Dirkjan at 20:14, 27 August 2020

2020-08-27T20:14:32Z

Dirkjan: Created page with "For a square image of size NxN the two dimensional DFT is given by:

`F(k,l)=1/(N^2)sum_(i=0)^(N-1)sum_(j=0)^(N-1)f(i,j)e^(-i2..."

2020-08-25T08:18:31Z

Created page with "<html>For a square image of size NxN the two dimensional DFT is given by: `F(k,l)=1/(N^2)sum_(i=0)^(N-1)sum_(j=0)^(N-1)f(i,j)e^(-i2..."

New page

<html>For
a square image of size NxN the two dimensional DFT is given by:
 `F(k,l)=1/(N^2)sum_(i=0)^(N-1)sum_(j=0)^(N-1)f(i,j)e^(-i2pi((ki)/N
+ (ij)/N)``` where `f(i,j)` = is the image in the [[spatial domain|spatial domain]] and
the exponential term is the basic function corresponding for each point
F(k,l) in the
Fourier space. The equatation can be interpreted as: the value of each
point F(k,l) is
obtained by multiplying the spatial image with the corresponding base
function and summing the result. The picture below shows 2
images with their Fourier Transforms directly underneath. The images
are a pure horizontal cosine of 8 cycles and a pure vertical cosine of
32 cycles. Notice that the FT for each just has a single component,
represented by 2 bright spots symmetrically placed about the enter of
the FT image. The center of the image is the origin of the frequency
coordinate system. The u-axis runs left to right through the center and
represents the horizontal component of frequency. The v-axis runs
bottom to top through the center and represents the vertical component
of frequency. In both cases there is a dot at the center that

represents the (0,0) frequency term or average value of the image.
Images usually have a large average value (like 128) and lots of low
frequency information so FT images usually have a bright blob of
components near the center. Notice that high frequencies in the
vertical direction will cause bright dots away from the center in the
vertical direction. And that high frequencies in the horizontal
direction will cause bright dots away from the center in the horizontal
direction. <img src="http://www.cs.unm.edu/%7Ebrayer/vision/cosines.gif"
/> Here
are 2 images of more general Fourier components. They are images of 2D
cosines with both horizontal and vertical components. The one on the
left has 4 cycles horizontally and 16 cycles vertically. The one on the
right has 32 cycles horizontally and 2 cycles vertically. (Note: You
see a gray band when the function goes through gray = 128 which happens
twice/cycle.) You may begin to notice there is a lot of symmetry. For
all REAL (as opposed to IMAGINARY or COMPLEX) images, the FT is
symmetrical about the origin so the 1st and 3rd quadrants are the same
and the 2nd and 4th quadrants are the same. If the image is symmetrical
about the x-axis (as the cosine images are) 4-fold symmetry
results. <img
src="http://www.cs.unm.edu/%7Ebrayer/vision/basis.gif" /> Sources: http://homepages.inf.ed.ac.uk/rbf/HIPR2/fourier.htm http://www.cs.unm.edu/~brayer/vision/fourier.html http://research.stowers-institute.org/efg/Report/FourierAnalysis.pdf </html>
[[Category:microscopy]]

[[Category:image]]

[[Category:processing]]

@@ Line 8: / Line 8: @@
 size="2">where <span class="AM" style="font-style:
 italic;">`f(i,j)`</span> = <span
-class="AM"></span>is the image in the [[spatial domain|spatial domain]] and
+class="AM"></span>is the image in the [[spatial domain]] and
 the exponential term is the basic function corresponding for each point
 <span style="font-style: italic;">F(k,l)</span> in the

Fourier transform - Revision history

Dirkjan at 20:14, 27 August 2020

Dirkjan: Created page with "For a square image of size NxN the two dimensional DFT is given by: `F(k,l)=1/(N^2)sum_(i=0)^(N-1)sum_(j=0)^(N-1)f(i,j)e^(-i2..."

Dirkjan: Created page with "For a square image of size NxN the two dimensional DFT is given by:

`F(k,l)=1/(N^2)sum_(i=0)^(N-1)sum_(j=0)^(N-1)f(i,j)e^(-i2..."