To see this single pixel more clearly lets also magnify that area of the
image...
convert constant_magnitude.png gravity center extent 5x5 \
scale 2000% constant_dc_zoom.gif

Note that the color of the DC point is the same as the original image.
Actually, it is a good idea to remember that what you are seeing is three
values. That is the FFT image generated is actually three separate Fast
Fourier transforms, one for each of the three red, green and blue image
channels. The FFT itself has no real knowledge about colors, only the channel
values or 'graylevels'.
Effects of the DC Color
In a more typical nonconstant image, the DC value is the average color of the
image. The color you should generally get if you had completely blurred,
averaged, or resized the image down to a single pixel or color.
For example lets extract the DC pixel from the FFT of the "Lena" image.
convert lena.png fft +delete lena_magnitude.png
convert lena_magnitude.png gravity center extent 1x1 \
scale 60x60 lena_dc_zoom.gif

As you can see the average color for the image is a sort of 'dark pink' color.
Another way of thinking about this special pixel is that it represents the
center 'bias' level for all the sine waves.
For example lets replace that 'dark pink' DC pixel with some other color
such as the more orange color 'tomato'...
convert size 60x60 xc:tomato tomato_swatch.gif
convert lena.png fft \
\( clone 0 fill tomato draw "point 128,128" alpha off \) \
delete 0 +swap ift lena_dc_replace.png

What is really happening is that by changing the DC value in the FFT magnitude,
you are changing the whole image in that same way. This is how changing the
average color of the image gets replicated from the 'frequency domain' back
into the 'spatial domain'.
This is not a recomended method of color tinting an image as there are other
more direct means. But it is an effective approach.
Spectrum Of A Sine Wave Image
Next, lets take a look at the spectrum from a single sine wave image with 4
cycles. We see that it has 3 dots. The center dot is at zero x and y
frequencies, i.e. the DC point, and in the original magnitude has a value equal
to the average over the whole input image channel. The other two dots are at
the positive and negative frequencies of the sine wave. As the sine wave has
n=4 cycles along the x direction only, the position of the two sides dots will
be 4 pixels to the left and right of the center dot.
convert size 128x129 gradient: chop 0x1 rotate 90 evaluate sine 4 sine4.png
convert sine4.png fft +delete \
contraststretch 0 evaluate log 100 sine4_spectrum.png

If we repeat this with a sine wave with 16 cycles, then again we see that it
has 3 dots, but the dots are further apart, since the waves are closer together.
In this case the side dots are spaced 16 pixels to the left and right sides of
the center dot.
convert size 128x129 gradient: chop 0x1 rotate 90 evaluate sine 16 sine16.png
convert sine16.png fft +delete \
contraststretch 0 evaluate log 100 sine16_spectrum.png

Spectrum Of A Rectangle Pattern Image
Next, lets look at the spectrum of white rectangle of width 8 and height 16
inside a black background.
convert size 8x16 xc:white gravity center background black extent 128x128 \
write rect8x16.png fft delete 1 \
contraststretch 0 evaluate log 100 rect8x16_spectrum.png

Notice that this looks like the image that we directly obtained from the
absolute value of the sinc function earlier. If we graph the center row and
column, we get
im_profile s rectangle_spectrum.png[128x1+0+64] rectangle_spec_row_profile.png
convert rect8x16_spectrum.png[1x128+64+0] rotate 90 miff: \
im_profile s  rect8x16_spec_col_profile.png

This clearly demonstrates that the transform of the rectangular shape is
indeed a sinc function and that the narrower dimension of the object in the
image produces a transform with features that are more spread out and wider.
According to the properties listed above, the spacing in each dimension
between the troughs is determined by N/a and N/b. Thus, when the width and
height of the rectangle are a=8 and b=16, then the spacings should be 128/8=16
horizontally and 128/16=8 vertically. This is indeed what one measures for the
spacing between the dark troughs in this image.
Now, lets rotate the rectangle by 45 degrees. We find that the spectrum is
also rotated in the same direction by 45 degrees.
convert rect8x16.png rotate 45 gravity center crop 128x128+0+0 +repage \
write rect8x16_rot45.png fft delete 1 \
contraststretch 0 evaluate log 100 rect8x16_rot45_spectrum.png

Spectrum Of A Flat Circular Pattern Image
Next, lets look at the spectrum from an image with a white, flat circular
pattern, in one case with diameters of 12 (radius 6) and in another case
with diameter of 24 (radius 12).
convert size 128x128 xc:black fill white \
draw "circle 64,64 64,70" alpha off write circle6.png fft delete 1 \
contraststretch 0 evaluate log 100 circle6_spectrum.png
convert size 128x128 xc:black fill white \
draw "circle 64,64 64,76" alpha off write circle12.png fft delete 1 \
contraststretch 0 evaluate log 100 circle12_spectrum.png

Note that the first image is very close to what we generated for the jinc
example further above. It is however a little broken up. This artifacting
occurs due to the small size of the circle. Since it is represented
digitally, its perimeter is not perfectly circular. The transform of the
larger circle is better as its perimeter is a closer approximation of a true
circle. We therefore conclude that indeed the transform of the flat circular
shape is a jinc function and that the image containing the smaller diameter
circle produces transform features that are more spread out and wider.
According to the properties listed above, the distance from the center to
the middle of the first dark ring in the spectrum will be 1.22*N/d. When the
diameter of the circle is d=12, we get a distance of 1.22*128/12=13.
Likewise when the diameter of the circle is d=24, we get a distance of
1.22*128/24=6.5. These are the values that one measures in those images.
Spectrum Of A Gaussian Pattern Image
Next, lets look at the spectrum from two images, each with a white Gaussian
circular pattern having sigmas of 8 and 16, respectively
convert size 128x128 xc:black fill white \
draw "point 64,64" gaussianblur 0x8 contraststretch 0 \
alpha off write gaussian8.png fft delete 1 \
contraststretch 0 evaluate log 100 gaussian8_spectrum.png
im_profile s gaussian8.png[128x1+0+64] gaussian8_profile.png
im_profile s gaussian8_spectrum.png[128x1+0+64] gaussian8_spectrum_profile.png

convert size 128x128 xc:black fill white \
draw "point 64,64" gaussianblur 0x16 contraststretch 0 \
alpha off write gaussian16.png fft delete 1 \
contraststretch 0 evaluate log 100 gaussian16_spectrum.png
im_profile s gaussian16.png[128x1+0+64] gaussian16_profile.png
im_profile s gaussian16_spectrum.png[128x1+0+64] gaussian16_spectrum_profile.png

This shows that the transform of the Gaussian circular shape is another
Gaussian and that the larger the sigma in the original image, the smaller the
sigma will be in the spectrum. From the properties stated above, we know that
the sigma in the spectrum will be just N/(2*sigma), where sigma is from the
original image. So for an image of size N and sigma=8, the sigma in the
spectrum will be 128/16=8. Similarly if the image's sigma is 16, then the
sigma in the spectrum will be 128/32=4.
Spectrum Of A Grid Pattern Image
Next, lets transform an image containing just a set of grid lines spaced 16x8
pixels apart. The resulting spectrum is just an array of dots where the grid
lines that are more closely spaced produce dots further apart and vice versa.
According to the properties above, since the grid lines are spaced 16x8 pixels
apart, then the dots should be spaced N/a=128/16=8 and M/b=128/8=16, which is
just what is measured in this image.
convert size 16x8 xc:white fill black \
draw "line 0,0 15,0" draw "line 0,0 0,7" \
alpha off write mpr:tile +delete \
size 128x128 tile:mpr:tile \
write grid16x8.png fft delete 1 \
contraststretch 0 evaluate log 100000 grid16x8_spectrum.png

Image Reconstructed From Magnitude Or Phase Only
Both the magnitude and the phase are important in the Fourier Transform.
However, is one more important than the other? First let's look at the
statistics of the two components.
convert lena.png fft format "channel=%s; min=%[min]; max=%[max]" info:
channel=0; min=0; max=46319.2
channel=1; min=0.01993; max=65535

As expected, we see that both components are always nonnegative. The magnitude
(first channel) is by definition nonnegative. The phase (second channel) has
been scaled intentionally from a range between pi and pi to full dynamic range
between 0 and quantumrange, which is 65535 for Q16. Thus both components look
equally important.
To get further insight, lets try reconstructing an image from either just
its magnitude component or just its phase component.
convert lena.png fft lena_mp_fft.png
convert lena_mp_fft0.png \( size 256x256 xc:gray \) ift lena_mag_ift.png
convert \( size 256x256 xc:gray \) lena_mp_fft1.png ift lena_phase_ift.png

What we find is that the reconstruction from the phase contains the edge
features in the original image, whereas the magnitude image contains most of
the coloring from the original image. You can recognize the girl in the
phase reconstructed image, but you cannot see anything resembling her in the
magnitude reconstructed image.
Image Reconstructed From Real Or Imaginary Only
Finally, lets see about the relative importance of the real and imaginary
components. Again we use hdrienabled Q16 ImageMagick compilation as both
the real and imaginary components contain positive and negative values.
First lets look at the statistics of the two components.
convert lena.png +fft format "channel=%s; min=%[min]; max=%[max]; mean=%[mean]" info:
channel=0; min=2521.81; max=46319.2; mean=0.63663
channel=1; min=2770.91; max=2770.91; mean=0.000439742

