Laplace transform of piecewise functions

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The discussion focuses on calculating the Laplace transform of a piecewise function defined as f(t) = 0 for t=2 and f(t) = t for t≠2. Participants clarify that the function should equal t after t=2, but the initial approach does not yield this result until t=3. It is suggested to evaluate the Laplace transform directly by plugging f(t) into the integral with appropriate limits, rather than using unit step functions. The Laplace transform of the unit step function is noted to be e^(-2s), which applies to similar step functions. The conversation emphasizes that the single point of discontinuity does not affect the overall evaluation of the transform.
Xyius
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I do not have trouble doing these types of problems if the domains are not equalities. However for this type of problem..

f(t)=
0 , t=2
t , t\neq 2

(I do not know how to do piecewise function in latex)

I figured this is on the right track...
t-tu_{2}(t)+2tu_{3}

I have a problem with this however, it needs to be equal to t after 2 and my answer isn't equal to t until 3. Any help would be appreciated. :\
 
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Did you mean for that function to differ from g(t)=t at only the one point t=2?
 
Yes exactly! :)
 
Don't bother writing it in terms of unit steps and the like. Just plug f(t) into the integral and figure out how to evaluate it using appropriate limits. You'll find the single point doesn't make a difference in the end.
 
I would but the question is asking for the laplace transform of the piecewise function. :\
 
I don't know what you mean by that. What specifically do you think the question is asking you to do?
 
Xyius said:
I would but the question is asking for the laplace transform of the piecewise function. :\

Which is e^(-2s) because the laplace transform of the unit step function with a step at c is e^(-sc).

It works the same way with all step functions, and you can even find an approximate method for the dirac delta function.
 

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