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Thread: Shifted fcn...

  1. #1
    Junior Member
    Sep 2007

    Shifted fcn...


    x(t) = u(t) - u(t-10)



    Am I thinking about this correctly... u(t) is 0 from neg. infinity to 0, and 1 after. u(t-10) is the same thing just shifted to the right to "turn on" at 10. So in effect we have a rectangle of amplitude 1 with bounds from 0 to 10. So our integral becomes:

    int (1)^2 dt (with bounds of 0 and 10), which = 10?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Feb 2007
    New York, USA
    your thinking won't be true in general (and we don't even have a definite integral here). Take $\displaystyle u(t) = \sin t$, for instance. I think we need more info here, if you have some definite answer to get to, as opposed to an expression.
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  3. #3
    Senior Member
    Jan 2010
    I'm assuming that $\displaystyle u(t)$ is intended to notate the Heaviside step function, not an arbitrary function.

    In any case, could you clarify whether this is a definite or indefinite integral? If we take it as written, it would be an indefinite integral, i.e.,

    $\displaystyle \int x^2(t) \, dt$

    As you observed, you could define $\displaystyle x^2(t)$ as a piecewise function like so:

    $\displaystyle x^2(t) = \begin{cases} 0, & \mbox{if } t<0 \\ 1, & \mbox{if } 0 \le t < 10 \\ 0, & \mbox{if } t \ge 10 \end{cases}$

    Then we could find the anti-derivative using piecewise techniques.

    If instead you are calculating a definite integral whose bounds cover the 0 to 10 range, then your work is correct.
    Last edited by drumist; Jul 1st 2010 at 03:34 AM.
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