Problem 14 - Advanced Engineering Mathematics

Question

Fill in the blanks or answer true/false. $$ \mathscr{L}^{-1}\left\\{\frac{1}{3 s-1}\right\\}= $$_____

Step-by-Step Solution

Answer

\( \frac{1}{3} e^{\frac{1}{3}t} \)

Step 1: Understanding the Given Expression

The expression given is \( \mathscr{L}^{-1} \left\{ \frac{1}{3s-1} \right\} \). This indicates that we need to find the inverse Laplace transform of the function \( \frac{1}{3s-1} \).

Step 2: Recall the General Formula for Inverse Laplace Transform

The inverse Laplace transform of \( \frac{1}{s-a} \) is \( e^{at} \). Our goal is to express \( \frac{1}{3s-1} \) in this form.

Step 3: Rewrite the Expression

Rewrite the expression \( \frac{1}{3s-1} \) in the form \( \frac{1}{s-a} \) by factoring out the 3: \( \frac{1}{3(s - \frac{1}{3})} \). We now identify \( a = \frac{1}{3} \).

Step 4: Apply the Inverse Laplace Transform Formula

Using the formula from step 2, substitute \( a = \frac{1}{3} \). Thus, the inverse Laplace transform, \( \mathscr{L}^{-1} \left\{ \frac{1}{3s-1} \right\} \), is \( \frac{1}{3} e^{\frac{1}{3}t} \).

Step 5: Verify the Solution

Ensure the solution is consistent by noting that inversely transforming a multiplication by a constant \( \frac{1}{3} \) outside the transformation results in the same constant multiplied to the time function, confirming \( \frac{1}{3} e^{\frac{1}{3}t} \) is accurate.

Key Concepts

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to transform complex functions from the time domain into a simpler algebraic form in the frequency domain. This makes solving differential equations more manageable. By applying this transform, continuous time signals are expressed as polynomials using the variable \( s \), which represents complex frequency.

• Laplace Transform is essential in engineering, particularly for control systems and signal processing.
• The notation for the Laplace Transform of a function \( f(t) \) is \( \, \mathscr{L}\{f(t)\} = F(s) \).
• Standard formulas and properties, such as linearity and time shifts, make the integration process straightforward.

Understanding how to convert a time-domain function using Laplace enables engineers and scientists to analyze system stability or behavior efficiently.

Exponential Function

Exponential functions play a critical role in the analysis and understanding of dynamic systems, especially when dealing with Laplace Transforms. An exponential function is of the form \( e^{at} \), where \( e \approx 2.718 \), the base of the natural logarithm, acting as a constant.

• These functions represent growth or decay processes, like radioactive decay or population growth.
• In the context of Laplace and inverse Laplace transforms, they help describe system responses over time.
• When transforming functions back to the time domain, the reappearance of an exponential function symbolizes the impact of initial conditions or input functions.

In this exercise, the inverse Laplace transform of \( \frac{1}{3s-1} \) reflects as \( \frac{1}{3} e^{\frac{1}{3}t} \), showcasing the exponential decay of a process starting at an initial value.

Inverse Transformation

Inverse Transformation refers to reverting functions from the frequency domain back to the time domain using techniques such as the Inverse Laplace Transform. This process allows the interpretation of physical phenomena in terms of time behavior based on frequency characteristics.

• The general formula for inverse transformation, \( \mathscr{L}^{-1}(F(s)) = f(t) \), helps translate frequency algebra back into functional time descriptions.
• In applying the inverse Laplace Transform, recognizing patterns in terms similar to standard forms, e.g., \( \frac{1}{s-a} \), enables efficient calculation.
• The task in this exercise involves re-identifying the constants and rewriting the expression to utilize known inverse Laplace pairs.

By achieving an inverse transformation, students can map theoretical solutions to real-world data, providing insight into how systems evolve over time.