Competing definitions of the Fourier transform

In summary, the Fourier transforms are two equivalent definitions for the same mathematical operation, which takes a function and transforms it into a series of "points" or "values." The factor of 1/\sqrt{2\pi} is added into the exponents of the two definitions to make sure that the transforms and their inverse don't multiply the result by a constant factor other than 1. If you leave the factors of 2\pi out of the definitions, they appear in the inverse transform.
  • #1
AxiomOfChoice
533
1
Just began a serious study of the Fourier transform with a couple of books. One of them defines the Fourier transform on [itex]\mathbb R[/itex] as

[tex]
\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{-i\xi x}dx.
[/tex]

Another defines it as

[tex]
\hat f(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i \xi x} dx.
[/tex]

A few questions:

(1) Are these definitions somehow equivalent? I cannot seem to obtain the second from the first by making a simple change of variables.

(2) Why worry about the factors of [itex]2\pi[/itex] in the definitions? What does that do for us? Why not leave those out altogether?
 
Physics news on Phys.org
  • #2
See M.L.Boas - "Mathematical Methods" for disambiguation. I had the same problem and it really helped me.
 
  • #3
The definitions are equivalent, and the factor of [itex]1/\sqrt{2\pi}[/itex] is added into ensure that applying the transform and its inverse doesn't mulitply the result by a constant factor other than 1. You need to know whether the [itex]2\pi[/itex] is in the exponential or not to figure out this normalization constant.
 
  • #4
AxiomOfChoice said:
(2) Why worry about the factors of [itex]2\pi[/itex] in the definitions? What does that do for us? Why not leave those out altogether?

If you leave them out of the definition of the Fourier transform, they appear in the inverse transform. If you try to leave them all out, then IFT ( FT [ f(x) ] ) doesn't equal f(x).

Unfortunately there isn't an "obvious" place to put the ##2\pi## that everybody agrees on, so you have to check what convention any particular book or paper is using.

Similar issues apply to the discrete Fourier transform, and computer software routines that calculate it. There you also have to watch out for factors of n and 1/n, where n is the number of samples in the DFT.
 
  • #5
Oh. that is easily solved. Make change of variables in the familiar transform you understand clearly to include the 2[itex]\pi[/itex] in the exponent, THEN see if the product of coefficients of Fourier and inverse Fourier transforms gives the factor claimed. if they are the same, this is correct, if it is not, it got to be wrong.
 

Related to Competing definitions of the Fourier transform

1. What is the Fourier transform?

The Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It is used to analyze signals and images in various fields, including physics, engineering, and mathematics.

2. What are the competing definitions of the Fourier transform?

The two competing definitions of the Fourier transform are the continuous Fourier transform and the discrete Fourier transform. The continuous Fourier transform is used for continuous signals and functions, while the discrete Fourier transform is used for discrete signals and functions.

3. What is the difference between the continuous and discrete Fourier transform?

The main difference between the continuous and discrete Fourier transform is the type of input signal they can process. The continuous Fourier transform can handle continuous signals and functions, while the discrete Fourier transform is designed for discrete signals and functions.

4. Which definition of the Fourier transform is more commonly used?

The answer to this question depends on the application. In fields such as signal processing and image analysis, the discrete Fourier transform is more commonly used due to its ability to handle discrete signals and functions. However, in mathematical analysis and theoretical physics, the continuous Fourier transform is often preferred.

5. Can the continuous and discrete Fourier transform be used interchangeably?

No, the continuous and discrete Fourier transform cannot be used interchangeably. They are two distinct mathematical operations with different inputs and outputs. However, in some cases, the discrete Fourier transform can be approximated to the continuous Fourier transform by using a large number of samples.

Similar threads

I On deriving the (inverse) Fourier transform from Fourier series
    • Topology and Analysis
Replies
4
Views
382
I Regarding Wirtinger's inequality
    • Topology and Analysis
Replies
4
Views
359
I Fourier coefficients of convolution
    • Topology and Analysis
Replies
1
Views
462
I On differentiability and Fourier coefficients (Vretblad's text)
    • Topology and Analysis
Replies
4
Views
441
I Proof check for Plancherel's theorem (Fourier Transform version)
    • Topology and Analysis
Replies
1
Views
1K
I On limit of convolution of function with a summability kernel
    • Topology and Analysis
Replies
2
Views
452
I What's the definition of "periodic extension of a function"?
    • Topology and Analysis
Replies
14
Views
3K
A Polar Fourier transform of derivatives
    • Differential Equations
Replies
4
Views
2K
I How to derive the Fourier transform of a comb function
    • Topology and Analysis
Replies
4
Views
2K
MHB Can the Fourier Transform Prove the Fundamental Solution for a Heat Equation?
    • Topology and Analysis
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top