Fourier mistake

  • Thread starter jennyjones
  • Start date
  • #1
35
0
can someone spot my mistake, i'm stuck

thanks,

jenny
 

Attachments

Answers and Replies

  • #2
BruceW
Homework Helper
3,611
120
I don't think step 3 is correct. the original function is not zero for |x| > a. But it looks like you assume that so that you can make the integral over this range, instead of from minus infinity to infinity. At least, that looks like what you had in mind...
 
  • Like
Likes 1 person
  • #3
BruceW
Homework Helper
3,611
120
Also, what definition are you using for the fourier transform? Because I get the feeling they have used a different convention.
 
  • Like
Likes 1 person
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,874
1,449
When ##\lvert\omega\rvert < a##, you have
$$\int_0^\infty \frac{\sin ax\cos \omega x}{x}\,dx = \frac{\pi}{2}.$$ When ##\lvert\omega\rvert > a##, you have
$$\int_0^\infty \frac{\sin ax\cos \omega x}{x}\,dx = 0.$$ Somehow, you have to figure out what happens when ##\lvert\omega\rvert = a##.
 
  • Like
Likes 1 person
  • #5
35
0
bruce W i'm using the cosine transform, i made a picture of this formula for my textbook.

Vela, do you know if i can than say |ω|= (∏/2+0)/2=∏/4 ?

the average?
 

Attachments

  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,874
1,449
Seems like it might, like the Fourier series does at discontinuities. Do you know of a theorem that establishes the same result for the Fourier transform?
 
  • Like
Likes 1 person
  • #7
BruceW
Homework Helper
3,611
120
I don't think step 3 is correct. the original function is not zero for |x| > a. But it looks like you assume that so that you can make the integral over this range, instead of from minus infinity to infinity. At least, that looks like what you had in mind...
Ah whoops, ignore this. I thought you wrote an ##a## for the upper limit, but it is an ##\infty##, as it should be.

jennyjones said:
bruce W i'm using the cosine transform, i made a picture of this formula for my textbook.
right. yes, you are using the same definition of the Fourier transform as they are. Which is good :)

jennyjones said:
Vela, do you know if i can than say |ω|= (∏/2+0)/2=∏/4 ?

the average?
I think you meant to say the average of the values of the Fourier transform on either side of the point ##\omega=a##. If this is what you meant, then yes that's right. Was it a guess? You have good intuition if it was. Yeah, there is a specific theorem (which is pretty hard to find on the internet), as vela is hinting at. This theorem works for certain kinds of function, like the rectangular function.
 
  • #8
35
0
Yey! thank you, than i solve the problem now!

Do you maybe know the name of this theorem?

jenny
 
  • #9
BruceW
Homework Helper
3,611
120
No worries :) I think wikipedia said it is a form of the fourier inversion theorem. They didn't give a direct link for it though. It's on the wikipedia page "fourier inversion theorem", about halfway down under the subtitle "piecewise smooth; one dimension", if you are interested. It looks like the book "Fourier Analysis and its Applications" by Folland, G. B. might have some more information.
 

Related Threads on Fourier mistake

  • Last Post
Replies
1
Views
982
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
833
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
6
Views
2K
Top