Problem 46 - Advanced Engineering Mathematics

Question

Suppose \(f(t)\) is a function for which \(f^{\prime}(t)\) is piece wise continuous and of exponential order \(c\). Use results in this section and Section \(4.1\) to justify $$ f(0)=\lim _{s \rightarrow \infty} s F(s), $$ where \(F(s)=\mathscr{L}\\{f(t)\\}\). Verify this result with \(f(t)=\cos k t\).

Step-by-Step Solution

Answer

The limit shows that \(f(0) = 1\), verifying the expression for \(f(t) = \cos kt\).

Step 1: Understanding the Exponential Order

If a function is of exponential order \(c\), it means there exists a constant \(M\) such that \(|f(t)| \leq M e^{ct}\) for all \(t\). This condition helps with the convergence of the Laplace transform \(F(s)\).

Step 2: Compute the Laplace Transform

For a function \(f(t)\), the Laplace Transform \(F(s) = \mathscr{L}\{f(t)\}\) is given by \(F(s) = \int_0^{\infty} e^{-st} f(t) \, dt\). This result is used for transforming the time domain function \(f(t)\) into the frequency domain function \(F(s)\).

Step 3: Evaluate Limit s→∞ for General Case

Evaluate the limit \( \lim_{s \rightarrow \infty} s F(s) \). Under the assumption that \(f(t)\) is of exponential order, the polynomial growth of \(s\) will dominate, leading to \( f(0) = \lim_{s \rightarrow \infty} s F(s) \).

Step 4: Verify with Specific Function \(f(t) = \cos k t\)

First find the Laplace Transform of \(f(t) = \cos kt\): the Laplace transform is \(F(s) = \frac{s}{s^2 + k^2}\).

Step 5: Evaluate Limit s→∞ for \(F(s) = \frac{s}{s^2 + k^2}\)

Substitute \(F(s)\) into the limit expression: \( \lim_{s \rightarrow \infty} s \cdot \frac{s}{s^2 + k^2} = \lim_{s \rightarrow \infty} \frac{s^2}{s^2 + k^2}\).

Step 6: Simplify the Expression

As \(s\) approaches infinity, \( k^2 \) becomes negligible compared to \(s^2\). Thus, \( \lim_{s \rightarrow \infty} \frac{s^2}{s^2 + k^2} = \lim_{s \rightarrow \infty} \frac{s^2}{s^2} = 1\).

Step 7: Final Step: Conclusion and Verification

Since the limit yields \(1\), this matches \(f(0) = \cos(0) = 1\). Thus, the expression \(f(0) = \lim_{s \rightarrow \infty} s F(s)\) is verified for the function \(f(t) = \cos kt\).

Key Concepts

Exponential Order

Understanding a function's exponential order is crucial when it comes to Laplace Transforms. A function is said to be of exponential order if there exists a constant, say \( M \), such that \(|f(t)| \leq M e^{ct}\) for all time \( t \).
This notion assures us that the function doesn't "blow up" too quickly as time moves forward. Instead, it grows at a rate that doesn't exceed a certain exponential function.
This condition greatly aids the convergence of a Laplace Transform.

• It ensures the integral involved in the Laplace Transform, \( \int_0^{\infty} e^{-st} f(t) \, dt \), converges.
• It allows us to work with the function in the frequency domain by transforming it from the time domain.

In essence, understanding the exponential order simplifies dealing with complex functions when using the Laplace Transform.

Piecewise Continuous Function

A piecewise continuous function is not as complicated as it sounds. Imagine it as a function that may "jump" or "break" but still behaves predictably within certain intervals.
This means the function can have different rules or definitions within different parts of its domain, but there shouldn't be any "wild" behavior.
The only requirement is that these jumps or breaks are finite and can be easily accounted for.

• This is essential for ensuring the proper computation of the Laplace Transform.
• Despite having discontinuities, the function should remain reasonably well-behaved within each section.

This quality makes it manageable to integrate and transform into the frequency domain, as required in Laplace Transform analysis.

Frequency Domain

The frequency domain is a perspective shift from analyzing signals and functions in time, to understanding them in terms of frequency.
When you use the Laplace Transform on a function, you are essentially translating it from the time domain into the frequency domain.
This new representation helps simplify many problems, especially those involving differential equations.

• It converts complex operations like differentiation into simpler algebraic forms like multiplication.
• By viewing functions in terms of frequency, you can more easily analyze their behavior and stability.

Overall, transitioning to the frequency domain provides a powerful toolset for engineering and physics, making complex signal behavior easier to understand and manipulate.