Chapter 2

2.3 Sincem is not a prime, it can be factored as the product of two integersa andb,

m = a · b

with 1 < a, b < m. It is clear that botha andb are in the set{1, 2, · · · , m − 1}. It follows

from the definition of modulo-m multiplication that

a ¡ b = 0.

Since0 is not an element in the set{1, 2, · · · ,m−1}, the set is not closed under the modulo-m

multiplication and hence can not be a group.

2.5 It follows from Problem 2.3 that, ifm is not a prime, the set{1, 2, · · · ,m − 1} can not be a

group under the modulo-m multiplication. Consequently, the set{0, 1, 2, · · · ,m− 1} can not

be a field under the modulo-m addition and multiplication.

2.7 First we note that the set of sums of unit element contains the zero element0. For any1 ≤` < λ,

∑i=1

1 +λ−∑i=1

1 =λ∑

i=1

1 = 0.

Hence every sum has an inverse with respect to the addition operation of the field GF(q). Since

the sums are elements in GF(q), they must satisfy the associative and commutative laws with

respect to the addition operation of GF(q). Therefore, the sums form a commutative group

under the addition of GF(q).

Next we note that the sums contain the unit element 1 of GF(q). For each nonzero sum

∑i=1

1

with 1 ≤ ` < λ, we want to show it has a multiplicative inverse with respect to the multipli-

cation operation of GF(q). Sinceλ is prime,` andλ are relatively prime and there exist two

1

integersa andb such that

a · ` + b · λ = 1, (1)

wherea andλ are also relatively prime. Dividinga by λ, we obtain

a = kλ + r with 0 ≤ r < λ. (2)

Sincea andλ are relatively prime,r 6= 0. Hence

1 ≤ r < λ

Combining (1) and (2), we have

` · r = −(b + k`) · λ + 1

Consider

∑i=1

1 ·r∑

i=1

1 =`·r∑i=1

1 =

−(b+k`)·λ∑i=1

+1

= (λ∑

i=1

1)(

−(b+k`)∑i=1

1) + 1

= 0 + 1 = 1.

Hence, every nonzero sum has an inverse with respect to the multiplication operation of GF(q).

Since the nonzero sums are elements of GF(q), they obey the associative and commutative

laws with respect to the multiplication of GF(q). Also the sums satisfy the distributive law.

As a result, the sums form a field, a subfield of GF(q).

2.8 Consider the finite field GF(q). Letn be the maximum order of the nonzero elements of GF(q)

and letα be an element of ordern. It follows from Theorem 2.9 thatn dividesq − 1, i.e.

q − 1 = k · n.

Thusn ≤ q − 1. Let β be any other nonzero element in GF(q) and lete be the order ofβ.

2

Suppose thate does not dividen. Let (n, e) be the greatest common factor ofn ande. Then

e/(n, e) andn are relatively prime. Consider the element

β(n,e)

This element has ordere/(n, e). The element

αβ(n,e)

has orderne/(n, e) which is greater thann. This contradicts the fact thatn is the maximum

order of nonzero elements in GF(q). Hencee must dividen. Therefore, the order of each

nonzero element of GF(q) is a factor ofn. This implies that each nonzero element of GF(q)

is a root of the polynomial

Xn − 1.

Consequently,q − 1 ≤ n. Sincen ≤ q − 1 (by Theorem 2.9), we must have

n = q − 1.

Thus the maximum order of nonzero elements in GF(q) is q-1. The elements of orderq − 1

are then primitive elements.

2.11 (a) Suppose thatf(X) is irreducible but its reciprocalf ∗(X) is not. Then

f ∗(X) = a(X) · b(X)

where the degrees ofa(X) andb(X) are nonzero. Letk andm be the degrees ofa(X) and

b(X) respectivly. Clearly,k + m = n. Since the reciprocal off ∗(X) is f(X),

f(X) = Xnf ∗(1

X) = Xka(

1

X) ·Xmb(

1

X).

This says thatf(X) is not irreducible and is a contradiction to the hypothesis. Hencef ∗(X)

must be irreducible. Similarly, we can prove that iff ∗(X) is irreducible,f(X) is also

irreducible. Consequently,f ∗(X) is irreducible if and only iff(X) is irreducible.

3

(b) Suppose thatf(X) is primitive butf ∗(X) is not. Then there exists a positive integerk less

than2n − 1 such thatf ∗(X) dividesXk + 1. Let

Xk + 1 = f ∗(X)q(X).

Taking the reciprocals of both sides of the above equality, we have

Xk + 1 = Xkf ∗(1

X)q(

1

X)

= Xnf ∗(1

X) ·Xk−nq(

1

X)

= f(X) ·Xk−nq(1

X).

This implies thatf(X) divides Xk + 1 with k < 2n − 1. This is a contradiction to the

hypothesis thatf(X) is primitive. Hencef ∗(X) must be also primitive. Similarly, iff ∗(X) is

primitive, f(X) must also be primitive. Consequentlyf ∗(X) is primitive if and only iff(X)

is primitive.

2.15 We only need to show thatβ, β2, · · · , β2e−1are distinct. Suppose that

β2i

= β2j

for 0 ≤ i, j < e andi < j. Then,

(β2j−i−1)2i

= 1.

Since the orderβ is a factor of2m − 1, it must be odd. For(β2j−i−1)2i= 1, we must have

β2j−i−1 = 1.

Since bothi andj are less thane, j − i < e. This is contradiction to the fact that thee is the

smallest nonnegative integer such that

β2e−1 = 1.

4

Henceβ2i 6= β2jfor 0 ≤ i, j < e.

2.16 Let n′ be the order ofβ2i. Then

(β2i

)n′ = 1

Hence

(βn′)2i

= 1. (1)

Since the ordern of β is odd,n and2i are relatively prime. From(1), we see thatn dividesn′

and

n′ = kn. (2)

Now consider

(β2i

)n = (βn)2i

= 1

This implies thatn′ (the order ofβ2i) dividesn. Hence

n = `n′ (3)

From (2) and (3), we conclude that

n′ = n.

2.20 Note thatc ·v = c · (0+v) = c ·0+ c ·v. Adding−(c ·v) to both sides of the above equality,

we have

c · v + [−(c · v)] = c · 0 + c · v + [−(c · v)]

0 = c · 0 + 0.

Since0 is the additive identity of the vector space, we then have

c · 0 = 0.

2.21 Note that0 · v = 0. Then for anyc in F ,

(−c + c) · v = 0

5

(−c) · v + c · v = 0.

Hence(−c) · v is the additive inverse ofc · v, i.e.

−(c · v) = (−c) · v (1)

Sincec · 0 = 0 (problem 2.20),

c · (−v + v) = 0

c · (−v) + c · v = 0.

Hencec · (−v) is the additive inverse ofc · v, i.e.

−(c · v) = c · (−v) (2)

From (1) and (2), we obtain

−(c · v) = (−c) · v = c · (−v)

2.22 By Theorem 2.22,S is a subspace if (i) for anyu andv in S, u + v is in S and (ii) for anyc

in F andu in S, c · u is in S. The first condition is now given, we only have to show that the

second condition is implied by the first condition forF = GF (2). Letu be any element inS.

It follows from the given condition that

u + u = 0

is also inS. Let c be an element in GF(2). Then, for anyu in S,

c · u =

0 for c = 0

u for c = 1

Clearlyc · u is also inS. HenceS is a subspace.

2.24 If the elements of GF(2m) are represented bym-tuples over GF(2), the proof that GF(2m) is

6

a vector space over GF(2) is then straight-forward.

2.27 Let u andv be any two elements inS1 ∩ S2. It is clear theu andv are elements inS1, andu

andv are elements inS2. SinceS1 andS2 are subspaces,

u + v ∈ S1

and

u + v ∈ S2.

Hence,u + v is in S1 ∩ S2. Now letx be any vector inS1 ∩ S2. Thenx ∈ S1, andx ∈ S2.

Again, sinceS1 andS2 are subspaces, for anyc in the fieldF , c · x is in S1 and also inS2.

Hencec · v is in the intersection,S1 ∩ S2. It follows from Theorem 2.22 thatS1 ∩ S2 is a

subspace.

7

Chapter 3

3.1 The generator and parity-check matrices are:

G =

0 1 1 1 1 0 0 0

1 1 1 0 0 1 0 0

1 1 0 1 0 0 1 0

1 0 1 1 0 0 0 1

H =

1 0 0 0 0 1 1 1

0 1 0 0 1 1 1 0

0 0 1 0 1 1 0 1

0 0 0 1 1 0 1 1

From the parity-check matrix we see that each column contains odd number of ones, and no

two columns are alike. Thus no two columns sum to zero and any three columns sum to a 4-

tuple with odd number of ones. However, the first, the second, the third and the sixth columns

sum to zero. Therefore, the minimum distance of the code is 4.

3.4 (a) The matrixH1 is an(n−k+1)×(n+1) matrix. First we note that then−k rows ofH are

linearly independent. It is clear that the first(n− k) rows ofH1 are also linearly independent.

The last row ofH1 has a′′1′′ at its first position but other rows ofH1 have a′′0′′ at their first

position. Any linear combination including the last row ofH1 will never yield a zero vector.

Thus all the rows ofH1 are linearly independent. Hence the row space ofH1 has dimension

n− k + 1. The dimension of its null space,C1, is then equal to

dim(C1) = (n + 1)− (n− k + 1) = k

HenceC1 is an(n + 1, k) linear code.

(b) Note that the last row ofH1 is an all-one vector. The inner product of a vector with odd

weight and the all-one vector is′′1′′. Hence, for any odd weight vectorv,

v ·HT1 6= 0

andv cannot be a code word inC1. Therefore,C1 consists of only even-weight code words.

(c) Let v be a code word inC. Thenv ·HT = 0. Extendv by adding a digitv∞ to its left.

8

This results in a vector ofn + 1 digits,

v1 = (v∞,v) = (v∞, v0, v1, · · · , vn−1).

Forv1 to be a vector inC1, we must require that

v1HT1 = 0.

First we note that the inner product ofv1 with any of the firstn−k rows ofH1 is 0. The inner

product ofv1 with the last row ofH1 is

v∞ + v0 + v1 + · · ·+ vn−1.

For this sum to be zero, we must require thatv∞ = 1 if the vectorv has odd weight and

v∞ = 0 if the vectorv has even weight. Therefore, any vectorv1 formed as above is a code

word inC1, there are2k such code words. The dimension ofC1 is k, these2k code words are

all the code words ofC1.

3.5 Let Ce be the set of code words inC with even weight and letCo be the set of code words in

C with odd weight. Letx be any odd-weight code vector fromCo. Addingx to each vector in

Co, we obtain a set ofC ′e of even weight vector. The number of vectors inC ′

e is equal to the

number of vectors inCo, i.e. |C ′e| = |Co|. Also C ′

e ⊆ Ce. Thus,

|Co| ≤ |Ce| (1)

Now addingx to each vector inCe, we obtain a setC ′o of odd weight code words. The number

of vectors inC ′o is equal to the number of vectors inCe and

C ′o ⊆ Co

Hence

|Ce| ≤ |Co| (2)

From (1) and (2), we conclude that|Co| = |Ce|.

9

3.6 (a) From the given condition onG, we see that, for any digit position, there is a row inG

with a nonzero component at that position. This row is a code word inC. Hence in the code

array, each column contains at least one nonzero entry. Therefore no column in the code array

contains only zeros.

(b) Consider the-th column of the code array. From part (a) we see that this column contains

at least one′′1′′. Let S0 be the code words with a′′0′′ at the`-th position andS1 be the

codewords with a′′1′′ at the`-th position. Letx be a code word fromS1. Addingx to each

vector inS0, we obtain a setS ′1 of code words with a′′1′′ at the`-th position. Clearly,

|S ′1| = |S0| (1)

and

S ′1 ⊆ S1. (2)

Adding x to each vector inS1, we obtain a set ofS ′0 of code words with a′′0′′ at the`-th

location. We see that

|S ′0| = |S1| (3)

and

S ′0 ⊆ S0. (4)

From (1) and (2), we obtain

|S0| ≤ |S1|. (5)

From (3) and (4) ,we obtain

|S1| ≤ |S0|. (6)

From (5) and (6) we have|S0| = |S1|. This implies that the-th column of the code array

consists2k−1 zeros and2k−1 ones.

(c) Let S0 be the set of code words with a′′0′′ at the`-th position. From part (b), we see that

S0 consists of2k−1 code words. Letx andy be any two code words inS0. The sumx + y

also has a zero at the-th location and hence is code word inS0. ThereforeS0 is a subspace

of the vector space of alln-tuples over GF(2). SinceS0 is a subset ofC, it is a subspace ofC.

The dimension ofS0 is k − 1.

10

3.7 Let x, y andz be any threen-tuples over GF(2). Note that

d(x,y) = w(x + y),

d(y, z) = w(y + z),

d(x, z) = w(x + z).

It is easy to see that

w(u) + w(v) ≥ w(u + v). (1)

Let u = x + y andv = y + z. It follows from (1) that

w(x + y) + w(y + z) ≥ w(x + y + y + z) = w(x + z).

From the above inequality, we have

d(x,y) + d(y, z) ≥ d(x, z).

3.8 From the given condition, we see thatλ < bdmin−12

c. It follows from the theorem 3.5 that all

the error patterns ofλ or fewer errors can be used as coset leaders in a standard array. Hence,

they are correctable. In order to show that any error pattern of` or fewer errors is detectable,

we need to show that no error patternx of ` or fewer errors can be in the same coset as an

error patterny of λ or fewer errors. Suppose thatx andy are in the same coset. Thenx + y

is a nonzero code word. The weight of this code word is

w(x + y) ≤ w(x) + w(y) ≤ ` + λ < dmin.

This is impossible since the minimum weight of the code isdmin. Hencex andy are in

different cosets. As a result, whenx occurs, it will not be mistaken asy. Thereforex is

detectable.

3.11 In a systematic linear code, every nonzero code vector has at least one nonzero component in

its information section (i.e. the rightmostk positions). Hence a nonzero vector that consists of

only zeros in its rightmostk position can not be a code word in any of the systematic code inΓ.

11

Now consider a nonzero vectorv = (v0, v1, · · · , vn−1) with at least one nonzero component

in its k rightmost positions,sayvn−k+i = 1 for 0 ≤ i < k. Consider a matrix of the following

form which hasv as itsi-th row:

p00 p01 · · · p0,n−k−1 1 0 0 0 · · · 0

p10 p11 · · · p1,n−k−1 0 1 0 0 · · · 0

......

v0 v1 · · · vn−k−1 vn−k vn−k+1 · · · · · vn−1

pi+1,0 pi+1,1 · · · pi+1,n−k−1 0 0 · · 1 · · 0

......

pk−1,0 pk−1,1 · · · pk−1,n−k−1 0 0 0 0 · · · 1

By elementary row operations, we can putG into systematic formG1. The code generated

by G1 containsv as a code word. Since eachpij has 2 choices,0 or 1, there are2(k−1)(n−k)

matricesG with v as thei-th row. Each can be put into systematic formG1 and eachG1

generates a systematic code containingv as a code word. Hencev is contained in2(k−1)(n−k)

codes inΓ.

3.13 The generator matrix of the code is

G = [P1 Ik P2 Ik]

= [G1 G2]

Hence a nonzero codeword inC is simply a cascade of a nonzero codewordv1 in C1 and a

nonzero codewordv2 in C2, i.e.,

(v1,v2).

Sincew(v1) ≥ d1 andw(v2) ≥ d2, hencew[(v1,v2)] ≥ d1 + d2.

3.15 It follows from Theorem 3.5 that all the vectors of weightt or less can be used as coset leaders.

There are (n

0

)+

(n

1

)+ · · ·+

(n

t

)

12

such vectors. Since there are2n−k cosets, we must have

2n−k ≥(

n

0

)+

(n

1

)+ · · ·+

(n

t

).

Taking logarithm on both sides of the above inequality, we obtain the Hamming bound ont,

n− k ≥ log2{1 +

(n

1

)+ · · ·+

(n

t

)}.

3.16 Arrange the2k code words as a2k × n array. From problem 6(b), each column of this code

array contains2k−1 zeros and2k−1 ones. Thus the total number of ones in the array isn ·2k−1.

Note that each nonzero code word has weight (ones) at leastdmin. Hence

(2k − 1) · dmin ≤ n · 2k−1

This implies that

dmin ≤ n · 2k−1

2k − 1.

3.17 The number of nonzero vectors of lengthn and weightd− 1 or less is

d−1∑i=1

(n

i

)

From the result of problem 3.11, each of these vectors is contained in at most2(k−1)(n−k) linear

systematic codes. Therefore there are at most

M = 2(k−1)(n−k)

d−1∑i=1

(n

i

)

linear systematic codes contain nonzero codewords of weightd− 1 or less. The total number

of linear systematic codes is

N = 2(k(n−k)

If M < N , there exists at least one code with minimum weight at leastd. M < N implies

13

that

2(k−1)(n−k)

d−1∑i=1

(n

i

)< 2k(n−k)

d−1∑i=1

(n

i

)< 2(n−k).

3.18 Let dmin be the smallest positive integer such that

dmin−1∑i=1

(n

i

)< 2(n−k) ≤

dmin∑i=1

(n

i

)

From problem 3.17, the first inequality garantees the existence of a systematic linear code

with minimum distancedmin.

14

Chapter 4

4.1 A parity-check matrix for the(15, 11) Hamming code is

H =

1 0 0 0 1 0 0 1 1 0 1 0 1 1 1

0 1 0 0 1 1 0 0 0 1 1 1 0 1 1

0 0 1 0 0 1 1 0 1 0 1 1 1 0 1

0 0 0 1 0 0 1 1 0 1 0 1 1 1 1

Let r = (r0, r1, . . . , r14) be the received vector. The syndrome ofr is (s0, s1, s2, s3) with

s0 = r0 + r4 + r7 + r8 + r10 + r12 + r13 + r14,

s1 = r1 + r4 + r5 + r9 + r10 + r11 + r13 + r14,

s2 = r2 + r5 + r6 + r8 + r10 + r11 + r12 + r14,

s3 = r3 + r6 + r7 + r9 + r11 + r12 + r13 + r14.

Set up the decoding table as Table 4.1. From the decoding table, we find that

e0 = s0s1s2s3, e1 = s0s1s2s3, e2 = s0s1s2s3,

e3 = s0s1s2s3, e4 = s0s1s2s3, e5 = s0s1s2s3,

. . . , e13 = s0s1s2s3, e14 = s0s1s2s3.

15

Table 4.1: Decoding Tables0 s1 s2 s3 e0 e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12 e13 e14

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 00 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 00 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 01 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 01 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 00 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 01 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 00 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 01 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 01 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 01 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

r 0

r 1

Buffer Register r 14

+

...

s 0

+

s 1

+

s 2

+

s 3

... s 0 s 1 s 2 s 3

s 0 s 1 s 2 s 3

s 0 s 1 s 2 s 3

s 0 s 1 s 2 s 3

+ e 0

r 0 +

e 1

r 1 +

e 2

r 2 +

e 14

r 14

Decoded bits

... ... ...

16

4.3 From (4.3), the probability of an undetected error for a Hamming code is

Pu(E) = 2−m{1 + (2m − 1)(1− 2p)2m−1} − (1− p)2m−1

= 2−m + (1− 2−m)(1− 2p)2m−1 − (1− p)2m−1. (1)

Note that

(1− p)2 ≥ 1− 2p, (2)

and

(1− 2−m) ≥ 0. (3)

Using (2) and (3) in (1), we obtain the following inequality:

Pu(E) ≤ 2−m + (1− 2−m)[(1− p)2

]2m−1 − (1− p)2m−1

= 2−m + (1− p)2m−1{(1− 2−m)(1− p)− 1}

= 2−m − (1− p)2m−1{(1− (1− p)(1− 2−m)}. (4)

Note that0 ≤ 1− p ≤ 1 and0 ≤ 1− 2−m < 1. Clearly0 ≤ (1− p) · (1− 2−m) < 1, and

1− (1− p) · (1− 2−m) ≥ 0. (5)

Since(1− p)2m−1 ≥ 0, it follows from (4) and (5) thatPu(E) ≤ 2−m.

4.6 The generator matrix for the 1st-order RM code RM(1,3) is

G =

v0

v3

v2

v1

=

1 1 1 1 1 1 1 1

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

(1)

The code generated by this matrix is a(8, 4) code with minimum distance 4. Let(a0, a3, a2, a1)

17

be the message to be encoded. Then its corresponding codeword is

b = (b0, b1, b2, b3, b4, b5, b6, b7)

= a0v0 + a3v3 + a2v2 + a1v1. (2)

From (1) and (2), we find that

b0 = a0, b1 = a0 + a1, b2 = a0 + a2, b3 = a0 + a2 + a1,

b4 = a0 + a3, b5 = a0 + a3 + a1, b6 = a0 + a3 + a2,

b7 = a0 + a3 + a2 + a1. (3)

From (3), we find that

a1 = b0 + b1 = b2 + b3 = b4 + b5 = b6 + b7,

a2 = b0 + b2 = b1 + b3 = b4 + b6 = b5 + b7,

a3 = b0 + b4 = b1 + b5 = b2 + b6 = b3 + b7,

a0 = b0 = b1 + a1 = b2 + a2 = b3 + a2 + a1 = b4 + a3

= b5 + a3 + a1 = b6 + a3 + a2 = b7 + a3 + a2 + a1.

Let r = (r0, r1, r2, r3, r4, r5, r6, r7) be the received vector. The check-sum for decodinga1, a2

anda3 are:

(a1) (a2) (a3)

A(0)1 = r0 + r1 B

(0)1 = r0 + r2 C

(0)1 = r0 + r4

A(0)2 = r2 + r3 B

(0)2 = r1 + r3 C

(0)2 = r1 + r5

A(0)3 = r4 + r5 B

(0)3 = r4 + r6 C

(0)3 = r2 + r6

A(0)4 = r6 + r7 B

(0)4 = r5 + r7 C

(0)4 = r3 + r7

18

After decodinga1, a2 anda3, we form

r(1) = r− a1v1 − a2v2 − a3v3

= (r(1)0 , r

(1)1 , r

(1)2 , r

(1)3 , r

(1)4 , r

(1)5 , r

(1)6 , r

(1)7 ).

Thena0 is equal to the value taken by the majority of the bits inr(1).

For decoding(01000101), the four check-sum for decodinga1, a2 anda3 are:

(1) A(0)1 = 1, A

(0)2 = 0, A

(0)3 = 1, A

(0)4 = 1;

(2) B(0)1 = 0, B

(0)2 = 1, B

(0)3 = 0, B

(0)4 = 0;

(3) C(0)1 = 0, C

(0)2 = 0, C

(0)3 = 0, C

(0)4 = 1.

Based on these check-sums,a1, a2 anda3 are decoded as 1, 0 and 0, respectively.

