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Berechnen Sie die FOURIER-Transformierte der Funktion \(f(t)=\frac{1}{a^{2}+t^{2}}, a>0\).

Short Answer

Expert verified
The Fourier transform of \( f(t)=\frac{1}{a^{2}+t^{2}} \) is \( F(\omega) = \pi \frac{e^{-\omega a}}{a} \).

Step by step solution

01

Understand the Problem

We are tasked with finding the Fourier transform of the function \( f(t) = \frac{1}{a^2 + t^2} \). This is an integration problem where we use the formula for the continuous Fourier transform of a function \( f(t) \).
02

Write the Formula for Fourier Transform

The formula for the Fourier transform \( F(\omega) \) of a function \( f(t) \) is given by:\[ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \, dt \] We will substitute \( f(t) = \frac{1}{a^2 + t^2} \) into this integral.
03

Set Up the Integral

Substitute \( f(t) = \frac{1}{a^2 + t^2} \) into the Fourier transform formula:\[ F(\omega) = \int_{-\infty}^{\infty} \frac{1}{a^2 + t^2} e^{-i \omega t} \, dt \] This is an improper integral that can be solved using contour integration methods.
04

Solve the Integral Using Contour Integration

To solve this integral, consider the complex function \( f(z) = \frac{e^{-i \omega z}}{a^2 + z^2} \) and integrate over a semicircular contour in the upper half-plane. The poles are located at \( z = ai \) and \( z = -ai \). Using the residue theorem, only the pole \( z = ai \) contributes for integration along the real line.
05

Calculate the Residue at the Pole

The residue at the pole \( z = ai \) can be calculated as:\[ \text{Res}(f, ai) = \lim_{z \to ai} (z - ai) f(z) \]Simplifying, this leads to the residue being \( \frac{e^{\omega a}}{2ai} \).
06

Apply the Residue Theorem

Use the Residue theorem which states \[ \oint_C f(z) \, dz = 2\pi i \sum \text{Res}(f, \text{poles in } C) \]Here, along the real line, this simplifies to:\[ F(\omega) = 2 \pi i \left( \frac{e^{-\omega a}}{2ai} \right) \]
07

Simplify the Expression

Simplify the result from the residue theorem:\[ F(\omega) = \pi \frac{e^{-\omega a}}{a} \] This gives us the Fourier transform of the function.

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Key Concepts

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

Contour Integration
When solving integrals, especially those that extend from negative to positive infinity, contour integration provides a powerful tool. This method leverages complex analysis to evaluate integrals by considering them over certain paths in the complex plane. Unlike straightforward real integrals, contour integration allows us to circumvent challenging or even undefined parts along the real line by considering a path that extends into the complex plane.
The contour is often shaped like a semicircle, surrounding regions that include the poles of the integrand. In the exercise of finding the Fourier transform of the function \( f(t) = \frac{1}{a^2 + t^2} \), the integral is first defined over the real line and then extended into the upper half of the complex plane to form a closed contour. This method is crucial for simplifying the calculations by focusing on contributions from singularities within the closed path.
Residue Theorem
The Residue Theorem is a core component of complex analysis and is extremely useful in computing complex integrals, especially in the context of contour integration. The theorem essentially relates the integral of a complex function over a closed contour to the sum of residues inside the contour. A residue is a particular complex number that provides insights into the behavior of a function near its singularities (poles). In the exercise example, the poles of the function \( f(z) = \frac{e^{-i \omega z}}{a^2 + z^2} \) are located at \( z = ai \) and \( z = -ai \). For integration involving the upper half-plane, only the pole at \( z = ai \) affects the contour integral outcome.
By applying the Residue Theorem, we focus on this singularity, calculating the residue, which leads to the final solution of the original Fourier transform problem. This approach not only simplifies the complex integral but also ensures that the calculated values adhere to mathematical rigor and accuracy.
Improper Integral
An improper integral is defined as an integral with one or both of its limits extending to infinity. It can also refer to an integral with a singularity in its interval. These integrals arise often in mathematical physics and engineering. Evaluating them typically involves extending the domain into the complex plane, making tools like contour integration and the Residue Theorem essential.In this specific exercise, evaluating the Fourier transform involves such an improper integral given by \( \int_{-\infty}^{\infty} \frac{1}{a^2 + t^2} e^{-i \omega t} \ dt \). Due to the infinite limits, one cannot simply compute this by standard techniques of definite integrals. Instead, by transforming the integral using contour integration, we handle the difficulties presented by infinity and singularities systematically.
This conversion helps not only in describing the original problem in a more approachable format but also in finding solutions that are simple and elegant.
Complex Analysis
Complex analysis is an elegant branch of mathematics that studies complex numbers and the functions that operate on them. It opens doors to solving real integrals that are otherwise difficult to tackle, such as those encountered in Fourier transform examples. Key concepts include complex differentiability and integrals, power series expansions, and particularly important theorems like Cauchy's integral theorem and the Residue Theorem.In the scenario of calculating Fourier transforms, complex analysis provides sophisticated techniques through which real-world problems are transformed into manageable ones. For instance, translating the Fourier transform of \( f(t) = \frac{1}{a^2 + t^2} \) into a contour integral simplifires dealing with infinite limits and singularities, making the solution tractable.
Through such methods, complex analysis not only reveals powerful mathematical insights but also significantly broadens the scope of solvable problems beyond what's possible with real analysis alone. This field ultimately serves as a foundational cornerstone for advancements in engineering, physics, and applied mathematics.

