EE4253 Digital Communications
Department of Electrical and Computer Engineering - University of New Brunswick, Fredericton, NB, Canada
The use of simple parity allows detection of single bit errors in a received message. However, a second error will go undetected, and further errors will not be noticed whenever the total number of bit errors is EVEN. Unless the probability of a error is very low and a message is very short (the case when a parity bit is added to a 7-bit ASCII character) the chances of some error event going undetected are high. One of the most useful techniques is called the Cyclic Redundancy Check (CRC).
In general, use of parity bits or the CRC are examples of "systematic (n,k) block codes", where a block of k input bits is taken to create an n-bit codeword. The number of extra bits is p = n-k. So the commonplace 7-bits plus parity is called an (8,7) code for which p= (8 - 7) =1. The code is systematic in that the extra bits are appended to unmodified data bits. In the case of CRC, the block size is often variable, while the constant p depends on the order of a defining polynomial P(x).
Consider a message having many data bits which are to be transmitted reliably by appending several check bits as shown below.
The exact number of extra bits and their makeup depends on a generating polynomial. For example one such polynomial is:
The number of CRC bits corresponds to the order of the generating polynomial. The above polynomial of order 16 generates a 16-bit CRC. Typically, the CRC bits are used for error detection only.
Consider a message represented by some polynomial G(x), and a generating polynomial P(x).
In this example, let G(x) represent the binary message 110010, and let P(x) = x3 + x2 + 1; (binary 1101).
The polynomial P(x) will be used to generate a (3-bit) CRC called C(x) which will be appended to G(x). Note that P(x) is prime.
Step 1 - Multiply the message G(x) by x3, where 3 is the number of bits in the CRC.
Add 3 three zeros to the binary G(x).
Step 2 - Divide the product x3 [G(x)] by the generating polynomial P(x).
We wish to find "the remainder, modulo P(x)"
Compute the following:
100100 (ignore this quotient) ------------ 1101 ) 110010000 1101 ---- 1100 1101 ---- 100 = remainder = C(x)Observe that if C(x) were in place of the appended zeros, the remainder would become 000.
(Try the Online Polynomial Calculator)
Consider the decimal number 41 and the prime divisor 13.
Q: What can be done to 41 to make the result divisible by 13?
A: It is clear that 41 = 2 mod 13. On the other hand, (41-2) = 0 mod 13.
In general, if the division X/Y gives a remainder Z, then (X-Z)/Y gives remainder zero.
Step 3 - Disregard the quotient and add the remainder C(x)to the product x3 [G(x)] to yield the code message polynomial F(x), which is represented as:
Put the remainder C(x)=100 in place of the three zeros added in Step 1.
Upon reception, the entire received F(x) = "message + crc" can be checked simply by dividing F(x)/P(x) using the same generatingpolynomial. If theremainder after division equals zero, then no error was found.
100100 (ignore this quotient) ------------ 1101 ) 110010100 1101 ---- 1101 1101 ---- 000 = remainder (no error)
A single bit error in bit position K in a message F(x) can be represented by adding the term E(x) = xK, (binary 1 followed by K-zeros).
sent: 110010100 = F(x) error: 000001000 = E(x) = x3 --------- received: 110011100 = F(x) + E(x) (error in bit 3)
The above error would be detected when the CRC division is performed:
100101 (ignore this quotient) ------------ 1101 ) 110011100 = F(x) + E(x) 1101 ---- 1111 1101 ---- 1000 1101 ---- 101 = remainder (error!)
Note that division by P(x) revealed the error. On the other hand, since F(x)/P(x) = 0 by definition, the remainder is a function only of the error. An error in this same bit would give the same non-zero remainder regardless of the message bits.
F(x) + E(x) F(x) E(x) E(x) ----------- = ----- + ----- = ----- P(x) P(x) P(x) P(x)
|The remainder is a function only of the errored bits E(x).|
1 (ignore this quotient) ------------ 1101 ) 000001000 = E(x) alone 1101 ---- 101 = remainder (error!)
Since E(x) = xK has no factors other than x, a single bit error will never produce a term exactly divisible by P(x). All single bit errors will be detected.
Similarly if E(x) = xK + xK+1, the error pattern includes two adjacent bits (e.g. E(x) = 000011000 for K=3).
Since this E(x) has no factors other than (x+1) and x, a double bit error will never produce a term exactly divisible by P(x). All double bit (adjacent-bit) errors will be detected
The above argument can be extended to other types of errors, and the error performance can be fully described by examining E(x) independently of any data messages.
Observe that a CRC calculation based on the polynomial P(x) = x + 1 leaves a one-bit remainer and is a simple parity generator (EVEN).
From the above discussion, any bit error term E(x) which is an exact multiple of P(x) will not be detected.
This is the case, in particular, for the two bit error 10000001, where the two bad bits are 7-bits apart. Note that 10000001 =(1011)(1101)(11). The allowable separation between two bad bits is related to the choice of P(x).
In general, bit errors and bursts up to N-bits long will be detected for a prime P(x) of order N. For arbitrary bit errors longer than N-bits, the odds are one in 2N than a totally false bit pattern will nonetheless lead to a zero remainder.
In essence, 100% detection is assured for all errors E(x) not an exact multiple of P(x). For a 16-bit CRC, this means:
Tue Jun 18 21:33:24 ADT 2013
Last Updated: 12 OCT 2001
|Richard Tervo [ firstname.lastname@example.org ]||Back to the course homepage...|