Neeed help with a laplace transform

  • Thread starter Lorens
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  • #1
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What is the laplace transform of

http://img150.imageshack.us/img150/8145/laplacetransform6wk.jpg [Broken]
 
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  • #2
Tom Mattson
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According to the section entitled "Homework Help" in the Physics Forums Global Guidelines which you agreed to:

NOTE: You MUST show that you have attempted to answer your question in order to receive help.

So whatcha got?
 
  • #3
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Tom Mattson said:
According to the section entitled "Homework Help" in the Physics Forums Global Guidelines which you agreed to:
So whatcha got?

It isn't so much to test, jsut to watch a tabel of formula, i know the transform for cos(x), but i can't find any rule which would let me multiplicat it with O(x).

I mean for problems like this you jsut think, and try to figure out how to do it.
I would say my problem is O(x).

Anyway i got to go to bed now 00:08.. lol...
 
  • #4
Tom Mattson
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Do you know what [itex]\theta(t)[/itex] is?
 
  • #5
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Tom Mattson said:
Do you know what [itex]\theta(t)[/itex] is?


for [itex]\theta(t)[/itex] t<0 gives t=0 and t>0 gives t=1 and the transform is 1/s, but that knowledge don't help me much :( ...
 
  • #6
Tom Mattson
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You are right about [itex]\theta(t)[/itex], but that knowledge should help you a great deal.

You have a function that is defined piecewise:

[tex]\theta(t) = \left\{ \begin{array}{cc}0 & t<0\\1 & t \geq 0\end{array}[/tex]

Now, if you multiply [itex]\theta(t)[/itex] by [itex]\cos(t)[/itex], then you just have to multiply both pieces by [itex]\cos(t)[/itex].

So...

[tex]\cos(t)\theta(t) = \left\{ \begin{array}{cc}0 & t<0\\\cos(t) & t \geq 0\end{array}[/tex]

Can you take it from there?
 
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  • #7
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You missed that [itex]\theta(o)[/itex]=1/2 my text book say so, but it don't matter.

Can I ignore [itex]\theta(t)[/itex] seen the laplace transform isn't defined for the second quadrant for the x-axis.

Also i must ask how do you get access to all the special signs?

Thx for your time Lorenz
 
  • #8
Galileo
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Lorens said:
Can I ignore [itex]\theta(t)[/itex] seen the laplace transform isn't defined for the second quadrant for the x-axis.
It's not a matter of being undefined, but you got the right idea. It's more precise to state that the laplace transform of f doesn't care what values f takes on for x<0.

Also i must ask how do you get access to all the special signs?
Thx for your time Lorenz
What special signs?
 
  • #9
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Like [itex]\theta(t)[/itex] i just copyed him there :smile:
 
  • #10
Galileo
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