To decodea0, we form

r(1) = (01000101)− a1v1 − a2v2 − a3v3

= (00010000).

From the bits ofr(1), we decodea0 to 0. Therefore, the decoded message is(0001).

4.14

RM(1, 3) = { 0, 1, X1, X2, X3, 1 + X1, 1 + X2,

1 + X3, X1 + X2, X1 + X3, X2 + X3,

1 + X1 + X2, 1 + X1 + X3, 1 + X2 + X3,

X1 + X2 + X3, 1 + X1 + X2 + X3}.

4.15 The RM(r,m− 1) and RM(r − 1,m− 1) codes are given as follows (from (4.38)):

RM(r,m− 1) = {v(f) : f(X1, X2, . . . , Xm−1) ∈ P(r,m− 1)},RM(r − 1,m− 1) = {v(g) : g(X1, X2, . . . , Xm−1) ∈ P(r − 1,m− 1)}.

Then

19

RM(r,m)={v(h) : h = f(X1, X2, . . . , Xm−1) + Xmg(X1, X2, . . . , Xm−1)

with f ∈ P(r,m− 1) andg ∈ P(r − 1,m− 1)}.

20

Chapter 5

5.6 (a) A polynomial over GF(2) with odd number of terms is not divisible byX + 1, hence it

can not be divisible byg(X) if g(X) has(X + 1) as a factor. Therefore, the code contains no

code vectors of odd weight.

(b) The polynomialXn + 1 can be factored as follows:

Xn + 1 = (X + 1)(Xn−1 + Xn−2 + · · ·+ X + 1)

Sinceg(X) dividesXn +1 and sinceg(X) does not haveX +1 as a factor,g(X) must divide

the polynomialXn−1 + Xn−2 + · · · + X + 1. Therefore1 + X + · · · + Xn−2 + Xn−1 is a

code polynomial, the corresponding code vector consists of all1′s.

(c) First, we note that noX i is divisible byg(X). Hence, no code word with weight one.

Now, suppose that there is a code wordv(X) of weight 2. This code word must be of the

form,

v(X) = X i + Xj

with 0 ≤ i < j < n. Putv(X) into the following form:

v(X) = X i(1 + Xj−i).

Note thatg(X) andX i are relatively prime. Sincev(X) is a code word, it must be divisible

byg(X). Sinceg(X) andX i are relatively prime,g(X) must divide the polynomialXj−i+1.

However,j − i < n. This contradicts the fact thatn is the smallest integer such thatg(X)

dividesXn + 1. Hence our hypothesis that there exists a code vector of weight 2 is invalid.

Therefore, the code has a minimum weight at least 3.

5.7 (a) Note thatXn + 1 = g(X)h(X). Then

Xn(X−n + 1) = Xng(X−1)h(X−1)

21

1 + Xn =[Xn−kg(X−1)

] [Xkh(X−1)

]

= g∗(X)h∗(X).

whereh∗(X) is the reciprocal ofh(X). We see thatg∗(X) is factor ofXn + 1. Therefore,

g∗(X) generates an(n, k) cyclic code.

(b) Let C andC∗ be two(n, k) cyclic codes generated byg(X) andg∗(X) respectively. Let

v(X) = v0 + v1X + · · · + vn−1Xn−1 be a code polynomial inC. Thenv(X) must be a

multiple ofg(X), i.e.,

v(X) = a(X)g(X).

ReplacingX by X−1 and multiplying both sides of above equality byXn−1, we obtain

Xn−1v(X−1) =[Xk−1a(X−1)

] [Xn−kg(X−1)

]

Note thatXn−1v(X−1), Xk−1a(X−1) andXn−kg(X−1) are simply the reciprocals ofv(X),

a(X) andg(X) respectively. Thus,

v∗(X) = a∗(X)g∗(X). (1)

From (1), we see that the reciprocalv∗(X) of a code polynomial inC is a code polynomial in

C∗. Similarly, we can show the reciprocal of a code polynomial inC∗ is a code polynomial in

C. Sincev∗(X) andv(X) have the same weight,C∗ andC have the same weight distribution.

5.8 Let C1 be the cyclic code generated by(X + 1)g(X). We know thatC1 is a subcode ofC

andC1 consists all the even-weight code vectors ofC as all its code vectors. Thus the weight

enumeratorA1(z) of C1 should consists of only the even-power terms ofA(z) =∑n

i=0 Aizi.

Hence

A1(z) =

bn/2c∑j=0

A2jz2j (1)

Consider the sum

A(z) + A(−z) =n∑

i=0

Aizi +

n∑i=0

Ai(−z)i

22

=n∑

i=0

Ai

[zi + (−z)i

].

We see thatzi + (−z)i = 0 if i is odd and thatzi + (−z)i = 2zi if i is even. Hence

A(z) + A(−z) =

bn/2c∑j=0

2A2jz2j (2)

From (1) and (2), we obtain

A1(z) = 1/2 [A(z) + A(−z)] .

5.10 Let e1(X) = X i + X i+1 ande2(X) = Xj + Xj+1 be two different double-adjacent-error

patterns such thati < j. Suppose thate1(X) ande2(X) are in the same coset. Thene1(X) +

e2(X) should be a code polynomial and is divisible byg(X) = (X + 1)p(X). Note that

e1(X) + e2(X) = X i(X + 1) + Xj(X + 1)

= (X + 1)X i(Xj−i + 1)

Sinceg(X) dividese1(X) + e2(X), p(X) should divideX i(Xj−i + 1). Howeverp(X) and

X i are relatively prime. Thereforep(X) must divideXj−i + 1. This is not possible since

j − i < 2m − 1 andp(X) is a primitive polynomial of degreem (the smallest integern such

thatp(X) dividesXn + 1 is 2m − 1). Thuse1(X) + e2(X) can not be in the same coset.

5.12 Note thate(i)(X) is the remainder resulting from dividingX ie(X) by Xn + 1. Thus

X ie(X) = a(X)(Xn + 1) + e(i)(X) (1)

Note thatg(X) dividesXn + 1, andg(X) andX i are relatively prime. From (1), we see that

if e(X) is not divisible byg(X), thene(i)(X) is not divisible byg(X). Therefore, ife(X) is

detectable,e(i)(X) is also detectable.

23

5.14 Suppose that does not dividen. Then

n = k · ` + r, 0 < r < `.

Note that

v(n)(X) = v(k·`+r)(X) = v(X) (1)

Sincev(`)(X) = v(X),

v(k·`)(X) = v(X) (2)

From (1) and (2), we have

v(r)(X) = v(X).

This is not possible since0 < r < ` and` is the smallest positive integer such thatv(`)(X) =

v(X). Therefore, our hypothesis that` does not dividen is invalid, hence must dividen.

5.17 Let n be the order ofβ. Thenβn = 1, andβ is a root ofXn + 1. It follows from Theorem

2.14 thatφ(X) is a factor ofXn + 1. Henceφ(X) generates a cyclic code of lengthn.

5.18 Let n1 be the order ofβ1 andn2 be the order ofβ2. Let n be the least common multiple ofn1

andn2, i.e. n = LCM(n1, n2). ConsiderXn + 1. Clearly,β1 andβ2 are roots ofXn + 1.

Sinceφ1(X) andφ2(X) are factors ofXn + 1. Sinceφ1(X) andφ2(X) are relatively prime,

g(X) = φ1(X) · φ2(X) dividesXn + 1. Henceg(X) = φ1(X) · φ2(X) generates a cyclic

code of lengthn = LCM(n1, n2).

5.19 Since every code polynomialv(X) is a multiple of the generator polynomialp(X), every root

of p(X) is a root ofv(X). Thusv(X) hasα and its conjugates as roots. Supposev(X) is

a binary polynomial of degree2m − 2 or less that hasα as a root. It follows from Theorem

2.14 thatv(X) is divisible by the minimal polynomialp(X) of α. Hencev(X) is a code

polynomial in the Hamming code generated byp(X).

5.20 Let v(X) be a code polynomial in bothC1 andC2. Thenv(X) is divisible by bothg1(X) and

g2(X). Hencev(X) is divisible by the least common multipleg(X) of g1(X) andg2(X),

i.e. v(X) is a multiple ofg(X) = LCM(g1(X),g2(X)). Conversely, any polynomial of

degreen − 1 or less that is a multiple ofg(X) is divisible byg1(X) andg2(X). Hence

v(X) is in bothC1 andC2. Also we note thatg(X) is a factor ofXn + 1. Thus the code

24

polynomials common toC1 andC2 form a cyclic code of lengthn whose generator polynomial

is g(X) = LCM(g1(X),g2(X)). The codeC3 generated byg(X) has minimum distance

d3 ≥ max(d1, d2).

5.21 See Problem 4.3.

5.22 (a) First, we note thatX2m−1 + 1 = p∗(X)h∗(X). Since the roots ofX2m−1 + 1 are the

2m − 1 nonzero elements in GF(2m) which are all distinct,p∗(X) andh∗(X) are relatively

prime. Since every code polynomialv(X) in Cd is a polynomial of degree2m − 2 or less,

v(X) can not be divisible byp(X) (otherwisev(X) is divisible byp∗(X)h∗(X) = X2m−1+1

and has degree at least2m − 1). Suppose thatv(i)(X) = v(X). It follows from (5.1) that

X iv(X) = a(X)(X2m−1 + 1) + v(i)(X)

= a(X)(X2m−1 + 1) + v(X)

Rearranging the above equality, we have

(X i + 1)v(X) = a(X)(X2m−1 + 1).

Sincep(X) dividesX2m−1 + 1, it must divide(X i + 1)v(X). Howeverp(X) andv(X) are

relatively prime. Hencep(X) dividesX i + 1. This is not possible since0 < i < 2m − 1 and

p(X) is a primitive polynomial(the smallest positive integern such thatp(X) dividesXn +1

is n = 2m−1). Therefore our hypothesis that, for0 < i < 2m−1, v(i)(X) = v(X) is invalid,

andv(i)(X) 6= v(X).

(b) From part (a), a code polynomialv(X) and its2m − 2 cyclic shifts form all the2m − 1

nonzero code polynomials inCd. These2m − 1 nonzero code polynomial have the same

weight, sayw. The total number of nonzero components in the code words ofCd isw·(2m−1).

Now we arrange the2m code words inCd as an2m× (2m− 1) array. It follows from Problem

3.6(b) that every column in this array has exactly2m−1 nonzero components. Thus the total

nonzero components in the array is2m−1 · (2m − 1). Equatingw.(2m − 1) to 2m−1 · (2m − 1),

we have

w = 2m−1.

25

5.25 (a) Any error pattern of double errors must be of the form,

e(X) = X i + Xj

wherej > i. If the two errors are not confined ton− k = 10 consecutive positions, we must

have

j − i + 1 > 10,

15− (j − i) + 1 > 10.

Simplifying the above inequalities, we obtain

j − i > 9

j − i < 6.

This is impossible. Therefore any double errors are confined to 10 consecutive positions and

can be trapped.

(b) An error pattern of triple errors must be of the form,

e(X) = X i + Xj + Xk,

where0 ≤ i < j < k ≤ 14. If these three errors can not be trapped, we must have

k − i > 9

j − i < 6

k − j < 6.

If we fix i, the only solutions forj andk arej = 5 + i andk = 10 + i. Hence, for three errors

not confined to 10 consecutive positions, the error pattern must be of the following form

e(X) = X i + X5+i + X10+i

26

for 0 ≤ i < 5. Therefore, only 5 error patterns of triple errors can not be trapped.

5.26 (b) Consider a double-error pattern,

e(X) = X i + Xj

where0 ≤ i < j < 23. If these two errors are not confined to 11 consecutive positions, we

must have

j − i + 1 > 11

23− (j − i− 1) > 11

From the above inequalities, we obtain

10 < j − i < 13

For a fixedi, j has two possible solutions,j = 11+i andj = 12+i. Hence, for a double-error

pattern that can not be trapped, it must be either of the following two forms:

e1(X) = X i + X11+i,

e1(X) = X i + X12+i.

There are a total of 23 error patterns of double errors that can not be trapped.

5.27 The coset leader weight distribution is

α0 = 1, α1 =

(23

1

), α2 =

(23

2

), α3 =

(23

3

)

α4 = α5 = · · · = α23 = 0

The probability of a correct decoding is

P (C) = (1− p)23 +

(23

1

)p(1− p)22 +

(23

2

)p2(1− p)21

27

+

(23

3

)p3(1− p)20.

The probability of a decoding error is

P (E) = 1− P (C).

5.29(a) Consider two single-error patterns,e1(X) = X i ande2(X) = Xj, wherej > i. Suppose

that these two error patterns are in the same coset. ThenX i + Xj must be divisible by

g(X) = (X3 + 1)p(X). This implies thatXj−i + 1 must be divisible byp(X). This is

impossible sincej − i < n andn is the smallest positive integer such thatp(X) divides

Xn + 1. Therefore no two single-error patterns can be in the same coset. Consequently, all

single-error patterns can be used as coset leaders.

Now consider a single-error patterne1(X) = X i and a double-adjacent-error patterne2(X) =

Xj + Xj+1, wherej > i. Suppose thate1(X) and e2(X) are in the same coset. Then

X i +Xj +Xj+1 must be divisible byg(X) = (X3 +1)p(X). This is not possible sinceg(X)

hasX + 1 as a factor, howeverX i + Xj + Xj+1 does not haveX + 1 as a factor. Hence no

single-error pattern and a double-adjacent-error pattern can be in the same coset.

Consider two double-adjacent-error patterns,X i +X i+1 andXj +Xj+1 wherej > i. Suppose

that these two error patterns are in the same cosets. ThenX i + X i+1 + Xj + Xj+1 must be

divisible by(X3 + 1)p(X). Note that

X i + X i+1 + Xj + Xj+1 = X i(X + 1)(Xj−i + 1).

We see that forX i(X + 1)(Xj−i + 1) to be divisible byp(X), Xj−i + 1 must be divisible

by p(X). This is again not possible sincej − i < n. Hence no two double-adjacent-error

patterns can be in the same coset.

Consider a single error patternX i and a triple-adjacent-error patternXj + Xj+1 + Xj+2. If

these two error patterns are in the same coset, thenX i +Xj +Xj+1 +Xj+2 must be divisible

by (X3 + 1)p(X). But X i + Xj + Xj+1 + Xj+2 = X i + Xj(1 + X + X2) is not divisible by

X3 +1 = (X +1)(X2 +X +1). Therefore, no single-error pattern and a triple-adjacent-error

pattern can be in the same coset.

Now we consider a double-adjacent-error patternX i+X i+1 and a triple-adjacent-error pattern

28

Xj + Xj+1 + Xj+2. Suppose that these two error patterns are in the same coset. Then

X i + X i+1 + Xj + Xj+1 + Xj+2 = X i(X + 1) + Xj(X2 + X + 1)

must be divisible by(X3+1)p(X). This is not possible sinceX i +X i+1+Xj +Xj+1+Xj+2

does not haveX+1 as a factor butX3+1 hasX+1 as a factor. Hence a double-adjacent-error

pattern and a triple-adjacent-error pattern can not be in the same coset.

Consider two triple-adjacent-error patterns,X i + X i+1 + X i+2 andXj + Xj+1 + Xj+2. If

they are in the same coset, then their sum

X i(X2 + X + 1)(1 + Xj−i)

must be divisible by(X3 + 1)p(X), hence byp(X). Note that the degree ofp(X) is 3 or

greater. Hencep(X) and(X2 + X + 1) are relatively prime. As a result,p(X) must divide

Xj−i + 1. Again this is not possible. Hence no two triple-adjacent-error patterns can be in the

same coset.

Summarizing the above results, we see that all the single-, double-adjacent-, and triple-

adjacent-error patterns can be used as coset leaders.

29

Chapter 6

6.1 (a) The elementsβ, β2 andβ4 have the same minimal polynomialφ1(X). From table 2.9, we

find that

φ1(X) = 1 + X3 + X4

The minimal polynomial ofβ3 = α21 = α6 is

φ3(X) = 1 + X + X2 + X3 + X4.

Thus

g0(X) = LCM(φ1(X), φ2(X))

= (1 + X3 + X4)(1 + X + X2 + X3 + X4)

= 1 + X + X2 + X4 + X8.

(b)

H =

1 β β2 β3 β4 β5 β6 β7 β8 β9 β10 β11 β12 β13 β14

1 β3 β6 β9 β12 β15 β18 β21 β24 β27 β30 β33 β36 β39 β42

H =

1 1 1 0 1 0 1 1 0 0 1 0 0 0 1

0 1 0 0 0 1 1 1 1 0 1 0 1 1 0

0 0 0 1 1 1 1 0 1 0 1 1 0 0 1

0 1 1 1 1 0 1 0 1 1 0 0 1 0 0

1 0 1 0 0 1 0 1 0 0 1 0 1 0 0

0 0 1 0 1 0 0 1 0 1 0 0 1 0 1

0 1 1 0 0 0 1 1 0 0 0 1 1 0 0

0 1 1 1 1 0 1 1 1 1 0 1 1 1 1

.

30

(c) The reciprocal ofg(X) in Example 6.1 is

X8g(X−1) = X8(1 + X−4 + X−6 + X−7 + X−8

= X8 + X4 + X2 + X + 1 = g0(X)

6.2 The table forGF (s5) with p(X) = 1 + X2 + X5 is given in Table P.6.2(a). The minimal

polynomials of elements inGF (2m) are given in Table P.6.2(b). The generator polynomials

of all the binary BCH codes of length 31 are given in Table P.6.2(c)

Table P.6.2(a) Galois Field GF(25) with p(α) = 1 + α2 + α5 = 0

0 (0 0 0 0 0)

1 (1 0 0 0 0)

α (0 1 0 0 0)

α2 (0 0 1 0 0)

α3 (0 0 0 1 0)

α4 (0 0 0 0 1)

α5 = 1 + α2 (1 0 1 0 0)

31

Table P.6.2(a) Continued

α6 = α + α3 (0 1 0 1 0)

α7 = α2 + α4 (0 0 1 0 1)

α8 = 1 + α2 + α3 (1 0 1 1 0)

α9 = α + α3 + α4 (0 1 0 1 1)

α10 = 1 + α4 (1 0 0 0 1)

α11 = 1 + α + α2 (1 1 1 0 0)

α12 = α + α2 + α3 (0 1 1 1 0)

α13 = α2 + α3 + α4 (0 0 1 1 1)

α14 = 1 + α2 + α3 + α4 (1 0 1 1 1)

α15 = 1 + α + α2 + α3 + α4 (1 1 1 1 1)

α16 = 1 + α + α3 + α4 (1 1 0 1 1)

α17 = 1 + α + α4 (1 1 0 0 1)

α18 = 1 + α (1 1 0 0 0)

α19 = α + α2 (0 1 1 0 0)

α20 = α2 + α3 (0 0 1 1 0)

α21 = α3 + α4 (0 0 0 1 1)

α22 = 1 + α2 + α4 (1 0 1 0 1)

α23 = 1 + α + α2 + α3 (1 1 1 1 0)

α24 = α + α2 + α3 + α4 (0 1 1 1 1)

α25 = 1 + α3 + α4 (1 0 0 1 1)

α26 = 1 + α + α2 + α4 (1 1 1 0 1)

α27 = 1 + α + α3 (1 1 0 1 0)

α28 = α + α2 + α4 (0 1 1 0 1)

α29 = 1 + + α3 (1 0 0 1 0)

α30 = α + α4 (0 1 0 0 1)

32

Table P.6.2(b)

Conjugate Roots φi(X)

1

α, α2, α4, α8, α16

α3, α6, α12, α24, α17

α5, α10, α20, α9, α18

α7, α14, α28, α25, α19

α11, α22, α13, α26, α21

α15, α30, α29, α27, α23

1 + X

1 + X2 + X5

1 + X2 + X3 + X4 + X5

1 + X + X2 + X4 + X5

1 + X + X2 + X3 + X5

1 + X + X3 + X4 + X5

1 + X3 + X5

Table P.6.2(c)

n k t g(X)

31 26 1 g1(X) = 1 + X2 + X5

21 2 g2(X) = φ1(X)φ3(X)

16 3 g3(X) = φ1(X)φ3(X)φ5(X)

11 5 g4(X) = φ1(X)φ3(X)φ5(X)φ7(X)

6 7 g5(X) = φ1(X)φ3(X)φ5(X)φ7(X)φ11(X)

6.3 (a) Use the table forGF (25) constructed in Problem 6.2. The syndrome components of

r1(X) = X7 + X30 are:

S1 = r1(α) = α7 + α30 = α19

S2 = r1(α2) = α14 + α29 = α7

S3 = r1(α3) = α21 + α28 = α12

33

S4 = r1(α4) = α28 + α27 = α14

The iterative procedure for finding the error location polynomial is shown in Table P.6.3(a)

Table P.6.3(a)

µ σ(µ)(X) dµ `µ 2µ− `µ

-1/2 1 1 0 -1

0 1 α19 0 0

1 1 + α19X α25 1 1(ρ = −1/2)

2 1 + α19X + α6X2 – 2 2(ρ = 0)

Henceσ(X) = 1 + α19X + α6X2. Substituting the nonzero elements ofGF (25) into σ(X),

we find thatσ(X) hasα andα24 as roots. Hence the error location numbers areα−1 = α30

andα−24 = α7. As a result, the error polynomial is

e(X) = X7 + X30.

The decoder decodesr1(X) into r1(X) + e(X) = 0.

(b) Now we consider the decoding ofr2(X) = 1 + X17 + X28. The syndrome components of

r2(X) are:

S1 = r2(α) = α2,

S2 = S21 = α4,

S4 = S22 = α8,

S3 = r2(α3) = α21.

The error location polynomialσ(X) is found by filling Table P.6.3(b):

34

Table P.6.3(b)

µ σ(µ)(X) dµ `µ 2µ− `µ

-1/2 1 1 0 -1

0 1 α2 0 0

1 1 + α2X α30 1 1(ρ = −1/2)

2 1 + α2X + α28X2 – 2 2(ρ = 0)

The estimated error location polynomial is

σ(X) = 1 + α2X + α28X2

This polynomial does not have roots inGF (25), and hencer2(X) cannot be decoded and must

contain more than two errors.

6.4 Let n = (2t + 1)λ. Then

(Xn + 1) = (Xλ + 1)(X2tλ + X(2t−1)λ + · · ·+ Xλ + 1

The roots ofXλ + 1 are 1,α2t+1, α2(2t+1), · · · , α(λ−1)(2t+1). Hence,α, α2, · · · , α2t are roots

of the polynomial

u(X) = 1 + Xλ + X2λ + · · ·+ X(2t−1)λ + X2tλ.

This implies thatu(X) is code polynomial which has weight2t + 1. Thus the code has

minimum distance exactly2t + 1.

