Problem 47 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The inverse Laplace transform is \( f(t) = u(t-1)(1 - e^{-(t-1)}) \).

Step 1: Identify the Inverse Laplace Transform

We need to find the inverse Laplace transform of \( \frac{e^{-s}}{s(s+1)} \). This requires recognizing the shift theorem and partial fraction decomposition.

Step 2: Apply the Shift Theorem

The term \( e^{-s} \) suggests a time shift, so first, identify \( F(s) = \frac{1}{s(s+1)} \). According to the first shifting theorem, \( \mathscr{L}^{-1} \left\{ e^{-s} F(s) \right\} = u(t-1) f(t-1) \), where \( u(t-1) \) is the unit step function.

Step 3: Partial Fraction Decomposition

Decompose \( \frac{1}{s(s+1)} \) into simpler fractions: \( \frac{1}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1} \). Solve for \( A \) and \( B \) by multiplying through and equating coefficients.

Step 4: Solve for A and B

Multiplying both sides by \( s(s+1) \) gives \( 1 = A(s+1) + Bs \). By setting \( s = 0 \), we find \( A = 1 \). By setting \( s = -1 \), we find \( B = -1 \). Thus, \( \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1} \).

Step 5: Find Inverse Laplace of Each Term

Find \( \mathscr{L}^{-1} \left\{ \frac{1}{s} \right\} = 1 \) and \( \mathscr{L}^{-1} \left\{ \frac{1}{s+1} \right\} = e^{-t} \). So, the inverse Laplace \( \mathscr{L}^{-1} \left\{ \frac{1}{s(s+1)} \right\} = 1 - e^{-t} \).

Step 6: Apply the Time Shift

Using the result from Step 2, apply the time shift to \( (1 - e^{-t}) \). Thus, \( f(t) = u(t-1)((1 - e^{-(t-1)})) \). This means \( f(t) = u(t-1)(1 - e^{-(t-1)}) \), for \( t \geq 1 \).

Key Concepts

Shift Theorem

The Shift Theorem, also known as the First Shifting Theorem or Time Shifting, is a vital tool in the Laplace Transform domain. It helps simplify the calculation of inverse Laplace transforms by shifting functions in time.When confronted with an expression like \( e^{-as}F(s) \), the theorem tells us that the inverse Laplace transform involves a unit step function. Specifically, the theorem states:

• If \( F(s) \) is the Laplace Transform of \( f(t) \),
• then \( \,\mathscr{L}^{-1}\{ \, e^{-as} F(s) \} = u(t-a)f(t-a) \, \).

In our problem, the presence of \( e^{-s} \) indicates a shift of one unit in the time domain. This means we will calculate \( f(t-1) \) and then multiply it by the unit step function \( u(t-1) \).Understanding this theorem is crucial because it provides a straightforward approach to account for delays or advancements in a system's response.

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to break down complex fractions into simpler parts, making it easier to apply inverse Laplace transforms. When we have a fraction like \( \frac{1}{s(s+1)} \), our goal is to express it as a sum of simpler fractions.Here’s how it works:

• Start by assuming our fraction can be decomposed: \( \frac{1}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1} \).
• Multiply through by the denominator \( s(s+1) \) to clear the fraction: \( 1 = A(s+1) + Bs \).
• Use strategic values of \( s \) to solve for \( A \) and \( B \). Setting \( s = 0 \) gives \( A = 1 \) and \( s = -1 \) gives \( B = -1 \).

By breaking the fraction down into \( \frac{1}{s} - \frac{1}{s+1} \), each part can be transformed back into the time domain more easily. This makes it more manageable to find the inverse Laplace Transform.

Unit Step Function

The Unit Step Function, denoted as \( u(t-a) \), is a piecewise function that changes value at a specified point in time. Its role in solving Laplace Transforms is primarily to handle time shifts or postponed system responses.Here’s what you need to know:

• For \( t < a \), the function value is 0.
• For \( t \geq a \), the function value is 1.

In practice, a function like \( u(t-1) \) essentially "turns on" another function at \( t = 1 \). By multiplying another function by \( u(t-1) \), we effectively delay it by 1 unit of time in the response.In our problem, the Unit Step Function \( u(t-1) \) was used to shift the expression from the Laplace domain to effectively model the delayed response in the time domain. It’s a great tool for modeling situations where outputs don’t initiate until after some delay.

Laplace Transform Properties

Laplace Transform Properties are a set of rules that help simplify the process of transforming functions back and forth between the time and s-domain. Knowing these properties makes it easier to manipulate and solve complex transformations.Some key properties include:

• Linearity: \( \,\mathscr{L}\{ af(t) + bg(t) \} = a\,\mathscr{L}\{ f(t) \} + b\,\mathscr{L}\{ g(t) \} \, \).
• Shift Theorem: Helps in situations with \( e^{-as} \), as discussed in the first section.
• Transform of Derivatives: Relates to transforming derivatives of functions from the time domain.
• Initial and Final Value Theorems: Provide insights into the behavior of the function as time approaches zero or infinity.

These properties, when combined, enable us to tackle even the most complex differential equations and systems by breaking them down into manageable parts. Each property contributes a key piece to the broader puzzle of understanding and solving Laplace Transforms.