Laplace Transforms

Question

Laplace Transforms

malijhaa · Answer

Yhur Question?

idku · Answer

Note: $u(t)$ is the unit step-function.

problem 1:

$\color{black}{\mathcal{L}\left\{4u(t-3)-3u(t-5)\right\}=}$

$\color{black}{4\mathcal{L}\left\{u(t-3)\right\}-3\mathcal{L}\left\{u(t-5)\right\}=}$

$\color{black}{\displaystyle \frac{-4e^{3s}}{s}+\frac{3e^{5s}}{s}}$.

idku · Answer

I am just trying to check my work.

idku · Answer

and $\mathcal{L}$ denotes, obviously, the laplace transform

irishboy123 · Answer

you've got the signs on your exponents wrong, and i think that's because you didn't implement $\mathcal{L} \{ u(t-c)  \} = \dfrac{e^{-c~s}}{s}$  

look at the sign on the c maybe !

idku · Answer

oh, that is the rule? didn't know, thank you for giving me the correct one.

idku · Answer

So, here, unlike in other cases, the exponent is with he same sign as the shift c.