1 / 12

4-2 binary fields and binary vector spaces

4-2 binary fields and binary vector spaces. Special Thanks to Dr. Samir Al-Ghadhban & EE430 Students. Binary fields. To describe codes, we use a mathematical structure called a field.

Download Presentation

4-2 binary fields and binary vector spaces

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 4-2 binary fields and binary vector spaces Special Thanks to Dr. Samir Al-Ghadhban & EE430 Students

  2. Binary fields • To describe codes, we use a mathematical structure called a field. • A field is defined to be a set of elements A and two arithmetic operation called addition (+) and multiplication (.) ,such that; 1- Closure For all a,b ϵ A → a+b ϵ A and a.b ϵ A 2- Associative properties of addition and multiplication a+b +c = (a+b)+c = a+(b+c) and a.b.c = (a.b).c = a.(b.c) 3- Identity elements a+0 = 0+a=a ; 0 is the additive identity a.1 = 1.a=a ; 1 is the multiplicative identity

  3. Cont. 4- Additive inverse For any a ϵ A, there exist b such that a+b =0 → b is the additive inverse b→ -a 5- Multiplicative inverse For any element a ϵ A ,except 0, there is a multiplicative inverse ,such that a.b=1 b → a-1 6- Addition is commutative a+b = b+a 7- Distributive property a.(b+c) = a.b + a.c (a+b).c = a.c + b.c

  4. Boolean arithmetic • For binary source, A={0,1} • The field is called Galois field with two elements, GF(2). • Addition and multiplication are defined as: • 0+0 = 0 ; 0+1 = 1+0 = 1 ; 1+1 = 0 • 0.0 = 1.0 = 0.0 = 0 ; 1.1 = 1 → Addition is the XOR "Exclusive or" Multiplication is the AND function

  5. Binaryvectors • We can construct binary vectors from the binary field GF(2) • Define An to be a set with elements a= (a0, a1 ,…, an-1) with each ai ϵ A= {0,1}. • Define two arithmetic operators 1. Vector addition: if a,bϵ An then vector addition (+) is defined as a+b ≡ ((a0+b0), (a1+b1),…, (an-1+bn-1)) 2. Scalar multiplication: If a ϵ An and b ϵ Ais a binary scalar , then b.a = a.b = (ba0, ba1 ,…, ban-1)

  6. Vector space • A vector space is a structure made of: • - vectors An • - scalars A • - and two arithmetic operations. • The sets and the arithmetic operations satisfy the following constraints: 1. Closure: For every a,b ϵ An,the sum a+bϵ An 2. Addition is commutative 3. Addition is associative 4. An contains a vector 0 such that a+0 = a 5. Additive inverses: for every a ϵ An,there is some vector bϵ An such that a+b = 0.

  7. Cont. 6. For every scalar a ϵ A and every vector b ϵ An there is a vector a b ϵ An→ closure of scalar multiplication. 7. Scalar multiplication is associative 8. Scalar multiplication is distributive with respect to vector addition 9. Scalar multiplication is distributive with respect to scalar addition 10. If 1 ϵ A is the scalar multiplicative identity, then for every a ϵ An , 1 a= a .

  8. Example: 2D Code. • A 2- D code is a simple error correcting code with additional error detection capability. The rate in this example is 9/16. • Let the 9 bit massage be • M=[m0, m1, m2,….,m8] , mi ϵ {0,1} • Let m8 be the first bit transmitted, then the codeword is • C = [c0,c1,c2 ,m0,c4,m1,c6,m2,c8,m3,m5,c12,m6,m7,m8] • Where

  9. Cont. • Where c12 = m8 + m7 + m6 similarly c1,c2,c4,c6 and c8 c0 = m8 + m7 + …+ m1 + m0 • During transmission , bit error may occur, the received word is v= c + e • where e = [ e0,e1,….,e15]; if ei = 1 → error ei = 0 → no error

  10. Cont. • At the receiver, we calculate seven error check bits, which are called syndrome bits. this code can correct single bit errors ,and can detect two bit errors.

  11. Cont. • Notice that, 1. If no errors occur, all si are zero. 2. If an odd number of errors occur , s0 = 1 3. If a single error occurs in one of the mimessage bits, one of the row and one of the column syndrome bits will equal to 1. In this case, we can correct the error bit by adding 1 to the vibit corresponding to the intersection of the row and column. 4. If a single error occurs in one of the parity bits (ci), then either one row or one column syndrome bit (but not both) will be equal to one and the s0 bit will be 1. in this case, no correction is necessary because none of the message bits are in error.

  12. Cont. 5. If two errors occur, s0=0, and multiple syndrome bits in a row or column will be 1 → we can detect two errors. but we can't guarantee to correct these errors For example, assume the errors are in v14 and v13, then s2 and s1 = 1 but s6 =0 So, we can't tell the error location. 6. If the number of errors are greater than two: Odd number → we can't guarantee that we will not miscorrect the error. Even number → we can't guarantee to detect the error. See example 4.2.3 (Retransmission Probability)

More Related