Math Problem Statement
Find the Laplace transform of the piecewise-defined function f(t), where f(t) = t/τ for 0 < t < τ and f(t) = 1 for t > τ.
Solution
The problem asks to find the Laplace transform of the piecewise-defined function where:
\begin{cases} t/\tau & \text{for } 0 < t < \tau, \\ 1 & \text{for } t > \tau.$$ Let’s proceed step by step: --- ### **Step 1: Definition of Laplace Transform** The Laplace transform of a function $$f(t)$$ is given by: $$\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty f(t) e^{-st} \, dt.$$ --- ### **Step 2: Handling Piecewise Functions** For a piecewise function, we split the Laplace transform as: $$F(s) = \int_0^\tau \frac{t}{\tau} e^{-st} \, dt + \int_\tau^\infty 1 \cdot e^{-st} \, dt.$$ We will evaluate both integrals separately. --- ### **Step 3: First Integral** Evaluate: $$\int_0^\tau \frac{t}{\tau} e^{-st} \, dt.$$ Let $$a = 1/\tau$$, so: $$\int_0^\tau \frac{t}{\tau} e^{-st} \, dt = \int_0^\tau a t e^{-st} \, dt.$$ Using integration by parts where $$u = t$$ and $$dv = e^{-st} dt$$: - $$u = t \implies du = dt$$, - $$v = \int e^{-st} dt = \frac{-1}{s} e^{-st}.$$ Thus: $$\int t e^{-st} \, dt = \left[ \frac{-t}{s} e^{-st} \right]_0^\tau + \int_0^\tau \frac{1}{s} e^{-st} \, dt.$$ The second term simplifies to: $$\int e^{-st} \, dt = \frac{-1}{s^2} e^{-st}.$$ After combining and simplifying, the first integral becomes: $$\int_0^\tau \frac{t}{\tau} e^{-st} \, dt = \frac{1}{\tau} \left[ \frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} \left( 1 - e^{-s\tau} \right) \right].$$ --- ### **Step 4: Second Integral** Evaluate: $$\int_\tau^\infty e^{-st} \, dt = \left[ \frac{-1}{s} e^{-st} \right]_\tau^\infty.$$ At $$t = \infty$$, $$e^{-st} \to 0$$. At $$t = \tau$$: $$\int_\tau^\infty e^{-st} \, dt = \frac{1}{s} e^{-s\tau}.$$ --- ### **Step 5: Combine Results** Summing both integrals, we get: $$F(s) = \frac{1}{\tau} \left[ \frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} \left( 1 - e^{-s\tau} \right) \right] + \frac{1}{s} e^{-s\tau}.$$ Simplify to get the final Laplace transform. --- Would you like detailed simplification or additional steps explained? --- ### **Related Questions** 1. How do you perform integration by parts for piecewise functions? 2. What is the role of the Heaviside function in handling piecewise Laplace transforms? 3. How does the Laplace transform handle discontinuities? 4. Can the transform be applied directly to generalized piecewise functions? 5. What are the applications of Laplace transforms in engineering and physics? --- ### **Tip** When dealing with piecewise functions, always separate the integrals carefully and simplify each part before combining.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Laplace transform
Piecewise functions
Integration by parts
Formulas
Laplace transform: L{f(t)} = ∫[0,∞] f(t)e^(-st) dt
Integration by parts: ∫u dv = uv - ∫v du
Theorems
Fundamental theorem of calculus for definite integrals
Suitable Grade Level
Undergraduate - Engineering or Applied Mathematics