3 ms·

Oh wow, I literally just finished my quadrature sinusoid DDS generator using MyHDL last night. I didn't use CORDIC but rather a LUT. I found out I can optimize

Oh wow, I literally just finished my quadrature sinusoid DDS generator using MyHDL last night. I didn't use CORDIC but rather a LUT. I found out I can optimize generating quadrature sinusoids by having two separate LUTs where each one stores from 0 to pi/2 and the other from pi/2 to pi, and this has an advantage because when the sine output takes inputs from the first LUT, the cos output takes inputs from the second LUT and vice versa, thus saving duplicates. I'm still cleaning up the testbench code and I plan to put out a blog post here if y'all are interested: hatsunearu.github.io

gugagore 8y ago
I'm not sure I followed, but I thought that the standard approach was to only store a quarter of a cycle (0 to pi/2) in the LUT, and use symmetry to generate the rest of the cycle. Is the advantage of having two LUTs that you can index into each of them separately?
- hatsunearu 8y ago
  edit: and yeah, the two LUTs allow me to index into them separately, and also guarantees that every quadrature combination never indexes into one LUT. OK so the problem is kinda twofold with that... First of all my application needs two sinusoids at quadrature (it's for downconversion in my SDR), which using that approach I believe I need dual read port RAM which takes up two RAM slices in my FPGA so I'm not losing anything by taking my approach. Secondly I tried doing that but I seem to get an "off by one" error--basically if you subdivide 0 to pi/2 with let's say 2^10 points, you either: a) the last point (index=2^10-1) corresponds to sin(pi/2) b) the last point + 1 corresponds to sin(pi/2), and the actual last point is sin(pi/2 - (pi/2) / (2^10)) the a) approach that means when you sweep through 0 to 2^10 - 1, the next index isn't 2^10 - 1, but rather 2^10 - 2, which means you'd need some annoying logic rather than simple bit slicing. the b) approach you'd have you make a special case and it's all too annoying. Maybe I'm just not creative enough but I haven't been able to come up with an elegant way of doing it with 0 to pi/2 LUTs (just by using simple bit slices) I also haven't looked at any actual implementations (actually I couldn't even find them lol) other than the basic description on wikipedia. Maybe that's why.
rshashik 8y ago
Couldn't find that post on your blog. You sure you didn't forget to commit it?
- rshashik 8y ago
  My bad I misread it.

gugagore 8y ago

I'm not sure I followed, but I thought that the standard approach was to only store a quarter of a cycle (0 to pi/2) in the LUT, and use symmetry to generate the rest of the cycle. Is the advantage of having two LUTs that you can index into each of them separately?

hatsunearu 8y ago
edit: and yeah, the two LUTs allow me to index into them separately, and also guarantees that every quadrature combination never indexes into one LUT. OK so the problem is kinda twofold with that... First of all my application needs two sinusoids at quadrature (it's for downconversion in my SDR), which using that approach I believe I need dual read port RAM which takes up two RAM slices in my FPGA so I'm not losing anything by taking my approach. Secondly I tried doing that but I seem to get an "off by one" error--basically if you subdivide 0 to pi/2 with let's say 2^10 points, you either: a) the last point (index=2^10-1) corresponds to sin(pi/2) b) the last point + 1 corresponds to sin(pi/2), and the actual last point is sin(pi/2 - (pi/2) / (2^10)) the a) approach that means when you sweep through 0 to 2^10 - 1, the next index isn't 2^10 - 1, but rather 2^10 - 2, which means you'd need some annoying logic rather than simple bit slicing. the b) approach you'd have you make a special case and it's all too annoying. Maybe I'm just not creative enough but I haven't been able to come up with an elegant way of doing it with 0 to pi/2 LUTs (just by using simple bit slices) I also haven't looked at any actual implementations (actually I couldn't even find them lol) other than the basic description on wikipedia. Maybe that's why.

rshashik 8y ago

Couldn't find that post on your blog. You sure you didn't forget to commit it?

rshashik 8y ago
My bad I misread it.