View source for Applied Mathematics/Laplace Transforms

The Laplace transform is an integral transform which is widely used in physics and engineering. 

Laplace Transforms involve a technique to change an expression into another form that is easier to work with using an improper integral.  We usually introduce Laplace Transforms in the context of differential equations, since we use them a lot to solve some differential equations that can't be solved using other standard techniques.      However, Laplace Transforms require only improper integration techniques to use. So you may run across them in first year calculus.

Notation: The Laplace Transform is denoted as <math>\displaystyle\mathcal{L} \left\{f(t)\right\}</math>.

The Laplace transform is named after mathematician and astronomer Pierre-Simon Laplace.

{| role="presentation" class="wikitable mw-collapsible mw-collapsed" style="margin: 1em auto 1em auto; width:90%"
| Topics You Need to Understand For This Page
|-
| Improper Integrals
|}

==Definition==
For a function <math>f(t)</math>, using Napier's constant <math>e</math> and a complex number <math>s</math>, the Laplace transform <math>F(s)</math> is defined as follows:   
: <math>F(s) = \mathcal{L} \left\{f(t)\right\}(s)=\int_0^{\infty} e^{-st} f(t) \,dt </math>

The parameter <math>s</math> is a complex number.

: <math>s = \sigma + i \omega, \, </math> with real numbers <math>\sigma</math> and <math>\omega</math>.

This <math>F(s)</math> is the '''Laplace transform''' of <math>f(t)</math>.

==Explanation==
Here is what is going on.  

==Examples of Laplace transform==
{| class="wikitable"
|+ Examples of Laplace transform
! <math>f(t)</math> !!  <math>F(s) = \mathcal{L}\{f(t)\} </math>

|- style="text-align: center;"
| <math>C</math>
| <math>\frac{C}{s} </math>  
|- style="text-align: center;"
|<math>t</math> 
| <math>\frac{1}{s^2}</math>
|- style="text-align: center;"
| <math>t^n</math> 
| <math>\frac{n!}{s^{n+1}} </math> 
|- style="text-align: center;"
|<math>\frac{t^{n-1}}{(n-1)!}</math> 
| <math>\frac{1}{s^{n}}</math>
|- style="text-align: center;"
| <math>e^{at}</math>
| <math>\frac{1}{s-a} </math>
|- style="text-align: center;"
| <math>e^{-at}</math>
| <math>\frac{1}{s+a} </math>
|- style="text-align: center;"
| <math>{\rm cos}\ \omega t</math>
| <math>\frac{s}{s^2 + {\omega}^2} </math>
|- style="text-align: center;"
| <math>{\rm sin}\ \omega t</math>
| <math>\frac{\omega}{s^2 + {\omega}^2} </math>
|- style="text-align: center;"
| <math> \frac{t^{n-1}}{\Gamma(n)}</math> 
| <math>\frac{1}{s^{n}}</math>   (n>0)
|- style="text-align: center;"
| <math> \delta (t-a) </math> 
| <math>e^{-as} </math>
|- style="text-align: center;"
| <math> H(t-a) </math>
| <math> \frac {e^{-as}}{s} </math>
|}

In the above table,
# <math>C</math> and <math>a</math> are constants
# <math>n</math> is a natural number
# <math>\delta(t-a)</math> is the Delta function
# <math>H(t-a)</math> is the Heaviside function



:{{:Engineering Tables/Laplace Transform Table 2}}


== Examples ==
1. Calculate <math>\mathcal{L}\{C\} </math> (where <math>C</math> is a constant) using the integral definition.<br><br>
<math>
\begin{array}{rcl}
\displaystyle{\int_0^{\infty} e^{-st} C \,dt} & = &  C \displaystyle{\int_0^{\infty} e^{-st} \,dt} \\
& = & \displaystyle{ C \lim_{b \to \infty}{ \int_0^{b} e^{-st}  \,dt} } \\
& = & \displaystyle{ \left. C \lim_{b \to \infty}{ \frac{e^{-st}}{-s} } \right|_{t=0}^{t=b} } \\
& = & \displaystyle{ -\frac{C}{s}\left[ \lim_{b \to \infty} {e^{-bs}} - e^0 \right] } \\
& = & \displaystyle{ -\frac{C}{s} [ 0 - 1 ] } \\
& = & \displaystyle{ \frac{C}{s} }
\end{array}
</math>
<br /><br />
<math>\therefore \displaystyle\mathcal{L} \left\{C\right\} = \frac{C}{s}</math>
<br />
<hr>
2. Calculate <math>\mathcal{L}\{e^{-at}\} </math> using the integral definition.
<br><br />
<math>
\begin{array}{rcl}
\displaystyle\int_0^{\infty} e^{-st} \cdot e^{-at} \,dt & = & \displaystyle\int_0^{\infty} e^{-(s+a)t} \,dt \\
& = & \displaystyle\lim_{b \to \infty}{
\displaystyle\int_0^{b} e^{-(s+a)t} \,dt
} \\
& = & \displaystyle\lim_{b \to \infty} \left[\displaystyle\frac{-e^{-(s+a)t}}{s+a}\right]_{t=0}^{t=b} \\
& = & \displaystyle\lim_{b\to\infty} 
\left[\displaystyle\frac{-e^{-(s+a)b}}{s+a} - \displaystyle\frac{-e^{-(s+a)0}}{s+a}\right] \\
& = & \displaystyle\frac{1}{s+a}

\end{array}
</math><br /><br />
<math> \therefore \mathcal{L}\{e^{-at}\} = \displaystyle\frac{1}{s+a} </math>

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