# Laplace Transform of a Continuous Piecewise Function

• Nov 27th 2010, 12:52 PM
Kasper
Laplace Transform of a Continuous Piecewise Function
Good afternoon, the question in question is;

Quote:

Consider the piecewise function;

$f(t) = \left\{
\begin{array}{c l}
-4t+8 & t \leq 9\\
5 & t>9\\
\end{array}
\right.
$

This can be written using step functions as :
$f(t)=$ ______ + $u_9(t)$_______ .

The Laplace Transform is then: $L(f(t))=F(s)$where
$F(s)=________$
Now for my work, I started by writing f(t) as a sum of functions.

$f(t) = -4t+8+H(t-9)(5)$
Where $H$ is the Heaviside Step function shifted by $+9$.

Now to take the Laplace transform of this, I am trying to use the Second Shift theorem on the second term,

$L(f(t-9)H(t-9)) = e^{-9s}F(s)$

But in my case, $f(t-9)=5$, so shifting it to the right by $+9$ won't do anything to a horizontal line, so f(t)=5. So for my answer, I obtained:

$L(f(t))=(-\frac{4}{s^2} + \frac{8}{s}) + e^{-9s}\frac{5}{s}$

Which is not showing as correct, I am not entirely sure where I am going wrong! Any help would be appreciated!
• Nov 27th 2010, 10:45 PM
CaptainBlack
Quote:

Originally Posted by Kasper
Good afternoon, the question in question is;

Now for my work, I started by writing f(t) as a sum of functions.

$f(t) = -4t+8+H(t-9)(5)$
Where $H$ is the Heaviside Step function shifted by $+9$.

No!

$f(t) = -4t+8+H(t-9)(5-(-4t+8))$

CB
• Nov 28th 2010, 03:27 PM
Kasper
Quote:

Originally Posted by CaptainBlack
No!

$f(t) = -4t+8+H(t-9)(5-(-4t+8))$

CB

Err, I feel foolish. Thanks!