Problem 18 - Advanced Engineering Mathematics

Question

Fill in the blanks or answer true/false. $$ \mathscr{L}^{-1}\left\\{\frac{e^{-5 s}}{s^{2}}\right\\}= $$____

Step-by-Step Solution

Answer

\((t-5)u(t-5)\)

Step 1: Analyze the given Laplace expression

You are given the inverse Laplace expression \( \mathscr{L}^{-1}\left\{\frac{e^{-5 s}}{s^{2}}\right\} \). This expression involves two components: \( e^{-5s} \) which represents a time delay and \( \frac{1}{s^{2}} \) which relates to a standard Laplace transform.

Step 2: Identify standard Laplace transform

The Laplace transform \( \frac{1}{s^{2}} \) corresponds to \( t \) in the time domain. This is because \( \mathscr{L}\{t\} = \frac{1}{s^2} \).

Step 3: Apply time-shift property

The term \( e^{-5s} \) in a Laplace transform represents a time delay of 5 units. The time-shift property states that if \( \mathscr{L}\{f(t)\} = F(s) \), then \( \mathscr{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \). Here, \( a = 5 \) and \( f(t) = t \).

Step 4: Combine results to find inverse

Applying the time-shift property, the inverse transform becomes \( t-5 \) times the unit step function \( u(t-5) \). Thus, the inverse Laplace transform is \( (t-5)u(t-5) \).

Key Concepts

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to convert functions from the time domain into the frequency domain. This transformation simplifies solving differential equations by turning them into algebraic ones. This is particularly useful for system analysis and control engineering.

Consider a time-domain function, usually expressed as \( f(t) \). The Laplace Transform of this function, denoted by \( F(s) \), is defined as:

• \( \mathscr{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t) \, dt \)

In this expression:

• \( s \) is a complex number that comprises a real part and an imaginary part.
• The exponential term \( e^{-st} \) acts as a dampening factor.

The process of transforming \( f(t) \) using this integral allows us to handle complex time functions easily, making it a cornerstone in signal processing and control systems. When using it, always keep standard Laplace transforms and properties, like linearity and time-scaling, in mind. These can greatly simplify the process.

Time-Shift Property

The time-shift property in Laplace Transforms helps manage functions that experience a delay or shift in time. This is particularly useful when dealing with delayed responses or control systems that activate after a certain time period.

For a given function \( f(t) \), if its Laplace Transform is \( F(s) \), then the time-shifted function \( f(t-a) \) with a delay of \( a \) units, has a Laplace Transform \( e^{-as} F(s) \).

This adjustment is crucial when modifying functions that start not at zero but at some later time. In an equation:

• \( \mathscr{L}\{f(t-a)u(t-a)\} = e^{-as} F(s) \)

In this formula:

• \( a \) is the shift amount.
• The term \( u(t-a) \) is a unit step function ensuring the function is zero for \( t < a \).

Understanding and applying this property allows seamless transitions between time-delayed functions, crucial for accurate system modeling and analysis.

Unit Step Function

The Unit Step Function, often noted as \( u(t) \), is a fundamental building block in mathematical analysis of control systems and signal processing. It acts like a switch that turns on a function at a specified point in time.

This function is defined as:

• \( u(t) = \begin{cases} 0, & t < 0 \ 1, & t \geq 0 \end{cases} \)

The unit step function can "enable" other functions to operate exclusively from a particular time onward. For example, if a function \( f(t) \) is multiplied by \( u(t-a) \), it becomes zero for all \( t < a \) and essentially starts or activates at \( t = a \).

When combined with the Laplace Transform, the unit step function helps model systems that begin operation after a time delay. This makes it an indispensable tool for control systems, offering a way to mirror physical processes accurately. Proper handling of unit step functions in transformations is key to problem-solving.