+18/(s^3)[/asciimath] for [asciimath]s>0[/asciimath]
Try an Example
C. First Shifting (Exponential) Theorem.
If [asciimath]F(s)= \mathcal{L}{f(t)}[/asciimath], then
[asciimath]\mathcal{L}[/asciimath] [asciimath]{e^(at)f(t)}=F(s-a)[/asciimath]
This theorem is valuable when solving differential equations with exponential terms or in analyzing systems with exponential inputs.
Use the first shifting theorem and the linearity property to determine
[asciimath]\mathcal{L}{2e^(9t)sin(7t)+8t^3 e^(-6t)}[/asciimath].
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Using the first shifting theorem, we have
[asciimath]\mathcal{L}[/asciimath] [asciimath]{e^(at)f(t)}=F(s-a)[/asciimath]
1. In [asciimath]\mathcal{L}{e^(9t)sin(7t)}[/asciimath], [asciimath]f(t)=sin(7t)[/asciimath] and the coefficient in the exponential term’s exponent is [asciimath]a=9[/asciimath].
[asciimath]F(s)=\mathcal{L}{sin(7t)}[/asciimath] [asciimath]=7/(s^2+7^2)[/asciimath] for [asciimath]s>0[/asciimath]
Shifting [asciimath]F(s)[/asciimath], we substitute [asciimath]s[/asciimath] with [asciimath]s-9.[/asciimath]
[asciimath]\mathcal{L}{e^(9t)sin(7t)}=F(s-9)[/asciimath] [asciimath]=7/((s-9)^2+7^2)[/asciimath] for [asciimath]s>9[/asciimath]
2. In [asciimath]\mathcal{L}{t^3e^(-6t))}[/asciimath], [asciimath]f(t)=t^3[/asciimath] and the coefficient in the exponential term’s exponent is [asciimath]a=-6[/asciimath].
[asciimath]F(s)=\mathcal{L}{t^3}[/asciimath] [asciimath]=(3!)/(s^(3+1))=(3!)/(s^4)[/asciimath] for [asciimath]s>0[/asciimath]
Shifting [asciimath]F(s)[/asciimath] , we substitute [asciimath]s[/asciimath] with [asciimath]s-(-6).[/asciimath]
[asciimath]\mathcal{L}{t^3e^(-6t))} =F(s+6)[/asciimath] [asciimath]=6/((s+6)^4)[/asciimath] for [asciimath]s> -6[/asciimath]
3. From the linearity theorem, we have
[asciimath]\mathcal{L}{2e^(9t)sin(7t)+8t^3 e^(-6t)}[/asciimath] [asciimath]=2\mathcal{L}{e^(9t)sin(7t)}[/asciimath] [asciimath]+8\mathcal{L}{t^3 e^(-6t)}[/asciimath]
[asciimath]=2(7/((s-9)^2+49) )[/asciimath] [asciimath]+8(6/(s+6)^4)[/asciimath]
[asciimath]=14/((s-9)^2+49)[/asciimath] [asciimath]+48/(s+6)^4[/asciimath] for [asciimath]s>9[/asciimath]
Try an Example
D. Differentiation in the Time Domain
Understanding how to transform derivatives is crucial for effectively solving differential equations. This property allows us to express the Laplace Transform of a function’s derivative in terms of the original function’s transform. For a function [asciimath]f(t)[/asciimath] with continuous derivatives up to [asciimath]n^{th}[/asciimath] order,
[asciimath]\mathcal{L}{f'(t)} = sF(s) - f(0)[/asciimath]
[asciimath]\mathcal{L}{f''(t)\} = s^2F(s) - s f(0) - f'(0)[/asciimath]
[asciimath]vdots[/asciimath]
[asciimath]\mathcal{L}{f^{(n)}(t)\} =[/asciimath] [asciimath]s^nF(s) - s^{n-1}f(0) - \cdots - f^{(n-1)}(0)[/asciimath]
Since we will mostly deal with second-order differential equations, we will focus on the Laplace Transform of the first and second derivatives.
For function [asciimath]f(t)=sin(3t)[/asciimath] show that [asciimath]\mathcal{L}{f'(t)} = sF(s) - f(0)[/asciimath].
