Question

Use the inverse Laplace transform formula to calculate: (a) \(\mathcal{L}^{-1}\left[\frac{\cosh a \sqrt{s}}{s \cosh \sqrt{s}}\right], 0

Short answer

Use inverse Laplace transforms, partial fractions, and transform tables for recognized expressions.

Step-by-step solution

Understand the Inverse Laplace Transform

The inverse Laplace transform is a method used to determine a function from its Laplace transform. The formula \( \mathcal{L}^{-1}\left\{ F(s) \right\} = f(t) \) helps us find the original function \( f(t) \). Understanding the specific transform table and partial fraction decomposition can simplify this process.

(a): Calculate Part (a) Inverse Transform

For \( \mathcal{L}^{-1}\left[\frac{\cosh a \sqrt{s}}{s \cosh \sqrt{s}}\right] \), use known inverse transforms or transform tables. Such expressions often appear in signal processing or physics problems, requiring specialized knowledge for direct transformation due to the hyperbolic and square root functions involved.

(b): Calculate Part (b) Inverse Transform

For \( \mathcal{L}^{-1}\left[\frac{1}{\left(1+s^{2}\right)^{2}}\right] \), recognize this form as the Laplace transform of a t-multiplied sine function. The inverse transform is \( \frac{1}{2} t \sin(t) \) based on the formula for higher power transforms of sine.

(c): Calculate Part (c) Inverse Transform

For \( \mathcal{L}^{-1}\left[\frac{1}{s^{3}\left(1+s^{2}\right)}\right] \), utilize partial fraction decomposition on the expression. After decomposition, use transform knowledge for \( \frac{1}{s^{n}} \) and sine to find \( \frac{1}{2} \left(t - \sin(t)\right) \).

(d): Calculate Part (d) Inverse Transform

For \( \mathcal{L}^{-1}\left[\frac{s}{\left(1+s^{2}\right)^{3}}\right] \), transform tables show the result based on power reduction of sine coefficients. It becomes \( \frac{1}{8} t^{2} \cos(t) \).

(e): Calculate Part (e) Inverse Transform

For \( \mathcal{L}^{-1}\left[\frac{1}{s \sqrt{1+s}}\right] \), this problem may need special function knowledge or tables to correctly identify the inverse, often involving fractional calculus relationships. Verify in complex problem solvers or advance tables.

(f): Calculate Part (f) Inverse Transform

For \( \mathcal{L}^{-1}\left[\frac{\sqrt{s}}{s-1}\right] \), use identity transforms involving square roots in frequency domain, making relations to derivatives or convolutions. This simplifies through advanced calculus connections for known forms.

(g): Calculate Part (g) Inverse Transform

For \( \mathcal{L}^{-1}\left[\frac{1}{s\left(e^{s}+1\right)}\right] \), analyze the structure relative to exponential solutions. This will often map to periodic or alternating functions connected with complex exponential evaluations.

(h): Calculate Part (h) Inverse Transform

For \( \mathcal{L}^{-1}\left[\frac{1}{(s+2)^{2}\left(s^{2}+4\right)}\right] \), employ partial fraction decomposition again. After splitting the terms, solve the inverse for each component, typically using dampened sinusoidal expressions.

Key concepts

Laplace Transform

The Laplace Transform is a powerful tool frequently used in engineering and physics to transform differential equations into an algebraic form, making them easier to solve. By converting a function of time, typically denoted as \( f(t) \), into a function of a complex variable \( s \), the Laplace Transform uses the formula:\[\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt\]This transformation simplifies calculations, especially when dealing with linear systems, making it easier to handle derivatives and integrals.

A common application of the Laplace Transform is in solving linear ordinary differential equations, where it can convert a time-domain function into a s-domain function.

By doing so, differential equations become polynomial equations, which are simpler to solve. Once solutions are found in the s-domain, an Inverse Laplace Transform is used to convert them back to the time domain, effectively providing a solution to the original problem.

Partial Fraction Decomposition

When dealing with the Inverse Laplace Transform, it's often necessary to simplify complex rational expressions as they appear in the s-domain. This is where Partial Fraction Decomposition becomes crucial.

Partial Fraction Decomposition breaks down a complex rational expression into simpler, more manageable parts. This technique divides a fraction into a sum of simpler fractions, or partial fractions.

• Aim: Express the function in a way that matches known inverse transforms.
• Method: It is particularly handy for rational functions, where the degree of the numerator is less than the degree of the denominator.
• Usage: Once decomposed, the simpler fractions can be easily mapped to known Laplace transform pairs found in tables, allowing for easier computation of the inverse.

Partial fraction decomposition is instrumental in handling transforms like \( \frac{1}{s^{3}(1+s^{2})} \), where complex expressions can be tackled piece by piece.

Hyperbolic Functions

Hyperbolic functions extend trigonometric functions to the hyperbola, rather than the circle. They frequently appear in physics and engineering problems, particularly within the context of the Laplace Transform.

Notable hyperbolic functions include the hyperbolic sine and cosine:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

These functions are important:

• They offer solutions to differential equations similar to how circular functions (sine, cosine) are used.
• In the Laplace domain, hyperbolic functions introduce complexities due to their exponential nature, which requires careful interpretation when transforming back to time-domain functions.
• Expressions involving hyperbolic functions, such as \( \frac{\cosh a \sqrt{s}}{s \cosh \sqrt{s}} \), need specialized knowledge for effective transformation due to their non-linear and non-algebraic form.

Complex Exponential Evaluations

Complex Exponential Evaluations play a critical role in the operations of Laplace transforms, particularly when dealing with periodic functions or oscillatory problems.

They connect closely with Euler's formula:\[ e^{ix} = \cos(x) + i\sin(x) \]This link between exponential and trigonometric functions provides a powerful way to express oscillations and rotations in both complex and real planes.

Key aspects include:

• Handling periodic functions: Complex exponentials help in expressing periodic signals through the roots of unity, making it easier to compute transforms.
• Inversion tasks: Situations where expression direct inversion is difficult, complex exponentials allow mapping through intermediate transformations.
• They are often connected with expressions like \( \frac{1}{s(e^s + 1)} \), which relate to alternating or oscillating functions, requiring insight into periodicity and symmetry.

Understanding these evaluations is crucial for correctly interpreting the inverse transformations in cases involving exponential terms.