Math Problem Statement
Find laplace transform of f(t)=0,t<4 t^2-8t+22,t>=4
Solution
To find the Laplace transform of the given piecewise function:
\begin{cases} 0, & t < 4 \\ t^2 - 8t + 22, & t \geq 4 \end{cases}$$ We can express $$f(t)$$ using the Heaviside step function $$u(t)$$: $$f(t) = (t^2 - 8t + 22)u(t-4)$$ ### Step 1: Compute the Laplace Transform The Laplace transform of a function multiplied by the Heaviside step function can be calculated using the formula: $$\mathcal{L}\{f(t)u(t-a)\} = e^{-as}F(s)$$ where $$F(s)$$ is the Laplace transform of $$f(t)$$ without the step function. ### Step 2: Calculate $$F(s)$$ First, we need to compute the Laplace transform of $$g(t) = t^2 - 8t + 22$$: $$F(s) = \mathcal{L}\{t^2 - 8t + 22\} = \mathcal{L}\{t^2\} - 8\mathcal{L}\{t\} + 22\mathcal{L}\{1\}$$ Using the standard transforms: - $$\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$$ - $$\mathcal{L}\{1\} = \frac{1}{s}$$ Calculating each term: 1. $$\mathcal{L}\{t^2\} = \frac{2!}{s^3} = \frac{2}{s^3}$$ 2. $$\mathcal{L}\{t\} = \frac{1}{s^2}$$ 3. $$\mathcal{L}\{1\} = \frac{1}{s}$$ Now substituting these back into $$F(s)$$: $$F(s) = \frac{2}{s^3} - 8\left(\frac{1}{s^2}\right) + 22\left(\frac{1}{s}\right) = \frac{2}{s^3} - \frac{8}{s^2} + \frac{22}{s}$$ ### Step 3: Laplace Transform of $$f(t)$$ Now applying the Heaviside step function: $$\mathcal{L}\{f(t)\} = e^{-4s}F(s) = e^{-4s}\left(\frac{2}{s^3} - \frac{8}{s^2} + \frac{22}{s}\right)$$ ### Final Result Thus, the Laplace transform of $$f(t)$$ is: $$\mathcal{L}\{f(t)\} = e^{-4s} \left( \frac{2}{s^3} - \frac{8}{s^2} + \frac{22}{s} \right)$$ Would you like more details on any specific step or further questions about Laplace transforms? Here are some related questions to consider: 1. What are the properties of the Laplace transform? 2. How does the Heaviside step function affect the Laplace transform? 3. Can you give examples of other piecewise functions and their Laplace transforms? 4. What applications do Laplace transforms have in solving differential equations? 5. How do you find the inverse Laplace transform? Tip: Always remember to express piecewise functions using the Heaviside function for easier Laplace transformation!Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Laplace Transforms
Piecewise Functions
Heaviside Step Function
Formulas
\mathcal{L}\{f(t)u(t-a)\} = e^{-as}F(s)
\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}
\mathcal{L}\{1\} = \frac{1}{s}
Theorems
Laplace Transform Properties
Suitable Grade Level
Grades 11-12
Related Recommendation
Laplace Transform of a Piecewise Function Using Heaviside and Time Shift
Laplace Transform of a Piecewise Function Involving the Heaviside Function
Laplace Transform of Piecewise Function with Heaviside Step Function
Laplace Transform of Piecewise Function f(t) = 3t for 0 < t < 2 and f(t) = 7 for 2 < t < 4
Laplace Transform of a Piecewise Function Using Heaviside Step Functions