Question

Use the results of Problem 19 to find the inverse Laplace transform of the given function. $$ F(s)=\frac{2 s+1}{4 s^{2}+4 s+5} $$

Short answer

Question: Find the inverse Laplace transform of the following function: $$ F(s) = \frac{2s+1}{4s^2 + 4s + 5} $$ Answer: The inverse Laplace transform of the given function is: $$ f(t) = \frac{1}{2}e^{-\frac{1}{2}t}\cos\left(\frac{\sqrt{3}t}{2}\right) - \frac{1}{2}e^{-\frac{1}{2}t}\sin\left(\frac{\sqrt{3}t}{2}\right) $$

Step-by-step solution

Perform partial fraction decomposition

To perform partial fraction decomposition, we can rewrite the given function as: $$ F(s) = \frac{2s+1}{4s^2 + 4s + 5} = \frac{A}{s - s_1} + \frac{B}{s - s_2} $$ where \(s_1\) and \(s_2\) are the roots of the denominator, and \(A\) and \(B\) are constants to be determined.

Determine the roots of the denominator

To find the roots of the denominator, we can rewrite the denominator in terms of its quadratic formula: $$ 4s^2 + 4s + 5 = 0 $$ Using the quadratic formula, we have: $$ s_{1,2} = \frac{-4 \pm \sqrt{4^2 - 4(4)(5)}}{8} = -\frac{1}{2} \pm \frac{\sqrt{3}i}{2} $$ The two roots are: $$ s_1 = -\frac{1}{2} + \frac{\sqrt{3}i}{2}, \quad s_2 = -\frac{1}{2} - \frac{\sqrt{3}i}{2} $$

Determine the constants A and B

To find the constants \(A\) and \(B\), we can use the residues theorem or by substituting back into the original equation: $$ 2s+1 = A(s - s_2) + B(s - s_1) $$ Substituting \(s = s_1\), we have: $$ 2s_1 + 1 = A(s_1 - s_2) $$ Substituting \(s = s_2\), we have: $$ 2s_2 + 1 = B(s_2 - s_1) $$ Solving for \(A\) and \(B\), we find: $$ A = \frac{1}{2}, \quad B = -\frac{1}{2} $$

Inverse Laplace transform

Now we rewrite \(F(s)\) with the obtained constants and roots: $$ F(s) = \frac{1/2}{s - (-\frac{1}{2} + \frac{\sqrt{3}i}{2})} - \frac{1/2}{s - (-\frac{1}{2} - \frac{\sqrt{3}i}{2})} $$ We know the inverse Laplace transform of \(\frac{1}{s+a}\) is \(e^{-at}\), and by linearity, the inverse Laplace transform of a sum of terms is the sum of the inverse Laplace transforms of each term. Therefore, we can write the inverse Laplace transform of \(F(s)\) as: $$ f(t) = \mathcal{L}^{-1} \{F(s)\} = \frac{1}{2}e^{(-\frac{1}{2}+\frac{\sqrt{3}i}{2})t} - \frac{1}{2}e^{(-\frac{1}{2}-\frac{\sqrt{3}i}{2})t} $$ This is the inverse Laplace transform of the given function: $$ f(t) = \frac{1}{2}e^{-\frac{1}{2}t}\cos\left(\frac{\sqrt{3}t}{2}\right) - \frac{1}{2}e^{-\frac{1}{2}t}\sin\left(\frac{\sqrt{3}t}{2}\right) $$

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in algebra to break down complex rational expressions into simpler fractions that are easier to work with, especially when performing integration or finding inverse Laplace transforms. The process involves rewriting a given fraction as a sum of fractions with simpler denominators. This is particularly useful when the denominator includes polynomials and is essential when handling higher-order differential equations in engineering and physics.

For example, to decompose a fraction with a quadratic denominator, you might express it as the sum of two or more fractions, each with a different linear factor in the denominator. Finding the coefficients of these 'partial fractions' typically requires setting the numerators equal and solving for the unknown constants. This step is crucial before moving onto finding the inverse Laplace transform, as it simplifies the function into a form where standard transformation rules can be applied.

Quadratic Formula

The quadratic formula is a staple in algebra for finding the roots of any quadratic equation. A quadratic equation is any equation in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). The quadratic formula is given by:

\[ x = \frac{-b \: \pm \: \sqrt{b^2 - 4ac}}{2a} \]
Using the formula, you can find both the real and complex roots of the quadratic equation. In the context of inverse Laplace transforms, identifying the roots of the denominator polynomial is essential to apply partial fraction decomposition accurately. These roots also give insight into the behavior of the solution to a differential equation, indicating oscillatory or exponential solutions depending on whether they are complex or real.

Complex Roots

Complex roots arise in equations when the discriminant (the part under the square root) in the quadratic formula is negative, leading to a square root of a negative number. Since the square root of a negative number doesn't have a real value, this introduces the imaginary unit \( i \), where \( i^2 = -1 \). Complex roots typically come in conjugate pairs, such as \( a + bi \) and \( a - bi \), where \( a \) and \( b \) are real numbers.

When dealing with inverse Laplace transforms, complex roots lead to solutions that involve trigonometric functions because of Euler's formula, which relates exponentials and complex numbers. This formula states that \( e^{ix} = \cos(x) + i \sin(x) \), a fundamental concept in understanding oscillatory behavior in systems described by differential equations.

Residues Theorem

In complex analysis, the residue theorem is a powerful tool used to evaluate complex integrals, often encountered in advanced calculus and differential equations. It states that the integral of a function around a closed loop in the complex plane is equal to \( 2\pi i \) times the sum of the residues contained inside the path of integration.

When applying the residue theorem to inverse Laplace transforms, it becomes invaluable for finding the coefficients in partial fraction decomposition. A residue, in this sense, is the coefficient of the \( \frac{1}{s-a} \) term in the decomposed function when \( a \) is a pole (a value where the function becomes unbounded or undefined). By calculating residues, you can efficiently determine the constants needed to reconstruct the original function in terms of its simpler components, which can then be inverted to find the time-domain solution.