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Derivative of unit step function

  1. May 3, 2003 #1
    [SOLVED] Derivative of unit step function

    How does one do this, for example x= e^(-3t)u(t-4); how do you get x' ??
     
  2. jcsd
  3. May 4, 2003 #2

    enigma

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    Do laplace transforms on it.
     
  4. May 4, 2003 #3

    Hurkyl

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    Write out the definition of the unit step function and it might be easier to see.
     
  5. May 4, 2003 #4
    I think I got it now. I used the property L{f'}(s) = sL{f}(s) - f(0)
    Is that correct?
     
  6. May 4, 2003 #5
    The Fourier Transform

    can also be used. It can be used for many unbounded functions.
     
  7. May 4, 2003 #6

    Hurkyl

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    You could just differentiate it directly.

    x(t) = e^(-3t)u(t-4)

    is equivalent to:

    Code (Text):

    x(t) =  e^(-3t)   (for t > 4)
               0      (for t < 4)
     
    with x(4) depending on the precise definition of u.

    Differentiating on each piece gives:

    Code (Text):

    x'(t) = (-3) e^(-3t)   (for t > 4)
              0            (for t < 4)
     
    And x'(4) does not exist because x(t) is discontinuous at t = 4

    IOW:

    x'(t) = (-3) e^(-3t) u(t - 4) for t [x=] 4
     
    Last edited: May 4, 2003
  8. May 6, 2003 #7

    ahrkron

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    Sometimes you can safely assume the derivative of a step to be a delta function (for instance, when you integrate a delta, you get a step).

    They need to be used as distributions, and there may be some requirements on the functions you use along with them (integrability, continuity,...).

    I'm sorry I don't remember much about it.
     
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