By Walker G.

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3. (a/g)21 mod n/g. The complete set of solutions within Zn is xk ¼ x0 þ k:(n=g), 8k ¼ 0, 1, . . , g À 1: Observe that if k , g and x0 , (n/g), then xk ¼ n 2 1. 4 (Chinese Remainder Theorem) Consider s pairwise relatively prime integers m1, m2, . . , ms whose product is equal to M. Then the system N ; r1 (mod m1 ), N ; r2 (mod m2 ), ... N ; rs (mod ms ), (2:3) has a unique solution N within ZM (jajm stands for a mod m): N ¼ S 1 Ã Ã i s mi :jri =mi jmi M , (2:4) where M ¼ P1 i s mi ; mÃi ¼ M=mi : (2:5) The ri are called residues modulo mi.

1 1. The set of natural numbers1 N ¼ f0, 1, 2, 3, . g. 2. The set of integers Z ¼ f . . , 23, 22, 21, 0, 1, 2, 3, . . g. y. 1 For convenience, the element zero has been included in N. Synthesis of Arithmetic Circuits: FPGA, ASIC, and Embedded Systems By Jean-Pierre Deschamps, Ge´ry J. A. Bioul, and Gustavo D. Sutter Copyright # 2006 John Wiley & Sons, Inc. 3 Given two integers x and y, with y . 0, there exist two integers q (the quotient) and r (the remainder) such that x ¼ q:y þ r, where 0 r , y: It can be proved that q and r are unique.

X ; d (mod n) has a solution x if and only if g divides d. 2. x ; (d/g) (mod n/g). 3. There are g solutions, all of them congruent modulo n/g. Proof 1. n. As g divides both a and n, it also divides d. g. n. b is a solution. 2. x ; (d/g) (mod n/g). x ; d (mod n). 3. (a/g)21 mod n/g. The complete set of solutions within Zn is xk ¼ x0 þ k:(n=g), 8k ¼ 0, 1, . . , g À 1: Observe that if k , g and x0 , (n/g), then xk ¼ n 2 1. 4 (Chinese Remainder Theorem) Consider s pairwise relatively prime integers m1, m2, .

### An isolated MOSFET gate driver by Walker G.

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