Question

Find the inverse Laplace transform of the given function. $$ \frac{1-2 s}{s^{2}+4 s+5} $$

Short answer

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

Step-by-step solution

Perform Partial Fraction Decomposition

To perform partial fraction decomposition on $$\frac{1-2s}{s^2 + 4s + 5}$$, we first need to identify the factors of the denominator. The denominator $$s^2 + 4s + 5$$ cannot be factored into linear terms, but it can be written in the standard quadratic form as $$(s+2)^2 + 1$$, with $$\alpha = -2$$ and $$\omega = 1$$. Now, let's perform partial fraction decomposition: $$ \frac{1-2s}{(s+2)^2 + 1} = \frac{A(s+2) + B}{(s+2)^2 + 1} $$ For the coefficients A and B, multiply both sides by the denominator to clear fractions: $$1-2s = A(s+2) + B$$ Now, we need to find the values for A and B. Let's solve for B first by setting s equal to 0: $$1 - 2(0) = A(0+2) + B \Rightarrow B + 2A = 1$$ Next, let's determine A. Since there is only one equation left, typically we would differentiate both sides. But in this case, let's substitute $$s = -3$$ and solve for A directly: $$1 - 2(-3) = A(-3+2) + B \Rightarrow 7 = -A + B$$ Now we have two equations: 1. $$B + 2A = 1$$ 2. $$7 = -A + B$$ Now we can solve for A and B: $$A = 2$$ $$B = -3$$ So, our function after partial fraction decomposition is: $$ \frac{1-2s}{(s+2)^2 + 1} = \frac{2(s+2) - 3}{(s+2)^2 + 1} $$

Find the Inverse Laplace Transform

Now that we have the function in the form $$\frac{2(s+2) - 3}{(s+2)^2 + 1}$$, we can find the inverse Laplace transform of each term separately using the Laplace transform pairs: 1. Inverse Laplace transform of $$\frac{2(s+2)}{(s+2)^2 + 1} = 2e^{-2t}\cos(t)$$ 2. Inverse Laplace transform of $$\frac{-3}{(s+2)^2 + 1} = -3e^{-2t}\sin(t)$$ Now we add them together to find the inverse Laplace transform of the original function: $$ L^{-1}\left\{\frac{1-2s}{s^2 + 4s + 5}\right\} = 2e^{-2t}\cos(t) - 3e^{-2t}\sin(t) $$ So, the inverse Laplace transform of the given function is: $$ 2e^{-2t}\cos(t) - 3e^{-2t}\sin(t) $$

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a key method used to simplify complex expressions, especially when dealing with inverse Laplace transforms. In our exercise, we need to decompose the function \( \frac{1-2s}{s^2 + 4s + 5} \). The first step involves rewriting the denominator. It is a quadratic that doesn't factor into linear terms easily. Instead, we rewrite it in a completed square form: \((s+2)^2 + 1\). This form allows us to express the original expression as a sum or difference that can be more easily inverse-transformed.

Once in this form, we represent the fraction as \( \frac{A(s+2) + B}{(s+2)^2 + 1} \). The next step is to determine the constants \(A\) and \(B\). We do this through algebraic manipulation, clearing the fraction, and equating the coefficients of corresponding powers of \(s\) from both sides of the equation. By setting up a system of equations, we solve for \(A\) and \(B\). This allows us to deconstruct our original expression into simpler, handleable parts: \( \frac{2(s+2) - 3}{(s+2)^2 + 1} \).

This step is fundamental because it converts a complex rational function into simpler components that are more straightforward to handle with Laplace transform pairs.

Differential Equations

Differential equations form the backbone of systems that model real-world scenarios. They describe how a system changes over time and are pervasive in fields like engineering and physics. The use of Laplace transforms in solving differential equations is fundamental because it converts differential equations into algebraic equations, which are usually easier to solve.

In our context, finding the inverse Laplace transform \(L^{-1}\) converts the function back into the time domain. This step is essential when solving initial value problems associated with differential equations. The end solution often reflects a real-world phenomenon such as oscillations or exponential decay.

• This conversion helps in comprehending complex systems.
• Inverse transforms reveal the behavior of the system with respect to time.
• Solutions often use standard Laplace transform pairs listed in tables for simplification purposes.

Inverse transforming complex rational functions, like \( \frac{1-2s}{s^{2}+4s+5} \), allows us to directly understand the behavior of the underlying processes modeled by the differential equation.

Complex Exponentials

Complex exponentials play a vital role in simplifying the processes of solving differential equations and performing operations like inverse Laplace transforms. Functions expressed with complex exponentials, such as \( e^{-2t}\cos(t) \), are derived from Euler's formula, which relates complex exponentials and trigonometric functions.

In this exercise, each part of the decomposed fraction can be linked with derivatives and integrals of these complex expressions. For instance, the inverse Laplace of \( \frac{2(s+2)}{(s+2)^2+1} \) yields \( 2e^{-2t}\cos(t) \), and \( \frac{-3}{(s+2)^2 +1} \) transforms to \(-3e^{-2t}\sin(t) \).

• They help in visualizing systems as combinations of simple waveforms.
• These are standard solutions to homogeneous linear differential equations.
• Sine and cosine terms indicate oscillatory motion in the system.

The interplay of the exponential function with trigonometric functions often describes systems with oscillatory outputs damped by exponential decay, a common characteristic in electrical circuits and mechanical systems as suggested by these function forms.