As we expected, both channels contain both positive and negative values. But
The real component has a relatively small negative minimum and a relatively
large positive maximum. On the other hand, the imaginary component has both a
relatively small negative minimum and a relatively small positive maximum. Both
are about equal in absolue value and its mean is about zero. This suggests
that the imaginary component is perhaps less important than the real component.
So, lets try reconstructing an image from either just its real component or
just its imaginary component.
convert lena.png +fft +adjoin lena_ri_fft.tif
convert lena_ri_fft0.tif \( size 256x256 xc:black \) +ift lena_real_ift.png
convert \( size 256x256 xc:white \) lena_ri_fft1.tif +ift lena_imag_ift.png

Note that the real and imaginary results of the +fft are saved in TIFF format,
since that supports 32bit values. However, most browsers (Safari being the
apparent exception) will not display them. So they have been converted to PNG
format for display. Consequency, those images cannot be used for further
processing as PNG does not support 32bits per channel. The proper TIFF files
are needed.
So it would appear that in this case, neither does a good job. Since both
components have relatively similar strengths for min and max, they both are
important. But as the imaginary component has a much smaller max value, its
reconstruction is poorer. For example, the ratio of max values for imaginary vs
real is about (2770.91/46319.2)=0.6 or 60%.
However, might it be possible under the right circumstances that only the real
component might be sufficient. The answer is yes. In particular when using
and image that is mostly black with just a little white in it. This is the case
of the blurring filters as exampled by the simple white rectangle or white
circle shown earlier. So lets repeat this for one of these cases, say the
circle of diameter 12.
convert circle6.png +fft format "channel=%s; min=%[min]; max=%[max]; mean=%[mean]" info:
channel=0; min=499.375; max=505.504; mean=0.000602056
channel=1; min=0.271139; max=0.271139; mean=9.94399e07
convert circle6.png +fft +adjoin circle6_ri_fft.tif
convert circle6_ri_fft0.tif \( size 128x128 xc:black \) +ift circle6_real_ift.png
compare metric rmse circle6.png circle6_real_ift.png null:
63.2857 (0.000965678)
convert \( size 128x128 xc:white \) circle6_ri_fft1.tif +ift circle6_imag_ift.png

Here the ratio of max values for imaginary vs real is about
(0.271139/505.504)=0.0005 or about 0.05%. So in the case of the circle,
the real component alone virtually reproduces the original. This will be
important later as it shows that for such simple filter images, in practice,
the imaginary component (or phase component) of the filter can often be
ignored.
Practical Applications
OK, now that we have covered the basics, what are the practical applications
of using the Fourier Transform?
Some of the things that can be done include: 1) increasing or decreasing the
contrast of an image, 2) blurring, 3) sharpening, 4) edge detection and
5) noise removal.
Changing The Contrast Of An Image  Coefficient Rooting
One can adjust the contrast in an image by performing the forward Fourier
transform, raising the magnitude image to a power and then using that with the
phase in the inverse Fourier transform. To increase, the contrast, one uses an
exponent slightly less than one and to decrease the contrast, one uses an
exponent slightly greater than one. So lets first increase the contrast on the
Lena image using an exponent of 0.9 and then decrease the contrast using an
exponent of 1.1.
convert lena.png fft \
\( clone 0 evaluate pow 0.9 \) delete 0 \
+swap ift lena_plus_contrast.png
convert lena.png fft \
\( clone 0 evaluate pow 1.1 \) delete 0 \
+swap ift lena_minus_contrast.png

Blurring An Image  Low Pass Filtering
One of the most important properties of Fourier Transforms is that convolution
in the spatial domain is equivalent to simple multiplication in the frequency
domain. In the spatial domain, one uses small, squaresized, simple
convolution filters (kernels) to blur an image with the convolve option. This is called a low pass filter. The
simplest filter is just a an equallyweighted, square array. That is all the
values are ones, which are normalized by dividing by their sum before applying
the convolution. This is equivalent to a local or neighborhood average.
Another low pass filter is the Gaussianweighted, circularly shaped filter
provided by either gaussianblur or blur.
In the frequency domain, one type of low pass blurring filter is just a
constant intensity white circle surrounded by black. This filter corresponds
to the magntitude (or real) component and there is no (or a zero) phase (or
imaginary) component as we discussed above. This filter would be similar to
a circularly shaped averaging convolution filter in the spatial domain.
However, since convolution in the spatial domain is equivalent to
multiplication in the frequency domain, all we need do is perform a forward
Fourier transform of the image, then multiply the filter with the magnitude
image and finally perform the inverse Fourier transform of the product. We
note that a small sized convolution filter will correspond to a large circle
in the frequency domain. Multiplication is carried out via composite with a compose multiply setting.
So let's try doing this with two sizes of circular filters, one of diameter 48
(radius 24) and the other of diameter 32 (radius 16). Note that we display the
spectrum and masked spectrum, but we actually use the masked magnitude to
create the blurred result.
convert lena.png fft \
\( size 256x256 xc:black fill white draw "circle 128,128 152,128" alpha off write circle24.png \
clone 0 compose multiply composite \) \
\( +clone contraststretch 0 evaluate log 10000 write lena_circle24_spec.png \) \
delete 0,3 +swap ift lena_circle24_blur.png
convert lena.png fft \
\( size 256x256 xc:black fill white draw "circle 128,128 144,128" alpha off write circle16.png \
clone 0 compose multiply composite \) \
\( +clone contraststretch 0 evaluate log 10000 write lena_circle16_spec.png \) \
delete 0,3 +swap ift lena_circle16_blur.png

A bash script, fftfilter, is available to make this easier. The simplified
equivalent without all the extra processing to the above is given by:
convert size 256x256 xc:black fill white draw "circle 128,128 152,128" \
alpha off circle24.png
fftfilter lena.png circle24.png lena_circle24_blur.png
convert size 256x256 xc:black fill white draw "circle 128,128 144,128" \
alpha off circle16.png \
fftfilter lena.png circle16.png lena_circle16_blur.png

So we see that the image that used the smaller diameter filter produced more
blurring. We also note the 'ringing' or 'ripple' effect near edges in the
resulting images. This occurs because the Fourier Transform of a circle, as we
saw earlier, is a jinc function, which has decreasing oscillations as it
progresses outward from the center. Here however, the jinc function and the
oscillations are in the spatial domain rather than in the frequency domain, as
we had demonstrated earlier above.
So what can we do about this? The simplest thing is to taper the edges of the
circles using various windowing functions. Alternately, one can use a filter
such as a Gaussian shape that is already by definition tapered. So lets do
the latter and use two Gaussian, circularly symmetric filters with sigma of
24 and 16, respectively. Here again we display the spectrum and masked
spectrum, but we actually use the masked magnitude to create the blurred result.
convert lena.png fft \
\( size 256x256 xc:black fill white draw "point 128,128" \
alpha off blur 0x24 contraststretch 0 write gaussian24.png \
clone 0 compose multiply composite \) \
\( +clone contraststretch 0 evaluate log 10000 write lena_gaussian24_spec.png \) \
delete 0,3 +swap ift lena_gaussian24_blur.png
convert lena.png fft \
\( size 256x256 xc:black fill white draw "point 128,128" \
alpha off blur 0x16 contraststretch 0 write gaussian16.png \
clone 0 compose multiply composite \) \
\( +clone contraststretch 0 evaluate log 10000 write lena_gaussian16_spec.png \) \
delete 0,3 +swap ift lena_gaussian16_blur.png

Using the bash script, fftfilter, the simplified equivalent without all the extra
processing to the above is given by:
convert size 256x256 xc:black fill white draw "point 128,128" \
alpha off blur 0x24 contraststretch 0 gaussian24.png
fftfilter lena.png gaussian24.png lena_gaussian24_blur.png
convert size 256x256 xc:black fill white draw "point 128,128" \
alpha off blur 0x16 contraststretch 0 gaussian16.png
fftfilter lena.png gaussian16.png lena_gaussian16_blur.png

