Generation of spectral nulls with permutation sequences

Int. J. Electron. Commun. (AEÜ) 67 (2013) 1–9

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International Journal of Electronics and Communications (AEÜ) journal homepage: www.elsevier.com/locate/aeue

Generation of spectral nulls with permutation sequences夽 Khmaies Ouahada ∗ , Hendrik C. Ferreira Department of Electrical and Electronic Engineering Science, University of Johannesburg, South Africa

a r t i c l e

i n f o

Article history: Received 13 June 2011 Accepted 25 May 2012 Keywords: Spectral nulls Permutation sequence

a b s t r a c t This paper describes the construction of classes of spectral null codes, which are obtained by shaping permutation sequences. The properties and the construction techniques of binary spectral null codes are used. The power spectral densities (PSD) of the proposed codes show nulls at the zero frequency and at submultiples of the frequencies of the transmitted symbols. The channel symbols or voltage levels mapped to each transmitted permutation symbol are chosen to satisfy the spectral null equation. Interesting results show the elimination of the DC component at the lower frequencies. The well-shaped power spectral densities of these codes may overcome some communications problem like zero frequency components. The proposed technique is generalized and the corresponding spectral null permutation sequences that generate spectral null codes are presented. © 2012 Elsevier GmbH. All rights reserved.

1. Introduction Spectral null codes are codes with simultaneous nulls and have shown great importance in certain applications like in the case of transmission systems employing pilot tones for synchronization and that of track-following servos in digital recording [1,2]. The techniques of designing binary codes to have a spectrum with nulls occurring at certain frequencies were extended to multilevel alphabets by Gorog [3]. Tallini and Vaccaro [4] presented efficient methods, better than other constructions previously published, to design m-ary balanced codes for all alphabet sizes m ≥ 2. Their basic idea is the construction of anm-ary balanced  code as a code of length n over the alphabet Zm = 1, 2, . . . , m with m ∈ N such that each codeword is balanced when the real sum of its components (or weights) is equal to (m− 1)n/2 . The m-ary balanced codes have many applications [5–7] where the case of m = 2 is considered as a special case as it presents the binary form of these sequences. The definitions presented in [4] for balanced codes coincide with definitions in other published works [8–11]. Mascella and Tallini [12,13] have defined a symbol permutation invariant balanced code over the m-ary alphabet Zm = {1, 2, . . . , m} with m ∈ N as a block code over Zm such that each alphabet symbol occurs as many times as any other symbol in every codeword. Earlier

夽 This paper was presented in part to the IEEE Information Theory Workshop, Paraty, Brazil, 2011. ∗ Corresponding author. E-mail addresses: [email protected] (K. Ouahada), [email protected] (H.C. Ferreira). 1434-8411/$ – see front matter © 2012 Elsevier GmbH. All rights reserved. http://dx.doi.org/10.1016/j.aeue.2012.05.010

Tallini [14] presented an efficient recursive method to encode k information m-ary digits, with m ≥ 2, into an m-ary second-order spectral-null code. Knuth [11] presented an attractive solution for binary balanced codes which constituted a very simple algorithm for constructing bipolar sequences with an equal number of 1s and −1s. His work was explored and generalized by Alon et al. [15], in which they have given an algebraic proof of the optimality of their construction. Authors in [12,13,16] extended the results by considering balanced codes such that the alphabet is a set of vectors of Rn . In this paper we consider the design of non-binary spectral null codes based on permutation sequences of non-repetitive symbols of length M selected from the set {1, 2, . . ., M}. It is important to mention that these codes have an advantage compared to binary spectral null codes in that they are also DC free codes. Using the same concept in constructing binary spectral null codes as in [3], we consider the vector Y = (y1 , y2 , . . ., yM ), yi ∈ { −1, + 1} with 1 ≤ i ≤ M, to be an element of a set S, which is called a codebook of codewords with elements in { − 1, + 1}. We investigate codewords of length, M, as an integer multiple of N, thus let

