# Thread: I can't understand Heaviside

1. ## I can't understand Heaviside

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

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

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

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

3. it's easy bro,

i got

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

4. 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
$\displaystyle \int_0^\infty f(x)e^{-sx}dx$

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

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

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

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

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

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

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

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