Problem 46 - Advanced Engineering Mathematics

Question

Find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}^{-1}\left\\{\frac{s e^{-\pi s / 2}}{s^{2}+4}\right\\} $$

Step-by-Step Solution

Answer

The inverse is \(f(t) = u(t-\frac{\pi}{2})(-\cos(2t))\).

Step 1: Identify the problem

We need to find the inverse Laplace transform of the given function \(\mathscr{L}^{-1}\left\{\frac{s e^{-\pi s / 2}}{s^{2}+4}\right\}\). This involves recognizing the component of the Laplace transform and the effect of the exponential term \(e^{-\pi s / 2}\).

Step 2: Find the basic inverse Laplace without the exponential

First, consider the function without the exponential, \(\frac{s}{s^2+4}\). This resembles the Laplace transform of \(\cos(2t)\), since \(\mathscr{L}\{\cos(kt)\} = \frac{s}{s^2+k^2}\). Therefore, \(\mathscr{L}^{-1}\left\{\frac{s}{s^2+4}\right\} = \cos(2t)\).

Step 3: Apply the shift theorem

The term \(e^{-\pi s / 2}\) indicates a time delay in the time domain. According to the shift property of Laplace transforms, \(\mathscr{L}\{u(t-a)f(t-a)\} = e^{-as}F(s)\). Therefore, \(\mathscr{L}^{-1}\left\{e^{-\pi s / 2}G(s)\right\} = u(t-\frac{\pi}{2})g(t-\frac{\pi}{2})\), where \(g(t) = \cos(2t)\).

Step 4: Formulate the final inverse Laplace transform

Apply the shift theorem result: the function is \(u(t-\frac{\pi}{2}) \cos\left(2(t-\frac{\pi}{2})\right)\), where \(u(t-\frac{\pi}{2})\) is the unit step function. This means \(f(t)\) is zero for \(t < \frac{\pi}{2}\) and \(\cos(2t-\pi)\) or \(-\cos(2t)\) for \(t \geq \frac{\pi}{2}\).

Step 5: Write the final function

Thus, the final function is \(f(t) = u(t-\frac{\pi}{2}) \cdot \cos(2t - \pi)\). When \(t \geq \frac{\pi}{2}\), this simplifies to \(-u(t-\frac{\pi}{2}) \cdot \cos(2t)\).

Key Concepts

Laplace Transform

The Laplace transform is a crucial tool in the field of engineering and mathematics, especially for solving differential equations and systems dynamics. It transforms a time-domain function, usually denoted as \(f(t)\), into a complex frequency-domain representation, \(F(s)\). This transformation makes it easier to manipulate and solve equations. Here are some essential points to understand the Laplace Transform:

• The Laplace transform is defined as \( \mathscr{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt \).
• This transformation is particularly useful for linear time-invariant systems, allowing us to convert differential equations into algebraic ones.
• The inverse Laplace transform, denoted as \( \mathscr{L}^{-1} \), allows you to convert back from the frequency domain to the time domain, retrieving the original function.

Understanding the Laplace transform helps simplify complex problems involving integrals and derivatives, providing a pathway to finding solutions more efficiently.

Shift Theorem

The shift theorem, or shifting property, is highly significant within the context of the Laplace Transform. It generally relates to how a function behaves over time, indicating delays and other temporal modifications. Here's a closer look:

• The Laplace transform shift theorem is expressed as \( \mathscr{L}\{u(t-a)f(t-a)\} = e^{-as}F(s) \), where \(a\) is the shift or delay.
• This property states that multiplying a Laplace transform \(F(s)\) by \(e^{-as}\) corresponds to shifting the original function \(f(t)\) by \(a\) units in the time domain.
• In practice, it means you can account for time delays directly within the Laplace domain, without reconstructing the entire time-based function.

This property is beneficial for systems analysis where timing and delay are significant factors, providing insight into how signals are modified as they traverse a system.

Unit Step Function

The unit step function, often denoted as \(u(t)\), is an essential concept in control systems and signal processing. It represents an instantaneous change in value and is crucial when describing systems that engage after a specific point in time. Here's why it's critical to understand:

• The unit step function is defined as \(u(t) = 0\) for \(t < 0\) and \(u(t) = 1\) for \(t \geq 0\).
• It is often used to model functions that "turn on" at a certain time \(a\), typically written as \(u(t-a)\) to represent a shift.
• In Laplace transforms, \(u(t-a)\) is essential for depicting functions with a delay or start time.

This function plays a significant role in analyzing systems, especially when they involve switches, controllers, and other mechanisms triggered at specific times. Combined with the shift theorem, it helps describe complex temporal behaviors in a very structured way.