Laplace Transform of Piecewise Continuous Functions Calculator

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Example

Created on 2024-06-20Asked by Evelyn Smith (Solvelet student)

Find the Laplace transform of a piecewise continuous function

f(t)

defined as:

f(t) = \begin{cases} 1, & 0 \leq t < 1 \\ 2, & t \geq 1 \end{cases}

Solution

To find the Laplace transform of the piecewise continuous function

f(t)

, we break it into two intervals and use the definition of the Laplace transform: For

0 \leq t < 1

\mathcal{L}\{1\} = \int_{0}^{1} e^{-st} \, dt = \left[ \frac{e^{-st}}{-s} \right]_{0}^{1} = \frac{1 - e^{-s}}{s}

For

t \geq 1

\mathcal{L}\{2\} = \int_{1}^{\infty} 2e^{-st} \, dt = 2 \left[ \frac{e^{-st}}{-s} \right]_{1}^{\infty} = \frac{2e^{-s}}{s}

Therefore, the Laplace transform of

f(t)

is:

\mathcal{L}\{f(t)\} = \frac{1 - e^{-s}}{s} + \frac{2e^{-s}}{s} = \frac{1 + e^{-s}}{s}

Solved on Solvelet with Basic AI Model

Some of the related questions asked by Luna Rodriguez on Solvelet

1. Compute the Laplace transform of the piecewise function

f(t) = \begin{cases} e^{-t}, & 0 \le t < 1 \\ 0, & t \ge 1 \end{cases}

,2. Find the inverse Laplace transform of the function

F(s) = \frac{s + 2}{s^2 + 4s + 13}

DefinitionPiecewise Continuous Functions: under this transformation, the functions being transformed are continuous in pieces with finite discontinuities. Many real-world applications have such functions. The Laplace transform is computed by summing the contributions from each continuous segment, and handling discontinuities properly. Example For the piecewise function \( f(t) \) given as \( f(t) = \left\{ \begin{array} { l l l } { 0, } & { 0 \leq t < 1 } \\ { 1, } & { t \geq 1 } \end{array} \right\), the Laplace transform is computed by summing the Laplace contributions obtained from each segment. For the detailed lecture about the classics Laplace Transform of Piecewise Continuous Functions

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