Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2011-12-20 04:26:51

model
Member
Registered: 2011-08-10
Posts: 142

Fourier Descriptors : reduce 2D to 1D problem .How ?

Fouier Descriptor representation has an great advantage . It reduce a 2D to 1D problem ./ How ?  what does that mean actually ?

Does it mean Fourier Descriptor ( that are actually  defined by complex numbers ) define 1D function ?

smaller the fourier co-efficient : lose of details and global  shape of boundary  of an object .
higher the Fourier co-efficient : more details and high frequency  components .

these descriptor retain the principal shape feature of the original boundary .  How can last figure( see image in below link )  retain the original shape although the number of Fourier co-efficient  are less  say M < N where N = total boundary  points . ?

Note
Attached Image Link : http://imageshack.us/photo/my-images/85/44022547.png/   

Thanks


Maths is Doctor that make our life easy .. smile smile

https://www.facebook.com/groups/Maths.GIS/

Offline

#2 2011-12-20 06:28:09

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Fourier Descriptors : reduce 2D to 1D problem .How ?

Hi model;

Doesn't that use a DFT? I only know of a FFT and a DFT which are transforms of list of numbers and functions. The videos on Fourier Descriptors are too sketchy for me to understand what you want. Can you put your question in the form of a list of digital values?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Offline

#3 2011-12-20 06:49:36

model
Member
Registered: 2011-08-10
Posts: 142

Re: Fourier Descriptors : reduce 2D to 1D problem .How ?

Well i came to know that

My 1st  question'  ans :
1D Function : weighted sum of many different complex exponents
So rotated image is may be difficult to recognize . while in 1D  since Fourier descriptor as  1D ( i-e weighted sum of different complex  exponent and DC == average  sum of points  ) that mean rotated or scaling will not effect in recognizing the image.  so problem solved from 2D  to 1D  as  a function .

So smaller the fourier co-efficient : lose of details and global  shape of boundary  of an object .
higher the Fourier co-efficient : more details and high frequency  components .

So  less Fourier descriptor , loss of details  but  since " The boundary in the complex plane is approximated by an ellipse with minimum Fourier descriptor 2 . Although resultant shape will be lose of details but approximated  to original  shape .
As  a result , Fourier descriptors are invariant to scaling  and rotation.

yes , last figure shown in the linked image ,retain the original shape
because boundary in complex numbers  are approximated to an ellipse ( but why ellipse may be because circle have no boundary  and after circle we get ellipse ; )

That's  all my thinking and understanding ,.
and i do't know more  than that

Last edited by model (2011-12-20 06:54:23)


Maths is Doctor that make our life easy .. smile smile

https://www.facebook.com/groups/Maths.GIS/

Offline

Board footer

Powered by FluxBB