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Laplace Transform of t.H(t-a)

  1. Jul 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the Laplace Transform of t.H(t-a) where H is the heavyside (unit step) function.

    2. Relevant equations

    Properties of Laplace Transforms

    L{t.f(t)} = -Y'(s)

    L{f(t-a).H(t-a)} = e-as.F(s)

    Maybe another one I dont know about?


    3. The attempt at a solution

    I'm not sure how to apply the two properties since they both match one half of the relevant information, but neither of them match the situation exactly. ie. one only refers to "t" on it's own as the input, while the other only refers to t-a as the input.

    What do I do? o_O
     
  2. jcsd
  3. Jul 29, 2011 #2
    Start by using

    [itex] \mathcal{L}\{ f(t-a)\cdot H(t-a)\} = e^{-as}\cdot F(s) [/itex]

    Just substitute [itex] f(t-a) [/itex] with [itex] 1 [/itex] and this should give you the laplace transform of [itex] H(t-a) [/itex].

    Then you could proceed by using the first of your two properties of the Laplace transform

    [itex] \mathcal{L}\{ t\cdot f(t)\} = -F'(s) [/itex]

    ... to evaluate [itex] \mathcal{L}\{ t\cdot H(t-a)\} [/itex]
     
    Last edited: Jul 29, 2011
  4. Jul 29, 2011 #3
    If it's still unclear let me know, however I have deliberately restrained myself from giving you a complete solution as this is not permitted here on PF.
     
  5. Jul 29, 2011 #4
    hrm thanks for helping but still unclear.

    How am I allowed to substitute "t-a" = 1 ? isnt that just assigning it an arbitrary value? shouldnt I only be substituting it for another variable, if anything?

    Even once I get the Laplace of H(t-a), how do I then use that transform to find t.f(t) ?
     
  6. Jul 29, 2011 #5
    You are not substituting [itex] t-a [/itex] with [itex] 1 [/itex], you are substituting (or I am :P) [itex] f(t-a) [/itex] with [itex] 1 [/itex] such that, according to

    [itex] \mathcal{L}\{ f(t−a)⋅H(t−a)\} =e^{−as}\cdot F(s) [/itex]

    one gets

    [itex] \mathcal{L}\{ 1⋅H(t−a)\} =e^{−as}\cdot\mathcal{L}\{ 1 \} [/itex]

    If you don't remember what [itex] \mathcal{L} \{ 1 \} [/itex] is, then just use the definition:

    [itex] \mathcal{L} \{ 1 \} = \int_0^\infty 1\cdot e^{-st} \; dt [/itex]

    From this you should be able to find

    [itex] \mathcal{L} \{ H(t-a) \} [/itex]

    You also asked how one would proceed using the

    [itex] \mathcal{L}\{ t\cdot f(t)\} =−F'(s) [/itex]

    to find

    [itex] \mathcal{L}\{ t\cdot H(t-a)\} [/itex]

    Well, all i can do is hint you in the right direction. You need to realize that [itex] F'(s) [/itex] is the derivative of the transform of [itex] f(t) [/itex]. And thus when you replace [itex] f(t) [/itex] with [itex] H(t-a) [/itex] you also replace [itex] F'(s) [/itex] with the derivative of [itex] \mathcal{L} \{ H(t-a) \} [/itex].

    Point out which parts of what I've written so far you don't understand and I'll try to help you as much as I can without spelling it out for you :P
     
    Last edited: Jul 29, 2011
  7. Jul 29, 2011 #6
    Why do I care what the laplace of H(t-a) is?

    How do I use e^(-as).L{1} to find H(t-a) ?

    The Laplace of H(t-a) is e^(-as).(1/s)

    But I dont see how that helps me?

    This isnt for homework by the way its for an exam I have in a few days time. I generally learn best from seeing looking at the solution and deconstructing it into the different steps and looking at how each property is applied, as opposed to trying to cryptically figure out something that already confuses me :P
     
  8. Jul 29, 2011 #7

    Ray Vickson

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    Apply the formula [itex] L\{t f(t)\}(s) = -F'(s)[/itex] to the function f(t) = H(t-a).

    RGV
     
    Last edited: Jul 29, 2011
  9. Jul 29, 2011 #8
    [itex] \mathcal{L}\{ f(t−a)\cdot H(t−a)\} = e^{−as}\cdot F(s) [/itex]

    This is the identity that gives you [itex] \mathcal{L} \{ H(t-a) \} [/itex].

    Just substitute [itex] f(t-a) [/itex] with [itex] 1 [/itex]... the identity now looks as follows

    [itex] \mathcal{L}\{ 1\cdot H(t−a)\} = e^{−as}\cdot \mathcal{L}\{ 1 \} [/itex]

    Clearly the left hand side is just the transform of [itex] H(t-a) [/itex] and [itex] F(s) [/itex] is the transform of [itex] f(t) = 1 [/itex] . Since [itex] \mathcal{L}\{ 1\} = \frac{1}{s} [/itex], we have that;

    [itex] \mathcal{L}\{ H(t−a)\} = e^{−as}\cdot \mathcal{L}\{ 1 \} = e^{−as}\cdot \frac{1}{s}[/itex].

    From here on you proceed in the same manner using the other identity that you have been given...

    [itex] \mathcal{L}\{ t\cdot f(t)\} =−F'(s) [/itex]

    Now you replace [itex] f(t) [/itex] with [itex] H(t-a) [/itex] and thus, since [itex] F(s) [/itex] is always the transform of f(t), it is now the transform of [itex] H(t-a) [/itex] which we have found above.

    The identity therefore yields...

    [itex] \mathcal{L}\{ t\cdot H(t-a)\} =−\frac{d}{ds}e^{−as}\cdot \frac{1}{s} = \cdots [/itex]
     
    Last edited: Jul 29, 2011
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