Necessary and sufficient conditions: Fourier Transform

In summary, the Fourier transform of a function that is "absolutely integrable" is not necessarily a function that exists on a space of absolutely integrable functions. However, the Fourier transform of a function that is not absolutely integrable can exist as a tempered distribution.
  • #1
klondike
125
2
I don’t know if the question belongs to engineering or math but here it goes.
I was taught that a sufficient (not necessary) condition for existence of Fourier transform of f(t) is f(t) is absolutely integratble. I was wondering what are the “necessary and sufficient conditions” for FT of f(t) to exist. Some textbook said “it can be very involved” but what is that?
 
Physics news on Phys.org
  • #2
This depends on what you mean by "Fourier transform." Usually (but there are many exceptions), the Fourier transform F is defined on the space L^1 of absolutely integrable functions. Are you trying to find the 'largest' domain for F?
 
  • #3
morphism, thank you for your reply.
morphism said:
This depends on what you mean by "Fourier transform."

The Fourier Transform which I was taught is:

[tex]
F(\omega)=\int _{-\infty}^{+\infty}f(t)e^{-j\omega t}dt
[/tex]
the one that I use to convert signals from timedomain to analyze and process in freq domain. All physically realizable signals that I have to deal with are "absolutely integrable".

Other forms of FT that I'm familiar with are those 2D or 3D FT that are used to solve linear multi-dimensional PDEs.

morphism said:
Usually (but there are many exceptions), the Fourier transform F is defined on the space L^1 of absolutely integrable functions.

Can you please show an example of exceptions or point me to some URLs for further reading?

morphism said:
Are you trying to find the 'largest' domain for F?

No. I was trying to find a signal (function of time) that doesn't satisfy "absolutely integrable" condition, but some sort of FT exists for such signal, for the sake of curiosity. As an analytical electrical engineer, I love engineering math in general but certainly can't afford much time digging into hardcore math.

thanks again.
 
Last edited:
  • #4
Hmm. I don't know if this is what you're looking for, but try to look into Fourier transforms of distributions.
 
  • #5
Hmm, let's see. My problem is to find a function that is NOT "absolutely integrable" but its Fourier transform exists. If I can find such function, that means "absolutely integrable" is not necessary condition for FT to exist.

Given the problem, let's say

[tex]
F(\omega)=\int_{-\infty}^{+\infty}f(t)e^{-j\omega t}dt
[/tex]
[tex]
f(t)=\dfrac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)e^{j\omega t}d\omega
[/tex]

let
[tex]
f(t)=\cos{\Omega t}
[/tex]
And clearly, f(t) is not absolutely integrable, but its FT does exist:

[tex]
F(\omega)=\pi \delta (\omega -\Omega)+ \pi \delta (\omega+\Omega)
[/tex]

Make sense?
 
  • #6
klondike said:
And clearly, f(t) is not absolutely integrable, but its FT does exist:

[tex]
F(\omega)=\pi \delta (\omega -\Omega)+ \pi \delta (\omega+\Omega)
[/tex]

Make sense?
That depends very much on what you mean by 'exist'! At the most elementary level (i.e. the math you learned in your basic calculus sequence in college), that Fourier transform doesn't exist.

To be able to use 'generalized functions' (such as the Dirac delta), you have to set up some additional mathematical scaffolding. And then to use the Fourier transform on them (or even things like derivatives), that's a bit more scaffolding.

If Wikipedia is to be trusted, if the Fourier transform is of primary interest, then the right scaffolding is the space of tempered distributions. The natural definition for the Fourier transform of a tempered distribution is implicitly through the following integral equation:

[tex]\int_{-\infty}^{+\infty} \mathcal{F} \{ f \} \phi = \int_{-\infty}^{+\infty} f \mathcal{F} \{ \phi \}[/tex]

where f denotes any tempered distribution and [itex]\phi[/itex] denotes any 'well-behaved' function. It turns out that the Fourier transform of a tempered distribution is a tempered distribution -- so in this context, Fourier transforms always exist.


Every 'somewhat well-behaved' function can be viewed as a tempered distribution -- the basic requirement is that it doesn't grow 'too fast' at infinity. And every tempered distribution can be written as a limit of distributions that came from somewhat well-behaved functions. (e.g. you do this when you write the Dirac delta as a 'limit' of ordinary functions)
 

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 convert a signal from its original domain (often time or space) to a representation in the frequency domain, where it can be analyzed in terms of its individual frequency components.

2. What are the necessary conditions for using the Fourier Transform?

The necessary conditions for using the Fourier Transform are that the function must be continuous and have a finite integral. In other words, the function must be well-defined and have a finite amount of energy or information.

3. What are the sufficient conditions for using the Fourier Transform?

The sufficient conditions for using the Fourier Transform are that the function must be absolutely integrable and have a finite number of discontinuities. This means that the function must be well-behaved and not have any abrupt changes or jumps.

4. How is the Fourier Transform calculated?

The Fourier Transform is calculated using an integral that sums up the contributions of each frequency component in the function. This integral, known as the Fourier Integral, is defined by a complex exponential function and can be evaluated using various mathematical techniques such as integration by parts.

5. What are the applications of the Fourier Transform?

The Fourier Transform has numerous applications in many fields, including signal processing, image processing, audio processing, and data compression. It is also used in solving differential equations and studying the behavior of physical systems. Additionally, the Fourier Transform is an essential tool in understanding the properties of waves and oscillatory phenomena.

Similar threads

Replies
3
Views
964
  • Calculus
Replies
2
Views
1K
Replies
7
Views
2K
Replies
11
Views
813
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
165
  • Calculus
Replies
3
Views
1K
Back
Top