4

fewer check digits, these codes operate at a higher efficiency.

3.2 LINEAR BLOCK CODES

Let a code word consists of a digits C1, C2, .... Cn , and a data word consists of k digits d1, d2, ....dk . Since the code word and the data word are an n-tuple and k-typle respectively. Hence, they are n- and k- dimensional vectors. using row matrices to represent these words, we have;

C= (C1,C2... Cn) ; d= (d1, d2... dk)

For linear block codes, all the n digits of c are formed by linear combinations (1modulo-2 a addition) of k data digits.

But, in a special case where C1=d1,C2=d2, Ck=dk and the remaining digits from Ck+1 to Cn are linear combinations of d1, d2.. dk is known as a systematic code. In a systematic code, the first k digits of a code word are the data digits and the last m=n-k digits are known as parity- check digits. Which are formed by linear combinations of data digits d1,d2,dk

Hence,

C1=d1

C2=d2

.

.

.

Ck=dk

CK+1=h11d1 (h12 d2 (.(h1kdk

Ck+1=h21d1 (h22d2 ( ( h2kdk

.

.

.

(3.3a)

Cn=hm1d1 ( hm2d2((hmkdk

Or

C=dG.(3.3b)

Where

G is kxm matrix and is known as the generator matrix, which can be partitioned into a k x k identity matrix Ik and a k x m matrix p. All the elements of P are either 0 or 1. Then, the code word can be expressed as

C=dg

= d [ Ik,p]

= [d, dp]

= [d,cp].. (3.5)

where

Cp=dp (3.6)

Example e1: For a (6,3) code the generator matrix G is

For all eight possible data words, find the corresponding code words, and verify that the code is a single error correcting code.

Solution

Applying,

C=dp

The table 3.1 below shows the eight data words and the corresponding code words.

Data word dCode word C

111

110

101

100

011

010

001

000111000

110110

101011

100101

011101

010011

001110

000000

Table 3.1

The distance between any two code words is at least 3. Hence, the code can correct at least one error . The fig 3.1 below shows a possible encoder for this code, using a three digit shift register and three modoulo -2 adders

Fig 3.1 Encoder for linear block codes

Decoding:

From equ. (3.6) and the fact that the modulo-2 sum of any sequence with iself is zero. we have;

dp( Cp= .....(3.7)

Where:

Im = Identity matrix of order m x m (m=n-k); Hence,

CHT=0 ................ (3.8)

Where:

......(3.8b)

and its transpose is

H= [ PT Im ] ......(3.8C)

is known as the parity check matrix.

In decoding, every code word must satisty equ. 3.8 .

Let us consider, ther received word r and because of possible errors caused by channel noise, r differs from the transmitted code word C,

( = C ( e

where,

e= error word ( or error vector)

Example 2: If the data word 100 in example 1 above is transmitted as a code word 10010, and the channel noise causes a detection error in the third digit;

Solution

Apply

(= 101101

C= 100101

( e = 001000

Thus, an element 1 in e indicates an eror in the corresponding position, and 0 indicates no error. The Hamming distance between r and c is simply the number of 1s in e, which is one in thirs case.

Supposed the transmitted code word is Code word is Ci and the channel noise causes an error li, then,

(=Ci+ei

where ( = the received word

if ei=000000, which implies no error, then (HT=0

But, as a result of possible channel errors, (HT is in general a non zero row vector s which is cared the syndrome.

S= (HT--------- (3.9)

= (Ci ( eiHT

=eiHT-----------(3.9b)

( can also be expressed in terms of code words other than li. Hence,

(= Cj( ej j#i

This implies that ;

S= (Cj (ej) HT+ejHTSince , there are 2k possible code words,

S= eHTis satisfied by 2k error vectors.

Example 3: If a data word d=100 is transmitted by a code word 100101 in example e1 above, and if a detection error is caused in the third digit, then the received word is 101101.

In this case, we have;

C=100101

e=001000

But , the same word could have been received if

C=101011 and e=000110, or if C= 010011 and e= 111110, etc.

