Transcript DIGITAL SYSTEMS

DIGITAL COMMUNICATION
Packet error detection (CRC)
November 2011
A.J. Han Vinck
Some books
Cyclic Redundancy Check
• Add n-k bits of extra data (the CRC field) to an k-bit message
to provide error detection function
8
Beginning
sequence
16
Header
16
Body
8
CRC Ending
sequence
– For efficiency, want n-k << n
– e.g., n-k = 32 for Ethernet and n = 12,000 (1500 bytes)
Automatic Repeat reQuest
IDEA: ask for retransmission if error detected
channel
Pass data if no error
ERROR ?
If Yes: ask retransmission
Some properties of ARQ
•
Automatic Repeat-reQuest (ARQ)
send ACK: packet OK
send NACK: packet in error
•
Some modes:
•
Some ARQ protocols are
– Resend the packet if no ACK received before timeout
– Use a predetermined maximum # of retransmissions until it is
correctly received
–
–
–
–
Stop-and-wait ARQ
Go-Back-N ARQ
Selective Repeat ARQ.
Hybrid ARQ is a combination of ARQ and forward error correction.
A pictorial view of coding
Replace k information bits by a unique n bit code word
2n vectors
2k code words (code)
Error detection
received vector not equal to one of the 2k code words.
2n vectors
transmitted word
received
Misdetection
2n vectors
transmitted word
received
Consider the following example
transmitted
00000
received
00001
00010
00100
01000
10000
00110
11000
01011
10101
11110
01010
01001
01111
00011
11011
10100
10111
10001
11101
00101
11111
11100
11010
10110
01110
11001
00111
10011
01101
01100
10010
Property: one error can be corrected
two errors can be detected
CHECK!
one error
Error detection capability
code words differ in at least dmin positions
2k vectors
0xx1x0x0
4 differences
1xx0x1x1
up to dmin –1
errors are detectable
< = 3 errors can be
detected
Misdetection
2n vectors
2k code words
for a random vector:
the probability of false
acceptance is (2k-1)/2n
= probability of hitting
the wrong code word
A polynomial description of the procedure
Let a = (a0 a1,..., ak-1) be k information bits
the information is represented by A(X) = Xn-k (a0+a1X +...+ak-1 Xk-1 )
the check polynomial is
F(X) = f0+f1X +...+fn-k Xn-k.
The transmitted code polynomial or encoding procedure is
C(X) = ( A(X) mod F(X) + A(X) )
Note that C(X) mod F(X)  0, and thus is C(X) a multiple of F(X).
A non-systematic procedure
• Transmit C(X) = F(X) (a0+a1X +...+ak-1 Xk-1 )
then, C(X) modulo F(X) = 0!
Some examples
• All coefficient operations are modulo-2
– ( 1+ X + X3 ) (X + X2 )
= X+ X2 + X4
+ X2+ X3 + X5
= X + X3 + X4 + X5
• errors change 0
1 and 1
0
• Ex: ( X + X4 + X5 ) modulo-(1 + X2 ) = 1
– Subtract as often as possible ( 1 + X2 ), but keep calculating mod-2
Ex: ( X + X4 + X5 ) Modulo- (1 + X2 )
X + X4 + X5
X3 + X5
X3
X + X3 + X4
X2
X + X2 + X3
X+
X3
X2
1 + X2
1
+ X4
X2
X
1
detection procedure
Assume that we receive the polynomial R(X) = C(X) + E(X)
where
E(X) is a binary error polynomial
i.e. an error vector ( 0,0,0,1,0)  C(X) = X3
the decoder calculates R(X) mod F(X)
- if the result = 0 no error detected
- if the result ≠ 0, then an error is detected.
The polynomial R(X) mod F(X) = 0 if and only if E(X) is a multiple of F(X).
performance
ASSUME: the polynomial F(X) has the form F(X) = 1 + ... + Xn-k
- degree n-k;
- and a nonzero constant term,
THEN: any error burst of length  n-k has a polynomial representation
that looks like E(X) = Xi (1 +... + Xn-k-1 )
and can thus never be a multiple of F(X).
