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Magazine Chapter 12: Problem 7
Show that \(\mathcal{L}^{-1}\left(\frac{1}{s^{2}(s+1)}\right)=t-1+e^{t}\).
Short Answer
Expert verified
The inverse Laplace transform is \( t - 1 + e^{t} \).
Step by step solution
01
Recognize the Inverse Laplace Transform
The problem asks to find the inverse Laplace transform of \( \frac{1}{s^{2}(s+1)} \). The solution involves rewriting the expression in a form that allows us to use known inverse Laplace transforms.
02
Decompose the Fraction
We use partial fraction decomposition to rewrite \( \frac{1}{s^{2}(s+1)} \). Assume \( \frac{1}{s^{2}(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} \). Multiply throughout by \( s^2(s+1) \) to obtain \( 1 = A s(s+1) + B(s+1) + Cs^2 \).
03
Solve for Coefficients
Compare coefficients from the equation in Step 2:1. Set \( s=0 \): \( 1 = B \).2. Set \( s=-1 \): \( 0 = A(-1) \) which simplifies to \( A = -1 \).3. Substitute \( A = -1 \) and \( B = 1 \) into the equation for \( s \) terms to solve for \( C \):\( 1 = -s(s+1) + 1(s+1) + Cs^2 \).Equating coefficients of \( s^2 \) gives \( C = 1 \).
04
Write Decomposed Expression
The partial fractions give:\[ \frac{1}{s^{2}(s+1)} = \frac{-1}{s} + \frac{1}{s^2} + \frac{1}{s+1} \].
05
Find Inverse Laplace for Each Term
Apply known inverse Laplace transforms:1. \( \mathcal{L}^{-1}\left(\frac{-1}{s}\right) = -1 \).2. \( \mathcal{L}^{-1}\left(\frac{1}{s^2}\right) = t \).3. \( \mathcal{L}^{-1}\left(\frac{1}{s+1}\right) = e^{t} \).
06
Combine results
The final inverse Laplace transform is:\[ t - 1 + e^{t} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler parts that are easier to handle, especially when trying to find inverse Laplace transforms. It involves expressing a fraction as a sum of simpler fractions. This technique is particularly useful when dealing with polynomial denominators that cannot be simplified through straightforward factorization. Here's how it works:
- Identify the poles of the denominator to determine the form of the decomposition.
- Set up the decomposition by assigning coefficients to each partial fraction term. For example, for the expression \( \frac{1}{s^2(s+1)} \), we write it as \( \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} \).
- Clear the common denominator by multiplying through by the denominator, and then solving for the coefficients (A, B, and C) by substituting convenient values of \( s \) or equating coefficients.
Laplace Transform Properties
Laplace transform properties are essential tools in solving differential equations and systems, offering a way to shift an analysis from the time domain to the complex frequency domain. Understanding these properties helps to efficiently find inverse Laplace transforms. Some important properties include:
- Linearity Property: This property states that the Laplace transform of a sum of functions is the sum of their Laplace transforms, making it easy to deal with linear combinations of terms. Mathematically, \( \mathcal{L}(af(t) + bg(t)) = aF(s) + bG(s) \) where \( a \) and \( b \) are constants.
- First Shifting Theorem (s-shifting): The Laplace transform of \( e^{at}f(t) \) is \( F(s-a) \). This is particularly useful when dealing with terms like \( \frac{1}{s+1} \).
- Derivative Property: The transform of a derivative is \( s \mathcal{L}(f(t)) - f(0) \). This property helps simplify the process of transforming higher-order derivatives.
Complex Analysis
Complex analysis is a branch of mathematics that studies functions of complex numbers and plays a critical role in understanding transformations like the Laplace transform. Dealing with complex functions involves ensuring that functions are analytic across a region, which means they are differentiable everywhere in that region.
- Complex Poles: In Laplace transforms, we often encounter complex poles. These poles help determine the form of the inverse function. For instance, a pole at \( s = -1 \) leads to an exponential term \( e^t \) in the inverse transform.
- Residue Theorem: Though not typically used in basic inverse transforms, this theorem helps evaluate complex integrals, especially when dealing with circuit systems and signals.
- Analytic Continuation: Ensures that the complex function behaves well in the defined domain, which is crucial for applying the Laplace transform to real-world problems.
