M KIM TRA LI CRCGii thiu. Cch xc nh CRC-n Mt s a thc sinh. V d. CRC-4 trong PCM-30

GII THIUCRC l mt phng php pht hin li bng cch gn thm mt khi bit pha sau khi d liu. Thut ton to ra khi bit CRC l da trn php cng modulo 2 (GF(2)). CRC l phn d ca php chia nh phn khng nh.

PHP TON MODULO 2 (mod 2)Trong trng GF(2), cc h s ca a thc l cc s 1 v 0. V d cng hai a thc:

( x 3 + x ) + ( x + 1) = x 3 + 2 x + 1 = x 3 + 1 (mod 2)

2 x = x + x = x(1 + 1) = x.0 = 0 (mod 2)

Php nhn cng vy:

( x 2 + x).( x + 1) = x 3 + 2 x 2 + x = x 3 + x (mod 2)Chng ta cng c th chia a thc theo mod 2 tm thng (quotient) v s d (remainder):

x +x +x 1 1 = ( x + 1) = ( x + 1) + (mod 2) x +1 x +1 x +13 2

CHUYN I GIA CHUI S NH PHN V A THCChui bit S bit: m a thc nh phn biu din chui bit trn l: b(m-1).x(m-1) + b(m-2).x(m-2) + + b2.x2 + b1.x1 + b0.x0 V d: Chui bit 101 c th c biu din di dng a thc nh phn nh sau:b(m-1) b(m-2) b2 b1 b0

1. x 2 + 0. x1 + 1. x 0 = x 2 + 1

CCH XC NH CRC-nCc bc thc hin:Biu din chui bit thnh a thc nh phn M(x). Nhn M(x) vi xn: M(x).xn. Chia M(x).xn cho a thc sinh G(x) ca CRC-n. Nh vy ta c thng Q(x) v s d R(x). S d R(x) chnh l CRC-n.

Nh vy tng qut ta c thvit:

M ( x).x = Q( x).G ( x) + R ( x)n

MT S A THC SINH G(x)CRC-n CRC-1 CRC-4 CRC-5 - CCITT CRC-5 USB CRC-7 CRC-8 CRC-12 CRC-32 - MPEG2 x+1 x4 + x + 1 x5 + x3+ x+1 x5 + x2+ 1 x7 + x3+ 1 x8 + x7 + x6 + x4 + x2 + 1 x12 + x11 + x3 + x2 + x + 1 x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1 use: telecom systems G(x) USE Hardware (parity bit) PCM-30 ITU-G.704 USB token packets Telecom systems, MMC

MT S V D TNH CRC-n V d 1: Tm CRC-1 ca chui s nh phn sau: 1101001010101010

MT S V D TNH CRC-n (tt)p s v d 1: M(x) = x15+x14+x12+x9+x7+x5+x3+x G(x) = x+1 Q(x) = x15+x12+x11+x10+x7+x6+x3+x2 R(x) = 0 CRC-1 = 0

MT S V D TNH CRC-n (tt) V d 2: Tm CRC-7 ca chui s nh phn sau: 1101001010101010

MT S V D TNH CRC-n (tt)p s v d 2: M(x) = x15+x14+x12+x9+x7+x5+x3+x G(x) = x7 + x3 + 1 Q(x) = x15+x14+x12+x11+x10+x9+x7+x6+x5+x4+x3 R(x) = x5+x4+x3 CRC-7 = 0111000

FRAME #1

CRC-4 TRONG PCM-30TS0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 FRAME TS0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 #8 C1 0 #9 #10 C2 0 #11 #12 C3 0 #13 #14 C4 0 #15 Submultiframe#1 Submultiframe#2

#0 C1 0 #2 C2 0 #3 #4 C3 0 #5 #6 C4 0 #7

Multi-frame

CRC-4 TRONG PCM-30 (tt)Gi tr CRC-4 ca a khung con #n c tnh t cc bit tin ca a khung con #(n-1) (tr cc bit CRC). m tng s bit 1 trong mt a khung con #(n-1). i s ny thnh s nh phn. Sau tnh CRC-4 ca s nh phn ny. Truyn CRC-4 ny a khung con #n.

V D TNH CRC-4V d 3: Gi s b pht PCM-30 to cc a khung con th (n-1) c tng s bit 1 l 1210.Hy xc nh gi tr CRC-4 ca a khung con ny?

V D TNH CRC-4 (tt)p s v d 3: CRC-4 = 0101

V D TNH CRC-4 (tt)Gii v d 3: i s 1210 ra s nh phn: 10010111010 M(x) = x10+x7+x5+x4+x3+x G(x) = x4+x+1 Q(x) = x10+x6+x5+x4+x+1 R(x) = x2+1

V D TNH CRC-4 (tt)V d 4: Gi s b pht PCM-30 to cc a khung con th (n-1) vi xc sut xut hin bit 1 l 50%. Hy xc nh gi tr CRC-4 ca a khung con ny?

V D TNH CRC-4 (tt)p s v d 4: CRC-4 = 1011

V D TNH CRC-4 (tt)Gii v d 4: Tng s bit 1: 2048*50% = 1024. S nh phn: 10000000000 M(x) = x10 G(x) = x4 + x + 1 Q(x) = x10 + x7 + x6+x4 + x2 + x + 1 R(x) = x3+ x + 1