1 / 14

基于 FFT 的频域 RS 译码

基于 FFT 的频域 RS 译码. 季文洁 2014.3.21. 频域译码. FFT. Prime factor Cyclostome FFT. Prime factor. e.g. GF(2 4 ) N=15=3×5. CFFT. CFFT. e.g. GF(2 3 ). polynomial. F= ALf. CFFT. 选择一组正规基 ( γ , γ 2 , γ 4 ) ， 此处取 γ = α 3 ， 满足 γ+γ 2 + γ 4 = 1 ）. F= ALf. Winograd Algorithm.

aquene
Télécharger la présentation

基于 FFT 的频域 RS 译码

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 基于FFT的频域RS译码 季文洁 2014.3.21

  2. 频域译码

  3. FFT • Prime factor • Cyclostome FFT

  4. Prime factor • e.g. GF(24) • N=15=3×5

  5. CFFT

  6. CFFT e.g. GF(23) polynomial F=ALf

  7. CFFT 选择一组正规基(γ, γ2, γ4)，此处取γ = α3，满足γ+γ2+γ4 = 1） F=ALf Winograd Algorithm

  8. CFFTGF(23) 3-point cyclic convolution

  9. CFFT GF(23) n=7 Lr=2 GF(25) n=31 Lr=6 GF(26) n=63 Lr=12 GF(28) n=255 Lr=34 …… GF(24) n=15 Lr=4

  10. 预估结果 F=ALf 计算n个校验子的硬件需求

  11. TF

  12. CFFT i mod m2 = 0，0≤i<m，→{0, 3} • e.g. GF(26) • m = 6=2×3 • m1 = 2, m2 = 3 [y0, y1, y2, y3, y4, y5]T = [a0, a1, a2, a3, a4,a5]C·[y0, y1, y2, y3, y4, y5]T

  13. 预估结果 • e.g. GF(26) 常数乘法器 相同的例子 如： 10=2×5 15=3×5

  14. THE END

More Related