F(X) is capable of detecting any burst of length  n-k !
Performance cont‘d
A nonzero burst of length  (n-k) can never give a zero result.
This can also be seen as follows
E(X) =
n-k
0 0 0 0 0 0 ... 0 1 0 ... 1 0 ... 1 0 0 0
Xi F(X) =
0 0 ...
0 ... 1
...
n-k+1
1
WHY: the binary representation of F(X) has length (n - k + 1)
cancellation of the highest degree nonzero component in E(X)
always gives a non-zero result.
CRC standard polynomials
•
•
•
•
•
•
CRC-8: x8+x2+x1+1
CRC-10: x10+x9+x5+x4+x1+1
CRC-12: x12+x11+x3+x2+1
CRC-16: x16+x15+x2+1
CRC-CCITT: x16+x12+x5+1
CRC-32: x32+x26+x23+x22+ x16+x12+x11+x10+x8 x7+x5+x4+x2+1
• Ethernet uses CRC-32
• HDLC: CRC-CCITT
• ATM: CRC-8, CRC-10, and CRC-32
Why these polynomials (CRC-16 and CRC-CCITT) ?
•
1st property: Multiples of these polynomials have even weight
Hence: error vectors of odd weight cannot be a multiple!
•
2nd property: error vectors of the form E(X) = (1 + Xi )
are not divisible by F(X) for i < L,
L larger than word length
Try to check this property!
•
Conclusion: 1, 2 and 3 errors are detectable. CHECK!
SUMMARY for ERROR DETECTION
Remark: We do all operations modulo 2 (XOR). F(X) = 1 + .... + Xn-k
A packet has the form
C(X) = A(X) mod F(X)
checksum
C
=
checksum
n-k bits
A(X) = Xn-k (a0 + a1X + ... + ak-1 Xk-1
information packet of length k.
a0 a1 a2
...
k bits data
ak-1
To calculate A(X) mod F(X), cancel highest nonzero term with F(X)
until result has degree less than n-k
Cont‘d
invert C at positions where the binary error vector E has ones: R =
C = checksum a0 a1 a2 ...
ak-1
C = c0 c1 c2
... 0
...
1 ... 1 ... 0 cn-1
E = e0 e1 e2
... 1
...
1 ... 0 ... 0 en-1
R = r0 r1 r2
If no errors
If errors
1 ...
0 ... 1 ... 0 rn-1
C E
EXAMPLE
000
1111
1011
000 0100
10 11
010 1
101
1
1 1 1 0 check sum
C(X) mod F(X) = A(X) mod F(X)  A(X) mod F(X) = 0!
{C(X)  E(X) } mod F(X) = E(X) mod F(X).
Shift-register
The following shift register can be used to
calculate A(X) modulo (1 + X + X3 )
a0 a1 a2 ... ak-1
Homework:
give a description of the shift control to obtain the result
Cont‘d
•
Example: calculate ( 1 +X2 +X4 +X5 ) modulo ( 1 +X + X3 )
– In binary: ( 1 0 1 0 1 1 ) modulo ( 1 1 0 1 )
The operations in binary are as follows
RESULT:
101011
001101
100110
011010
111100
110100
001000
Homework: draw the corresponding shift register
1 +X2 +X4 +X5
X2 ( 1 +X + X3 )
X ( 1 +X + X3 )
( 1 +X + X3 )
X2
CRC cont‘d
• The following „clocked“ shift register can be used to
implement CRC16:
F(X) = 1 +X2 + X15 + X16
Data in
13 flip-flops
Example: X16 modulo F(X) = 1 + X2 + X15
Question: When is the result A(X) mod F(X) equal to 0?
What does this mean?
A simple method
L
0110011001 1
0001100110 0
0000000000 0
Fill row wise
0110011000 0
0001100111 1
0000000001 1
RESULT: any burst of length L
can be detected
Transmit column wise
row parity
A simple code: the single parity check code
•
Make the number of ones in a code word even:
– Method 1
• Take F(X) = 1 + X, i.e. C(X) = F(X) A(X).
