Register to reply

Fourier Transform Convergence

by nabeel17
Tags: convergence, fourier, transform
Share this thread:
nabeel17
#1
Dec8-13, 06:59 PM
P: 50
From -infinity to infinity at the extreme ends do Fourier transforms always converge to 0? I know in the case of signals, you can never have an infinite signal so it does go to 0, but speaking in general if you are taking the fourier transform of f(x)

If you do integration by parts, you get a term (f(x)e^ikx evaluated from -infinity to infinity why does this always = 0?
Phys.Org News Partner Physics news on Phys.org
Step lightly: All-optical transistor triggered by single photon promises advances in quantum applications
The unifying framework of symmetry reveals properties of a broad range of physical systems
What time is it in the universe?
DaleSpam
#2
Dec8-13, 07:56 PM
Mentor
P: 17,318
Quote Quote by nabeel17 View Post
From -infinity to infinity at the extreme ends do Fourier transforms always converge to 0?
No, not always. If a signal is periodic in one domain then it is discrete in the other domain. So if you have a signal which is discrete in time, then it is periodic in frequency. Since it is periodic in frequency it does not converge to 0 at infinity.
nabeel17
#3
Dec8-13, 08:07 PM
P: 50
Quote Quote by DaleSpam View Post
No, not always. If a signal is periodic in one domain then it is discrete in the other domain. So if you have a signal which is discrete in time, then it is periodic in frequency. Since it is periodic in frequency it does not converge to 0 at infinity.
Ok then why is it that we the first term in the integration by parts goes to 0 then regardless of the function (Whether it is periodic or not)? For example when finding the fourier transform of a derivative F[d/dx] = ∫d/dxf(x)e^ikx= f(x)e^ikx evaluated -infinity to infinity -ik∫f(x)e^ikx

the first term = 0, why is that? If it were a wave function like in QM then it makes sense because the area under the wave function must be finite and converge to 0 at the extremes for it to have a probability density, but why here?

DaleSpam
#4
Dec8-13, 08:53 PM
Mentor
P: 17,318
Fourier Transform Convergence

Quote Quote by nabeel17 View Post
Ok then why is it that we the first term in the integration by parts goes to 0 then regardless of the function (Whether it is periodic or not)? For example when finding the fourier transform of a derivative F[d/dx] = ∫d/dxf(x)e^ikx= f(x)e^ikx evaluated -infinity to infinity -ik∫f(x)e^ikx

the first term = 0, why is that? If it were a wave function like in QM then it makes sense because the area under the wave function must be finite and converge to 0 at the extremes for it to have a probability density, but why here?
I think that the various properties of the Fourier transform all assume that f satisfies the Dirichlet conditions.
AlephZero
#5
Dec8-13, 09:20 PM
Engineering
Sci Advisor
HW Helper
Thanks
P: 7,169
The OP is asking about Fourier transforms, not Fourier series (of periodic functions) which is what #2 and #4 appear to be about.

A reasonable condition for Fourier transforms to behave sensibly is that ##\int_{-\infty}^{+\infty}|f(x)|dx## is finite. Note that if you use Lebesque measure to define integration, that does not imply ##f(x)## converges to 0 as x tends to infinity. ##f(x)## can take any values on a set of measure zero.

(Also note, "reasonable" does not necessarily mean either "necessary" or "sufficient"!)

The mathematical correspondence between Fourier series and Fourier transforms is not quite "obvious", since the Fourier transform of a periodic function (defined by an integral with an infinte range) involves Dirac delta functions, and indeed the Fourier transform of a periodic function is identically zero except on a set of measure zero (i.e. the points usually called the "Fourier coefficients").

On the other hand if you integrate over one period of a periodic function, it is a lot simpler to get to some practical results, even if you have to skate over why the math "really" works out that way.


Register to reply

Related Discussions
Fourier Transform using Transform Pair and Properties Calculus & Beyond Homework 0
Computing the Hilbert transform via Fourier transform Math & Science Software 2
Fourier transform, Fourer Integral transform Calculus & Beyond Homework 5
Convergence problem for Fourier Transform of 1/t Calculus & Beyond Homework 0
The difference between Fourier Series, Fourier Transform and Laplace Transform General Physics 1