Find Laplace Transform of the following function
You must break the integral up on the discontinuities to take the Laplace Transform of a discontinuous function: $$\begin{align} \mathcal{Lf}(s) &= \int_0^\infty f(t)e^{st}dt \\ &=\int_0^1\underbrace{t}_{\text{f(t) for t < 1}}e^{st}dt + \int_1^\infty\underbrace{(2t)}_{\text{f(t) for t > 1}}e^{st}\;dt\\ &=\cdots \end{align}$$
Please let me know if this needs more explaining.
More explanation:
The function $f$ is defined by: $$f(t) = \begin{cases}\color{red}{t} & \text{if }0\le t \le1\\ \color{blue}{2t} & \text{if } 1\le t\end{cases}$$
To compute the Laplace transform, we compute the integral: $$\int_0^\infty f(t)e^{st}dt$$
Based on the property of integrals that says $\int_a^b f(x)\;dx + \int_b^cf(x)\;dx = \int_a^c f(x)\;dx$, we can write: $$\int_0^\infty f(t)e^{st}dt = \int_0^1f(t)e^{st}\;dt + \int_1^\infty f(t)e^{st}\;dt$$
We now replace $f(t)$ based on the definition: $$\int_0^1\color{red}{f(t)}e^{st}\;dt + \int_1^\infty \color{blue}{f(t)}e^{st}\;dt = \int_0^1\color{red}{t}e^{st}\;dt + \int_1^\infty \color{blue}{(2t)}e^{st}\;dt$$
The rest is simple integration. (The first you do byparts, the second you distribute, then a $u$sub and byparts again.)
Related videos on Youtube
A A
Updated on August 01, 2022Comments

A A over 1 year
How do I find the Laplace transform for the function:
$f(t)=t, 0 \leq t \leq 1$ and $2t, t \geq 1$
I tried looking up the process online, but it remains unclear to me.
Thanks in advance!

David H over 9 yearsThe function you have defined is not a step function...

A A over 9 years@DavidH Oh okay, I got a bit confused. I'll fix the question, thanks!


A A over 9 yearsCan you please further explain? Thank you

apnorton over 9 years@AA Is that better?

A A over 9 yearsGreat, I understand now. Thank you!