Problem 9 - Advanced Engineering Mathematics

Question

In Problems, find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}\left\\{\left(1-e^{t}+3 e^{-4 t}\right) \cos 5 t\right\\} $$

Step-by-Step Solution

Answer

The Laplace transform is \( F(s) = \frac{s}{s^2 + 25} - \frac{s - 1}{(s - 1)^2 + 25} + 3\frac{s + 4}{(s + 4)^2 + 25} \).

Step 1: Identify the Function to Transform

We're given the function \(f(t) = (1 - e^{t} + 3e^{-4t}) \cos(5t)\) and need to find its Laplace transform \(F(s)\).

Step 2: Use the Linearity of Laplace Transform

Apply the linearity of the Laplace transform, which means we can break down the transformation: \( \mathscr{L}\{(1) \cos(5t)\} - \mathscr{L}\{e^{t} \cos(5t)\} + 3\mathscr{L}\{e^{-4t} \cos(5t)\}\)

Step 3: Apply Laplace Transform Formula for \(\cos(at)\)

The Laplace transform of \(\cos(at)\) is \(\frac{s}{s^2 + a^2}\). So, for each component of the form \(b \cos(5t)\): \- For \(\mathscr{L}\{\cos(5t)\} = \frac{s}{s^2 + 25}\- For \(\mathscr{L}\{e^{t} \cos(5t)\} = \frac{s - 1}{(s - 1)^2 + 25}\)- For \(\mathscr{L}\{e^{-4t} \cos(5t)\} = \frac{s + 4}{(s + 4)^2 + 25}\)\)

Step 4: Combine the Transformed Components

Combine the results from Step 3 using linearity: \[F(s) = \frac{s}{s^2 + 25} - \frac{s - 1}{(s - 1)^2 + 25} + 3\frac{s + 4}{(s + 4)^2 + 25}\]This is the Laplace transform of the original function.

Key Concepts

Linearity of Laplace Transform

The Linearity of Laplace Transform is an essential concept to grasp when transforming functions from the time domain to the frequency domain. It allows us to simplify the process of finding the Laplace transform of a sum of functions.

This property states that the Laplace transform of a linear combination of functions is equal to the linear combination of their individual Laplace transforms. In mathematical terms, if we have functions \(f(t)\) and \(g(t)\) with constants \(a\) and \(b\), the transform is given by:

• \(\mathscr{L}\{a \cdot f(t) + b \cdot g(t)\} = a \cdot \mathscr{L}\{f(t)\} + b \cdot \mathscr{L}\{g(t)\}\)

This makes the process both straightforward and efficient, as we can work with each component of a function separately rather than dismantling complex expressions all at once.

In the original solution, the function \((1 - e^{t} + 3e^{-4t}) \cos(5t)\) was broken down using linearity. Each part of the expression was dealt with individually, speeding up the process of transforming it completely.

Cosine Function in Laplace Transform

The transformation of cosine functions is a common task when working with the Laplace Transform. The cosine function, \(\cos(at)\), is one of the simplest periodic functions.

Understanding how cosine transforms help us in dealing with signals and oscillating functions. The Laplace transform for a standard cosine function \( \cos(at) \) is given by:

• \(\mathscr{L}\{\cos(at)\} = \frac{s}{s^2 + a^2}\)

This formula becomes a very powerful tool, especially when handling systems that naturally exhibit periodic behaviors like electrical circuits or mechanical oscillations.

In the outlined problem, the formula was applied to transform \(\cos(5t)\) and versions of it modified by exponential terms. It provides a way to translate a time-based cosine, contingent on an angular frequency, into a frequency-based form that is easier to manipulate for further analysis.

Exponentially Damped Cosine

An exponentially damped cosine function combines oscillation with exponential growth or decay. This function is of the form \( e^{bt} \cos(at) \), which indicates multiplication of a cosine by an exponential term.

This can model various real-world phenomena, such as damped electrical signals or mechanical vibrations that decrease over time.

• If \(b > 0\), it shows exponential growth (amplification) while if \(b < 0\), it indicates exponential decay (damping).

To find the Laplace transform, a combination of transformation formulas is used. For an exponentially damped cosine like \( e^{bt} \cos(at) \):

• \(\mathscr{L}\{e^{bt} \cos(at)\} = \frac{s - b}{(s - b)^2 + a^2}\)

This formula reflects changes in amplitude over time, incorporating both the damping effects and the frequency of the original cosine signal.

In the problem solution, this transformative property was used for both \(e^{t} \cos(5t)\) and \(e^{-4t} \cos(5t)\), accounting for their respective damping factors and allowing simplified manipulation in the s-domain.