0
$\begingroup$

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 $k_k$?
$\endgroup$
  • $\begingroup$ The two definitions of $k_k$ are equivalent in the range $0..N$ though (?) $\endgroup$ – A_A Jun 4 at 10:32
  • $\begingroup$ 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. $\endgroup$ – Eduardo Reis Jun 4 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.