Laplace Transform
4.3 Inverse Laplace Transform
In previous sections, we defined the Laplace Transform as an integral operator that can map a function [asciimath]f(t)[/asciimath] and its derivatives in a differential equation into an algebraic equation in terms of [asciimath]s[/asciimath] and function [asciimath]F(s)[/asciimath]. As part of solving differential equations, it is often necessary to obtain [asciimath]f(t)[/asciimath] from its transform [asciimath]F(s)[/asciimath] to solve the original initial value problem. This process is facilitated by the Inverse Laplace Transform.
The formal inversion formula is typically not used directly due to its complexity. Instead, we rely on tables of Laplace Transforms to find the inverse transforms of [asciimath]F(s)[/asciimath] obtained from the original problem. The Inverse Laplace Transform is denoted as
[asciimath]f=\mathcal{L}^-1{F}[/asciimath]
Linearity of the inverse Laplace Transform
Similar to the Laplace Transform, the inverse operation is also linear. If [asciimath]F_1[/asciimath] and [asciimath]F_2[/asciimath] are functions in the s-domain with constants [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath], then then the inverse Laplace Transform of a linear combination of [asciimath]F_1[/asciimath] and [asciimath]F_2[/asciimath] for [asciimath]s>s_0[/asciimath] is given by
[asciimath]\mathcal{L}^(-1){c_1F_1+c_2F_2}=c_1\mathcal{L}^(-1){F_1}+c_2\mathcal{L}^(-1){F_2}[/asciimath]
This property ensures that the process of finding the inverse transform of a complicated expression can often be broken down into simpler, more manageable parts.
Determine [asciimath]\mathcal{L}^(-1){5/(s+7)+(8s)/ (s^2+16)}[/asciimath].
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From Table 4.1
[asciimath]e^(-7t)\ harr\ 1/(s+7)[/asciimath] and [asciimath]cos(4t)\ harr\ s/(s^2+4^2)[/asciimath]
Thus from linearity, we obtain
[asciimath]\mathcal{L}^(-1){5/(s+7)+(8s)/ (s^2+16)}[/asciimath] [asciimath]=5\mathcal{L}^(-1){1/(s+7)}[/asciimath] [asciimath]+8\mathcal{L}^(-1){s/ (s^2+16)}[/asciimath]
From the table of Laplace Transform, we get
[asciimath]=5e^(-7t)+8cos(4t)[/asciimath]
Try an Example
Try an Example
In the process of finding the inverse Laplace transform, we often encounter rational function [asciimath]F(s)[/asciimath] in the form
[asciimath]F(s)=(P(s))/(Q(s))[/asciimath]
Here, [asciimath]P(s)[/asciimath] and [asciimath]Q(s)[/asciimath] are polynomials. To ensure that [asciimath]F(s)[/asciimath] represents a valid Laplace Transform, we typically consider cases where the degree of [asciimath]P(s)[/asciimath] is less than that of [asciimath]Q(s)[/asciimath], as it can be shown that [asciimath]F(s)[/asciimath] is a Laplace transform if [asciimath]lim_(s->oo)F=0[/asciimath]. This condition is often referred to as the condition for the properness of a rational function in the Laplace domain.
In such cases, finding the inverse may require completing the square in the denominator or performing a partial fraction expansion, a technique similar to one used in integral calculus. These techniques are particularly necessary when attempting to match [asciimath]F(s)[/asciimath] to a known inverse transform from standard tables. The choice between completing the square and partial fraction decomposition depends on the nature and composition of the denominator [asciimath]Q(s)[/asciimath].
- Partial Fraction Decomposition is often the first approach considered. It is effective when the denominator [asciimath]Q(s)[/asciimath] is factorable into linear or irreducible quadratic factors. This technique breaks down complex rational expressions into simpler parts, making it easier to find the inverse Laplace Transform for each term individually.
- Completing the Square is used when the denominator [asciimath]Q(s)[/asciimath] contains quadratic terms that do not factor into real linear terms, often indicating complex roots.
To illustrate these methods, let’s proceed with a few examples demonstrating how to apply these techniques to find the inverse Laplace Transform of various functions.
