Math Problem Statement
Find the Laplace transform of the given piecewise function f(t).
Solution
The image contains a mathematical problem that involves finding the Laplace transform of a piecewise-defined function . Here's the transcription and detailed steps to solve it.
Function Definition
The function is defined as:
\begin{cases} 1, & 0 < t \leq 1, \\ t, & 1 < t \leq 2, \\ 0, & t > 2. \end{cases}$$ The goal is to compute the Laplace transform of $$f(t)$$. --- ### Laplace Transform Formula The Laplace transform of a function $$f(t)$$ is given by: $$F(s) = \int_0^\infty f(t) e^{-st} \, dt.$$ For this piecewise function, we compute the integral over each interval separately and sum the results. --- ### Step-by-Step Solution #### Interval 1: $$0 < t \leq 1$$ Here, $$f(t) = 1$$. The Laplace transform contribution is: $$F_1(s) = \int_0^1 1 \cdot e^{-st} \, dt = \int_0^1 e^{-st} \, dt.$$ The integral of $$e^{-st}$$ is: $$\int e^{-st} \, dt = \frac{-e^{-st}}{s}.$$ Evaluate from 0 to 1: $$F_1(s) = \left[ \frac{-e^{-st}}{s} \right]_0^1 = \frac{-e^{-s}}{s} - \frac{-1}{s} = \frac{1 - e^{-s}}{s}.$$ --- #### Interval 2: $$1 < t \leq 2$$ Here, $$f(t) = t$$. The Laplace transform contribution is: $$F_2(s) = \int_1^2 t \cdot e^{-st} \, dt.$$ Using integration by parts ($$u = t, dv = e^{-st} dt$$): $$u = t, \quad du = dt, \quad v = \frac{-e^{-st}}{s}.$$ The integral becomes: $$F_2(s) = \left[ \frac{-t e^{-st}}{s} \right]_1^2 + \int_1^2 \frac{e^{-st}}{s} \, dt.$$ The first term evaluates to: $$\left[ \frac{-t e^{-st}}{s} \right]_1^2 = \frac{-2 e^{-2s}}{s} - \frac{-1 e^{-s}}{s} = \frac{-2 e^{-2s} + e^{-s}}{s}.$$ For the second term: $$\int_1^2 \frac{e^{-st}}{s} \, dt = \frac{1}{s} \int_1^2 e^{-st} \, dt = \frac{1}{s} \left[ \frac{-e^{-st}}{s} \right]_1^2 = \frac{1}{s} \left( \frac{-e^{-2s}}{s} - \frac{-e^{-s}}{s} \right).$$ Simplify: $$\int_1^2 \frac{e^{-st}}{s} \, dt = \frac{1}{s^2} \left( e^{-s} - e^{-2s} \right).$$ Combine the results for $$F_2(s)$$: $$F_2(s) = \frac{-2 e^{-2s} + e^{-s}}{s} + \frac{1}{s^2} \left( e^{-s} - e^{-2s} \right).$$ --- #### Interval 3: $$t > 2$$ Here, $$f(t) = 0$$. The Laplace transform contribution is: $$F_3(s) = \int_2^\infty 0 \cdot e^{-st} \, dt = 0.$$ --- ### Combine Results The total Laplace transform is: $$F(s) = F_1(s) + F_2(s) + F_3(s).$$ Substitute: $$F(s) = \frac{1 - e^{-s}}{s} + \frac{-2 e^{-2s} + e^{-s}}{s} + \frac{1}{s^2} \left( e^{-s} - e^{-2s} \right).$$ --- ### Simplified Expression Combine terms and simplify: $$F(s) = \frac{1}{s} - \frac{e^{-s}}{s} - \frac{2 e^{-2s}}{s} + \frac{e^{-s}}{s} + \frac{e^{-s}}{s^2} - \frac{e^{-2s}}{s^2}.$$ --- Would you like me to further simplify or explain any part of the process? Let me know! --- ### Related Questions 1. How is the Laplace transform useful for piecewise functions in engineering applications? 2. What is the role of the Heaviside step function in simplifying Laplace transform computations? 3. How do you handle Laplace transforms of more complex piecewise functions? 4. Can this method be extended to compute Laplace transforms of periodic functions? 5. What are the common challenges in solving Laplace transforms with integration by parts? --- ### Tip To simplify piecewise Laplace transforms, always consider expressing the function using step functions for easier integration!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Laplace Transform
Piecewise Functions
Integration
Formulas
F(s) = ∫₀⁺∞ f(t) e^(-st) dt
Theorems
Integration by Parts
Properties of Laplace Transform
Suitable Grade Level
Undergraduate Level
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