# Implement Fast Hartley Transform

I'm trying to implement a simple FHT on .NET platform myself. I follow this document: Optimized fast Hartley transform for the MC68000 with applications in image processing.

In the page 20th, equation (20):

H(k) = He(k) + [Ho(k)cos(2πk/N) + Ho(N/2-k)sin(2πk/N)]
H(k+N/2) = He(k) - [Ho(k)cos(2πk/N) + Ho(N/2-k)sin(2πk/N)]

He(k) is the N/2 point DHT of the even indexed elements of H(k) and
Ho(k) is the N/2 point DHT of the odd indexed elements of H(k)

$k$ is zero index, so where is the item $N/2$ if $N/2$ is the length? I know I must have misunderstood something, but I realy don't know what exactly the Ho and He are. I have succeeded implement the FFT equation (in the page 17th). But the FHT, I can not.

Can somebody please help me explain the above equation details?

First of all, I'm pretty sure that $H_e(k)$ is the $N/2$ point DHT of the even elements of the original sequence to be transformed, not of $H(k)$. The same is true for $H_o(k)$. As for the index $k$, when used with $H_e(k)$ and $H_o(k)$, it is evaluated modulo $N/2$, so it ranges from $0$ to $N/2-1$. So the index $k=N/2$ is mapped to the index $0$ when evaluating $H_e(k)$ and $H_o(k)$.