Math Problem Statement
In correct and simple way to solve
Find the Laplace Transform of f(t) defined as: f(t) = t/τ, when 0 < t < τ; 1, when t > τ.
Solution
To solve this problem and find the Laplace Transform of the given , we will proceed step by step:
Problem Analysis
The function is defined as:
\begin{cases} \frac{t}{\tau}, & 0 < t < \tau, \\ 1, & t > \tau. \end{cases}$$ We want to find the Laplace transform of $$f(t)$$, defined as: $$\mathcal{L}[f(t)] = \int_0^\infty f(t)e^{-st} dt$$ ### Step 1: Split the Function Since $$f(t)$$ is piecewise, we need to break the Laplace integral into two parts: 1. From $$0$$ to $$\tau$$, where $$f(t) = \frac{t}{\tau}$$, 2. From $$\tau$$ to $$\infty$$, where $$f(t) = 1$$. Thus, the Laplace transform becomes: $$\mathcal{L}[f(t)] = \int_0^\tau \frac{t}{\tau} e^{-st} dt + \int_\tau^\infty e^{-st} dt$$ ### Step 2: Evaluate Each Integral #### (a) First Integral: $$\int_0^\tau \frac{t}{\tau} e^{-st} dt$$ Let $$I_1 = \int_0^\tau \frac{t}{\tau} e^{-st} dt$$. We will use integration by parts: - Set $$u = t$$ and $$dv = e^{-st} dt$$, - Then, $$du = dt$$ and $$v = \frac{-1}{s} e^{-st}$$. Applying integration by parts: $$I_1 = \frac{1}{\tau} \left[ \frac{-t}{s} e^{-st} \bigg|_0^\tau + \int_0^\tau \frac{1}{s} e^{-st} dt \right]$$ Evaluate each term: - The first term is: $$\frac{-t}{s} e^{-st} \bigg|_0^\tau = \frac{-\tau}{s} e^{-s\tau} - \frac{0}{s} e^{-s \cdot 0} = \frac{-\tau}{s} e^{-s\tau}.$$ - The second integral $$\int_0^\tau e^{-st} dt$$ evaluates to: $$\int_0^\tau e^{-st} dt = \frac{1}{s} \left( 1 - e^{-s\tau} \right).$$ Thus: $$I_1 = \frac{1}{\tau} \left[ \frac{-\tau}{s} e^{-s\tau} + \frac{1}{s} \left( 1 - e^{-s\tau} \right) \right].$$ Simplify: $$I_1 = \frac{1}{s\tau} \left[ -\tau e^{-s\tau} + 1 - e^{-s\tau} \right] = \frac{1}{s\tau} \left( 1 - (\tau + 1)e^{-s\tau} \right).$$ #### (b) Second Integral: $$\int_\tau^\infty e^{-st} dt$$ Let $$I_2 = \int_\tau^\infty e^{-st} dt$$. This is a standard exponential integral: $$I_2 = \frac{1}{s} e^{-st} \bigg|_\tau^\infty = \frac{1}{s} \left( 0 - e^{-s\tau} \right) = -\frac{1}{s} e^{-s\tau}.$$ ### Step 3: Combine Results The total Laplace transform is: $$\mathcal{L}[f(t)] = I_1 + I_2.$$ Substitute $$I_1$$ and $$I_2$$: $$\mathcal{L}[f(t)] = \frac{1}{s\tau} \left( 1 - (\tau + 1)e^{-s\tau} \right) - \frac{1}{s} e^{-s\tau}.$$ Simplify further: $$\mathcal{L}[f(t)] = \frac{1}{s\tau} - \frac{(\tau + 1)}{s\tau} e^{-s\tau} - \frac{1}{s} e^{-s\tau}.$$ Combine like terms: $$\mathcal{L}[f(t)] = \frac{1}{s\tau} - \frac{\tau + 2}{s\tau} e^{-s\tau}.$$ ### Final Answer: $$\mathcal{L}[f(t)] = \frac{1}{s\tau} \left( 1 - (\tau + 2)e^{-s\tau} \right).$$ Let me know if you'd like further clarification or have additional questions! 😊 --- ### Related Questions: 1. How does integration by parts simplify Laplace transforms of linear terms? 2. What are some common Laplace transform properties used for piecewise functions? 3. How do you handle discontinuities in the Laplace transform? 4. Can you apply the shifting property to solve this problem more efficiently? 5. What are the applications of Laplace transforms in solving differential equations? **Tip:** When solving problems involving piecewise functions, always break the integral into sections corresponding to the definition of the function.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 definition: \( \mathcal{L}[f(t)] = \int_0^\infty f(t)e^{-st} dt \)
Integration by parts: \( \int u dv = uv - \int v du \)
Theorems
Laplace Transform linearity
Integration by parts theorem
Suitable Grade Level
Undergraduate - Mathematics or Engineering