Why is Fourier Integral Meaningless for f(x)=A*cos(ax)?

[caduceus]

I couldn't understand that why the Fourier integral is meaningless for f(x)=A*cos(ax) ?

Any comments will be appreciated.

 

[mathman]

|f(x)| is not integrable over (-oo,oo). so the Fourier integral cannot be properly defined.

 

[DeepSeeded]

I couldn't understand that why the Fourier integral is meaningless for f(x)=A*cos(ax) ?

Any comments will be appreciated.

Because that is a sinusoidal function. What is there to analyze?

You want to find the sinusoidal functions that make up a non sinusoidal function.

 

[caduceus]

Oh, I guess I can see the point now. You mean square integrability. That is why I should use Delta function. Thank you.

 

[Peeter]

Prof Brad Osgood has some excellent Fourier transform lectures on iTunesU
(The Fourier Transform and its Applications).

In particular, there's a few lectures (~lectures 10-14 I think) on distributions and Schwartz functions that help show how Fourier transforms of sines, exponentials, deltas, ... can be better justified. Suprisingly (to me after having seen and given up trying to understand Functional analysis), the basics required for application are not actually all that difficult, mostly requiring a change in approach, and an extra level of indirection.

Some lecture notes to go with the lectures can be found here:

http://www.stanford.edu/class/ee261/book/all.pdf

(chapter 4 covers distributions).

 

[Meir Achuz]

I couldn't understand that why the Fourier integral is meaningless for f(x)=A*cos(ax) ?

Any comments will be appreciated.

The Fourier integral gives two delta functions. That is good enough for physicists.

 

[akshaythegeek]

well i am a computer guy :) so i don't know...ok here is my question
following are two equations


$fk(k1) = \frac{1}{nj} \sum_{j=1}^{nj} c_j(j) \exp (i k1 x_j(j))$\\

$where \frac {-ms}{2} <k1 < \frac{ms-1}{2}$

a) what is k1
b) what is x_j(j)
c) what is c_j(j)

Is this a forward transform

d) What will be the inverse transform, and is inverse tranform means we are evaluating Fourier series

 

[Peeter]

that's not really readable as is. Can you edit with [ tex ] [ / tex ] (no spaces), replacing the dollar signs.

 

[akshaythegeek]

I am experimenting with non uniform sampling, I applied Fast Fourier transform on the non uniform sampling in MATLAB it has given me some results. I can't understand how FFT runs on non uniform sampling. What i am getting after applying FFT on Non uniform samples is what...is it a errorfull value if yes then why