M = Nz, where N represents the number of groupings, Ai , in the spectral null equation and z represents the number of elements in each grouping. The values of f = r/N, are the frequencies values of the nulls at the rational submultiples r/N [17] as appearing in the power spectral density of the codebook of the code. In this paper we investigate codewords of length M, where yi represents the permutation symbol’s channel level or voltage level as yi ∈ { . . . , −3, − 2, − 1, 0, + 1, + 2, + 3, . . . }. To ensure the presence

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K. Ouahada, H.C. Ferreira / Int. J. Electron. Commun. (AEÜ) 67 (2013) 1–9

of these nulls in the continuous component at the spectrum, it is sufficient to satisfy the following spectral null equation [18], A1 = A2 = · · · = AN ,

(1)

where

PS :

 z−1

Ai =

yi+N ,

i = 1, 2, . . . , N,

(2)

=0

which also can be rewritten as:



z





A1

= y1

+

y1+N

+

y1+2N

+

···

+

y1+(z−1)N ,

A2

= y2

+

y2+N

+

y2+2N

+

···

+

y2+(z−1)N ,

A3

= y3

+

y3+N

+

y3+2N

+

···

+

y3+(z−1)N ,

.. . = yN

+

y2N

+

y3N

+

···

+

yzN .

= A2 ,

x1 + x3 = x2 + x4 . The spectral null codebook is:

⎧ ⎫ 0 0 0 0⎪ ⎪ ⎪ ⎪0 0 1 1⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨0 1 1 0⎪ ⎬ . 1 0 0 1⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪1 1 0 0⎪ ⎪ ⎩ ⎭ 1

M−1 2

···

1

↓ M−1 CS : − 2

PS :

Example 1. As an example, if N = 2 and M = 4, the following relationship must hold, A1

0

↓ M−3 − 2

···

↓ ···

···

0

M−2

M−1

↓ M−3 + 2

↓ M−1 + , 2

For even values of M, the symbol mapping could be

(3)

.. .

AN

Each permutation symbol (PS) is mapped to a channel symbol (CS) [19], which could represent the voltage level of a signal. In general, for odd values of M, the symbol mapping is

1 1 1

It is important to mention that if all codewords in a codebook satisfying (1), then the codebook will exhibit nulls at the desired frequencies. This can still hold for non-binary sequences as it will later be detailed in following sections. The paper is organized as follows. Section 2 presents the technique of spectral shaping of permutation sequences and the generation of spectral null sequences. Few properties related to each case are also presented. Section 3 generalizes the design of spectral null sequences. In Section 4, magic squares related to spectral null sequences are discussed. Shaping permutation sequences with repeated symbols are discussed in Section 5. Conclusions are offered in Section 6. 2. Spectral shaping sequence The difference between a binary and a non-binary permutation sequences of same length, M, is that the binary sequences have only two symbols and the permutation sequences have M different symbols if take the case of non-repetitive symbols. Shaping a non-binary sequence may present more complexity than shaping a binary one because of the weight of each of the M symbols. This complexity is more serious in the case of spectral null codes since we have more constraints on our permutation sequences represented by the spectral null equation related to the groupings that should be taken into consideration. Our challenge in this work is to find the non-binary symbols for our permutation sequence and their corresponding channel symbols that we need to design our non-binary spectral null codes. The sequence that we start with to satisfy the spectral null equation in (1) is called SN sequence.

0

↓ M CS : − 2

1 ↓ M−2 − 2

···

M−2 2

M 2

↓

↓

−1

+1

···

···

···

M−2

M−1

↓ M−2 + 2

↓ M + . 2

From the channel levels mapping to permutation symbols, it is just a matter of placing the symbols in such a way that SN equations are satisfied. Even though these sequences can be found by trialand-error or computer searches, in this paper we aim to find a general construction approach for sequences to satisfy the spectral null equation. As we know in the standard channel symbol assignment for optimum detection performance in AWGN, symbols have uniform spacing. This is not the case for even lengths of the permutation sequences, where we can see that middle pair of channel symbols have twice the distance of all others. This is is made to simplify some of the algebra in some of the subsequent proofs and most importantly is to guarantee the generation DC-free codes. The generalized SN sequences are presented in the following propositions and definitions based on special cases related to the values of N and z. Example 2. We consider an SN sequence for M = 8 and N = 4, with channel symbols assigned as follows: PS :

0

1

2

3

4

5

6

7

↓

↓

↓

↓

↓

↓

↓

↓

−3

−2

−1

+1

+2

+3

+4.

