Question

Express the given function \(f\) in terms of unit step functions and use Theorem 8.4 .1 to find \(\mathcal{L}(f) .\) Where indicated by, graph \(f\). $$ f(t)=\left\\{\begin{array}{cl} t, & 0 \leq t<1 ,\\\ 2-t, & 1 \leq t<2, \\ 6, & t>2. \end{array}\right. $$

Short answer

#tag_title#Short Answer#tag_content# The Laplace Transform of the given piecewise function is: $$\mathcal{L}(f) = \frac{1}{s^2} + \frac{2e^{-s}}{s} - \frac{e^{-s}}{s^2} + \frac{6e^{-2s}}{s}$$

Step-by-step solution

Express the function in terms of unit step functions

First, we can break the function into three separate terms based on the given conditions: 1. \(t\) multiplied by a unit step function that is activated when \(0\leq t<1\). 2. \((2-t)\) multiplied by a unit step function that is activated when \(1\leq t<2\). 3. \(6\) multiplied by a unit step function that is activated when \(t>2\). We can represent the unit step functions as: 1. \(u(t)\) for \(0\leq t<1\). 2. \(u(t-1)\) for \(1\leq t<2\). 3. \(u(t-2)\) for \(t>2\). Now, we can write the given function \(f(t)\) in terms of unit step functions: \(f(t) = tu(t) + (2-t)u(t-1) + 6u(t-2)\)

Apply Laplace Transform Using Theorem 8.4.1

Following the theorem 8.4.1 (Laplace transform of a function multiplied by a delayed unit step function), we have: $$\mathcal{L}\{f(t)u(t-a)\} = e^{-as}F(s)$$ Now, let's apply the Laplace Transform to each term in \(f(t)\): 1. \(\mathcal{L}\{t u(t)\} = \mathcal{L}\{t\} = \frac{1}{s^2}\) (since \(u(t)\) has no delay) 2. \(\mathcal{L}\{(2-t) u(t-1)\} = e^{-s}\mathcal{L}\{(2-t) \}\) 3. \(\mathcal{L}\{6 u(t-2)\} = 6e^{-2s}\mathcal{L}\{1\}\) For the second term, we need to apply the Laplace Transform to the function \((2-t)\). We can split that into two parts and apply the properties of the Laplace Transform: $$\mathcal{L}\{(2-t)\} = \mathcal{L}\{(2 - t)\} = \mathcal{L}\{2\} - \mathcal{L}\{t\} = \frac{2}{s} - \frac{1}{s^2}$$ Now, let's substitute this back to the second term's Laplace Transform: $$e^{-s}\left(\frac{2}{s} - \frac{1}{s^2}\right) = \frac{2e^{-s}}{s} - \frac{e^{-s}}{s^2}$$ For the third term, the Laplace Transform of a constant, in this case, \(1\) is: $$\mathcal{L}\{1\} = \frac{1}{s}$$ Let's substitute this back into the third term's Laplace Transform: $$6e^{-2s}\frac{1}{s} = \frac{6e^{-2s}}{s}$$ Now, let's sum up all terms: $$\mathcal{L}(f) = \frac{1}{s^2} + \frac{2e^{-s}}{s} - \frac{e^{-s}}{s^2} + \frac{6e^{-2s}}{s}$$ So the Laplace Transform of the given function is: $$\mathcal{L}(f) = \frac{1}{s^2} + \frac{2e^{-s}}{s} - \frac{e^{-s}}{s^2} + \frac{6e^{-2s}}{s}$$

Key concepts

Unit Step Functions

Let's delve into understanding unit step functions, a concept critical in handling piecewise functions within the realm of Laplace transforms. Unit step functions, often denoted as u(t-a), serve as 'switches' that activate at a particular moment a. If you imagine plotting their graph, they shoot from zero to one at the time a, creating a step-like discontinuity. This jump has significant implications, as it allows us to model functions that change behavior or activate at specific times.

In our given problem, these unit step functions are employed to express the different intervals of the function f(t): whenever the condition for the interval becomes true, the unit step 'switches on', indicating that the function in that interval should now take effect. Such modeling is tremendously useful as it simplifies the otherwise complex analysis of piecewise functions when subjected to operations like the Laplace transform.

Laplace Transform Properties

Exploring the properties of Laplace transforms uncovers the toolkit for transforming complex time-domain functions into a more manageable frequency-domain representation. These properties include linearity, time-shift, and frequency-shift, amongst others. Linearity allows us to take transforms of sums term by term, effectively dissecting a complicated function into more elementary parts.

In the exercise, we use the time-shift property when handling unit step functions. This property tells us how a time-delayed function, such as (2-t)u(t-1), affects the transform. The formula e^-asF(s), where a is the delay and F(s) is the original transform, provides a straightforward approach to calculating transforms of delayed functions. Mastery of these properties serves to make the Laplace transform a powerful and flexible tool for solving differential equations.

Piecewise Continuous Functions

Piecewise continuous functions are like a puzzle, composed of distinct function 'pieces' that apply over specific intervals. Visualizing graphs of these functions reveals that they can change abruptly at certain points, which can challenge traditional analytical methods.

However, unit step functions pair beautifully with them, turning each 'piece' into a function multiplied by a step function that indicates its interval of relevancy. Our provided example f(t) is a classic case, with three segments each defined over different intervals. Through the ingenuity of step functions and the transforming power of Laplace, what initially seems fragmented can be expressed and operated on as a unified whole.

Theorem Application in Laplace Transforms

The application of theorems in Laplace transforms is akin to a chef using a recipe: it guides us how to mix and match properties and functions to achieve our desired transform. Theorem 8.4.1, used in our exercise, lays out the method for transforming a function multiplied by a delayed unit step function. Essentially, this theorem is a specific application of the time-shift property, packaging it in a form that's directly applicable to piecewise functions.

By using this theorem, we conveniently deal with each segment of the piecewise function f(t) on its own, transforming it as if it were the entire function active over all time, just scaled and shifted according to its unit step. This piecemeal approach leads to a comprehensive Laplace Transform of the entire function, simplified to manageable terms that can be tackled with further mathematical or computational tools.