Problem 19 - Advanced Engineering Mathematics

Question

Fill in the blanks or answer true/false. $$ \mathscr{L}^{-1}\left\\{\frac{s+\pi}{s^{2}+\pi^{2}} e^{-s}\right\\}= $$____

Step-by-Step Solution

Answer

\( u(t-1) \cos(\pi(t-1)) \)

Step 1: Identify the Problem Type

The problem involves finding the inverse Laplace transform of a given function. The function provided is \( \frac{s+\pi}{s^{2}+\pi^{2}} e^{-s} \).

Step 2: Decompose the Expression

The given function can be understood as a Laplace transform that includes a time-shift factor. The exponential term \( e^{-s} \) suggests a shift in the time domain.

Step 3: Handle the Time Shift

The time shift, represented by \( e^{-as} \), indicates that the inverse transform of the function without the exponential should be shifted by \( a = 1 \) in the time domain. This means that once we find \( \mathscr{L}^{-1}\left\{\frac{s+\pi}{s^{2}+\pi^{2}}\right\} \), we will shift the found function by 1 unit in time.

Step 4: Find the Inverse Transform of the Main Function

Consider the inverse Laplace transform of \( \frac{s+\pi}{s^{2}+\pi^{2}} \). This resembles the form \( \frac{s-a}{(s-a)^2+b^2} \). The standard inverse Laplace of \( \frac{s}{s^2+\omega^2} \) is \( \cos(\omega t) \) and of \( \frac{\omega}{s^2+\omega^2} \) is \( \sin(\omega t) \). Here, \( \omega = \pi \). Thus, the inverse is \( \cos(\pi t) \).

Step 5: Apply the Time Shift

Now, apply the time shift. The inverse function \( \cos(\pi t) \) is shifted by 1 unit to give \( u(t-1) \cos(\pi(t-1)) \), where \( u(t-1) \) is the unit step function applied to ensure the function is valid for \( t \geq 1 \).

Step 6: Write the Final Solution

The final expression after applying the time shift is \( u(t-1) \cos(\pi(t-1)) \).

Key Concepts

Time Shift in Laplace Transform

The time shift property in the Laplace Transform is an important tool in electrical engineering and mathematics. This property allows us to handle functions that start, stop, or change their behavior at a specific time. When you see an exponential term like \( e^{-as} \) in the Laplace Transform, it signals that there is a time shift involved in the function's inverse transformation.
To dissect this, the Laplace Transform of a function \( f(t) \) shifts in time by replacing it with another function \( f(t-a) \), valid for \( t \geq a \). The shift indicated by \( e^{-as} \) moves the entire response of the system to a later time frame. In essence, you compute the inverse Laplace of the function without the exponential term, and then ensure that the result is valid starting from \( t = a \).
This means if you find \( \, \mathscr{L}^{-1}\{ F(s) \} = f(t) \), then \( \, \mathscr{L}^{-1}\{ F(s) e^{-as} \} = u(t-a)f(t-a) \). Here, the unit step function \( u(t-a) \) ensures everything starts at \( t = a \). It effectively ‘turns on’ the shifted function only from this new time forward.

Unit Step Function

The unit step function, often denoted as \( u(t) \), is a simple yet powerful tool used in Laplace Transforms to define functions that start at a specific time. It is represented graphically as a step that jumps from 0 to 1 at a given point on the time axis, and remains at 1 thereafter.
In the context of inverse Laplace transformations, the unit step function is crucial for applying time shifts, especially when combined with transforms that incorporate the \( e^{-as} \) term. The role of the unit step is to 'switch on' the transformed function at the correct time. Thus for a shifted function \( f(t-a) \), the unit step \( u(t-a) \) defines the onset at \( t = a \).

• Before the shift time \( t
• After the shift time \( t \geq a \): \( u(t-a) = 1 \).

This step function thus ensures that the inverse Laplace transform aligns perfectly in the time domain, becoming active just when needed.

Cosine Function Transformation

Transforming trigonometric functions such as the cosine function using Laplace techniques allows us to analyze oscillatory systems, common in physics and engineering domains.
Invariant forms, such as \( \frac{s}{s^2 + \omega^2} \), directly translate to cosine functions in the time domain through the inverse Laplace transform. Specifically, \( \mathscr{L}^{-1}\left\{\frac{s}{s^2 + \omega^2}\right\} = \cos(\omega t) \).
When tackling problems like the one given, where \( \omega = \pi \), the expression \( \frac{s+\pi}{s^2+\pi^2} \) can be broken down, with the inverse transform yielding \( \cos(\pi t) \).

• Adding terms directly related to the sine and cosine identities is common in real-world signals.
• Such transformations help describe periodic phenomena and their shifts.

This cosine function transformation describes smooth oscillations, prevalent in systems like AC circuits or any periodic motion analyses. Understanding this concept is therefore pivotal for interpreting and designing time-based waveforms and solutions in complex engineering problems.