Math Problem Statement
correct and simple way
Find the Laplace transform of f(t) defined as f(t) = t/τ when 0 < t < τ, and f(t) = 1 when t > τ.
Solution
The given problem is to find the Laplace Transform of a function defined as:
\begin{cases} \frac{t}{\tau}, & \text{when } 0 < t < \tau \\ 1, & \text{when } t > \tau \end{cases}$$ Here is a step-by-step solution: --- ### Step 1: Definition of Laplace Transform The Laplace Transform of a function $$f(t)$$ is given by: $$\mathcal{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} \, dt$$ --- ### Step 2: Split the Integral Based on $$f(t)$$ Since $$f(t)$$ is piecewise-defined, we split the Laplace Transform as: $$\mathcal{L}\{f(t)\} = \int_0^\tau \frac{t}{\tau} e^{-st} \, dt + \int_\tau^\infty 1 \cdot e^{-st} \, dt$$ --- ### Step 3: Solve the First Integral ($$0 < t < \tau$$) The first term is: $$\int_0^\tau \frac{t}{\tau} e^{-st} \, dt$$ Let $$\frac{1}{\tau}$$ be a constant factor: $$= \frac{1}{\tau} \int_0^\tau t e^{-st} \, dt$$ Using integration by parts: $$\int t e^{-st} \, dt = \left[ \frac{-t e^{-st}}{s} \right]_0^\tau + \int_0^\tau \frac{e^{-st}}{s} \, dt$$ 1. Evaluate the boundary term: $$\left[ \frac{-t e^{-st}}{s} \right]_0^\tau = \frac{-\tau e^{-s\tau}}{s} - 0 = \frac{-\tau e^{-s\tau}}{s}$$ 2. Solve the remaining integral: \[ \int_0^\tau \frac{e^{-st}}{s} \, dt = \frac{1}{s} \left[ \frac{-e^{-st}}{s} \right]_0^\tau = \frac{1}{s^2} \left[ -e^{-s\tau} + 1 \right] \] Combine both terms: $$\int_0^\tau t e^{-st} \, dt = \frac{-\tau e^{-s\tau}}{s} + \frac{1 - e^{-s\tau}}{s^2}$$ Factor out $$e^{-s\tau}$$: $$= \frac{1}{\tau} \left[ \frac{-\tau e^{-s\tau}}{s} + \frac{1 - e^{-s\tau}}{s^2} \right]$$ --- ### Step 4: Solve the Second Integral ($$t > \tau$$) The second term is: $$\int_\tau^\infty e^{-st} \, dt = \left[ \frac{-e^{-st}}{s} \right]_\tau^\infty = 0 - \left( \frac{-e^{-s\tau}}{s} \right)$$ $$= \frac{e^{-s\tau}}{s}$$ --- ### Step 5: Combine the Results The Laplace Transform is: $$\mathcal{L}\{f(t)\} = \frac{1}{\tau} \left[ \frac{-\tau e^{-s\tau}}{s} + \frac{1 - e^{-s\tau}}{s^2} \right] + \frac{e^{-s\tau}}{s}$$ Simplify further if required. --- Do you want further simplifications or step-by-step explanations? ### Expanding Questions: 1. Can you apply this method to a different piecewise function? 2. How does this transform behave as $$s \to \infty$$? 3. How would you interpret this solution in terms of signals or systems? 4. Can you derive the inverse Laplace Transform from this result? 5. How would the Laplace Transform change if $$f(t)$$ were scaled differently? ### Tip: Always identify the boundaries of a piecewise function before solving its Laplace Transform. It simplifies integration and avoids errors.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: \( \mathcal{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} dt \)
Integration by Parts: \( \int u dv = uv - \int v du \)
Theorems
Laplace Transform Definition
Piecewise Integration
Suitable Grade Level
Undergraduate Engineering or Mathematics