Math Problem Statement
Find the Laplace transform of f(t) defined as f(t) = t/T when 0 ≤ t ≤ T and f(t) = 1 when t > T.
Solution
The given problem involves finding the Laplace transform of a piecewise function . Based on the image, the function is defined as:
\begin{cases} \frac{t}{T}, & 0 \leq t \leq T, \\ 1, & t > T. \end{cases}$$ To find the Laplace transform $$\mathcal{L}\{f(t)\}$$, we need to evaluate the Laplace transform for the two parts of $$f(t)$$ and sum them: ### Definition of Laplace Transform: The Laplace transform of a function $$f(t)$$ is: $$\mathcal{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} dt.$$ ### Step-by-Step Solution: #### 1. For $$0 \leq t \leq T$$ ($$f(t) = \frac{t}{T}$$): The Laplace transform is: $$\mathcal{L}\left\{\frac{t}{T}\right\} = \int_0^T \frac{t}{T} e^{-st} dt.$$ Substitute $$\frac{1}{T}$$ outside the integral: $$\mathcal{L}\left\{\frac{t}{T}\right\} = \frac{1}{T} \int_0^T t e^{-st} dt.$$ Using integration by parts where $$u = t$$ and $$dv = e^{-st} dt$$: $$\int t e^{-st} dt = \frac{-t e^{-st}}{s} + \frac{1}{s} \int e^{-st} dt.$$ $$\int t e^{-st} dt = \frac{-t e^{-st}}{s} - \frac{1}{s^2} e^{-st}.$$ Evaluating the limits from $$0$$ to $$T$$: $$\int_0^T t e^{-st} dt = \left[\frac{-t e^{-st}}{s} - \frac{1}{s^2} e^{-st} \right]_0^T.$$ At $$t = T$$: $$\frac{-T e^{-sT}}{s} - \frac{1}{s^2} e^{-sT}.$$ At $$t = 0$$: $$\frac{-0 \cdot e^{0}}{s} - \frac{1}{s^2} e^{0} = 0.$$ Thus: $$\int_0^T t e^{-st} dt = \frac{-T e^{-sT}}{s} - \frac{1}{s^2} e^{-sT} + \frac{1}{s^2}.$$ Substituting back: $$\mathcal{L}\left\{\frac{t}{T}\right\} = \frac{1}{T} \left[\frac{-T e^{-sT}}{s} - \frac{1}{s^2} e^{-sT} + \frac{1}{s^2} \right].$$ Simplify: $$\mathcal{L}\left\{\frac{t}{T}\right\} = \frac{-e^{-sT}}{s} - \frac{e^{-sT}}{s^2 T} + \frac{1}{s^2 T}.$$ #### 2. For $$t > T$$ ($$f(t) = 1$$): The Laplace transform is: $$\mathcal{L}\{1\} = \int_T^\infty e^{-st} dt.$$ Integrate: $$\int e^{-st} dt = \frac{-e^{-st}}{s}.$$ Evaluate from $$T$$ to $$\infty$$: At $$t = \infty$$, $$e^{-st} \to 0$$; at $$t = T$$, $$e^{-sT}$$: $$\mathcal{L}\{1\} = \left[0 - \left(\frac{-e^{-sT}}{s}\right)\right] = \frac{e^{-sT}}{s}.$$ #### 3. Combine Results: The total Laplace transform is: $$\mathcal{L}\{f(t)\} = \mathcal{L}\left\{\frac{t}{T}\right\} + \mathcal{L}\{1\}.$$ Substitute the results: $$\mathcal{L}\{f(t)\} = \left(\frac{-e^{-sT}}{s} - \frac{e^{-sT}}{s^2 T} + \frac{1}{s^2 T}\right) + \frac{e^{-sT}}{s}.$$ Simplify: $$\mathcal{L}\{f(t)\} = \frac{1}{s^2 T} - \frac{e^{-sT}}{s^2 T}.$$ This is the final Laplace transform. Would you like me to verify the steps, simplify further, or provide more explanation? --- ### Related Questions: 1. How is the Laplace transform used in solving differential equations? 2. What happens to the Laplace transform when $$T \to \infty$$? 3. Can this function $$f(t)$$ be expressed in terms of the Heaviside step function? 4. What are the properties of Laplace transform for piecewise functions? 5. How would the Laplace transform change if the definition of $$f(t)$$ had different intervals? **Tip:** Always double-check the piecewise intervals and boundaries when solving Laplace transforms for segmented functions.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
Definition of Laplace Transform
Properties of Linearity in Laplace Transforms
Suitable Grade Level
Undergraduate