6.5 Consider the Galois fieldGF (22m). Note that22m − 1 = (2m − 1) · (2m + 1). Let α be

a primitive element inGF (22m). Thenβ = α(2m−1) is an element of order2m + 1. The

elements1, β, β2, β2, β3, β4, · · · , β2m are all the roots ofX2m+1+1. Letψi(X) be the minimal

35

polynomial ofβi. Then at-error-correcting non-primitive BCH code of lengthn = 2m + 1 is

generated by

g(X) = LCM {ψ1(X), ψ2(X), · · · , ψ2t(X)} .

6.10 Use Tables 6.2 and 6.3. The minimal polynomial forβ2 = α6 andβ4 = α12 is

ψ2(X) = 1 + X + X2 + X4 + X6.

The minimal polynomial forβ3 = α9 is

ψ3(X) = 1 + X2 + X3.

The minimal polynomial forβ5 = α15 is

ψ5(X) = 1 + X2 + X4 + X5 + X6.

Hence

g(X) = ψ2(X)ψ3(X)ψ5(X)

The orders ofβ2, β3 andβ5 are 21,7 and 21 respectively. Thus the length is

n = LCM(21, 7, 21),

and the code is a double-error-correcting (21,6) BCH code.

6.11 (a) Letu(X) be a code polynomial andu∗(X) = Xn−1u(X−1) be the reciprocal ofu(X).

A cyclic code is said to be reversible ifu(X) is a code polynomial thenu∗(X) is also a code

polynomial. Consider

u∗(βi) = β(n−1)iu(β−i)

Sinceu(β−i) = 0 for−t ≤ i ≤ t, we see thatu∗(βi) hasβ−t, · · · , β−1, β0, β1, · · · , βt as roots

36

and is a multiple of the generator polynomialg(X). Thereforeu∗(X) is a code polynomial.

(b) If t is odd,t+1 is even. Henceβt+1 is the conjugate ofβ(t+1)/2 andβ−(t+1) is the conjugate

of β−(t+1)/2. Thusβt+1 andβ−(t+1) are also roots of the generator polynomial. It follows from

the BCH bound that the code has minimum distance2t + 4 (Since the generator polynomial

has(2t + 3 consecutive powers ofβ as roots).

37

Chapter 7

7.2 The generator polynomial of the double-error-correcting RS code over GF(25) is

g(X) = (X + α)(X + α2)(X + α3)(X + α4)

= α10 + α29X + α19X2 + α24X3 + X4.

The generator polynomial of the triple-error-correcting RS code over GF(25) is

g(X) = (X + α)(X + α2)(X + α3)(X + α4)(X + α5)(X + α6)

= α21 + α24X + α16X2 + α24X3 + α9X4 + α10X5 + X6.

7.4 The syndrome components of the received polynomial are:

S1 = r(α) = α7 + α2 + α = α13,

S2 = r(α2) = α10 + α10 + α14 = α14,

S3 = r(α3) = α13 + α3 + α12 = α9,

S4 = r(α4) = α + α11 + α10 = α7,

S5 = r(α5) = α4 + α4 + α8 = α8,

S6 = r(α6) = α7 + α12 + α6 = α3.

The iterative procedure for finding the error location polynomial is shown in Table P.7.4. The

error location polynomial is

σ(X) = 1 + α9X3.

The roots of this polynomial areα2, α7, andα12. Hence the error location numbers areα3, α8,

andα13.

From the syndrome components of the received polynomial and the coefficients of the error

1

Table P.7.4

µ σµ(X) dµ lµ µ− lµ

−1 1 1 0 −1

0 1 α13 0 0

1 1 + α13X α10 1 0 (takeρ = −1)

2 1 + αX α7 1 1 (takeρ = 0)

3 1 + α13X + α10X2 α9 2 1 (takeρ = 1)

4 1 + α14X + α12X2 α8 2 2 (takeρ = 2)

5 1 + α9X3 0 3 2 (takeρ = 3)

6 1 + α9X3 − − −

location polynomial, we find the error value evaluator,

Z0(X) = S1 + (S2 + σ1S1)X + (S3 + σ1S2 + σ2S1)X2

= α13 + (α14 + 0α13)X + (α9 + 0α14 + 0α13)X2

= α13 + α14X + α9X2.

The error values at the positionsX3, X8, andX13 are:

e3 =−Z0(α

−3)

σ′(α−3)=

α13 + α11 + α3

α3(1 + α8α−3)(1 + α13α−3)=

α7

α3= α4,

e8 =−Z0(α

−8)

σ′(α−8)=

α13 + α6 + α8

α8(1 + α3α−8)(1 + α13α−8)=

α2

α8= α9,

e13 =−Z0(α

−13)

σ′(α−13)=

α13 + α + α13

α13(1 + α3α−13)(1 + α8α−13)=

α

α13= α3.

Consequently, the error pattern is

e(X) = α4X3 + α9X8 + α3X13.

and the decoded codeword is the all-zero codeword.

2

7.5 The syndrome polynomial is

S(X) = α13 + α14X + α9X2 + α7X3 + α8X4 + α3X5

Table P.7.5 displays the steps of Euclidean algorithm for finding the error location and error

value polynomials.

Table P.7.5

i Z(i)0 (X) qi(X) σi(X)

−1 X6 − 0

0 α13 + α14X + α9X2 + α7X3 + α8X4 + α3X5 − 1

1 1 + α8X + α5X3 + α2X4 α2 + α12X α2 + α12X

2 α + α13X + α12X3 α12 + αX α3 + αX + α13X2

3 α7 + α8X + α3X2 α8 + α5X α9 + α3X3

The error location and error value polynomials are:

σ(X) = α9 + α3X3 = α9(1 + α9X3)

Z0(X) = α7 + α8X + α3X2 = α9(α13 + α14X + α9X2)

From these polynomials, we find that the error location numbers areα3, α8, andα13, and error

values are

e3 =−Z0(α

−3)

σ′(α−3)=

α7 + α5 + α12

α9α3(1 + α8α−3)(1 + α13α−3)=

α

α12= α4,

e8 =−Z0(α

−8)

σ′(α−8)=

α7 + 1 + α2

α9α8(1 + α3α−8)(1 + α13α−8)=

α11

α2= α9,

e13 =−Z0(α

−13)

σ′(α−13)=

α7 + α10 + α7

α9α13(1 + α3α−13)(1 + α8α−13)=

α10

α7= α3.

3

Hence the error pattern is

e(X) = α4X3 + α9X8 + α3X13.

and the received polynomial is decoded into the all-zero codeword.

7.6 From the received polynomial,

r(X) = α2 + α21X12 + α7X20,

we compute the syndrome,

S1 = r(α1) = α2 + α33 + α27 = α27,

S2 = r(α2) = α2 + α45 + α47 = α,

S3 = r(α3) = α2 + α57 + α67 = α28,

S4 = r(α4) = α2 + α69 + α87 = α29,

S5 = r(α5) = α2 + α81 + α107 = α15,

S6 = r(α6) = α2 + α93 + α127 = α8.

Therefore, the syndrome polynomial is

S(X) = α27 + αX + α28X2 + α29X3 + α15X4 + α8X5

Using the Euclidean algorithm, we find

σ(X) = α23X3 + α9X + α22,

Z0(X) = α26X2 + α6X + α18,

as shown in the following table: The roots ofσ(X) are: 1 = α0, α11 andα19. From these

roots, we find the error location numbers:β1 = (α0)−1

= α0, β2 = (α11)−1 = α20, and

4

i Z(i)0 (X) qi(X) σi(X)

-1 X6 - 0

0 S(X) - 1

1 α5X4 + α9X3 + α22X2 + α11X + α26 α23X + α30 α23X + α30

2 α8X3 + α4X + α6 α3X + α5 α24X2 + α30X + α10

3 α26X2 + α6X + α18 α28X + α α23X3 + α9X + α22

β3 = (α19)−1 = α12. Hence the error pattern is

e(X) = e0 + e12X12 + e20X

20.

The error location polynomial and its derivative are:

σ(X) = α22(1 + X)(1 + α12X)(1 + α20X),

σ′(X) = α22(1 + α12X)(1 + α20X) + α3(1 + X)(1 + α20X) + α11(1 + X)(1 + α12X).

The error values at the 3 error locations are given by:

e0 =−Z0(α

0)

σ′(α0)=

α26 + α6 + α8

α22(1 + α12)(1 + α20)= α2,

e12 =−Z0(α

−12)

σ′(α−12)=

α2 + α25 + α18

α3(1 + α19)(1 + α8)= α21,

e20 =−Z0(α

−20)

σ′(α−20)=

α17 + α17 + α18

α11(1 + α11)(1 + α23)= α7.

Hence, the error pattern is

e(X) = α2 + α21X12 + α7X20

and the decoded codeword is

v(X) = r(X)− e(X) = 0.

5

7.9 Let g(X) be the generator polynomial of at-symbol correcting RS codeC over GF(q) with α,

α2, . . . , α2t as roots, whereα is a primitive element of GF(q). Sinceg(X) dividesXq−1 − 1,

then

Xq−1 − 1 = g(X)h(X).

The polynomialh(X) hasα2t+1, . . . , αq−1 as roots and is called the parity polynomial. The

dual codeCd of C is generated by the reciprocal ofh(X),

h∗(X) = Xq−1−2th(X−1).

We see thath∗(X) hasα−(2t+1) = αq−2t−2, α−(2t+2) = αq−2t−3, . . . , α−(q−2) = α, and

α−(q−1) = 1 as roots. Thush∗(X) has the following consecutive powers ofα as roots:

1, α, α2, . . . , αq−2t−2.

HenceCd is a(q − 1, 2t, q − 2t) RS code with minimum distanceq − 2t.

7.10 The generator polynomialgrs(X) of the RS codeC hasα, α2, . . . , αd−1 as roots. Note that

GF(2m) has GF(2) as a subfield. Consider those polynomialv(X) over GF(2) with degree

2m−2 or less that hasα, α2, . . . , αd−1 (also their conjugates) as roots. These polynomials over

GF(2) form a primitive BCH codeCbch with designed distanced. Since these polynomials are

also code polynomials in the RS codeCrs, henceCbch is a subcode ofCrs.

7.11 Supposec(X) =∑2m−2

i=0 ciXi is a minimum weight code polynomial in the(2m − 1, k) RS

codeC. The minimum weight is increased tod + 1 provided

c∞ = −c(1) = −2m−2∑i=0

ci 6= 0.

We know thatc(X) is divisible byg(X). Thusc(X) = a(X)g(X) with a(X) 6= 0. Consider

c(1) = a(1)g(1).

Since 1 is not a root ofg(X), g(1) 6= 0. If a(1) 6= 0, thenc∞ = −c(1) 6= 0 and the vector

(c∞, c0, c1, . . . , c2m−2) has weightd+1. Next we show thata(1) is not equal to 0. Ifa(1) = 0,

6

thena(X) hasX − 1 as a factor andc(X) is a multiple of(X − 1)g(X) and must have a

weight at leastd + 1. This contradicts to the hypothesis thatc(X) is a minimum weight code

polynomial. Consequently the extended RS code has a minimum distanced + 1.

7.12 To prove the minimum distance of the doubly extended RS code, we need to show that no2t

or fewer columns ofH1 sum to zero over GF(2m) and there are2t + 1 columns inH1 sum

to zero. Suppose there areδ columns inH1 sum to zero andδ ≤ 2t. There are 4 case to be

considered:

(1) All δ columns are from the same submatrixH.

(2) Theδ columns consist of the first column ofH1 andδ − 1 columns fromH.

(3) Theδ columns consist of the second column ofH1 andδ − 1 columns fromH.

(4) Theδ columns consist of the first two columns ofH1 andδ − 2 columns fromH.

The first case leads to aδ × δ Vandermonde determinant. The second and third cases lead to

a (δ − 1) × (δ − 1) Vandermonde determinant. The 4th case leads to a(δ − 2) × (δ − 2)

Vandermonde determinant. The derivations are exactly the same as we did in the book. Since

Vandermonde determinants are nonzero,δ columns ofH1 can not be sum to zero. Hence the

minimum distance of the extended RS code is at least2t + 1. However,H generates an RS

code with minimum distance exactly2t + 1. There are2t + 1 columns inH (they are also in

H1), which sum to zero. Therefore the minimum distance of the extended RS code is exactly

2t + 1.

7.13 Consider

v(X) =2m−2∑i=0

a(αi)X i =2m−2∑i=0

(k−1∑j=0

ajαij)X i

Let α be a primitive element in GF(2m). ReplacingX by αq, we have

v(αq) =2m−2∑i=0

k−1∑j=0

ajαijαiq

=k−1∑j=0

aj(2m−2∑i=0

αi(j+q)).

7

We factor1 + X2−1 as follows:

1 + X2m−1 = (1 + X)(1 + X + X2 + · · ·+ X2m−2)

Since the polynomial1 + X + X2 + · · · + X2m−2 hasα, α2, . . . , α2m−2 as roots, then for

1 ≤ l ≤ 2m − 2,2m−2∑i=0

αli = 1 + αl + α2l + · · ·+ α(2m−2)l = 0.

Therefore,

∑2m−2i=0 αi(j+q) = 0 when1 ≤ j + q ≤ 2m − 2.

This implies that

v(αq) = 0 for 0 ≤ j < k and1 ≤ q ≤ 2m − k − 1.

Hencev(X) hasα, α2, . . . , α2m−k−1 as roots. The set{v(X)} is a set of polynomial over

GF(2m) with 2m−k−1 consecutive powers ofα as roots and hence it forms a(2m−1, k, 2m−k) cyclic RS code over GF(2m).

8

Chapter 8

8.2 The order of the perfect difference set{0, 2, 3} is q = 2.

(a) The length of the coden = 22 + 2 + 1 = 7.

(b) Let z(X) = 1 + X2 + X3. Then the parity-check polynomial is

h(X) = GCD{1 + X2 + X3, X7 + 1} = 1 + X2 + X3.

(c) The generator polynomial is

g(X) =X7 + 1

h(X)= 1 + X2 + X3 + X4.

(d) Fromg(X), we find that the generator matrix in systematic form is

G =

1 0 1 1 1 0 0

1 1 1 0 0 1 0

0 1 1 1 0 0 1

.

The parity-check matrix in systematic form is

H =

1 0 0 0 1 1 0

0 1 0 0 0 1 1

0 0 1 0 1 1 1

0 0 0 1 1 0 1

.

The check-sums orthogonal on the highest order error digite6 are:

A1 = s0 + s2,

A2 = s1,

A3 = s3.

Based on the above check-sum, a type-1 decoder can be implemented.

8.4 (a) Since all the columns ofH are distinct and have odd weights, no two or three columns

can sum to zero, hence the minimum weight of the code is at least4. However, the first,

1

the second, the third and the6th columns sum to zero. Therefore the minimum weight,

hence the minimum distance, of the code is 4.

(b) The syndrome of the error vectore is

s = (s0, s1, s2, s3, s4) = eHT

with

s0 = e0 + e5 + e6 + e7 + e8 + e9 + e10,

s1 = e1 + e5 + e6 + e8,

s2 = e2 + e5 + e7 + e9,

s3 = e3 + e6 + e7 + e10,

s4 = e4 + e8 + e9 + e10.

(c) The check-sums orthogonal one10 are:

A1,10 = s0 + s1 + s2 = e0 + e1 + e2 + e5 + e10,

A2,10 = s3 = e3 + e6 + e7 + e10,

A3,10 = s4 = e4 + e8 + e9 + e10.

The check-sums orthogonal one9 are:

A1,9 = s0 + s1 + s3,

A2,9 = s2,

A3,9 = s4.

The check-sums orthogonal one8 are:

A1,8 = s0 + s2 + s3,

A2,8 = s1,

A3,8 = s4.

2

The check-sums orthogonal one7 are:

A1,7 = s0 + s1 + s4,

A2,7 = s2,

A3,7 = s3.

The check-sums orthogonal one6 are:

A1,6 = s0 + s2 + s4,

A2,6 = s1,

A3,6 = s3.

The check-sums orthogonal one5 are:

A1,5 = s0 + s3 + s4,

A2,5 = s1,

A3,5 = s2.

(d) Yes, the code is completely orthogonalizable, since there are 3 check-sums orthogonal

on each message bit and the minimum distance of the code is 4.

8.5 Form = 6, the binary radix-2 form of 43 is

43 = 1 + 2 + 23 + 25.

The nonzero-proper descendants of43 are:

1, 2, 8, 32, 3, 9, 33, 10, 34, 40, 11, 35, 41, 42.

8.6

α∞ α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 α12 α13 α14

u = ( 1 1 1 0 0 1 0 1 0 1 1 0 0 0 0 1 )

Applying the permutationZ = α3Y + α11 to the above vector, the component at the location

αi is permute to the locationα3αi+α11. For example, the1-component at the locationY = α8

is permuted to the locationα3α8 + α11 = α11 + α11 = α∞. Performing this permutation to

3

the each component of the above vector, we obtain the following vector

α∞ α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 α12 α13 α14

v = ( 1 1 0 1 0 0 1 0 0 1 1 0 1 0 1 0 )

8.7 ForJ = 9 andL = 7, X63 + 1 can be factor as follows:

X63 + 1 = (1 + X9)(1 + X9 + X18 + X27 + X36 + X45 + X54).

Thenπ(X) = 1+X9+X18+X27+X36+X45+X54. Letα be a primitive element in GF(26)

whose minimal polynomial isφ1(X) = 1+X +X6. Becauseα63 = 1, the polynomial1+X9

hasα0, α7, α14, α21, α28, α35, α42, α49, andα56 as all it roots. Therefore, the polynomialπ(X)

hasαh as a root whenh is not a multiple of7 and0 < h < 63. From the conditions (Theorem

8.2) on the roots ofH(X), we can findH(X) as:H(X) = LCM{minimal polynomialsφi(X)

of the roots ofH(X)}. As the result,

H(X) = φ1(X)φ3(X)φ5(X)φ9(X)φ11(X)φ13(X)φ27(X),

whereφ1 = 1 + X + X6, φ3 = 1 + X + X2 + X4 + X6, φ5 = 1 + X + X2 + X5 + X6,

φ9 = 1 + X2 + X3, φ11 = 1 + X2 + X3 + X5 + X6, φ13 = 1 + X + X3 + X4 + X6, and

φ27 = 1 + X + X3. Then, we can findG(X),

G(X) =X63 + 1

H(X)=

(1 + X9)π(X)

H(X)

= (1 + X9)(1 + X2 + X4 + X5 + X6)(1 + X + X4 + X5 + X6)(1 + X5 + X6).

For type-1 DTI code of length 63 andJ = 9, the generator polynomial is:

g1(X) =X27G(X−1)

1 + X=

(1 + X9)(1 + X + X2 + X4 + X6)(1 + X + X2 + X5 + X6)(1 + X + X6)1 + X

= (1 + X + X2 + X3 + X4 + X5 + X6 + X7 + X8)(1 + X + X2 + X4 + X6)(1 + X + X2 + X5 + X6)(1 + X + X6).

Represent the polynomialsπ(X), Xπ(X), X2π(X), X3π(X), X4π(X), X5π(X), X6π(X),

X7π(X), andX8π(X) by 63-tuple location vectors. Add an overall parity-check digit and

apply the affine permutation,Y = αX + α62, to each of these location vectors. Then, remove

the overall parity-check digits of all location vectors. By removing one vector with odd weight,

4

we can obtain the polynomials orthogonal on the digit positionX62. They are:

X11 + X16 + X18 + X24 + X48 + X58 + X59 + X62,

X1 + X7 + X31 + X41 + X42 + X45 + X57 + X62,

X23 + X33 + X34 + X37 + X49 + X54 + X56 + X62,

X2 + X14 + X19 + X21 + X27 + X51 + X61 + X62,

X0 + X3 + X15 + X20 + X22 + X28 + X52 + X62,

X9 + X10 + X13 + X25 + X30 + X32 + X38 + X62,

X4 + X6 + X12 + X36 + X46 + X47 + X50 + X62,

X5 + X29 + X39 + X40 + X43 + X55 + X60 + X62.

8.8 ForJ = 7 andL = 9,

X63 + 1 = (1 + X7)(1 + X7 + X14 + X21 + X28 + X35 + X42 + X49 + X56)

andπ(X) = 1+X7+X14+X21+X28+X35+X42+X49+X56. Letα be a primitive element in

GF(26) whose minimal polynomial isφ1(X) = 1+X +X6. Becauseα63 = 1, the polynomial

1 + X7 hasα0, α9, α18, α27, α36, α45, andα54 as all it roots. Therefore, the polynomialπ(X)

hasαh as a root whenh is not a multiple of9 and0 < h < 63. From the conditions (Theorem

8.2) on the roots ofH(X), we can findH(X) as:H(X) = LCM{minimal polynomialsφi(X)

of the roots ofH(X)}. As the result,

H(X) = φ1(X)φ3(X)φ5(X)φ7(X)φ21(X),

whereφ1 = 1 + X + X6, φ3 = 1 + X + X2 + X4 + X6, φ5 = 1 + X + X2 + X5 + X6,

φ7 = 1 + X3 + X6, andφ21 = 1 + X + X2. Then, we can findG(X),

G(X) =X63 + 1

H(X)=

(1 + X7)π(X)

H(X)

= (1 + X7)(1 + X2 + X3 + X5 + X6)(1 + X + X3 + X4 + X6)

(1 + X2 + X4 + X5 + X6)(1 + X + X4 + X5 + X6)(1 + X5 + X6).

5

For type-1 DTI code of length 63 andJ = 7, the generator polynomial is:

g1(X) =X37G(X−1)

1 + X

=(1 + X7)(1 + X + X3 + X4 + X6)(1 + X2 + X3 + X5 + X6)

1 + X

(1 + X + X2 + X4 + X6)(1 + X + X2 + X5 + X6)(1 + X + X6)= (1 + X + X2 + X3 + X4 + X5 + X6)(1 + X + X3 + X4 + X6)(1 + X2 + X3 + X5 + X6)

(1 + X + X2 + X4 + X6)(1 + X + X2 + X5 + X6)(1 + X + X6).

Represent the polynomialsπ(X), Xπ(X), X2π(X), X3π(X), X4π(X), X5π(X), and

X6π(X) by 63-tuple location vectors. Add an overall parity-check digit and apply the affine

permutation,Y = αX + α62, to each of these location vectors. By removing one vector with

odd weight, we can obtain the polynomials orthogonal on the digit positionX62. They are:

X11 + X14 + X32 + X36 + X43 + X44 + X45 + X52 + X56 + X62,

X3 + X8 + X13 + X19 + X31 + X33 + X46 + X48 + X60 + X62,

X2 + X10 + X23 + X24 + X26 + X28 + X29 + X42 + X50 + X62,

X1 + X6 + X9 + X15 + X16 + X35 + X54 + X55 + X61 + X62,

X0 + X4 + X7 + X17 + X27 + X30 + X34 + X39 + X58 + X62,

X5 + X21 + X22 + X38 + X47 + X49 + X53 + X57 + X59 + X62.

