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A Structured Dual Split-Radix Algorithm for the Discrete Hartley Transform of Length \(2^{N}\)

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Abstract

This paper presents a new split-radix algorithm for DHT of length \({N}=2^{n}\), called the Dual Split-Radix DHT (DSR DHT), that allows an efficient parallel implementation using a dual core system. Moreover, as it is different from existing split-radix algorithms for DHT, it offers an efficient hardware implementation similar to that for FFT. It avoids the so-called retrograde indexing specific to existing DHT algorithms that do not allow an efficient pipeline implementation.

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Correspondence to Doru Florin Chiper.

Appendix

Appendix

1.1 Algorithms for Small DHTs

An efficient implementation of a fast DHT algorithm, with a simple and low complexity software or hardware structure, closely depends on efficient algorithms for a small DHTs. At the following link http://www.etti.tuiasi.ro/index.php?option=com_content&view=article&id=118:dfchiper&catid=37:cadre-didactice&Itemid=78, there are some efficient DHT algorithms for length \(N=4\) and \(N=8\) [7].

1.2 MATLAB Script for the Proposed Split-Radix DHT Algorithm

figure a
figure b

1.3 Solving the Recurrence Relations for the Arithmetic Complexity

(a) Solving \(M(N)=M(N/2)+2\cdot M(N/4)+7N/8-4\)

with \(M(4)=0\), \(M(8)=2\)

We can write

$$\begin{aligned} \left[ \begin{array}{l} {M(N)} \\ *\\ \end{array} \right] =A^{n-3}\cdot b+\left[ I+A/2+\cdots +(A/2)^{n-4}\right] \cdot cN-\left[ I+A+\cdots +A^{n-4}\right] d \end{aligned}$$

where

$$\begin{aligned} A=\left[ {{\begin{array}{ll} 1&{} 2 \\ 1&{} 0 \\ \end{array} }} \right] ; \quad b=\left[ {{\begin{array}{l} 2 \\ 0 \\ \end{array} }} \right] ; \quad c=\left[ {{\begin{array}{l} {7/8} \\ 0 \\ \end{array} }} \right] ; \quad d=\left[ {{\begin{array}{l} 4 \\ 0 \\ \end{array} }} \right] . \end{aligned}$$

We note \(a_{00} (n)\), \(a_{01} (n)\) the elements \(a_{00}\) and \(a_{01}\) of \(A^{n}\).

We have

$$\begin{aligned} a_{00}(n)= & {} \frac{2}{3}2^{n}+(-1)^{n}\frac{1}{3}\\ a_{01}(n)= & {} \frac{2}{3}2^{n}-(-1)^{n}\frac{2}{3}. \end{aligned}$$

We can further write

$$\begin{aligned} M(N)= & {} \left[ \frac{2}{3}\cdot 2^{n-3}+(-1)^{n-3}\frac{1}{3}\right] \cdot 2+\left\{ 1+(n-4)\frac{2}{3}\right. \nonumber \\&\left. +\frac{1}{3}\left[ -\frac{1}{2}+\frac{1}{2^{2}}-\frac{1}{2^{3}}+\cdots +(-1)^{n-4}\frac{1}{2^{n-4}}\right] \right\} \frac{7}{8}N \\&-\left[ 1+\frac{2}{3}2-\frac{1}{3}+\frac{2}{3}2^{2}+\frac{1}{3}+\cdots +\frac{2}{3}2^{n-4}+(-1)^{n}\frac{1}{3}\right] \cdot 4, \end{aligned}$$

where \(n=\log _2 N\).

Results

$$\begin{aligned} M(N)= & {} \frac{1}{3}\cdot \left[ 2^{n-2}+(-1)^{n-3}\right] \cdot 2+\left\{ {\frac{2}{3}(n-3)+\frac{1}{9}\frac{2^{n-3}-(-1)^{n-3}}{2^{n-4}}} \right\} \frac{7}{8}N \\&-\frac{1}{3}\left[ 1+2(2^{n-3}-1)\right] \cdot 4+\frac{4}{3}\cdot <\log _2 N>_2 \\ \end{aligned}$$

and finally

$$\begin{aligned} M(N)=\frac{7}{12}N\cdot \log _2 N-\frac{31}{18}N+\frac{4}{3}+\frac{8}{9}(-1)^{\log _2 N}+\frac{1}{3}<\log _2 N>_2 \cdot 4. \end{aligned}$$