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Most popular questions from this chapter

Berechnen Sie das Faltungsprodukt \((f * g)(t)=\int_{-\infty}^{\infty} f(t-\tau) g(\tau) d \tau\) für \(f(t)=\) \(e^{-c|t|}\) und \(g(t)=\cos (\omega t)\), wobei \(c>0\) gilt.

Sei \(U: \mathbb{R} \times[0, \infty[\rightarrow \mathbb{R}\) die Lösung des Anfangswertproblems $$ \begin{gathered} \frac{\partial U}{\partial t}(x, t)=4 \frac{\partial^{2} U}{\partial x^{2}}(x, t) \\ \lim _{x \rightarrow \pm \infty} U(x, t)=\lim _{x \rightarrow \pm \infty} \frac{\partial U}{\partial x}(x, t)=0, \text { f.a. } t>0 \\ U(x, 0)=\frac{1}{x^{2}+9}, \text { f.a. } x \in \mathbb{R} \end{gathered} $$ Verwenden Sie die FOURIER-Transformierte bezüglich \(x\), um \(U(x, t)\) zu berechnen.

Lösen Sie mit Hilfe der LAPLACE-Transformation das Anfangswertproblem $$ \begin{gathered} u^{\prime}(t)=u(t)+4 v(t)+e^{t} \\ v^{\prime}(t)=u(t)+v(t)+e^{t} \\ u(0)=-\frac{1}{2}, v(0)=1 . \end{gathered} $$

Berechnen Sie die FOURIER-Transformierten \(F=\mathcal{F}[f]\) und \(G=\mathcal{F}[g]\), wobei $$ f(t)=\chi_{[-1,1]}(t)\left(1-t^{2}\right)= \begin{cases}\left(1-t^{2}\right) & \text { falls }-1 \leq t \leq 1 \\ 0 & \text { sonst }\end{cases} $$ und \(^{1}\) $$ g(t)=\chi_{[-3,-1]}(t)+\chi_{[1,3]}(t)= \begin{cases}1 & \text { falls }-3 \leq t \leq-1 \text { oder } 1 \leq t \leq 3 \\ 0 & \text { sonst. }\end{cases} $$ Nutzen Sie die Ergebnisse zur Bestimmung der FOURIER-Transformierten von \(H(x)=\frac{\sin x \cos 2 x}{x}\)

Zeigen Sie für eine ungerade Funktion \(f(t)\) die Beziehung $$ \mathcal{F}[f](\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(t) e^{-i \omega t} d t=-\frac{i}{\pi} \int_{0}^{\infty} f(t) \sin (\omega t) d t $$

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