Problem 18 - A First Course in Complex Analysis with Applications

Question

Use the theory of residues to compute the inverse Laplace transform \(\mathscr{L}^{-1}\\{F(s)\\}\) for the given function \(F(s)\). $$ \frac{s+4}{s^{2}+6 s+11} $$

Step-by-Step Solution

Answer

The inverse Laplace transform is \(f(t) = \frac{2}{\sqrt{8}}e^{-3t}(\cos(\sqrt{2}t) + \sin(\sqrt{2}t))\).

Step 1: Identify the Laplace Transform Function

We start by identifying the given Laplace transform function to be \[F(s) = \frac{s+4}{s^2 + 6s + 11}.\]

Step 2: Find the Poles of F(s)

To find the poles of the function, solve the quadratic equation in the denominator,\[s^2 + 6s + 11 = 0.\]By using the quadratic formula, \[s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]where \(a = 1\), \(b = 6\), and \(c = 11\), we have:\[s = \frac{-6 \pm \sqrt{36 - 44}}{2} = \frac{-6 \pm \sqrt{-8}}{2}.\]This gives us\[s = -3 \pm i\sqrt{2}.\]

Step 3: Use Residue Theory to Find Inverse Laplace Transform

To find the inverse Laplace transform, use the residue theorem:The inverse Laplace transform, \( \mathscr{L}^{-1}\{F(s)\}\), is given by the sum of the residues of \(e^{st} F(s)\) at its poles.Compute the residues at \(s = -3 + i\sqrt{2}\) and \(s = -3 - i\sqrt{2}\):For the pole at \(s = -3 + i\sqrt{2}\):\[\text{Residue} = \lim_{s \to -3 + i\sqrt{2}} (s - (-3 + i\sqrt{2})) \cdot \frac{s+4}{s^2 + 6s + 11}.\]The expression simplifies directly, as the term \((s - (-3 + i\sqrt{2}))\) cancels out the denominator leaving the numerator evaluated at the pole.For the pole at \(s = -3 - i\sqrt{2}\), compute similarly.

Step 4: Compute Residues

For the pole at \(s = -3 + i\sqrt{2}\):The residue is calculated as the derivative of the denominator evaluated at the pole:\[ \frac{d}{ds}\left(s - (-3 + i\sqrt{2})\right) = 1,\]Thus,\[\frac{s+4}{2s + 6}\bigg|_{s = -3 + i\sqrt{2}} = \frac{(-3 + i\sqrt{2}) + 4}{2(-3 + i\sqrt{2}) + 6} \implies \text{Residue 1} = \frac{1 + i\sqrt{2}}{-6 + i\sqrt{8}}.\]For the pole at \(s = -3 - i\sqrt{2}\):\[\frac{s+4}{2s + 6}\bigg|_{s = -3 - i\sqrt{2}} = \frac{1 - i\sqrt{2}}{-6 - i\sqrt{8}},\]\(\text{Residue 2}\) can be determined similarly, using conjugates to simplify.

Step 5: Calculate Inverse Laplace Transform

The inverse Laplace transform is given by combining the residues:\[f(t) = e^{-3t} \left[ \cos(\sqrt{2}t) + i\sin(\sqrt{2}t) \right] R_1 + e^{-3t} \left[\cos(-\sqrt{2}t) + i\sin(-\sqrt{2}t)\right] R_2,\]with \(R_1\) and \(R_2\) as computed residues.By symmetry of complex pairs, simplify it:\[f(t) = 2e^{-3t}\left[\frac{1}{\sqrt{8}}\cos(\sqrt{2}t) + \frac{1}{\sqrt{8}}\sin(\sqrt{2}t) \right].\]

Key Concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis, particularly in evaluating integrals and computing inverse transforms. When dealing with functions like Laplace transforms, the residue theorem simplifies calculating inverse transforms by focusing on the poles of a complex function. Here's how it works in simple terms:

• Identify the poles of the function: These are the values of the complex variable where the denominator of the function becomes zero.
• Compute the residue at each pole: This involves simplifying the expression to a point where you evaluate the numerator at the pole, often after eliminating factors that would otherwise make the denominator zero.
• Apply the inverse transform: Once you have the residues, you can sum them up to find the inverse Laplace transform, each residue contributing to the function based on its corresponding pole.

By breaking the problem into manageable parts, the residue theorem helps transform complex integrals into simpler calculations.

Complex Analysis

Complex analysis is a branch of mathematics that focuses on functions of complex numbers. It's crucial in many fields, especially in engineering and physics, where it is used for signal processing and control theory. In the context of inverse Laplace transforms, complex analysis allows us to handle the mathematical behavior of functions in the complex plane. It helps us analyze:

• Poles: Specific points in the complex plane where a function becomes undefined (infinite), but are critical for calculating integrals.
• Contours: Paths in the complex plane used to evaluate integrals. The residue theorem applies mainly to contours around poles.

Poles and residues play a vital role as they simplify the evaluation of otherwise difficult integrals through contour integration. Using this, engineers can devise systems that are more predictable and manageable under various conditions.

Laplace Transform Poles

Poles of a Laplace transform are essential for determining the behavior of a system described by the transform. These poles are values of the Laplace variable, typically denoted by \(s\), that make the denominator zero, resulting in an undefined output for the function.To identify these poles:

• Solve for values where the denominator equals zero, often using algebraic methods such as factoring or the quadratic formula.
• Understand that these poles correspond to exponential terms in the inverse Laplace transform, influencing system stability and transient response.

For example, in the exercise, the denominator \(s^2 + 6s + 11 = 0\) was solved using the quadratic formula to find poles at \(s = -3 \pm i\sqrt{2}\). These complex poles indicate an oscillatory response that decays over time. Recognizing pole patterns can greatly aid in predicting the dynamic behavior of systems modeled by Laplace transforms.