91Ó°ÊÓ

Partial fractions are a helpful way to simplify complex rational functions, making it easier to apply the inverse Laplace transform. By expressing a complicated fraction as a sum of simpler fractions, you can address each part individually. For instance, with the given function \( \frac{2s^2 + 1}{s(s+1)^2} \), partial fraction decomposition allows us to write this as \( \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2} \). Here’s how you find coefficients like \( A, B, \) and \( C \):

• Multiply through by the common denominator to eliminate the fractions.
• Expand and combine like terms.
• Set the resulting polynomial equal to the original numerator, \( 2s^2 + 1 \), and solve for the coefficients.

Then, equate coefficients of like terms from both sides of the equation. This step simplifies each term of the original fraction so that later, it is easier to find their inverse Laplace transforms.

The inverse Laplace transform is a method that converts a function from the \( s \)-domain back to the time-domain. When dealing with partial fractions, each term can be inverted separately, which makes the process straightforward thanks to known transforms.

• The inverse Laplace transform of \( \frac{1}{s} \) is \( 1 \).
• The inverse Laplace transform of \( \frac{1}{s+1} \) results in the exponential function \( e^{-t} \).
• For \( \frac{3}{(s+1)^2} \), the inverse transform is \( 3t e^{-t} \), a model for systems with exponential decay and response over time.

Combining these results provides the time-domain function that matches the behavior of the original \( s \)-domain function, demonstrating how these smaller, simpler component transforms can be summed to achieve an overall solution.

In solving for coefficients in partial fraction decomposition, you'll often encounter a system of equations. This arises when equating coefficients of the polynomial equation obtained after clearing the denominators. For our given problem:1. We decomposed the rational expression into terms \( \frac{1}{s} + \frac{1}{s+1} - \frac{3}{(s+1)^2} \).2. Upon multiplying out and collecting terms, the equation \( (A + B)s^2 + (2A + B + C)s + A \) was formed.3. Equating each coefficient with those from the original polynomial (\( 2s^2 + 1 \)) led to the system:

• \( A + B = 2 \)
• \( 2A + B + C = 0 \)
• \( A = 1 \)

Solving these linear equations requires substitution or elimination methods. Substituting known values simplifies finding unknowns, allowing us to quickly determine \( B = 1 \) and \( C = -3 \). Understanding this process is crucial, as it's a standard procedure anytime you decompose a rational function for Laplace transformation.