Math Problem Statement
Find the Laplace Transform of the piecewise function f(t) defined as f(t) = t/τ for 0 < t < τ and f(t) = 1 for t > τ.
Solution
The problem asks to find the Laplace Transform of the function defined as:
t/\tau, & 0 < t < \tau, \\ 1, & t > \tau. \end{cases}$$ ### Solution: The Laplace Transform of $$f(t)$$ is given by: $$F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} dt.$$ We split the integral into two parts based on the piecewise definition of $$f(t)$$: $$F(s) = \int_0^\tau \frac{t}{\tau} e^{-st} dt + \int_\tau^\infty e^{-st} dt.$$ #### 1. Evaluate the first integral: $$\int_0^\tau \frac{t}{\tau} e^{-st} dt.$$ Let $$A = \frac{1}{\tau}$$, so the integral becomes: $$A \int_0^\tau t e^{-st} dt.$$ Using integration by parts ($$u = t$$ and $$dv = e^{-st} dt$$): $$\int t e^{-st} dt = \left[ -\frac{t}{s} e^{-st} \right]_0^\tau + \int_0^\tau \frac{1}{s} e^{-st} dt.$$ For the second term, $$\int e^{-st} dt = -\frac{1}{s} e^{-st}$$. Substituting: \[ \int t e^{-st} dt = \left[ -\frac{t}{s} e^{-st} \right]_0^\tau + \left[ -\frac{1}{s^2} e^{-st} \right]_0^\tau. \] Evaluating at the limits $$t = \tau$$ and $$t = 0$$: $$\int t e^{-st} dt = -\frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} e^{-s\tau} - \left(0 - \frac{1}{s^2}\right).$$ Simplify: $$\int t e^{-st} dt = -\frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} e^{-s\tau} + \frac{1}{s^2}.$$ Multiplying by $$A = \frac{1}{\tau}$$: $$\int_0^\tau \frac{t}{\tau} e^{-st} dt = \frac{1}{\tau} \left(-\frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} e^{-s\tau} + \frac{1}{s^2}\right).$$ Simplify: $$\int_0^\tau \frac{t}{\tau} e^{-st} dt = -\frac{1}{s} e^{-s\tau} + \frac{1}{s^2 \tau} e^{-s\tau} + \frac{1}{s^2 \tau}.$$ #### 2. Evaluate the second integral: $$\int_\tau^\infty e^{-st} dt.$$ This is straightforward: $$\int_\tau^\infty e^{-st} dt = \left[ -\frac{1}{s} e^{-st} \right]_\tau^\infty.$$ At $$t = \infty$$, $$e^{-st} \to 0$$. At $$t = \tau$$: $$\int_\tau^\infty e^{-st} dt = 0 - \left(-\frac{1}{s} e^{-s\tau}\right).$$ Simplify: $$\int_\tau^\infty e^{-st} dt = \frac{1}{s} e^{-s\tau}.$$ #### 3. Combine results: $$F(s) = \left(-\frac{1}{s} e^{-s\tau} + \frac{1}{s^2 \tau} e^{-s\tau} + \frac{1}{s^2 \tau}\right) + \frac{1}{s} e^{-s\tau}.$$ Simplify terms: $$F(s) = \frac{1}{s^2 \tau} + \frac{1}{s^2 \tau} e^{-s\tau}.$$ This is the Laplace Transform of $$f(t)$$. Would you like a deeper explanation or further simplification? Here are some related questions to consider: 1. How is the Laplace Transform used in engineering applications? 2. Can you extend this result to other piecewise functions? 3. How would this result change if $$f(t)$$ was defined differently for $$t > \tau$$? 4. Can you verify this solution using numerical integration? 5. How do you handle Laplace Transforms of functions with discontinuities? **Tip:** Always check your piecewise definitions carefully while setting up integrals for Laplace Transforms!Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Laplace Transform
Piecewise Functions
Integration by Parts
Exponential Functions
Formulas
Laplace Transform formula: F(s) = ∫_0^∞ f(t)e^(-st) dt
Integration by parts: ∫u dv = uv - ∫v du
Theorems
Properties of Laplace Transform
Evaluation of piecewise integrals
Suitable Grade Level
Undergraduate (Engineering or Mathematics Courses)
Related Recommendation
Laplace Transform of Piecewise Function f(t) with Time Intervals
Laplace Transform of a Piecewise Function: Step-by-Step Solution
Laplace Transform of a Piecewise Function with Exponential Terms
Laplace Transform of a Piecewise Function with t/τ and Constant 1
Find the Laplace Transform of a Piecewise Function