CS : −4

From (1) and (2) we must satisfy y1 + y5 = y2 + y6 = y3 + y7 = y4 + y8 .

(4)

It is clear that each grouping has two elements with a difference of indices of a 4. This means that if yi is mapped to a negative channel symbol, the corresponding positive channel symbol must be mapped to yi+4 , as follows: PS :

0

1

2

3

7

6

5

4

↓

↓

↓

↓

↓

↓

↓

↓

CS : −4

−3

−2

−1

+4

+3

+2

+1

y1

y2

y3

y4

y5

y6

y7

y8

Clearly this satisfies (4) and the required SN sequence that generates nulls at normalized frequencies 0, 1/4, 1/2, 3/4 and 1 is therefore (0)(1)(2)(3)(7)(6)(5)(4). Starting with the obtained SN equation and using the property of commutativity by permuting the groupings and also by swapping the CS in each one of them, we

P. S. D.

K. Ouahada, H.C. Ferreira / Int. J. Electron. Commun. (AEÜ) 67 (2013) 1–9

16 14 12 10 8 6 4 2 0

0

0.2

0.4

0.6

3

0.8

1

Normalized Frequency Fig. 1. PSD of spectral nulls codebook for M = 8 and N = 4.

can obtain all different permutation sequences that satisfy (4). The final codebook is presented below:

The following equation depicts this concept: invert



⎧ ⎫ 01237654, 01327645, 10236754, 10326745, 01247653, 01427635, ⎪ ⎪ ⎪ ⎪ 10246753, 10426735, 01537624, 01357642, 10536724, 10356742, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 01547623, 01457632, 10546723, 10456732, 06237154, 06327145, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 60231754, 60321745, 06247153, 06427135, 60241753, 60421735, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 06537124, 06357142, 60531724, 60351742, 06547123, 06457132, ⎪ ⎬

0

1

···

↓

↓

.. .

0

1

···

PS :

···

M−2

.. .

↓ ↓ M−2 M−1 2

60541723, 60451732, 71230654, 71320645, 17236054, 17326045,

⎪ ⎪ ⎪ 71240653, 71420635, 17246053, 17426035, 71530624, 71350642, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 17536024, 17356042, 71540623, 71450632, 17546023, 17456032, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 76320145, 67231054, 67321045, 76240153, 76420135, 76230154, ⎪ ⎪ ⎪ ⎪ ⎪ 67241053, 67421035, 76530124, 76350142, 67531024, 67351042, ⎪ ⎪ ⎪ ⎩ ⎭



M−2 M 2 2

↓ M+2 ··· 2

M−1 ↓ M 2

arrow M−2 − 2

M CS : − 2

76540123, 76450132, 67541023, 67451032

···

−1

+

M 2

···

+2

+1

↓

↓

↓

↓

↓

↓

↓

↓

y1

y2

···

yM

yM + 2

···

yM−1

yM ,

2 Since the sum of each grouping in the equation is zero, there will be no discrete component at the zero frequency as depicted in Fig. 1, which represents the power spectral density of the corresponding designed codebook. In the case when we calculate the SN sequence directly as in Example 2, we call this, the basic SN sequence or the start SN sequence (SNS). In the case where the SN equation is derived from the start SN sequence, then we can call it as the general SN sequences.



2

resulting in, A1

=

y1 + y M + 2

=

A2

=

2 y2 + y(M−1)

=

.. . AN

.. . =

y M + yM

M M + 2 2

=

0,

M−2 M−2 + 2 2

=

0,

− −

(5)

.. . =

−1 + 1

.. . =

0.

