Efficient RNS Reverse Converters for Moduli Sets with Dynamic Ranges Up to $(10n+1)$-bit

Abstract

This paper presents reverse converters for dynamic ranges (DRs) up to $10n+1$ achieved with horizontal and vertical extensions of the traditional three-moduli set $\{2^{n},2^n-1, 2^n+1\}$. The proposed approach results in increased parallelism with a smaller number of bits per channel. An arithmetic application is considered to show that the proposed converter scales better with an increasing number of residue operations per channel, confirming the practical applications of the moduli set and reverse converter. Experimental results suggest that the proposed architecture outperforms the best reverse converters in the related state of the art, with equivalent DR, presenting speedup of 2.09 times, and average reductions of 62.00 and 45.20% in area and power consumption, respectively.

Appendices

Appendices

1.1 Proof of Lemmas in the Paper

Proof

In order to prove the value of multiplicative inverse of $(2^{9n}-2^{3n})$ modulo $M_{b}=2^{n-1}-1$ for $n=12k_1+6, k_1 \in {\mathbb {N}}$ given in Eq. (4), we need to distinguish two subcases $n=6$ and $n>6$. For $n=6$ we have to show that:

$$\begin{aligned} \left| (2^{9n}-2^{3n}) \times \left| {(2^{9n}-2^{3n})}^{-1} \right| _{M_b} \right| _{M_b} = \left| (2^{9n}-2^{3n}) \times 2^{n-4}\right| _{M_b} = 1. \end{aligned}$$

(11)

Therefore substituting the value $n=6$:

$$\begin{aligned} | (\overbrace{|2^{54}|_{31}}^{16}-\overbrace{|2^{18}|_{31}}^{8}) \times (2^{6-4})|_{31} = |8\times 4|_{31} = 1. \end{aligned}$$

(12)

For $n>6$ we have to show that:

$$\begin{aligned}&\left| (2^{9n}-2^{3n}) \times \left| (2^{9n}-2^{3n})^{-1} \right| _{M_b} \right| _{M_b}\nonumber \\&\quad =\left| (2^{9n}-2^{3n}) \times \left( 2^{n-4}+\frac{2^{n}}{504}-\frac{64}{504}\right) \right| _{M_b} = 1, \end{aligned}$$

(13)

where $| 2^{9n}-2^{3n} |_{M_b} = | \overbrace{|2^{9n}|_{M_b}}^{2^{9}}-\overbrace{|2^{3n}|_{M_b}}^{2^{3}} |_{M_b} = 504$. Therefore, Eq. (13) can be written as:

$$\begin{aligned}&\left| (2^{9}-2^{3}) \times \left( 2^{n-4}+\frac{2^{n}}{504}-\frac{64}{504} \right) \right| _{M_b} \nonumber \\&\quad = \left| 2^{n+5}-2^{n-1}+2^{n}-2^{6}\right| _{M_b}= \left| 2^{6}-1+2-2^{6}\right| _{M_b}=1. \end{aligned}$$

(14)

It is important to note that $\left| (2^{9n}-2^{3n})^{-1} \right| _{M_b}$ is also expressed for $n>6$ in terms of powers of two in (4). Thus, to finish the proof we have to show that:

$$\begin{aligned} \left| (2^{9n}-2^{3n})^{-1} \right| _{M_b} = 2^{n-4}+\displaystyle {\sum _{i=0}^{\frac{n-12}{6}}} 2^{6i+3}. \end{aligned}$$

(15)

By using the sum of n terms of a geometric series with common ratio r:

$$\begin{aligned} \sum _{i=m}^{n}r^{i}=\frac{r^{m}-r^{n+1}}{1-r} . \end{aligned}$$

(16)

we can obtain:

$$\begin{aligned} 2^{n-4}+\displaystyle {\sum _{i=0}^{\frac{n-12}{6}}} 2^{6i+3}= & {} 2^{n-4}+2^{3} \times \frac{1-2^{n-6}}{1-2^6} \nonumber \\= & {} 2^{n-4}+2^{3} \times \left( \frac{2^{n-6}}{63}- \frac{1}{63}\right) =2^{n-4}+\frac{2^{n}}{504}-\frac{64}{504}. \end{aligned}$$

