# How To Calculate Inverse Laplace Transform?

The Inverse Laplace Transform is one of the useful equations for turning a Laplace transform into a differential equation. The process of Calculating the Laplace Transform theoretically varies. However, there is one major way that mathematicians and students have accepted, and that is by using the Laplace Transform Table.

The Inverse Laplace Transform table allows students to compute an inverse Laplace transform by using the comparing of equations method. There are many theorems and proves that establish both the LT and ILT. However, the scope of this article is to calculate the ILT and highlight the necessary equations that make it possible.

Inverse Laplace Transform Formula

The Inverse Laplace Transform of a function denoted by F(s) is is the exponentially-restricted and piecewise-continuous real function f(t) with the property

$L\left&space;\{&space;f&space;\right&space;\}(s)=L\left&space;\{&space;f&space;t\right&space;\}(s)=F(s)$

Where $L$ denotes the Laplace function.

In simpler terms, the ILT is the transformation of an LT F(s) into a time function f(t). So since $L\left&space;\{&space;f&space;\right&space;\}(s)=L\left&space;\{&space;f&space;t\right&space;\}(s)=F(s)$

Then the ILT is the

The above Fnc (1) is the standard ILT formula.

There are two major formulas that explicitly (but mostly theoretically) define the ILT, and they are the

The Mellin’s Inverse Formula

The Melllin’s Inverse formula for finding an ILT is

The i in the formula denotes complex numbers which means that the formula depends on understanding a complex plane. Theoretically, Mellin’s Inverse Formula is not the best option for calculating ILT.

Post’s Inversion Formula

The Post Inversion Formula is an impractical formula that cannot be used to determine an ILT due to the need to calculate very high orders.

The formula states;

Let f(t) be a continuous function on the interval [0, ∞) of exponential order i.e $sup_{_{t>0}}&space;\frac{f(t)}{e^{ht}}&space;<\infty$

of a real number b. For all s > b, the Laplace Transform for F(t) exists & can be infinitely differentiable with respect to S. If F(s) is the Laplace transform of f(t), the ILT of F(s) is  $f(t)=L^{-1}\left&space;\{&space;f&space;\right&space;\}(t)=(\frac{k}{t})^{k-1}&space;F^{k}&space;(\frac{k}{t})$

The above formula is impracticable as K, which is the order of the equation can contain extremely high values and render the entire equation extremely confounding.

The Laplace Transform Table.

Another way to calculate the inverse Laplace Transform Formula is by using the Laplace Transform Table. The Table for popular Laplace Transforms is outlined below

The Laplace transform tables provide the most practical method to compute ILTs because it does not require a complex variable or very long calculations.

Useful Theorems for calculating the ILT using the Laplace Tables

Theorem 1: The Linearity Property

If $F_{1},F_{2},...F_{n}$  are Laplace transforms &  $C_{1},C_{2},...C_{n}$ then

$L^{-1}(C_{1}C_{1}F_{1},C_{2}F_{2}+...+C_{n}F_{n})&space;=&space;C_{1}L^{-1}F_{1},&space;C_{2}L^{-1}F_{2}+...+C_{n}L^{-1}F_{n}$

Theorem 2:

How to Calculate the Inverse Laplace Transform Tables?

As already stated, it will be very difficult to compute the inverse Laplace transform using Mellin’s Inverse Formula or Post’s Inversion Formula. So the best method to calculate the Inverse Laplace Transform is to use the Laplace Transform Table. Below are the steps to calculate the Laplace Transform Table.

Step 2:  Compare the style of the function for which the Inverse Laplace is being computed with the stylization in the Laplace Transform Table.

Step 3: choose the pattern that matches that in the Table and perform your calculation

Step 4: Always apply the theory on which the Inverse Laplace transform is based for a quick answer.

The Inverse Laplace Transform Calculation

Example 1

Find the Inverse Laplace Transform for  $F(s)=L^{-1}\left&space;(\frac{3}{s^{2}-9}\right&space;)$

Solution

We will use the Table already listed above

Step 2:  Compare the style of the function for which the Inverse Laplace is being computed with the stylization in the Laplace Transform Table.