From equ 3.9b, since we have eight possible error vectors (2k error vectors), hence the problem will be which one to choope. Hence, the reasonable criterion is the maximum-likelihood rule where, if we receiver (, then we decide in favour of that C for which ( is most likely to be received. Thus, we decide Ci transmitted if

p((\Ci)>p((/Ck) add k#i

For a BSC, suppose the Hamming distance b/w ( and Ci is d, i, the channel noise causes errors in d digits, Then if Pe is the digit error probability of a BSc,

P((\Ci)=Ped(1-pe)n-d=(1-pe)n

If Pe < 0.5, then P((\Ci)is a monotonically decreasing function of d because Pc/(1-Pe)k. Thus, every k-bit input frame produces an n-bit out put frame. Redundancy in the output, since n>k .There is also memory in the coder, since the output frame depends on the previous k input fromes where k>1 and the code rate is R=k/n, which is in this case. The constraint length, k, is the number of input frames that are held in the kK-bit Shift register. Data from the kK stages of the shift register are added (modulo 2) and used to set the bits in the n-stage output register.

Fig 3.6 Convolution encoding (k=3,n=4,k=5,and R=3/4)

Let as consider the convolution coder shown in fig 3.7 where k=1 , n=d2 , k=3 , and a commentator whit two inputs performs the function of a two stage output shift register. Hence, the convolution code is generated by inputting a bit of data and then giving the commutator a complete revolution. This process is repeated for successive in put bits to produce the convolution ally encoded output . In this case, each k=1 in put bit produces n=2 out put bits, thus, the code Rate (R) is . Let is illustrate this with an input digits of 11010. Firstly, all the stages of the registor is clear; ie they are in 0 state. When the first data digit 1 enters the register, the stage S1 shows 1 and all the other stages (S2 and S3) are unchanged; ie at 0 state. The two modulo -2 adders V1=1 and v2=1 . The commutator samples this output, given the coder output to 11. However , when the 2nd message bit1 enters the register, it enters the stage s1, and the previous 1 is S1 is shifted to S2=1 . Hence, the decoder output is 01. Also, when the 3rd message digits 0enters the register, then S1=0 , S2=1 and S3=1 and the decoder output is 01 . The process continues until the last data digits enters the stage S1.

In general, the convolution coder operates on a continuous basis . fewer check digits, these codes operate at a higher efficiency.

3.2 LINEAR BLOCK CODES

Let a code word consists of a digits C1, C2, .... Cn , and a data word consists of k digits d1, d2, ....dk . Since the code word and the data word are an n-tuple and k-typle respectively. Hence, they are n- and k- dimensional vectors. using row matrices to represent these words, we have;

C= (C1,C2... Cn) ; d= (d1, d2... dk)

For linear block codes, all the n digits of c are formed by linear combinations (1modulo-2 a addition) of k data digits.

But, in a special case where C1=d1,C2=d2, Ck=dk and the remaining digits from Ck+1 to Cn are linear combinations of d1, d2.. dk is known as a systematic code. In a systematic code, the first k digits of a code word are the data digits and the last m=n-k digits are known as parity- check digits. Which are formed by linear combinations of data digits d1,d2,dk

Hence,

C1=d1

C2=d2

.

.

.

Ck=dk

CK+1=h11d1 (h12 d2 (.(h1kdk

Ck+1=h21d1 (h22d2 ( ( h2kdk

.

.

.

(3.3a)

Cn=hm1d1 ( hm2d2((hmkdk

Or

C=dG.(3.3b)

Where

G is kxm matrix and is known as the generator matrix, which can be partitioned into a k x k identity matrix Ik and a k x m matrix p. All the elements of P are either 0 or 1. Then, the code word can be expressed as

C=dg

= d [ Ik,p]

= [d, dp]

= [d,cp].. (3.5)

where

Cp=dp (3.6)

Example e1: For a (6,3) code the generator matrix G is

For all eight possible data words, find the corresponding code words, and verify that the code is a single error correcting code.

Solution

Applying,

C=dp

The table 3.1 below shows the eight data words and the corresponding code words.