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Identifying the derivative and initial value:
[asciimath]f(t)=sin(3t)[/asciimath] [asciimath]->[/asciimath] [asciimath]f'(t)=3cos(3t)[/asciimath] and
[asciimath]f(0)=sin(0)=0[/asciimath]
Finding the Laplace Transforms:
From the Laplace Transform table, we have
[asciimath]\mathcal{L}{cos(3t)}=s/(s^2+3^2)[/asciimath]
[asciimath]\mathcal{L}{sin(3t)}=3/(s^2+3^2)[/asciimath]
Applying the Differentiation Property:
We need to show
[asciimath]\mathcal{L}{3cos(3t) } = s \mathcal{L}{sin(3t) } - sin(0)[/asciimath]
Plugging in the transforms and initial value yields
[asciimath]3(s/(s^2+3^2))=s(3/(s^2+3^2))-0[/asciimath]
Simplifying both sides gives
[asciimath](3s)/(s^2+3^2)=(3s)/(s^2+3^2[/asciimath]
This equality confirms the differentiation property as the two sides match.
Find the Laplace Transform of [asciimath]y''[/asciimath] given the initial conditions [asciimath]y(0)=-3[/asciimath] and [asciimath]y'(0)=1[/asciimath]. Use [asciimath]Y[/asciimath]for [asciimath]\mathcal{L}{y}[/asciimath].
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From the differentiation property, we have
[asciimath]mathcal{ L}{y''}= s^2Y-sy(0)-y(0)[/asciimath]
Plugging in initial conditions [asciimath]y(0)=-3[/asciimath] and [asciimath]y'(0)=1[/asciimath], we obtain
[asciimath]\mathcal{L}{y''}=s^2Y+3s-1[/asciimath]
Try an Example
Table 4.2.1 summarizes the above properties of the Laplace Transform. These properties are crucial for simplifying computations and effectively utilizing the Laplace Transform in solving initial value problems.
Table 4.2.1: Properties of Laplace Transform
| Property | Example |
| [asciimath]\mathcal{L}{f+g}[/asciimath] [asciimath]=\mathcal{L}{f} +\mathcal{L}{g}[/asciimath] | [asciimath]\mathcal{L}{t+cos(2t)}[/asciimath] [asciimath]=\mathcal{L}{t}[/asciimath] [asciimath]+\mathcal{L}{cos(2t)}[/asciimath] [asciimath]=1/s^2+s/(s^2+2^2)[/asciimath] |
| [asciimath]\mathcal{L}{cf}[/asciimath] [asciimath]=c \mathcal{L}{f}[/asciimath] for any constant [asciimath]c[/asciimath] | [asciimath]\mathcal{L}{4t}[/asciimath] [asciimath]=4\mathcal{L}{t}[/asciimath] [asciimath]=4(1/s^2)[/asciimath] |
| [asciimath]\mathcal{L}{e^(at)f}(s)[/asciimath] [asciimath]=\mathcal{L}{f} (s-a)[/asciimath] | [asciimath]\mathcal{L}{e^(3t)sin(5t) }[/asciimath] [asciimath]=5/((s-3)^2+5^2)[/asciimath] |
| [asciimath]\mathcal{L}{f'}[/asciimath] [asciimath]=s\mathcal{L}{f} -f(0)[/asciimath] | |
| [asciimath]\mathcal{L}{f''}[/asciimath] [asciimath]=s^2\mathcal{L}{f} -s f(0)-f'(0)[/asciimath] | |
| [asciimath]\mathcal{L}{t^n f(t)}[/asciimath] [asciimath]=(-1)^n (d^n)/(ds^n)(\mathcal{L}{f})[/asciimath] | [asciimath]\mathcal{L}{t^1 sin(7t)}[/asciimath] [asciimath]=(-1)^1 (d)/(ds)(\mathcal{L}{sin(7t)})[/asciimath] [asciimath]=-d/(ds)(7/(s^2+7^2))[/asciimath] [asciimath]=(14s)/(s^2+49)^2[/asciimath] |
Section 4.2 Exercises
- Find the Laplace transform of the function [asciimath]f(t)=-3t^5+9sin(t), \ t gt 0[/asciimath].
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[asciimath]F(s)=-360/s^(6)+9/(s^2+1),\ sgt0[/asciimath]
- Find the Laplace transform, [asciimath]F(s)[/asciimath], of the function
[asciimath]f(t)=[/asciimath] [asciimath]10 e^(t)sin(t), \ t gt 0[/asciimath].
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[asciimath]F(s)=10/((s - 1)^2+1),\ sgt1[/asciimath]
- Find the Laplace Transform of [asciimath]y''[/asciimath] given the initial conditions [asciimath]y(0)=4[/asciimath] and [asciimath]y'(0)=-2[/asciimath].
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[asciimath]\mathcal{L}{y''}=s^2Y-4s+2[/asciimath]
