Chapter 9: Problem 11
Use Table \(9.1\) to find \(\mathcal{I}^{-1}\\{F(s)\\}\) for each of the functions \(F\) defined \(F(s)=\frac{s-2}{s^{2}+5 s+6}\)
Short Answer
Expert verified
The inverse Laplace transform of the given function \(F(s) = \frac{s-2}{s^2 + 5s + 6}\) is \(\mathcal{I}^{-1}\\{F(s)\\} = 4e^{-2t} - e^{-3t}\).
Step by step solution
01
Perform Partial Fraction Decomposition
To find the inverse Laplace transform, we will first decompose the given function \(F(s)\) into simpler fractions. Factoring the denominator, we have:
\(s^2 + 5s + 6 = (s+2)(s+3)\)
Now, we decompose the function using partial fraction decomposition:
\(F(s) = \frac{A}{s+2} + \frac{B}{s+3}\), where A and B are constants to be determined.
02
Combine the Fractions and Solve for A and B
To combine the fractions and solve for A and B, we set:
\(\frac{s-2}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3}\)
Clearing the denominators, we have:
\(s - 2 = A(s + 3) + B(s + 2)\)
Now, solve for A and B by substituting the roots of the denominators.
For A:
\(s = -2\)
\(-2 - 2 = A(-2 + 3) + B(-2 + 2)\)
\(-4 = -A => A = 4\)
For B:
\(s = -3\)
\(-3 - 2 = A(-3 + 2) + B(-3 + 3)\)
\(-5 = -4 + B => B = -1\)
Thus, our decomposed function becomes:
\(F(s) = \frac{4}{s + 2} - \frac{1}{s + 3}\)
03
Taking Inverse Laplace Transform
Now, we use Table 9.1 and take the inverse Laplace transform of our decomposed function. We know that:
\( \mathcal{I}^{-1}\\{\frac{1}{s+a}\\} = e^{-at}\)
Applying this to our decomposed function, we have:
\(\mathcal{I}^{-1}\\{F(s)\\} = 4\mathcal{I}^{-1}\\{\frac{1}{s+2}\\} - \mathcal{I}^{-1}\\{\frac{1}{s+3}\\}\)
Now, apply the inverse Laplace transform to each term:
\(4e^{-2t} - e^{-3t}\)
04
Final Answer
We have found the inverse Laplace transform of the given function \(F(s)\):
\(\mathcal{I}^{-1}\\{F(s)\\} = 4e^{-2t} - e^{-3t}\)
This is the solution to the given problem.
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 Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions, which can be more easily integrated or, in the case of Laplace transforms, inverted. This technique is particularly useful when dealing with polynomials as denominators.
Imagine you're trying to separate a multi-layered cake into single flavors to savor each one distinctly. Similarly, partial fraction decomposition separates a complex fraction into a sum of simpler fractions with linear factors or irreducible quadratic factors in the denominator.
For the given function \(F(s) = \frac{s-2}{s^{2}+5 s+6}\), the first step is to factor the denominator, revealing its simpler components, \(s+2\) and \(s+3\). Then, to find constants A and B, we assume that \(F(s)\) can be expressed as a sum of fractions: \(\frac{A}{s+2} + \frac{B}{s+3}\). Subsequently, we solve for A and B by equating coefficients or by using the values that make each denominator zero, known as the roots of the denominators. This way, we transform the complex expression into a simpler one that is more manageable for inverse Laplace transformations.
Imagine you're trying to separate a multi-layered cake into single flavors to savor each one distinctly. Similarly, partial fraction decomposition separates a complex fraction into a sum of simpler fractions with linear factors or irreducible quadratic factors in the denominator.
For the given function \(F(s) = \frac{s-2}{s^{2}+5 s+6}\), the first step is to factor the denominator, revealing its simpler components, \(s+2\) and \(s+3\). Then, to find constants A and B, we assume that \(F(s)\) can be expressed as a sum of fractions: \(\frac{A}{s+2} + \frac{B}{s+3}\). Subsequently, we solve for A and B by equating coefficients or by using the values that make each denominator zero, known as the roots of the denominators. This way, we transform the complex expression into a simpler one that is more manageable for inverse Laplace transformations.
Laplace Transform
The Laplace transform is a powerful tool in mathematics, particularly for engineers and physicists who often have to solve differential equations. Think of it like a mathematical magnifying glass that can transform complex, time-domain functions into a clearer, frequency-domain representation.
Essentially, the Laplace transform converts differential equations into algebraic equations, which are generally easier to manipulate and solve. The transform is defined by the integral \(\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)dt\), where \( f(t) \) is a given time-domain function, \( s \) is a complex frequency parameter, and \( e^{-st} \) is an exponential decay factor.
In the context of our exercise, the inverse Laplace transform is what we're interested in. It allows us to revert from the frequency domain back to the time domain. In simpler terms, if the Laplace transform helps you to simplify and solve the equation, the inverse Laplace transform helps you to translate the solution back into a function of time, \( f(t) \), by using a table of known transforms.
Essentially, the Laplace transform converts differential equations into algebraic equations, which are generally easier to manipulate and solve. The transform is defined by the integral \(\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)dt\), where \( f(t) \) is a given time-domain function, \( s \) is a complex frequency parameter, and \( e^{-st} \) is an exponential decay factor.
In the context of our exercise, the inverse Laplace transform is what we're interested in. It allows us to revert from the frequency domain back to the time domain. In simpler terms, if the Laplace transform helps you to simplify and solve the equation, the inverse Laplace transform helps you to translate the solution back into a function of time, \( f(t) \), by using a table of known transforms.
Exponential Function
The exponential function, often noted as \(e^x\) or \(\exp(x)\), is ubiquitous in mathematics due to its unique properties and its appearance in naturally occurring processes like growth and decay. Picture it as the growth pattern of bacteria; just as they multiply rapidly, the exponential function increases quickly as well.
An exponential function has the form \( f(t) = e^{at} \), where \( a \) is a constant. When you see \( e \), it's referring to Euler's number, approximately equal to 2.71828. What makes the exponential function so special is its rate of growth: it is proportional to its current value, which leads to the exponential increase or decrease in applications like radioactive decay or compound interest calculations.
In our inverse Laplace transformation, the exponential's role is captured when we apply the transform table to find the time-domain equivalent of expressions like \(\frac{1}{s+a}\). The table tells us that this corresponds to \(e^{-at}\) in the time domain, which directly factors into our final solution for the inverse Laplace transform.
An exponential function has the form \( f(t) = e^{at} \), where \( a \) is a constant. When you see \( e \), it's referring to Euler's number, approximately equal to 2.71828. What makes the exponential function so special is its rate of growth: it is proportional to its current value, which leads to the exponential increase or decrease in applications like radioactive decay or compound interest calculations.
In our inverse Laplace transformation, the exponential's role is captured when we apply the transform table to find the time-domain equivalent of expressions like \(\frac{1}{s+a}\). The table tells us that this corresponds to \(e^{-at}\) in the time domain, which directly factors into our final solution for the inverse Laplace transform.