Data word dCode word C

111

110

101

100

011

010

001

000111000

110110

101011

100101

011101

010011

001110

000000

Table 3.1

The distance between any two code words is at least 3. Hence, the code can correct at least one error . The fig 3.1 below shows a possible encoder for this code, using a three digit shift register and three modoulo -2 adders

Fig 3.1 Encoder for linear block codes

Decoding:

From equ. (3.6) and the fact that the modulo-2 sum of any sequence with iself is zero. we have;

dp( Cp= .....(3.7)

Where:

Im = Identity matrix of order m x m (m=n-k); Hence,

CHT=0 ................ (3.8)

Where:

......(3.8b)

and its transpose is

H= [ PT Im ] ......(3.8C)

is known as the parity check matrix.

In decoding, every code word must satisty equ. 3.8 .

Let us consider, ther received word r and because of possible errors caused by channel noise, r differs from the transmitted code word C,

( = C ( e

where,

e= error word ( or error vector)

Example 2: If the data word 100 in example 1 above is transmitted as a code word 10010, and the channel noise causes a detection error in the third digit;

Solution

Apply

(= 101101

C= 100101

( e = 001000

Thus, an element 1 in e indicates an eror in the corresponding position, and 0 indicates no error. The Hamming distance between r and c is simply the number of 1s in e, which is one in thirs case.

Supposed the transmitted code word is Code word is Ci and the channel noise causes an error li, then,

(=Ci+ei

where ( = the received word

if ei=000000, which implies no error, then (HT=0

But, as a result of possible channel errors, (HT is in general a non zero row vector s which is cared the syndrome.

S= (HT--------- (3.9)

= (Ci ( eiHT

=eiHT-----------(3.9b)

( can also be expressed in terms of code words other than li. Hence,

(= Cj( ej j#i

This implies that ;

S= (Cj (ej) HT+ejHTSince , there are 2k possible code words,

S= eHTis satisfied by 2k error vectors.

Example 3: If a data word d=100 is transmitted by a code word 100101 in example e1 above, and if a detection error is caused in the third digit, then the received word is 101101.

In this case, we have;

C=100101

e=001000

But , the same word could have been received if

C=101011 and e=000110, or if C= 010011 and e= 111110, etc.

From equ 3.9b, since we have eight possible error vectors (2k error vectors), hence the problem will be which one to choope. Hence, the reasonable criterion is the maximum-likelihood rule where, if we receiver (, then we decide in favour of that C for which ( is most likely to be received. Thus, we decide Ci transmitted if

p((\Ci)>p((/Ck) add k#i

For a BSC, suppose the Hamming distance b/w ( and Ci is d, i, the channel noise causes errors in d digits, Then if Pe is the digit error probability of a BSc,

P((\Ci)=Ped(1-pe)n-d=(1-pe)n

If Pe < 0.5, then P((\Ci)is a monotonically decreasing function of d because Pc/(1-Pe)k. Thus, every k-bit input frame produces an n-bit out put frame. Redundancy in the output, since n>k .There is also memory in the coder, since the output frame depends on the previous k input fromes where k>1 and the code rate is R=k/n, which is in this case. The constraint length, k, is the number of input frames that are held in the kK-bit Shift register. Data from the kK stages of the shift register are added (modulo 2) and used to set the bits in the n-stage output register.

Fig 3.6 Convolution encoding (k=3,n=4,k=5,and R=3/4)

Let as consider the convolution coder shown in fig 3.7 where k=1 , n=d2 , k=3 , and a commentator whit two inputs performs the function of a two stage output shift register. Hence, the convolution code is generated by inputting a bit of data and then giving the commutator a complete revolution. This process is repeated for successive in put bits to produce the convolution ally encoded output . In this case, each k=1 in put bit produces n=2 out put bits, thus, the code Rate (R) is .

Let is illustrate this with an input digits of 11010. Firstly, all the stages of the registor is clear; ie they are in 0 state. When the first data digit 1 enters the register, the stage S1 shows 1 and all the other stages (S2 and S3) are unchanged; ie at 0 state. The two modulo -2 adders V1=1 and v2=1 . The commutator samples this output, given the coder output to 11. However , when the 2nd message bit1 enters the register, it enters the stage s1, and the previous 1 is S1 is shifted to S2=1 . Hence, the decoder output is 01. Also, when the 3rd message digits 0enters the register, then S1=0 , S2=1 and S3=1 and the decoder output is 01 . The process continues until the last data digits enters the stage S1.

