How do I convert f(x) into its Fourier Transform?

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Homework Help Overview

The discussion revolves around converting a function involving Dirac delta functions into its Fourier Transform, specifically in the context of electrodynamics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to integrate the function multiplied by an exponential term to find the Fourier Transform. Some participants question the integration steps and properties of the Dirac delta function.

Discussion Status

Participants are exploring the integration of the Dirac delta functions and discussing properties related to the Fourier Transform. Some guidance has been provided regarding the properties of the delta function, but there is no explicit consensus on the approach yet.

Contextual Notes

The original poster expresses confusion about the problem and lacks clear equations or methods to proceed, indicating a need for clarification on the Fourier Transform process.

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Homework Statement



I am really confused in my electrodynamics class. I have the following function.

f(x) = \delta (x + \alpha ) + \delta(x -\alpha)

How do i convert this into Fourier Tranform ?
Those are dirac delta functions on either sides of the origin.


Homework Equations



no clue

The Attempt at a Solution



I multiplied f(x) by 1/sqrt(2*pi)*e-ikt and integrated the whole thing.
 
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And what did you get?
 
1/\sqrt{2\pi}\int\delta(x+\alpha)e-ikx + \delta(x-\alpha)e-ikx dx

I tried to integrate those terms by parts (uv substitution).

\int\delta(x+\alpha) = 1. Isn't that right ?
Also I think f(x)\int\delta(x+\alpha) = f(\alpha). Am i right ?
 
All you need is the following property:

\int\limits_{-\infty}^{+\infty} f(x)\delta\left(x-a\right)=f(a)
 
so

\inte-ikx \delta (x - a) = e-ika ?
 
Exactly.
 

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