Solving Laplace transforms from a given, linear graph

**zepto-** · February 26th 2010, 02:04 PM

Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?

**TheEmptySet** · February 26th 2010, 02:10 PM

Originally Posted by zepto-

Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?

How discontinous? if it is a finite set $t_0,t_1,...t_n$
then just break up the integral to get

$\int_{0}^{t_0}e^{-st}f(t)dt+\int_{t_0}^{t_1}e^{-st}f(t)dt+...+\int_{t_{n-1}}^{t_n}e^{-st}f(t)dt$

**zepto-** · February 26th 2010, 02:18 PM

I think my probelm might be writing the Heaviside / Unit step functions in U(t) format.

**TheEmptySet** · February 26th 2010, 02:31 PM

Originally Posted by zepto-

I think my probelm might be writing the Heaviside functions in U(t) format.

Okay well Here is an example I guess

$f(t)=\begin{cases} e^t, 0 \le t\le 1 \\ e, t > 1\end{cases}$

The Lapalce transform is

$\mathcal{L}(f)=\int_{0}^{\infty}e^{-st}dt=\int_{0}^{1}e^{-st}f(t)dt+\int_{1}^{\infty}e^{-st}f(t)dt$

Now using the definition of f gives

$\mathcal{L}(f)=\int_{0}^{1}e^{-st}e^{t}dt+\int_{1}^{\infty}e^{-st}\cdot 1dt$

Now just evaluate each of these integrals to get

$\mathcal{L}(f)=\int_{0}^{1}e^{-(s-1)t}dt+\int_{1}^{\infty}e^{-st}dt=-\frac{1}{s-1}e^{-(s-1)t}\bigg|_{0}^{1}-\frac{1}{s}e^{-st}\bigg|_{1}^{\infty}=$
$-\frac{e^{-(s-1)}}{s-1}+\frac{1}{s-1}+\frac{1}{s}e^{-s}$

**CaptainBlack** · February 26th 2010, 11:28 PM

Originally Posted by zepto-

Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?

Can you post it?

CB

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