In general, the convolution coder operates on a continuous basis .

4.0 Noise in Amplitude modulation systems Introduction: Now we have to look at the be haviour of analog communication systems in the presence of noise. When a signal power ST is transmitted over a channel as showed in fig. 4.0 . The transmitted signal is corrupted by channel noise during transmission. The Channel will altercate the signal. Hence, at the receiver input, a signal will nixed with noise, and the signal and noise powers at this point are Si and Ni respectively. The receiver processes the signal to yield the desire dsignal plys noise. The signal and noise power at the receiver out put are So and No respectively. Thus, in analog communication systems, the quality of the received signal is determined by raion . So/No which is th output SNR. This rationcan be increased by increasing the transmitted power ST. other factors which can affect the maximum value of ST are transmitter cost, channels etc. Although, it is more convenient to deal with the received power rather than the transmitted power ST. 4.1 Base band systems

This is where a signal is transmitted directly with out any modulation. This mode of ommunication is suitable over a pair of wires, optical fiber, or coaxial cables. S0, it is mainly used in short have links. Generally, base band systems serves as a basis against which other systems may be compared. Hence the transmitter and the receiver are ideal baseband filters as shown in fig 4.1. We have two low-pass filters. The first one is Hp(w) which is at the transmitter limits the input signal spectrum to a given bandwidth, while the other ,Hd(w) which is at the receiver eliminate the out-of-band noise and channel interference. These filter can also serve as an optimize the signal-to-noise ratio (SNR) at the receiver.Fig.4.1.A baseband system

If we assume the baseband signal m(t) to be a zero mean ,wide sense stationary random process band limited to B Hz. Let us consider , an ideal low-pass (baseband) filter with bandwidth B at the transmitter and the receiver . If the channel is assumed to be distortion less. Thus,

So = Si ---------------------------------------------- (4.0) And

No = 2Sn(w)df --------------------------(4.1)

Note: 1 ( 1 = 0 ( 0 =0

0 ( 1 = 1 ( 0 =1

1 0 0 0

0 1 0 0

.

.

0 0 0.0

h11 h21 hm1

h12 h22 hm2

.

.

h1k h2k .hmk

G=

Ik (k x k)

P(k x m)

.3.4

1 0 0

0 1 0

0 0 1

1 0 1

0 1 1

1 1 0

G=

Ik

P

Commutator

C1

C2

C3

C4

C5

C6

Data input

d1

d3

d2

d Cp

P

Im

=

0

C

HT=

P

Im

100011

1 0 1

0 1 1

1 1 0

1 0 0

0 1 0

0 0 1

=

e1 e2 e3 e4 e5 e6

1 0 1

0 1 1

1 1 0

1 0 0

0 1 0

0 0 1

=

Constraint length= k input famine

Logic

Coded output data

n-bit frame

Input data

n-bit

Shifts

register

(n-bits)

k-bit frame

Shift register

(kK. bits)

Constraint length= k input famine

Logic

Coded output data

n-bit frame

Input data

n-bit

Shifts

register

(n-bits)

k-bit frame

Shift register

(kK. bits)

Data input

S1

S2

S3

Coded sequence output

commentator

Modulo 2

adders

Note: 1 ( 1 = 0 ( 0 =0

0 ( 1 = 1 ( 0 =1

1 0 0 0

0 1 0 0

.

.

0 0 0.0

h11 h21 hm1

h12 h22 hm2

.

.

h1k h2k .hmk

G=

Ik (k x k)

P(k x m)

.3.4

1 0 0

0 1 0

0 0 1

1 0 1

0 1 1

1 1 0

G=

Ik

P

Commutator

C1

C2

C3

C4

C5

C6

Data input

d1

d3

d2

d Cp

P

Im

=

0

C

HT=

P

Im

100011

1 0 1

0 1 1

1 1 0

1 0 0

0 1 0

0 0 1

=

e1 e2 e3 e4 e5 e6

1 0 1

0 1 1

1 1 0

1 0 0

0 1 0

0 0 1

=

_1101277449.unknown

_1101277706.unknown

_1099376621.unknown