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Chapter 8: Problem 11

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 \(f\). $$ (t)=\left\\{\begin{array}{cc} -t, & 0 \leq t<2 ,\\\ t-4, & 2 \leq t<3 ,\\\ 1, & t \geq 3. \end{array}\right. $$

Short Answer

Expert verified
Based on the given step by step solution, provide a short answer in the space below: The Laplace Transform of the given function is: $$ \mathcal{L}(f(t)) = -\frac{1}{s^2} + e^{-2s}\left(\frac{1}{s^2} - \frac{4}{s}\right) + e^{-3s}\frac{1}{s} $$ To graph the given function, follow these steps: 1. Draw a straight line with a negative slope (-t) for \(0 \leq t < 2\), starting at the origin and ending at the point \((2, -2)\). 2. Draw a straight line with a positive slope (t-4) for \(2 \leq t < 3\), starting at the point \((2, -2)\) and ending at the point \((3, -1)\). 3. Draw a horizontal line (y=1) for \(t \geq 3\), starting at the point \((3, 1)\) and extending towards positive infinity.

Step by step solution

01

Express the function in terms of unit step functions

We will express the given function \(f(t)\) in terms of unit step functions \(u_c(t)\). To do that, we'll write each of the three sections of the given function in the u_c(t) notation. $$ f(t) = -tu_0(t) + (t - 4)u_2(t) + 1u_3(t) $$
02

Apply Theorem 8.4.1 to find the Laplace Transform

Theorem 8.4.1 states the following property for Laplace Transforms of functions defined by unit step functions: $$ \mathcal{L}(f(t)u_c(t)) = e^{-cs}F(s) $$ Now, we will apply this theorem and find the Laplace Transform of the function \(\mathcal{L}(f(t))\). $$ \mathcal{L}(f(t)) = \mathcal{L}(-tu_0(t) + (t-4)u_2(t) + 1u_3(t)) $$ Now, we can calculate the Laplace Transform separately for each term and then combine their results. $$ \mathcal{L}(-tu_0(t)) = (-t)e^{0}F(s) = -\frac{1}{s^2} $$ $$ \mathcal{L}((t-4)u_2(t)) = e^{-2s}\left(\frac{1}{s^2} - \frac{4}{s}\right) $$ $$ \mathcal{L}(1u_3(t)) = e^{-3s}\frac{1}{s} $$ Now, combine the results, and we get the Laplace Transform of the given function: $$ \mathcal{L}(f(t)) = -\frac{1}{s^2} + e^{-2s}\left(\frac{1}{s^2} - \frac{4}{s}\right) + e^{-3s}\frac{1}{s} $$
03

Graph the given function

To graph the given function, we will draw each section of the function separately on the same graph, keeping in mind the domain restrictions: 1. The first section is \(-t\) for \(0 \leq t < 2\). This is a straight line with a negative slope, starting at the origin and ending at the point \((2, -2)\). 2. The second section is \(t - 4\) for \(2 \leq t < 3\). This is a straight line with a positive slope, starting at the point \((2, -2)\) and ending at the point \((3, -1)\). 3. The last section is \(1\) for \(t \geq 3\). This is a horizontal line at \(y = 1\), starting at the point \((3, 1)\) and extending towards positive infinity. Draw these 3 sections on the graph (make sure to use appropriate arrows to indicate the domain restrictions) and this will represent the given function \(f(t)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Step Functions
Unit step functions, often denoted as \( u_c(t) \), are an essential component in mathematical analysis and engineering. They are used to describe functions that switch on or off at particular times. The unit step function \( u_c(t) \) is defined as:
  • \( u_c(t) = 0 \) when \( t < c \)
  • \( u_c(t) = 1 \) when \( t \geq c \)
These functions can be used to express piecewise functions more conveniently, especially in contexts like differential equations or signal processing. Using unit step functions allows us to leverage the properties of Laplace Transforms smoothly, making complex operations replaceable by simpler, formulaic approaches.
In the given exercise, the function \( f(t) \) is expressed using unit step functions as \( f(t) = -tu_0(t) + (t - 4)u_2(t) + 1u_3(t) \). This expression helps to capture the behavior of \( f(t) \) at different intervals using a unified form.
Piecewise Functions
Piecewise functions are mathematical expressions defined by different expressions based on the variable's specific intervals. They offer a flexible way to define functions that have various behaviors in different regions of their input.
The function in our exercise is given as a piecewise function:
  • \(-t\) for \( 0 \leq t < 2 \)
  • \(t - 4\) for \( 2 \leq t < 3 \)
  • \(1\) for \( t \geq 3 \)
Each segment represents a distinct behavior of the function over the specified interval. These intervals illustrate how the function "steps" through various definitions. Breaking down complex functions into piecewise segments allows easier manipulation and analysis.
You can convert these into a form compatible with unit step functions to apply tools like the Laplace Transform. This is crucial in systems where inputs change over time or in response to certain conditions.
Graphing Functions
Graphing functions, especially piecewise ones, is a visual way to understand their structure and behavior. It involves plotting each segment of the piecewise function over its corresponding interval and ensuring the transitions between these segments are noted correctly.
For the given function \( f(t) \), the graph consists of three segments:
  • A line with a negative slope, represented by \(-t\) from \((0, 0)\) to \((2, -2)\).
  • A line with a positive slope, represented by \(t-4\) from \((2, -2)\) to \((3, -1)\).
  • A horizontal line at \(y = 1\), starting at \((3, 1)\) continuing to infinity.
Graphing these segments helps in visualizing and understanding how the function changes across different values of \( t \). This can clarify how to apply transformations or evaluate the function at given points. Making sure to mark changes clearly can provide insights on continuity and limits, which are vital when applying mathematical theories or solving real-world problems.

