# Context

I am reading this book Discrete Cosine Transform [...]. K.R. Rao and P.YIP. 1990.

Given That

From Equation (2.4.11.a)

$$X^{C(2)}(m) = \left ( \frac{2}{N} \right )^{1/2} k_m \sum^{N-1}_{n=0} x(n) \cos \left [ \frac{(2n+1)m \pi}{2N} \right ]$$

$$m=0, ..., N-1$$

Here, we have

$$k_p = \begin{cases} \frac{1}{\sqrt{2}} & \text{when } p = 0 \text{ or } N \\ 1 & \text{otherwise.} \end{cases}$$

From Book's Equation (2.8.7)

$$F^{C(2)}(k) = k_k \exp \left [ \frac{-j \pi k}{2N} \right ] \hat{F}_F(k)$$

Here, we have

$$k_p = \begin{cases} \frac{1}{\sqrt{2}} & p = 0 \\ 1 & p = 1, ..., N-1 \\ 0 & \text{otherwise} \\ \end{cases}$$

Now Comes the following

It is not difficult to show that the sequence $$f(n)$$ can be recovered from the transformed sequence $$F^{C(2)}(k)$$, as as a result we have

From (Equation 2.8.8)

$$f(n) = \sqrt{\frac{2}{n}} \Re \left \{ \sum^{2N-1}_{k=0} k_k F^{C(2)} \exp \left [ \frac{j \pi k}{2N} \right ] \exp \left [ \frac{2j \pi k n}{2N} \right ] \right \}$$

$$n=0, ..., N-1$$

# Question

1. I do not understand how that $$k_k$$ is present on the (2.8.8). If I had to deduct this equation myself from (2.8.7) $$k_k$$ would pop on the other side of the equality inverted, as $$k_k$$, though it would yield a division by zero. But I could pretend (2.8.7) uses the same definition for as (2.4.11.a) resulting on a $$F^{C(2)}(k)$$ that is defined for $$k > N-1$$, then $$k_k^{-1}$$ wouldn't yield division by zero. How is (2.8.8) actually obtained?
2. Why having two different definitions for $$k_k$$?
• The two definitions of $k_k$ are equivalent in the range $0..N$ though (?) – A_A Jun 4 at 10:32
• Indeed they are. However, I am kind of using a different version of the iDCT that uses the first definition, the second one wouldn't be suitable as I said in the question. Though I am not sure why the book does what it does. – Eduardo Reis Jun 4 at 16:44