离散余弦变换公式视差
嗯，我正在编写一些需要使用 DCT 的程序。我找到了 DCT 公式的 2 个资源：
最初我使用维基百科版本的 DCT-II。在 wiki 页面的 DCT-II 部分中，写道一些作者进一步将 X0 项乘以 1/√2，并将所得矩阵乘以整体比例因子，这使得 DCT-II 矩阵正交，但打破了直接与半移位输入的实偶 DFT 对应。 mathworks 网站只做这个。
正在讨论的这个属性是什么？
Well, I was programming something that required the use of DCT. I found 2 resources for the DCT formula:
Initially I used the wikipedia version of DCT-II. In the DCT-II section of wiki page, it is written that some authors further multiply the X0 term by 1/√2 and multiply the resulting matrix by an overall scale factor, which makes the DCT-II matrix orthogonal, but breaks the direct correspondence with a real-even DFT of half-shifted input. And the mathworks site does this only.
What is this property being talked about?
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我相信他们试图说的是关心使 DCT-II 变换矩阵成为 酉矩阵< /a>.从信号处理的角度来看，有一个酉矩阵是很好的，因为当我们将信号变换回其原始域时，我们不会向信号添加任何更多的功率。
但是，一维 DFT：
可以用正弦和 consies 重写（使用 欧拉恒等式）。如果输入是实偶信号，则 DFT 的偶项将对应于 DCT 的项。有些人喜欢通过简单地对信号进行 DFT 来简化他们的算法，并且只关注偶数项。
I beleive that they are trying to say that that are concerned about making the DCT-II transform matrix a unitary matrix. It is nice from a signal processing standpoint to have a unitary matrix because when we transform the signal back to its original domain, we are not adding any more power into the signal.
However, the 1-D DFT:
can be rewritten in terms of sines and consies (using Euler's Identity). If the input is a real-even signal, the even terms of the DFT will correspond to the terms of the DCT. Some people like to simplify their algorithms by simply taking the DFT of a signal, and only concentrating on the even terms.