2 2.1. For all N and z is even We fist start with the case of z = 2 and then we generalize the sequence. Proposition 1. The SN sequence of any permutation sequence with any value of N and z = 2 is the same permutation sequence with its second half inverted.

It is clear that all sums of elements in Ai for 1 ≤ i ≤ N are equal and this satisfies (1).  Example 3. We take the case of M = 16, with N = 4 and z = 4. This sequence will be generalized for all values of N and even values of z. We arrange the permutation sequence [20] in the same way as depicted in Fig. 2. The final SN permutation sequence is the cascaded sequence or the concatenation

Proof. The codeword length in this case is M = 2N. With z = 2, the number of elements in each grouping Ai is two. It is clear from (2) that symbols in each grouping are separated from one another in their positions by a value of N, which means that the two symbols in each groupings are yi and yi+N . Then:

yi + yi+N = yi + yi+ M = yi + y 2i+M . 2

2

As the channel symbols are opposite in signs, we can say that yi and yi+N should have opposite channel symbol signs to guarantee that the sums of all the Ai are equal. Hence, we need to invert the second half of the codeword sequence to guarantee that the channel symbols of yi and yi+N have the opposite value.

Fig. 2. Snake sequence.

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K. Ouahada, H.C. Ferreira / Int. J. Electron. Commun. (AEÜ) 67 (2013) 1–9

Table 1 Snake sequence for ∀ N and z even. 2N − 1 2N − 2 2N − 3 . . . N+1 N

0 1 2 . . . N−2 N−1

2N 2N + 1 2N + 2 . . . 3N − 2 3N − 1

... ... ... ..

.

... ...

Nz − 1 Nz − 2 Nz − 3 . . . N(z − 1) + 1 N(z − 1)

of the columns. The SN permutation sequence generating nulls at the normalized frequencies 0, 1/4, 1/2, 3/4 and 1, is (0)(1)(2)(3)(7)(6)(5)(4)(8)(9)(10)(11)(15)(14)(13)(12). Table 1 is completed by placing the symbols column by column. The first column starts with 0 at the top and continues to N − 1 at the bottom. In the second column we start from the bottom, proceeding to the top for N up to 2N − 1. We do the same for the rest of the columns by alternating with sequences going down and going up, producing a sequence which has a curving shape like a snake. Proposition 2. The SN sequence of any permutation sequence for any value of N and any even value of z is the result of the concatenation of the columns of the snake sequence as depicted in Table 1. Proof. For any even value of z, there exists an integer r where z = 2r. We can group all elements in each grouping Ai in pairs as shown as follows: z=2

  

z=2







= y1 + y1+N

+ y1+2N + y1+3N

+

···

+ y1+(z−1)N ,

A2

= y2 + y2+N

+ y2+2N + y2+3N

+

···

+ y2+(z−1)N ,

. . .

AN

. . .

= yN + y2N

  

. . .

+ y3N + y4N

  

z=2

+

(6)

. . .

···

+ yzN .

= y1 + y2N

+

A2

= y2 + y (2N−1)

+

AN

. . .

   y1+2N + y4N

+ ···

+ yzN ,

y2+2N + y4N−1

+ ···

+ y(zN−1) ,

. . .

= yN + y1+N

  

1

2

3

4

5

6

7

8

9

↓ shift 0

4

8

2

6

1

5

9

3

7

↓

↓

↓

↓

↓

↓

↓

↓

↓

↓

−5

−1

+4

−3

+2

−4

+1

+5

−2

+3

y1

y2

y3

y4

y5

y6

y7

y8

y9

y10

The SN sequence that generates nulls at frequencies 0, 1/2 and 1 then is (0)(4)(8)(2)(6)(1)(5)(9)(3)(7).

 yi =

(i − 1)(z − 1) (mod M),

1 ≤ i ≤ z,

yi−z + 1,

z + 1 ≤ i ≤ 2z,

z=2

A1

. . .