These results with the Gaussian filter are of course much better.
The bad thing about filtering in the frequency domain is that to produce a
small amount of (blurring) filtering, you need a large filter, which may take
longer to create. The good thing about filtering in the frequency domain is
that to produce a large amount of (blurring) filtering, you only need a small
filter. Therefore, the delay you pay for doing the forward and inverse
Fourier transforms, can often be made up by the shorter time it takes to
create and apply the filter. However, when the filter gets very small, in
this case a circle, it may degrade into some other shape not intended, due to
pixelization from having a digital image. However, antialiasing the boundary
of the filter shape should help.
Detecting Edges In An Image  High Pass Filtering
In the spatial domain, high pass filters that extract edges from an image are
often implemented as convolutions with positive and negative weights such that
they sum to zero. Things are much simpler in the frequency domain. Here a high
pass filter is just the negated version of the low pass filter. That is where
the low pass filter is bright, the high pass filter is dark and vice versa. So
in ImageMagick, all we need do is apply the
negate option to the low pass filter image. So lets apply
high pass filters to the Lena image that correspond to the constant intensity
and gaussian rolloff circularly symmetric filters of diameter=24 and sigma=24,
respectively. Here again we display the spectrum and masked spectrum, but we
actually use the masked magnitude to create the edge result.
convert lena.png fft \
\( size 256x256 xc:black fill white draw "circle 128,128 152,128" negate alpha off write circle24n.png \
clone 0 compose multiply composite \) \
\( clone 0 contraststretch 0 evaluate log 100000 circle24n.png \
compose multiply composite write lena_circle24n_spec.png \) \
delete 0,3 +swap ift normalize lena_circle24n_edge.png
convert lena.png fft \
\( size 256x256 xc:black fill white draw "point 128,128" \
alpha off blur 0x24 contraststretch 0 negate write gaussian24n.png \
clone 0 compose multiply composite \) \
\( clone 0 contraststretch 0 evaluate log 100000 gaussian24n.png \
compose multiply composite write lena_gaussian24n_spec.png \) \
delete 0,3 +swap ift normalize lena_gaussian24n_edge.png

Using the bash script, fftfilter, the simplified equivalent without all the extra
processing to the above is given by:
convert size 256x256 xc:black fill white draw "circle 128,128 152,128" \
negate alpha off write circle24n.png
fftfilter lena.png circle24n.png lena_circle24n_edge.png
convert size 256x256 xc:black fill white draw "point 128,128" \
alpha off blur 0x24 contraststretch 0 negate gaussian24n.png
fftfilter lena.png gaussian24n.png lena_gaussian24n_edge.png

Carefully examining these two results, we see that the simple circle is not
quite as good as the gaussian, as it has 'ringing' artifacts and is not quite
as sharp.
Sharpening An Image  High Boost Filtering
The simplest way to sharpen an image is to high pass filter it (without
the normalization stretch) and then blend it with the original image. The
high pass filtering, in this case, is done in the frequency domain and the
result transformed back to the spatial domain where it is blended with the
original image.
convert lena.png fft \
\( clone 0 \( size 256x256 xc:black fill white \
draw "point 128,128" blur 0x24 contraststretch 0 negate \) \
alpha off compose multiply composite \) \
delete 0 +swap ift \
lena.png compose blend set "option:compose:args" 100x100 composite \
lena_gauss24_sharp.png

Noise Removal  Notch Filtering
Many noisy images contain some kind of patterned noise. This kind of noise is
easy to remove in the frequency domain as the patterns show up as either a
pattern of a few dots or lines. Recall a simple sine wave is a repeated
pattern and shows up as only 3 dots in the spectrum.
In order to remove this noise, one simply, but unfortunately, has to manually
mask (or notch) out the dots or lines in the magnitude image. We do this by
transforming to the frequency domain, create a grayscale version of the
spectrum, mask the dots or lines, threshold it, multiply the binary mask image
with the magnitude image and then transform back to the spatial domain.
Lets try this on the clown image, which contains a diagonally striped
ditherlike pattern. First we transform the clown image to create its
magnitude and phase images.
convert clown.jpg fft clown_fft.png

Next, we create the spectrum image from the magnitude, but we add an extra
step to push any image values at graylevel zero to graylevel 1, so that we
can preserve graylevel zero for the masked areas.
convert clown_fft0.png contraststretch 0 evaluate log 100000 \
fill "gray(1)" opaque "gray(0)" clown_spectrum.png

A bash script, spectrum, is also available to make this easier. One can
use either the magnitude image for input or just provide the original image
and it will convert that to the magnitude before generating the spectrum.
In fact, the script does an adequate job of estimating the log scaling
constant, so in fact, you may leave off the s argument. Thus, the equivalent
to the above is either of the following:
spectrum s 100000 clown_fft0.png clown_spectrum.png
or
spectrum m s 100000 clown.jpg clown_spectrum.png

We see that the spectrum contains four bright starlike dots, one in each
quadrant. We ignore the bright dot and lines in the middle as they represent
the DC or zero frequency components in the image, i.e. the constant intensity
regions. So now we need to mask out those starlike dots. We could display
this image in ImageMagick and use the display functions to measure the
locations and sizes of the starlike dots; then use
draw to cover them with black circles or rectangles.
But it is easier to use some other tool such as GIMP or Photoshop to do this
interactively and then threshold the result to a binary image using
ImageMagick. The results are shown below.
convert clown_spectrum_mask.png threshold 0 clown_mask.png

You can even draw in color and then extract the mask as follows:
convert clown_spectrum_edited.png fill white +opaque red \
fill black opaque red clown_mask2.png

Now we simply multiply the mask with the magnitude and use the result with
the phase image to transform back to the spatial domain. We display the
original image next to it for comparison. Here again we display the spectrum,
but we actually mask the magnitude to create the filtered result.
We note, however, that there is some 'ringing' artifacts remaining in the
result. This can be removed by tapering the sharp whitetoblack transitions
in the mask. One can use a Gaussian shaped taper as we did before or even
simply a linear taper of 5 pixels. Note that blur radiusx65000 will produce
such a linear taper.
Both examples above become simpler using the script,
fftfilter (after creating the mask).
fftfilter clown.jpg clown_mask.png clown_filtered.png
fftfilter clown.jpg clown_mask_taper.png clown_filtered_taper.png

We can even take the difference between the original and the result to create
an image that is just the noise that was removed.
convert clown.jpg clown_filtered_taper.png compose difference \
composite normalize clown_noise.png

Lets try this on an another example. This time on the twig image,
which contains an irregular horizontal and vertical striped patter. First we
transform the twig image to create its magnitude and phase images. Then, we
create a graylevel spectrum image from the magnitude with graylevel zero
preserved for the masked areas.
convert twig.jpg fft twig_fft.png
spectrum c 100000 twig_fft0.png twig_spectrum.png

In this case, as the noise in the image is horizontally and vertically
oriented, this shows up as vertical and horizontal bright areas along the
center of the spectrum. So we mask those out, again manually, using GIMP
or Photoshop. Then we threshold the result in ImageMagick to form a binary
mask.
convert twig_spectrum_mask.png threshold 0 twig_mask.png

Now we again multiply the mask with the magnitude and use the result with
the phase image to transform back to the spatial domain. The result is
presented next to the original image for comparison. Here again we display the
spectrum, but we actually mask the magnitude to create the filtered result.
Again just using the script (after creating the mask), fftfilter, we get:
fftfilter twig.jpg twig_mask.png twig_filtered.png

Now we can take the difference between the original and the result to create
an image that is just the noise that was removed
convert twig.jpg twig_filtered.png compose difference \
composite normalize twig_noise.png

