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Laplace transform of f(t-b)

  1. Sep 21, 2011 #1
    1. The problem statement, all variables and given/known data
    is it possible to find the laplace transform of f(t-b) ? i don't know if its possible, i am just trying.

    3. The attempt at a solution

    so, where integral from 0 to infinity,

    [itex]\int[/itex] f(t-b) e-stdt

    let t-b = z

    =[itex]\int[/itex]f(z) e-s(z+b)dz

    =[itex]\int[/itex]f(z) e-sze-sbdz

    =e-sb f(s) , where f(s) is the laplace transform?

    if this is correct, then

    question) is f(s) the laplace transform of f(z)? i.e of f(t-b) ? OR is it the laplace transform of f(t) only?

    because, from above, the expressions are telling me that f(s) is the laplace transform of f(z) which is f(t-b)

    but if that is the case, then it wouldn't make any sense to even compute the laplace transform by the above method. because it would then mean L { f(t-b) } = f (s) ?



    if the entire above is wrong, then how do i compute it?


    because i know the inverse laplace transform of e-bsf(s) is f(t-b), so the reverse has to be true
     
  2. jcsd
  3. Sep 21, 2011 #2

    LCKurtz

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    That last statement is not correct, which I will get to in a moment. You usually use slightly different techniques when doing the direct transform vs. the inverse involving the unit step function. Let's look at the direct first

    The unit step function is useful when you are given a problem to find the LT of a function that is 0 for t < a and some formula f(t) for t > a. You would write that function as f(t)u(t-a). When you take its transform you get:

    [tex]\mathcal L(f(t)u(t-a)) = \int_a^\infty e^{-st}f(t)\,dt[/tex]
    since u(t) = 1 only for t > a. Now using your substitution u = t-a gives
    [tex]\int_0^\infty e^{-s(u+a)}f(u+a)\,du=e^{-as}\int_0^\infty e^{-st}f(t+a)\,dt=
    e^{-as}\int_0^\infty e^{-st}F(t)\,dt=e^{-as}F(s)[/tex]
    where F(t) = f(t+a). This is the formula you would normally use for taking transforms. One way of using this, but not the only way, is to replace t by t+a in f(t), take transform of that, and multiply the result by e-as.

    Now let's look at what we would get if we wanted the transform of f(t-a)u(t-a), which isn't the same thing as above.
    [tex]\mathcal L(f(t-a)u(t-a)) = \int_a^\infty e^{-s(t-a)}f(t-a)\,dt[/tex]
    as before. Now using your substitution u = t-a gives
    [tex]e^{-as}\int_0^\infty e^{-su}f(u)\,du=e^{-as}f(s)[/tex]

    This says if you have e-asf(s) and you want to invert it, you can inverse f(s) to get f(t) and then translate and truncate to give the answer f(t-a)u(t-a), which is why that quote of yours above is wrong.

    This also gives an alternate method to take the transform in the first case where you want the transform of f(t)u(t-a). You can express f(t) in terms of t-a, perhaps by Taylor expansion, and use the bottom formula.
     
    Last edited: Sep 21, 2011
  4. Sep 22, 2011 #3
    er how did you get from the first expression to the 2nd expression here? are you just changing the variable of integration from u to t? we can just do this? but i thought u =t - a?
    is "u" a step function? i haven't really learnt that yet.



    also, in the LHS of the last equation, where did the extra e-as come from? if i sub u = t-a, shouldn't it give me
    [tex]\int_0^\infty e^{-su}f(u)\,du=f(s)[/tex] ?

    is it due to a special property of 'u' or something?





    i think i should rephrase my original question, i need to learn by examples :( i can't learn well through theory.

    assuming we have a function that is 0 for t < a and some formula f(t) for t > a ,i.e, it behaves nicely that i don't have to introduce the step function?

    assuming this function is sin(t-[itex]\pi[/itex])

    when i laplace transform sin(t-[itex]\pi[/itex]) ,

    do i get e-[itex]\pi[/itex]s { 1 / (s2+1) } ?

    since the laplace transform of sin(t) is { 1 / (s2+1) }


    so now if i inverse e-[itex]\pi[/itex]s { 1 / (s2+1) },

    do i get sin(t) or do i get sin(t-[itex]\pi[/itex]) ?

    if i understood what you wrote previously, are you saying i will get sin(t), but i have to 'translate and truncate' to get sin(t-[itex]\pi[/itex]) ??? what is 'translate and truncate' ?

    thanks!
     
  5. Sep 22, 2011 #4

    LCKurtz

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    u(t) is the unit step function. I guess I shouldn't have used the substitution u = t-a since it causes confusion. I should have used v = t-a so the integrals have v's in them. Still, the variable of integration is a dummy variable and once the substitution is done, the dummy variable can be anything as in
    [tex]\int_a^b g(v)\,dv=\int_a^b g(t)\, dt[/tex]

    Unfortunately it won't let me edit that.
    Sorry, that is a typo. Those last two lines should read

    [tex]\mathcal L(f(t-a)u(t-a)) = \int_a^\infty e^{-st}f(t-a)\,dt[/tex]
    as before. Now using your substitution u = t-a gives
    [tex]e^{-as}\int_0^\infty e^{-su}f(u)\,du=e^{-as}f(s)[/tex]

    and, again, it won't let me correct it.

    Yes. because you would be taking the transform of sin(t-π)u(t-π) by the last formula. But you would have to use the first formula if you were doing sin(t)u(t-π)
    Translate means sin(t-π), truncate means multiply by u(t-π) to make it 0 when t < π. So the inverse would be sin(t-π)u(t-π).
     
  6. Sep 22, 2011 #5
    oh i think i understand most, except, the above,

    shouldn't my variable in e-stbe t-a, and not t now? since i am laplace transforming a function that is f(t-a)?

    so it should be [tex]\mathcal L(f(t-a)u(t-a)) = \int_a^\infty e^{-s(t-a)}f(t-a)\,dt[/tex] ? as per your first post?
     
  7. Sep 22, 2011 #6

    LCKurtz

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    No. No matter what function you are transforming, the transform always uses e-st times the function in the integrand. Remember there was a typo in that first post.
     
  8. Sep 23, 2011 #7
    oh, so even if i have f(t-21) , it still uses e-st and not e-(t-21)?

    because i was reminded by for e.g, when i use e-(s-a)t , the transform will be in terms of (s-a) instead of the usual s.

    so say transform of e-at 5 = 5/ (s+a) ?
     
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