Problem 28 - Differential Equations and Linear Algebra

Question

Sketch \(f(t),\) express \(f(t)\) in terms of \(u_{a}(t),\) and determine \(L\\{f(t)\\}\). $$f(t)=\left\\{\begin{array}{cc} t, & 0 \leq t < 1 \\ 1 & 1 < t \leq 2 \\ 3-t, & 2 < t \leq 3 \\ 0, & t > 3 \end{array}\right.$$

Step-by-Step Solution

Answer

The short answer for this question is: 1. The sketch of the function, \(f(t)\), consists of four segments: a diagonal line from (0,0) to (1,1), a horizontal line at y=1 for \(1 < t \leq 2\), a diagonal line from (2,1) to (3,0), and \(f(t) = 0\) for \(t > 3\). 2. We can express \(f(t)\) in terms of the unit step function as: \(f(t) = t[u_0(t) - u_1(t)] + 1[u_1(t) - u_2(t)] + (3-t)[u_2(t) - u_3(t)]\) 3. Lastly, the Laplace Transform of the given function is: \(L\\{f(t)\\} = \frac{1}{s^2} - \frac{e^{-s} - e^{-2s}}{s} + \frac{e^{-3s}}{s^2}\)

Step 1: Sketch the function f(t)

To sketch the function, we will draw each segment of the piecewise function on the same graph. The function is defined as follows: - \(t\) when \(0 \leq t < 1\) - \(1\) when \(1 < t \leq 2\) - \(3-t\) when \(2 < t \leq 3\) - \(0\) when \(t > 3\) Next, plot each segment on the graph, making sure to show the appropriate intervals for each piece.

Step 2: Express f(t) in terms of the unit step function u_a(t)

The unit step function u_a(t) is defined as: $$ u_a(t) = \left\\{ \begin{array}{cc} 0, & t < a \\\ 1, & t \geq a \end{array} \right. $$ Now, we will express the given function f(t) in terms of the unit step function u_a(t): $$ f(t) = t[u_0(t) - u_1(t)] + 1[u_1(t) - u_2(t)] + (3-t)[u_2(t) - u_3(t)] + 0 \cdot u_3(t) $$

Step 3: Determine the Laplace Transform of f(t)

To find the Laplace Transform of f(t), we will apply the Laplace Transform to each term in the expression from Step 2: $$ L\\{f(t)\\} = L\\{t(u_0(t) - u_1(t))\\} + L\\{1(u_1(t) - u_2(t))\\} + L\\{(3-t)(u_2(t) - u_3(t))\\} + L\\{0 \cdot u_3(t)\\} $$ Now, we can use the properties of the Laplace Transform to evaluate of each term: 1. \(L\\{t(u_0(t) - u_1(t))\\} = L\\{t (u_0(t) - u_1(t)) e^{-0s}\\} = \frac{1}{s^2} - \frac{e^{-s}}{s^2}\) 2. \(L\\{1(u_1(t) - u_2(t))\\} = \frac{e^{-s} - e^{-2s}}{s}\) 3. \(L\\{(3-t)(u_2(t) - u_3(t))\\} = L\\{(3 - t) (u_2(t) - u_3(t)) e^{-2s}\\} = \frac{-e^{-2s} + e^{-3s}}{s^2}\) 4. Since the last term is multiplied by 0, the Laplace Transform will also be 0. Finally, add all the Laplace Transforms together to obtain the final result: $$ L\\{f(t)\\} = \left( \frac{1}{s^2} - \frac{e^{-s}}{s^2} \right) + \left( \frac{e^{-s} - e^{-2s}}{s} \right) + \left( \frac{-e^{-2s} + e^{-3s}}{s^2} \right) = \frac{1}{s^2} - \frac{e^{-s} - e^{-2s}}{s} + \frac{e^{-3s}}{s^2} $$ This is the final Laplace Transform of the given function f(t): $$ L\\{f(t)\\} = \frac{1}{s^2} - \frac{e^{-s} - e^{-2s}}{s} + \frac{e^{-3s}}{s^2} $$

Key Concepts

Unit Step Function

The Unit Step Function, commonly denoted as \( u_a(t) \), is an essential concept in mathematical analysis and control systems. It is sometimes called the Heaviside step function. This function plays a crucial role in switching operations because it can "turn on" a function at a specified time. The unit step function is defined as follows:

• \( u_a(t) = 0 \) when \( t < a \)
• \( u_a(t) = 1 \) when \( t \geq a \)

This step function allows us to model scenarios where a process or system starts at a particular time. In practical applications, it helps to define functions over specific intervals, which is exactly what is done in the problem we are examining. By expressing functions in terms of the unit step function, we gain the flexibility to handle piecewise functions efficiently with transforms like the Laplace Transform.In the context of this exercise, the unit step function helps us to convert a piecewise function into an expression that can be more easily transformed, especially for further analysis like solving differential equations. It is a powerful tool in signal processing and in the analysis of systems control, where time-based switching is common.

Piecewise Functions

Piecewise functions are mathematical expressions that define a function by different expressions over separate intervals of the independent variable. These functions are particularly useful for describing systems and phenomena that change behavior at certain points in time or space. This means that within different portions or pieces of its domain, the function will have a different form or rule.In our exercise, the function \( f(t) \) is a piecewise function with distinct expressions for each time interval:

• \( f(t) = t \) for \( 0 \leq t < 1 \)
• \( f(t) = 1 \) for \( 1 < t \leq 2 \)
• \( f(t) = 3-t \) for \( 2 < t \leq 3 \)
• \( f(t) = 0 \) for \( t > 3 \)

Such functions are highly versatile and can model real-life situations like traffic flow, material properties varying with temperature, or signal processing. By using techniques like the Laplace Transform in conjunction with unit step functions, we can effectively analyze and simplify these functions to solve complex mathematical and engineering problems seamlessly.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are fundamental in describing various phenomena in physics, engineering, biology, economics, and beyond. These equations represent how a quantity changes in relation to another and are essential for modeling dynamic systems.When dealing with differential equations, the Laplace Transform becomes a vital tool. It transforms differential equations from the time domain into the frequency domain, where they often become algebraic and simpler to solve. After solving in the frequency domain, the inverse Laplace Transform is used to switch back to the time domain.The exercise under discussion explores expressing a piecewise function in terms of the unit step function. This is a technique that frequently appears when formulating or solving differential equations using Laplace Transforms. By simplifying the representation of our function \( f(t) \) into terms involving \( t \) and the step function \( u_a(t) \), we set the stage for easier manipulation and solution of associated differential equations. Understanding how to handle such transformations is key to mastering the resolution of differential equations in applied mathematics.