/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Use the result of Problem 28 to ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 30

Use the result of Problem 28 to find the Laplace transform of the given function. $$ f(t)=\left\\{\begin{aligned} 1, & 0 \leq t<1 \\\\-1, & 1 \leq t<2 \end{aligned}\right. $$ $$ f(t+2)=f(t) $$

Short Answer

Expert verified
Question: Find the Laplace transform of the given piecewise-defined function: $$ f(t) = \begin{cases} 1, & 0 \leq t < 1 \\ -1, & 1 \leq t < 2 \end{cases} $$ where the function is periodic with period 2. Answer: The Laplace transform of the given function is $$ F(s) = \frac{1}{s} \left(1 - 2e^{-s} + e^{-2s} \right). $$

Step by step solution

01

Write down the definition of the Laplace transform

Recall that the Laplace transform of a function f(t) is defined as $$ F(s) = \mathcal{L}\{f(t)\} = \int_0^{\infty}e^{-st}f(t)dt. $$ We will apply this definition to our given piecewise function.
02

Separate the function into its two pieces

Since the given function is defined differently for \(0 \leq t < 1\) and \(1 \leq t < 2\), we will separate the function into these two parts and find the Laplace transform of each part individually: $$ F(s) = \int_0^{1} e^{-st} \cdot 1 dt + \int_1^{2} e^{-st} \cdot (-1) dt. $$
03

Calculate the integrals

Now we evaluate the two integrals separately. First, the integral of \(e^{-st}\) with respect to \(t\): $$ \int e^{-st} dt = -\frac{1}{s}e^{-st} + C. $$ Plugging this back into the previous equation, we get $$ F(s) = -\frac{1}{s}e^{-s(1)} - \left(-\frac{1}{s}e^{-s(0)}\right) - \left[-\frac{1}{s}e^{-s(2)} - \left(-\frac{1}{s}e^{-s(1)}\right)\right]. $$
04

Simplify the expression

Now, we will simplify the expression for \(F(s)\): $$ F(s) = -\frac{1}{s}e^{-s} - \left(-\frac{1}{s}\right) - \left[-\frac{1}{s}e^{-2s} - \left(-\frac{1}{s}e^{-s}\right)\right]. $$ Simplifying further, $$ F(s) = \frac{1}{s} \left(1 - e^{-s} + e^{-2s} - e^{-s} \right). $$
05

Apply the periodic condition

Now, we will make use of the information that the function is periodic with a period of 2, which means that: $$ f(t+2) = f(t). $$ This allows us to represent the Laplace transform more compactly: $$ F(s) = \frac{1}{s} \left(1 - 2e^{-s} + e^{-2s} \right). $$ Finally, we have found the Laplace transform of the given function: $$ F(s) = \frac{1}{s} \left(1 - 2e^{-s} + e^{-2s} \right). $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Piecewise Functions
When studying mathematics, one often encounters piecewise functions, which are defined by different expressions over various intervals. In the context of the Laplace Transform, piecewise functions require particular attention because each segment must be transformed separately.

The example function provides a perfect illustration of how a piecewise function can be handled. In this case, the function switches from 1 to -1 at the point t=1. To find the Laplace Transform, we consider each piece in the interval it's defined. We integrate the corresponding function segment within its specific interval and then aggregate the results.

Understanding piecewise functions is crucial for analyzing complex signals in engineering and physics, as real-world phenomena often behave differently under various conditions. By mastering the approach to these functions, one gains the versatility to tackle a wide range of problems where standard functions fall short.
Periodic Functions
Another significant concept in mathematics and engineering is that of periodic functions. A periodic function is one that repeats its values at regular intervals, known as the period. The function from the exercise demonstrates periodicity with a period of 2. This repetition simplifies many operations, including the Laplace Transform.

For periodic functions, once the Laplace Transform for one period is determined, the entire transform can be constructed due to the function's repetitive nature. This characteristic is particularly useful when analyzing signals in electronics or understanding harmonic motions in mechanics. By recognizing periodicity, one can streamline many complex calculations to a simpler analysis of a single period, saving time and reducing the potential for errors.
Integral Transforms
Lastly, the core of the exercise lies within the realm of integral transforms. Integral transforms, such as the Laplace Transform, are powerful mathematical tools used to convert functions from the time domain to the complex frequency domain. This alter-ego of the function opens doors to easier manipulation and analysis, particularly for differential equations and system responses in engineering.

In the case of the Laplace Transform, we apply an integral operation to the function over the entire range of time, using the kernel e-st. It is vital for students to understand that integral transforms can greatly simplify the process of solving complex equations, making them a cornerstone in the fields of control theory, signal processing, and more.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the Laplace transform of the given function. $$ f(t)=(t-3) u_{2}(t)-(t-2) u_{3}(t) $$

Find the Laplace transform of the given function. $$ f(t)=t-u_{1}(t)(t-1), \quad t \geq 0 $$

Consider the initial value problem $$ y^{\prime \prime}+\gamma y^{\prime}+y=\delta(t-1), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ where \(\gamma\) is the damping coefficient (or resistance). (a) Let \(\gamma=\frac{1}{2} .\) Find the value of \(k\) for which the response has a peak value of \(2 ;\) call this value \(k_{1} .\) (b) Repeat part (a) for \(\gamma=\frac{1}{4}\). (c) Determine how \(k_{1}\) varies as \(\gamma\) decreases. What is the value of \(k_{1}\) when \(\gamma=0 ?\)

The Tautochrone. A problem of interest in the history of mathematics is that of finding the tautochrone-the curve down which a particle will slide freely under gravity alone, reaching the bottom in the same time regardless of its starting point on the curve. This problem arose in the construction of a clock pendulum whose period is independent of the amplitude of its motion. The tautochrone was found by Christian Huygens \((1629-\) \(1695)\) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using analytical arguments. Bernoulli's solution (in \(1690)\) was one of the first occasions on which a differential equation was explicitly solved. The Tautochrone. A problem of interest in the history of mathematics is that of finding the tautochrone-the curve down which a particle will slide freely under gravity alone, reaching the bottom in the same time regardless of its starting point on the curve. This problem arose in the construction of a clock pendulum whose period is independent of the amplitude of its motion. The tautochrone was found by Christian Huygens \((1629-\) \(1695)\) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using analytical arguments. Bernoulli's solution (in \(1690)\) was one of the first occasions on which a differential equation was explicitly solved. The geometrical configuration is shown in Figure \(6.6 .2 .\) The starting point \(P(a, b)\) is joined to the terminal point \((0,0)\) by the arc \(C .\) Arc length \(s\) is measured from the origin, and \(f(y)\) denotes the rate of change of \(s\) with respect to \(y:\) $$ f(y)=\frac{d s}{d y}=\left[1+\left(\frac{d x}{d y}\right)^{2}\right]^{1 / 2} $$ Then it follows from the principle of conservation of energy that the time \(T(b)\) required for a particle to slide from \(P\) to the origin is $$ T(b)=\frac{1}{\sqrt{2 g}} \int_{0}^{b} \frac{f(y)}{\sqrt{b-y}} d y $$

Determine whether the given integral converges or diverges. $$ \int_{0}^{\infty} t e^{-t} d t $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.