Advanced Applications
Some of the other more advanced applications of using the Fourier Transform
include: 1) deconvolution (deblurring) of motion blurred and defocused images
and 2) normalized cross correlation to find where a small image best matches
within a larger image.
However, as these applications generally require the use of the real and
imaginary components of the Fourier Transform and/or very high precision, we
must do the processing using Q32 or an HDRI compiled version of IM 6.5.47 or
later until proper ImageMagick functions can be made for them. When that
occurs, the HDRI restriction can be lifted. The examples below were
processed using a Q16 HDRI compilation of ImageMagick.
Convolution And Deconvolution
In the spatial domain, the process of blurring or edge detection of an image
is usually accomplished with convolutions. That is, the image is convolved
with a small kernel or filter. For example, a 3x3 array of ones, is a small
blurring (averaging) filter. In ImageMagick, this 2D array of ones is simply
supplied as a 1D list of ones to the convolve option. However, in other systems, the filter is represented
as a small image, for example, of size 3x3 containing only values of 1 or
white pixels for the averaging filter. However, in principal, it could be
padded out with black pixels on all sides to any size desired and one would
get the same result, only slower.
One of the Fourier Transform principles that was listed earlier is that in
the frequency domain, the equivalent of convolution is multiplication.
Therefore, blurring or edge detection in the frequency domain is simply the
multiplication, pixelbypixel, between the Fourier Transform of the image
and the Fourier Transform of the appropriate filter image.
Likewise, the process of deconvolution (or deblurring), which is hard if not
impossible in the spatial domain, becomes simply a division in the frequency
domain. That is, one divides the Fourier Transform of the blurred image by
the Fourier Transform of the filter image that created the blurring.
So the way to convolve or deconvolve an image (digital picture) using the
Fourier Transform is to create a filter in the form of a small, centered
image of appropriate values and then pad it with black all around to fill it
out to the same size as the picture. Then transform both this filter and the
picture to the frequency domain, multiply or divide the two together
pixelbypixel, depending upon whether one is convolving or deconvolving,
respectively, and then transform back to the spatial domain.
But what do we really mean by multiplication and division in the frequency
domain?
In order to understand the meaning, we recall that the Fourier Transform
converts spatial intensity (graylevel) values in each image channel into
complex numbers that represent the amplitudes of the exponential sinusoidal
waves. We recall that these exponential sinusoidal waves are described by
the following equation.
(7)
This has two components, the cosine, which is the real part and the sine,
which is the imaginary part (because it is multiplied by i, the symbol for
square root of 1). Thus we are fundamentally dealing with a complex number
for both the filter (F) and the picture (P) of the form.
(8)
Therefore multiplication and division in the frequency domain means that for
each pixel, we must multiply or divide two complex numbers. So when doing
blurring in the frequency domain, we want the product of each complex filter
value multiplied by the corresponding complex picture value. The product
then becomes just another complex number with a real part and an imaginary
part as follows. (Note that i*i=1). See also multiplication and division of
two Complex Numbers.
(9)
For deblurring, what we would like to do is divide each complex picture
value by the corresponding complex filter value, namely, P/F. But there is
actually no formal division by a complex number in real and imaginary form.
However, we can multiply both the numerator and denominator by the complex
conjugate of the filter, denoted as F^{*}. A complex conjugate
number is the same as the complex number, except there is a negative sign
rather than a positive sign before the imaginary symbol, i. Thus the
division of the complex picture by the complex filter becomes again just
another complex number with a real part and an imaginary part as follows.
(Note F^{2} is just a real number. It has no imaginary
component).
(10)
One can also do complex multiplication and division using magnitude and phase
components, which turns out actually to be simpler mathematics. Here, A and C
are the magnitude components and B and D are the phase components.
(11)
Here multiplication is achieved by multiplying the two magnitude components and
by adding the two phase components. See again,
Complex Numbers.
(12)
Thus the new magnitude is just AC and the new phase is just (B+D).
For division, we can divide the two magnitude components and subtract the two
phase components.
(13)
Thus the new magnitidue is just (C/A) and the new phase is just (DB).
Motion Blur
Motion blur in a photograph occurs when either the object
being photographed or the camera itself moves while the camera shutter is
open. If this motion is linear and uniform over time, then in the spatial
domain, the blurring filter may be considered simply a white straight line
in a black background, since the motion is linear in one dimension and
uniform. Of course the motion may be at an angle relative to horizontal and
thus the line would be oriented at some angle. The length of the line
corresponds directly to the amount of motion blurring.
Lets now create a 15 pixel, horizontal motion blur filter and use it to blur
a picture using both multiplication approaches, first using real and
imaginary components and then using magnitude and phase components. Both
images will be 256x256 pixels in size, but we will display them at half
size. You may not be able to see the white line in the middle of the black
filter image at half size. However, you can view the full size images by
clicking on the small ones.
There are two other things that must be done in practice in addition to
applying the equation for complex multiplication. First, we need to "roll"
the picture using the
roll
option so that what is now the center of the 15 pixel long white line
at coordinates (128,128) are moved to the top left at coordinates (0,0). We
must do this, since that is the real origin for the Fourier Transform and the
filter must be properly centered at the origin before it can be multiplied
with the Fourier Transform of the image. Otherwise the opposite occurs and
the resulting image will end up rolled so that its top left corner will end
up at the center and vice versa. Second, we must linearly stretch the Fourier
Transform of the filter or the result will be too dark. Since its maximum
value is at the DC point and is equal by the mean of its spatial image, we
simply need to stretch the mean value to full white. We can do this by
computing the mean value and using the
level option. Alternately, we could simply use an automatic stretch such
as
linearstretch 0x1. This forces the stetch to occur on the white side
from its maximum value found at the first nonempty bin (i.e. having a count
of 1 or more) in the histogram to full white and leaves the black side
unchanged.
To create the spatial domain motion blur filter, we simply draw a 15 pixel
horizontal white line in the center of a black background using the
draw
option. So if the center is 128, we draw a line from (121,128) to
(135,128) so that the difference between the left and right ends is
(135121+1)=15.
Once the filter image is created in the first section below, we start the
second section, which blurs the image, by first computing the mean of the
filter image and save that value. Then we transform both the filter and image
from the spatial domain to the frequency domain using the +fft
option to create the real and imaginary components of each. Then we
roll the filter so that its center is moved to the upper left corner and
apply the mean to stretch both its real and imaginary components using the
level
option. Once that is done, we apply equation (9) to do the complex
multiply of the real and imaginary components from both the filter and image.
This is accomplished with a sequence of
compose and
composite options. Finally the real and imaginary component products
are returned to the spatial domain using the +ift
option.
In the third section, we do the same blurrinig, but use the fft
option to create the magnitude and phase components of each. We do the
same roll and stretch to the magnitude and phase of the filter and use a
different set of
compose and
composite options to do the complex multiplication according to
equation (12). Then we use the ift
option to return the magnitude and phase products back to the spatial
domain.
As there is a lot to remember to do here and it involves a formidable amount
of
compose and
composite options, a bash script, fftconvol, is available to do all this work. The script
works as follows to achieve the same thing as what is presented in the
second section above using real and imaginary fft components.
fftconvol cameraman.png motion15.png cameraman_ri_motion15.png

This script produces the same results as depicted above.
You will need to click on the filter image to enlarge it so that you can see
the white line. Then if you enlarge the original image and the motion
blurred image and compare them you will notice several things indicative of
motion blur. First, the blurring is only horizontally. You can see this, for
example, by examining the bright spot on the top of the far, tall, narrow
bright building. You will see in the motion blurred image that it is only
blurred horizontally and has a nice uniform brightness to it. There is also
a bright spot about 3/4 of the way down the dark part of the front two legs
of the tripod. They also get blurred only horizontally. The final
interesting thing to notice is that the bottom of the two front tripod legs
get doubled by the large amount of motion blur.
Now lets use the same blurring filter image to deblur this picture using the
division equations (10) and (13) above using, respectively, the real and
imaginary division and the magnitude and phase division approaches. These
are shown in the second and third sections after recreating the same filter
image in the first section.
Again, as there is a lot to remember to do here and it involves an even more
formidable amount of
compose and
composite options, a bash script, fftdeconvol, is available to do all this work. The script
works as follows to achieve the same thing as what is presented in the
second section above using real and imaginary fft components.
fftdeconvol cameraman_ri_motion15.png motion15.png cameraman_ri_motion15_deblur_0.png

This script produces the same results as depicted above.
Now a comparison of the results with the original image shows a very small rmse
value. Consequently, you might say, this is too good to be true. And you are
right. In the real world, photographic pictures are not so conducive.
In particular, noise is introduced into the pictures when it is recorded to
whatever is the relevant media. In filmbased photography, the silver grains
contribute to film grain noise. In digital photography, the sensors that
define the pixel array have electronic noise. For more details on either,
see Image
Noise. Also see Additive White Noise. The important factor is that the
noise is introduced after the blurring.
Therefore, to make a more realistic simulation of a blurred image, we need
to add noise to our motion blurred image from above. This is accomplished by
creating a random grayscale noise pattern and adding it to the image. Here
we create a noise image that has full dynamic range and subtract midgray
from it and then add a factor of 0.01 times the noise image to the blurred
image. In doing so we have subtracted midgray first so as to have both
positive and negative noise changes in the image. As you can see the amount
of noise is so small that you cannot detect it with the naked eye.
convert size 256x256 xc:gray +noise random \
channel green separate noise_gray.png
convert cameraman_ri_motion15.png noise_gray.png +swap \
compose mathematics set option:compose:args "0,1,0.01,0.005" \
composite cameraman_motion15_noise.png
compare metric rmse cameraman_motion15.png cameraman_motion15_noise.png null:
190.033 (0.00289972)

So now what do we get when we try to deblur this image? So that we do not
repeat a large amount of duplication and to make life easier, we simply
use the fftdeconvol script.
fftdeconvol cameraman_motion15_noise.png motion15.png cameraman_motion15_noise_deblur_0.png

As you can see it has been deblurred, but at the cost of much more noise than
we originally introduced into the blurred image. Why is this? The noise
amplification occurs from the divide we do when the denominator is very
small and approaches zero. This tends to amplify the noise in the image.
So what can we do about it, if anything? The answer is we can do something
very simple. Just add a small number to the denominator in our complex
division equation. So when the denominator gets very small, the constant will
be enough to keep from dividing by something too close to zero. But when the
denominator is not small, the small constant will not contribute significantly.
(14)
There is in fact some basis for doing this and some real meaning to the small
number used in the denominator. This idea was first proposed by
Norbert Wiener and the resulting deblurring approach has
been named the Wiener Filter after him. The approach produces a minimum
mean squared error optimal filter and the small number is determined from the
noise to signal power spectrum density ratio. This is just a fancy way of
expressing the ratio of the noise variance to the signal variance and can be
estimated directly from the image. The signal variance can be computed from
overall image variance (squared standard deviation) statistic and a noise
variance can be estimated from the variance of a subsection of the image that
has little signal variation, namely a region that has a constant graylevel.
The equivalent using magnitude and phase simply becomes:
(15)
So let's now try this on the noisy blurred image. The image below shows where
the noise variance is estimated.
convert cameraman_motion15_noise.png fill none stroke red \
draw "rectangle 240,40 249,89" alpha off cameraman_motion15_noise_area.png
convert cameraman_motion15_noise.png[10x50+240+40] format "%[standard_deviation]" info:
379.925
convert cameraman_motion15_noise.png format "%[standard_deviation]" info:
16865.8
convert xc: format "%[fx:(379.925/16865.8.9)^2]" info:
0.000507437

