Math Problem Statement
In correct and detail explanation
Find the Laplace transform of the given piecewise function f(t) defined as:
- f(t) = t, when 0 < t < τ
- f(t) = 1, when t > τ
Solution
To solve this problem, we will compute the Laplace transform of the given function , which is defined as:
\begin{cases} t, & \text{when } 0 < t < \tau \\ 1, & \text{when } t > \tau \end{cases}$$ ### Definition of Laplace Transform: The Laplace transform of $$f(t)$$ is given by: $$\mathcal{L}\{f(t)\} = \int_{0}^{\infty} f(t)e^{-st} dt$$ ### Step-by-Step Solution: The function $$f(t)$$ is piecewise, so the Laplace transform must be split into two integrals based on the intervals: $$\mathcal{L}\{f(t)\} = \int_{0}^{\tau} t e^{-st} dt + \int_{\tau}^{\infty} 1 \cdot e^{-st} dt$$ #### 1. First Integral: $$\int_{0}^{\tau} t e^{-st} dt$$ Let’s compute this integral: $$\int_{0}^{\tau} t e^{-st} dt$$ Use integration by parts, where: - Let $$u = t$$ and $$dv = e^{-st} dt$$. - Then $$du = dt$$ and $$v = \frac{e^{-st}}{-s} = -\frac{1}{s}e^{-st}$$. Now apply the integration by parts formula: $$\int u dv = uv - \int v du$$ $$\int_{0}^{\tau} t e^{-st} dt = \left[ -\frac{t}{s} e^{-st} \right]_0^{\tau} + \int_{0}^{\tau} \frac{1}{s} e^{-st} dt$$ 1. Evaluate the boundary term $$\left[ -\frac{t}{s} e^{-st} \right]_0^{\tau}$$: $$= -\frac{\tau}{s} e^{-s\tau} + 0 = -\frac{\tau}{s} e^{-s\tau}$$ 2. Compute the second term $$\int_{0}^{\tau} \frac{1}{s} e^{-st} dt$$: $$\int_{0}^{\tau} \frac{1}{s} e^{-st} dt = \frac{1}{s} \left[ \frac{e^{-st}}{-s} \right]_0^{\tau}$$ $$= \frac{1}{s} \left( -\frac{e^{-s\tau}}{s} + \frac{1}{s} \right) = \frac{1}{s^2} \left( 1 - e^{-s\tau} \right)$$ Combine the results: $$\int_{0}^{\tau} t e^{-st} dt = -\frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} \left( 1 - e^{-s\tau} \right)$$ Simplify: $$\int_{0}^{\tau} t e^{-st} dt = \frac{1}{s^2} - \frac{\tau + 1/s}{s} e^{-s\tau}$$ #### 2. Second Integral: $$\int_{\tau}^{\infty} e^{-st} dt$$ Now compute the second part: $$\int_{\tau}^{\infty} e^{-st} dt$$ The integral of $$e^{-st}$$ is: $$\int e^{-st} dt = \frac{e^{-st}}{-s}$$ Evaluate from $$\tau$$ to $$\infty$$: $$\int_{\tau}^{\infty} e^{-st} dt = \left[ \frac{e^{-st}}{-s} \right]_{\tau}^{\infty}$$ 1. At $$t = \infty$$, $$e^{-st} \to 0$$. 2. At $$t = \tau$$, $$e^{-st} = e^{-s\tau}$$. So: $$\int_{\tau}^{\infty} e^{-st} dt = 0 - \left( \frac{e^{-s\tau}}{-s} \right)$$ $$= \frac{e^{-s\tau}}{s}$$ #### 3. Combine Both Integrals: The total Laplace transform is: $$\mathcal{L}\{f(t)\} = \int_{0}^{\tau} t e^{-st} dt + \int_{\tau}^{\infty} e^{-st} dt$$ Substitute the results: $$\mathcal{L}\{f(t)\} = \left( \frac{1}{s^2} - \frac{\tau + 1/s}{s} e^{-s\tau} \right) + \frac{e^{-s\tau}}{s}$$ Simplify: $$\mathcal{L}\{f(t)\} = \frac{1}{s^2} + \frac{-\tau e^{-s\tau}}{s}$$ ### Final Answer: $$\mathcal{L}\{f(t)\} = \frac{1}{s^2} - \frac{\tau}{s} e^{-s\tau}$$ Would you like me to clarify any step further or verify alternative methods? --- ### Related Questions: 1. How does integration by parts help solve Laplace transforms of polynomial functions? 2. What is the physical significance of the Laplace transform in engineering systems? 3. How do piecewise functions influence the Laplace transform calculation? 4. Can this problem be solved using the Heaviside function instead? Why or why not? 5. What are the common errors to avoid when solving Laplace transforms for piecewise functions? ### Tip: When dealing with piecewise functions, consider expressing the function using the Heaviside step function to streamline the Laplace transform calculation.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Laplace Transform
Integration by Parts
Piecewise Functions
Formulas
Laplace transform formula: \( \mathcal{L}\{f(t)\} = \int_{0}^{\infty} f(t)e^{-st} dt \)
Integration by parts formula: \( \int u dv = uv - \int v du \)
Theorems
Linearity of Laplace Transforms
Integration by Parts Theorem
Suitable Grade Level
Undergraduate
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