Issue in a fourier transform of a function

[claudiocamera]

Fourier transform

The Fourier transform of a given function f(x) is F(jw) , if f(x)>0 for every x, Is it correct to say that F(jw) has a maximum at the origin ? If so, why ?

 

[LouisSheffield]

Fourier transform

Is f(x) allowed to be complex (with real component > 0)? If so, what are the rules for the imaginary component?

 

[claudiocamera]

Fourier transform

For instance consider f(x) the gaussian function. If f(x) = Exp( (x^2)/2) it is always positive, in this case,F(jw) has its maximum at w=0, if the gaussian function is shifted, for instance f(x) = exp{ ((x-2)^2)/2} it continues to be always positive and its fourier transform continues having a maximum at w=0.

Another instance f(x) = a , where a is a constant and a>0. The fourier transform has a maximum at w=0, so it comes my question:

if f(x)>0 for every x, Is it correct to say that F(jw) has a maximum at the origin ? Is it always true ? if so , what is the mathematical proof ?

 

[LouisSheffield]

Re: Fourier transform

For f(x) > 0 (real only) it must be maximum at zero-frequency.

It can be visualized by considering the Fourier transform as an integrating filterbank.
Any non-zero frequency solved for by the Fourier transform will modulate some of its finite energy into the +/- imaginary planes (due to the Exp(i...) term).
Integrating, the total energy will be reduced, since its phase was not zero everywhere at that frequency.

In the limit (so as to not have f(x) = 0) of f(x) becoming a delta function (for example Exp(-1000000 x^2)), the energy of the non-zero frequencies will approach that of the zero frequency, but can never exceed it.

 

[thecreator]

Re: Fourier transform

VISUALISING THE FOURIER TRANSFORM CAN HELP YOU A BIT
SINCE f(X)>0 for x>0
we can say that the average value of x is going to be positive,
actually the zero frequency magnitude is the average value of the
signal in consideration. It has to be maximum sine freqencies other than 0 have
negative areas.

 

[claudiocamera]

Re: Fourier transform

It has to be maximum sine freqencies other than 0 have
negative areas.

I didn't understand this part, why they have negative areas ? I can't visualize it.

 

[matchasm]

Re: Fourier transform

For a real-valued input signal, it may be easier to consider the linear property of the Fourier transform, i.e. if h(x)=f(x)+g(x), then H(w)=F(w)+G(w).

Now look at the table on

https://mathworld.wolfram.com/FourierTransform.html


The transform of any DC or positive constant function is a delta function. You will see that any other waveform with an amplitude equal to this constant function (in order to make a minimum of zero), will have also a negative frequency component and thus at most a peak half the size of the constant function.

(ignore the ramp function!! )

 

[Sal]

Fourier transform

Hi

I agree with the last post, if f(x) > 0, the you can make f(x)=A + g(x), where g(x) is function which has positive and negative values, now applying the Fourier transform you will find the delta function for F(A), just you need to go further and prove that this value is greater that the other frequency components

Cheers

Sal

 

[DugalMc]

Re: Fourier transform

Hi, this is not a rigorous proof ...

Max(F(ω)) = ∫|f(x)exp(-jωx)dx| = ∫f(x)dx = F(0) (given f(x) >0)