Now lets deblur the noisy blurred image using the Wiener Filter approach with
n=0.0005. The image in the center is this result. For comparison and to see
how sensitive is the value used for n, the image to its left is
the result using n=0.005 and the image to its right is the result using
n=0.00005, thus using a factor of 10 larger and smaller than the center image.
fftdeconvol n 0.00005 cameraman_motion15_noise.png motion15.png \
cameraman_motion15_noise_deblur_0p00005.png

Upon examination of the full size images, we see that the image on the left is
still somewhat blurred and has some 'ghosting' artifacts, but has little
amplified noise. Whereas, the image on the right is deblurred well, but has
noise amplification. The center image with our measure value of n=0.0005
seems to be a reasonable compromise between residual blurring and noise
amplification.
Defocus
Defocus in a picture occurs when the lens is moved
from its optimal location and causes a deviation in the plane or surface
of best focus. In an ideal case, the defocus filter corresponds simply to
a constant intensity (whitefilled) circle, although other shapes are possible
and depend upon the shape of the shutter mechanism. The amount of blurring that
is introduced then depends upon the diameter of the circle. When using such a
spatial filter for blurring or deblurring in the frequency domain, we again
center the circle and pad it out with black to fill an image the same size as
the image that we want to blur or deblur.
Lets now create a 15 pixel diameter, circular defocus filter and use it to
blur the same picture. Both images will be 256x256 pixels in size, but we
will again display them at half size. However, you can view the full size
images by clicking on the small ones. We use the exact same blurring and
deblurring techniques (and scripts) as above for the case of motion blur.
The only difference is the filter image that we create and apply.
To create the spatial domain defocus filter, we simply draw a 15 pixel
diameter white circle in the center of a black background using the
draw
option. So if the center is 128, we draw a circle whose center is at
(128,128) and having a point on its perimeter to its right at (135,128).
This means that the point on the perimeter to its left will be at (121,128)
so that the diameter becomes (135121+1)=15. We use the same fftonvolve script, but use the t defocus argument to create
the defocus type blurring.
convert size 256x256 xc:black fill white \
draw "circle 128,128 135,128" alpha off defocus15.png
fftconvol cameraman.png defocus15.png cameraman_defocus15.png

Upon examining the full size images, we see that blurring occurs uniformly
in all directions. Now let's deblur this image using the fftdeconvol script.
convert size 256x256 xc:black fill white \
draw "circle 128,128 135,128" alpha off defocus15.png
fftdeconvol cameraman_defocus15.png defocus15.png cameraman_defocus15_deblur_0.png
compare metric rmse cameraman.png cameraman_defocus15_deblur_0.png null:
693.721 (0.0105855)

And a comparison with the original image shows a small rmse value, but about
3 times larger than we got from the 15 pixel motion blur example above. We
now add noise to the blurred image using the same random noise image that
we created before.
convert cameraman_defocus15.png noise_gray.png +swap \
compose mathematics set option:compose:args "0,1,0.01,0.005" \
composite cameraman_defocus15_noise.png
compare metric rmse cameraman_defocus15.png cameraman_defocus15_noise.png null:
190.033 (0.00289972)

And now estimate the signal and noise variance and thus the number to add to
the denominator term.
convert cameraman_defocus15_noise.png fill none stroke red \
draw "rectangle 240,40 249,89" alpha off cameraman_defocus15_noise_area.png
convert cameraman_defocus15_noise.png[10x50+240+40] format "%[standard_deviation]" info:
351.106
convert cameraman_defocus15_noise.png format "%[standard_deviation]" info:
16489.1
convert xc: format "%[fx:(351.106/16489.1)^2]" info:
0.000453401

Now we use the number n=0.0005 to deblur the image along with the blurring
filter as we did before. The image in the center is this result. For
comparison the image to its left is the result using n=0.005 and the image to
its right is the result using n=0.00005, thus using a factor of 10 larger and
smaller than the center image.
convert size 256x256 xc:black fill white \
draw "circle 128,128 135,128" alpha off defocus15.png
fftdeconvol n 0.0005 cameraman_defocus15_noise.png defocus15.png \
cameraman_defocus15_noise_deblur_0p0005.png

Upon examination of the full size images, we see that the image on the left is
still somewhat blurred and has some 'ghosting' artifacts, but has little
amplified noise. Whereas, the image on the right is deblurred well, but has
noise amplification. The center image with our measure value of n=0.0005
seems to be a reasonable compromise between residual blurring and noise
amplification.
Ideal Blurring And Deblurring
The approach above has one specific advantange and one specific
disadvantage. The advantage is that is permits almost any kind of spatial
filter that one can create in the form of an image. It is not limited to
motion blur or defocus. For example, if the shutter on your camera is
hexagonal rather than round, one can create a hexagonal shape for the
filter. Or if you want to produce an edge result you can, for example, make
a black circle in a white image for the filter. The disadvantage that it has
when applying motion blur and defocus is that it transform the line and circle
filter images from the spatial domain to the frequency domain. As we saw above,
when one does this, for example, for a circle, the result in the frequency
domain should be an image of the jinc function. But as we saw earlier, in
practice, it ends up with a less than ideal form and is somewhat broken up.
Here is the comparison again for the case of a 12 pixel diameter circle. On
the left is the Fourier Transform spectrum of the circle image and on the right
is the absolute value of the log of the ideal jinc function for the same
diameter.
Therefore another approach to blurring and deblurring, at least for motion
blur and defocus, would be to create the ideal Fourier Transform filter
directly without having to transform the equivalent spatial filter. We can
do this, since, as we saw above, we know the form and equations that are
needed to create the Fourier Transform filters for these two types of
blurring. For motion blur, it is the 1D (possibly rotated) sinc function and
for defocus, it is the jinc function. We note however, that we must calculate
and use the actual sinc and jinc functions rather than just their absolute
values. Consequently, both will have positive and negative values, which is
why these filter images must be saved in HDRI compatible image types.
We also know from our experiments with real only reconstruction of the circle
image that its imaginary component is relatively insignificant and so we do
not need an imaginary component for the filter.
The downside to this approach is that there is no current ImageMagick option
to create either the sinc or jinc function. Therefore they currently must be
created with the generally slower fx
option, although we can speed things up for special cases where the sinc
function's orientation is along the coordinate axes and also by using a 1D
look up table to implement the radially symmetric jinc function.
On the upside side, these filters can be stored and reused later, if desired.
But more importantly, the filtering operation is much simpler, since these
filters have no imaginary component. Therefore, for blurring and deblurring,
the corresponding multiplication and division math (with real and imaginary
components) becomes:
(16)
(17)
(18)
Ideal Motion Blur
Now let's repeat the same motion blur by 15 pixels horizontally, but create
the blurring filter in the form of the sinc function as an image for use
directly in the frequency domain. To blur the image, we then use the
modified multiplication equation (17) above between that sinc filter and the
Fourier Transform of the image. We do not transform the filter image as it
is already the equivalent filter for use in the frequency domain. Finally,
we transform the product back to the spatial domain. Since the actual sinc
filter has both positive and negative values, it cannot be displayed
directly. Therefore, the absolute value of the log of the filter is also
created for viewing purposes.
length=15
xfreq=`convert xc: format "%[fx:$length*pi/256]" info:`
convert size 256x1 xc: \
fx "zz=$xfreq*(iw/2); zz?sin(zz)/(zz):1" \
scale 256x256\! \
\( clone 0 clone 0 compose multiply composite \
evaluate pow 0.5 contraststretch 0 \
evaluate log 100 write sincfilter15_abs_log100.png \) \
+delete sincfilter15.pfm
convert \( cameraman.png +fft \) \
\( clone 0 sincfilter15.pfm compose multiply composite \) \
\( clone 1 sincfilter15.pfm compose multiply composite \) \
delete 0,1 +ift cameraman_idealmotion15.png

As before, a script version, camerablur, is available and its use to create the same
motion bur as above is as follows.
camerablur t motion a 15 r 0 cameraman.png cameraman_idealmotion15.png

Note there is no filter image, since the script creates the sinc function
internally. The argument r 0 indicates that the motion blur will be
horizontal. This is the default orientation, but other values may be supplied
to produce motion blur at any angle.
And now we deblur this result using equation (18) for the complex divide
operation. We can either use the filter that we created or create it again
exactly as above. We will do the former. However, when one has some actual
blurred image that was not artificially created, we would do the latter.
noise=0
qnoise=`convert xc: format "%[fx:quantumrange*$noise]" info:`
convert \( cameraman_idealmotion15.png +fft \) \
\( sincfilter15.pfm sincfilter15.pfm compose multiply composite evaluate add $qnoise \) \
\( clone 0 sincfilter15.pfm compose multiply composite \) \
\( clone 1 sincfilter15.pfm compose multiply composite \) \
\( clone 3 clone 2 +swap compose divide composite \) \
\( clone 4 clone 2 +swap compose divide composite \) \
delete 04 +ift cameraman_idealmotion15_deblur_0.png
compare metric rmse cameraman.png cameraman_idealmotion15_deblur_0.png null:
188.748 (0.00288011)