Now the ILT to be solved is $F(s)=L^{-1}\left&space;(\frac{3}{s^{2}-9}\right&space;)$

Now, if we replace 3 with c, we will have  $\inline&space;F(s)=L^{-1}\left&space;(\frac{3}{s^{2}-e^{2}}\right&space;)$

Now number 17 in the Laplace table above shows that

Sinhat $\rightarrow&space;\left&space;(\frac{a}{s^{2}-a^{2}}&space;\right&space;)$

The screenshot below highlights this

Step 3: choose the pattern that matches that in the Table and perform your calculation

Clearly; $\left&space;(&space;\frac{c}{s^{2}-c^{2}}&space;\right&space;)&space;=&space;\left&space;(&space;\frac{a}{s^{2}-a^{2}}&space;\right&space;)$

And this subsequently means that

$L^{-1}\left&space;(&space;\frac{e}{^{s^{2}-e^{2}}}&space;\right&space;)&space;=&space;Sinhct$

Now since c=3, Then

$F(s)=L^{-1}\left&space;(&space;\frac{3}{^{s^{2}-9}}&space;\right&space;)&space;=&space;Sinh3t$

Example 2

Find the Inverse Laplace Transform for

$F(s)=L^{-1}\left&space;(&space;\frac{s}{^{s^{2}+16}}&space;\right&space;)$

Solution

Let d = 4, then;

$F(s)=L^{-1}\left&space;(&space;\frac{s}{^{s^{2}+d^{2}}}&space;\right&space;)$

From the number 18 of the Table;    $Cosat=L^{-1}\left&space;(&space;\frac{s}{^{s^{2}+a^{2}}}&space;\right&space;)$

$\left&space;(&space;\frac{s}{^{s^{2}+d^{2}}}&space;\right&space;)&space;=&space;\left&space;(&space;\frac{s}{^{s^{2}+a^{2}}}&space;\right&space;)$

Therefore, the ILT of F(s)

$\large&space;F(s)=L^{-1}\left&space;(&space;\frac{s}{^{s^{2}+d^{2}}}&space;\right&space;)&space;=&space;Cos\mathbf{\textbf{}st}$

Since d=4

$\large&space;F(s)=L^{-1}\left&space;(&space;\frac{s}{^{s^{2}+16}}&space;\right&space;)&space;=&space;Cos\mathbf{\textbf{}4t}$

Example 3

Calculate the Inverse Laplace Transform of

$\large&space;F(s)=L^{-1}&space;(&space;\frac{9}{s+6}&space;+&space;\frac{11}{s^{}+7})$

SOLUTION

Checking the Table there is no equation that compares to the LT above, so we will need to apply the linearity property and have

$\large&space;L^{-1}&space;(&space;\frac{9}{s+6}&space;+&space;\frac{11}{s^{}+7})&space;=&space;9L^{-1}&space;\frac{1}{s+6}+11L^{-1}&space;\frac{1}{s^{2+7}}$

We will work on  $\large&space;L^{-1}&space;\frac{1}{s+6}$ this first part  of the new equation

Now from the Table, No 2 shows that $\large&space;e^{at}=\frac{1}{s-a}$ which is similar to that of $\large&space;\frac{1}{s+6}$

in the left side of the equation apart from the minus sign. So let a = -6

And we will have $\large&space;F(s)&space;=9L^{-1}&space;\frac{1}{s-a}+11L^{-1}\frac{1}{s^{2}+7}$

$\large&space;F(s)&space;=9e^{at}&space;+11L^{-1}\frac{1}{s^{2}+7}$

For the second part of the equation $\large&space;L^{-1}\frac{1}{s^{2}+7}$

Now from the same Table, $\large&space;\frac{1}{s^{2}+7}$ wil match with $\large&space;\frac{a}{s^{2}+7}=Sinat$ only if $\large&space;a=\sqrt{7}$ and if the numeratore can be made to look like 1.

To accomplish this, the next step will be to have something like this $\large&space;\frac{11}{\sqrt{7}}$ x $\large&space;\frac{\sqrt{7}}{^{s^{2}+7}}$

And subsequently, have;

$\large&space;\frac{11}{\sqrt{7}}$ x $\large&space;\frac{\sqrt{7}}{^{s^{2}+a^{2}}}$

$\large&space;\frac{11}{\sqrt{7}}$ Sinat

As such; the new equation will look  $\large&space;F(s)=9e^{at}+\frac{11}{\sqrt{7}}$ Sinat

Now replace the respective a’s with their value, and the ILT will be $\large&space;F(s)=93^{-6t}+\frac{11}{\sqrt{7}}&space;sin\sqrt{7t}$

Statistical software for Computing Inverse Laplace Transforms

The computations determined in this article are not very complex and, as such, are manually practicable. There are complex functions that require bigger and vaster calculations and, as such, are not expedient to be manually computed. In most cases, using math or statistical software to compute them is the best way to arrive at the answer. The top software for calculating ILT are

Mathematica: Inverse Laplace Transform using complex variable

MATLAB: Numerical Inversion of Laplace Transforms

Conclusion

Obviously, calculating the Laplace Transform does not require a lot of technicalities when you use the Laplace Table. This article considered three examples to highlight how the ILT calculation process work. For students to get it easy, they need to always seek to simplify a complex equation until it matches with one in the Laplace table.

There are many complex ILTs that may require the need of software and two major mathematical software that allow these computations were highlighted in this article.

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