Laplace Transforms

Posted by Jeff Schuler on Sep 6, 2009 in Uncategorized |

What is the Laplace Transform method?

The Laplace Transform is a method that simplifies integral and differential equations into algebraic equations. This practice is commonly used to solve for a function out of a differential equation, which otherwise may have been unsolvable or very difficult. The following integrals can be used to transform between F(s) \Leftrightarrow f(t) where \mathcal{L} denotes Laplace and \mathcal{L}^{-1} denotes Inverse Laplace:

\mathcal{L} [f(t)] = F(s) = \int\limits_0^\infty {f(t)*e ^ {-st}d} t \mathcal{L}^{-1}[F(s)] = f(t) = \frac{1}{2 \pi j} \int\limits_{c - j \omega}^{c + j \omega} {F(s)*e ^ {-st}d}s

Table of Laplace Transforms

Since these integrals can be tedious and certain functions tend to reoccur, a table of Laplace Transforms has been linked:

Laplace Transforms Table

A Laplace Transform example

This table can be a little complex to use at first so an example is provided below to get you started. In this problem we implement the Laplace Transform and Inverse Laplace Transform to solve for y(t) .

\frac{d^2y}{dt^2} + 7 \frac{dy}{dt} + 12y = 10 \quad y(0)=3, y'(0)

The first step is to take the Laplace Transform of both sides of the equation. Use element 1 of our table for the right side and element 18 for the left side. Note that the initial conditions are necessary to take the Laplace Transform of the left side.

s^2 Y(s) - s y(0) - y'(0) + 7 (s Y(s) - y(0))  + 12 Y(s) = \frac{10}{s}

Inputting our initial conditions:

s^2 Y(s) - 3s - 0 + 7 (s Y(s) - 3)  + 12 Y(s) = \frac{10}{s}

Assuming you are an engineering student and can do a little alegebra, our next step is to find the terms that have Y(s) in common and factor it out. Our goal is to find Y(s) = F(s) because, after all, \mathcal{L}^{-1}[Y(s)] = y(t) :

(s^2 +7s + 12)Y(s) - 3s -21 = \frac{10}{s}

After solving for Y(s) and factoring the denominator:

Y(s) = \frac{3s^2 + 21s + 10}{s(s+4)(s+3)}

Taking the Inverse Laplace Transform

Now we arrive at the trickier part of this procedure. We must take the Inverse Laplace of Y(s) to find y(t) . If our function Y(s) does not match anything in the table, such as this case, factoring is a good place to start. This problem can easily be factored using the expand function on your TI-89. Just go to catalog \rightarrow expand and enter your function in parenthesis. Using this function:

Y(s) = \frac{\frac{5}{6}}{s} - \frac{\frac{13}{2}}{s+4} + \frac{\frac{26}{3}}{s+3}

Noting \mathcal{L} [f_1(t) + f_2(t)] = \mathcal{L} [f_1(t)] + \mathcal{L} [f_2(t)] and \mathcal{L} [k*f(t)] = k*\mathcal{L} [f(t)] , we can take the Laplace Transform of each term independently and also manipulate the constant terms if necessary or just pull them out. Using the 2nd property in our Laplace Transform table:

y(t) = \frac{5}{6} - \frac{13}{2}e^{-4t} + \frac{26}{3}e^{-3t}

To check your work you can plug y(t), \frac{dy}{dt}, \frac{d^2y}{dt^2} into the original, differential equation and at t=0 we find that 10=10.

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3 Comments

Jeff
Sep 7, 2009 at 4:58 pm

Awesome post Luke, thank you


 
Tahir Aljazeera
Oct 1, 2010 at 7:49 pm

Not bad at all but you need to do some more complex problems involving laplace transformation


 
Luke
Oct 2, 2010 at 7:40 am

Thank you, that is a good idea. I will look into it.


 

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