Find the inverse Laplace transform
[asciimath]3/(s^2+2s+17)[/asciimath]
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The denominator is not factorable. Therefore, we try to complete the square:
[asciimath]s^2+2s+17[/asciimath] [asciimath]=s^2+2s+1+16[/asciimath] [asciimath]=(s+1)^2+16[/asciimath] [asciimath]=(s+1)^2+4^2[/asciimath]
From Table 4.1, we see that
[asciimath]b/((s-a)^2+b^2)[/asciimath] [asciimath]harr \ \e^(at) sin(bt)[/asciimath]
Thus, [asciimath]a=-1[/asciimath] and [asciimath]b=4[/asciimath] . To be able to use the above inverse transform we need to create a 4 in the numerator. So we multiply both the numerator and the denominator of the original function by 4. We obtain
[asciimath]3/(s^2+2s+17)[/asciimath] [asciimath]=3/4(4/((s+1)^2+4^2))[/asciimath]
Now, we can use the inverse from the table
[asciimath]\mathcal{L}^-1{3/(s^2+2s+17) }[/asciimath] [asciimath]=3/4\mathcal{L}^-1{4/((s+1)^2+4^2)}[/asciimath] [asciimath]=3/4 \ e^(-t) \ sin(4t)[/asciimath]
Try an Example
Find the inverse Laplace Transform
[asciimath](s^2-s-5)/((s-2)^2(s+1))[/asciimath]
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In the denominator, we have a repeated linear factor [asciimath]s-2[/asciimath] with multiplicity two and a non-repeated linear factor [asciimath]s-1[/asciimath]. This composition leads us to structure the partial fraction expansion as:
[asciimath](s^2-s-5)/((s-2)^2(s+1))[/asciimath] [asciimath]=A/(s-2)+B/(s-2)^2+C/(s+1)[/asciimath]
One way to find constants A, B, and C is to multiply both sides of the equality by [asciimath](s-2)^2(s+1)[/asciimath] to eliminate denominators:
[asciimath]s^2-s-5=A(s-2)(s+1)+B(s+1)+C(s-2)^2[/asciimath]
We can then solve for the constants by equating coefficients of like terms on both sides. This forms a system of equations.
An alternative and often simpler method is strategically choosing values for [asciimath]s[/asciimath] that simplify the equation and isolate each constant. For instance. For example
For B: Set [asciimath]s=2[/asciimath], which nullifies the terms with A and C, leading to:
[asciimath]2^2-(2)-5=A(2-2)(2+1)+B(2+1)+C(2-2)^2[/asciimath]
[asciimath]-3=3B[/asciimath] [asciimath]->[/asciimath] [asciimath]B=-1[/asciimath]
For C: Set [asciimath]s=-1[/asciimath] , simplifying the equation to solve for C:
[asciimath](-1)^2-(-1)-5=A(-1-2)(-1+1)-(-1+1)+C(-1-2)^2[/asciimath]
[asciimath]-3=9C[/asciimath] [asciimath]->[/asciimath] [asciimath]C=-1/3[/asciimath]
For A: Choose a different [asciimath]s[/asciimath] , say [asciimath]s=0[/asciimath] to isolate and solve for A:
[asciimath]-5=A(-2)(1)-(1)-1/3(-2)^2[/asciimath] [asciimath]->[/asciimath] [asciimath]2A=5-1-4/3[/asciimath] [asciimath]->[/asciimath] [asciimath]2A=8/3[/asciimath]
[asciimath]A=4/3[/asciimath]
With [asciimath]A=4/3[/asciimath], [asciimath]B=-1[/asciimath], and [asciimath]C=-1/3[/asciimath], the partial fraction becomes
[asciimath]\mathcal{L}^-1{(s^2-s-5)/((s-2)^2(s+1)) }[/asciimath] [asciimath]=\mathcal{L}^-1{4/3(1/(s-2))-1/(s-2)^2-1/3(1/(s+1))}[/asciimath]
Using linearity, the inverse Laplace Transform is
[asciimath]=4/3 \mathcal{L}^-1{1/(s-2)}-\mathcal{L}^-1{1/(s-2)^2}[/asciimath] [asciimath]-1/3\mathcal{L}^-1{1/(s+1)}[/asciimath]
Referring to the table of inverse transforms
[asciimath]1/(s-a)[/asciimath] [asciimath]harr \ \e^(at)[/asciimath] and [asciimath](n!)/(s-a)^(n+1)[/asciimath] [asciimath]harr \ \t^n e^(at)[/asciimath]
Applying these with [asciimath]a=2[/asciimath] for the first two terms and [asciimath]a=-1[/asciimath] for the last term, we obtain
[asciimath]=4/3e^(2t)-t e^(2t)-1/3e^(-t)[/asciimath]
Try an Example
Section 4.3 Exercises
- Find the inverse Laplace transform of the function [asciimath]F(s)=(-s-6)/(s^2+49),\ s gt 0[/asciimath].
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[asciimath]f(t)=-cos(7t)-6/7sin(7t)[/asciimath]
- Find the inverse Laplace transform of [asciimath]F(s) = (-7s-2)/(s^2+s-2).[/asciimath]
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[asciimath]f(t)=-4 e^(-2t)-3 e^(t)[/asciimath]
- Solving a differential equation using the Laplace transform, you find [asciimath]Y(s) = \mathcal{L}{y}[/asciimath] to be
[asciimath]Y(s) = 16/((s-7)^2+16)+(-5s)/(s^2+9)+8/(s^2+16)[/asciimath]
Find the inverse Laplace transform [asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath].
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[asciimath]y(t)=4 e^(7t)sin(4t)-5cos(3t)+2sin(4t)[/asciimath]
