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Input/output question

  1. Aug 12, 2005 #1
    if y(x)=e^(-h)*(integral *(e^h)*rdx) + ce^(-h)
    then why can you look at that equation from this view:
    total output = response to the input+response to the initial data
     
  2. jcsd
  3. Aug 15, 2005 #2

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    The term [tex]e^{-h}\int{e^{h}rdx}[/tex] is the response to the input and the term [tex]ce^{-h}[/tex] is the response to the initial conditions.
     
  4. Aug 16, 2005 #3
    i know, but why?
     
  5. Aug 16, 2005 #4

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    In the expression, [tex]r(x)[/tex] is the input and [tex]c[/tex] is the initial state. So, the integral is a function of the input and the second term is a function of the initial state.
     
  6. Aug 17, 2005 #5
    @@a
    i thought c was a constant? why is that particular one the initial state?
    r(x) is the input because usually x is the input, right?
     
  7. Aug 17, 2005 #6

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    [tex]x[/tex] is the independent variable. [tex]r(x)[/tex] is an arbitrary function of the independent variable and is the input to your system. [tex]c = y(x_0)[/tex] is the initial value of the dependent variable [tex]y[/tex]. [tex]h = a.(x - x_0)[/tex].
    The integral is in reality a definite one. So:
    [tex]y(x) = e^{-h}\int_{x_0}^{x}e^hr(x')dx' + c e^{-h}[/tex]
    At [tex]x = x_0[/tex] [tex]e^{-h} = e^{-a.(x_0 - x_0)} = e^0 = 1[/tex] and the integral from [tex]x_0[/tex] to [tex]x_0[/tex] is zero. So,
    [tex]y(x_0) = 0 + c.1 = c[/tex]
     
  8. Aug 18, 2005 #7
    thank you very much! :)
     
  9. Aug 18, 2005 #8
    by the way, do you mean that a "definite integral" is one that converges?
     
  10. Aug 18, 2005 #9

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    No, a definite integral is one that has specified limits, so its result is a number and not a function of the integration variable. As you can see, in my example, I used x' as integration variable and x as one of the limits. Relative to x', x is a constant.
     
  11. Aug 18, 2005 #10
    thanks! :)
     
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