Again a script, cameradeblur, is available and the equivlant to the above is:
cameradeblur t motion a 15 r 0 cameraman_idealmotion15.png cameraman_idealmotion15_deblur_0.png

We now add noise to the blurred image using the same random noise image
that we generated before.
convert cameraman_idealmotion15.png noise_gray.png +swap \
compose mathematics set option:compose:args "0,1,0.01,0.005" \
composite cameraman_idealmotion15_noise.png
compare metric rmse cameraman_idealmotion15.png cameraman_idealmotion15_noise.png null:
190.033 (0.00289972)

And now get the noise and signal variances to estimate the noise parameter.
convert cameraman_idealmotion15_noise.png fill none stroke red \
draw "rectangle 240,40 249,89" alpha off cameraman_idealmotion15_noise_area.png
convert cameraman_idealmotion15_noise.png[10x50+240+40] format "%[standard_deviation]" info:
379.22
convert cameraman_idealmotion15_noise.png format "%[standard_deviation]" info:
16862
convert xc: format "%[fx:(379.22/16862)^2]" info:
0.000505783

So we again deblur using n=0.0005 and also n=0.005 and n=0.00005 for
comparison. The sinc function filter image is not explicity supplied, but is
created internally. We simply display it here for reference.
cameradeblur t motion a 15 r 0 n 0.0005 cameraman_idealmotion15_noise.png \
cameraman_idealmotion15_noise_deblur_0p005.png

Ideal Defocus
Now let's repeat the same defocus with a 15 pixel diameter, but create the
blurring filter as the jinc function as an image in the frequency domain. To
blur the image, we then use the modified multiplication equation (17) above
between the jinc filter and the Fourier Transform of the image. We do not
transform the filter image as it is already the equivalent filter for use in
the frequency domain. Finally, we transform the product back to the spatial
domain. Since the actual jinc filter has both positive and negative values,
it cannot be displayed directly. Therefore, the absolute value of the log of
the filter is also created for viewing purposes.
diam=15
diag=`convert xc: format "%[fx:sqrt(2)*256]" info:`
hdiag=`convert xc: format "%[fx:sqrt(2)*128]" info:`
rfreq=`convert xc: format "%[fx:$diam*pi/256]" info:`
a0=0.5; a1=.56249985; a2=.21093573; a3=.03954289; a4=.00443319; a5=.00031781; a6=.00001109
uu="(zz/3)"
jinc1="($a0+$a1*pow($uu,2)+$a2*pow($uu,4)+$a3*pow($uu,6)+$a4*pow($uu,8)+$a5*pow($uu,10)+$a6*pow($uu,12))"
b0=.79788456; b1=.00000156; b2=.01659667; b3=.00017105; b4=.00249511; b5=.00113653; b6=.00020033
c0=2.35619; c1=.12499612; c2=.00005650; c3=.00637879; c4=.00074348; c5=.00079824; c6=.00029166
iuu="(3/zz)"
vv="($b0+$b1*$iuu+$b2*pow($iuu,2)+$b3*pow($iuu,3)+$b4*pow($iuu,4)+$b5*pow($iuu,5)+$b6*pow($iuu,6))"
ww="(zz+$c0+$c1*$iuu+$c2*pow($iuu,2)+$c3*pow($iuu,3)+$c4*pow($iuu,4)+$c5*pow($iuu,5)+$c6*pow($iuu,6))"
jinc2="$vv*cos($ww)/(zz*sqrt(zz))"
convert size 1x${hdiag} gradient: rotate 90 \
fx "zz=hypot($rfreq*i,$rfreq*j); (zz<=3)?2*$jinc1:2*$jinc2" \
jincfilter15_lut.pfm
convert \( size ${diag}x${diag} radialgradient: negate \
gravity center crop 256x256+0+0 +repage \) \
jincfilter15_lut.pfm clut \
\( clone 0 clone 0 compose multiply composite \
evaluate pow 0.5 contraststretch 0 \
evaluate log 100 write jincfilter15_abs_log100.png \) \
+delete jincfilter15.pfm
convert \( cameraman.png +fft \) \
\( clone 0 jincfilter15.pfm compose multiply composite \) \
\( clone 1 jincfilter15.pfm compose multiply composite \) \
delete 0,1 +ift cameraman_idealdefocus15.png

The script version of the above is just:
camerablur t defocus a 15 m fast cameraman.png cameraman_idealdefocus15.png

The m argument allows for either a fast, but less accurate computation of the
jinc or a much slower, but more accurate computation.
One point to note about the blurred result. If one looks carefully at the
image, one sees some mottled darkening along the top edge and some brightening
along the bottom edge. This is a result of the Fourier transform assumption
that the image is periodic. In ImageMagick terms, it is the equivalent to
virtualpixel
tile. Thus the blurring along the one edge mixes those pixels with the opposite
edge. The only way to avoid this is to pad the image first with mirrored pixels
by an amount somewhat larger than the blur amount. So for example, let's
mirror unfold this image by 22 pixels (which is larger than 15) to a size of
300x300.
convert cameraman.png virtualpixel mirror \
set option:distort:viewport 300x3002222 \
distort SRT 0 +repage \
cameraman_mirror.png

Now lets do the same defocus blurring and crop the result back to its original
size of 256x256.
camerablur t defocus a 15 m fast cameraman_mirror.png cameraman_idealdefocus15_mirror.png
convert cameraman_idealdefocus15_mirror.png[256x256+22+22] cameraman_idealdefocus15_mirror_crop.png

This results in a higher quality image without the artifacts at the top and
bottom.
And now we go back and deblur the earlier result using equation (18) for the
complex divide operation. We can either use the filter that we created or
create it again exactly as above. We will do the former. However, when one
has some actual blurred image that was not artificially created, we would do
the latter.
noise=0
qnoise=`convert xc: format "%[fx:quantumrange*$noise]" info:`
convert \( cameraman_idealdefocus15.png +fft \) \
\( jincfilter15.pfm jincfilter15.pfm compose multiply composite evaluate add $qnoise \) \
\( clone 0 jincfilter15.pfm compose multiply composite \) \
\( clone 1 jincfilter15.pfm compose multiply composite \) \
\( clone 3 clone 2 +swap compose divide composite \) \
\( clone 4 clone 2 +swap compose divide composite \) \
delete 04 +ift cameraman_idealdefocus15_deblur_0.png
compare metric rmse cameraman.png cameraman_idealdefocus15_deblur_0.png null:
970.639 (0.014811)

And it even reverses the artifacts that were introduced along the top and
bottom edges from the Fourier transform assumption of image periodicity.
The script version of the above is just:
cameradeblur t defocus a 15 m fast cameraman_idealdefocus15.png \
cameraman_idealdefocus15_deblur_0.png

We now add noise to the blurred image using the same random noise image
that we generated before.
convert cameraman_idealdefocus15.png noise_gray.png +swap \
compose mathematics set option:compose:args "0,1,0.01,0.005" \
composite cameraman_idealdefocus15_noise.png
compare metric rmse cameraman_idealdefocus15.png cameraman_idealdefocus15_noise.png null:
190.033 (0.00289972)

And now we get the noise and signal variances to estimate the noise parameter.
convert cameraman_idealdefocus15_noise.png fill none stroke red \
draw "rectangle 240,40 249,89" alpha off cameraman_idealmotion15_noise_area.png
convert cameraman_idealdefocus15_noise.png[10x50+240+40] format "%[standard_deviation]" info:
372.646
convert cameraman_idealdefocus15_noise.png format "%[standard_deviation]" info:
16558.5
convert xc: format "%[fx:(372.646/16558.5)^2]" info:
0.000506467

So we again deblur using n=0.0005 and also n=0.005 and n=0.00005 for
comparison. The jinc function filter image is not explicity supplied, but is
created internally. We simply display it here for reference.
cameradeblur t defocus a 15 m fast n 0.0005 cameraman_idealdefocus15_noise.png \
cameraman_idealdefocus15_noise_deblur_0p0005.png

