The two dimensional DCT can be generated using the equation

$F=CfC’ \space\space (eq.1)$

Where f = Given image

$C(u,v) = \sqrt{1/N}\space\space\space u=0,0\lt=v\lt=N-1 \\ C(u,v) = \sqrt{2/N} \cos[\dfrac{π(2v+1)u}{2N}]1\lt=u\lt=N-1, 0\lt=v\lt=N-1$

DCT of the given image

The given image is

$\begin{bmatrix} 2&4&4&2\\ 4&6&8&3 \\ 2&8&10&4 \\ 3&8&6&2 \end{bmatrix}$

Using eq.1 i.e. $F=CfC’$

Calculating C(u,v)

For $u=0, 0\lt=v\lt=N-1\\ C(u,v)= \sqrt{1/N} \\ = \sqrt{¼} = 0.5$

Therefore $C(0,0) = C(0,1) = C(0,2) = C(0,3) = 0.5$

$C(0,0) = C(0,1) = C(0,2) = C(0,3) = 0.5 \\ For\space\space u=1, v=0\\ C(u,v) = \sqrt{2/N}\cos[\dfrac {π(2v+1)u}{2N}] \\ C(1,0) = \sqrt{2/4} \cos[\dfrac {π(0+1)1}8] = 0.653 \\ For \space\space u=1, v=1 \\ C(1,1) = \sqrt{2/4} \cos[\dfrac {π(2+1)1}8] = 0.2705 \\ For\space \space u=1, v=2 \\ C(1,2) = \sqrt{2/4}\cos[\dfrac {π(4+1)1}8] = -0.2705 \\ For \space\space u=1, v=3\\ C(1,3) = \sqrt{2/4}\cos[\dfrac {π(6+1)1}8] = -0.653 \\ For \space \space u=2, v=0\\ C(2,0) = \sqrt{2/4}\cos[\dfrac {π(0+1)2}8] = 0.5 \\ For \space\space u=2, v=1 \\ C(2,1) = \sqrt{2/4} \cos[\dfrac {π(2+1)2}8] = -0.5 \\ For\space\space u=2, v=2 \\ C(2,2) = \sqrt{2/4}\cos[\dfrac {π(4+1)2}8] = -0.5 \\ For \space\space u=2, v=3 \\ C(2,3) = \sqrt{2/4} \cos[\dfrac {π(6+1)2}8] = 0.5 \\ For\space\space u=3, v=0 \\ C(3,0) = \sqrt{2/4} \cos[\dfrac {π(0+1)3}8] = 0.2705 \\ For \space\space u=3, v=1 \\ C(3,1) = \sqrt{2/4}\cos[\dfrac {π(2+1)3}8] = -0.653 \\ For\space\space u=3, v=2 \\ C(3,2) = \sqrt{2/4} \cos[\dfrac {π(4+1)3}8] = 0.653 \\ For \space\space u=3, v=3\\ C(3,3) = \sqrt{2/4} \cos[\dfrac {π(6+1)3}8] = -0.2705$