关于OpenCV中DCT函数方程及Scilab JPEG压缩实现的疑问

Hey there! Let's break down how OpenCV's dct function maps to the equation you're using in your Scilab code for JPEG compression—since you're comparing results between Scilab's built-in imdct and your custom my_function on the Lenna image, this should clear up the confusion.

First: Your Scilab Equation Context #

From what you've shared, your my_function uses the standard 8x8 2D DCT-II (the industry standard for JPEG), which looks like this (assuming you're working with inverse DCT, since you mentioned imdct):

For an 8x8 block of DCT coefficients ( X(u,v) ), the inverse transform to recover the pixel block ( x(i,j) ) is:
[
x(i,j) = \frac{1}{4} \sum_{u=0}^{7} \sum_{v=0}^{7} C(u)C(v) X(u,v) \cos\left( \frac{(2i+1)u\pi}{16} \right) \cos\left( \frac{(2j+1)v\pi}{16} \right)
]
Where the scaling factors ( C(k) ) are:

• ( C(0) = \frac{1}{\sqrt{2}} )
• ( C(k) = 1 ) for ( k = 1,2,...,7 )

OpenCV's Corresponding DCT Equation #

OpenCV's dct function implements the exact same JPEG-compliant DCT-II, but with one key difference: default scaling. Here's the breakdown for both forward and inverse transforms:

Forward DCT (Compression Step) #

OpenCV's forward DCT (called with cv::dct(src, dst, DCT_FORWARD)) calculates coefficients as:
[
dst(u,v) = C(u)C(v) \sum_{i=0}^{7} \sum_{j=0}^{7} x(i,j) \cos\left( \frac{(2i+1)u\pi}{16} \right) \cos\left( \frac{(2j+1)v\pi}{16} \right)
]
Notice this omits the ( \frac{1}{4} ) scaling factor present in your Scilab equation. To match Scilab's forward DCT output, divide OpenCV's result by 4.

Inverse DCT (Decompression Step) #

For inverse transforms (what imdct handles), OpenCV's inverse DCT (called with cv::dct(src, dst, DCT_INVERSE)) outputs:
[
dst(i,j) = \sum_{u=0}^{7} \sum_{v=0}^{7} C(u)C(v) X(u,v) \cos\left( \frac{(2i+1)u\pi}{16} \right) \cos\left( \frac{(2j+1)v\pi}{16} \right)
]
Again, no ( \frac{1}{4} ) scaling—so to get results matching Scilab's imdct, divide OpenCV's inverse DCT output by 4.

Quick Verification Steps for Lenna Image #

To confirm alignment between your code and OpenCV:

• Grab an 8x8 pixel block from Lenna, shift values to the JPEG standard range (-128 to 127, since JPEG centers pixels around 0).
• Compute forward DCT in Scilab and OpenCV, then divide OpenCV's output by 4—you should get identical coefficient values.
• For inverse DCT: take the same DCT coefficients, run Scilab's imdct and OpenCV's inverse DCT, then divide OpenCV's result by 4—pixel values should match.

Key Takeaway #

OpenCV's dct uses the exact same mathematical core as your Scilab equation; the only gap is the default scaling factor. Add a division by 4 to OpenCV's forward/inverse results, and you'll get perfect alignment with your my_function and Scilab's imdct.

内容的提问来源于stack exchange，提问作者Chris Tine