8.9 The generator polynomial of the maximum-length sequence code of lengthn = 2m − 1 is

g(X) = (Xn + 1)/p(X) = (X + 1)(1 + X + X2 + . . . + Xn−1)/p(X),

wherep(X) is a primitive polynomial of degreem over GF(2). Sincep(X) and(X + 1) are

relatively prime,g(X) has1 as a root. Since the all-one vector1 + X + X2 + . . . + Xn−1

does not have1 as a root, it is not divisible byg(X). Therefore, the all-one vector is not a

codeword in a maximum-length sequence code.

8.17 There are five 1-flats that pass through the pointα7. The 1-flats passing throughα7 can be

represented byα7 + βa1, wherea1 is linearly independent ofα7 andβ ∈ GF (22). They are

6

five 1-flats passing throughα7 which are:

L1 = {α7, α9, α13, α6},L2 = {α7, α14, α10, α8},L3 = {α7, α12, 0, α2},L4 = {α7, α4, α11, α5},L5 = {α7, α3, 1, α}.

8.18 (a) There are twenty one 1-flats that pass through the pointα63. They are:

L1 = {α63, 0, α42, α21}L2 = {α63, α6, α50, α39}L3 = {α63, α12, α15, α37}L4 = {α63, α32, α4, α9}L5 = {α63, α24, α11, α30}L6 = {α63, α62, α7, α17}L7 = {α63, α1, α18, α8}L8 = {α63, α26, α41, α38}L9 = {α63, α48, α60, α22}

L10 = {α63, α45, α46, α53}L11 = {α63, α61, α34, α14}L12 = {α63, α25, α3, α51}L13 = {α63, α2, α16, α36}L14 = {α63, α35, α31, α40}L15 = {α63, α52, α13, α19}L16 = {α63, α23, α54, α58}L17 = {α63, α33, α44, α57}L18 = {α63, α47, α49, α20}

7

L19 = {α63, α27, α43, α29}L20 = {α63, α56, α55, α10}L21 = {α63, α59, α28, α5}

(b) There are five 2-flats that intersect on the 1-flat,{α63 + ηα}, whereη ∈ GF (22). They

are:

F1 = {1, α6, α50, α39, 0, α, α22, α43, α42, α29, α18, α48, α21, α60, α27, α8}F2 = {1, α6, α50, α39, α12, α26, α10, α46, α15, α53, α41, α56, α37, α55, α45, α38}F3 = {1, α6, α50, α39, α32, α35, α20, α57, α4, α33, α31, α47, α9, α49, α44, α40}F4 = {1, α6, α50, α39, α24, α16, α34, α17, α11, α62, α36, α14, α30, α61, α7, α2}F5 = {1, α6, α50, α39, α25, α5, α54, α52, α3, α13, α59, α58, α51, α23, α19, α28}

8.19 The 1-flats that pass through the pointα21 are:

L1 = {α21, α42, α11, α50, α22, α44, α25, α37}L2 = {α21, α60, α35, α31, α47, α54, α32, α52}L3 = {α21, α58, α9, α26, α6, α5, α28, α34}L4 = {α21, α30, α57, 0, α3, α48, α39, α12}L5 = {α21, α51, α61, α27, α19, α14, α62, α2}L6 = {α21, α38, α40, α33, α46, α17, α18, α7}L7 = {α21, α29, α16, α53, α23, α0, α4, α1}L8 = {α21, α59, α15, α45, α56, α8, α55, α13}L9 = {α21, α43, α41, α20, α10, α36, α24, α49}

8.20 a. The radix-23 expansion of47 is expressed as follows:

47 = 7 + 5 · 23.

8

Hence, the23-weight of47

W23(47) = 7 + 5 = 12.

b.

W23(47(0)) = W23(47) = 7 + 5 = 12,

W23(47(1)) = W23(31) = 7 + 3 = 10,

W23(47(2)) = W23(62) = 6 + 7 = 13.

Hence,

max0≤l<3

W23(47(l)) = 13.

c. All the positive integersh less than 63 such that

0 < max0≤l<3

W23(h(l)) ≤ 23 − 1.

are

1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 17, 20,

21, 24, 28, 32, 33, 34, 35, 40, 42, 48, 49, 56.

9

Chapter 11

Convolutional Codes

11.1 (a) The encoder diagram is shown below.

u

v( 1 )

v( 2 )

v ( 0 )

(b) The generator matrix is given by

G =

111 101 011111 101 011

111 101 011. . . . . .

.

(c) The codeword corresponding to u = (11101) is given by

v = u · G = (111, 010, 001, 110, 100, 101, 011).

11.2 (a) The generator sequences of the convolutional encoder in Figure 11.3 on page 460 are given in(11.21).

1

2

(b) The generator matrix is given by

G =

1111 0000 00000101 0110 00000011 0100 0011

1111 0000 00000101 0110 00000011 0100 0011

. . . . . .

.

(c) The codeword corresponding to u = (110, 011, 101) is given by

v = u ·G = (1010, 0000, 1110, 0111, 0011).

11.3 (a) The generator matrix is given by

G(D) =[

1 + D 1 + D2 1 + D + D2].

(b) The output sequences corresponding to u(D) = 1 + D2 + D3 + D4 are

V(D) =[v(0)(D),v(1)(D),v(2)(D)

]

=[1 + D + D2 + D5, 1 + D3 + D5 + D6, 1 + D + D4 + D6

],

and the corresponding codeword is

v(D) = v(0)(D3) + Dv(1)(D3) + D2v(2)(D3)= 1 + D + D2 + D3 + D5 + D6 + D10 + D14 + D15 + D16 + D19 + D20.

11.4 (a) The generator matrix is given by

G(D) =[

1 + D D 1 + DD 1 1

]

and the composite generator polynomials are

g1(D) = g(0)1 (D3) + Dg(1)

1 (D3) + D2g(2)1 (D3)

= 1 + D2 + D3 + D4 + D5

and

g2(D) = g(0)2 (D3) + Dg(1)

2 (D3) + D2g(2)2 (D3)

= D + D2 + D3.

(b) The codeword corresponding to the set of input sequences U(D) =[1 + D + D3, 1 + D2 + D3

]is

v(D) = u(1)(D3)g1(D) + u(2)(D3)g2(D)= 1 + D + D3 + D4 + D6 + D10 + D13 + D14.

3

11.5 (a) The generator matrix is given by

G =

111 001 010 010 001 011111 001 010 010 001 011

111 001 010 010 001 011. . . . . .

.

(b) The parity sequences corresponding to u = (1101) are given by

v(1)(D) = u(D) · g(1)(D)= (1 + D + D3)(1 + D2 + D3 + D5)= 1 + D + D2 + D3 + D4 + D8,

and

v(2)(D) = u(D) · g(2)(D)= (1 + D + D3)(1 + D + D4 + D5)= 1 + D2 + D3 + D6 + D7 + D8.

Hence,

v(1) = (111110001)v(2) = (101100111).

11.6 (a) The controller canonical form encoder realization, requiring 6 delay elements, is shown below.

v( 0 )

v( 1 )

v

u( 1 )

u( 2 )

( 2 )

4

(b) The observer canonical form encoder realization, requiring only 3 delay elements, is shown below.

v(2)

v(1)

v(0)

u(2)

u(1)

11.14 (a) The GCD of the generator polynomials is 1.

(b) Since the GCD is 1, the inverse transfer function matrix G−1(D) must satisfy

G(D)G−1(D) =[1 + D2 1 + D + D2

]G−1(D) = I.

By inspection,

G−1(D) =[

1 + DD

].

11.15 (a) The GCD of the generator polynomials is 1 + D2 and a feedforward inverse does not exist.

(b) The encoder state diagram is shown below.

S2 S5

1/10

0/00 S7

S6

S3

S0

S4

S1

0/01

0/10

1/10

0/11

0/011/11

0/00

1/00

1/11

1/01

1/00

1/01

0/10

0/11

(c) The cycles S2S5S2 and S7S7 both have zero output weight.

(d) The infinite-weight information sequence

u(D) =1

1 + D2= 1 + D2 + D4 + D6 + D8 + · · ·

5

results in the output sequences

v(0)(D) = u(D)(1 + D2

)= 1

v(1)(D) = u(D)(1 + D + D2 + D3

)= 1 + D,

and hence a codeword of finite weight.

(e) This is a catastrophic encoder realization.

11.16 For a systematic (n, k, ν) encoder, the generator matrix G(D)is a k × n matrix of the form

G(D) = [Ik|P(D)] =

1 0 · · · 0 g(k)1 (D) · · · g(n−1)

1 (D)0 1 · · · 0 g(k)

2 (D) · · · g(n−1)2 (D)

......

...0 0 · · · 1 g(k)

k (D) · · · g(n−1)k (D)

.

The transfer function matrix of a feedforward inverse G−1(D) with delay l = 0 must be such that

G(D)G−1(D) = Ik.

A matrix satisfying this condition is given by

G−1(D) =[

Ik

0(n−k)×k

]=

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 10 0 · · · 0...

......

0 0 · · · 0

.

6

11.19 (a) The encoder state diagram is shown below.

S2 S1

S3

S0

0/000

1/001

1/1110/011

1/100

0/110 1/010

0/101

(b) The modified state diagram is shown below.

S3

X

X

X

S1 S2 S0S0

X2

X2X3X2

(c) The WEF function is given by

A(X) =

∑i

Fi∆i

∆.

7

There are 3 cycles in the graph:

Cycle 1: S1S2S1 C1 = X3

Cycle 2: S1S3S2S1 C2 = X4

Cycle 3: S3S3 C3 = X .

There is one pair of nontouching cycles:

Cycle pair 1: (Cycle 1, Cycle 3) C1C3 = X4.

There are no more sets of nontouching cycles. Therefore,

∆ = 1 −∑

i

Ci +∑i′,j′

Ci′Cj′

= 1 − (X + X3 + X4) + X4

= 1 − X − X3.

There are 2 forward paths:

Forward path 1: S0S1S2S0 F1 = X7

Forward path 2: S0S1S3S2S0 F2 = X8.

Only cycle 3 does not touch forward path 1, and hence

∆1 = 1 − X.

Forward path 2 touches all the cycles, and hence

∆2 = 1.

Finally, the WEF is given by

A(X) =X7(1 − X) + X8

1 − X − X3=

X7

1 − X − X3.

Carrying out the division,

A(X) = X7 + X8 + X9 + 2X10 + · · · ,indicating that there is one codeword of weight 7, one codeword of weight 8, one codeword ofweight 9, 2 codewords of weight 10, and so on.

(d) The augmented state diagram is shown below.

(e) The IOWEF is given by

A(W, X, L) =

∑i

Fi∆i

∆.

There are 3 cycles in the graph:

Cycle 1: S1S2S1 C1 = WX3L2

Cycle 2: S1S3S2S1 C2 = W 2X4L3

Cycle 3: S3S3 C3 = WXL.

8

S1 S2S0 S0

W X3L X2LX2L

X2L

S3

WXL

WXL

WXL

There is one pair of nontouching cycles:

Cycle pair 1: (Cycle 1, Cycle 3) C1C3 = W 2X4L3.

There are no more sets of nontouching cycles. Therefore,

∆ = 1 −∑

i

Ci +∑i′,j′

Ci′Cj′

= 1 − WX3L2 + W 2X4L3 + WXL + W 2X4L3.

There are 2 forward paths:

Forward path 1: S0S1S2S0 F1 = WX7L3

Forward path 2: S0S1S3S2S0 F2 = W 2X8L4.

Only cycle 3 does not touch forward path 1, and hence

∆1 = 1 − WXL.

Forward path 2 touches all the cycles, and hence

∆2 = 1.

Finally, the IOWEF is given by

A(W, X, L) =WX7L3(1 − WXL) + W 2X8L4

1 − (WX3L2 + W 2X4L3 + WXL) + X4Y 2Z3=

WX7L3

1 − WXL − WX3L2.

Carrying out the division,

A(W, X, L) = WX7L3 + W 2X8L4 + W 3X9L5 + · · · ,indicating that there is one codeword of weight 7 with an information weight of 1 and length 3,one codeword of weight 8 with an information weight of 2 and length 4, and one codeword ofweight 9 with an information weight of 3 and length 5.

9

11.20 Using state variable method described on pp. 505-506, the WEF is given by

A(X)=−X4(−1−2X4+X3+9X20−42X18+78X16+38X12+3X17+5X13+9X11−9X15−2X7−74X14−14X10−9X9+X6+6X8)

1−X+2X4−X3−X2+3X24+X21−3X20+32X18−8X22−X19−45X16−8X12−5X17−6X11+9X15+2X7+27X14+3X10−2X9−4X6+X8 .

Performing the division results in

A(X) = X4 + X5 + 2X6 + 3X7 + 6X8 + 9X9 + · · · ,

which indicates that there is one codeword of weight 4, one codeword of weight 5, two codewords ofweight 6, and so on.

11.28 (a) From Problem 11.19(c), the WEF of the code is

A(X) = X7 + X8 + X9 + 2X10 + · · · ,

and the free distance of the code is therefore dfree = 7, the lowest power of X in A(X).

(b) The complete CDF is shown below.

0 1 2 3 4

1

2

3

4

5

6

7

5 6 7

d

dmin = 5

dfree =7

(c) The minimum distance isdmin = dl|l=m=2 = 5.

10

11.29 (a) By examining the encoder state diagram in Problem 11.15 and considering only paths that beginand end in state S0 (see page 507), we find that the free distance of the code is dfree = 6. Thiscorresponds to the path S0S1S2S4S0 and the input sequence u = (1000).

(b) The complete CDF is shown below.

0 1 2 3 4

1

2

3

4

5

6

d

dfree = 6

dmin = 3

(c) The minimum distance isdmin = dl|l=m=3 = 3.

11.31 By definition, the free distance dfree is the minimum weight path that has diverged from and remergedwith the all-zero state. Assume that [v]j represents the shortest remerged path through the statediagram with weight free dfree. Letting [dl]re be the minimum weight of all remerged paths of lengthl, it follows that [dl]re = dfree for all l ≥ j. Also, for a noncatastrophic encoder, any path that remainsunmerged must accumulate weight. Letting [dl]un be the minimum weight of all unmerged paths oflength l, it follows that

liml→∞

[dl]un → ∞.

Therefore

liml→∞

dl = min

{liml→∞

[dl]re, liml→∞

[dl]un

}= dfree.

Q. E. D.

Chapter 12

Optimum Decoding of ConvolutionalCodes

12.1 (Note: The problem should read “ for the (3,2,2) encoder in Example 11.2 ”rather than “ for the (3,2,2)code in Table 12.1(d)”.) The state diagram of the encoder is given by:

01/111

11/101

00/000

01/011

00/100

10/101

S3

S1S2

S0

00/0

11

11/1

10

00/111

10/010

10/001

01/100

11/001

10/110

11/010

01/000

1

2

From the state diagram, we can draw a trellis diagram containing h + m + 1 = 3 + 1 + 1 = 5 levels asshown below:

00/000

10/10101

/011

11/110

10/01000/111

01/10011

/001

01/11110/001

11/010

00/100

11/101

01/000

10/110

00/011

S3

S1

S2

S0S000/000

10/10101

/01111/110

10/010

00/111

01/10011

/001

01/111

10/001

11/010

00/100

11/101

01/000

10/110

S0

S3

S1

S2

00/011

S0

S3

S1

S2

00/000

10/10101

/011

11/110

00/000

00/111

00/10000/011

S0

Hence, for u = (11, 01, 10),v(0) = (1001)v(1) = (1001)v(2) = (0011)

andv = (110, 000, 001, 111),

agreeing with (11.16) in Example 11.2. The path through the trellis corresponding to this codeword isshown highlighted in the figure.

3

12.2 Note that

N−1∑l=0

c2 [log P (rl|vl) + c1] =N−1∑l=0

[c2 log P (rl|vl) + c2c1]

= c2

N−1∑l=0

log P (rl|vl) + Nc2c1.

Since

maxv

{c2

N−1∑l=0

log P (rl|vl) + Nc2c1

}= c2 max

v

{N−1∑l=0

log P (rl|vl)

}+ Nc2c1

if C2 is positive, any path that maximizes∑N−1

l=0 log P (rl|vl) also maximizes∑N−1

l=0 c2[log P (rl|vl) +c1].

12.3 The integer metric table becomes:

01 02 12 11

0 6 5 3 01 0 3 5 6

The received sequence is r = (111201, 111102, 111101, 111111, 011201, 120211, 120111). The decoded se-quence is shown in the figure below, and the final survivor is

v = (111, 010, 110, 011, 000, 000, 000),

which yields a decoded information sequence of

u = (11000).

This result agrees with Example 12.1.

4

0/000

1/111

0/000 0/000 0/000 0/000 0/000 0/000

1/111

1/111

1/111

1/111

0/101

1/010

1/010

1/010

1/010

0/101

0/101

0/101

0/101

1/001

0/110

1/001 1/001

0/110

0/110

0/1101/100

0/011

1/100

1/100

0/011

0/011

0/011

0/011

S3

S1

S0S0 S0 S0 S0 S0 S0

S1 S1 S1 S1

S2 S2 S2

S3

S2

S3 S3

S2

S0

0 9 13 26 52 67 75 84

11 24 32 46 57

22 36 42 63

20 40 48 53 73

5

12.4 For the given channel transition probabilities, the resulting metric table is:

01 02 03 04 14 13 12 11

0 −0.363 −0.706 −0.777 −0.955 −1.237 −1.638 −2.097 −2.6991 −2.699 −2.097 −1.638 −1.237 −0.955 −0.777 −0.706 −0.363

To construct an integer metric table, choose c1 = 2.699 and c2 = 4.28. Then the integer metric tablebecomes:

01 02 03 04 14 13 12 11

0 10 9 8 7 6 5 3 01 0 3 5 6 7 8 9 10

12.5 (a) Referring to the state diagram of Figure 11.13(a), the trellis diagram for an information sequenceof length h = 4 is shown in the figure below.

(b) After Viterbi decoding the final survivor is

v = (11, 10, 01, 00, 11, 00).

This corresponds to the information sequence

u = (1110).

6

0/000

0/00 0/00 0/00 0/00 0/00 0/00

0/00

0/00

0/00

1/10

1/10

1/10

S1

S0

0/00

0/01

S0 S0 S0 S0 S0

S4 S4 S4 S4

S2 S2S2S2

S6 S6 S6

S5 S5

S1 S1 S1

0/01

0/01

0/01

S3 S3 S3

3

19 12 21 42

22

38

16

1/00

0/110/11

0/110/11

1/00

1/01

1/01

0/10

0/10

0/10

46 51

S7 S7

20 48

1/10

1/01

0/01

0/01

0/10

0/10

40 61

1/00

1/11

0/11

0/00

0/00

0/00

53 71 66

20 45 79

0/110/11

0/11

27 68 83 94

34 49 80 98 115

S0 S0

7

12.6 Combining the soft decision outputs yields the following transition probabilities:

0 10 0.909 0.0911 0.091 0.909

For hard decision decoding, the metric is simply Hamming distance. For the received sequence

r = (11, 10, 00, 01, 10, 01, 00),

the decoding trellis is as shown in the figure below, and the final survivor is

v = (11, 10, 01, 01, 00, 11, 00),

which corresponds to the information sequence

u = (1110).

This result matches the result obtained using soft decisions in Problem 12.5.

8

0/000

0/00 0/00 0/00 0/00 0/00 0/00

0/00

0/00

0/00

1/10

1/10

1/10

S1

S0

0/00

0/01

S0 S0 S0 S0 S0

S4 S4 S4 S4

S2 S2S2S2

S6 S6 S6

S5 S5

S1 S1 S1

0/01

0/01

0/01

S3 S3 S3

2

0 3 5 4

2

0

3

1/00

0/110/11

0/110/11

1/00

1/01

1/01

0/10

0/10

0/10

1 3

S7 S7

4 2

1/10

1/01

0/01

0/01

0/10

0/10

2 3

1/00

1/11

0/11

0/00

0/00

0/00

1 1 2

4 4 3

0/110/11

0/11

4 2 2 3

3 4 3 3 3

S0 S0

9

12.9 Proof: For d even,

Pd =12

(d

d/2

)pd/2(1 − p)d/2 +

d∑e=(d/2)+1

(de

)pe(1 − p)d−e

<

d∑e=(d/2)

(de

)pe(1 − p)d−e

<d∑

e=(d/2)

(de

)pd/2(1 − p)d/2

= pd/2(1 − p)d/2d∑

e=(d/2)

(de

)

< 2dpd/2(1 − p)d/2

and thus (12.21) is an upper bound on Pd for d even. Q. E. D.

12.10 The event error probability is bounded by (12.25)

P (E) <

∞∑d=dfree

AdPd < A(X)|X=2

√p(1−p)

.

From Example 11.12,

A(X) =X6 + X7 − X8

1 − 2X − X3= X6 + 3X7 + 5X8 + 11X9 + 25X10 + · · · ,

which yields

(a) P (E) < 1.2118× 10−4 for p = 0.01,(b) P (E) < 7.7391× 10−8 for p = 0.001.

The bit error probability is bounded by (12.29)

Pb(E) <∞∑

d=dfree

BdPd < B(X)|X=2

√p(1−p)

=1k

∂A(W, X)∂W

∣∣∣∣X=2

√p(1−p),W=1

.

From Example 11.12,

A(W, X) =WX7 + W 2(X6 − X8)

1 − W (2X + X3)= WX7+W 2

(X6 + X8 + X10

)+W 3

(2X7 + 3X9 + 3X11 + X13

)+· · · .

Hence,∂A(W, X)

∂W=

X7 + 2W (X6 − 3X8 − X10) − 3W 2(2X7 − X9 − X11)(1 − 2WX − WX3)2

+ · · ·and

∂A(W, X)∂W

∣∣∣∣W=1

=2X6 − X7 − 2X8 + X9 + X11

(1 − 2X − X3)2= 2X6 + 7X7 + 18X8 + · · · .

10

This yields

(a) Pb(E) < 3.0435× 10−4 for p = 0.01,

(b) Pb(E) < 1.6139× 10−7 for p = 0.001.

12.11 The event error probability is given by (12.26)

P (E) ≈ Adfree

[2√

p(1 − p)]dfree ≈ Adfree

2dfreepdfree/2

and the bit error probability (12.30) is given by

Pb(E) ≈ Bdfree

[2√

p(1 − p)]dfree ≈ Bdfree

2dfreepdfree/2.

From Problem 12.10,dfree = 6, Adfree

= 1, Bdfree= 2.

(a) For p = 0.01,P (E) ≈ 1 · 26 · (0.01)6/2 = 6.4 × 10−5

Pb(E) ≈ 2 · 26 · (0.01)6/2 = 1.28 × 10−4.

