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

model · 2011-12-20 04:26:51

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

bobbym · 2011-12-20 06:28:09

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?

model · 2011-12-20 06:49:36

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)

Math Is Fun Forum

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

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

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

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

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

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

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