Understanding gain of the DCT transform

Here is the small experiment I was able to conduct on my Debian system. I crafted two small $8 \times 8$ JPEG files again:

• white: ()
• black: ()

I used the following commands:

% convert -size 8x8 -depth 8 xc:white white.pgm % convert -size 8x8 -depth 8 xc:black black.pgm 

then:

% cjpeg -sample 1x1 -quality 100 -grayscale -outfile white.jpg white.pgm % cjpeg -sample 1x1 -quality 100 -grayscale -outfile black.jpg black.pgm 

Note that in this case, it is a single component JPEG (no color space transform):

% file white.jpg white.jpg: JPEG image data, JFIF standard 1.01, aspect ratio, density 1x1, segment length 16, baseline, precision 8, 8x8, components 1 

Upon inspection, one can check that the DC coefficient stored in white.jpg is 1016, while the DC coefficient stored in black.jpg is -1024. Which can also be represented as:

$$\begin{bmatrix} 1016&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0 \end{bmatrix}$$

One can also check that the quantization table is simply a a bunch of ones:

% djpeg -verbose -verbose -outfile dummy.ppm white.jpg [...] Define Quantization Table 0 precision 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 

So taking the entry-for-entry product with the quantization matrix from above results simply in the same matrix.

Going back to the equations on wikipedia page:

$$f_{x,y} = \frac{1}{4} \sum_{u=0}^7 \sum_{v=0}^7 \alpha(u) \alpha(v) F_{u,v} \cos \left[\frac{(2x+1)u\pi}{16} \right] \cos \left[\frac{(2y+1)v\pi}{16} \right]$$

Simply gives on our case (no rounding error):

$$f_{x,y} = \frac{1}{4}\left(\frac{1}{\sqrt{2}\sqrt{2}}1016\times 1+63\times 0\right) = 127$$

So adding 128 to each entries lead to:

$$\left[ \begin{array}{rrrrrrrr} 255 & 255 & 255 & 255 & 255 & 255 & 255 & 255 \\ 255 & 255 & 255 & 255 & 255 & 255 & 255 & 255 \\ 255 & 255 & 255 & 255 & 255 & 255 & 255 & 255 \\ 255 & 255 & 255 & 255 & 255 & 255 & 255 & 255 \\ 255 & 255 & 255 & 255 & 255 & 255 & 255 & 255 \\ 255 & 255 & 255 & 255 & 255 & 255 & 255 & 255 \\ 255 & 255 & 255 & 255 & 255 & 255 & 255 & 255 \\ 255 & 255 & 255 & 255 & 255 & 255 & 255 & 255 \end{array} \right]$$

Repeating the above for the black.jpg image with DC value -1024 gives:

$$f_{x,y} = \frac{1}{4}\left(\frac{1}{\sqrt{2}\sqrt{2}}\times -1024\times 1+63\times 0\right) = -128$$

Which is what we expected.