(b) For p = 0.001,P (E) ≈ 1 · 26 · (0.001)6/2 = 6.4 × 10−8

Pb(E) ≈ 2 · 26 · (0.001)6/2 = 1.28 × 10−7.

12.12 The (3, 1, 2) encoder of (12.1) has dfree = 7 and Bdfree= 1. Thus, expression (12.36) becomes

Pb(E) ≈ Bdfree2dfree/2e−(Rdfree/2)·(Eb/No) = 2(7/2)e−(7/6)·(Eb/No)

and (12.37) remains

Pb(E) ≈ 12e−Eb/No .

These expressions are plotted versus Eb/No in the figure below.

11

0 5 10 1510

-16

10-14

10-12

10-10

10-8

10-6

10-4

10-2

100

102

Eb/N

o (dB)

Pb(E

)

12.36 (Coded)12.37 (Uncoded)

12

Equating the above expressions and solving for Eb/No yields

2(7/2)e−(7/6)·(Eb/No) =12e−Eb/No

1 = 2(9/2)e−(1/6)(Eb/No)

e(−1/6)(Eb/No) = 2−(9/2)

(−1/6)(Eb/No) = ln(2−(9/2))Eb/No = −6 ln(2−(9/2)) = 18.71,

which is Eb/No = 12.72dB, the coding threshold. The coding gain as a function of Pb(E) is plottedbelow.

10-10

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

-6

-4

-2

0

2

4

6

8

10

12

14

Pb

Eb/N

o (d

B)

Reguired SNR, CodedRequired SNR, UncodedGain

( E )

Note that in this example, a short constraint length code (ν = 2) with hard decision decoding, theapproximate expressions for Pb(E) indicate that a positive coding gain is only achieved at very smallvalues of Pb(E), and the asymptatic coding gain is only 0.7dB.

12.13 The (3, 1, 2) encoder of Problem 12.1 has dfree = 7 and Bdfree= 1. Thus, expression (12.46) for the

unquantized AWGN channel becomes

Pb(E) ≈ Bdfreee−RdfreeEb/No = e−(7/3)·(Eb/No)

and (12.37) remains

Pb(E) ≈ 12e−Eb/No .

These expressions are plotted versus Eb/No in the figure below.

13

0 5 10 1510

-35

10-30

10-25

10-20

10-15

10-10

10-5

100

Eb/N

o (dB)

Pb(E

)

12.46 (Coded)12.37 (Uncoded)

Equating the above expressions and solving for Eb/No yields

e−(7/3)·(Eb/No) =12e−Eb/No

1 =12e(4/3)(Eb/No)

e(4/3)(Eb/No) = 2(4/3)(Eb/No) = ln(2)

Eb/No = (3/4) ln(2) = 0.5199,

which is Eb/No = −2.84dB, the coding threshold. (Note: If the slightly tighter bound on Q(x) from(1.5) is used to form the approximate expressin for Pb(E), the coding threshold actually moves to−∞ dB. But this is just an artifact of the bounds, which are not tight for small values of Eb/No.)The coding gain as a function of Pb(E) is plotted below. Note that in this example, a short constraintlength code (ν = 2) with soft decision decoding, the approximate expressions for Pb(E) indicate thata coding gain above 3.0 dB is achieved at moderate values of Pb(E), and the asymptotic coding gainis 3.7 dB.

14

10-10

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

-2

0

2

4

6

8

10

12

14

Pb

Eb/N

o (d

B)

Reguired SNR, CodedRequired SNR, UncodedGain

( E )

15

12.14 The IOWEF function of the (3, 1, 2) encoder of (12.1) is

A(W, X) =WX7

1 − WX − WX3

and thus (12.39b) becomes

Pb(E) < B(X)|X=Do =1k

∂A(W, X)∂W

∣∣∣∣X=D0,W=1

=X7

(1 − WX − WX3)2

∣∣∣∣X=D0,W=1

.

For the DMC of Problem 12.4, D0 = 0.42275 and the above expression becomes

Pb(E) < 9.5874× 10−3.

If the DMC is converted to a BSC, then the resulting crossover probability is p = 0.091. Using (12.29)yields

Pb(E) < B(X)|X=Do =1k

∂A(W, X)∂W

∣∣∣∣X=2

√p(1−p),W=1

=X7

(1 − WX − WX3)2

∣∣∣∣X=2

√p(1−p),W=1

= 3.7096×10−1,

about a factor of 40 larger than the soft decision case.

12.16 For the optimum (2, 1, 7) encoder in Table 12.1(c), dfree = 10, Adfree= 1, and Bdfree

= 2.

(a) From Table 12.1(c)γ = 6.99dB.

(b) Using (12.26) yieldsP (E) ≈ Adfree

2dfreepdfree/2 = 1.02 × 10−7.

(c) Using (12.30) yieldsPb(E) ≈ Bdfree

2dfreepdfree/2 = 2.04 × 10−7.

(d) For this encoder

G−1 =[

D2

1 + D+D2

]

and the amplification factor is A = 4.

For the quick-look-in (2, 1, 7) encoder in Table 12.2, dfree = 9, Adfree= 1, and Bdfree

= 1.

(a) From Table 12.2γ = 6.53dB.

(b) Using (12.26) yieldsP (E) ≈ Adfree

2dfreepdfree/2 = 5.12 × 10−7.

(c) Using (12.30) yieldsPb(E) ≈ Bdfree

2dfreepdfree/2 = 5.12 × 10−7.

16

(d) For this encoder

G−1 =[

11

]

and the amplification factor is A = 2.

12.17 The generator matrix of a rate R = 1/2 systematic feedforward encoder is of the form

G =[1 g(1)(D)

].

Letting g(1)(D) = 1 + D + D2 + D5 + D7 achieves dfree = 6 with Bdfree= 1 and Adfree

= 1.

(a) The soft-decision asymptotic coding gain is

γ = 4.77dB.

(b) Using (12.26) yieldsP (E) ≈ Adfree

2dfreepdfree/2 = 6.4 × 10−5.

(c) Using (12.30) yieldsPb(E) ≈ Bdfree

2dfreepdfree/2 = 6.4 × 10−5.

(d) For this encoder (and all systematic encoders)

G−1 =[

10

]

and the amplification factor is A = 1.

12.18 The generator polynomial for the (15, 7) BCH code is

g(X) = 1 + X4 + X6 + X7 + X8

and dg = 5. The generator polynomial of the dual code is

h(X) =X15 + 1

X8 + X7 + X6 + X4 + 1= X7 + X6 + X4 + 1

and hence dh ≥ 4.

(a) The rate R = 1/2 code with composite generator polynomial g(D) = 1 + D4 + D6 + D7 + D8 hasgenerator matrix

G(D) =[1 + D2 + D3 + D4 D3

]and dfree ≥ min(5, 8) = 5.

(b) The rate R = 1/4 code with composite generator polynomial g(D) = g(D2)+Dh(D2) = 1+D+D8 + D9 + D12 + D13 + D14 + D15 + D16 has generator matrix

G(D) =[1 + D2 + D3 + D4 1 + D2 + D3 D3 D3

]and dfree ≥ min(dg + dh, 3dg, 3dh) = min(9, 15, 12) = 9.

17

The generator polynomial for the (31, 16) BCH code is

g(X) = 1 + X + X2 + X3 + X5 + X7 + X8 + X9 + X10 + X11 + X15

and dg = 7. The generator polynomial of the dual code is

h(X) =X15 + 1g(X)

= X16 + X12 + X11 + X10 + X9 + X4 + X + 1

and hence dh ≥ 6.

(a) The rate R = 1/2 code with composite generator polynomial g(D) = 1 + D + D2 + D3 + D5 +D7 + D8 + D9 + D10 + D11 + D15 has generator matrix

G(D) =[1 + D + D4 + D5 1 + D + D2 + D3 + D4 + D5 + D7

]

and dfree ≥ min(7, 12) = 7.

(b) The rate R = 1/4 code with composite generator polynomial g(D) = g(D2)+Dh(D2) = 1+D+D2 +D3 +D4 +D6 +D9 +D10 +D14 +D16 +D18 +D19 +D20 +D21 +D22 +D23 has generatormatrix

G(D) =[1 + D + D4 + D5 1 + D2 + D5 + D6 + D8 1 + D + D2 + D3 + D4 + D5 + D7 1 + D5

]

and dfree ≥ min(dg + dh, 3dg, 3dh) = min(13, 21, 18) = 13.

12.20 (a) The augmented state diagram is shown below.

S1 S2

S3

S0 S0WX 3L X2L

X2LWXL

WXL

X2L

WXL

The generating function is given by

A(W, X, L) =

∑i

Fi∆i

∆.

There are 3 cycles in the graph:

18

Cycle 1: S1S2S1 C1 = WX3L2

Cycle 2: S1S3S2S1 C2 = W 2X4L3

Cycle 3: S3S3 C3 = WXL.

There is one pair of nontouching cycles:

Cycle pair 1: (loop 1, loop 3) C1C3 = W 2X4L3.

There are no more sets of nontouching cycles. Therefore,

∆ = 1 −∑

i

Ci +∑i′,j′

Ci′Cj′

= 1 − (WX3L2 + W 2X4L3 + WXL) + W 2X4L3.

There are 2 forward paths:

Forward path 1: S0S1S2S0 F1 = WX7L3

Forward path 2: S0S1S3S2S0 F2 = W 2X8L4.

Only cycle 3 does not touch forward path 1, and hence

∆1 = 1 − WXL.

Forward path 2 touches all the cycles, and hence

∆2 = 1.

Finally, the WEF A(W, X, L) is given by

A(W, X, L) =WX7L3(1 − WXL) + W 2X8L4

1 − (WX3L2 + W 2X4L3 + WXL) + W 2X4L3=

WX7L3

1 − WXL − WX3L2

and the generating WEF’s Ai(W, X, L) are given by:

A1(W, X, L) =WX3L(1 − WXL)

∆=

WX3L(1 − WXL)1 − WXL − WX3L2

= W X3 L + W 2 X6 L3 + W 3 X7 L4 + (W 3X9 + W 4X8 )L5 + (2 W 4, X10 + W 5X9 )L6 + · · ·A2(W, X, L) =

WX5L2(1 − WXL) + W 2X6L3

∆=

WX5L2

1 − WXL − WX3L2

= W X5 L2 + W 2 X6 L3 + (W 2 X8 + W 3 X7)L4 + (2 W 3 X9 + W 4 X8)L5

+(W 3 X11 + 3 W 4 X10 + W 5 X9)L6 + · · ·A3(W, X, L) =

W 2X4L2

∆=

W 2X4L2

1 − WXL − WX3L2

= W 2 X4 L2 + W 3 X5 L3 + (W 3 X7 + W 4 X6)L4 + (2 W 4 X8 + W 5 X7)L5

+(W 4 X10 + 3 W 5 X9 + W 6 X8)L6 · · ·

(b) This code has dfree = 7, so τmin is the minimum value of τ for which d(τ ) = dfree + 1 =8. Examining the series expansions of A1(W, X, L), A2(W, X, L), and A3(W, X, L) above yieldsτ min = 5.

19

(c) A table of d(τ ) and Ad(τ ) is given below.

τ d(τ ) Ad(τ )

0 3 11 4 12 5 13 6 14 7 15 8 1

(d) From part (c) and by looking at the series expansion of A3(W, X, L), it can be seen that

limτ→∞

d(τ ) = τ + 3.

12.21 For a BSC, the trellis diagram of Figure 12.6 in the book may be used to decode the three possible 21-bit subequences using the Hamming metric. The results are shown in the three figures below. Since ther used in the middle figure (b) below has the smallest Hamming distance (2) of the three subsequences,it is the most likely to be correctly synchronized.

0/000

1/111

0/000 0/000 0/000 0/000 0/000 0/000

1/111

1/111

1/111

1/111

0/101

1/010

1/010

1/010

1/010

0/101

0/101

0/101

0/101

1/001

0/110

1/001 1/001

0/110

0/110

0/1101/100

0/011

1/100

1/100

0/011

0/011

0/011

0/011

S3

S1

S0S0 S0 S0 S0 S0 S0

S1 S1 S1 S1

S2 S2

S3

S2

S3 S3

S2

S0

0 2 3 4 4 6 4 6

1 4 3 5 5

3 5 3 6

2 3 6 3 7S2

r 011 100 110 010 110 010 001

(a)

20

0/000

1/111

0/000 0/000 0/000 0/000 0/000 0/000

1/111

1/111

1/111

1/111

0/101

1/010

1/010

1/010

1/010

0/101

0/101

0/101

0/101

1/001

0/110

1/001 1/001

0/110

0/110

0/1101/100

0/011

1/100

1/100

0/011

0/011

0/011

0/011

S3

S1

S0S0 S0 S0 S0 S0 S0

S1 S1 S1 S1

S2 S2

S3

S2

S3 S3

S2

S0

0 3 4 4 5 4 5 2

0 5 1 4 1

2 4 4 6

1 3 1 5 2S2

r 111 001 100 101 100 100 011

(b)

21

0/000

1/111

0/000 0/000 0/000 0/000 0/000 0/000

1/111

1/111

1/111

1/111

0/101

1/010

1/010

1/010

1/010

0/101

0/101

0/101

0/101

1/001

0/110

1/001 1/001

0/110

0/110

0/1101/100

0/011

1/100

1/100

0/011

0/011

0/011

0/011

S3

S1

S0S0 S0 S0 S0 S0 S0

S1 S1 S1 S1

S2 S2

S3

S2

S3 S3

S2

S0

0 2 4 4 4 5 5 6

1 3 5 5 6

2 2 3 3

3 4 4 6 5S2

r 110 011 001 011 001 000 111

(c)

Chapter 13

Suboptimum Decoding ofConvolutional Codes

13.1 (a) Referring to the state diagram of Figure 11.13a, the code tree for an information sequence oflength h = 4 is shown below.

(b) For u = (1001), the corresponding codeword is v = (11, 01, 11, 00, 01, 11, 11), as shown below.

11

10

01

00

11

00

01

10

10

01

11

00

00

11

10 01 00 11

01 00 11 10

11

00

10

11 00 00

11 11

01

10

10

11

11

11

00

00

11 110100

11 00 00 00

01 00 1101

00 11 1010

10

11 00 00

11 1100

11

10

11

11

11

00

00

10

01

00 00 00 00

11 11 1101

1

2

13.2 Proof: A binary-input, Q-ary-output DMC is symmetric if

P (r` = j|0) = P (r` = Q− 1− j|1) (a-1)

for j = 0, 1, . . . , Q− 1. The output probability distribution may be computed as

P (r` = j) =1∑

i=0

P (r` = j|i)P (i),

where i is the binary input to the channel. For equally likely input signals, P (1) = P (0) = 0.5 and

P (r` = j) = P (r` = j|0)P (0) + P (r` = j|1)P (1)= 0.5 [P (r` = j|0) + P (r` = j|1)]= 0.5 [P (r` = Q− 1− j|0) + P (r` = Q− 1− j|1)] [using (a-1)]= P (r` = Q− 1− j).

Q. E. D.

13.3 (a) Computing the Fano metric with p = 0.045 results in the metric table:

0 10 0.434 −3.9741 −3.974 0.434

Dividing by 0.434 results in the following integer metric table:

0 10 1 −91 −9 1

(b) Decoding r = (11, 00, 11, 00, 01, 10, 11) using the stack algorithm results in the following eightsteps:

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 81(-2) 11(-6) 10(-6) 100(-4) 1001(-2) 10010(0) 100100(-8) 1001000(-6)0(-18) 10(-6) 111(-14) 111(-14) 111(-14)

0(-18) 110(-14) 110(-14) 110(-14)0(-18) 0(-18) 0(-18)

101(-24) 1000(-22)101(-24)

The decoded sequence is v = (11, 01, 11, 00, 01, 11, 11) and u = [1001]. The Viterbi algorithmrequires fifteen steps to decode the same sequence.

(c) Decoding r = (11, 10, 00, 01, 10, 01, 00) using the stack algorithm results in the following sixteensteps:

3

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 81(-2) 11(4) 111(-4) 1110(-2) 110(-4) 11100(-10) 1100(-12) 1101(-12)0(-18) 10(-16) 110(-4) 110(-4) 11100(-10) 1100(-12) 1101(-12) 10(-16)

0(-18) 10(-16) 10(-16) 10(-16) 1101(-12) 10(-16) 111000(-18)0(-18) 0(-18) 0(-18) 10(-16) 111000(-18) 0(-16)

1111(-22) 1111(-22) 0(-18) 0(-18) 11000(-20)Step 9 Step 10 Step 11 Step 12 Step 13 Step 14 Step 15 Step 16

11010(-10) 10(-16) 101(-14) 1011(-12) 10110(-10) 101100(-18) 111000(-18) 1110000(-16)10(-16) 111000(-18) 111000(-18) 111000(-18) 111000(-18) 111000(-18) 110100(-18) 110100(-18)

111000(-18) 110100(-18) 110100(-18) 110100(-18) 110100(-18) 110100(-18) 0(-18) 0(-18)0(-18) 0(-18) 0(-18) 0(-18) 0(-18) 0(-18) 11000(-20) 11000(-20)

11000(-20) 11000(-20) 11000(-20) 11000(-20) 11000(-20) 11000(-20) 1010(-32) 1010(-32)100(-34) 1010(-32) 1010(-32) 1010(-32) 100(-34) 100(-34)

100(-34) 100(-34) 100(-34) 1011000(-36) 1011000(-36)

The decoded sequence is v = (11, 10, 01, 01, 00, 11, 00) and u = (1110). This agrees with the resultof Problem 12.6.

13.4 (a) Computing the Fano metric using

M(r`|v`) = log2

P (r`|v`)P (r`)

−R

with R = 1/2 andP (01) = P (11) = 0.218P (02) = P (12) = 0.1025P (03) = P (13) = 0.095P (04) = P (14) = 0.0845

results in

01 02 03 04 14 13 12 11

0 0.494 0.443 0.314 −0.106 −1.043 −2.66 −4.07 −7.2681 −7.268 −4.07 −2.66 −1.043 −0.106 0.314 0.443 0.494

Multiplying by 3/0.106 yields the following integer metric table:

01 02 03 04 14 13 12 11

0 14 12 9 −3 −30 −75 −115 −1671 −167 −115 −75 −30 −3 9 12 14

4

(b) Decoding r = (1211, 1201, 0301, 0113, 1202, 0311, 0302) using the stack algorithm results in thefollowing thirteen steps:

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 71(26) 11(52) 110(-9) 1100(-70) 111(-106) 1110(-83) 1101(-167)

0(-282) 10(-256) 111(-106) 111(-106) 1101(-167) 11000(-173) 11100(-186)0(-282) 10(-256) 1101(-167) 11000(-173) 11000(-173) 11100(-186)

0(-252) 10(-256) 10(-256) 10(-256) 10(-256)0(-282) 0(-282) 0(-282) 0(-282)

1111(-348) 1111(-348)Step 8 Step 9 Step 10 Step 11 Step 12 Step 13

11010(-143) 11000(-173) 11100(-186) 110100(-204) 111000(-247) 1110000(-226)11000(-173) 11100(-186) 110100(-204) 111000(-247) 10(-256)11100(-186) 110100(-204) 10(-256) 10(-256) 0(-282)

10(-256) 10(-256) 0(-282) 0(-282) 110000(-348)0(-282) 0(-282) 110000(-331) 110000(-331) 1111(-348)

1111(-348) 1111(-348) 1111(-348) 1111(-348) 1101000(-394)

The decoded sequence is v = (11, 10, 01, 01, 00, 11, 00) and u = (1110). This agrees with the resultof Problem 12.5(b).

13.6 As can be seen from the solution to Problem 13.3, the final decoded path never falls below the secondstack entry (part (b)) or the fourth stack entry (part(c)). Thus, for a stacksize of 10 entries, the finaldecoded path is not effected in either case.

13.7 From Example 13.5, the integer metric table is

0 10 1 −51 −5 1

5

(a) Decoding the received sequence r = (010, 010, 001, 110, 100, 101, 011) using the stack bucket algo-rithm with an interval of 5 results in the following decoding steps:

Interval 1 Interval 2 Interval 3 Interval 4 Interval 5 Interval 6 Interval 79 to 5 4 to 0 -1 to -5 -6 to -10 -11 to -15 -16 to -20 -21 to -25

Step 1 0(-3) 1(-9)Step 2 00(-6) 01(-12)

1(-9)Step 3 1(-9) 000(-12)

001(-15)01(-12)

Step 4 11(-6) 000(-12) 10(-24)001(-15)01(-12)

Step 5 111(-3) 000(-12) 110(-21)001(-15) 10(-24)01(-12)

Step 6 1110(0) 000(-12) 1111(-18) 110(-21)001(-15) 10(-24)01(-12)

Step 7 11101(3) 11100(-15) 1111(-18) 110(-21)000(-12) 10(-24)001(-15)01(-12)

Step 8 111010(6) 11100(-15) 1111(-18) 110(-21)000(-12) 10(-24)001(-15)01(-12)

Step 9 1110100(6) 11100(-15) 1111(-18) 110(-21)000(-12) 10(-24)001(-15)01(-12)

The decoded sequence is v = (111, 010, 001, 110, 100, 101, 011) and u = (11101), which agrees withthe result of Example 13.5.

6

(b) Decoding the received sequence r = (010, 010, 001, 110, 100, 101, 011) using the stack bucket algo-rithm with an interval of 9 results in the following decoding steps.

Interval 1 Interval 2 Interval 3 Interval 48 to 0 -1 to -9 -10 to -18 -19 to -27

Step 1 0(-3)1(-9)

Step 2 00(-6) 01(-12)1(-9)

Step 3 1(-9) 000(-12)001(-15)01(-12)

Step 4 11(-6) 000(-12) 10(-24)001(-15)01(-12)

Step 5 111(-3) 000(-12) 110(-21)001(-15) 10(-24)01(-12)

Step 6 1110(0) 1111(-18) 110(-21)000(-12) 10(-24)001(-15)01(-12)

Step 7 11101(3) 11100(-15) 110(-21)1111(-18) 10(-24)000(-12)001(-15)01(-12)

Step 8 111010(6) 11100(-15) 110(-21)1111(-18) 10(-24)000(-12)001(-15)01(-12)

Step 9 1110100(9) 11100(-15) 110(-21)1111(-18) 10(-24)000(-12)001(-15)01(-12)

The decoded sequence is v = (111, 010, 001, 110, 100, 101, 011) and u = (11101), which agrees withthe result of Example 13.5 and part (a).