• C(X= 1) adds up the number of ones and since X = 1 is a root of
F(X), the result is 0.
– Method 2.
• Use the n-th bit to make the parity of a word even
•
CONCLUSION: detection of any number of odd errors
– Efficiency: R = (n-1)/n
ISBN checksum: checksum for 10-digit ISBN
- Given 9 digit product code. Starting at leftmost digit:
- multiply corresponding digit by 10, 9, 8, ... down to 2 inclusive
- add the resulting numbers: add digit 10 such that the result is divisible by 11
the number 10 is written as X
e.g., 0-201-61586-X is valid. The last digit has to be 10 (= X).
10*0 + 9*2 + 8*0 + 7*1 + 6*6 + 5*1 + 4*5 + 3*8 + 6*2 + 1*10 = 122 + 10 = 132 = 12*11
• detects transpositions and single digit errors
- an error e at position I gives as result the value eI modulo 11  0
- detection of an transposition error of A and B:
A I + B (I-1)  BI + A(I-1) = AI + B(I-1) – A + B
the value ( – A + B )  0 modulo 11 for A  B
Combining error detection-correction
• G1(X) G2(X) generates a code with minimum distance D2
• G1(X) generates a code with minimum distance D1 < D2
• C(X) = I(X) G1(X) G2(X) = I’(X)G1(X)
decoding:
step 1: correct a maximum of ( D1 – 1 )/2 errors with G1(X)
step 2: detect the remaining errors with G1(X)G2(X)
properties:
1. t  ( D1 – 1 )/2 errors correctable
2. t  D2 - 1 - ( D1 – 1 )/2 errors detectable
Homework: construct some examples
A.J. Han Vinck
29
Combining error detection-correction
• Example:
n
k
dmin
generator polynomial (octal)
63
57
51
45
3
5
7
G1 = 103 (octal)
G2 x G1 = G1 x (127)
G3 x G2 x G1= G2 x G1 x (147)
A.J. Han Vinck
30
Example: bluetooth
• Three error correction schemes are defined for Bluetooth:
– 1/3 rate FEC, a simple repetition
– 2/3 rate FEC, a shortened Hamming Code (10,15)
– ARQ scheme for data (automatic retransmission request)
• FEC schemes reduce the number of retransmissions
• A CRC code decides whether a packet/header contains errors, i.e.
• transmit
• receive
• check
C(X) = I(X) g(X)
R(X) = C(X) + E(X)
R(X) modulo g(X) = 0 ?
A.J. Han Vinck
31
Bookland EAN Symbol
Proper codes
Linear block codes do not necessarily obey the 2-(n-k) bound. In other words, a code is proper if PU is
monotonically increasing in p, for 0.0  p  0.5. Hamming codes and primitive double error-correcting BCH
codes are proper.
To illustrate the above behavior, we compare the standard CRC-16 for a length 80 packet with a double error
correcting BCH code of length 81 and 17 redundant digits.
Detection error probability as a function of the channel error probability.
Unidirectional errors detection: Berger codes
application: memory systems
Errors: unidirectional, i.e. In a word only from 0 to 1 (or from 1 to 0)
n binary data bits
Example: n = 7; log2 (n+1) = 3
log2 (n+1) check bits
indicate the # of zeros in the first part
1000111 011
Suppose errors from 1 to 0
in the first part only: number of zeros does not match check
in the second part only: idem dito
in the first and second part: the first part gets more zeros
the second part # gets smaller
CONCLUSION: any number of errors detectable!
Some practical info
INTEL website:
http://datasheets.chipdb.org/Intel/x86/Pentium/Embedded%20Pentium%AE%20Processor/ERRORDET.PDF
Xilinx website:
http://www.xilinx.com/support/documentation/application_notes/xapp645.pdf
Altera website:
http://www.altera.com/literature/an/an357.pdf