Problem 45 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

\( f(t) = \sin(t-\pi)u(t-\pi) \)

Step 1: Identify the Laplace Transform

The problem asks for the inverse Laplace transform of \( \mathscr{L}^{-1}\left\{\frac{e^{-\pi s}}{s^{2}+1}\right\} \). Here, the expression looks similar to \( \frac{1}{s^2 + 1} \), which matches the form for the Laplace transform of a sine function.

Step 2: Recall the Basic Inverse Transform

Recall that the inverse Laplace transform of \( \frac{1}{s^2 + 1} \) is \( \sin(t) \). Using this knowledge, we understand that if the \( \frac{1}{s^2 + 1} \) corresponds to \( \sin(t) \), then the given problem only differs by the exponential factor \( e^{-\pi s} \).

Step 3: Address the Exponential Shift

The term \( e^{-\pi s} \) represents a time delay in the inverse Laplace domain. The effect of \( e^{-as} \) in the Laplace domain results in a delay of \( a \) units in the time domain. Therefore, multiplying by \( e^{-\pi s} \) results in a time shift of \( \pi \) units.

Step 4: Apply the Time Shift Property

According to the time shift property, if \( \mathscr{L}^{-1}\left\{ F(s) \right\} = f(t) \), then \( \mathscr{L}^{-1}\left\{ e^{-as}F(s) \right\} = f(t-a)u(t-a) \), where \( u(t-a) \) is the unit step function. Here \( f(t) = \sin(t) \) and \( a = \pi \).

Step 5: Write the Final Solution

Using the time shift property from Step 4, the inverse Laplace transform is \( f(t) = \sin(t-\pi)u(t-\pi) \). This is the function in the time domain that corresponds to the given Laplace transform.

Key Concepts

Time Domain Analysis

Time domain analysis involves studying functions with respect to time. It is a crucial way to understand how systems behave over time, particularly in engineering and physics. One big reason we use time domain analysis is to examine how a system responds to different inputs. This response shows up as a function of time, giving us insightful information about the system’s dynamics.

When we talk about a function's behavior in the time domain, it's essentially how the system will react at various moments in time. Analyzing a system in the time domain can help us work out things like how long it will take the system to reach a steady state or how it will react to sudden changes. It's a direct way to look at the performance of a system as it naturally occurs over time. This is very useful when using inverse Laplace transforms, like converting back from the frequency domain to time domain, which can reveal timed-based behaviors we want to study.

In the context of the original problem, converting the expression from the Laplace domain back into the time domain allows us to see how the system behaves considering factors such as time delay.

Laplace Transform Properties

The Laplace transform is a powerful tool that helps us move between the time and frequency domains. It’s particularly celebrated for its ability to simplify differential equations into algebraic equations, which are much easier to solve. One of the key properties of the Laplace transform is linearity. This means you can take the Laplace transform of a sum of functions by summing the Laplace transforms of each function individually.

• Linearity: Utilizes the fact that the transform of a sum is the sum of the transforms.
• Shifting: Helps in dealing with delays and advances in systems.
• Frequency domain manipulation: Simplifies the handling of complex systems.

In the exercise at hand, one important property to highlight is the exponential shift property. When you multiply a function by an exponential factor in the Laplace domain, it corresponds to a time shift in the time domain. This is crucial to understanding how time delays are represented and managed when dealing with systems in the frequency domain. Recognizing this shift property enables us to properly translate the behavior of delayed systems back to the time domain.

Unit Step Function

The unit step function, often denoted as \( u(t) \), plays a significant role in time domain analysis. Essentially, it's a function that "turns on" at a specific time. It is predominantly used to model switches or sudden changes in systems.

The function is defined as:

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

This function is integral when dealing with the time shifts mentioned earlier. It acts to "activate" the delayed portion of a time-shifted function. By incorporating the unit step function, expressions can properly reflect real-world behaviors where actions begin or end at a discrete point in time.

In the original exercise solution, the unit step function was used to conveniently manage the delay effect introduced by \( e^{-\pi s} \) in the Laplace domain. Adding \( u(t-\pi) \) ensured that the sine function \( \sin(t-\pi) \) would only be observed from time \( t=\pi \) onwards, accurately reflecting the system's delayed response in the time domain.