Transform of a Piecewise-Continuous Function Evaluate [latex]\mathscr{L}\{f(t)\}[/latex] for [latex]f(t) \quad\left\{\begin{array}{lr}0, & 0 \leq t<3 \\ 2, & t \geq 3 .\end{array}\right.[/latex]

This piecewise-continuous function appears in FIGURE 4.1.5. Since [latex]f[/latex] is defined in two pieces, [latex]\mathscr{L}\{f(t)\}[/latex] is expressed as the sum of two integrals: [latex] \begin{aligned} \mathscr{L}\{f(t)\} \quad \int_{0}^{\infty} e^{-s t} f(t) d t & \int_{0}^{3} e^{-s t}(0) d t+\int_{3}^{\infty} e^{-s t}(2) d t \\ = & -\left.\frac{2 e^{-s t}}{s}\right|_{3} ^{\infty} \\ & \frac{2 e^{-3 s}}{s}, \quad s>0 . \end{aligned} [/latex]

Question 4.1.6: Transform of a Piecewise-Continuous Function Evaluate L{f(t)......

Advanced Engineering Mathematics [1367744]

Transform of a Piecewise-Continuous Function

Evaluate $\mathscr{L}\{f(t)\}$ for $f(t) \quad\left\{\begin{array}{lr}0, & 0 \leq t<3 \\ 2, & t \geq 3 .\end{array}\right.$

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This piecewise-continuous function appears in FIGURE 4.1.5. Since $f$ is defined in two pieces, $\mathscr{L}\{f(t)\}$ is expressed as the sum of two integrals:

$\begin{aligned} \mathscr{L}\{f(t)\} \quad \int_{0}^{\infty} e^{-s t} f(t) d t & \int_{0}^{3} e^{-s t}(0) d t+\int_{3}^{\infty} e^{-s t}(2) d t \\ = & -\left.\frac{2 e^{-s t}}{s}\right|_{3} ^{\infty} \\ & \frac{2 e^{-3 s}}{s}, \quad s>0 . \end{aligned}$