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Find the inverse transforms of the given functions of \(s\). $$F(s)=\frac{s+2}{\left(s^{2}+9\right)^{2}}$$

Short Answer

Expert verified
The inverse transform is \( f(t) = \frac{1}{54}(\sin(3t) - 3t \cos(3t)) + \frac{t}{3} \sin(3t) \).

Step by step solution

01

Identify the form of the given function

The function provided is \( F(s) = \frac{s+2}{(s^2+9)^2} \). We need to transform this from the Laplace domain back to the time domain, often using inverse Laplace transform tables.
02

Break down the function

Notice that \( F(s) \) is a fraction involving \( s+2 \) in the numerator and \( (s^2+9)^2 \) in the denominator. This suggests using partial fraction decomposition or known inverse Laplace pairs related to sine and cosine functions.
03

Recognize components for inverse transforms

Recognizing \( s^2 + 9 = (s^2 + 3^2) \), this aligns with inverse Laplace transforms of \( \sin(3t) \) and \( \cos(3t) \). The denominator \( (s^2+9)^2 \) suggests the need for a special rule for repeated terms, specifically forms such as \( \frac{1}{(s^2+a^2)^2} \).
04

Use inverse Laplace formula for sine and cosine

The inverse Laplace transform of \( \frac{s}{(s^2+a^2)^2} \) is known to be \( \frac{1}{2a^3}(\sin(at) - at \cos(at)) \). In our case \( a = 3 \), so the term \( \frac{s}{(s^2+9)^2} \) leads to \( \frac{1}{54}(\sin(3t) - 3t \cos(3t)) \).
05

Handle constant shift

The \( +2 \) component requires handling \( \frac{2}{(s^2+9)^2} \), which transforms separately. The inverse transformation of \( \frac{1}{(s^2+a^2)^2} \) for our \( a = 3 \) becomes \( \frac{t}{6} \sin(3t) \). Therefore, \( \frac{2}{(s^2+9)^2} \) becomes \( \frac{t}{3} \sin(3t) \).
06

Combine components

Combine results from Steps 4 and 5: the inverse Laplace transform combines to give \( f(t) = \frac{1}{54}(\sin(3t) - 3t \cos(3t)) + \frac{t}{3} \sin(3t) \). Simplify this step as needed.

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Key Concepts

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

Laplace Transform
The Laplace Transform is a powerful mathematical tool used to move from the time domain to the frequency domain by converting time-based functions into a new domain of complex frequency, denoted as \( s \). This transform simplifies the analysis of systems, particularly for solving linear differential equations.
  • The basic integral form of the Laplace Transform for a function \( f(t) \) is given by:
\[L\{f(t)\} = \int_{0}^{\infty} e^{-st}f(t) \, dt\]This equation effectively transforms the function dependent on time, \( t \), into a function of \( s \).
The beauty of the Laplace Transform lies in how it changes differentiation into algebraic operations.
For example, the derivative of a function in the time domain becomes simple multiplication by \( s \) in the Laplace domain.
Partial Fraction Decomposition
Partial Fraction Decomposition is a method used to break down complex rational expressions into simpler fractions. This is particularly effective when working with the Laplace Transform to simplify inverse calculations.
  • In the context of this exercise, the aim was to separate the rational expression:
\[\frac{s+2}{(s^2+9)^2}\]into parts that can be efficiently managed and transformed back to the time domain.
By breaking this expression into recognizable components, the process of finding its inverse becomes more straightforward, often matching known inverse Laplace Transform pairs.
This method is crucial especially when the denominator involves polynomial terms that are perfect candidates for decomposition, as seen in this exercise.
Sine and Cosine Functions
Sine and cosine functions often appear in processes related to oscillatory or periodic behavior, such as waves. In Laplace Transforms, the expressions:
  • \( \frac{s}{s^2 + a^2} \) and \( \frac{a}{s^2 + a^2} \)
correspond to \( \cos(at) \) and \( \sin(at) \) in the time domain.
Recognizing these patterns allows for the straightforward calculation of inverse transforms in exercises like this.
For repetitive terms like \((s^2+a^2)^n\), special rules apply:
  • For example, the inverse of \( \frac{s}{(s^2+a^2)^2} \) results in an expression involving both sine and cosine functions, reflecting both amplitude and phase changes in time.
Time Domain
The time domain represents functions and signals as they occur over time, depicted with a variable \( t \). It is the natural domain for most real-world phenomena.
  • When using the Laplace Transform, functions are initially moved out of the time domain to simplify specific calculations.
However, the ultimate goal is to return to this domain via the inverse Laplace Transform, as seen in this exercise.
Inverse transformations convert expressions from the frequency domain back:
  • Allow these time-domain representations to reflect dynamics like oscillations, decays, and growths linked with physical systems or signals.
Understanding this transition is key in applying mathematical models effectively back to real-world interpretations, as shown when computing \( f(t) \), the time-based outcome.

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Most popular questions from this chapter

The displacement \(y\) (in \(\mathrm{cm}\) ) of an object at the end of a spring is described by the equation \(d^{2} y / d t^{2}+2 d y / d t+4 y=0,\) where \(t\) is the time (in \(\mathrm{s}\) ). Find \(y=f(t)\) if \(f(0)=0\) and \(f(1)=2.00 \mathrm{cm}\).

Solve the given problems by solving the appropriate differential equation. In a town of \(N\) persons, during a flu epidemic, it was determined that the rate \(d S / d t\) at which persons were being infected was proportional to the product of the number \(S\) of infected persons and the number \(N-S\) of healthy persons. Find \(S\) as a function of \(t\)

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Find the inverse transforms of the given functions of \(s\). $$F(s)=\frac{3 s^{4}+3 s^{3}+6 s^{2}+s+1}{s^{5}+s^{3}}$$ (Explain your method of solution.)

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