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6.8: Step Functions

  • Page ID
    395
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    In this discussion, we will investigate piecewise defined functions and their Laplace Transforms. We start with the fundamental piecewise defined function, the Heaviside function.

    Definition: The Heaviside Function

    The Heaviside function, also called the unit step function, is defined by

    \[ u_c(t) = \left\{\begin{aligned}
    &1 && t <c \\
    &0 && t \ge c
    \end{aligned}
    \right.\]

    for \(c \ge 0\).

    The Heaviside function \(y = u_c(t)\) and \(y = 1 - u_c(t)\) are graphed below.

    stepFu2.gif stepFu3.gif

    Example \(\PageIndex{1}\)

    We can write the function

    \[ f(t) = \left\{\begin{aligned}
    &3 && 0 \le x < 2 \\ &e^x && 2 \le x < 5 \\ &0 && 5 \le x
    \end{aligned}
    \right.\]

    in terms of Heaviside functions.

    Solution

    We tackle the functions in parts. The function that is 1 from 0 to 2, and 0 otherwise is

    \[1 - u_2(x) .\]

    Multiplying by 3 gives

    \[ 3\left(1 - u_2(x)\right) = 3 - 3u_2(x) .\]

    To get the function that is 1 between 2 and 5 and 0 otherwise, we subtract

    \[ u_2(x) - u_5(x) .\]

    Now multiply by \(e^x\) to get

    \[ e^x(u_2(x) - u_5(x)) = e^x u_2(x) - e^x u_5(x). \]

    Adding these together gives

    \[\begin{align} f(x) &= 3 - 3u_2(x) + e^x\, u_2(x) - e^x \,u_5(x) \\ &= 3 + (e^x - 3)\,u_2(x) - e^x\, u_5(x) . \end{align}\]

    We can find the Laplace transform of \(u_c(t)\) by integrating

    \[ \mathcal{L}\{u_c(t)\} = \int_0^{\infty} e^{st} u_c(t) \, dt \]

    \[ = \int _c^{\infty} e^{st} dt = \dfrac{e^{-cs}}{s} \]

    \[ L \{ u_c(t)\} = \dfrac{e^{-cs}}{s}. \]

    In practice, we want to find the Laplace transform of a more general piecewise defined function such as

    \[ f(t) = \left\{\begin{aligned}
    &0 && x < \pi \\ &\sin x && x \ge \pi
    \end{aligned}
    \right.\]

    This type of function occurs in electronics when a switch is suddenly turned on after one second and a forcing function is applied. We can write

    \[ f(x) = u_p(x) \sin x.\]

    We will be interested in the Laplace transform of a product of the Heaviside function with a continuous function. The result that we need is

    \[ \mathcal{L}\{(u_c(t) f(t - c)\} = e^{-cs} \mathcal{L}\{f(t)\} .\]

    By taking inverses we get that if \(F(s) = \mathcal{L}\{f(t)\}\), then

    \[ L^{-1}\{e^{-cs}F(s)\} = u_c(t)f(t - c) .\]

    Proof

    We use the definition to get

    \[ \mathcal{L}\{u_c(t)f(t-c)\} = \int _0^{\infty} e^{-st}u_c(t)f(t-c)\,dt\]

    substitute \(\nu=t-c\)

    \[ = \int _c^{\infty} e^{-st}f(t-c)\,dt = \int _0^{\infty} e^{-(\nu +c)s}f(\nu)\,d\nu \]

    \[ e^{-cs}\int _0^{\infty} e^{-s\nu}f(\nu)\,d\nu = e^{-cs}F(s).\]

    Example \(\PageIndex{2}\)

    Find the Laplace transform of

    \[ f(t) = \left\{\begin{aligned}
    &0 && x < \pi \\ &\sin x && x \ge \pi.
    \end{aligned}
    \right.\]

    Solution

    We use that fact that

    \[ f(x) = u_p(x) \sin x = -u_p(x) \sin[(x - p)] . \]

    Now we can use the formula to get that

    \[ \mathcal{L}\{f(x)\} = -\mathcal{L}\{u_p(x) \sin[(x - p)]\} = -e^{-cs} \mathcal{L}\{\sin x\}. \]

    By the table of Laplace Transforms, we get

    \[ = \dfrac{-e^{-cs}}{s^2 + 1} .\]

    Contributors and Attributions


    This page titled 6.8: Step Functions is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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