Question

In each exercise, graph the function \(f(t)\) for \(0 \leq t<\infty\), and use Table \(5.1\) to find the Laplace transform of \(f(t)\). \(f(t)=(2-t)[h(t-1)-h(t-3)]\)

Short answer

The Laplace transform of the function \(f(t)\) can be found using the Laplace transform properties, resulting in the expression: \(F(s) = -\frac{3}{s^2}(e^{-s} - e^{-3s})\). In this expression, the Laplace transform helps describe f(t) in terms of exponentials and the parameter s. This can be useful in solving differential equations and analyzing system responses in engineering applications.

Step-by-step solution

Understand Heaviside step function behavior

Heaviside step function, also known as the unit step function, is defined as follows: $h(t) = \begin{cases} 0, & \text{if } t < 0 \\ 1, & \text{if } t \ge 0 \end{cases}$ In our problem, we have \(h(t-1)\) and \(h(t-3)\). These functions are just a shift in the original function. So, for \(h(t-1)\), it will be \(0\) when \(t < 1\), and \(1\) when \(t\ge 1\). Similarly, for \(h(t-3)\), it will be \(0\) when \(t < 3\), and \(1\) when \(t\ge 3\).

Graph the function \(f(t)\)

To graph the function, we'll use the given expression: \(f(t) = (2-t)[h(t-1)-h(t-3)]\). We can split the function into different intervals based on the Heaviside functions present: 1. For \(t < 1\), both \(h(t-1)\) and \(h(t-3)\) are \(0\), so \(f(t) = 0\) 2. For \(1 \le t < 3\), \(h(t-1) = 1\) and \(h(t-3) = 0\), so \(f(t) = (2-t)(1) = 2 - t\) 3. For \(t \ge 3\), both \(h(t-1)\) and \(h(t-3)\) are \(1\), so \(f(t) = (2 - t)(1-1) = 0\) So the graph of the function shows a linear segment starting at point \((1, 1)\) and decreasing to point \((3, -1)\), with the function being equal to \(0\) everywhere else.

Use Laplace transform properties

Now we need to find the Laplace transform of the function \(f(t)\). We can rewrite the function as follows and apply the Laplace transform to each term separately: $$F(s) = \mathcal{L}\{f(t)\} = \mathcal{L}\{(2-t)h(t-1)\} - \mathcal{L}\{(2-t)h(t-3)\}$$ For the Laplace transform of the product of a function and a delayed Heaviside function, we have: $$\mathcal{L}\{(f(t-a)h(t-a)\} = e^{-as}F(s)$$ So we have: $$F(s) = e^{-s}\mathcal{L}\{(2-t)\} - e^{-3s}\mathcal{L}\{(2-t)\}$$

Calculate the Laplace transforms of the individual terms

We can find the Laplace transforms of the individual terms using the properties of Laplace transforms: $$\mathcal{L}\{(2-t)\} = \frac{d}{ds}\left(\frac{1}{s}\right)\mathcal{L}\{2\} - \mathcal{L}\{t\} = \frac{d}{ds}\left(\frac{2}{s}\right) - \frac{1}{s^2} = -\frac{2}{s^2} - \frac{1}{s^2} = -\frac{3}{s^2}$$ So, substituting this result back into the expression for \(F(s)\), we get: $$F(s) = e^{-s}\left(-\frac{3}{s^2}\right) - e^{-3s}\left(-\frac{3}{s^2}\right)$$

Simplify the final expression

The final Laplace transform of the function \(f(t)\) is: $$F(s) = -\frac{3}{s^2}(e^{-s} - e^{-3s})$$ So, the Laplace transform of the given function \(f(t) = (2-t)[h(t-1) - h(t-3)]\) is \(F(s) = -\frac{3}{s^2}(e^{-s} - e^{-3s})\) for \(0 \le t < \infty\).

Key concepts

Heaviside Step Function

The Heaviside step function is a mathematical function that represents a sudden change or transition in the value of a function at a specific point. It's commonly used in control theory and signal processing to model switches and on/off events. In the realm of Laplace transforms, the Heaviside function facilitates the breakdown of complex, time-dependent signals into simpler segments for analysis.

It is defined as 0 for negative inputs and 1 for non-negative inputs, providing a clear 'step' at the transition from negative to positive. Remember, when the Heaviside function includes a shift, like in our exercise with h(t-1) and h(t-3), it doesn't flip to 1 until the input exceeds the offset value. This behavior is precisely what allows us to graph piecewise functions accurately, as it defines clear breakpoints for different functional behaviors.

Graphing Piecewise Functions

Graphing piecewise functions, like our exercise's f(t), involves understanding how the function behaves on different intervals. Because these functions have different expressions based on the input value, we need to consider each segment separately.

Key Steps for Graphing Piecewise Functions:

• Identify the intervals and corresponding function expressions.
• Plot each interval's function behavior on the graph.
• Ensure continuity or denote discontinuity (jumps) at the interval boundaries.

These steps lead to a complete visual representation of the function, such as a line segment for 1 ≤ t < 3 in our exercise, which then reverts to zero outside this interval. Understanding this visual aspect is crucial for interpreting the behavior of piecewise functions in time-dependent systems.

Differential Equations

Differential equations form the cornerstone for modeling continuous, dynamic systems in engineering, physics, and other sciences. They express the relationship between a function and its derivatives, indicating how a particular quantity changes over time or another variable.

In practice, solving differential equations allows you to predict system behavior. This requires initial or boundary conditions that help determine a specific solution from a family of possible solutions. Differential equations can be ordinary (ODEs), with a single variable and its derivatives, or partial (PDEs), involving multiple variables and partial derivatives. The Laplace transform is a valuable tool for solving linear ODEs with given initial conditions, as it reduces the problem of solving a differential equation to an algebraic one - often making the solution process more straightforward.

Boundary Value Problems

Boundary value problems (BVPs) are a type of differential equation problem where you seek a solution that satisfies certain conditions at the boundaries of the domain. Unlike initial value problems, boundary conditions are given at more than one point and are essential for determining the unique solution of a differential equation on a bounded domain.

In the context of our Laplace transform exercise, while we don't directly deal with a traditional BVP, understanding boundary conditions is still instrumental. The Heaviside function’s offset shifts act as 'synthetic' boundaries within the piecewise function, stipulating its behavior within certain intervals. When we eventually work our way back from the Laplace transform to the time domain, the inverse process must respect these boundary-like conditions to yield an accurate representation of the original function.