Problem 43 - Advanced Engineering Mathematics

Question

Find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}^{-1}\left\\{\frac{e^{-2 s}}{s^{3}}\right\\} $$

Step-by-Step Solution

Answer

The inverse Laplace transform is \(f(t) = \frac{(t-2)^2}{2} \cdot u(t-2)\).

Step 1: Recall the Inverse Laplace Transform Formula

The inverse Laplace transform of a function is denoted by \[\mathscr{L}^{-1}\{ F(s) \} = f(t)\.\]Given the function, we recognize \[\mathscr{L}^{-1}\{ \frac{e^{-as}}{s^n} \} = \frac{(t-a)^{n-1}}{(n-1)!} u(t-a)\]where \(u(t-a)\) is the Heaviside step function.

Step 2: Identify Parameters

In the given problem, \( \frac{e^{-2s}}{s^3} \), we have \(a = 2\) and \(n = 3\). This means the formula becomes\[ (t-2)^{3-1} = (t-2)^2 \].

Step 3: Apply Inverse Laplace Transform

Using the identified parameters and the inverse Laplace formula, the inverse Laplace transform is:\[\mathscr{L}^{-1}\{\frac{e^{-2s}}{s^3}\} = \frac{(t-2)^{2}}{2!} \cdot u(t-2) = \frac{(t-2)^{2}}{2} \cdot u(t-2)\.\]Here, the \(u(t-2)\) ensures the function is shifted to start at \(t=2\).

Step 4: Simplify the Expression

Simplifying \(\frac{(t-2)^2}{2}\), we maintain the expression as it is with the Heaviside step function:\[f(t) = \frac{(t-2)^2}{2} \cdot u(t-2)\.\]

Key Concepts

Heaviside step function

The Heaviside step function, often denoted as \(u(t-a)\), is essential in handling functions that have delays or shifts in time. It effectively represents a function that turns on, or "steps up," at a specific point, \(t = a\). This step function is given by \(u(t-a)\), where:

• \(u(t-a) = 0\) for \(t < a\)
• \(u(t-a) = 1\) for \(t \geq a\)

The utility of the Heaviside step function lies in its ability to model real-world phenomena where an action starts abruptly at a given time. For instance, when utilizing the inverse Laplace transform, the Heaviside function helps in accounting for the delays in function response, ensuring that the solution is accurate past specific time points. In our exercise, \(u(t-2)\) indicates that the function behavior starts at \(t = 2\). Without it, the expression \(\frac{(t-2)^2}{2}\) would inaccurately represent the function starting from \(t = 0\).

Laplace Transform properties

The Laplace Transform is a powerful integral transform used to convert functions from the time domain to the frequency domain, making them easier to manipulate and solve. It comes with several critical properties, making it versatile for analyzing linear systems, such as linear circuits and control systems. Here are a few essential properties of the Laplace Transform:

• Linearity: \(\mathscr{L}\{a f(t) + b g(t)\} = a F(s) + b G(s)\)
• Shifting Theorem: Shifting by \(a\) in time results in multiplication by \(e^{-as}\) in the Laplace domain.
• Derivative in Time Domain: \(\mathscr{L}\{f'(t)\} = sF(s) - f(0)\)

In the inverse Laplace context, these properties help revert the frequency domain back to the time domain by linking each term strategically. For our exercise, the shifting property, \(\mathscr{L}^{-1}\left\{\frac{e^{-as}}{s^n}\right\}\), demonstrates the time delay and aligns with applying the Heaviside function. Recognizing and correctly applying these properties can simplify complex inverse transform problems.

Partial fraction decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions, making them easier to solve or integrate, especially within the context of Laplace Transforms. It is particularly helpful when dealing with composite fractions that are hard to manipulate in their original form. Here's how it typically works:

• Given a complicated fraction, the goal is to rewrite it as a sum of simpler fractions.
• Look for common factors in the denominator and express the fraction in terms of these factors.
• Set up the decomposition with unknown coefficients and solve for these coefficients by equating both sides of the equation.

In relation to the Laplace Transform, partial fraction decomposition assists in making the inverse transformation more straightforward. It allows complicated Laplace-transformed functions to be expressed as a sum of simpler ones, each corresponding to a term with a known inverse Laplace transform formula. While our current problem did not require explicit use of partial fractions, understanding this process is fundamental for handling more complicated inverse Laplace transforms that may arise in broader applications.