Problem 2 - Arbeitsbuch Mathematik für Ingenieure

Question

Bestimmen Sie jeweils die Originalfunktion zu den folgenden LaplaceTransformierten: a) \(F(s)=\frac{s+1}{s^{2}+s-6}\) (Partialbruchzerlegung), b) \(F(s)=\frac{1}{\left(s^{2}+1\right)^{2}}\) (Faltungssatz), c) \(F(s)=\frac{1}{s\left(1-e^{-s}\right)}\), d) \(F(s)=\ln \frac{s+1}{s-1}\) (Differentiationssatz).

Step-by-Step Solution

Answer

a) \[f(t) = \frac{2}{5} e^{-3t} + \frac{3}{5}e^{2t} \] b) \[t sin(t) \] c) \[\frac{sin(t)}{t} \]d)\[\log 2(s)\]

Step 1: Teilaufgabe a: Partialbruchzerlegung

Die Laplacetransformierte ist gegeben als \[F(s) = \frac{s+1}{s^2 + s - 6} \] Der Nenner kann faktorisiert werden zu \[ s^2 + s - 6 = (s+3)(s-2) \] Daraus folgt: \[F(s) = \frac{s+1}{(s+3)(s-2)} \] Nun führt man eine Partialbruchzerlegung durch: \[\frac{s+1}{(s+3)(s-2)} = \frac{A}{s+3} + \frac{B}{s-2} \] Zunächst bestimmt man die Koeffizienten A und B, indem man beide Seiten der Gleichung multipliziert mit (s+3)(s-2) und die resultierende Gleichung vergleicht. \[(s+1) = A(s - 2) + B(s + 3) \] Setze s = 2: \[ 2 + 1 = A(2 - 2) + B(2 + 3) \ 3 = 5B \Rightarrow B = \frac{3}{5} \] Setze s = -3: \[-3 + 1 = A(-3 - 2) + B(-3 + 3) \ -2 = -5A \Rightarrow A = \frac{2}{5} \] Somit gilt: \[\frac{s+1}{(s+3)(s-2)} = \frac{\frac{2}{5}}{s+3} + \frac{\frac{3}{5}}{s-2} \] Nun wendet man den Laplace-Transformation an (rückwärts): \[ F(t) = \frac{2}{5} e^{-3t} + \frac{3}{5} e^{2t} \]

Step 2: Teilaufgabe b: Faltungssatz

Die Laplacetransformierte ist gegeben als: \[F(s) = \frac{1}{(s^2+1)^2} \] Diese Funktion entspricht der Laplace-Transformation des Faltungssatzes: \[\mathcal{L}[t \times \frac{\text{sin}(t)}{t}] = \frac{1}{(s^2+1)^2} \] Also ist die Originalfunktion: \[f(t) = t \times \frac{\text{sin}(t)}{t} = t \text{sin}(t) \]

Step 3: Teilaufgabe c

Die Laplacetransformierte ist gegeben als: \[F(s) = \frac{1}{s (1-e^{-s})} \] Diese Funktion entspricht der Laplace-Transformation der Funktion: \[f(t) = \frac{\text{sin}(t)}{t}\] Deshalb ist: \[f(t) = \frac{\text{sin}(t)}{t} \]

Step 4: Teilaufgabe d: Differentiationssatz

Die Laplacetransformierte ist gegeben als: \[F(s) = \ln{\frac{s+1}{s-1}} \] Laut dem Differentiationssatz ergibt sich: \[F(s) = s + 1 = H(s + 1 - s) + \frac{1}{s + 1} - s + \frac{1}{s-1}-\frac{1}{(s-1)}}\]

Key Concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a method used to break down a complex fraction into simpler fractions that are easier to work with. It is particularly useful in finding the inverse Laplace Transform. For example, consider the function \( F(s) = \frac{s+1}{s^{2}+s-6} \). To simplify, start by factoring the denominator to get \( s^{2}+s-6 = (s+3)(s-2) \). Then, we can represent it as \( F(s) = \frac{s+1}{(s+3)(s-2)} \).To decompose it into partial fractions, we write: \( \frac{s+1}{(s+3)(s-2)} = \frac{A}{s+3} + \frac{B}{s-2} \). Solving for A and B, we get \( A = \frac{2}{5} \) and \( B = \frac{3}{5} \). Substituting these back, we have \( \frac{s+1}{(s+3)(s-2)} = \frac{\frac{2}{5}}{s+3} + \frac{\frac{3}{5}}{s-2} \). Each part can then be inverted back to the time domain using the inverse Laplace transform, giving us \( F(t) = \frac{2}{5} e^{-3t} + \frac{3}{5} e^{2t} \).

Convolution Theorem

The Convolution Theorem is a powerful tool in Laplace transforms which simplifies dealing with products of functions. It translates a multiplication in the s-domain into a convolution in the time domain. The equation for the Convolution Theorem is: \[ \mathcal{L}{f(t) * g(t)} = F(s) \cdot G(s) \].For example, suppose we have \( F(s) = \frac{1}{(s^{2}+1)^{2}} \). We recognize this as the Laplace transform of a convolution: \[ \mathcal{L}{t \cdot \frac{\sin(t)}{t}} = \frac{1}{(s^{2}+1)^{2}} \]. This means the original function f(t) is: \[ f(t) = t \cdot \frac{\sin(t)}{t} = t \sin(t) \].Convolution makes it straightforward to handle complex products by breaking them down to simpler individual parts.

Differentiation Theorem

The Differentiation Theorem deals with the Laplace transforms of derivatives. It states that if we know the Laplace transform of a function, we can find the transform of its derivative, and vice versa.For instance, given \( F(s) = \ln{\frac{s+1}{s-1}} \), applying the differentiation theorem, we get \[ \mathcal{L}{f'(t)} = s \cdot F(s) - f(0) \]. Here, \( F(s) \) is the Laplace transform of some function whose derivative we need to consider. Rewriting the above example as a differential relationship helps solve it step-by-step.This theorem can greatly simplify calculations in solving differential equations using Laplace transforms.

Inverse Laplace Transform

The Inverse Laplace Transform is the process of finding the original time-domain function, given its Laplace transform. It is often the final step after using various Laplace transform properties and theorems.For instance, after performing partial fraction decomposition on \( F(s) = \frac{s+1}{s^{2}+s-6} \), we simplify it to \[ \frac{\frac{2}{5}}{s+3} + \frac{\frac{3}{5}}{s-2} \]. This directly translates to \( f(t) = \frac{2}{5} e^{-3t} + \frac{3}{5} e^{2t} \) in the time domain by applying the inverse Laplace transform on each term.One often makes use of Laplace transform tables or known transforms to effectively carry out the inversion process for various functions.