7

13.8 (i) 4 = 5

Step Look MF MB Node Metric T0 - - - X 0 01 LFB -3 −∞ X 0 -52 LFB -3 - 0 -3 -53 LFB -6 0 X 0 -54 LFNB -9 −∞ X 0 -105 LFB -3 - 0 -3 -106 LFB -6 - 00 -6 -107 LFB -9 - 000 -9 -108 LFB -12 -9 00 -6 -109 LFNB -15 -3 0 -3 -1010 LFNB -12 0 X 0 -1011 LFNB -9 - 1 -9 -1012 LFB -6 - 11 -6 -1013 LFB -3 - 111 -3 -514 LFB 0 - 1110 0 015 LFB 3 - 11101 3 016 LFB 6 - 111010 6 517 LFB 9 - 1110100 9 Stop

The decoded sequence is v = (111, 010, 001, 110, 100, 101, 011) and u = [11101] which agrees with theresult of Example 13.5.

(ii) 4 = 10

Step Look MF MB Node Metric T0 - - - X 0 01 LFB -3 −∞ X 0 -102 LFB -3 - 0 -3 -103 LFB -6 - 00 -6 -104 LFB -9 - 000 -9 -105 LFB -12 -6 00 -6 -106 LFNB -15 -3 0 -3 -107 LFNB -12 0 X 0 -10*8 LFNB -9 - 1 -9 -109 LFB -6 - 11 -6 -1010 LFB -3 - 111 -3 -1011 LFB 0 - 1110 0 012 LFB 3 - 11101 3 013 LFB 6 - 111010 6 014 LFB 9 - 1110100 9 Stop

* Not the first visit. Therefore no tightening is done.

Note: The final decoded path agrees with the results of Examples 13.7 and 13.8 and Problem 13.7.The number of computations in both cases is reduced compared to Examples 13.7 and 13.8.

8

13.20 (a)

g(1)(D) = 1 + D + D3 + D5 + D8 + D9 + D10 + D11

Therefore:

H =

1 1

1 0 1 1

0 0 1 0 1 1

1 0 0 0 1 0 1 1

0 0 1 0 0 0 1 0. . .

1 0 0 0 1 0 0 0. . .

0 0 1 0 0 0 1 0. . .

0 0 0 0 1 0 0 0. . .

1 0 0 0 0 0 1 0. . .

1 0 1 0 0 0 0 0. . .

1 0 1 0 1 0 0 0 1 1

1 0 1 0 1 0 1 0 . . . . . . . . . . . . . . . . . . 1 0 1 1

(b) Using the parity triangle of the code, we obtain:

s0 = e(0)0 +e

(1)0

s1 = e(0)0 +e

(0)1 +e

(1)1

s2 = +e(0)1 +e

(0)2 +e

(1)2

s3 = e(0)0 +e

(0)2 +e

(0)3 +e

(1)3

s4 = +e(0)1 +e

(0)3 +e

(0)4 +e

(1)4

s5 = e(0)0 +e

(0)2 +e

(0)4 +e

(0)5 +e

(1)5

s6 = +e(0)1 +e

(0)3 +e

(0)5 +e

(0)6 +e

(1)6

s7 = +e(0)2 +e

(0)4 +e

(0)6 +e

(0)7 +e

(1)7

s8 = e(0)0 +e

(0)3 +e

(0)5 +e

(0)7 +e

(0)8 +e

(1)8

s9 = e(0)0 +e

(0)1 +e

(0)4 +e

(0)6 +e

(0)8 +e

(0)9 +e

(1)9

s10 = e(0)0 +e

(0)1 +e

(0)2 +e

(0)5 +e

(0)7 +e

(0)9 +e

(0)10 +e

(1)10

s11 = e(0)0 +e

(0)1 +e

(0)2 +e

(0)3 +e

(0)6 +e

(0)8 +e

(0)10 +e

(0)11 +e

(1)11

9

(c) The effect of error bits prior to time unit ` are removed. Thus, the modified syndrome bits aregiven by the following equations:

s′` = e(0)`

s′`+1 = e(0)` +e

(0)`+1 +e

(1)`+1

s′`+2 = +e(0)`+1 +e

(0)`+2 +e

(1)`+2

s′`+3 = e(0)` +e

(0)`+2 +e

(0)`+3 +e

(1)`+3

s′`+4 = +e(0)`+1 +e

(0)`+3 +e

(0)`+4 +e

(1)`+4

s′`+5 = e(0)` +e

(0)`+2 +e

(0)`+4 +e

(0)`+5 +e

(1)`+5

s′`+6 = +e(0)`+1 +e

(0)`+3 +e

(0)`+5 +e

(0)`+6 +e

(1)`+6

s′`+7 = +e(0)`+2 +e

(0)`+4 +e

(0)`+6 +e

(0)`+7 +e

(1)`+7

s′`+8 = e(0)` +e

(0)`+3 +e

(0)`+5 +e

(0)`+7 +e

(0)`+8 +e

(1)`+8

s′`+9 = e(0)` +e

(0)`+1 +e

(0)`+4 +e

(0)`+6 +e

(0)`+8 +e

(0)`+9 +e

(1)`+9

s′`+10 = e(0)` +e

(0)`+1 +e

(0)`+2 +e

(0)`+5 +e

(0)`+7 +e

(0)`+9 +e

(0)`+10 +e

(1)`+10

s′`+11 = e(0)` +e

(0)`+1 +e

(0)`+2 +e

(0)`+3 +e

(0)`+6 +e

(0)`+8 +e

(0)`+10 +e

(0)`+11 +e

(1)`+11

13.28 (a) Using either of the methods described in Examples 13.15 and 13.16, we arrive at the conclusionthat dmin = 7.

(b) From the parity triangle, we can see that this code is not self-orthogonal (for instance, s3 and s5

both check information error bit e(0)2 ).

(c) From the parity triangle shown below, we see that the maximum number of orthogonal paritychecks that can be formed on e

(0)0 is 4: {s0, s1, s2 + s11, s5}.

110101001111

11010100111

1101010011

110101001

11010100

1101010

110101

11010

1101

110

11 1

(d) So, because of (c), this code is not completely orthogonalizable.

10

13.32 (a) According to Table 13.2(a), there is only one rate R = 1/2 self orthogonal code with dmin = 9:the (2,1,35) code with

g(1)(D) = 1 + D7 + D10 + D16 + D18 + D30 + D31 + D35and m(a) = 35.

(b) According to Table 13.3(a), there is only one rate R = 1/2 orthogonalizable code with dmin = 9:the (2,1,21) code with

g(1)(D) = 1 + D11 + D13 + D16 + D17 + D19 + D20 + D21and m(b) = 21.

(c) The best rate R = 1/2 systematic code with dmin = 9 is the (2,1,15) code with

g(1)(D) = 1 + D + D3 + D5 + D7 + D8 + D11 + D13 + D14 + D15and m(c) = 15∗.

(d) According to Table 12.1(c), the best rate R = 1/2 nonsystematic code with dfree = 9 (actually,10 in this case) is the (2,1,6) code with

g(0)(D) = 1 + D + D2 + D3 + D6, g(1)(D) = 1 + D2 + D3 + D5 + D6, and m(d) = 6.

Thus m(d) < m(c) < m(b) < m(a).

* This generator polynomial can be found in Table 13.1 of the first edition of this text.

13.34 (a) This self orthogonal code with g(1)(D) = 1+D2 +D7 +D13 +D16 +D17 yields the parity triangle:

→ 10 1

→ 1 0 10 1 0 10 0 1 0 10 0 0 1 0 10 0 0 0 1 0 1

→ 1 0 0 0 0 1 0 10 1 0 0 0 0 1 0 10 0 1 0 0 0 0 1 0 10 0 0 1 0 0 0 0 1 0 10 0 0 0 1 0 0 0 0 1 0 10 0 0 0 0 1 0 0 0 0 1 0 1

→ 1 0 0 0 0 0 1 0 0 0 0 1 0 10 1 0 0 0 0 0 1 0 0 0 0 1 0 10 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1

→ 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1→ 1 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1

The orthogonal check sums on e(0)0 are {s0, s2, s7, s13, s16, s17}.

11

(b) The block diagram of the feedback majority logic decoder for this code is:

+

+ + + + + +

+

MAJORITY GATE

)1(17r

)0(17r

)0(e

s

u)0(r

17s

13.37 (a) These orthogonalizable code with g(1)(D) = 1 + D6 + D7 + D9 + D10 + D11 yields the paritytriangle:

100000110111 1

10000011011

1000001101

100000110

10000011

1000001

100000

10000

1000

100

10

The orthogonal check sums on e(0)0 are {s0, s6, s7, s9, s1 + s3 + s10, s4 + s8 + s11}.

12

(b) The block diagram of the feedback majority logic decoder for this code is:

+

+ + + + +

+

MAJORITY GATE

+

++

)0(11r

)1(11r 11s s

)0(r u

)0(e

Chapter 16

Turbo Coding

16.1 Let u = [u0, u1, . . . , uK−1] be the input to encoder 1. Then

v(1) = [v(1)0 , v

(1)1 , . . . , v

(1)K−1] = u G(1)

is the output of encoder 1, where G(1) is the K ×K parity generator matrix for encoder 1, i.e.,v(1) = uG(1) is the parity sequence produced by encoder 1. For example, for the (2,1,2) systematicfeedbacK encoder with

G(1)(D) =[1

1 + D2

1 + D + D2

],

the parity generator matrix is given by

G(1) =

1 1 1 0 1 1 0 1 1 . . . . . . . . . . . . . . .1 1 1 0 1 1 0 1 1 . . . . . . . . . . . .

1 1 1 0 1 1 0 1 1 . . . . . . . . .. . .

. (K ×K)

Now let u′ = [u′0, u′1, . . . , u

′K−1] be the (interleaved) input to encoder 2. Then we can write

u′ = u T,

where T is a K×K permutation matrix with a single one in each row and column. (If T is the identitymatrix, then u′ = u.)

The output of encoder 2 is then given by

v(2) = [v(2)0 , v

(2)1 , . . . , v

(2)K−1] = u′ G(2) = u T G(2),

where G(2) is the K ×K parity generator matrix for encoder 2.

1

2

Now consider two input sequences u1 and u2, both of length K. The corresponding encoder outputsin each case are given by the 3-tuples

[u1,u1 G(1),u′1 G(2)] = [u1, u1 G(1),u1 T G(2)]4= y1

and

[u2,u2 G(1),u′2 G(2)] = [u2, u2 G(1),u2 T G(2)]4= y2.

Now consider the input sequence u = u1 + u2. The encoder output 3-tuple in this case is given by

[u,u G(1),u′ G(2)] = [u, u G(1),u T G(2)]

= [u1 + u2,u1 G(1) + u2 G(2),u1T G(2) + u2T G(2)]

= [u1,u1 G(1),u1T G(2)] + [u2,u2 G(2),u2T G(2)]= y1 + y2.

Thus superposition holds and the turbo encoder is linear.

16.3 Define C1 to be the (7,4,3) Hamming code and C2 to be the (8,4,4) extended Hamming code.

C1 Codeword list:

0000000110100001110011010001101101001100101100011000101111101000011100100110110001010101110100011010001111111111

The CWEF for C1 is given by

AC11 (Z) = 3Z2 + Z3

AC12 (Z) = 3Z + 3Z2

AC13 (Z) = 1 + 3Z

AC14 (Z) = Z3

3

C2 Codeword list:

0000 0000110100010111001010100011101101000110010111000110000101111110100000111001100110101000101101011100100011011000111011111111

The CWEF for C2 is given by

AC21 (Z) = 4Z3

AC22 (Z) = 6Z2

AC23 (Z) = 4Z

AC24 (Z) = Z4

The CWEF for the PCBC is given by equation (16.23) as

APCw (Z) = AC1

w (Z) ∗ AC2w (Z)/

(Kw

)

APC1 (Z) = (3Z2 + Z3)(4Z3)/4 = 3Z5 + Z6

APC2 (Z) = (3Z + 3Z2)(6Z2)/6 = 3Z3 + 3Z4

APC3 (Z) = (1 + 3Z)(4Z)/4 = Z + 3Z2

APC4 (Z) = (Z3)(Z4)/1 = Z7

The bit CWEF’s are given by

BPCw (Z) =

( w

K

)APC

w (Z)

BPC1 (Z) =

(14

)(3Z5 + Z6) = .75 Z5 + .25 Z6

BPC2 (Z) =

(24

)(3Z3 + 3Z4) = 1.5Z3 + 1.5Z4

BPC3 (Z) =

(34

)(Z + 3Z2) = .75Z3 + 2.25Z2

4

BPC4 (Z) =

(44

)(Z7) = Z7

The IRWEF’s are given by

APC(W,Z) = W (3Z5 + Z6) + W 2(3Z3 + 3Z4) + W 3(Z + 3Z2) + W 7Z7

BPC(W,Z) = W (.75Z5 + .25Z6) + W 2(1.5Z3 + 1.5Z4) + W 3(.75Z + 2.25Z2) + W 4Z7

To find the WEF’s, set W = Z = X:

APC(X) = X4 + 6X5 + 6X6 + X7 + X11

BPC(X) = .75X4 + 3.75X5 + 2.25X6 + .25X7 + X11

16.4 The codeword IRWEF can be found from equation (16.34) as follows:

APC(W,Z) = W (4.5Z4 + 3Z5 + 0.5Z6) +W 2(1.29Z2 + 2.57Z3 + 1.29Z4 + 3.86Z5 + 6.43Z6 + 3Z7 +3.32Z8 + 3.86Z9 + 1.93Z10 + 0.43Z11 + 0.04Z12) +W 3(0.07 + 0.43Z + 0.64Z2 + 1.29Z3 + 5.57Z4 + 5.57Z5 +7.07Z6 + 15.43Z7 + 14.14Z8 + 5.14Z9 + 0.64Z10) +W 4(3.21Z4 + 17.14Z5 + 29.29Z6 + 17.14Z7 + 3.21Z8) +W 5(0.64Z2 + 5.14Z3 + 14.14Z4 + 15.43Z5 + 7.07Z6 + 5.57Z7 +5.57Z8 + 1.29Z9 + 0.64Z10 + 0.43Z11 + 0.07Z12) +W 6(0.04 + 0.43Z + 1.93Z2 + 3.86Z3 + 3.32Z4 + 3Z5 +6.43Z6 + 3.86Z7 + 1.29Z8 + 2.57Z9 + 1.29Z10) +W 7(0.05Z6 + 3Z7 + 4.5Z8) + W 8Z12

To find the bit IRWEF, take each term in the expression for APC(W,Z) and divide by w/8, where wis the exponent of W.

BPC(W,Z) = W (0.56Z4 + 0.38Z5 + 0.06Z6) +W 2(0.32Z2 + 0.64Z3 + 0.32Z4 + 0.97Z5 + 1.61Z6 + .75Z7 +0.83Z8 + 0.97Z9 + 0.48Z10 + 0.11Z11 + 0.01Z12) +W 3(0.003 + 0.16Z + 0.24Z2 + 0.48Z3 + 2.09Z4 + 2.09Z5 +2.65Z6 + 5.79Z7 + 5.30Z8 + 1.93Z9 + 0.24Z10) +W 4(1.61Z4 + 8.57Z5 + 14.65Z6 + 8.57Z7 + 1.61Z8) +W 5(0.40Z2 + 3.21Z3 + 8.84Z4 + 9.64Z5 + 4.42Z6 + 3.48Z7

3.48Z8 + 0.81Z9 + 0.40Z10 + 0.27Z11 + 0.04Z12) +W 6(0.03 + 0.32Z + 1.45Z2 + 2.90Z3 + 2.49Z4 + 2.25Z5

4.82Z6 + 2.90Z7 + 0.97Z8 + 1.93Z9 + 0.97Z10

5

W 7(0.04Z6 + 2.63Z7 + 3.94Z8) + W 8Z12

To find the WEF’s, substitute X=W=Z into the equations for the IRWEF’s, as shown below.

APC(X) = 0.07X3 + 1.71X4 + 7.71X5 + 5.61X6 + 11X7 + 22.29X8 + 45.21X9 +66.79X10 + 44.79X11 + 20.14X12 + 9.71X13 + 9.46X14 + 5.14X15 + 3X16 +0.07X17 + 1.29X18 + X20

BPC(X) = 0.03X3 + 0.48X4 + 1.45X5 + 1.21X6 + 3.84X7 + 9.96X8 + 23.71X9 +33.39X10 + 21.19X11 + 10.71X12 + 5.82X13 + 5.04X14 + 0.96X15 + 2.20X16 +0.04X17 + 0.96X18 + X20

16.5 Consider using an h-repeated (n, k, dmin) = (7, 4, 3) Hamming code as the constituent code, formingan (h[2n − k], k) = (10h, 4h) PCBC. The output consists of the original input bits u, the parity bitsfrom the non-interleaved encoder v(1), and the parity bits from the interleaved encoder v(2). For theh = 4-repeated PCBC, the IRWEF of the (28,16) constituent code is

A(W,Z) = [1 + W (3Z2 + Z3) + W 2(3Z + 3Z2) + W 3(1 + 3Z) + W 4Z3]4 − 1= W (12Z2 + 4Z3) +

W 2(12Z + 12Z2 + 54Z4 + 36Z5 + 6Z6) +W 3(4 + 12Z + 108Z3 + 144Z4 + 36Z5 + 108Z6 +

108Z7 + 36Z8 + 4Z9) +W 4(90Z2 + 232Z3 + 90Z4 + 324Z5 + 540Z6 +

252Z7 + 117Z8 + 108Z9 + 54Z10 + 12Z11 + Z12) +W 5(36Z + 144Z2 + 108Z3 + 432Z4 + 1188Z5 + 780Z6 +

468Z7 + 648Z8 + 432Z9 + 120Z10 + 12Z11) +W 6(6 + 36Z + 54Z2 + 324Z3 + 1296Z4 + 1296Z5 +

918Z6 + 1836Z7 + 1620Z8 + 556Z9 + 66Z10) +W 7(144Z2 + 780Z3 + 1188Z4 + 1080Z5 + 2808Z6 + 3456Z7 +

1512Z8 + 324Z9 + 108Z10 + 36Z11 + 4Z12) +W 8(36Z + 252Z2 + 540Z3 + 783Z4 + 2592Z5 + 4464Z6 +

2592Z7 + 783Z8 + 540Z9 + 252Z10 + 36Z11) +

6

W 9(4 + 36Z + 108Z2 + 324Z3 + 1512Z4 + 3456Z5 +2808Z6 + 1080Z7 + 1188Z8 + 780Z9 + 144Z10) +

W 10(66Z2 + 556Z3 + 1620Z4 + 1836Z5 + 918Z6 + 1296Z7 +1296Z8 + 324Z9 + 54Z10 + 36Z11 + 6Z12) +

W 11(12Z + 120Z2 + 432Z3 + 648Z4 + 468Z5 + 780Z6 +1188Z7 + 432Z8 + 108Z9 + 144Z10 + 36Z11) +

W 12(1 + 12Z + 54Z2 + 108Z3 + 117Z4 + 252Z5 +540Z6 + 324Z7 + 90Z8 + 232Z9 + 90Z10) +

W 13(4Z3 + 36Z4 + 108Z5 + 108Z6 + 36Z7 +144Z8 + 108Z9 + 12Z11 + 4Z12) +

W 14(6Z6 + 36Z7 + 54Z8 + 12Z10 + 12Z11) +W 15(4Z9 + 12Z10) +W 16Z12.

The IRWEF’s of the (40,16) PCBC are found, using APCw (Z) =

(Kw

)−1

[Aw(Z)]2, where K = hk = 16

is the interleaver (and input) length, to be:

APC1 (Z) = 9Z4 + 6Z5 + Z6

APC2 (Z) = 1.2Z2 + 2.4Z3 + 1.2Z4 + 10.8Z5 + 18Z6 + 8.4Z7 + 25.5Z8 + 32.4Z9 + 16.2Z10 +

3.6Z11 + 0.3Z12

APC3 (Z) = 0.029 + 0.171Z + 0.257Z2 + 1.543Z3 + 6.686Z4 + 6.686Z5 + 23.914Z6 + 61.714Z7 +

56.057Z8 + 61.771Z9 + 99.686Z10 + 83.314Z11 + 54.771Z12 + 48.343Z13 +

35.229Z14 + 15.429Z15 + 3.857Z16 + 0.514Z17 + 0.029Z18

APC4 (Z) = 4.451Z4 + 22.945Z5 + 38.475Z6 + 54.989Z7 + 140.459Z8 + 194.637Z9 + 186.903Z10 +

257.697Z11 + 294.389Z12 + 216.831Z13 + 151.273Z14 + 117.156Z15 + 73.844Z16 +

36.316Z17 + 17.268Z18 + 8.229Z19 + 3.155Z20 + 0.831Z21 +

0.138Z22 + 0.013Z23 + 0.001Z24

APC5 (Z) = 0.297Z2 + 2.374Z3 + 6.527Z4 + 14.242Z5 + 50.736Z6 + 112.549Z7 + 160.615Z8 +

315.099Z9 + 550.385Z10 + 579.363Z11 + 551.505Z12 + 611.802Z13 + 540.89Z14 +

360.791Z15 + 238.088Z16 + 158.176Z17 + 80.901Z18 + 27.297Z19 +

5.670Z20 + 0.659Z21 + 0.033Z22

7

APC6 (Z) = 0.004 + 0.054Z + 0.243Z2 + 0.971Z3 + 5.219Z4 + 17.964Z5 + 43.615Z6 +

133.355Z7 + 345.929Z8 + 533.928Z9 + 682.391Z10 + 1030.59Z11 + 1269.74Z12 +

1130.6Z13 + 993.686Z14 + 891.674Z15 + 597.803Z16 + 255.219Z17 + 65.307Z18 +

9.165Z19 + 0.544Z20

APC7 (Z) = 1.813Z4 + 19.636Z5 + 83.089Z6 + 189.189Z7 + 341.333Z8 + 694.221Z9 +

1194.5Z10 + 1462.3Z11 + 1702.7Z12 + 2064.99Z13 + 1874.92Z14 + 1101.01Z15 +

456.243Z16 + 169.326Z17 + 61.439Z18 + 18.05Z19 + 4.116Z20 + 0.906Z21 +

0.189Z22 + 0.025Z23 + 0.001Z24

APC8 (Z) = 0.101Z2 + 1.41Z3 + 7.955Z4 + 25.527Z5 + 67.821Z6 + 192.185Z7 +

454.462Z8 + 795.877Z9 + 1316.39Z10 + 2201.74Z11 + 2743.06Z12 + 2201.74Z13 +

1316.39Z14 + 795.877Z15 + 454.462Z16 + 192.185Z17 + 67.821Z18 + 25.527Z19 +

7.955Z20 + 1.41Z21 + 0.101Z22

APC9 (Z) = 0.001 + 0.025Z + 0.189Z2 + 0.906Z3 + 4.116Z4 + 18.05Z5 + 61.439Z6 +

169.326Z7 + 456.243Z8 + 1101.01Z9 + 1874.92Z10 + 2064.99Z11 + 1702.7Z12 +

1462.3Z13 + 1194.5Z14 + 694.221Z15 + 341.333Z16 + 189.189Z17 +

83.089Z18 + 19.636Z19 + 1.813Z20

APC10 (Z) = 0.544Z4 + 9.165Z5 + 65.307Z6 + 255.219Z7 + 597.803Z8 + 891.674Z9 +

993.686Z10 + 1130.6Z11 + 1269.74Z12 + 1030.59Z13 + 682.391Z14 + 533.928Z15 +

345.929Z16 + 133.355Z17 + 43.615Z18 + 17.964Z19 + 5.219Z20 + 0.971Z21 +

0.243Z22 + 0.054Z23 + 0.004Z24

APC11 (Z) = 0.033Z2 + 0.659Z3 + 5.67Z4 + 27.297Z5 + 80.901Z6 + 158.176Z7 +

238.088Z8 + 360.791Z9 + 540.89Z10 + 611.802Z11 + 551.505Z12 + 579.363Z13 +

550.385Z14 + 315.099Z15 + 160.615Z16 + 112.549Z17 + 50.736Z18 + 14.242Z19 +

6.527Z20 + 2.374Z21 + 0.297Z22

APC12 (Z) = 0.001 + 0.013Z + 0.138Z2 + 0.831Z3 + 3.155Z4 + 8.229Z5 + 17.268Z6 +

36.316Z7 + 73.844Z8 + 117.156Z9 + 151.273Z10 + 216.831Z11 + 294.389Z12 +

257.697Z13 + 186.903Z14 + 194.637Z15 + 140.459Z16 + 54.989Z17 +

38.475Z18 + 22.945Z19 + 4.451Z20

APC13 (Z) = 0.029Z6 + 0.514Z7 + 3.857Z8 + 15.429Z9 + 35.229Z10 + 48.343Z11 + 54.771Z12 +

83.314Z13 + 99.686Z14 + 61.771Z15 + 56.057Z16 + 61.714Z17 + 23.914Z18 +

8

6.686Z19 + 6.686Z20 + 1.543Z21 + 0.257Z22 + 0.171Z23 + 0.029Z24

APC14 (Z) = 0.3Z12 + 3.6Z13 + 16.2Z14 + 32.4Z15 + 25.5Z16 + 8.4Z17 + 18Z18 +

10.8Z19 + 1.2Z20 + 2.4Z21 + 1.2Z22

APC15 (Z) = Z18 + 6Z19 + 9Z20

APC16 (Z) = Z24.