Determining The Blurring Parameters Using The Cepstrum
So far we have generated artificial examples where we knew exactly the type
of blur as either motion blur or defocus and the amount and orientation of
the blurring. But what if we don't know any of this. What can we do?
If we are lucky and the type of blurring is due to motion blur or defocus,
then one can extract the type, amount and orientation of the blurring from
something called the
Cepstrum
image. For defocus, the image will contain a white annular ring in the center
of the image whose radius is equal to the diameter of the defocus. For
motion blur, the image will contain an array of white dots at the orientation
of the motion blur. The center most dots, one on either side of the center
of the image will be separated from the origin by a distance equal to the
amount of motion blur (or the distance between them will be twice the amount
of motion blur). Of course there will also be noise in the image that may
obscure the dots, especially if the original image is grayscale.
The cepstrum is generate by first computing a running average of the
magnitude of the Fourier Transform from a large set of subsections of the
image each of which is windowed using typically about a 40% cosinebell
rolloff to black. The windowing is done to help assure that each subsection
conforms to the Fourier Transform assumption that the image is periodic. We
take the average result from many subsections to reduce the 'noise' from the
image signal and emphasise the effects from the blurring, since the blurring
will be the same in each subsection, but the image content will not. The
negative log is applied to this running average, which is then stretched to
span the full dynamic range. We use a form of the log that has a scaling
constant built in. The log is used to emphasize small values more and
deemphasize large values. An appropriate choice of this constant is
helpful in reducing the 'noise' from the image signal versus the
effect from the blurring. Finally, this result is treated as the real
component and combined with a black image for the imaginary component and
inverse transformed back to the spatial domain. The name cepstrum is derived
from the word spectrum, but with the first four letters reversed. It is in
some sense an inverse spectrum, because we create the spectrum (i.e.
log of the magnitude), but inverse transform it back to the spatial domain.
NOTE that this process requires the use of the real and imaginary components
of the Fourier Transform therefore must be done with ImageMagick compiled
with HDRI enabled.
Let's do this on the 15 pixel diameter ideal defocused image that we created
earlier. The image is 256x256 in size. We will use subsections that are 25%
in size (64x64) with a 50% overlap (i.e. offset of 32x32) and a 40% windowing
rolloff (i.e. 25.6x25.6). The windowing image will be created and shown next
to the defocused image. Once we create the cepstrum image, a red circle of
radius 15 relative to the center will be drawn on it.
convert \( size 64x1 xc: \
fx "(0.5*cos(pi*max(0,(1i/25.6)))+0.5)*(0.5*cos(pi*max(0,(1(wi)/25.6)))+0.5)" scale 64x64\! \) \
\( size 1x64 xc: \
fx "(0.5*cos(pi*max(0,(1j/25.6)))+0.5)*(0.5*cos(pi*max(0,(1(hj)/25.6)))+0.5)" scale 64x64\! \) \
compose multiply composite cos_bell64.png
convert size 64x64 xc:black magave.pfm
k=1
dh=0
for ((i=0;i<=192;i+=32)); do
dw=0
for ((j=0;j<=192;j+=32)); do
echo "Iteration: $k"
pct=`convert xc: format "%[fx:100/($k)]" info:`
convert cameraman_idealdefocus15.png[64x64+${dw}+${dh}] cos_bell64.png \
compose multiply composite fft delete 1 magave.pfm \
+swap compose blend set option:compose:args $pct composite magave.pfm
#echo "i=$i; j=$j; k=$k; pct=$pct"
dw=$(($dw+32))
((k++))
done
dh=$(($dh+32))
done
convert magave.pfm evaluate log 20000 negate \
\( size 64x64 xc:black \) +ift roll 3232 \
cameraman_idealdefocus15_cepstrum_20000.png
convert delay 100 cameraman_idealdefocus15_cepstrum_20000.png \
\( clone 0 fill none stroke red draw "circle 32,32 32,46" alpha off \) loop 0 \
cameraman_idealdefocus15_cepstrum_20000.gif

A script version, cepstrum, is available and its use to create the same
preanimation result as above is as follows. The windowing image does
not have to be pregenerated as it is created internally in the script.
cepstrum c 20000 cameraman_idealdefocus15.png cameraman_idealdefocus15_cepstrum_20000.png
convert delay 100 cameraman_idealdefocus15_cepstrum_20000.png \
\( clone 0 fill none stroke red draw "circle 32,32 32,46" alpha off \) loop 0 \
cameraman_idealdefocus15_cepstrum_20000.gif

Now let's do the same with motion blur, but let's create a motion blur of
length 15, but at a 45 degree angle. The same windowing image will be shown
next to the motion blurred image, but it is not actually a script input, as it
is created internally to the script. Once we create the cepstrum image,
a red line of length 30 will be drawn centered on the image at an angle
of 45 degrees.
length=15
angle=45
sinang=`convert xc: format "%[fx:sin($angle*pi/180)]" info:`
cosang=`convert xc: format "%[fx:cos($angle*pi/180)]" info:`
xfact=`convert xc: format "%[fx:$cosang*256/2]" info:`
yfact=`convert xc: format "%[fx:$sinang*256/2]" info:`
freq1=`convert xc: format "%[fx:$length*pi/256]" info:`
freq2=`convert xc: format "%[fx:$freq1/(2*pi)]" info:`
convert \( size 256x257 gradient: \
gravity center crop 256x256+0+1 +repage \
linearstretch 1x0 function polynomial "2,1" \) \
\( clone 0 rotate 90 evaluate multiply $xfact \) \
\( clone 0 rotate 180 evaluate multiply $yfact \) \
\( clone 1 clone 2 compose plus composite \) \
\( clone 3 function sinusoid "$freq2,0,1,0" clone 3 +swap compose divide composite \
linearstretch 0x1 fill white opaque black write sincfilter15r45.pfm \) \
\( clone 4 clone 4 compose multiply composite evaluate pow 0.5 \) \
delete 04 evaluate log 100 sincfilter15r45_abs_log100.png
convert \( cameraman.png +fft \) \
\( clone 0 sincfilter15r45.pfm compose multiply composite \) \
\( clone 1 sincfilter15r45.pfm compose multiply composite \) \
delete 0,1 +ift cameraman_idealmotion15r45.png
cepstrum c 5000 cameraman_idealmotion15r45.png cameraman_idealmotion15r45_cepstrum_5000.png
convert delay 100 cameraman_idealmotion15r45_cepstrum_5000.png \
\( clone 0 fill none stroke red draw "line 21.4,42.6 42.6,21.4" alpha off \) loop 0 \
cameraman_idealmotion15r45_cepstrum_5000.gif

