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Unit-step FT

  1. Mar 11, 2006 #1
    how can I calculate the fourier transform for unit step function:

    v(t)=1 where 0=<t<+infinity

    v(t)=0 otherwise

    I applied the general definition relation for FT:

    v(w)=integral(v(t)*e^-jwt) ; - infinity<t<+infinity

    but i had v(w)=infinity due to the term infinity-displaced e^(+jwt) so that's wrong of course!

    I think we could divide this function into two functions for example:

    v'(t)=1/2 ; t of all values

    v''(t)=1/2 ; 0=<t<+infinity
    v''(t)=-1/2; -infinity<t<0

    so we notice v(t)=v'(t)+v''(t)

    I don't know what to do , could anyone help!

  2. jcsd
  3. Mar 11, 2006 #2


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    Your v(t) does not have a Fourier transform. When functions like this are encountered in practice, Laplace transform is used instead (You need to describe the application you are working on that needs this transform).
  4. Mar 11, 2006 #3

    George Jones

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    Staff Emeritus
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    Gold Member

    The Fourier transform of the Heaviside step function doesn't exist as a function. It does, however, exist as a (somewhat nasty) distribution that involves the the Dirac delta "function".

  5. Mar 12, 2006 #4
    thanks.Anyway could you clarify more, how did you deduct that Heaviside step function exists as Dirac delta function and how can we describe that distribution practically on diagram
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