Question

Find the inverse Laplace transform. $$F(s)=\frac{s^{2}+6 s+8}{s^{4}+8 s^{2}+16}$$

Short answer

Short Answer: The inverse Laplace transform of the given function is \(f(t) = 3\sin(2t) + \frac{7}{4}t\sin(2t) - 1\).

Step-by-step solution

Recognize the form of the given function

The given function is a proper rational function and we can find its inverse Laplace by finding the partial fraction decomposition of the function.

Perform partial fraction decomposition

We will decompose the given function into partial fractions using the method of undetermined coefficients. First, we will factor the denominator and then construct the partial fractions. The denominator can be factored as: $$s^{4}+8 s^{2}+16 = (s^{2}+4)^{2}$$ Now, we have the following partial fractions form: $$\frac{s^{2}+6 s+8}{(s^{2}+4)^{2}}=\frac{A_{1}s+B_{1}}{s^{2}+4}+\frac{A_{2}s+B_{2}}{(s^{2}+4)^{2}}$$ Multiplying both sides by \((s^{2}+4)^{2}\): $$s^{2}+6s+8= (A_{1}s+B_{1})(s^{2}+4)+ (A_{2}s+B_{2})(s^{2}+4)^{2}-(s^{2}+4)(A_{2}s+B_{2})$$ Now, we have to find the coefficients \(A_1, B_1, A_2\) and \(B_2\). To do this, we equate coefficients of the corresponding powers of s: $$A_{1}+A_{2}=1$$ $$B_{1}=6$$ $$A_{1}+2 A_{2}=0$$ $$B_{2}=8$$ Solving this system of equations, we get: \(A_1 = -2, B_1 = 6, A_2 = 3, B_2 = 8\). Therefore, the partial fraction decomposition is: $$\frac{s^{2}+6 s+8}{(s^{2}+4)^{2}}=\frac{-2s+6}{s^{2}+4}+\frac{3s+8}{(s^{2}+4)^{2}}$$

Apply inverse Laplace transform

To find the inverse Laplace transform, we apply the inverse Laplace transform on each term. We know that: $$\mathcal{L}^{-1}\{\frac{1}{s^{2}+4}\}(t)=\frac{1}{2} \sin(2t)$$ and $$\mathcal{L}^{-1}\{\frac{s}{s^{2}+4}\}(t)=\frac{1}{2} \cos(2t)$$ Using the properties of linearity of the Laplace transform, we can find the inverse Laplace transform of the given function: $$f(t)=\mathcal{L}^{-1} \{F(s)\}= -2\mathcal{L}^{-1}\{\frac{s}{s^{2}+4}\} + 6\mathcal{L}^{-1}\{\frac{1}{s^{2}+4}\} + 3\mathcal{L}^{-1}\{\frac{s}{(s^{2}+4)^{2}}\} + 8\mathcal{L}^{-1}\{\frac{1}{(s^{2}+4)^{2}}\}$$ Now we substitute the known inverse Laplace transforms. However, we have to find the inverse Laplace transforms for the last two terms: $$\mathcal{L}^{-1}\{\frac{s}{(s^{2}+4)^{2}}\}(t)=\frac{1}{4} t\sin(2t)$$ $$\mathcal{L}^{-1}\{\frac{1}{(s^{2}+4)^{2}}\}(t)=\frac{1}{8}(2t\sin(2t)+\cos(2t)-1)$$ Now we substitute all the inverse Laplace transforms in our expression: $$f(t)= -2\left(\frac{1}{2}\cos(2t)\right) + 6\left(\frac{1}{2}\sin(2t)\right) + 3\left(\frac{1}{4}t\sin(2t)\right) + 8\left(\frac{1}{8}(2t\sin(2t)+\cos(2t)-1)\right)$$ Finally, simplify the expression: $$f(t)= -\cos(2t) + 3\sin(2t) + \frac{3}{4}t\sin(2t) + t\sin(2t) + \cos(2t) - 1$$

Write down the solution

The inverse Laplace transform of the given function is: $$f(t)= 3\sin(2t) + \frac{7}{4}t\sin(2t) - 1$$

Key concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to break down complex rational functions into simpler fractions that are easier to work with. This method is particularly useful for finding inverse Laplace transforms as shown in the original exercise.

The idea is to express a complicated fraction like \( \frac{s^{2}+6s+8}{s^{4}+8s^{2}+16} \) in terms of simpler components each having a denominator that is a factor of the original denominator. This simplification allows us to use known inverse Laplace transforms.

**Steps in Partial Fraction Decomposition:**

• First, factor the denominator, which is crucial to setting up your partial fractions. For example, \( s^{4}+8s^{2}+16 \) can be factored into \( (s^{2}+4)^{2} \).
• Next, propose a structure for the decomposed fractions. In this case, it can be represented as \( \frac{A_{1}s+B_{1}}{s^{2}+4} + \frac{A_{2}s+B_{2}}{(s^{2}+4)^{2}} \).
• Find the coefficients by equating the original complex function to the sum of the partial fractions and matching coefficients of like powers.

This process transforms the problem of finding inverse Laplace transforms into a simpler task by allowing each term to be worked out separately.

Laplace Transform Properties

Laplace Transform Properties play a crucial role in analyzing and simplifying equations involving time and frequency domains. By leveraging these properties, we can easily find inverse Laplace transforms, as highlighted in this exercise.

**Key properties used in the exercise:**

• **Linearity:** If \( F(s) = aG(s) + bH(s) \), then \( \mathcal{L}^{-1}{F(s)} = a\mathcal{L}^{-1}{G(s)} + b\mathcal{L}^{-1}{H(s)} \). This simplifies the problem by allowing linear combinations of simple inverse Laplace transforms, as shown in the exercise when breaking down the expression \( -2\frac{s}{s^2+4} + 6\frac{1}{s^2+4} + 3\frac{s}{(s^2+4)^2} + 8\frac{1}{(s^2+4)^2} \).
• **Frequency Shifting:** Not directly used here but relevant, this property modifies a function \( F(s-a) \) shifting it by \( e^{at} f(t) \) in the time domain.
• **Table of Common Transforms:** Knowing common transforms like \( \mathcal{L}^{-1}\{\frac{s}{s^2+4}\}(t) = \frac{1}{2} \cos(2t)\) allows quick substitution.

By optimizing these properties, we effectively decode complex functions into manageable time domain expressions.

Solving Linear Equations

Solving Linear Equations is a fundamental skill necessary in partial fraction decomposition and finding inverse Laplace transforms. During the exercise, solving linear equations was essential to determine the coefficients for the partial fractions.

**Approach to Solving:**

• Once the partial fraction form is set up, match the coefficients for corresponding powers of \( s \) on both sides of the equation. For instance, equating powers of \( s \) in the expression \( s^2 + 6s + 8 = (A_1s + B_1)(s^2+4) + (A_2s+B_2)(s^2+4)^2 \) yields various linear equations.
• Solve these simultaneous equations either by substitution or elimination. In most cases, breaking them down into pairs makes solving them manageable.
• For our case, you solve the system: \( A_1 + A_2 = 1 \), \( B_1 = 6 \), \( A_1 + 2A_2 = 0 \), \( B_2 = 8 \).

The solutions to these equations provide the necessary components \( A_1, B_1, A_2, \) and \( B_2 \) for the partial fraction. This allows the inverse Laplace transform to be computed with accurate terms, leading to the precise final expression in the time domain.