0

Proposition 3. The spectral null sequence for the case of N = 2 and any odd value of z, is

Using the same technique of inversion as used in (5), the system (6) can be written after inverting the subgrouping of z = 2 as follows:

  

PS :

The following proposition generalizes this class of SN sequences.

z=2

z=2

the first half, thus we get the following symbols: 1, 5, 9, 3 and 7, yielding:

CS :

A1

. . .

Fig. 3. Graph allocation of spectral null symbols for N = 2 and z odd.

+

. . .

y3N + y1+3N

  

z=2

+ ···

. . .

where mod results in the residue after division.

(7)

+ y1+(z−1)N .

z=2

Taking into consideration the channel symbol values, the values of Ai in (7) are zero and this satisfies the spectral shaping condition. 

2.2. N= even and z= odd 2.2.1. N = 2 and z = odd Example 4. We take the case of M = 10, with N = 2 and z = 5. The first half of the symbols starts with 0 and subsequent symbols are obtained by adding four (modulo 10), yielding the symbols 0, 4, 8, 2 and 6. The second half is obtained by adding 1 to the symbols of

Proof. This sequence is depicted in Fig. 3, where it is shown that every (z − 1)th symbol, the corresponding symbol is chosen for the SN sequence with taking into consideration the order and the consistency of the order as a result of the cyclic rotation through all symbols. We can start with any symbol but we need to use the start symbol as a reference. When the cycle is completed and returned to the reference symbol then we take the symbol next to it in the order. For example if we start with the symbol 0, the following symbol in the SN sequence is z − 1, then we carry on with the same analogy until we complete the first cycle of all symbols. This accounts for the first half of symbols, which are all even. The next half is the first half repeated but with 1 added to each symbol, which makes them all odd symbols.  Proposition 4. The sum of all odd symbols in the permutation sequence, denoted by Sodd is equal to the sum of all even symbols, denoted by Seven added to the value of z as follows: Sodd = Seven + z.

K. Ouahada, H.C. Ferreira / Int. J. Electron. Commun. (AEÜ) 67 (2013) 1–9 Table 2 SN sequence for N even, z = 3.

A

B

a

b

c

0 1 2 . . . N/2 −1

3N/2 3N/2 +1 3N/2 +2 . . . 2N − 1

3N − 2 3N − 4 3N − 6 . . . 2N

N/2 N/2 +1 . . . N−1

N N+1 . . . 3N/2 −1

3N − 1 3N − 3 . . . 2N + 1

The first column of Table 2B starts with the value of N/2 +1 until the value of N. The second column is a continuity of the last value of the first column, which is (N + 1) where we increment the values by 1 until the we reach the value N/2. The third column begins with 3N and is decremented in each row by two until the value becomes equal to 2N + 2.  Definition 1. The SN sequence of any permutation sequence with any even value of N and z = 3, is the concatenation of the columns starting from the first one in Table 2. Proposition 6. The sum of each row in Table 2A and-B denoted respectively by S1 and S2 is always constant as presented as follows: S1 =

Proof. The proof will be done by using induction. We start with a special case of M = 6, where N = 2 and z = 3. PS :

0

1

2

3

4

5

shift 0

2

4

↓

↓

1

3

5

↓

↓

↓

+2

−2

+1

+3

y3

y4

y5

y6

↓

CS :

−3 −1 y1

y2

Sodd = 1 + 3 + 5 = 9

(8)



