SUBJECT: DIFFERENTIAL EQUATION **Problem #5:**Given the function $ f(t) = -\frac{1}{2}t + 8 $, where $ 0 \leq t < 4 $, and it is periodic such that $ f(t+4) = f(t) $.Find $ F(s) = \mathcal{L} \{ f(t) \} $ of the Periodic Function. **Explanation:**- **Function Definition:** - $ f(t) = -\frac{1}{2}t + 8 $ - This is valid for the interval $ 0 \leq t < 4 $. - The function repeats every 4 units (periodic with period 4).- **Objective:** - Calculate the Laplace Transform $ F(s) = \mathcal{L} \{ f(t) \} $ for the periodic function described. No graphs or diagrams are provided.

Answered: Given f (t)=--1+8 +8,0

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Question

SUBJECT: DIFFERENTIAL EQUATION

**Problem #5:**

Given the function $ f(t) = -\frac{1}{2}t + 8 $, where $ 0 \leq t < 4 $, and it is periodic such that $ f(t+4) = f(t) $.

Find $ F(s) = \mathcal{L} \{ f(t) \} $ of the Periodic Function.

**Explanation:**

- **Function Definition:**
- $ f(t) = -\frac{1}{2}t + 8 $
- This is valid for the interval $ 0 \leq t < 4 $.
- The function repeats every 4 units (periodic with period 4).

- **Objective:**
- Calculate the Laplace Transform $ F(s) = \mathcal{L} \{ f(t) \} $ for the periodic function described.

No graphs or diagrams are provided.

Expert Solution

Step 1

According to the given information it is required to find

$F (s) = L \{f (t)\}$

Where, L represents Laplace Transformation and given:

$\begin{array}{l} f (t) = - \frac{1}{2} t + 8, 0 \leq t < 4 \\ f (t + 4) = f (t) \end{array}$

Step 2

Laplace transformation of a periodic function with period p>0 is given by

$\begin{array}{l} L \{f (t)\} = L \{f_{1} (t)\} \times \frac{1}{1 - e^{- s p}} \\ L \{f (t)\} = \frac{\int_{0}^{p} f (t) e^{- s t} d t}{1 - e^{- s p}} \\ here, f_{1} (t) is one period of the function. \end{array}$

Step by step

Solved in 4 steps

Knowledge Booster

$Differential Equation$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions