Question

Find the inverse Laplace transform of the given function. $$ \frac{2 s-3}{s^{2}-4} $$

Short answer

\) Answer: The inverse Laplace transform of the given function is \(\frac{7}{4} e^{-2t} + \frac{1}{4} e^{2t}\).

Step-by-step solution

Apply Partial Fraction Decomposition

To find the inverse Laplace transform, we first need to decompose the given function into simpler fractions. Given function: $$ \frac{2s-3}{s^2-4} $$ Since the denominator is a quadratic equation, we can rewrite it as: $$ \frac{2s-3}{(s+2)(s-2)} = \frac{A}{s+2} + \frac{B}{s-2} $$

Solve for A and B

To find the values of A and B, we can multiply both sides of the equation by \((s+2)(s-2)\) and use the properties of equal fractions. $$ (2s-3)= A(s-2) + B(s+2) $$ Now, we will solve for A and B: 1. Solve for A: Let \(s=-2\): $$ (-2 * 2) - 3 = -7 = A(-4) \Rightarrow A = \frac{7}{4} $$ 2. Solve for B: Let \(s=2\): $$ (2 * 2) - 3 = 1 = B(4) \Rightarrow B = \frac{1}{4} $$ Hence, the partial fraction decomposition is: $$ \frac{7}{4 (s+2)} + \frac{1}{4 (s-2)} $$

Find the Inverse Laplace Transform

Now, we can find the inverse Laplace transform for each term and sum them to find the final solution. 1. Inverse Laplace transform of the first term: $$ \mathcal{L}^{-1}\left\{\frac{7}{4(s+2)}\right\} = \frac{7}{4} e^{-2t} $$ 2. Inverse Laplace transform of the second term: $$ \mathcal{L}^{-1}\left\{\frac{1}{4(s-2)}\right\} = \frac{1}{4} e^{2t} $$ Summing both inverse Laplace transforms, we get the solution: $$ \frac{7}{4} e^{-2t} + \frac{1}{4} e^{2t} $$ So, the inverse Laplace transform of the given function is: $$ \mathcal{L}^{-1}\left\{\frac{2s-3}{s^2-4}\right\} = \frac{7}{4} e^{-2t} + \frac{1}{4} e^{2t} $$

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used in calculus to break down complex rational expressions into simpler ones that are easier to integrate or transform. It’s particularly useful when dealing with Laplace transforms, where breaking down a fraction into parts makes finding the inverse transform more manageable.

At its core, partial fraction decomposition involves expressing a fraction whose numerator and denominator are polynomials as a sum of simpler fractions, each with a simpler denominator. This process usually follows a few steps:

• Factoring the denominator into simpler polynomials, if possible.
• Writing out a sum of fractions corresponding to these factors, with unknown constants in the numerators.
• Figuring out these unknown constants by multiplying both sides by the factored denominator and comparing coefficients or by substituting convenient values for the variable.

Once the partial fractions are determined, each one can be inverted separately if working with Laplace transforms, ultimately simplifying the process of finding the inverse Laplace transform.

Laplace Transform Properties

The Laplace transform is a powerful tool used to solve differential equations by transforming them into algebraic equations, which are usually easier to solve. It transforms a function of time, often a physical signal or system response, into a function of complex frequency.

Understanding the properties of the Laplace transform is essential for effectively using it to analyze systems and signals. Some key properties include:

• Linearity: The transform of a sum of functions is the sum of their Laplace transforms.
• First and Second Shifting Theorems: These allow for the transformation of functions multiplied by an exponential term, which is common in solving inverse Laplace transforms.
• Convolution Theorem: This facilitates the calculation of the inverse Laplace transform of the product of two Laplace transforms, relating to the convolution of the original time functions.
• Initial and Final Value Theorems: These provide a way to determine the initial and final values of a function given its Laplace transform without having to calculate the entire inverse transform.

These properties are particularly useful when practicing partial fraction decomposition. In the context of our exercise, for instance, the shifting theorems apply directly to the exponential functions in the final solution.

Exponential Functions

Exponential functions play a crucial role in many areas of mathematics, including the study of Laplace transforms. In the context of Laplace transforms, they emerge naturally as solutions to differential equations and appear as part of inverse Laplace transforms.

An exponential function can generally be written as \( e^{at} \), where \( a \) is a constant. The sign of \( a \) affects the function's behavior:

• If \( a > 0 \), the function grows exponentially as \( t \) increases.
• If \( a = 0 \), the function is constant, equivalently \( e^{0} = 1 \).
• If \( a < 0 \), the function decays exponentially as \( t \) increases.

In the solution of our exercise, we find two exponential terms - \( \frac{7}{4} e^{-2t} \) and \( \frac{1}{4} e^{2t} \). The first is a decaying exponential because the exponent -2t is negative, indicative of a response that decreases over time. The second is a growing exponential since its exponent 2t is positive, pointing to a quantity that increases over time.

Understanding exponential functions is vital for mastering inverse Laplace transforms, as these functions are often present in the transformed solutions of linear time-invariant systems.