Question

Laplace Transforms

Answer 1

Yhur Question?

Answer 2

Note: \(u(t)\) is the unit step-function.

problem 1:

\(\color{black}{\mathcal{L}\left\{4u(t-3)-3u(t-5)\right\}=}\)

\(\color{black}{4\mathcal{L}\left\{u(t-3)\right\}-3\mathcal{L}\left\{u(t-5)\right\}=}\)

\(\color{black}{\displaystyle \frac{-4e^{3s}}{s}+\frac{3e^{5s}}{s}}\).

Answer 3

I am just trying to check my work.

Answer 4

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

Answer 5

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 !

Answer 6

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

Answer 7

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