(b) Solving \(A(N)=A(N/2)+2A(N/4)+21N/8-4\)

with \(A(4)=8\); \(A(8)=16.\)

We can write

$$\begin{aligned} \left[ {{\begin{array}{l} {A(N)} \\ *\\ \end{array} }} \right] =A^{n-3}\cdot b+\left[ I+A/2+\cdots +(A/2)^{n-4}\right] \cdot cN-\left[ I+A+\cdots +A^{n-4}\right] d \end{aligned}$$

where

$$\begin{aligned} A=\left[ {{\begin{array}{ll} 1&{} 2 \\ 1&{} 0 \\ \end{array} }} \right] ; \quad b=\left[ {{\begin{array}{l} {16} \\ 8 \\ \end{array} }} \right] ; \quad c=\left[ {{\begin{array}{l} {21/8} \\ 0 \\ \end{array} }} \right] ; \quad d=\left[ {{\begin{array}{l} 4 \\ 0 \\ \end{array} }} \right] . \end{aligned}$$

We note \(a_{00} (n)\), \(a_{01} (n)\) the elements \(a_{00}\) and \(a_{01} \) of \(A^{n}\).

We have

$$\begin{aligned} a_{00}(n)= & {} \frac{2}{3}2^{n}+(-1)^{n}\frac{1}{3}\\ a_{01}(n)= & {} \frac{2}{3}2^{n}-(-1)^{n}\frac{2}{3}. \end{aligned}$$

We can further write

$$\begin{aligned} A(N)= & {} \left[ \frac{2}{3}\cdot 2^{n-3}+(-1)^{n-3}\frac{1}{3}\right] \cdot 16+\left[ \frac{2}{3}\cdot 2^{n-3}-(-1)^{n-3}\frac{2}{3}\right] \cdot 8 \\&+\left\{ {1+(n-4)\frac{2}{3}+\frac{1}{3}\left[ -\frac{1}{2}+\frac{1}{2^{2}}-\frac{1}{2^{3}}+\cdots +(-1)^{n-4}\frac{1}{2^{n-4}}\right] } \right\} \frac{21}{8}N \\&-\left[ 1+\frac{2}{3}2-\frac{1}{3}+\frac{2}{3}2^{2}+\frac{1}{3}+\cdots +\frac{2}{3}2^{n-4}+(-1)^{n}\frac{1}{3}\right] \cdot 4, \end{aligned}$$

where \(n=\log _2 N\).

$$\begin{aligned} A(N)= & {} \frac{1}{3}\cdot \left[ 2^{n-2}+(-1)^{n-3}\right] \cdot 16+\frac{1}{3}\cdot \left[ 2^{n-2}-(-1)^{n-3}2\right] \cdot 8\\&+\left\{ {\frac{2}{3}(n-3)+\frac{1}{9}\frac{2^{n-3}-(-1)^{n-3}}{2^{n-4}}} \right\} \frac{21}{8}N \\&-\frac{1}{3}\left[ 1+2(2^{n-3}-1)\right] \cdot 4+\frac{4}{3}\cdot<\log _2 N>_2 \\ A(N)= & {} \frac{1}{3}\cdot \left[ \frac{N}{4}-(-1)^{\log _2 N}\right] 16+\frac{1}{3}\cdot \left[ \frac{N}{4}+(-1)^{\log _2 N}2\right] 8\\&+\frac{21}{12}N\cdot (\log _2 N-3)+\frac{7}{24}\left( {2N+(-1)^{\log _2 N}16} \right) \\&-\frac{1}{3}\left[ 1+2(N/8-1)\right] \cdot 4+\frac{1}{3}<\log _2 N>_2 \cdot 4 \\ \end{aligned}$$

and finally results

$$\begin{aligned} A(N)=\frac{21}{12}N\cdot \log _2 N-3N+\frac{4}{3}+\frac{14}{3}(-1)^{\log _2 N}+\frac{1}{3}<\log _2 N>_2 \cdot 4. \end{aligned}$$

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Chiper, D.F. A Structured Dual Split-Radix Algorithm for the Discrete Hartley Transform of Length \(2^{N}\) . Circuits Syst Signal Process 37, 290–304 (2018). https://doi.org/10.1007/s00034-017-0552-3

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