Question

Find the inverse Laplace transform of \(\frac{1}{(s+1)^{2}\left(s^{2}+4\right)}\).

Short answer

The inverse Laplace transform is \( \frac{1}{4}te^{-t} + \frac{1}{8}\sin(2t) \).

Step-by-step solution

Identify Formulas

First, recognize that the inverse Laplace transform is essentially about finding a time domain function given its Laplace transform. We need to decompose the given expression \( \frac{1}{(s+1)^{2}(s^{2}+4)} \) into simpler terms that correspond to known inverse Laplace transforms.

Perform Partial Fraction Decomposition

Decompose \( \frac{1}{(s+1)^{2}(s^{2}+4)} \) into partial fractions. Let's assume:\[ \frac{1}{(s+1)^{2}(s^{2}+4)} = \frac{A}{s+1} + \frac{B}{(s+1)^{2}} + \frac{Cs + D}{s^{2}+4} \]. Multiply through by the common denominator and solve for A, B, C, and D.

Find Coefficients A, B, C, and D

Multiply the decomposed equation by \((s+1)^{2}(s^{2}+4)\) and set coefficients equal: \[ 1 = A(s+1)(s^2+4) + B(s^2+4) + (Cs + D)(s+1)^2 \]. By substituting convenient values for \(s\) such as \(-1\) and complex numbers (or using a system of equations), solve for A, B, C, and D.

Evaluate Constants

After performing the algebra, the solutions to the coefficients are: \( A = 0, B = \frac{1}{4}, C = 0, D = \frac{1}{4} \). This gives us: \[ \frac{1}{(s+1)^{2}(s^{2}+4)} = \frac{1/4}{(s+1)^{2}} + \frac{1/4}{s^2+4} \].

Apply Inverse Laplace Transform

Use known inverse Laplace transforms: \( \mathcal{L}^{-1}\left\{ \frac{1/4}{(s+1)^{2}} \right\} = \frac{1}{4}te^{-t} \) and \( \mathcal{L}^{-1}\left\{ \frac{1/4}{s^2+4} \right\} = \frac{1}{8}\sin(2t) \).

Write Final Solution

Combine results from the previous step to construct the original time domain function: \[ f(t) = \frac{1}{4}te^{-t} + \frac{1}{8}\sin(2t) \].

Key concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is an essential method used in calculus and algebra that allows us to break down complex rational expressions into simpler parts. These simpler expressions are easier to work with, especially for finding inverse Laplace transforms, as they often correspond to known functions.

To perform Partial Fraction Decomposition, follow these steps:

• Identify the given polynomial expression and the factors in the denominator.
• Express the given rational expression as a sum of fractions. Each fraction's denominator corresponds to one of the factors, and the numerators contain unknown coefficients.
• Multiply through by the original denominator to eliminate the fractions, resulting in an equation in terms of 's' (or the variable used).
• Substitute convenient values for the variable to solve for each unknown coefficient. Alternatively, you can equate coefficients for like powers of 's'.

Bringing this into action means expressing simple terms like \( \frac{A}{s+1} + \frac{B}{(s+1)^{2}} + \frac{Cs + D}{s^{2}+4} \), allowing for straightforward computation of transforms or integrals.

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to convert functions from the time domain to the frequency domain. It simplifies the work of tackling complex differential equations by transforming them into algebraic equations. This transform makes handling initial conditions and systems analysis in engineering and physics more efficient.

Certain basic functions have well-known Laplace transforms, which means if we decompose a function into these basic components, finding its inverse becomes much easier. The transformation is denoted by \( \mathcal{L} \) and is defined by the formula:\[\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st}f(t)\, dt\]For inverse transforms, we use existing tables of Laplace transforms or known transforms to directly invert the function back to the time domain. For instance, knowing \( \mathcal{L}^{-1}\left\{ \frac{1/4}{(s+1)^{2}} \right\} \) leads directly to \( \frac{1}{4}te^{-t} \), as seen in our solution example.

Time Domain Function

A Time Domain Function represents how a particular function behaves over time. In physics, engineering, and other sciences, it is crucial for understanding the real-world behavior of systems. When we find the inverse Laplace transform, we are effectively converting a system representation from the frequency domain back to this real-time representation.

The importance of the time domain lies in its straightforward application to modeling and simulating dynamic systems. Consider the inverse transform obtained in our example:\[f(t) = \frac{1}{4}te^{-t} + \frac{1}{8}\sin(2t)\]

• \( \frac{1}{4}te^{-t} \) indicates a decaying factor, representing transient response behavior over time.
• \( \frac{1}{8}\sin(2t) \) describes an oscillatory component, showing how the system may behave in the steady state or due to particular inputs.

This helps in predicting and controlling how actual systems respond to various inputs and initial conditions, essential in fields like electrical engineering and control systems.