Inverse Laplace Transforms

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.4 – Inverse Laplace Transforms

Expected Educational Results

• Objective 21–1: I understand the Laplace transform as a map of a function onto a specific class of functions.
• Objective 21–2: I can find the inverse Laplace Transform of a given transform.

Inverse Laplace Transforms

Definition: Inverse Laplace Transform

Given $F(s)$, if there is a function $f(t)$ that is continuous on $[0,\infty)$ and satisfies $\mathcal{L}\left\{f(t)\right\}(s)=F(s)$ then $f(t)$ is the inverse Laplace transform of $F(s)$, then $\displaystyle \mathcal{L}^{-1}\left\{F(s)\right\}(t)=f(t)$

Integral Form for Inverse Laplace Transform

Here is the formula for calculating the inverse Laplace transform:

$\displaystyle\mathcal{L}^{-1}\left\{F(s)\right\}(t)=\frac{1}{2\pi}\lim_{Y\rightarrow +\infty}\int_{-Y}^Y{e^{t(\sigma+i\xi)}F(\sigma+i\xi)\,d\xi}$ where the integral is over a line in the complex plane for suitable $\sigma>0$.

NOTE: It will be easier to use the method of partial fractions and your knowledge of Laplace transforms to compute an inverse Laplace transform!

Linearity of the Inverse Laplace Transform

Theorem: Linearity of the Inverse Laplace Transform

Assume $\mathcal{L}^{-1}\left\{F(s)\right\}(t)$, $\mathcal{L}^{-1}\left\{F_1(s)\right\}(t)$, and $\mathcal{L}^{-1}\left\{F_2(s)\right\}(t)$ exist and are continuous on $[0,\infty)$. Let $c$ be a constant. Then

$\mathcal{L}^{-1}\left\{F_1(s)+F_2(s)\right\}(t)=\mathcal{L}^{-1}\left\{F_1(s)\right\}(t)+\mathcal{L}^{-1}\left\{F_2(s)\right\}(t)$ and

$\mathcal{L}^{-1}\left\{cF(s)\right\}(t)=c\mathcal{L}^{-1}\left\{F(s)\right\}(t)$.

NOTE: Often the Method of Partial Fractions will be needed to compute an inverse Laplace transform!

Example 01:

Find the inverse Laplace transform of $\displaystyle F(s)={\color{gsured}\frac{1}{s^2-3s-10}}$.

Solution:

$f(t)=\mathcal{L}^{-1}\left\{{\color{gsured}\dfrac{1}{s^2-3s-10}}\right\}(t)$

Using method of partial fractions:

Use Technology to use the method of partial fractions

Mathematica

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(* Separate into partial fractions: 1/(s^2-3s-10) *)
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Apart[1/(s^2-3s-10)]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.
2. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

$\Rightarrow f(t) =\mathcal{L}^{-1}\left\{\dfrac{\dfrac{1}{7}}{s-5}-\dfrac{\dfrac{1}{7}}{s+2}\right\}(t)$

Using the linearity property of inverse Laplace transforms:

$\Rightarrow f(t)=\frac{1}{7}\mathcal{L}^{-1}\left\{\dfrac{1}{s-5}\right\}(t)-\frac{1}{7}\mathcal{L}^{-1}\left\{\dfrac{1}{s+2}\right\}(t)$

Using our knowledge of Laplace transforms:

$\Rightarrow f(t)=\frac{1}{7}e^{5t} - \frac{1}{7}e^{-2t}$

Check Your Work

Use Technology to Verify Inverse Laplace Transform

Mathematica

xxxxxxxxxx
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(* Example from Example 01: Find inverse Laplace transform of 1/(s^2-3s-10) *)
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InverseLaplaceTransform[1/(s^2-3s-10), s, t]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

3. The arguments for InverseLaplaceTransform[ ] are:

   1. the function that the inverse transform is being applied;
   2. the independent variable of the function;
   3. the independent variable for the inverse transform.

Example 02:

Find the inverse Laplace transform of $\displaystyle F(s)={\color{gsured}\frac{s+3}{s^2+4}}$.

Solution:

$f(t)=\mathcal{L}^{-1}\left\{\dfrac{s+3}{s^2+4}\right\}(t)$

Rewrite using algebra:

$\Rightarrow f(t)=\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+4}+\dfrac{3}{s^2+4}\right\}(t)$

Use the linearity property of inverse Laplace transforms:

$\Rightarrow f(t)=\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+4}\right\}(t)+{3}\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+4}\right\}(t)$

Rewriting using algebra:

$\Rightarrow f(t)=\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+4}\right\}(t)+\dfrac{1}{2}3\mathcal{L}^{-1}\left\{\dfrac{2}{s^2+4}\right\}(t)$

Using our knowledge of Laplace transforms:

$\Rightarrow f(t)=\cos{(2t)} + \frac{3}{2}\sin{(2t)}$

Example 03:

Find the inverse Laplace transform of $\displaystyle F(s)={\color{gsured}\frac{s}{(s+2)^2-4}}$.

Solution:

$f(t)=\mathcal{L}^{-1}\left\{\dfrac{s}{(s+2)^2-4}\right\}(t)$

Rewrite using algebra:

$\Rightarrow f(t)=\mathcal{L}^{-1}\left\{\dfrac{s \,+\,2-2}{(s+2)^2-4}\right\}(t)$

Rewrite using algebra:

$\Rightarrow f(t)=\mathcal{L}^{-1}\left\{\dfrac{s\,+\,2}{(s+2)^2-4}+\dfrac{\,-\,2}{(s+2)^2-4}\right\}(t)$

Using the linearity property of inverse Laplace transform:

$\Rightarrow f(t)=\mathcal{L}^{-1}\left\{\dfrac{s\,+\,2}{(s+2)^2-4}\right\}(t) \,-\, \mathcal{L}^{-1}\left\{\dfrac{2}{(s+2)^2-4}\right\}(t)$

Using our knowledge of Laplace transforms:

$\Rightarrow f(t)=e^{-2t}\cosh{(2t)} + e^{-2t}\sinh{(2t)}$

Investigation 01

Determine the inverse Laplace transforms of the given functions:

1. $\displaystyle F(s)=\frac{3}{s^2+9}$
2. $\displaystyle F(s)=\frac{2}{s-5}$
3. $\displaystyle F(s)=\frac{s}{s^2-4}$
4. $\displaystyle F(s)=\frac{s-4}{(s-4)^2+1}$
5. $\displaystyle F(s)=\frac{1}{s^4}$
6. $\displaystyle F(s)=\frac{1}{s(s^2+3)}$
7. $\displaystyle F(s)=\frac{3s+2}{s^4+9s^3}$
8. $\displaystyle F(s)=\frac{s}{(s-3)^2+7}$
9. $\displaystyle F(s)=\frac{s-1}{s^3+4s^2+9s}$
10. $\displaystyle F(s)=\frac{1}{s^5+4s^3+4s}$

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Last Modified: Sunday, 8 November 2020 17:10 EDT