Laplace transformation

1. Nov 27, 2008

Karmel

1. The problem statement, all variables and given/known data
fing the laplace transformation of f(t) where
-1 if 0<t<2
f(t)= e^3t if 2<t<4
2t if 4<t

2. Relevant equations

3. The attempt at a solution

I can get to the part where f(t) = -1 + (e^3t +1)u(t-2) + (2t-e^3t)u(t-4)
then I get lost ...Any help is appreciated. Thank you all in advance

2. Nov 28, 2008

Unco

Hi Karmel,

First note that the -1 you have there should be -u(t); also, a minor point, but I'll assume u(t) is the Heaviside function restricted to t>0 (since f is only defined on t>0).

Your task now is to work with the formula $$g(t-a)u(t-a)\mapsto e^{-as}F(s)$$ to take the Laplace transform of f(t). For example, we may rewrite the second term as

$$(e^{3t} + 1)u(t-2) = (e^6e^{3(t-2)} + 1)u(t-2)$$,

and so by the above formula this transforms to $$e^{-2s}\left(e^6\frac{1}{s-3} + \frac{1}{s}\right)$$.

Last edited: Nov 28, 2008
3. Nov 28, 2008

Karmel

Okay so by using the examples from the book and the above example for the last part of the above problem I get and if Im right it will be luck!

(2t-e^3t)u(t-4) = (2t-(e^12)(e^3(t-4))u(t-4) which transforms into e^-4s(-e^12((1/s-3)+(2/s^2))

so assuming that this part is right. Do I leave the answer as is or do I need to multiply throughout and see if I can add anything and simplyfy it..

karmel

4. Nov 29, 2008

Unco

Almost, but you haven't completely made the factor multiplied by u(t-4) look like a function g(t-4) like in formula I gave above. We need to write it as

$$(2t - e^{3t})u(t - 4) = (2(t - 4) + 8 - e^{12}e^{3(t - 4)})u(t - 4)$$.

Once you have transformed each term of f(t), then you have found the Laplace transform of f(t), as required.

5. Nov 30, 2008

Karmel

Thank you so much for the help and assuming I havent made any errors I do believe I have finally arrivied at the right answer. Once again thank you karmel