Math Problem Statement
Solution
We are tasked with finding the Laplace transform of a periodic function , where the period is and the function is defined as follows in one period:
1, & 0 < t < \frac{a}{2} \\ -1, & \frac{a}{2} < t < a \end{cases}$$ To compute the Laplace transform of a periodic function, we can use the following formula for the Laplace transform of a periodic function with period $$T$$: $$\mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sT}} \int_0^T e^{-st} f(t) \, dt$$ In this case, the period $$T = a$$. So the Laplace transform becomes: $$\mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sa}} \int_0^a e^{-st} f(t) \, dt$$ We'll now compute the integral in two parts based on the piecewise definition of $$f(t)$$: 1. For $$0 < t < \frac{a}{2}$$, $$f(t) = 1$$: $$\int_0^{\frac{a}{2}} e^{-st} \, dt = \left[ \frac{-e^{-st}}{s} \right]_0^{\frac{a}{2}} = \frac{1 - e^{-\frac{sa}{2}}}{s}$$ 2. For $$\frac{a}{2} < t < a$$, $$f(t) = -1$$: $$\int_{\frac{a}{2}}^a e^{-st} \, dt = \left[ \frac{-e^{-st}}{s} \right]_{\frac{a}{2}}^a = \frac{e^{-\frac{sa}{2}} - e^{-sa}}{s}$$ Now, we can add these two integrals together: $$\int_0^a e^{-st} f(t) \, dt = \frac{1 - e^{-\frac{sa}{2}}}{s} + \frac{e^{-\frac{sa}{2}} - e^{-sa}}{s}$$ Simplifying: $$\int_0^a e^{-st} f(t) \, dt = \frac{1 - e^{-sa}}{s}$$ Thus, the Laplace transform of $$f(t)$$ is: $$\mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sa}} \cdot \frac{1 - e^{-sa}}{s} = \frac{1}{s}$$ Therefore, the Laplace transform of the given periodic function is: $$\boxed{\frac{1}{s}}$$ Let me know if you need more details or explanations. Here are 5 related questions to expand your understanding: 1. What is the general formula for the Laplace transform of a piecewise function? 2. How does the periodicity of a function affect its Laplace transform? 3. What are the conditions for a function to have a Laplace transform? 4. Can you find the inverse Laplace transform of $$\frac{1}{s}$$? 5. How does the choice of period $$a$$ influence the result of the Laplace transform? **Tip:** For periodic functions, the Laplace transform often simplifies significantly due to the periodicity property—always check if the function is periodic before applying standard formulas!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Laplace Transform
Periodic Functions
Piecewise Functions
Formulas
Laplace Transform of periodic function: \( \mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sT}} \int_0^T e^{-st} f(t) dt \)
Theorems
Laplace Transform Properties for Periodic Functions
Suitable Grade Level
Undergraduate Level (Calculus & Differential Equations)
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