Seven = 0 + 2 + 4 = 6

5

⇒ Sodd = Seven + z

From (8), we see how the resultant channel symbols satisfy the spectral null equation in (1). We have proven the proposition for the special case of M = 6 and now we assume that our proposition is satisfied for the case of any value of z and we just need to prove it for the case of z + 2. This value is chosen to keep the values of z odd since we know that it is already odd. As N = 2, the new sequence length is denoted by M , then M  = 2(z + 2) = 2z + 4 = M + 4. It is clear that we have four more symbols in our new sequence. As N = 2, the value of M is even, which means that the symbol with value M + 1 is odd. The new corresponding sums to this case are respectively denoted by Seven for the even sum and Sodd for the odd sum as presented as follows: Sodd = Sodd + (M + 1) + (M + 3) = Seven + z + (M + 1) + (M + 3) = Seven + z + M + (M + 2) + 2 = Seven + M + (M + 2) + (z + 2) = Seven + (z + 2). As we can see, Seven has the same configuration as Seven , which satisfies the general form of the spectral null sequences.  2.2.2. N = even and z = 3 Table 2 has N rows and three columns, where in the first column we have symbols from 1 to N. Table 2 is broken up into two subtables, each one has N/2 rows. Let us call them Table 2A and B. Proposition 5. The SN sequence of any permutation sequence with any even value of N and z = 3, is the concatenation of the columns starting from the first one in Table 2. Proof. The first column in Table 2A is filled with integers from 1 to N/2. The second column starts with the value of 3N/2 +1 where we increment this value by 1 until we reach the value of 2N corresponding to the row N/2. The third column starts with the value of 3N − 1 where we start decrementing by two until we reach the value of 2N + 1 corresponding to the row N/2.

32 N + 1, 2

32 N + 2. 2

S2 =

Proof. By induction We can prove the proposition for the case of M = 6 and then we can generalize it for the case of M = 3(N + 2), we choose the value of N + 2 to the keep the property of even satisfied. If M = 6, then N = 2. The sequence from Table 2 is (0)(1)(3)(2)(4)(5) SN : 0

1

3

2

4

5

↓

↓

↓

↓

↓

↓

−3

−2

+1

−1

+2

+3.

y1

y2

y3

y4

y5

y6

CS :

(9)

We can see then that the equation of spectral shaping is satisfied A1







A2







y1 + y3 + y5 y + y4 + y6 = 2 (−3) + (+1) + (+2) (−2) + (−1) + (+3).

This leads to A1 = A2 . This satisfies the spectral null Eq. (1) and the values of the sum for each row as follows: 32 × 2 + 1 = 10, 2 32 × 2 + 2 = 11, S2 = 2 + 3 + 6 = 2 S1 = 1 + 4 + 5 =

and prove the proposition for the first case of M = 6. We now suppose that our proposition is satisfied for the case of N and we can prove it for the case of N + 2. To prove this case we just verify the sums S1 and S2 . We denote by S1 , the sum of each row in Table 2A, which is, S1 =

N+2 + 2(N + 2) + (N + 2)z + (1 − (N + 2)) 2

=

N+2 + 2(N + 2) + (N + 2)z + (1 − (N + 2)) 2

= (N + 2) = (N + 2)

1 2

3 2



+2+z−1 +1 +z



+ 1 since z = 3 =

32 (N + 2) + 1. 2

It is clear that S1 has the same form as for S1 and this satisfies the sequences in Table 2A. The channel symbols corresponding to the values in the table also satisfy the Eq. (1).

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K. Ouahada, H.C. Ferreira / Int. J. Electron. Commun. (AEÜ) 67 (2013) 1–9

Table 3 SN sequence for N odd, z = 3.

A

B

Proposition 7. The sum of each row in Table 3 denoted by S is always constant as presented as follows:

a

b

c

0 1 2 . . . (N − 1)/2

(3N − 1)/2 (3N + 1)/2 (3N + 3)/2 . . . 2N − 1

3N − 1 3N − 3 3N − 5 . . . 2N

(N + 1)/2 (N + 3)/2 . . . N−1

N N+1 . . . (3N − 3)/2

3N − 2 3N − 4 . . . 2N + 1

We denote by S2 , the sum of each row in Table 2B, which is, S2 = (N + 2) +



3(N + 2) + (N + 2)z + (2 − (N + 2)) 2

= (N + 2) 1 +



3 + z − 1 + 2 = (N + 2) 2

+ 2 since z = 3 =

3 2

+z

3 (3N + 1). 2

S=

Proof. By induction We can prove the proposition for the case of M = 9 and then generalize it for the case of M = 3(N + 2). We make N = N + 2 for the sake of having an odd number every time. With M = 9 we have N = 3 and from Table 3 the corresponding sequence is (0)(1)(2)(4)(5)(3)(8)(6)(7), SN : CS :

