Determining the Inverse Laplace Transform of a Rational Function

In this article, we will explore the process of determining the inverse Laplace transform of a rational function. Specifically, we will focus on the example of finding the inverse Laplace transform of the function:

$$F(s) = \frac{3s + 5}{s^2 + 4s + 13}$$

The Laplace transform is a powerful tool in mathematics, particularly in the field of differential equations, and its inverse plays a crucial role in solving these equations. By finding the inverse Laplace transform, we can obtain the original function in the time domain, which is often more intuitive and useful for practical applications.

Step 1: Partial Fraction Expansion

The first step in determining the inverse Laplace transform of a rational function is to perform a partial fraction expansion. This process involves breaking down the rational function into a sum of simpler fractions, each of which can be individually transformed.

To perform the partial fraction expansion, we need to factor the denominator of the rational function. In this case, the denominator is a quadratic expression, $s^2 + 4s + 13$. We can factor this expression using the quadratic formula:

$$s^2 + 4s + 13 = (s + 2)^2 + 9$$

Now, we can express the original rational function as a sum of partial fractions:

$$F(s) = \frac{A}{s + 2} + \frac{B}{(s + 2)^2}$$

where $A$ and $B$ are constants to be determined.

Step 2: Determining the Constants A and B

To find the values of $A$ and $B$, we can use the method of undetermined coefficients. This involves multiplying both sides of the equation by the common denominator, $(s + 2)^2$, and then equating the coefficients of like powers of $s$.

Multiplying both sides by $(s + 2)^2$, we get:

$$3s + 5 = A(s + 2)^2 + B(s + 2)$$

Expanding the right-hand side and equating the coefficients of $s$ and the constant term, we obtain a system of two equations:

$$3 = 2A + B$$ $$5 = 2A + 2B$$

Solving this system of equations, we find that:

$$A = 1 \quad \text{and} \quad B = 1$$

Step 3: Inverse Laplace Transform

Now that we have the partial fraction expansion of the original rational function, we can apply the inverse Laplace transform to each term individually. The inverse Laplace transform of the partial fractions is given by:

$$\mathcal{L}^{-1}\left(\frac{1}{s + a}\right) = e^{-at}$$ $$\mathcal{L}^{-1}\left(\frac{1}{(s + a)^2}\right) = te^{-at}$$

Substituting the values of $A$ and $B$, we get:

$$\mathcal{L}^{-1}\left(\frac{3s + 5}{s^2 + 4s + 13}\right) = \mathcal{L}^{-1}\left(\frac{1}{s + 2} + \frac{1}{(s + 2)^2}\right) = e^{-2t} + te^{-2t}$$

Therefore, the inverse Laplace transform of the given rational function is:

$$f(t) = e^{-2t} + te^{-2t}$$

This result represents the solution to the corresponding differential equation in the time domain, which can be useful in various applications, such as circuit analysis, control systems, and signal processing.