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Chapter 6: Problem 1

Bestimmen Sie mit Hilfe der Definitionsgleichung der Laplace-Transformation die Bildfunktionen der folgenden Originalfunktionen: a) \(f(t)=\cos (\omega t)\) b) \(f(t)=2 t+\mathrm{e}^{-4 t}\) c) \(f(t)=\mathrm{e}^{-\phi t} \cdot \sin (\omega t)\) d) \(f(t)=\sinh (a t)\) e) \(f(t)=t^{3}\) f) \(f(t)=\sin ^{2} t\)

Short Answer

Expert verified
a) \frac{s}{s^2 + \omega^2}, b) \frac{2}{s^2} + \frac{1}{s+4}, c) \frac{\omega}{(s + \phi)^2 + \omega^2}, d) \frac{a}{s^2 - a^2}, e) \frac{6}{s^4}, f) 0.5\left(\frac{1}{s} - \frac{s}{s^2 + 4}\right)

Step by step solution

01

- Definition of the Laplace Transform

The Laplace Transform of a function f(t) is given by: \[ \text{L}\{f(t)\} = F(s) = \int_{0}^{\infty} f(t) \mathrm{e}^{-st} \ \mathrm{d}t \]
02

- Transformation of \(f(t)=\cos (\omega t)\)

We need to use the Laplace transform definition for the function \(f(t)=\cos (\omega t)\):\[ F(s) = \int_{0}^{\infty}\cos(\omega t) \mathrm{e}^{-st} \ \mathrm{d}t \]This evaluates to: \[ F(s)= \frac{s}{s^{2} + \omega^{2}} \]
03

- Transformation of \(f(t)=2t + \mathrm{e}^{-4t}\)

For the function \(f(t)=2t + \mathrm{e}^{-4t}\):Firstly, using linearity of Laplace transforms: \( \text{L}\{2t + \mathrm{e}^{-4t}\} = 2 \text{L}\{t\} + \text{L}\{\mathrm{e}^{-4t}\}\)Using known Laplace transforms:\[ \text{L}\{t\} = \frac{1}{s^2} \]\[ \text{L}\{\mathrm{e}^{-4t}\} = \frac{1}{s+4} \]Thus, the result is: \[ F(s) = 2 \frac{1}{s^2} + \frac{1}{s+4} = \frac{2}{s^2} + \frac{1}{s+4} \]
04

- Transformation of \(f(t)=\mathrm{e}^{-\phi t}\cdot\sin(\omega t)\)

Using the transformation definition for the product \(\mathrm{e}^{-\phi t}\cdot\sin(\omega t)\)\[ F(s) = \int_{0}^{\infty} \mathrm{e}^{-(s+\phi)t} \sin(\omega t) \ \mathrm{d}t \]This evaluates to: \[ F(s) = \frac{\omega}{(s+\phi)^{2} + \omega^{2}} \]
05

- Transformation of \(f(t)=\sinh(at)\)

We need to find Laplace transform of \(f(t)=\sinh(at)\):The Laplace transform of hyperbolic sine is: \[ \text{L}\{\sinh(at)\} = \frac{a}{s^{2} - a^{2}} \]
06

- Transformation of \(f(t)=t^{3}\)

Using the known Laplace Transform for \(t^n\), \( \text{L}\{t^n\} = \frac{n!}{s^{n+1}} \):For \(n=3:\) \[ F(s) = \text{L}\{t^{3}\} = \frac{3!}{s^{4}} = \frac{6}{s^{4}} \]
07

- Transformation of \(f(t)=\sin^{2}(t)\)

