Inverse Laplace transforms
The inverse Laplace transform undoes the Laplace transform
Normally when we do a Laplace transform, we start with a function ???f(t)??? and we want to transform it into a function ???F(s)???.
As you might expect, an inverse Laplace transform is the opposite process, in which we start with ???F(s)??? and put it back to ???f(t)???.
Hi! I'm krista.
I create online courses to help you rock your math class. Read more.
Ideally we want to simplify ???F(s)??? to the point where we can compare each part of it to a formula from a Laplace transform table.
If we can identify formulas in the table that match each part of ???F(s)???, then we can change each part back into something in terms of ???t??? and get a function for ???f(t)???.
How to calculate a Laplace transform step-by-step
Take the course
Want to learn more about Differential Equations? I have a step-by-step course for that. :)
Let’s do an inverse Laplace transform example with a partial fractions decomposition
Example
Use an inverse Laplace transform to find ???f(t)???.
???F(s)=\frac{s+7}{s^2-3s-10}???
Our goal is to simplify ???F(s)??? so that we can identify a Laplace transform formula from a table of Laplace transforms for every part of ???F(s)???. As-is, ???F(s)??? is too complicated to match to any formula from the table, so we’ll try to simplify it by factoring the denominator.
???F(s)=\frac{s+7}{(s+2)(s-5)}???
Now we’ll try a partial fractions decomposition.
???\frac{s+7}{(s+2)(s-5)}=\frac{A}{s+2}+\frac{B}{s-5}???
Multiplying both sides by the denominator from the left side, we get
???s+7=A(s-5)+B(s+2)???
If we set ???s=-2???, this will eliminate ???B??? and allow us to solve for ???A???.
???-2+7=A(-2-5)+B(-2+2)???
???5=A(-7)+B(0)???
???5=-7A???
???A=-\frac57???
If we set ???s=5???, this will eliminate ???A??? and allow us to solve for ???B???.
???5+7=A(5-5)+B(5+2)???
???12=A(0)+B(7)???
???12=7B???
???B=\frac{12}{7}???
an inverse Laplace transform is the opposite process, in which we start with F(s) and put it back to f(t).
Plugging the values we found for ???A??? and ???B??? back into the partial fractions decomposition gives
???F(s)=\frac{-\frac{5}{7}}{s+2}+\frac{\frac{12}{7}}{s-5}???
Factoring the constants out of each numerator, we can rewrite the function as
???F(s)=-\frac57\left(\frac{1}{s+2}\right)+\frac{12}{7}\left(\frac{1}{s-5}\right)???
In this form, we can see that both values inside the parentheses resemble the Laplace transform formula
???e^{at}=\frac{1}{s-a}???
We’ll just change the ???1/(s+2)??? so that it matches the transform formula exactly.
???F(s)=-\frac{5}{7}\left[\frac{1}{s-(-2)}\right]+\frac{12}{7}\left(\frac{1}{s-5}\right)???
Now we identify ???a_1=-2??? and ???a_2=5??? and use the transform formula to rewrite the equation as
???f(t)=-\frac57\left(e^{-2t}\right)+\frac{12}{7}\left(e^{5t}\right)???
???f(t)=\frac{12}{7}e^{5t}-\frac57e^{-2t}???
This is the original function ???f(t)??? that we found using an inverse Laplace transform. If we started with this function ???f(t)??? and applied a Laplace transform to it, we would find that we’d get back to the ???F(s)??? function we started with.
