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Step Function Help

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    The piecewise continuous function

    f(x) = { -8 for x ≤ 8
    -x – 1 for x > 8}

    Rewrite it in step functions.

    2. Relevant equations



    3. The attempt at a solution
    I think it's 1-(-8)(h(x-8)) + (-x-1)(h(x-8))

    Where h(x-a) is the Heaviside function (x-8) meaning it "jumps up" to 1 at x = a.

    Is that right? It seems wrong to me.
     
  2. jcsd
  3. Nov 24, 2009 #2

    berkeman

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    Staff: Mentor

    Why would you need a step function for x≤8?

    Oh, and the function doesn't look continuous at x=8, unless I'm missing something. Is there a typo in the definition?
     
  4. Nov 24, 2009 #3
    It's not continuous at x = 8. You have to write the entire function using step functions so that it's in the form

    f(x) = ___________ + u8(x)* __________

    "Fill in the blanks", basically. It will be two step functions added together.
     
  5. Nov 24, 2009 #4

    berkeman

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    But in your original post (OP), you said
     
  6. Nov 24, 2009 #5
    That's what the question says =/ Each piece of the function is continuous in pieces. The goal of this is to find the Laplace transform of f(x), if that helps.
     
  7. Nov 24, 2009 #6

    berkeman

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    Well, I'm no expert, but being continuous means no discontinuities:

    http://en.wikipedia.org/wiki/Piecewise_continuous

    But whatever. It seems like if you want to use a step funtion for x≤8, then that one should run backwards in x, no?
     
  8. Nov 24, 2009 #7
    Oh well I suppose it's not discontinuous at x = 8. The definition of the function just changes at x = 8, according to Wikipedia.

    Yes, I think so. The step function I got was -8 + u8(t)(7-t). (Still not sure if that's entirely correct...)

    So that's what I need to find the Laplace transform of. Laplace of -8 is -8/s

    Laplace transform of u8(t)(7-t) is what's troubling me.
     
  9. Nov 24, 2009 #8

    HallsofIvy

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    No, jumbogala, you were right here and, unfortunately, berkeman was wrong (I'll never let him live it down!). "Piecewise continuous" means "continuous in pieces"- that is that it is continuous everywhere except possibly at a finite number of point, separating the continuous "pieces".

    And, berkeman, he does NOT "want to use a step funtion for x≤8", he wants to use a step function for the entire function. That is, write a formula, including a step function so that the entire function is given by that step function.

    Let's see you suggested f(x)= 1-(-8)(h(x-8)) + (-x-1)(h(x-8)). Let's check to see if that is correct. h(x) is 0 if x< 0 so h(x-8) is 0 if x is less than 8. If x< 8 then f(x)= 1-(-8)(0)+ (-x-1)(0)= 1. No, that isn't what you want- you want -8. Also h(x-8) is equal to 1 if x is larger than or equal to 8. If x is larger than or equal to 8, f(x)= 1- (-8)(1)+(-x-1)(1)= 1+8- x-1= 8- x. No, that isn't it either.

    Think about this: If f(x)= u(x) for x< a and f(x)= v(x) for [itex]x\ge a[/itex], and we want to write it as f(x)= p(x)h(x-a)+ q(x), what must p and q equal? Well, if x< a, h(x-a)= 0 so f(x)= p(x). Obviously, we must have p(x)= u(x). If [itex]x\ge a[/itex], h(x-a)= 1 so f(x)= u(x)+ q(x)(1)= v(x) so q(x)= v(x)- u(x).

    In this problem u(x)= -8 and v(x)= -x-1
     
  10. Nov 25, 2009 #9

    berkeman

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    Thanks, Halls. Like I said, I'm no expert in this. Still, piecewise continuous used to mean something different to me. Guess I need to review some, eh? :blushing:
     
  11. Nov 25, 2009 #10

    LCKurtz

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