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Laplace transformation

  • Thread starter Karmel
  • Start date
  • #1
12
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Homework Statement


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

Homework Equations





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
 

Answers and Replies

  • #2
156
0
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 [tex]g(t-a)u(t-a)\mapsto e^{-as}F(s)[/tex] to take the Laplace transform of f(t). For example, we may rewrite the second term as

[tex](e^{3t} + 1)u(t-2) = (e^6e^{3(t-2)} + 1)u(t-2)[/tex],

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

Your turn. Show us your work if you get stuck.
 
Last edited:
  • #3
12
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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
156
0
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

[tex](2t - e^{3t})u(t - 4) = (2(t - 4) + 8 - e^{12}e^{3(t - 4)})u(t - 4)[/tex].

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..
Once you have transformed each term of f(t), then you have found the Laplace transform of f(t), as required.
 
  • #5
12
0
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
 

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