MIMO continued and Error Correction Code

2 by 2 MIMO

• Now consider we have two transmitting antennas and two receiving antennas.

• A simple scheme called ``V-BLAST:’’ Send independent data symbols over the transmitting antennas as well as over time.

MIMO

• MIMO receiver. Will receive two samples per time slot. hij: the channel coefficient from Tx ant j to Rx ant i.

• How to decode the data?

MIMO receiver

• The simplest receiver just do a matrix inversion:

• This is NOT the optimal decoder! The maximum likelihood decoder is better.

Beyond 2 by 2

• (Section 7.1.1 of Tse book.) Let’s say we have nt transmitting antennas and nr receiving antennas. The received signal is y=Hx+w, where y is a nr by 1 vector, H is an nt by nr matrix, and x is a nt by 1 vector.

• Again, hij: the channel coefficient from Tx ant j to Rx ant i.

Beyond 2 by 2

• Each matrix has an Singular Value Decomposition (SVD). Meaning that where U is an nr by nr unitary matrix, V is an nt by nt unitary matrix, and is an nr by nt matrix whose elements in the diagonal are nonnegative.

• unitary matrix: a complex matrix times its conjugate transpose is the identity matrix.

Beyond 2 by 2

• So, at the sender side, we send

• At the receiver side, we process the received vector :

• And we will get

• And the lambda matrix is diagonal. Meaning that you will have min{nr, nt} independent channels.

Error Control Code• Widely used in many areas, like communications, DVD,

data storage…• In communications, because of noise, you can never be

sure that a received bit is right• In physical layer, what you do is, given k data bits, add n-k

redundant bits and make it into a n-bit codeword. You send the codeword to the receiver. If some bits in the codeword is wrong, the receiver should be able to do some calculation and find out– There is something wrong– Or, these things are wrong (for binary codes, this is enough)– Or, these things should be corrected as this for non-binary

codes– (this is called Block Code)

Error Control Codes

• You want a code to – Use as few redundant bits as possible– Can detect or correct as many error bits as

possible

Error Control Code

• Repetition code is the simplest, but requires a lot of redundant bits, and the error correction power is questionable for the amount of extra bits used

• Checksum does not require a lot of redundant bits, but can only tell you “something is wrong” and cannot tell you what is wrong

(7,4) Hamming Code• The best example for introductory purpose and is also

used in many applications• (7,4) Hamming code. Given 4 information bits,

(i0,i1,i2,i3), code it into 7 bits C=(c0,c1,c2,c3,c4,c5,c6). The first four bits are just copies of the information bits, e.g., c0=i0. Then produce three parity checking bits c4, c5, and c6 as (additions are in the binary field)– c4=i0+i1+i2– c5=i1+i2+i3– c6=i0+i1+i3

• For example, (1,0,1,1) coded to (1,0,1,1,0,0,0).• Is capable of correcting one bit error

Generator matrix

• Matrix representation. C=IG where

• G is called the generator matrix

Parity check matrix

• It can be verified that any CH=(0,0,0) for all codeword C

Error Correction

• What you receive is R=C+E. You multiply R with H: S=RH=(C+E)H=CH+EH=EH. S is called the syndrome. If there is only one `1’ in E, S will be one of the rows of H. Because each row is unique, you know which bit in E is `1’.

• The decoding scheme is: – Compute the syndrome– If S=(0,0,0), do nothing. If S!=(0,0,0), output one

error bit.

How G is chosen

• How G is chosen such that it can correct one error?

• Any combinations of the row vectors of G has weight at least 3 (having at least three `1’s) – and codeword has weight at least 3.

• The sum of any two codeword is still a codeword, so the distance (number of bits that differ) is also at least 3.

• So if one bit is wrong, won’t confuse it with other codewords

The existence of H

• We didn’t compare a received vector with all codewords. We used H.

• The existence of H is no coincidence (need some basic linear algebra!) Let \Omega be the space of all 7-bit vectors. The codeword space is a subspace of \Omega spanned by the row vectors of G. There must be a subspace orthogonal to the codeword space spanned by 3 vectors which is the column vectors of H.

Linear Block Code

• Hamming Code is a Linear Block Code. Linear Block Code means that the codeword is generated by multiplying the message vector with the generator matrix.

• Minimum weight as large as possible. If minimum weight is 2t+1, capable of detecting 2t error bits and correcting t error bits.

Cyclic Codes

• Hamming code is useful but there exist codes that offers same (if not larger) error control capabilities while can be implemented much simpler.

• Cyclic code is a linear code that any cyclic shift of a codeword is still a codeword.

• Makes encoding/decoding much simpler, no need of matrix multiplication.

Cyclic code

• Polynomial representation of cyclic codes, here, assume all coefficients are either 0 or 1.

• That is, if your code is (1010011) (c6 first, c0 last), you write it as

• Addition and subtraction of polynomials --- Done by doing binary addition or subtraction on each bit individually, no carry and no borrow.

• Division and multiplication of polynomials. Try divide by .

Cyclic Code

• A (n,k) cyclic code can be generated by a polynomial g(x) which has degree n-k and is a factor of xn-1. Call it the generator polynomial.

• Given message bits, (mk-1, …, m1, m0), the code is generated simply as:

• In other words, C(x) can be considered as the product of m(x) and g(x).

Example

• A (7,4) cyclic code g(x) = x3+x+1.• If m(x) = x3+1, C(x) = x6+x4+x+1.

Cyclic Code

• One way of thinking it is to write it out as the generator matrix

• So, clearly, it is a linear code. Each row of the generator matrix is just a shifted version of the first row. Unlike Hamming Code.

• Why is it a cyclic code?

Example

• The cyclic shift of C(x) = x6+x4+x+1 is C1(x) = x5+x2+x+1.

• It is still a code polynomial, because it the code polynomial if m(x) = x2+1.

Cyclic Code

• Given a code polynomial

• We have

• C1(x) is the cyclic shift of C(x) and (1) has a degree of no more than n-1 and (2) divides g(x) (why?) hence is a code polynomial.

Cyclic Code

• So, to generate a cyclic code is to find a polynomial that (1) has degree n-k and (2) is a factor of xn-1.

Generating Systematic Cyclic Code

• A systematic code means that the first k bits are the data bits and the rest n-k bits are parity checking bits.

• To generate it, you let

where

• The claim is that C(x) must divide g(x) hence is a code polynomial. 33 mod 7 = 6. Hence 33-6=28 can be divided by 7.

Division Circuit• Division of polynomials can be done efficiently by the

division circuit. (just to know there exists such a thing, no need to understand it)

Remaining Questions for Those Really Interested

• Decoding. Divide the received polynomial by g(x). If there is no error you should get a 0 (why?). Make sure that the error polynomial you have in mind does not divide g(x).

• How to make sure to choose a good g(x) to make the minimum degree larger? Turns out to learn this you have to study more – it’s the BCH code.

Cyclic code used in IEEE 802

• g(x) = x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1– all single and double bit errors – all errors with an odd number of bits – all burst errors of length 32 or less

Other codes

• RS code. Block code. Used in CD, DVD, HDTV transmission.

• LDPC code. Also block code. Reinvented after first proposed 40 some years ago. Proposed to be used in 802.11n. Achieve close-to-Shannon bound

• Trellis code. Not block code. More closely coupled with modulation.

• Turbo code. Achieve close-to-Shannon bound.