Question

Berechnen Sie die LAPLACE-Transformierten der Funktionen (a) \(f(t)=e^{b t} \cos (a t)\), (b) \(g(t)=e^{b t} \sinh (a t)\), und weisen damit die Gültigkeit der entsprechenden Einträge in der Tabelle der LAPLACE-Transformierten nach.

Short answer

The Laplace transforms are consistent with the table: \( f(t) = e^{bt}\cos(at) \) yields \( \frac{s-b}{(s-b)^2 + a^2} \) and \( g(t) = e^{bt}\sinh(at) \) yields \( \frac{a}{(s-b)^2 - a^2} \).

Step-by-step solution

Understand the Laplace Transform Formula

The Laplace transform of a function \( f(t) \) is given by \( \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \). We will use this definition to find the Laplace transform of the given functions \( f(t) = e^{bt} \cos(at) \) and \( g(t) = e^{bt} \sinh(at) \).

Use the Laplace Transform Table Entries for Exponentials and Trigonometric Functions

The Laplace transform of \( f(t) = e^{bt} \cos(at) \) can be directly looked up using the table entry: \( \mathcal{L}\{e^{bt} \cos(at)\} = \frac{s-b}{(s-b)^2 + a^2} \). Similarly, for \( g(t) = e^{bt} \sinh(at) \), we use: \( \mathcal{L}\{e^{bt} \sinh(at)\} = \frac{a}{(s-b)^2 - a^2} \).

Verify by Deriving Laplace Transform of Part (a)

For \( f(t) = e^{bt} \cos(at) \), the Laplace transform is calculated as:\[\mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} e^{bt} \cos(at) \, dt = \int_{0}^{\infty} e^{-(s-b)t} \cos(at) \, dt\]Using integration by parts or the known result, this simplifies to:\[\mathcal{L}\{f(t)\} = \frac{s-b}{(s-b)^2 + a^2}\]

Verify by Deriving Laplace Transform of Part (b)

For \( g(t) = e^{bt} \sinh(at) \), the Laplace transform is calculated as:\[\mathcal{L}\{g(t)\} = \int_{0}^{\infty} e^{-st} e^{bt} \sinh(at) \, dt = \int_{0}^{\infty} e^{-(s-b)t} \sinh(at) \, dt\]Using the table or solving directly, this results in:\[\mathcal{L}\{g(t)\} = \frac{a}{(s-b)^2 - a^2}\]

Confirm Consistency with Laplace Table

Both derived expressions match the standard table results for Laplace transforms of functions \( f(t) = e^{bt} \cos(at) \) and \( g(t) = e^{bt} \sinh(at) \), demonstrating their validity and consistency in the Laplace transform table.

Key concepts

Integration

Integration is a key mathematical tool used to calculate areas under curves and more generally, to accumulate quantities. In the context of the Laplace Transform, integration helps us convert functions of time into functions of a complex variable, often denoted as "s". This conversion is very useful in solving differential equations which would otherwise be difficult to handle in their original form.

When we compute the Laplace Transform of a function, we perform integration from zero to infinity. This technique converts time domain functions into the "s" domain, simplifying the process of analyzing and solving equations. For example, evaluating:

• \(\int_{0}^{\infty} e^{-st} f(t) \, dt\), we effectively sum up all the effects of the function \(f(t)\) over time.
• Such integrations often involve techniques like integration by parts to simplify complex expressions.

Understanding how to properly perform integrations is crucial to derive the correct Laplace Transforms.

Differential Equations

Differential equations describe how variables change over time or space and are fundamental in modeling physical and biological systems. However, these equations can sometimes be hard to solve directly. Laplace Transforms come to the rescue by converting these differential equations into simpler algebraic equations.

Converting to the "s" domain via Laplace Transform can turn derivatives into multipliers of "s", significantly reducing the complexity of solving the equations. Once solved, inverse Laplace Transforms are applied to fetch solutions back into the time domain.

• The Laplace Transform simplifies the analysis of linear time-invariant systems by eliminating the need for solving differential equations directly.
• As seen in the original exercise, using this powerful tool allows students to transform complex functions and work towards solving them efficiently.

Knowing how to apply Laplace Transforms in solving differential equations is essential for tackling advanced engineering and physics problems.

Mathematical Functions

Mathematical functions form the backbone of many complex calculations. In the problem at hand, exponential and trigonometric functions combined to form the input to the Laplace Transform.

Different types of functions behave uniquely under Laplace Transform, providing specific results on how they alter when shifted into the s-domain.

• Exponential functions like \(e^{bt}\) involve a rate of growth or decay, often highlighting natural processes or decaying factors that can be observed in physics.
• Trigonometric functions, on the other hand, describe oscillatory behavior, such as that seen in waves.

Understanding the behavior of these functions and how their transforms appear in the Laplace table gives insight into how complex systems operate and how to analyze them effectively.