Using a 4x4 row-column (block) interleaver, the minimum distance of the PCBC is 5. This comesfrom a weight 1 input, resulting in wH(u) = 1, wH(v(1)) = 2, and wH(v(2)) = 2, where wH(x) is theHamming weight of x.

16.7 On page 791 in the text, it is explained that “...any multiple error event in a codeword belongingto a terminated convolutional code can be viewed as a succession of single error events separated bysequences of 0’s.” Another way to look at it is that for convolutional codewords, we start in the zerostate and end in the zero state and any multiple error event can be seen as a series of deviations fromthe zero state. These deviations from the zero state can be separated by any amount of time spentin the zero state. For example, the codeword could have a deviation starting and another deviationending at the same zero state, or the codeword could stay in the zero state for several input zeros.Likewise the first input bit could take the codeword out of the zero state or the last input bit could bethe one to return the codeword to the zero state.

So, if there are h error events with total length λ in a block codeword length K, then there must beK−λ times that state 0 occurs or, as explained in the text, K−λ 0’s. Now, in the codeword, there areK − λ + h places where an error event can start, since the remaining λ− h places are then determinedby the error events. Since there are h events, the number of ways to choose h elements from N −λ+hplaces gives the multiplicity of block codewords for h-error events.

16.8 The IRWEF and WEF for the PCCC in Example 16.6 is found in the same manner as for a PCBC.The following combines the equations in (16.49) for the CWEF’s:

APC(W,Z) = W 2(4Z8 + 4Z10 + Z12) + W 3(1.81Z4 + 6.74Z6 + 9.89Z8 + 6.74Z10 + 1.81Z12) +W 4(0.93Z4 + 5.93Z4 + 10.59Z8 + 4.67Z10 + 3.89Z12 + 0.67Z14 + 0.33Z16) +W 5(0.33Z4 + 2.22Z6 + 6.37Z8 + 9.33Z10 + 6.81Z12 + 1.78Z14 + 0.15Z16) +W 6(0.04Z4 + 1.04Z6 + 8.15Z8 + 12.44Z10 + 5.33Z12) + W 7(5.44Z8 + 3.11Z10 +0.44Z12) + W 9(Z12)

APC(X) = 1.81X7 + 0.93X8 + 7.07X9 + 9.97X10 + 12.11X11 + 15.63X12 + 13.11X13 +13.82X14 + 15.95X15 + 16.33X16 + 9.92X17 + 6X18 + 2.22X19 + 0.33X20 + 1.15X21

9

16.9 The bit IRWEF is given by:

BPC(W,Z) = W 2(0.89Z8 + 0.89Z10 + 0.22Z12) + W 3(0.60Z4 + 2.25Z6 + 3.30Z8 +2.25Z10 + 0.60Z12) + W 4(0.41Z4 + 2.63Z6 + 4.71Z8 + 2.07Z10 + 1.73Z12 +0.30Z14 + 0.15Z16) + W 5(0.19Z4 + 1.23Z6 + 3.54Z8 + 5.18Z10 + 3.79Z12 +0.99Z14 + 0.08Z16) + W 6(0.02Z4 + 0.69Z6 + 5.43Z8 + 8.30Z10 + 3.55Z12) +W 7(4.23Z8 + 2.42Z10 + 0.35Z12) + W 9Z12

The bit WEF is given by:

BPC(X) = 0.60X7 + 0.41X8 + 2.43X9 + 3.55X10 + 4.53X11 + 6.29X12 + 5.79X13 + 7.73X14

10.02X15 + 10.02X16 + 6.20X17 + 3.85X18 + 1.33X19 + 0.15X20 + 1.08X21

16.10 Redrawing the encoder block diagram and the IRWEF state diagram for the reversed generators, wesee that the state diagram is essentially the same except that W and Z are switched. Thus, we canuse equation (16.41) for the new IRWEF with W and Z reversed.

A(W,Z,L) = L3W 2Z3 + L4W 4Z2 + L5(W 4Z3 + W 2Z4) + L6(2W 4Z3 + W 4Z4) + L7(W 2Z5 +2W 4Z4 + W 4Z5 + W 6Z2) + L8(W 4Z5 + 3W 4Z4 + 2W 4Z5 + 2W 6Z3) + L9(W 2Z6 +3W 4Z5 + 2W 4Z6 + W 4Z7 + 3W 6Z3 + 3W 6Z4).

After dropping all terms of order larger than L9, the single-error event enumerators are obtained asfollows:

A(1)2,3 = Z3 A

(1)2,5 = Z4 A

(1)2,7 = Z5 A

(1)2,9 = Z6

A(1)4,4 = Z2 A

(1)4,5 = Z3 A

(1)4,6 = 2Z3 + Z4

A(1)4,7 = 2Z4 + Z5 A

(1)4,8 = 3Z4 + 2Z5 + Z6 A

(1)4,9 = 3Z5 + 2Z6 + Z7

A(1)6,7 = Z2 A

(1)6,8 = 2Z3 A

(1)6,9 = 3Z3 + 2Z4

The double-error event IRWEF is

A2(W,Z, L) = [A(W,Z, L)]2

= L6W 4Z6 + 2L7W 6Z5 + L8(2W 4Z7 + 2W 6Z6 + W 8Z4) + L9(6W 6Z6 + 2W 6Z7 +2W 8Z5) + . . .

Dropping all terms of order greater than L9 gives the double-error event enumerators A(2)2,`(Z):

A(2)4,6 = Z6 A

(2)4,8 = 2Z7

A(2)6,7 = 2Z5 A

(2)6,8 = 2Z6 A

(2)6,9 = 6Z6 + 2Z7

A(2)8,8 = Z4 A

(2)8,9 = 2Z5

10

Following the same procedure, the triple-error event IRWEF is given by A(3)(W,Z,L) = [A(W,Z, L)]3 =L9W 6Z9 + . . ., and the triple-error event enumerator is given by A

(3)6,9(Z) = Z9.

Before finding the CWEF’s we need to calculate hmax. Notice that there are no double-error events ofweight less than 4 and that the only triple-error event has weight 6. Also notice that, for the terminatedencoder, odd input weights don’t appear. So for weight 2, hmax = 1. For weight 4 and 8, hmax = 2.For weight 6, hmax = 3. Now we can use equations (16.37) and (16.39) to find the CWEF’s as follows:

A2(Z) = c[3, 1]A(1)2,3(Z) + c[5, 1]A(1)

2,5(Z) + c[7, 1]A(1)2,7(Z) + c[9, 1]A(1)

2,9(Z)

= 3Z2 + 7Z3 + 5Z4 + Z6

A4(Z) = c[4, 1]A(1)4,4(Z) + c[5, 1]A(1)

4,5(Z) + c[6, 1]A(1)4,6(Z) + c[7, 1]A(1)

4,7(Z) + c[8, 1]A(1)4,8(Z) +

c[9, 1]A(1)4,9(Z) + c[6, 2]A(2)

4,6(Z) + c[8, 2]A(2)4.8(Z)

= 6Z2 + 13Z3 + 16Z4 + 10Z5 + 14Z6 + 7Z7

A6(Z) = c[7, 1]A(1)6,7(Z) + c[8, 1]A(1)

6,8(Z) + c[9, 1]A(1)6,9(Z) + C[6, 2]A(2)

6,7(Z) + c[8, 2]A(2)6,8(Z) +

c[9, 2]A(2)6,9(Z) + c[9, 3]A(3)

6,9(Z)

= 3Z2 + 7Z3 + 3Z4 + 12Z5 + 12Z6 + 2Z7 + Z9

A8(Z) = c[8, 2]A(2)8,8(Z) + c[9, 2]A(2)

8,9(Z)

= 3Z4 + 2Z5

A quick calculation shows that the CWEF’s include a total of 127 nonzero codewords, the correctnumber for an (18,7) code.

The IRWEF and WEF are given by:

A(W,Z) = W 2(3Z2 + 7Z3 + 5Z4 + Z6) + W 4(6Z2 + 13Z3 + 16Z4 + 10Z5 + 14Z6 + 7Z7) +W 6(3Z2 + 7Z3 + 3Z4 + 12Z5 + 12Z6 + 2Z7 + Z9) + W 8(3Z4 + 2Z5)

A(X) = 3X4 + 7X5 + 11X6 + 13X7 + 20X8 + 17X9 + 17X10 + 20X11 + 15Z12 + 4X13 + X15

This convolutional code has minimum distance 4, one less than for the code of Example 6.6. As in theexample, we must modify the concept of the uniform interleaver because we are using convolutionalconstituent codes. Therefore, note that for weight 2, there are 16 valid input sequences. For weight 4,there are 66. For weight 6, there are 40, and for weight 8, there are 5. For all other weights, there areno valid input sequences.

11

The CWEF’s of the (27,7) PCCC can be found as follows:

APC2 (Z) = (3Z2 + 7Z3 + 5Z4 + Z6)2/16

= 0.56Z4 + 2.62Z5 + 4.94Z6 + 4.37Z7 + 1.94Z8 + 0.87Z9 + 0.62Z10 + 0.06Z12

APC4 (Z) = (6Z2 + 13Z3 + 16Z4 + 10Z5 + 14Z6 + 7Z7)2/66

= 0.55Z4 + 2.36Z5 + 5.47Z6 + 8.12Z7 + 10.36Z8 + 11.63Z9 + 11.06Z10 + 7.65Z11

5.09Z12 + 2.97Z13 + 0.74Z14

APC6 (Z) = (3Z2 + 7Z3 + 3Z4 + 12Z5 + 12Z6 + 2Z7 + Z9)2/40

= 0.22Z4 + 1.05Z5 + 1.67Z6 + 2.85Z7 + 6.22Z8 + 6.3Z9 + 6.1Z10 + 7.65Z11 +5.15Z12 + 1.35Z13 + 0.7Z14 + 0.6Z15 + 0.1Z16 + 0.02Z18

APC8 (Z) = (3Z4 + 2Z5)2/5

= 1.8Z8 + 2.4Z9 + 0.8Z10

16.11 As noted in Problem 16.10, odd input weights do not appear. For weight 2, hmax = 1 and equation(16.55) simplifies to

APC2 (Z) = 2[A(1)

2 (Z)]2 and BPC2 (Z) =

4K

[A(1)2 (Z)]2.

For weight 4 and 8, hmax = 2, so we have

APC4 (Z) = 6[A(2)

4 (Z)]2 and BPC4 (Z) =

24K

[A(2)4 (Z)]2.

APC8 (Z) =

10080K4

[A(2)8 (Z)]2 and BPC

4 (Z) =80640K5

[A(2)8 (Z)]2.

For weight 6, hmax = 3, so

APC6 (Z) = 20[A(3)

6 (Z)]2 and BPC6 (Z) =

120K

[A(3)6 (Z)]2.

Including only these terms in the approximate IRWEF’s for the PCCC gives

A(W,Z) = 2W 2[A2(Z)]2 + 6W 4[A4(Z)]2 + 20W 6[A6(Z)]2 +10080W 8

K4[A8(Z)]2

≈ 2W 2[A2(Z)]2 + 6W 4[A4(Z)]2 + 20W 6[A6(Z)]2

B(W,Z) =4W 2

K[A2(Z)]2 +

24W 4

K[A4(Z)]2 +

120W 6

K[A6(Z)]2 +

80640W 8

K5[A8(Z)]2

≈ 4W 2

K[A2(Z)]2 +

24W 4

K[A4(Z)]2 +

120W 6

K[A6(Z)]2.

12

The approximate CWEF’s from the previous problem are:

A2(Z) ≈ 3Z2 + 7Z3 + 5Z4 + Z6

A4(Z) ≈ 6Z2 + 13Z3 + 16Z4 + 10Z5 + 14Z6 + 7Z7

A6(Z) ≈ 3Z2 + 7Z3 + 3Z4 + 12Z5 + 12Z6 + 2Z7 + Z9

A8(Z) ≈ 3Z4 + 2Z5.

Now we obtain the approximate IRWEF’s as:

A(W,Z) ≈ 2W 2Z4(3 + 7Z + 5Z2 + Z4)2 + +6W 4Z4(6 + 13Z + 16Z2 + 10Z3 + 14Z4 + 7Z5)2 +20W 6Z4(3 + 7Z + 3Z2 + 12Z3 + 12Z4 + 2Z5 + Z7)2

B(W,Z) ≈ (4W 2Z4(3 + 7Z + 5Z2 + Z4)2 + 24W 4Z4(6 + 13Z + 16Z2 + 10Z3 + 14Z4 + 7Z5)2 +120W 6Z4(3 + 7Z + 3Z2 + 12Z3 + 12Z4 + 2Z5 + Z7)2)/K

and the approximate WEF’s are given by:

APC(X) ≈ 18X6 + 84X7 + 374X8 + 1076X9 + 2408X10 + 4084X11 + 5464X12 + 6888X13 +9362X14 + 8064X15 + 6896X16 + 7296X17 + 4414X18 + 1080X19 + 560X20 +480X21 + 80X22 + 20X24

BPC(X) ≈ 36X6/K + 168X7/K + 1180X8/K + 4024X9/K + 9868X10/K + 17960X11/K +24496X12/K + 32112X13/K + 47404X14/K + 42336X15/K + 37344X16/K +41424X17/K + 25896X18/K + 6480X19/K + 3360X20/K + 2880X21/K +480X22/K + 120X24/K

16.13 First constituent encoder: G(1)ff (D) = [1 + D + D2 1 + D2]

Second constituent encoder: G(2)ff (D) = [1 1 + D + D2]

For the first constituent encoder, the IOWEF is given by

AC1(W,X, L) = WX5L3 + W 2L4(1 + L)X6 + W 3L5(1 + L)2X7 + . . . ,

and for the second one the IRWEF can be computed as

AC2(W,Z,L) = WZ3L3 + W 2Z2L4 + W 2Z4L5 + . . . ,

The effect of uniform interleaving is represented by

APCw (X, Z) =

AC1w (X) ∗AC2

w (Z)(Kw

)

13

and for large K, this can be approximated as

APCw (X, Z) ≈

∑

1≤ h1≤ hmax

∑

1≤ h2≤ hmax

c[h1]c[h2](Kw

) A(h1)w (X)A(h2)

w (Z),

where A(h1)w (X) and A

(h2)w (Z) are the conditional IO and IR h-error event WEF’s of the nonsystematic

and systematic constituent encoders, C1 and C2, respectively. With further approximations, we obtain

APCw (X, Z) ≈ w!

h1max!h2max!K(h1max+h2max−w)A(h1max)

w (X)A(h2max)w (Z)

and

BPCw (X, Z) ≈ w

KAPC

w (X,Z).

Going directly to the large K case, for any input weight w, w ≥ 1, hmax = w, since the weight 1 singleerror event can be repeated w times. It follows that

A(w)w (X) = X5w, w ≥ 1

and

A(w)w (Z) = Z3w, w ≥ 1.

Hence

APCw (X, Z) ≈ w!

w!w!K(w+w−w)X5wZ3w =

Kw

w!X5wZ3w

and

BPCw (X,Z) ≈ K(w−1)

(w − 1)!X5wZ3w.

Therefore

APC(W,X) ≈∑1≤w≤KWw Kw

w!X8w = KWX8 +

K

2

2

W 2X16 +K

6

3

W 3X24 − . . . ,

BPC(W,X) ≈ WX8 + KW 2X16 +K2

2W 3X24 + . . . ,

the average WEF’s are

14

APC(X) = KX8 +K2

2X16 +

K3

6X24 + . . . ,

BPC(X) = X8 + KX16 +K2

2X24 + . . . ,

and the free distance of the code is 8.

16.14 The encoder and augmented modified state diagram for this problem are shown below:

S0 S2 S0

S3

S1

LWZ

LW

LWZLWL

LZ

LZ

(a) (b)

(a) Encoder and (b) Augmented Modified State Diagram for G(D) = 1+D+D2

1+D

Note that in order to leave from and return to state S0 (i.e., terminating a K − 2 bit input sequence),an even overall-input (including termination bits) weight is required. Examination of the state diagramreveals 6 paths that contain overall-input weight W < 6 (“· · ·” means an indefinite loop around S3):

Path Function{S0, S1, S2, S0} L3W 2Z3

{S0, S1, S2, S1, S2, S0} L5W 4Z4

{S0, S1, S3, · · · , S3, S2, S0} L4W 2Z2

1−ZL

{S0, S1, S2, S1, S3, · · · , S3, S2, S0} L6W 4Z3

1−ZL

{S0, S1, S3, · · · , S3, S2, S1, S2, S0} L6W 4Z3

1−ZL

{S0, S1, S3, · · · , S3, S2, S1, S3, · · · , S3, S2, S0} L7W 4Z2

(1−ZL)2

resulting in the single error event IRWEF for W < 6 of

A(1)(W,Z, L) = L3W 2Z3 + L5W 4Z4 +L4W 2Z2 + 2L6W 4Z3

1− ZL+

L7W 4Z2

(1− ZL)2,

and for both W < 6 and L < 10,

15

A(1)(W,Z, L) = L3W 2Z3 + L5W 4Z4 +

[5∑

n=0

L4+nW 2Z2+n

]+

[2

3∑n=0

L6+nW 4Z3+n

]+

[2∑

n=0

(n + 1)L7+nW 4Z2+n

].

The IRWEF for double error events for W < 6 is then

A(2)(W,Z,L) = [A(1)(W,Z, L)]2 = L6W 4Z6 +2L7W 4Z5

1− ZL+

L8W 4Z4

(1− ZL)2,

and for both W < 6 and L < 10,

A(2)(W,Z,L) = L6W 4Z6 +

[2

2∑n=0

L7+nW 4Z5+n

]+

[1∑

n=0

(n + 1)L8+nW 4Z4+n

].

Thus the single- and double-error event enumerators for W < 6 and L < 10 are

A(1)2,3(Z) = Z3 A

(1)2,4+n(Z) = Z2+n

A(1)4,5(Z) = Z4 A

(1)4,6+n(Z) = 2Z3+n A

(1)4,7(Z) = Z2 A

(1)4,8(Z) = 2Z3 A

(1)4,9(Z) = 3Z4

A(2)4,6(Z) = Z6 A

(2)4,7+n(Z) = 2Z5+n A

(2)4,8(Z) = Z4 A

(2)4,9(Z) = 2Z5,

and thus, up to Z7,

A(1)2 (Z) = Z3 +

Z2

1− Z= Z3 +

K−4∑n=0

Z2+n = Z2 + 2Z3 + Z4 + Z5 + Z6 + Z7 + · · ·+ Zn + · · ·

A(2)4 (Z) = Z6 +

2Z5

1− Z+

Z4

(1− Z)2= Z6 + 2

K−7∑n=0

Z5+n +K−8∑n=0

(n + 1)Z4+n

= Z4 + 4Z5 + 6Z6 + 6Z7 + · · ·+ (n− 1)Zn + · · · .

The multiple turbo code, as shown in Figure 16.2, contains 3 encoders, and it is assumed that theinterleavers are uniformly distributed, the block size K is large, and the code uses the same constituentencoders. Assume that a particular input sequence of weight w enters the first encoder, generatingthe parity weight enumerator APC

w (Z). Then, by the definition of the uniform interleaver, the secondand third encoders will generate any of the weights in APC

w (Z) with equal probability. The codewordCWEF of this (4K, K − 2) terminated multiple turbo code is given by

APCw (Z) =

[hmax∑

h1=1

c[h1, w]A(h1)w (Z)

][hmax∑

h2=1

c[h2, w](

Kw

)−1

A(h2)w (Z)

][hmax∑

h3=1

c[h3, w](

Kw

)−1

A(h3)w (Z)

]

=hmax∑

h1=1

hmax∑

h2=1

hmax∑

h3=1

(Kw

)−2

c[h1, w] c[h2, w] c[h3, w] A(h1)w (Z) A(h2)

w (Z) A(h3)w (Z) ,

16

where hmax is the maximum number of error events for the particular choice of w and c[h, w] =(K − h + w

w

). With K À h and K À w, c[h,w] ≈

(Kh

)≈ Kh

h! , and hmax = bw2 c. These

approximations, along with keeping only the highest power of K, i.e., the term corresponding toh1 = h2 = h3 = hmax, result in the codeword CWEF

APCw (Z) ≈ (w!)2

(hmax!)3K(3hmax−2w)

[A(hmax)

w (Z)]3

and the bit CWEF

BPCw (Z) =

w

KAPC

w (Z) ≈ w (w!)2

(hmax!)3K(3hmax−2w−1)

[A(hmax)

w (Z)]3

.

