I n f o r m a t i o n T r a n s m i s s i o n - PowerPoint PPT Presentation

I n f o r m a t i o n T r a n s m i s s i o n C h a p t e r 5 , B l o c k c o d e s OVE EDFORS ELECTRICAL AND INFORMATION TECHNOLOGY

L e a r n i n g o u t c o m e s ● A f t e r t h e l e c t u r e s t h e s t u d e n t s h o u l d : – U n d e r s t a n d a n d b e a b l e t o u s e m o d u l o - t w o a r i t h m e t i c , – k n o w t h a t a r e c e i v e d w o r d i s c o m p o s e d o f t h e t r a n s m i t t e d c o d e w o r d a n d a n e r r o r p a t t e r n , – k n o w w h a t t h e m i n i m u m d i s t a n c e o f a c o d e i s a n d h o w i t r e l a t e d t o e r r o r c o r r e c t i o n a n d e r r o r d e t e c t i o n p r o p e r t i e s , – b e a b l e t o p e r f o r m e n c o d i n g o f m e s s a g e s i n t o c o d e w o r d s , – u n d e r s t a n d h o w c o d e w o r d s a r e g e n e r a t e d u s i n g a g e n e r a t o r m a t r i x , a n d – u n d e r s t a n d h o w e r r o r s c a n b e d e t e c t e d u s i n g a p a r i t y c h e c k m a t r i x . O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 2

Wh e r e a r e w e i n t h e B I G P I C T U R E ? Block codes Lecture relates to pages 189-195 in textbook. O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 3

Me t h o d s o f c h a n n e l c o d i n g ● F o r c h a n n e l c o d i n g ( e r r o r c o r r e c t i o n ) w e h a v e t w o m a i n c l a s s e s o f c o d e s , n a m e l y : – b l o c k c o d e s , w h i c h w e fi r s t e n c o u n t e r e d w h e n w e d i s c u s s e d S h a n n o n ' s c h a n n e l c o d i n g t h e o r e m – c o n v o l u t i o n a l c o d e s . We s h a l l b r i e fl y d i s c u s s b o t h c l a s s e s . O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 4

D i g i t a l c h a n n e l – s y m b o l s i n a n d o u t Our code words Symbols X as input to channel Symbols Y as output from channel Our code words, corrupted by the noisy channel O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 5

Tie b i n a r y fj e l d For the following calculations we use the binary field , for which the rules of addition and multiplication are those of modulo-two arithmetic: Notice that since 1 + 1 = 0, subtraction is the same as addition, which is very convenient O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 6

Tie e r r o r p a t t e r n Suppose that the codeword is transmitted over the binary symmetric channel and that is the possibly erroneously received version of it, then the error pattern is defined to be the N -tuple that satisfies If we have one error, that is, e consists of one 1 and N - 1 0's, then one component in v is altered. Two errors cause two altered components in v . O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 7

Mi n i m u m d i s t a n c e The minimum distance , d min , of a block code B is the minimum of all distances between two non-identical codewords of the code. If the sum of any two codewords is a codeword, then the code is said to be linear . For a linear block code the minimum distance is simply equal to the least number of 1's in a nonzero codeword In general, a block code with minimum distance d min will correct up ( d min -1)/2 errors. Alternatively, it can be used to detect up to d min -1 errors. O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 8

Tie ( 7 , 4 ) H a m m i n g c o d e Hamming constructed a class of single-error-correcting linear block codes with minimum distance d min = 3. In the table we specify an encoder mapping for the (7,4) Hamming code with M = 2 4 = 16 code words. This code has rate O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 9

E x a m p l e Assume that we would like to transmit the information 4- tuple u = (1011) over a binary symmetric channel. Then we encode it, by using the mapping in the table, and obtain the code word v = (0110011). Let, for example, the sixth position be altered by the channel. Thus, we receive r = (0110001). O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 0

E x a m p l e ( c o n t . ) To correct the error we add position-wise modulo-two rows 2, 3, and 7 (the positions corresponding to the 1's in r ) from a given matrix H T (which we will explain later) and obtain that is, the binary representation of 6; we flip the sixth position in r = (0110001) and obtain the estimate of the code word which corresponds to the information 4-tuple . O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 1

H o w d o e s i t w o r k ? ( I ) Why does our scheme work? We can write the received 7- tuple as the sum of the code word and the error pattern, that is, r = v + e. Remember that 1 + 1 = 0! Due to this simple equality we can obtain the sum of the rows corresponding to the 1's in r by adding component- wise the sums of the rows corresponding to the 1's in v and e . O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 2

H o w d o e s i t w o r k ? ( I I ) Now we exploit that the mapping in the table is constructed such that the sum of the rows corresponding to the 1's in any code word is 000. Hence, we conclude that the sum of the rows corresponding to the 1's in r (this is the sum that the decoder computes) is equal to the sum of the rows corresponding to the 1's in e . O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 3

H o w d o e s i t w o r k ? ( I I I ) But assuming at most one error during the transmission we obtain in case of no errors the sum of zero rows which we interpret as 000 and then we accept r as our estimate In case of one error the sum contains one row, namely, precisely the row which is the binary representation of the position of the 1 in e . Hence, flip that position in r and we obtain our estimated codeword O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 4

Tie g e n e r a t o r m a t r i x How do we obtain the remarkable encoder mapping? Since the Hamming code is linear the code words corresponding to the information 4 tuples 1000, 0100, 0010, 0001 are of particular interest; these code words form a so- called generator matrix for the (7,4) Hamming code: O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 5

C o d e w o r d g e n e r a t i o n All code words can be obtained as the product of the corresponding information 4-tuples and the generator matrix: For example, the code word corresponding to u = (1011) is obtained as the position-wise modulo-two sum of the first, third and fourth rows in G , that is, FROM PREVIOUS SLIDE in agreement with the mapping. O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 6

G e n e r a t i o n o f t h e p a r i t y c h e c k m a t r i x Assume that we have a K x N generator G , then by the theory of matrices there exists an (N-K) x N matrix H such that It follows immediately that that is, we have the fundamental result Where H is the so-called parity check matrix . O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 7

Tie p a r i t y - c h e c k m a t r i x Parity check in different wording : Let v be a code word, then, if we add (position-wise modulo-two) the rows of H T corresponding to the 1s in v we obtain the all-zero (N-K) - tuple. This computation is a parity-checking procedure and thus we call the matrix H a parity-check matrix of our code. O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 8

G e n e r a t o r a n d p a r i t y c h e c k m a t r i c e s It is easily verified that Using linear algebra we can obtain the generator matrix G for a given parity-check matrix H and vice versa. O v e E d f o r s E I T A 3 0 - C h a p t e r 5 ( P a r t 4 ) 1 9