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Residue Generation

IP.com Disclosure Number: IPCOM000044712D
Original Publication Date: 1984-Apr-01
Included in the Prior Art Database: 2005-Feb-06
Document File: 1 page(s) / 11K

Publishing Venue

IBM

Related People

Vaghasia, B: AUTHOR

Abstract

A conventional way to generate residue codes for addition of two operands is: S(AI + BI) = S((AI)Mod M + (BI) Mod M)) Mod M Example: Mod 5 A = 1101 (Mod 5 = 3) B = 1001 (Mod 5 = 4) Mod 5 of the addition of these two operands is 2 In some arithmetic circuits there may not be enough time to generate a result in one cycle. A speed-up can be achieved by starting, in a previous cycle, the formation of propagate and generate terms for bits of two operands.

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Residue Generation

A conventional way to generate residue codes for addition of two operands is: S(AI + BI) = S((AI)Mod M + (BI) Mod M)) Mod M

Example: Mod 5

A = 1101 (Mod 5 = 3)

B = 1001 (Mod 5 = 4)

Mod 5 of the addition of these two operands is 2

In some arithmetic circuits there may not be enough time to generate a result in one cycle. A speed-up can be achieved by starting, in a previous cycle, the formation of propagate and generate terms for bits of two operands. Propagate (Pi) = Ai OR Bi

Generate (Gi) = Ai AND Bi

Residues can be generated using the P and G terms in accordance with the following: S(Ai + Bi) = S(Gi + Pi)

Proof:

Ai Bi Gi Pi

0 0 0 0

0 1 0 1

1 0 0 1

1 1 1 1

S(Ai + Bi) Mod M = S((Gi) Mod M + (Pi) Mod M)) Mod M

Example: Mod 5

From previous example

Pi = Ai OR Bi = 1101 ----T 3

Gi = Ai AND Bi= 1001 ----T 4 Mod 5 of the addition of these two operands is 2

Disclosed anonymously

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