(17)

Thus, Lemma 1 is proved. $\square$

Proof

In order to prove the value of the multiplicative inverse of $(2^{9n}-2^{3n})$ modulo $M_{b}=2^{n-1}-1$ for $n=12k_2+14, k_2 \in {\mathbb {N}}$ given in Eq. (5) we need to distinguish again two subcases $n=14$ and $n>14$. For $n=14$ we have to show that:

$$\begin{aligned}&\left| (2^{9n}-2^{3n}) \times \left| {(2^{9n}-2^{3n})}^{-1} \right| _{M_b} \right| _{M_b} \nonumber \\&\quad = \left| (2^{9n}-2^{3n}) \times (2^{n-2}+2^{n-3}-2^{n-9}+2^{4}-1) \right| _{M_b} = 1. \end{aligned}$$

(18)

Therefore, substituting the value $n=14$:

$$\begin{aligned} |(\overbrace{|2^{126}|_{8191}}^{512}-\overbrace{|2^{42}|_{8191}}^{8}) \times (6127)|_{8191} = |504 \times 6127|_{8191} = 1. \end{aligned}$$

(19)

For $n>14$ we have to show that:

(20)

It is important to note that $\left| (2^{9n}-2^{3n})^{-1} \right| _{M_b}$ is also expressed for $n>14$ in terms of powers of two in Eq. (5). Thus, to finish the proof we have to show that:

$$\begin{aligned} \left| (2^{9n}-2^{3n})^{-1} \right| _{M_b} = 2^4 - 1 + 2^{n-2} + 2^{n-3} - 2^{n-9} + \sum _{i=0}^{\frac{n-20}{6}} \left( 2^{6i + 10} - 2^{6i + 5}\right) . \nonumber \\ \end{aligned}$$

(21)

Applying Eq. (16), we can obtain:

$$\begin{aligned}&2^4 - 1 + 2^{n-2} + 2^{n-3} - 2^{n-9} + \frac{2^{n-4} - 2^{10}}{2^6 - 1} - \frac{2^{n-9} - 2^{5}}{2^6 - 1}\nonumber \\&\quad =\frac{(2^{n-1}+376(2^{n-1}-1))}{(2^{9n}-2^{3n})}. \end{aligned}$$

(22)

Thus, Lemma 2 is proved. $\square$

Proof

In order to prove the value of the multiplicative inverse of $(2^{9n}-2^{3n})$ modulo $M_{b}=2^{n+1}-1$ for $n=12k_3+6, k_3 \in {\mathbb {N}}$ given in (6), we need to distinguish two subcases $n=6$ and $n>6$. For $n=6$ we have to show that:

$$\begin{aligned} \left| (2^{9n}-2^{3n}) \times \left| {(2^{9n}-2^{3n})}^{-1} \right| _{M_b} \right| _{M_b} = \left| (2^{9n}-2^{3n}) \times 2^{3}\right| _{M_b} = 1. \end{aligned}$$

(23)

Therefore substituting the value $n=6$:

$$\begin{aligned} \left| (\overbrace{|2^{54}|_{127}}^{32}-\overbrace{|2^{18}|_{127}}^{16}) \times (2^{3})\right| _{127} = |16 \times 8|_{127} = 1. \end{aligned}$$

(24)

For $n>6$ we have to show that:

$$\begin{aligned} \left| (2^{9n}-2^{3n}) \times \left| (2^{9n}-2^{3n})^{-1} \right| _{M_b} \right| _{M_b} = \left| (2^{9n}-2^{3n}) \times \frac{8 \times 2^{n+1} - 520}{(2^{6}-1)} \right| _{M_b} = 1. \end{aligned}$$

(25)

where $| 2^{9n}-2^{3n} |_{M_b} = | \overbrace{|2^{3n}|_{M_b}}^{2^{n-2}}(\overbrace{|2^{6n}|_{M_b}}^{2^{n-5}}-1) |_{M_b} = 2^{n-2} (2^{n-5}-1)$. Therefore, Eq. (25) can be written as:

