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[Signals & Systems] Continuous-Time Unit Step

  1. Apr 13, 2008 #1
    hi all, i was wondering what is the most correct approach concerning this function, since sometimes i find that the function is discontinuous at t=0 and sometimes its not.

    why is that there are 2 interpretations for a single function? bit confusing...

  2. jcsd
  3. Apr 13, 2008 #2
    The signal (and not function, strictly speaking) is equal to 0 for t < 0 and equal to 1 for t > 0. It is discontinuous at t = 0, but some authors define it to be either 0 or 1 at t = 0, making it left or right continuous at that point, respectively.

    In either case, the signal is not continuous at t = 0 (continuity at a point requires (a) existence of left and right hand limits, (b) equality of the left and right hand limits, to the value of the function at that point)...it is either left continuous or right continuous or neither left nor right continuous.

    Sidenote: sometimes, the value at t = 0 is also defined to be the average of 0 and 1, i.e. 1/2. This is done because a theorem from Fourier Analysis states that the Fourier series of a function at a point of discontinuity [itex]\zeta[/itex] converges to the average of the values at [itex]\zeta-[/itex] and [itex]\zeta+[/itex].
  4. Apr 14, 2008 #3


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    it ends up being more elegant and useful for more reasons. like defining u(t) = (1+sgn(t))/2 (and then there is a meaningful result for the Fourier Transform of the unit step) and being able to say that u(t) + u(-t) = 1 for all t. i think, in any decent signals & systems book, it should be defined that way for continuous-time u(t).
  5. Apr 14, 2008 #4
    Right, you can use it to "define" the transform of a unit step.
  6. Apr 14, 2008 #5


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    no, to compute it from the definition of the unit step (in terms of the sgn(t) function, which is the premise i suggested) and from the definition of the Fourier Transform.
  7. Apr 15, 2008 #6
    Yes, I'm sorry thats what I meant too...but of course one has to figure out what sgn(t) transforms to, in the Fourier domain.

    [To the OP--have a look at Oppenheim/Wilsky/Nawab.]
  8. Apr 15, 2008 #7
    I you want to have a rigorous statement you might think it as the integral of the Dirac Delta function (or distribution as we should say) from [itex]-\infty[/itex] to [itex]\infty[/itex] . Then, the rest follows from the definitions of distributions, which mostly considered to be technical overkill for such a simple function (It is a function of time! [itex]\mathbf{1}(t)[/itex] ) and hence the confusion e.g. is it defined or not, does it have a value at zero or not, etc..
  9. Apr 16, 2008 #8


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    it's in the books. if you want to figure it out, try to transform this limiting function:

    [tex] \mbox{sgn}(t) = \lim_{\alpha \to 0} \begin{cases}
    e^{-\alpha t} & \mbox{if } t > 0 \\[3pt]
    0 & \mbox{if } t = 0 \\[3pt]
    -e^{\alpha t} & \mbox{if } t < 0
    \end{cases} [/tex]

    you can figure out the Fourier Transform of that if [itex]\alpha > 0[/itex].

    this has a lot to do with the Hilbert Transform.
    Last edited: Apr 16, 2008
  10. Apr 18, 2008 #9
    Thanks, thats an interesting way to look at sgn(t). I've only encountered trambolin's approach though...not too rigorously--during some brief digressions into distribution theory during an introductory signal analysis course.

    Anyway, so as not to confuse tko_gx, I want to point out that different books/authors and teachers will define the unit step differently. Its important to remember that the continuous time unit step is a discontinuous function at t = 0 no matter how you define it, because continuity requires three conditions to be met simulataneously:

    1. Left hand limit should exist
    2. Right hand limit should exist
    3. Left hand limit = right hand limit = value of function at that point

    Note that with a continuous time unit step, you cannot satisfy these 3 conditions simultaneously (first two are true).
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