Problem 13 - Analysis 1: Ein Lehr- und Arbeitsbuch für Studienanfänger

Question

Bestimmen Sie die Funktionen, deren Laplace-Transformierte gegeben sind durch: a) \(\mathscr{L}[f](s)=\frac{s+\alpha}{s\left(s^{2}+a^{2}\right)} \quad, \quad s>0\), b) \(\mathscr{L}[f](s)=\frac{s}{(s+\alpha)(s+\beta)}, \quad s>\max \\{-\alpha,-\beta\\}\).

Step-by-Step Solution

Answer

(a) \( f(t) = \frac{\alpha}{a^2} + \cos(at) \); (b) \( f(t) = \frac{\beta}{\beta - \alpha} e^{-\alpha t} - \frac{\alpha}{\beta - \alpha} e^{-\beta t} \).

Step 1: Identify the Type of Laplace Transform

We know that the Laplace Transform of a function \( f(t) \) is defined as \( \mathscr{L}[f](s) = \int_0^{\infty} e^{-st} f(t) \, dt \). Given the form of the transforms, we recognize these as standard types that can be transformed back using known inverse Laplace transformations or partial fraction decomposition.

Step 2: Apply Partial Fraction Decomposition (Part a)

For part a, decompose \( \frac{s+\alpha}{s(s^2 + a^2)} \) into partial fractions: \( \frac{A}{s} + \frac{Bs+C}{s^2+a^2} \). Equating coefficients, solve for \( A \), \( B \), and \( C \).

Step 3: Solve for Constants in Decomposition (Part a)

Set \( \frac{s+\alpha}{s(s^2+a^2)} = \frac{A}{s} + \frac{Bs+C}{s^2+a^2} \). Multiply through by the common denominator and equate coefficients of powers of \( s \) to get equations for \( A \), \( B \), and \( C \). From this, find \( A = \alpha/a^2 \), \( B = 1 \), and \( C = 0 \).

Step 4: Inverse Laplace Transform (Part a)

The inverse Laplace transform of \( \frac{1}{s} \) is \( 1 \) and of \( \frac{s}{s^2+a^2} \) is \( \cos(at) \). Applying these, get \( f(t) = \frac{\alpha}{a^2} + \cos(at) \).

Step 5: Apply Partial Fraction Decomposition (Part b)

For part b, decompose \( \frac{s}{(s+\alpha)(s+\beta)} \) into partial fractions: \( \frac{A}{s+\alpha} + \frac{B}{s+\beta} \). Solve for \( A \) and \( B \) using algebraic equations.

Step 6: Solve for Constants in Decomposition (Part b)

Set \( \frac{s}{(s+\alpha)(s+\beta)} = \frac{A}{s+\alpha} + \frac{B}{s+\beta} \). Find \( A = \frac{\beta}{\beta - \alpha} \) and \( B = \frac{-\alpha}{\beta - \alpha} \) by equating coefficients.

Step 7: Inverse Laplace Transform (Part b)

Use the inverse transforms: \( \frac{1}{s+\alpha} \) corresponds to \( e^{-\alpha t} \) and \( \frac{1}{s+\beta} \) corresponds to \( e^{-\beta t} \). Combine using the found constants to find \( f(t) = \frac{\beta}{\beta - \alpha} e^{-\alpha t} - \frac{\alpha}{\beta - \alpha} e^{-\beta t} \).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic technique used to simplify complex rational expressions into simpler fractions that are easier to work with. This is particularly useful in solving Laplace transforms, as the inverse operation requires the expression in a simplified form.
In this context, we usually take a complex function, like \( \frac{s+\alpha}{s(s^2 + a^2)} \), and express it as a sum of simpler fractions like \( \frac{A}{s} + \frac{Bs+C}{s^2+a^2} \).

• First, identify the form of the expression in the denominator. This helps determine the number of terms needed in the decomposition.
• Next, write out the expression in its decomposed form with unknown constants like \(A\), \(B\), and \(C\).
• Finally, solve for these constants by equating coefficients after multiplying through by the common denominator.

Partial fraction decomposition allows us to separate components of the transform which can then be individually inverted. This method is essential for bringing complex transforms back into simpler, recognizable inverse forms.

Inverse Laplace Transform

The inverse Laplace transform is a technique to find the original time-domain function from its Laplace-transformed version. Once an expression is decomposed into partial fractions or if it's already in a standard form, finding the inverse becomes straightforward.
For instance, given the expression \( \frac{1}{s+c} \), its inverse Laplace transform is simply \( e^{-ct} \). Similarly, the inverse transform for \( \frac{s}{s^2+a^2} \) is \( \cos(at) \).

• Identify each term in the decomposed fractions that correspond to known inverse Laplace forms.
• Use standard inverse Laplace transformation tables or formulas to convert each term back to its time-domain form.
• Sum up these individual terms to obtain the complete inverse function.

This process allows us to translate complex Laplace-transformed functions back into the time domain, revealing insights about the original functions.

Coefficient Equating

Coefficient equating is a method used to solve for unknown constants in partial fraction decompositions. Once a rational expression is written in its decomposed form, solving for the constants involves equating coefficients of corresponding powers of variables.
Here's how it generally works:

• Write the complex fraction as a sum of simpler fractions with unknown coefficients.
• Multiply through by the common denominator to eliminate fractions.
• Expand both sides of the equation and collect like terms.
• Set the coefficients of corresponding powers of \(s\) equal to each other creating a set of linear equations.

This technique enables us to determine the values of the unknown constants accurately and is crucial in simplifying expressions in preparation for inverse Laplace transforms.