Math Problem Statement
Find the Laplace transform of the piecewise function f(t) defined as: f(t) = t/τ when 0 < t < τ, and f(t) = 1 when t > τ.
Solution
The problem appears to be about finding the Laplace Transform of a piecewise function defined as:
\begin{cases} t/\tau, & 0 < t < \tau, \\ 1, & t > \tau. \end{cases}$$ Let's compute the Laplace Transform of this function. --- ### Laplace Transform Formula The Laplace Transform of a function $$f(t)$$ is defined as: $$\mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st}f(t) \, dt$$ We evaluate $$\mathcal{L}\{f(t)\}$$ in two parts, as the function $$f(t)$$ is piecewise. --- ### Step 1: Laplace Transform for $$0 < t < \tau$$ For $$0 < t < \tau$$, $$f(t) = t / \tau$$. The Laplace Transform for this interval is: $$\mathcal{L}_1 = \int_{0}^{\tau} e^{-st} \frac{t}{\tau} \, dt$$ Let $$I = \int_{0}^{\tau} t e^{-st} \, dt$$. Using integration by parts ($$u = t$$, $$dv = e^{-st} dt$$): $$I = \left[ -\frac{t}{s} e^{-st} \right]_0^\tau + \int_{0}^{\tau} \frac{1}{s} e^{-st} \, dt$$ The second term simplifies as: $$\int_{0}^{\tau} e^{-st} \, dt = \left[ \frac{-1}{s} e^{-st} \right]_0^\tau = \frac{-1}{s} (e^{-s\tau} - 1)$$ Thus, $$I = \left( -\frac{\tau}{s} e^{-s\tau} + 0 \right) + \frac{1}{s^2} (1 - e^{-s\tau}) = \frac{1}{s^2} - \frac{e^{-s\tau}}{s} \left( \frac{\tau}{s} + \frac{1}{s} \right)$$ Multiply by $$1/\tau$$ to complete $$\mathcal{L}_1$$. --- ### Step 2: Laplace Transform for $$t > \tau$$ For $$t > \tau$$, $$f(t) = 1$$. The Laplace Transform is: $$\mathcal{L}_2 = \int_{\tau}^{\infty} e^{-st} \, dt = \left[ \frac{-1}{s} e^{-st} \right]_\tau^\infty$$ This evaluates to: $$\mathcal{L}_2 = \frac{1}{s} e^{-s\tau}$$ --- ### Combine Results Adding $$\mathcal{L}_1$$ and $$\mathcal{L}_2$$ gives the final Laplace Transform of $$f(t)$$. Let me know if you would like this expressed in detail or fully simplified.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⁺∞ e⁻ˢᵗ f(t) dt
Integration by Parts: ∫u dv = uv - ∫v du
Theorems
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Suitable Grade Level
Undergraduate (Math/Engineering Courses)