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Most popular questions from this chapter

Use the Laplace transform to solve the initial value problem. \(y^{\prime \prime}+4 y=4, \quad y(0)=0, \quad y^{\prime}(0)=1\)

$$ y^{\prime \prime}+4 y^{\prime}+4 y=-\delta(t), \quad y(0)=1, \quad y_{-}^{\prime}(0)=5 $$

Show that if \(p(s)=a s^{2}+b s+c\) has a repeated real zero \(r_{1}\) then the solution of $$ a y^{\prime \prime}+b y^{\prime}+c y=f(t), \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1} $$ is $$ y(t)=k_{0}\left(1-r_{1} t\right) e^{r_{1} t}+k_{1} t e^{r_{1} t}+\frac{1}{a} \int_{0}^{t} \tau e^{r_{1} \tau} f(t-\tau) d \tau $$

Find the Laplace transform. (a) \(\int_{0}^{t} \sin a \tau \cos b(t-\tau) d \tau\) (b) \(\int_{0}^{t} e^{\tau} \sin a(t-\tau) d \tau\) (c) \(\int_{0}^{t} \sinh a \tau \cosh a(t-\tau) d \tau\) (d) \(\int_{0}^{t} \tau(t-\tau) \sin \omega \tau \cos \omega(t-\tau) d \tau\) (e) \(e^{t} \int_{0}^{t} \sin \omega \tau \cos \omega(t-\tau) d \tau\) (f) \(e^{t} \int_{0}^{t} \tau^{2}(t-\tau) e^{\tau} d \tau\) (g) \(e^{-t} \int_{0}^{t} e^{-\tau} \tau \cos \omega(t-\tau) d \tau\) (h) \(e^{t} \int_{0}^{t} e^{2 \tau} \sinh (t-\tau) d \tau\) (i) \(\int_{0}^{t} \tau e^{2 \tau} \sin 2(t-\tau) d \tau\) (j) \(\int_{0}^{t}(t-\tau)^{3} e^{\tau} d \tau\) (k) \(\int_{0}^{t} \tau^{6} e^{-(t-\tau)} \sin 3(t-\tau) d \tau\) (I) \(\int_{0}^{t} \tau^{2}(t-\tau)^{3} d \tau\) \((\mathbf{m}) \int_{0}^{t}(t-\tau)^{7} e^{-\tau} \sin 2 \tau d \tau\) (n) \(\int_{0}^{t}(t-\tau)^{4} \sin 2 \tau d \tau\)

Find a formula for the solution of the initial value problem. (a) \(y^{\prime \prime}+3 y^{\prime}+y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0\) (b) \(y^{\prime \prime}+4 y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0\) (c) \(y^{\prime \prime}+2 y^{\prime}+y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0\) (d) \(y^{\prime \prime}+k^{2} y=f(t), \quad y(0)=1, \quad y^{\prime}(0)=-1\) (e) \(y^{\prime \prime}+6 y^{\prime}+9 y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=-2\) (f) \(y^{\prime \prime}-4 y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=3\) (g) \(y^{\prime \prime}-5 y^{\prime}+6 y=f(t), \quad y(0)=1, \quad y^{\prime}(0)=3\) (h) \(y^{\prime \prime}+\omega^{2} y=f(t), \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1}\)
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