Problem 15 - Differential Equations and Linear Algebra

Question

Determine the inverse Laplace transform of \(F.\) $$F(s)=\frac{s e^{-s}}{s^{2}+4}$$.

Step-by-Step Solution

Answer

The inverse Laplace transform of the given function \(F(s)\) is: \(f(t) = \cos(2(t - 1)) u(t - 1)\).

Step 1: Find the inverse Laplace Transform of G(s)

First, let's find the inverse Laplace transform for G(s). We have: \[ G(s) = \frac{s}{s^2 + 4} \] This can be recognized as the Laplace transform of the derivative of the sine function, which is the cosine function. Therefore, \[ \mathcal{L}^{-1} \{ G(s) \} = g(t) = \cos(2t) \]

Step 2: Apply the Delay Property

Now let's apply the delay property to find the inverse Laplace transform of F(s). Given the delay property: \[ \mathcal{L}^{-1} \{ e^{-s} G(s) \} = g(t - 1) u(t - 1) \] Where \(u(t-1)\) is the unit step function. Therefore, \[ \mathcal{L}^{-1} \{ F(s) \} = \cos(2(t - 1)) u(t - 1) \]

Step 3: Final Answer

The inverse Laplace transform of the given function F(s) is: \[ f(t) = \cos(2(t - 1)) u(t - 1) \]

Key Concepts

Delay Property

The Delay Property is a powerful tool when working with inverse Laplace transforms. It allows us to shift functions in time without direct modification of the function's core properties. The delay property states that if you have a function multiplied by an exponential term like \( e^{-as} \), it translates to a shift in time by \( a \) seconds in the time domain. Here's the formula:\[\mathcal{L}^{-1} \{ e^{-as} G(s) \} = g(t - a) u(t - a)\]This means the original function \( g(t) \) is delayed by \( a \) units, effectively starting it later. This is very useful in modelling systems where operations or signals are delayed, such as communication systems or process control.
The Delay Property simplifies complex calculations by converting a multiplication operation in the Laplace domain to a simple shift in time in the original function, allowing more control and prediction of system behavior.

Laplace Transform

The Laplace Transform is a key concept in solving differential equations and analyzing systems. It transforms functions from the time domain into the s-domain (or complex frequency domain). This powerful method simplifies convolution into multiplication and differentiation into polynomial operations. Here's the fundamental formula:\[\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st}f(t) dt\]In practical terms, the Laplace transform, represented as \( F(s) \), captures all the frequencies of a time-based function \( f(t) \), turning complex equations simpler to manage.
Through the inverse Laplace transform, these frequency domain instructions are converted back to their time domain forms, which can be used to solve real-world problems.

Unit Step Function

The Unit Step Function, often denoted as \( u(t - a) \), is a fundamental concept in control systems and signal processing. It represents a signal that switches on at a certain time \( a \), remaining active from that point onwards. The function can be described by:\[u(t - a) = \begin{cases} 0, & \text{if } t < a \ 1, & \text{if } t \ge a \end{cases}\]This simple yet powerful function is key in describing signals and systems that have sudden starts or changes in their behavior. When combined with the delay property in Laplace transforms, as shown by shifting elements like \( g(t - a) u(t - a) \), it represents real-world problems like systems that "kick in" at a specific time.

• The Unit Step Function models system start-ups or blockages.
• It also assists in defining piecewise functions.

In many applications, it serves to activate portions of functions and to model switching actions within systems.