(26)

Note that $\left| (2^{9n}-2^{3n})^{-1} \right| _{M_b}$ is also expressed in terms of powers of two in (6). Thus, to finish the proof we have to show that:

$$\begin{aligned} \left| (2^{9n}-2^{3n})^{-1} \right| _{M_b} = 2^3 + \sum _{i=0}^{\frac{n-12}{6}}{2^{6i+10}}. \end{aligned}$$

(27)

Applying Eq. (16), we can obtain:

$$\begin{aligned} 2^3 + \left( \frac{2^{n-6}-1}{2^6-1} \right) \times 2^{10} = 2^3 + \left( \frac{2^{n+1}-2^7}{2^7 \times (2^6-1)} \right) \times 2^{10} = \frac{8 \times 2^{n+1} - 520}{(2^{6}-1)}.\nonumber \\ \end{aligned}$$

(28)

Thus, Lemma 3 is proved. $\square$

Proof

In order to prove the value of the multiplicative inverse of $(2^{9n}-2^{3n})$ modulo $M_{b}=2^{n+1}-1$ for $n=12k_4+10, k_4 \in {\mathbb {N}}$ given in (7), we need to distinguish two subcases $n=10$ and $n>10$. For $n=10$ we have to show that:

$$\begin{aligned}&\left| (2^{9n}-2^{3n}) \times \left| {(2^{9n}-2^{3n})}^{-1} \right| _{M_b} \right| _{M_b} \nonumber \\&\quad =\left| (2^{9n}-2^{3n}) \times ( 2^{n} + (2^9 - 2^5) + (2^4 - 2^0))\right| _{M_b} = 1. \end{aligned}$$

(29)

Therefore substituting the value $n=10$:

$$\begin{aligned}&|(\overbrace{|2^{90}|_{2047}}^{4}-\overbrace{|2^{30}|_{2047}}^{256}) \times \overbrace{(2^{10} + (2^9 - 2^5) + (2^4 - 2^0))}^{1519}|_{2047}\nonumber \\&\quad = |-252 \times 1519|_{2047} = 1. \end{aligned}$$

(30)

For $n>10$ we have to show that:

$$\begin{aligned} \begin{aligned} \left| (2^{9n}-2^{3n}) \times \left| (2^{9n}-2^{3n})^{-1} \right| _{M_b} \right| _{M_b} {=} \left| (2^{9n}-2^{3n}) \times \frac{(47\times 2^{n+1}-559)}{(2^{6}-1)} \right| _{M_b} = 1. \end{aligned} \end{aligned}$$

(31)

where $| 2^{9n}-2^{3n} |_{M_b} = | \overbrace{|2^{3n}|_{M_b}}^{2^{n-2}}(\overbrace{|2^{6n}|_{M_b}}^{2^{n-5}}-1) |_{M_b} = 2^{n-2} (2^{n-5}-1)$. Therefore, Eq. (31) can be written as:

(32)

It is important to note that $\left| (2^{9n}-2^{3n})^{-1} \right| _{M_b}$ is also expressed in terms of powers of two in (5). Thus, to finish the proof we have to show that:

$$\begin{aligned} \left| (2^{9n}-2^{3n})^{-1} \right| _{M_b} = 2^n + \displaystyle {\sum _{i=0}^{\frac{n-16}{6}} \left( 2^{6i + 15} - 2^{6i + 10}\right) } + (2^9 - 2^5) + (2^4 - 2^0). \end{aligned}$$

(33)

Applying Eq. (16), we can obtain:

$$\begin{aligned}&2^n + \frac{2^{15} + 2^{15} \times 2^{n-16}}{1 - 2^6} - \frac{2^{10} + 2^{10} \times 2^{n-16}}{1 - 2^6} + (2^9 - 2^5) + (2^4 - 2^0) \nonumber \\&\quad = \frac{(47 \times 2^{n+1} - 559)}{63}. \end{aligned}$$

(34)

Thus, Lemma 4 is proved. $\square$