Using trigonometric identity \(\sin^{2}t = 0.5(1-\cos(2t))\)Then, transform each term separately:\[ \text{L}\{0.5-0.5\cos(2t)\} = 0.5 \frac{1}{s} - 0.5 \frac{s}{s^{2} + 4} \]Thus: \[ F(s) = 0.5 \left(\frac{1}{s} - \frac{s}{s^2 + 4} \right) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laplace Transform definition
The Laplace Transform is a powerful tool in engineering mathematics. It transforms a time-domain function, often shown as \( f(t) \), into an s-domain function, denoted by \( F(s) \). This transformation is given by the integral: \[ \text{L}\{f(t)\} = F(s) = \int_{0}^{\infty} f(t) \mathrm{e}^{-st} \ \mathrm{d}t \]It is particularly useful for solving linear differential equations and for analyzing systems in control engineering.By converting differential equations into algebraic equations, it simplifies solving them. A solved Laplace transform provides insight into the behavior of systems in the s-domain, which is easier to work with than the time domain.
Transforming trigonometric functions
Transforming trigonometric functions like sine and cosine into the s-domain can show their behavior more clearly. For example, let's find the Laplace transform of \( \cos(\omega t) \):
Start with the definition: \[ \text{L}\{\cos(\omega t)\} = \int_{0}^{\infty} \cos(\omega t) \mathrm{e}^{-st} \ \mathrm{d}t \]Using integration properties, you get: \[ \text{L}\{\cos(\omega t)\} = \frac{s}{s^{2} + \omega^{2}} \]Similarly, for transforms involving \( \sin(\omega t) \), the process is similar but results in different denominators. These results help in understanding oscillatory systems and their behavior in the frequency domain.
Transforming exponential functions
Exponential functions are common in engineering, especially for modeling growth and decay. The Laplace Transform of \( \mathrm{e}^{-at} \) is straightforward:\[ \text{L}\{\mathrm{e}^{-at}\} = \frac{1}{s+a} \]This simplicity allows for easy manipulation of exponential models in the s-domain. When dealing with linear combinations like \( f(t) = 2t + \mathrm{e}^{-4t} \), use the linearity property:\[ \text{L}\{2t + \mathrm{e}^{-4t}\} = 2 \text{L}\{t\} + \text{L}\{\mathrm{e}^{-4t}\} = \frac{2}{s^2} + \frac{1}{s+4} \]This method allows for compact and clear representations of complex functions.
Transforming polynomial functions
Polynomial functions can be transformed using a standard transform for \( t^n \). The Laplace transform of a function like \( t^3 \) is:\[ \text{L}\{t^3\} = \frac{3!}{s^{4}} = \frac{6}{s^{4}} \]This rule generalizes for any \( t^n \), showing that the transform of a polynomial results in a factorial in the numerator and a power of s in the denominator. This simplification aids in working with higher-order differential equations, making them easier to solve.
Hyperbolic functions Laplace Transform
Hyperbolic functions such as \( \sinh(at) \) also have specific Laplace transforms. For \( \sinh(at) \), the transform is:\[ \text{L}\{\sinh(at)\} = \frac{a}{s^{2} - a^{2}} \]These transforms are useful in various physics and engineering applications, such as heat transfer and wave equations. They follow similar principles as the transforms for trigonometric functions but differ due to the distinct nature of hyperbolic functions.
Hyperbolic functions provide insights into various behaviors in engineering, enhancing the analysis of non-oscillatory but exponentially varying processes.

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Most popular questions from this chapter

Unterwerfen Sie die folgenden linearen Differentialgleichungen mit konstanten Koeffizienten der Laplace-Transformation und bestimmen Sie die Bildfunktion \(Y(s)=\mathscr{Z}\\{y(t)\\}\) der allgemeinen Lösung \(y(t):\) a) \(3 y^{\prime}+2 y=t\) b) \(y^{\prime \prime}+2 y^{\prime}+y=\cos (2 t)\)

Lösen Sie die folgenden Anfangsweriprobleme: a) \(y^{\prime}-y=\mathrm{e}^{\prime}, \quad y(0)=1\) b) \(y^{\prime}+3 y=-\cos t, \quad y(0)=5\) c) \(y^{\prime}-5 y=2 \cdot \cos t-\sin (3 t), \quad y(0)=0\)

Ein elektrischer Schwingkreis mit einer Induktivität \(L\) und einer Kapazität \(C\) wind durch die sinusförmige Spannung \(u_{a}=u_{0} \cdot \sin (\omega t)\) zu Schwingungen angeregt. Bei der mathematischen Behandlung dieses Problems mit Hilfe der Laplace-Transformation erh?lt man im sog. Resonanzfall für die Stromstärke \(i(t)\) eine Bildfunktion vom Typ $$ F(s)=\frac{s}{\left(s^{2}+a^{2}\right)^{2}}=\left(\frac{1}{s^{2}+a^{2}}\right) \cdot\left(\frac{s}{s^{2}+a^{2}}\right) $$ Bestimmen Sie durch Anwendung des Faltungssatzes die zugehörige Originalfunktion \(f(t)=\mathscr{L}^{-1}\\{F(s)\\}\)

Die Funktion \(f(t)=\sin (\omega t)\) ist eine Lösung der Schwingungsgleichung \(f^{\prime \prime}=-\omega^{2} \cdot f\). Bestimmen Sie hieraus die zugeh?rige Bildfunktion \(F(s)\). indem Sie die Laplace-Transformation auf diese Differentialgleichung anwenden und dabei den Ableitungssatz fir Originalfunktionen verwenden.

Lösen Sie die inhomogene lineare Differentialgleichung 2. Ordnung \(y^{\prime \prime}+2 y^{\prime}+y=\) \(t+\sin t\) für die Anfangswerte \(y(0)=1\), \(y^{\prime}(0)=0\) a) durch Aufsuchen einer partikulären Lösung, b) mit Hilfe der Laplace-Transformation.
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