Then, for w < 6, the approximate IRWEF’s for this PCCC are

APC(W,Z) ≈5∑

w=2

WwAPCw (Z) and BPC(W,Z) ≈

5∑w=2

WwBPCw (Z),

and the WEF’s are APC(X) = APC(X, X) and BPC(X) = BPC(X, X). From the above,[A

(1)2 (Z)

]3

= Z9 +3Z8

1− Z+

3Z7

(1− Z)2+

Z6

(1− Z)3

= Z6 + 6Z7 + 15Z8 + 23Z9 + 30Z10 + 39Z11 + · · ·+(

n2 − 3n− 102

)Zn + · · ·

and[A

(2)4 (Z)

]3

= Z18 +6Z17

1− Z+

15Z16

(1− Z)2+

20Z15

(1− Z)3+

16Z14

(1− Z)4+

6Z13

(1− Z)5+

Z12

(1− Z)6

= Z12 + 12Z13 + 66Z14 + 226Z15 + 561Z16 + 1128Z17 + 1995Z18 + 3258Z19 +

5028Z20 + · · ·+(−21720− 7046n + 3015n2 − 155n3 − 15n4 + n5

120

)Zn + · · · .

(a) Using the above, the approximate CWEF’s up to Z14 are

APC2 (Z) ≈ 4

K

[A

(1)2 (Z)

]3

= 4K Z6 + 24

K Z7 + 60K Z8 + 92

K Z9 + 120K Z10 + 156

K Z11+196K Z12 + 240

K Z13 + 288K Z14 + · · ·+ 2

(n2−3n−10

K

)Zn + · · ·

APC4 (Z) ≈ 72

K2

[A

(2)4 (Z)

]3

≈ 72K2 Z12 + 864

K2 Z13 + 4752K2 Z14 + · · ·

APC3 (Z) = APC

5 (Z) = 0

BPC2 (Z) = 2

K

[APC

2 (Z)] ≈ 8

K2 Z6 + 48K2 Z7 + 120

K2 Z8 + 184K2 Z9 + 240

K2 Z10 + 312K2 Z11+

392K2 Z12 + 480

K2 Z13 + 576K2 Z14 + · · ·+ 4

(n2−3n−10

K2

)Zn + · · ·

BPC4 (Z) = 4

K

[APC

4 (Z)] ≈ 288

K3 Z12 + 3456K3 Z13 + 19008

K3 Z14 + · · ·BPC

3 (Z) = BPC5 (Z) = 0

17

(b) Using the above, the approximate IRWEF’s up to W 5 and Z14 are

APC(W,Z) ≈ W 2[

4K Z6 + 24

K Z7 + 60K Z8 + 92

K Z9 + 120K Z10 + 156

K Z11+196K Z12 + 240

K Z13 + 288K Z14 + · · ·+ 2

(n2−3n−10

K

)Zn + · · ·

]+

W 4[

72K2 Z12 + 864

K2 Z13 + 4752K2 Z14 + · · ·]

andBPC(W,Z) ≈ W 2

[8

K2 Z6 + 48K2 Z7 + 120

K2 Z8 + 184K2 Z9 + 240

K2 Z10 + 312K2 Z11+

392K2 Z12 + 480

K2 Z13 + 576K2 Z14 + · · ·+ 4

(n2−3n−10

K2

)Zn + · · ·

]+

W 4[288K3 Z12 + 3456

K3 Z13 + 19008K3 Z14 + · · ·]

(c) Using the above, the approximate WEF’s up to X14, neglecting any higher power of K terms, are

APC(X) ≈ 4K

X8 +24K

X9 +60K

X10 +92K

X11 +120K

X12 +156K

X13 +196K

X14 + · · ·

andBPC(X) ≈ 8

K2X8 +

48K2

X9 +120K2

X10 +184K2

X11 +240K2

X12 +312K2

X13 +392K2

X14 + · · ·

(d) The union bounds on the BER and WER for K = 103 and K = 104, assuming a binary input,unquantized output AWGN channel are shown below. The plots were created using the loose bounds

WER ≤ APC(X)∣∣∣∣X=e

−R EbN0

and BER ≤ BPC(X)∣∣∣∣X=e

−R EbN0

where R = 1/4 is the code rate and Eb

N0

is the SNR.

10-2

10-3

10-1

10-4

10-5

10-6

10-7

1 2 3 4 5 6 7

- - - - - N = 1000 ______ N = 10000

Eb/N0 (dB)

(a) Word Error Rate

18

Eb/N0 (dB)

(b) Bit Error Rate

10-5

10-6

10-4

10-7

10-8

10-9

10-10

1 2 3 4 5 6 7

- - - - - N = 1000 ______ N = 10000

16.15 (a)[1 1

1+D

]

i. Weight 2Minimum parity weight generating information sequence = 1 + D.Corresponding parity weight = 1.In a PCCC (rate 1/3) we therefore have (assuming a similar input after interleaving):Information weight = 2; Parity weight = 1 + 1 = 2; Total weight = 4.

ii. Weight 3For this encoder a weight 3 input does not terminate the encoder and hence is not possible.

(b)[1 1+D2

1+D+D2

]

i. Weight 2Minimum parity weight generating information sequence = 1 + D3.Corresponding parity weight = 4.In a PCCC we therefore have:Information weight = 2; Parity weight = 4 + 4 = 8; Total weight = 10.

ii. Weight 3Minimum parity weight generating information sequence = 1 + D + D2.Corresponding parity weight = 2.In a PCCC we therefore have:Information weight = 3; Parity weight = 2 + 2 = 4; Total weight = 7.

(c)[1 1+D2+D3

1+D+D3

]

i. Weight 2Minimum parity weight generating information sequence = 1 + D7.Corresponding parity weight = 6.In a PCCC we therefore have:Information weight = 2; Parity weight = 6 + 6 = 12; Total weight = 14.

19

ii. Weight 3Minimum parity weight generating information sequence = 1 + D + D3.Corresponding parity weight = 4.In a PCCC we therefore have:Information weight = 3; Parity weight = 4 + 4 = 8; Total weight = 11.

(d)[1 1+D+D2+D4

1+D3+D4

]

i. Weight 2Minimum parity weight generating information sequence = 1 + D15.Corresponding parity weight = 10.In a PCCC we therefore have:Information weight = 2; Parity weight = 10 + 10 = 20; Total weight = 22.

ii. Weight 3Minimum parity weight generating information sequence = 1 + D3 + D4.Corresponding parity weight = 4.In a PCCC we therefore have:Information weight = 3; Parity weight = 4 + 4 = 8; Total weight = 11.

(e)[1 1+D4

1+D+D2+D3+D4

]

i. Weight 2Minimum parity weight generating information sequence = 1 + D5.Corresponding parity weight = 4.In a PCCC we therefore have:Information weight = 2; Parity weight = 4 + 4 = 8; Total weight = 10.

ii. Weight 3For this encoder a weight 3 input does not terminate the encoder and hence is not possible.

In all cases the input weight two sequence gives the free distance codeword for large K.

16.16 For a systematic feedforward encoder, for input weight w, the single error event for input weight 1 canoccur w times. Therefore hmax, the largest number of error events associated with a weight w inputsequence, is equal to w. That is, the maximum number of error events produced by a weight w inputsequence occurs by repeating the single error event for input weight 1 w times. Thus,

A(w)w (Z) = [A(1)

1 (Z)]w.

This is not true for feedback encoders. Feedback encoders require at least a weight 2 input sequenceto terminate. Thus, for an input sequence of weight 2w, hmax = w, i.e., we have a maximum of w errorevents. This occurs when an error event caused by a weight 2 input sequence is repeated w times (fora total input weight of 2w).

Therefore

A2w(w)(Z) = [AZ

(1)(Z)]w.

20

m0 m1 m2

u v(())

V(2)

V(n-1)

V(1)

m-1

16.19 An (n, 1, v) systematic feedback encoder can be realized by the following diagram:

In the figure, the dashed lines indicate potential connections that depend on the encoder realization.It can be seen from the encoder structure that the register contents are updated as

mi,t = mi−1,t−1 , i = 1, 2, . . . , v − 1

and

m0,t = mv−1,t−1 + ut +v−2∑

i=0

aimi,t−1,

where t is the time index, ut represents the input bit at time t, and the coefficients ai depend on theparticular encoder realization. When the encoder is in state S2v−1 = (0 0 . . . 1) at time t− 1, it followsfrom the given relations that

mi,t = mi−1,t−1 = 0, i = 1, 2, . . . , v − 1

and

m0,t = mv−1,t−1 + ut +v−2∑

i=0

aimi,t−1 = 1 + ut.

Thus if the input bit ut = 1, the first memory position at time t is also 0, and the encoder is in theall-zero state.

16.20 For the primitive encoder D with Gp(D) =[1 D4+D2+D+1

D4+D3+1

], the shortest terminating weight 2 input

sequence is

u(D) = 1 + D24−1 = 1 + D15,

21

which generates the parity sequence

v(D) = (1 + D3 + D4 + D6 + D8 + D9 + D10 + D11)(D4 + D2 + D + 1)= 1 + D + D2 + D3 + D4 + D8 + D11 + D12 + D14 + D15.

Thus

Zmin = 10

and

deff = 2 + 2 Zmin = 2 + 2× 10 = 22.

For the nonprimitive encoder E with Gnp(D) =[1 D4+1

D4+D3+D2+D+1

], the shortest terminating weight

2 input sequence is u(D) = 1 + D5, which generates the parity sequence

v(D) = (D + 1)(D4 + 1) = 1 + D + D4 + D5.

Thus Zmin = 4 and deff = 2 + 2Zmin = 10.

Chapter 20

20.1 If an (n, k) codeC is designed to correct all the bursts of lengthl or less and simultaneously to

detect all the bursts of lengthd or less withd ≥ l, then no burst of lengthl + d or less can be

a codeword. Suppose there is a burstx of lengthl + d or less that is a codeword inC. We can

decompose this burstx into two burstsy andz such that: (1)x = y + z; (2) the length ofy

is l or less and the length ofz is d or less. Since the code is designed to correct all the bursts

of lengthl or less,y must be a coset leader in a standard array for the code. Sincey + z = x

is codeword, thenz must be in the same coset asy. As a results,z has the same syndrome as

y. In this case, ifz occurs as an error burst during the transmission of a codeword inC, then

y will be taken as the error burst for correction. This will result in an incorrect decoding and

hence the decoder will fail to detectz which contradicts to the fact that the code is designed

to correct all the bursts of lengthsl or less and simultaneously to detect all the bursts of length

d or less withd ≥ l. Therefore, a burst of lengthl + d or less can not be a codeword. Then it

follows from Theorem20.2 thatn− k ≥ l + d.

20.2 An error-trapping decoder is shown in Figure P20.2.

Gate 1 Buffer Register Gate 2 +

Gate 4Gate 3 +

Syndrome Register

. . .

Test All 0’s

. . .l stages

Figure P20.2 An error-trapping decoder.

The decoding operation is given below:

1

Step 1.The received polynomialr(x) is shifted into the buffer and syndrome registers simulta-

neously with Gates 1 and 3 turned on and all the other gates turned off.

Step 2.As soon as the entirer(x) has been shifted into the syndrome register, the contents of the

syndrome form the syndromeS(n−k)(x) of r(n−k)(x). If the n− k− l leftmost stages of

the syndrome register contain all zeros, the error burst is confined to the positionsXn−l,

. . ., Xn− 1. Gate 2 and Gate 4 are activated. The received information digits are read

out of the buffer register and connected by the error digits shifted out from the syndrome

register. Ifn− k − l leftmost stages of the syndrome register does not contain all zeros,

go to step 3.

Step 3.The syndrome register starts to shift with Gate 3 on. As soon as itsn − k − l leftmost

stages contain only zeros, itsl rightmost stages contain the burst-error pattern. The error

correction begins. there are three cases to be considered.

Step 4. If the n− k − l leftmost stages of the syndrome register contains all zeros afterith shift

0 < i ≤ 2k−n (Assume2k−n ≥ 0, otherwise, no this step), the error burst is confined

to positionsXn−i−1−l, . . ., Xn−i−1. In this event, Gate 2 is first activated and thei high

position information digits are shifted out. Then Gate 4 is activated, the rest information

digits are read out and corrected by the error digit shifted out from the syndrome register.

Step 5. If the n − k − l leftmost stages of the syndrome register contains all zeros after theith

shift k ≤ i ≤ n− l, the errors of the burste(x) are confined to the parity-check positions

of r(x). In this event, thek received information digits in the buffer are error free. Gate

2 is activated and thek error-free information digits in the buffer are shifted out to the

data sink.

Step 6. if then−k−l leftmost stages of the syndrome register contains all zeros after(n−l+i)th

shift 0 < i < l, the error burst is confined to positionsXn−i, . . ., Xn−1, X0, X l−i−1. In

this event, thel− i digits contained in thel− i rightmost stages of the syndrome register

match the errors at the parity-check positions ofr(X), and thei digits contained int the

next i stages of the syndrome register match the errors at the positionsXn−i, . . ., Xn−1

of r(X). At this instant, a clock starts to count from(n − l + i + 1). The syndrome

register is then shifted with Gate 3 turned off. As soon as the clock has counted up ton,

thei rightmost digits in the syndrome register match the errors at the positionsXn−i, . . .,

Xn−l, Xn−1 of r(X). Gate 4 and Gate 2 are then activated. The received information

digits are read out of the buffer register and corrected by the error digits shifted out from

2

the syndrome register.

If the n − k − l stages of the syndrome never contain all zeros after all the steps above,

an uncorrectable burst of errors has been detected.

20.4 The generator polynomial of the desired Fire code is

g(X) = (X2l−1 + 1)p(X)

Since the code is designed to correct all the bursts of length4 or less,l = 4. Hence

g(X) = (X7 + 1)(1 + X + X4) = 1 + X + X4 + X7 + X8 + X11.

Sincep(X) is a primitive polynomial of degree4, its periodρ is 24 − 1 = 15. Hence the

length of the code isn = LCM(7, 15) = 105. Since the degree of the generator polynomial

of the code is11 which is the number of parity check digits of the code, the code is a(105, 94)

Fire code. The decoder of this code is shown in Figure P20.4.

+ + + ++

Test for all 0’s

Buffer Register

Gate 1

Gate 4 +

Gate 3

Gate 2

OutputInput r(x)

1 X X4 X7 X8 X11

Figure P20.4 A decoder for the(105, 94) Fire code.

3

20.5 The high-speed decoder for the(105, 94) Fire code designed in Problem20.4 is shown in

Figure P20.5.

+

Error Pattern Register

Comparator test for a match Test for all 0’s

++

Buffer Register

r(X) Input

Error Location RegisterComputation

for n-q

Counter 1

Counter 2

Figure P20.5 A high-speed decoder for the(105, 94) Fire code.

20.6 We choose the second code given in Table20.3 which is a(15, 9) code and capable of correct-

ing any burst of length3 or less. The burst-error-correcting efficiencyz = 1. The generator

polynomial of this code isg(X) = 1+X3 +X4 +X5 +X6. Suppose we interleave this code

by a degreeλ = 17. This results in a(255, 153) cyclic code that is capable of correcting any

single burst of length51 or less and has burst-error-correcting efficiencyz = 1. The generator

of this code is

g′(X) = g(X17) = 1 + X51 + X68 + X85 + X102.

A decoder for this code can be implemented with a de-interleaver and the decoder for the

(15, 9) code as shown in Figure P20.6. First the received sequence is de-interleaved and

arranged as a17 × 15 array. Then each row of the array is decoded based on the base(15, 9)

code.

De-interleaver Buffer

Decoder C

Figure P20.6 A decoder for the(255, 153) burst-error-correcting code designed above.

20.7 A codeword in the interleaved codeClλ is obtained by interleavingl codewords from the(n, k)

base codeC. Let v1(X), v2(X), . . . , vl(X) be l codewords inC. Then the codeword inClλ

4

obtained by interleaving thesel codewords inC in polynomial form is

v(X) = v1(Xl) + Xv2(X

l) + X2v3(Xl) + . . . + X l−1vl(X

l).

Sincevi(X) is divisible byg(X) for 1 ≤ i ≤ l, vi(Xl) is divisible byg(X l). Thereforev(X)

is divisible byg(X l). The degree ofg(X l) is (n − k)l. Sinceg(X l) is a factor ofX ln + 1,

g(X l) generates a(ln, lk) cyclic code.

20.8 In order to show that the Burton code is capable of correcting all phased bursts confined to

a single subblock ofm digits, it is necessary and sufficient to prove that no two such bursts

are in the same coset of a standard array for the code. LetXmiA(X) andXmjB(X) be the

polynomial representations of two bursts of lengthl1 andl2 with l1 ≤ m andl2 ≤ m confined

to ith andjth subblocks(0 ≤ i ≤ σ, 0 ≤ j ≤ σ), respectively, where

A(X) = 1 + a1X + . . . + al1−2Xl1−2 + X l1−1,

and

B(X) = 1 + b1X + . . . + bl2−2Xl2−2 + X l2−1.

SupposeXmiA(X) and XmjB(X) are in the same coset. ThenV(X) = XmiA(X) +

XmjB(X) must be a code polynomial and divisible by the generator polynomialg(X) =

(Xm + 1)p(X). Assumemj ≥ mi, thenmj −mi = qm and

V(X) = Xmi(A(X) + B(X)) + XmiB(X)(Xqm + 1).

SinceXm − 1 is a factor of the generatorg(X), V(X) must be divisible byXm − 1. Note

that Xqm + 1 is divisible by Xm + 1, andXmi and Xm + 1 are relatively prime. Since

V(X) andXmiB(X)(Xqm +1) are divisible byXm +1, A(X)+B(X) must be divisible by

Xm + 1. However both degreesl1 andl2 of A(X) andB(X) are less thanm, the degree of

A(X)+B(X) is less thanm. HenceA(X)+B(X) is not divisible byXm +1. ForV(X) to

be divisible byXm + 1, A(X) andB(X) must be identical, i.e.,A(X) = B(X). Therefore,

V(X) = XmiB(X)(Xqm + 1).

5

Since the degree ofB(X) is less thanm, B(X) andp(X) must be relatively prime. AlsoXmi

andp(X) are relatively prime. SinceV(X) is assumed to be a code polynomial,Xqm + 1

must be divisible by bothp(X) and Xm − 1. This implies thatXqm + 1 is divisible by

(Xm + 1)p(X) andqm must be a multiple ofn = σm. This is not possible, sinceq = j − i

is less thanσ. ThereforeV(X) can not be a code polynomial and the burstsXmiA(X) and

XmiB(X) can not be in the same coset. Consequently, all the phased bursts confined to a

single subblock ofm digits are in different cosets of a standard array for the Burton code

generated by(Xm + 1)p(X) and they are correctable.

20.9 Choosep(X) = 1 + X2 + X5 which is a primitive polynomial with periodρ = 25 − 1 = 31.

Then generator polynomial of the Burton codeC with m = 5 is

g(X) = (Xm + 1)p(X) = 1 + X2 + X7 + X10.

The length of the code isn = LCM(31, 5) = 155. Hence the code is a(155, 145) code

that is capable of correcting any phased burst confined to a single subblock of5 digits. If the

code is interleaved to a degreeλ = 6, we obtain a(930, 870) codeC6 generated byg(X6) =

1 + X12 + X42 + X60. This code is capable of correcting any burst of length5× 5 + 1 = 26

or less.

20.10 From Table9.3, code(164, 153) has burst-correcting-capabilityl = 4. If it is interleaved to

degreeλ = 6, thenλn = 6 × 164 = 984, λk = 6 × 153 = 918, andλl = 6 × 4 = 24. The

interleaved code is a(984, 918) code with burst-error-correcting capability24.

The interleaved Burton code of Problem 20.9 is a(930, 870) code with burst-error-correcting

capability 26. The burst-error-correcting efficiency of the(984, 918) code is

z =2λl

λn− λk=

48

984− 918=

8

11.

The efficiency of the interleave Burton code of Problem 20.9 is

z′ =2[(λ− 1)m + 1]

2λm=

52

60=

13

15.

Therefore,z′ ≥ z. The interleaved Burton code is more powerful and efficient.

6

20.11 The (15, 7) BCH code has random error-correcting capability 2. Interleaving this code to a

degree 7, we obtain a(105, 49) cyclic code that is capable of correcting any combination of

two random bursts, each of length 7 or less. Of course, it is capable of correcting any single

burst of length 14 or less. In decoding, a received sequence is de-interleaved and arranged into

a7× 15 array. Each row of the array is decoded based on the(15, 7) BCH code.

20.12 If each symbol of the(31, 15) RS code over GF(25) is replaced by its corresponding 5-bit

byte, we obtain a(155, 75) binary code. Since the(31, 15) RS code is capable of correcting

8 or fewer random symbol errors, the binary(155, 75) code is capable of correcting: (1) any

single burst of length(8− 1)× 5 + 1 = 36 or less; or (2) correcting any combination random

bursts that result in 8 or fewer random symbol errors for the(31, 15) RS code.

20.13 From Problem 20.4, the generator polynomial

g(X) = 1 + X + X4 + X7 + X8 + X11

generates a(105, 94) Fire code. If it is shortened by deleting the 15 high-order message digits,

it becomes a(90, 79) code.

Dividing Xn−k+15 = X26 by g(X), we obtain a remainderρ(X) = X10+X8+X5+X3+X.

Then a decoder for the(105, 94) shortened Fire code is shown in Figure P20.13

Test for all 0’s

Gate 1

Buffer Register

Gate 2

Gate 4

Gate 3

Output

1 x

x x3

x4

x5

x7

x8

x8

x11

Figure P20.13 A decoder for the(90, 79) Fire code.

20.14 Let α be a primitive element of the Galois field GF(26). The order ofα is 26− 1 = 63 and the

minimal polynomial ofα is

Φ(X) = 1 + X + X6.

7

Since63 = 7× 9, we can factorX63 + 1 as follows:

X63 + 1 = (X7 + 1)(1 + X7 + X14 + X21 + X28 + X35 + X42 + X49 + X56).

The code generated by the polynomial

g1(X) = (X7 + 1)(1 + X + X6)

is a Fire code of length 63 which is capable of correcting any single error burst of length4

or less. Letg2(X) be the generator polynomial of the double-error-correcting BCH code of

length 63. From Table 6.4, we find that

g2(X) = (1 + X + X6)(1 + X + X2 + X4 + X6).

The least common multiple ofg1(X) andg2(X) is

g(X) = (1 + X7)(1 + X + X6)(1 + X + X2 + X4 + X6).

Hence,g(X) generates a(63, 44) cyclic code which is a modified Fire code and is capable of

correcting any single error burst of length 4 or less as well as any combination of two or fewer

random errors.

8