Deblurring Real World Images
In practice, deblurring images with actual real world degradations is generally
much harder than those artificially degraded by ideal motion blur or ideal
defocus. With real world situations and actual cameras, the degradations do not
ideally conform to simple uniform linear motion blur or lens defocus. For
example, most motion blur is not uniform and not linear. It does not conform to
an ideal or pure straight line spatial filter (or sinc function in the frequency
domain). Likewise, most lens defocus does not conform ideally to an ideal
or perfect circular disk as the spatial filter (or jinc function in the
frequency domain. Furthermore there are other degradations that might be
included, which are not uniform over the image, such as depth of field
blurring.
Sometimes one may be successful with defocus, if the cepstrum shows a good
circular ring. For example, the following is an actual image (subsection)
taken with the lens purposely defocused. Image curtesy of Maxim Usatov.
The following is the result of running the cepstrum script followed by
contrastsstretch 0x1% and then a 9 pixel radius circle drawn about the center
of the image to match the ring of circular dots indicative of the diameter
of the defocus.
To get the appropriate noise constant, we measure the square of
the standard deviation in a flat region of the image divided by the square of
the standard deviation of the whole image. This concept is shown in the
following image and results in a noise constant value estimate of 0.005.
Then we use the cameradeblur script with t defocus d 9 n 0.005 to recover
the image.
On the other hand, true motion blur is almost never a simple situation that
conforms to the ideal models above.
Processing of real (as opposed to ideally blurred images) is very hard. If the
degradation is known or can be identified either in the spatial domain (as the
point spread function or PSF) or in the frequency domain (as the modulation
transfer function or MTF), then it can be used to deconvolve the image. Much
study has been reported in the literature on efforts to identify and use this
information. However, the techniques for obtaining such information are
difficult and the restorations are generally complicated and time consuming.
Another technique is called
blind deconvolution. It makes only a few assumptions
about the degradation. The solutions are often iterative, updating an estimate
for the degradation and then deconvolving the image. Then using that result to
improve the estimate for the degradation, etc. One such technique is the
RichardsonLucy technique, which is popular for
deblurring astronomical photography. Another technique that is not iterative was
first described by Jae S. Lim, TwoDimensional Signal and Image Processing,
PrenticeHall, 1990. Section 9.3.2, "Algorithms for Blind Deconvolution" and
summarized in a lecture by
Harvey Rhody. A variation on that theme was later
developed as the Quarktet Tria  SeDDaRA Method.
One point about these latter two techniques is that they cannot properly handle
motion blur or defocus, because they make an assumption that the degradation is
positive only. Both motion blur and defocus do not satisfy that assumption,
because they obey a sinc and jinc function, respectively, in the frequency
domain and therefore contain negative lobes. However, they do a nice job of
removing many other types of degradations and so can make significant
improvement in realworld degraded images. Another interesting point about
these latter techniques is that they can use a similar, but unblurred image
to help restore the blurred image.
Template Matching By Normalized Cross Correlation
In this section, we discuss the template matching or the matching of a small
image within a larger image to see where the small image best matches to the
same size subsection of the larger image. The technique that we use is called
Normalized Cross Correlation.
The cross correlation of two equal sized image, is basically the pointbypoint
multiplication of values within each image, totalling the products.
If the two images are not the same size, then the smaller image is moved
across the larger image and the cross correlation is computed for each shift
by one pixel and stored in resulting image whose origin represents the smaller
image located at the top left corner of the larger image.
The problem with this technique is that the two images may have similar
features at least at one shift position of the smaller image relative to the
larger image. However, the exposure differences between the two images may
make the match values too poor and in fact lead to false matches.
Consequently, we normalize the two image in two ways. First we subtract the mean
value of the template image from each pixel value in the template and also
subtract the mean value of each subsection of the larger image before doing
the correlation operation. Second we divide the template image by its standard
deviation and also divide each subsection of the larger image by its standard
deviation. The result of this is a normalized cross correlation surface image,
which varies from 1 to +1 at best, where +1 corresponds to a perfect match.
In general, we can truncate the correlation surface to only keep nonnegative
values as anticorrelation is not generally relevant in image matching.
The formula for the discrete normalized cross correlation surface (i.e. from
matching digital images) can be expressed as follows.
(19)
Here, C(h,k) is the normalized cross correlation value at each shift
position (h,k) of the smaller image's upper left corner relative the larger
images upper left corner. The values M_{S} and M_{L} are the
mean value of the small image and the mean value of the subsection of the
larger image for the current shift position. Likewise, σ_{S}
and σ_{L} are the standard deviations of the smaller image and
the standard deviation of the subsections of the larger image at shift
position (h,k). The mean and standard deviations introduce the
normalization of the images during the cross correlation process. The value
N is the total number of pixels contained in the smaller image.
Now, the standard deviation is just the square root of the
variance.
And computationally, the variance of an image can be computed as the
(average of the square of the image) minus the (square of the average of the
image). Thus, the standard deviation of the subsection of the large image can
be expressed as:
(20)
Therefore, the normalized cross correlation becomes:
(21)
The problem with doing the normalized cross correlation in the spatial domain
as defined by the above equation is that it is a very slow process.
Correlation in the spatial domain is very similar to convolution. However,
in the case of convolution, we generally use small kernels or filter images.
Consequently this can be done in a reasonable amount of time. As the small
kernel or in this case the small image attains any reasonable size, this
process slows down as the square of the small image size.
Consequently, it is more efficient to convert this equation to its frequency
domain equivalent and pay the time price of a Fourier Transform on both the
small and large images, but be able to do correlation much faster, because
correlation much like convolution in the frequency domain is simply a
multiply. Thus the time to do the processing increases approximately linearly
with the size of the larger image and is mostly independent of the size of
the smaller image.
Therefore, the following equation is presented without derivation as the
frequency domain equivalent. We introduce a special shorthand notation, an X
inscribed inside a circle, , to represent the Forward Fourier Transform of two images,
correlation of the two transformed images and finally the Inverse Fourier
Transform of the correlation product. For example, A B = I[F*(A) x F(B)]. More
specifically this means take the Forward Fourier Transform of A and B. Then
convert F(A) to its complex conjugate, F*(A). Then do complex multiplication
between F*(A) and F(B) using the real and imaginary components. Then finally
take the Inverse Fourier Transform of the product. Using this shorthand
notation, the equation for the normalize cross correlation surface or image,
becomes:
(22)
In order to multiply two images in the frequency domain, they must be
the same size. Therefore we pad the smaller image with zeroes (black) to fill
it out to the same size as the larger image. The padding is on the right and
bottom. Thus, S is the smaller image padded to the same size as the larger
image. L is the larger image and L^{2} is the square of the larger
image, i.e, multiplied by itself. U is a unit image (values=1) the size of
the original smaller image, but again padded with zeroes (black). This
image is a windowing image that is used to select every possible section of
the L image that is the size of the smaller (unpadded image). M_{S}
is an image the size of the smaller image, whose constant value is equal to
the mean value of the smaller image, but again it is padded with black. In
effect, SM_{S} is just the smaller image with its mean subtracted
and the difference is padded with black. Note, however, that it will have
both positive and negative values. But we can do that in HDRI ImageMagick
with a proper choice of image format that permits negative values. Here
again N is the total number of pixels in the original (unpadded) smaller
image and σ_{S} is the standard deviation of the original
(unpadded) smaller image.
In practice, we convert the two images to grayscale to do the normalized
cross correlation to avoid any possibility of getting different matching
locations from each channel.
Two examples will be presented. Each is produced by the script, normcrosscorr. The script is presented here rather than
an extremely long set of command line operations, because the implementation
requires 3 repetitions of complex multiply, each of which is accomplished with
a long set of composite with a compose operations.
Under ordinary circumstances, the above equation should normalize to a range
between 1 and +1. A perfect match would then become +1. ImageMagick is a
bit strange in that it normalizes the values in each image to a range of 0
to 1, internally, before doing any operation such as imagetoimage
multiplication. The above formulae are founded upon proper multiplication of
the unnormalized FFT (and normalized IFT) images. The result is that with
Imagemagick, one has to account for scaling of multiplied images by
quantumrange, scaling the square root by dividing by sqrt(quantumrange) and
also account for correction for the type of FFT normalization by multiplying
the two paddedimages, U and (SMs) by the total pixels of the large image, if
normalization is in the forward FFT rather than the inverse IFT. Due to
Imagemagick's internal normalization of images to the range 0 to 1, the U
image is actually a fully white region the size of the smaller image (rather
than a unit image), but padded with black to the size of the larger image.
See Fourier Transform Normlization
for a list of various FFT vs IFT normalizations.
In the first example, the smaller image is just a 64x64 pixel subsection of
the larger 128x128 imagee. The subsection was created with its upper left
corner coordinates at (32,27) in the larger image. A 3D plot of the
stretched correlation surface is also shown.
normcrosscorr zelda3_32_27.png zelda3.png zelda3_ncc.png zelda3_match.png
Match Coords: (32,27) And Score In Range 0 to 1: (0.999681)

The s option tells the script to stretch the correlation surface to full
dynamic range before output. The result shows that the correct match was
indeed found, and the match score was 0.665. This is the largest value in
the correlation surface image and it corresponds to the location of the top
left corner of the smaller image at the best match locatation in the larger
image.
In the second example, the smaller image is a 23x14 subsection of the larger
256x256 image. It corresponds to the eye of the mandril on the right side of
the image. The subsection was created with its upper left corner coordinates
at (156,22) in the larger image.
normcrosscorr c white mandril3_156_22.png mandril3.png mandril3_ncc.png mandril3_match.png
Match Coords: (156,22) And Score In Range 0 to 1: (0.998915)

The p argument tells the script to pseudocolor the correlation surface rather
than leave it grayscale. Low values are purple and high values are red.
The c argument tells the script to draw the box representing the match
location of the smaller image in the larger image in the desired color, in
this case white. The result again shows that the correct match, the right eye,
was indeed found, and the match score was 0.276.
However, one can see that the left eye also has a high score by its coloration
in the correlation surface. The selection of the small eye image was done
purposely to show such a case. So how can we determing the value at this
second peak in the correlation surface, since the script only gives us the
highest score. Furthermore, the area around the right eye is also a close
match.
We can process the stretched correlation surface by subtracting the local
mean value in a 3x3 neighborhood from the each pixel and threshold that at
an appropriate level to separate out the two highest values in the surface.
This difference is like an edge or slope extraction. The difference will be
positive where the image is higher than the mean and negative where the
image is lower than the mean. Since the peaks are very sharp near the eyes,
there will be a large slope there and an appropriate threshold should then
be able to locate just those two peak values. Once we get the two highest
values we can then draw boxes the size of the smaller image within the
larger image for where those two matches occur. A 3D plot of the stretched
correlation surface is also shown.
convert mandril3_ncc.png \
\( clone 0 convolve '1,1,1,1,1,1,1,1' \) \
\( clone 0 clone 1 +swap compose minus composite \) \
delete 0,1 threshold 15% txt:  grep white  \
tail n +2
156,22: (65535,65535,65535) #FFFFFFFFFFFF white
76,24: (65535,65535,65535) #FFFFFFFFFFFF white
convert mandril3.png fill none stroke white \
draw "rectangle 156,22 178,35 rectangle 76,24 98,37" alpha off mandril3_match2.png

A more flexible method is to locate the first peak, then mask it out,
locate the next highest peak and mask it out, etc. I have a new script
called maxima that will do just that.
maxima n 2 mandril3_ncc.png
156,22 gray=65535,255,100%
76,24 gray=54453,212,83.09%

Fast Template Matching Using Local Statistics In The Fourier Domain
The following link is to my unpublished paper describing the mathematical basis
for various types of correlationbased template matching computed fully via
Fourier Transforms, which are 1 to 2 orders of magnitude faster than doing the
same in the spatial domain. ACCELERATED TEMPLATE MATCHING USING LOCAL STATISTICS AND FOURIER TRANSFORMS.pdf
My thanks to Sean Burke for his coding of the original demo and to
ImageMagick's creator for integrating it into ImageMagick. Both were heroic
efforts.
 