Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?
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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 $\displaystyle t_0,t_1,...t_n$Quote:
Originally Posted by zepto-
then just break up the integral to get
$\displaystyle \int_{0}^{t_0}e^{-st}f(t)dt+\int_{t_0}^{t_1}e^{-st}f(t)dt+...+\int_{t_{n-1}}^{t_n}e^{-st}f(t)dt$
I think my probelm might be writing the Heaviside / Unit step functions in U(t) format.
Okay well Here is an example I guessQuote:
Originally Posted by zepto-
$\displaystyle f(t)=\begin{cases} e^t, 0 \le t\le 1 \\ e, t > 1\end{cases}$
The Lapalce transform is
$\displaystyle \mathcal{L}(f)=\int_{0}^{\infty}e^{-st}dt=\int_{0}^{1}e^{-st}f(t)dt+\int_{1}^{\infty}e^{-st}f(t)dt$
Now using the definition of f gives
$\displaystyle \mathcal{L}(f)=\int_{0}^{1}e^{-st}e^{t}dt+\int_{1}^{\infty}e^{-st}\cdot 1dt$
Now just evaluate each of these integrals to get
$\displaystyle \mathcal{L}(f)=\int_{0}^{1}e^{-(s-1)t}dt+\int_{1}^{\infty}e^{-st}dt=-\frac{1}{s-1}e^{-(s-1)t}\bigg|_{0}^{1}-\frac{1}{s}e^{-st}\bigg|_{1}^{\infty}=$
$\displaystyle -\frac{e^{-(s-1)}}{s-1}+\frac{1}{s-1}+\frac{1}{s}e^{-s}$
Can you post it?Quote:
Originally Posted by zepto-
CB
