I can't understand Heaviside

**Provoke** · November 25th 2009, 04:46 PM

Find the Laplace transformation:

Simply telling me what to do won't help me. I went through the definitions and more than 5 examples and I can't understand how they undergo the Heaviside calculation steps. If someone could be kind enough to explain each step of what he's doing would be really nice!

Thanx

**dedust** · November 25th 2009, 06:09 PM

the heaviside function is defined by
$u_{c}(t)=\left\{ \begin{array} [c]{cc}% 0, & t<c\\ 1, & t\geq c \end{array} \right.$

using this, we get
$f(t)=t\left[ u_{0}(t)-u_{1}(t)\right] +u_{1}(t)$

Calculate the Laplace transform of $f(t)$ using the identity for Laplace transform of heaviside function.

**dedust** · November 25th 2009, 10:26 PM

it's easy bro,

i got

$F(s)=\frac{1-e^{-s}}{s^{2}}$

**HallsofIvy** · November 26th 2009, 02:57 AM

Originally Posted by Provoke

Find the Laplace transformation:

Simply telling me what to do won't help me. I went through the definitions and more than 5 examples and I can't understand how they undergo the Heaviside calculation steps. If someone could be kind enough to explain each step of what he's doing would be really nice!

Thanx

The Laplace transform of a function, f(x) is
$\int_0^\infty f(x)e^{-sx}dx$

$f(t)=t\left[ u_{0}(t)-u_{1}(t)\right] +u_{1}(t)$
For t< 0 both $u_0$ and $u_1$ are 0 so f(x)= t(0- 0)+ 0= 0.

For $0\le t< 1$ $u_0$ is 1 and $u_1$ is 0 so f(x)= t(1- 0)+ 0= t.

for $1\le t$ both $u_0$ and $u_1$ are 1 so f(x)= t(1-1)+ 1= 1.

The Laplace transform of f is $\int_0^1 t e^{-st}dt+ \int_1^\infty e^{-st}dt$

Integrate the first by parts with u= t, $dv= e^{-st}dt$ so that du= dt and $v= -\frac{1}{s}e^{-st}$ .

$\int_0^1 te^{-st}dt= \left[-\frac{1}{s}te^{-st}\right]_0^1+ \frac{1}{s}\int_0^1 e^{-st}dt$
$= -\frac{1}{s}e^{-s}- \left[\frac{1}{s^2}(e^{-st}-1)\right]_0^1$
$= -\frac{1}{s}e^{-s}- \frac{1}{s^2}(e^{-s}-1)$

The second integrates easily
$\int_1^\infty e^{-st}dt= \left[-\frac{1}{s}e^{-st}\right]_1^\infty$
$= \frac{1}{s}e^{-s}$

The $\frac{1}{s}e^{-s}$ terms cancel when we add those integrals, leaving $-\frac{e^{-s}-1}{s^2}$ which is the same as dedust's result.

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