这些 2D 余弦函数与您的输入图像无关。它们只是“余弦波”，或在转换“ ”块时产生 64 个系数的 64 个基函数。给定 DCT-II 的矩阵和图像块，您将通过以下方式获得系数：$8\times 8$$D$ $8\times 8$$8\times 8$$I$$64$$C$

$$C=DID^T$$

[编辑] 如果您只想显示 2D DCT-II 函数，则必须为每个创建 2D 数组（和）一对水平/垂直整数索引，其中和，其中和。经典的 JPEG，。一个公式是（直到一个转置，可能还有一个比例因子）：$D_{h,v}(m,n)$$0\le m< M$$0\le n< N$$(h,v)$$0\le h< H$$0\le v< V$$H=M$$V=N$$M=N=8$

$$D_{h,v}(m,n) = 2\sum_{m=0}^{M-1}\sum_{n=0}^{N-1} \frac{1}{\sqrt{M}}\frac{1}{\sqrt{N}}\eta_h \eta_v \cos\left(\frac{\pi h}{2M}(2m+1)\right)\cos\left(\frac{\pi v}{2N}(2n+1)\right)$$

with $\eta_x = \frac{1}{\sqrt{2}}$ if $x=0$, $\eta_x = 1$ if $x\neq0$. There are a lot of clever implementations, I'll provide the most straightforward. The goal is to get a picture like the following:

A pseudo-code version (from Matlab) could be (with the issue of 0-based or 1-based indices):

nRow = 8;nCol = 8;
nFreqHoriz = 8;
nFreqVerti = 8;
for iFreqH = 1:nFreqHoriz
    for iFreqV = 1:nFreqVerti
        iFreqHoriz = iFreqH-1;
        iFreqVerti = iFreqV-1;
        normFreqDC = 1/sqrt(2);
        matFreq2D = zeros(nRow,nCol);
        for iRow = 1:nRow
            for iCol = 1:nCol
                iRow0 = iRow-1;
                iCol0 = iCol-1;
                matFreq2D(iRow,iCol) = 2/(sqrt(nFreqHoriz)*sqrt(nFreqVerti))*cos(pi*iFreqHoriz*(2*iRow0+1)/(2*nRow))*cos(pi*iFreqVerti*(2*iCol0+1)/(2*nCol));
                if ~iFreqHoriz ,  matFreq2D(iRow,iCol) =  matFreq2D(iRow,iCol)*normFreqDC;end
                if ~iFreqVerti ,  matFreq2D(iRow,iCol) =  matFreq2D(iRow,iCol)*normFreqDC;end
            end
        end
        sum(matFreq2D(:).^2)
        subplot(nFreqHoriz,nFreqVerti,8*(iFreqV-1)+iFreqH)
        imagesc(matFreq2D);axis off;axis square;colormap gray
    end
end