I have been trying to understand the definition of the DCT and found a really good website describing its definition.

Here is the link DCT.

What I don't understand is the part of the DFT. Why is the summation of the DFT multiplied by $\frac{1}{\sqrt{2N}}$?


1 Answer 1


Given $\{x_n, 0 \leq n \leq N - 1\}$, the conventional DFT is defined as

$$X_k = \sum_n x_n e^{-j2\pi kn/N}$$

The Parseval theorem gives

$$\sum_{n=0}^{N-1} |x_n|^2 = \frac{1}{N}\sum_{k=0}^{N-1} |X_k|^2$$

The DFT in your link is called Normalized DFT (NDFT): $$Y_k = \frac{1}{\sqrt{N}} X_k = \frac{1}{\sqrt{N}} \sum_n x_n e^{-j2\pi kn/N}$$


$$\sum_{n=0}^{N-1} |x_n|^2 = \sum_{k=0}^{N-1} |Y_k|^2$$

Let me cite the idea behind this Normalized DFT (please follow the link if you desire more details)

While this definition is much cleaner from a geometric signal theory point of view, it is rarely used in practice since it requires slightly more computation than the typical definition. However, note that the only difference between the forward and inverse transforms in this case is the sign of the exponent in the kernel. Advantages of the NDFT over the DFT in fixed-point implementations (Appendix G) are discussed in Appendix A.

It can be said that only the NDFT provides a proper change of coordinates from the time-domain (shifted impulse basis signals) to the frequency-domain (DFT sinusoid basis signals). That is, only the NDFT is a pure rotation in $\mathbb{C}^N$ , preserving both orthogonality and the unit-norm property of the basis functions. The DFT, in contrast, preserves orthogonality, but the norms of the basis functions grow to $\sqrt{N}$ . Therefore, in the present context, the DFT coefficients can be considered denormalized frequency-domain coordinates.


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