Inverse Laplace Transform
The Time-Domain Recovery Guide

So, the Laplace transform is an integral transformation that makes electrical and mechanical problems easier to solve.

How?

By taking a time-domain function with \(\mathrm{t}\) or \(\mathrm{x}\) as the variable and converting it into a frequency-domain function with \(\mathrm{s}\) as the variable.

But what if we are given the frequency-domain function in terms of s and we desire to know the time-domain function?

What do we do then?

We employ the Inverse Laplace Transform and reconstruct the original function in terms of t.

Laplace Transform Table

Let’s recall our table of Laplace Transform that we learned from our previous lesson.

\begin{equation}
\begin{array}{|c|c|}
\hline \text { If } f(x)= & \text { Then } L\{f(x)\}=F(s)= \\
\hline 1 & \frac{1}{s} \\
\hline k & \frac{k}{s}, s>0 \\
\hline e^{a x} & \frac{1}{s-a} \\
\hline x^n, n=1,2,3, \ldots & \frac{n !}{s^{n+1}} \\
\hline x^n e^{a x}, n=1,2,3, \ldots & \frac{n !}{(s-a)^{n+1}} s>a \\
\hline \sin (a x) & \frac{a}{s^2+a^2} \\
\hline \cos (a x) & \frac{s}{s^2+a^2} \\
\hline \sinh (a x) & \frac{a}{s^2-a^2}, s>|a| \\
\hline \cosh (a x) & \frac{s}{s^2-a} \\
\hline x \sin (a x) & \frac{2 a s}{\left(s^2+a^2\right)^2} \\
\hline x \cos (a x) & \frac{s^2-a^2}{\left(s^2+a^2\right)^2} \\
\hline e^{a x} \sin (b x) & \frac{b}{(s-a)^2+b^2} \\
\hline e^{a x} \cos (b x) & \frac{s-a}{(s-a)^2+b^2}\\
\hline
\end{array}
\end{equation}

Typically, we work from the left-side of the table to produce the right-side. But in this lesson, we will go in reverse direction by trying to match the Laplace Transform to one of the functions.

If \(\mathrm{F}\) is the Laplace transform of a continuous function \(\mathrm{f}\), such that \(L\{f\}=F\), then the inverse Laplace transform of \(\mathrm{F}\) is \(L^{-1}\{F\}=f\).

In other words, the inverse Laplace transformation recovers the continuous function \(\mathrm{f}\) when \(\mathrm{F}\) is given and is a well-defined, linear operation.

Example 1: Basic Inverse Laplace Transform

Let’s look at an example.

Evaluate \(L^{-1}\left\{\frac{5}{s^{2}+25}\right\}\)

First, we will match the form to the Laplace transforms table to be:

\begin{equation}
\frac{a}{s^2+a^2}
\end{equation}

This indicates to us that all we need to do is identify a, which is 5.

Next, we must recognize that the continuous function that produces such a transformation is of the form \(\sin (a x)\). So, to arrive at our desired result, we simply substitute our value of a into our known function

\begin{equation}
\text { If } L\{\sin (a x)\}=\frac{a}{s^2+a^2} \text {, then }
\end{equation}

\begin{equation}
L^{-1}\left\{\frac{5}{s^2+25}\right\}=\sin (5 x)
\end{equation}

So, \(f(x)=\sin (5 x)\)!

Let’s try another one.

Example 2: Inverse Laplace Transform with Adjustments

Evaluate \(L^{-1}\left\{\frac{1}{s^{4}}\right\}\)

First, we will match the form to the Laplace transforms table to be:

\begin{equation}
\frac{n !}{s^{n+1}}
\end{equation}

This indicates to us that we need to find \(\mathrm{n}\), which we identify to be:

\begin{equation}
n+1=4 \text {, so } n=3
\end{equation}

Next, we must recognize that the continuous function that produces such a transformation is of the form \(x^{n}\).

But because our numeartor in our tranform is not \(\mathrm{n}\) ! we must multiply and divide by 3 ! To arrive at our desired result.

\begin{equation}
\begin{aligned}
& \frac{3 !}{3 !} L^{-1}\left\{\frac{1}{s^4}\right\} \\
& \frac{1}{3 !} \underbrace{L^{-1}\left\{\frac{3 !}{s^4}\right\}}_{x^3}
\end{aligned}
\end{equation}

\begin{equation}
f(x)=\frac{1}{6} x^3
\end{equation}

See, not too bad, right?

Additional Techniques and Considerations

Now, sometimes the inverse transform will be straight forward, and other times we will need to have a little ingenuity and use techniques such as Completing the Square and Partial Fractions, for which this lesson will rely heavily upon, in order to recover the continuous function \(\mathrm{f}\).

In fact, Paul’s Online Notes even makes the statement, “Partial fractions are a fact of life when using Laplace transforms to solve differential equations.”

Next Steps

In the video, you’ll…

• Refresh your understanding of partial fractions to efficiently “fix up” the function of s to match the table
• Work through ten detailed examples to become proficient in evaluating inverse Laplace transforms
• Gain a deeper understanding of the concepts explained in the article by watching the video
• Visualize the step-by-step process with clear examples and explanations

Let’s go!

Video Tutorial w/ Full Lesson & Detailed Examples

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