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The standard type III discrete cosine transformation (DCT) is defined as follows:

$${X_k} = \frac{1}{2}{x_0} + \sum\limits_{n = 1}^{N - 1} {{x_n}} \cos \left[ {\frac{\pi }{N}n\left( {k + \frac{1}{2}} \right)} \right]\quad \quad k = 0, \ldots ,N - 1.$$

It transforms a list of $\{x_n\}$ into list of $\{X_k\}$

Now I encountered a summation in a paper like this:

$${X_k} = \frac{1}{2}{x_0} + \sum\limits_{n = 1}^{N - 1} {{x_n}} \cos \left[ {\frac{\pi }{{\tilde N}}n\left( {k + \frac{1}{2}} \right)} \right]\quad \quad k = 0, \ldots ,\tilde N - 1.$$

The only difference is that now $\tilde N = 2N$.

The paper says this summation could exploit the power of fast DCT. However, it tooks me days, still can't figure out how to relate it to the standard DCT-III. How to do it?

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    $\begingroup$ for a helpful response, you'll want to disclose in which paper you saw this alternative definition... $\endgroup$ Apr 4, 2016 at 12:23
  • $\begingroup$ @CarloBeenakker Thank you for reply. Actually, it is a physics paper. Here journals.aps.org/rmp/abstract/10.1103/RevModPhys.78.275 On page 284, section "1. Discrete cosine and Fourier transforms" $\endgroup$
    – user15964
    Apr 4, 2016 at 12:28

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in the second transformation the integer $k$ runs from $0$ to $\tilde{N}-1$, not to $N-1$; so you have twice as many data points $X_k$ than you have moments $x_n$, which is perfectly OK (oversampling); to use the conventional formulas for the DCT, just extend the list of moments $x_n$ by padding it with zeros ($x_n=0$ for $N\leq n\leq \tilde{N}-1$, so that you have $\tilde{N}$ moments $x_n$; these then give you $\tilde{N}$ values $X_n$.

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  • $\begingroup$ You are so smart !! Just padding zeros !! I feel stupid : ) But Can you explain the oversampling a little bit. Because, theoretically we could pad any number zero to make it 2N,3N,4N,... I am sure it becomes worse when zero is too much. However, is 2N better than N? why? $\endgroup$
    – user15964
    Apr 4, 2016 at 15:06

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