0

1

2

4

5

3

8

6

7

↓

↓

↓

↓

↓

↓

↓

↓

↓

−4

−3

−2

0

+1

−1 +4

+2

+3

y1

y2

y3

y4

y5

y6

y8

y9 .

y7

(10)

We can see then, that the equation of spectral shaping is satisfied as follows:



A1 = y1 + y4 + y7 = (−4) + (0) + (+4) = 0, A2 = y2 + y5 + y8 = (−3) + (+1) + (+2) = 0, A3 = y3 + y6 + y9 = (−2) + (−1) + (+3) = 0.

32 (N + 2) + 2 2

This leads to

Using the same observation as in the case of S1 , we can see that S2 has the same value for as S2 . This satisfies the sequences in Table 2B and their corresponding channel symbols satisfy (1). 

A1 = A2 = A3 . This will satisfy the spectral null Eq. (1) and the value of S for each row, 3 (3 × 3 + 1) = 15, 2 3 = 2 + 6 + 7 = (3 × 3 + 1) = 15, 2 3 = 3 + 4 + 8 = (3 × 3 + 1) = 15. 2

Srow1 = 1 + 5 + 9 =

2.3. N= odd and z= 3 Table 3 has N rows and 3 columns. The first column represents the symbols from 1 to N. Table 3 is divided into two tables, each one with (N + 1)/2 and (N − 1)/2 rows. We denote each one of them as Table 3A and B. The first column of Table 3A is filled from 1 to (N + 1)/2. At the second column, which starts with the value of (3N + 1)/2, we add 1 until the row reaches the value of (N + 1)/2. It corresponds to the value of N(3 + 1)/2. For the third column, it starts with 3N, where we decrement its values by 2 until we reach the row (N + 1)/2. The corresponding value is then 2N + 1. In Table 3B, the first column starts with the entry of (N + 3)/2 increasing the entries till it reaches the value of N. The first entry of the second column is a continuity of the last value of the first column, which is (N + 1). Then we carry on by adding 1 until we reach the value (3N − 1)/2. The third column starts with 3N − 1 and at each row we subtract by 2 until the entry reaches 2N + 2. Definition 2. The SN sequence of any permutation sequence with any odd value of N and z = 3, is the concatenation of the columns starting from the first one in Table 3. Although the construction of Table 3 is based on different constructions of Table 3A and B, the sum of all rows is always the same.

Srow2 Srow3

The above equations prove the proposition for the first case of M = 9. We can now suppose that our proposition is satisfied for the case of N and we try to prove it for the case of N + 2, which we have to prove and verify the sum S. We denote by S , the sum of the each row in Table 3, which is, S = (N + 2) +

3(N + 2) − 1 + 3(N + 2) + (2 − N − 2) 2

= (N + 2) +

3 1 (N + 2) − + 3(N + 2) − N 2 2

= (N + 2) +

3 1 (N + 2) + 3(N + 2) − (N + 2) + 2 − 2 2



= (N + 2) 1 + =



3 3 + 3 − 1 + = (N + 2) 2 2

9 2

+

3 2

3 [3(N + 2) + 1] 2

We can see that the sum S has the same form as S, which satisfies the sequences in Table 3 and then their corresponding channel symbols satisfy (1). 

Table 4 SN sequence for ∀N and ∀z. Sequence length

2N N(z + 2) N(z + 4) N(z + 6) . . . M

Index

2 z+2 z+4 z+6 . . . M/N

Final sequence

P2 Pz+2 Pz+4 Pz+6 . . . PM N

Basic sequence

 

Added sequence

 

 

P

P = P + P

P = P + P 

P = P + P 

...

P P P P . . . P

P P P . . . P

P P . . . P

P . . . P

. . . ...

K. Ouahada, H.C. Ferreira / Int. J. Electron. Commun. (AEÜ) 67 (2013) 1–9

7

Table 5 SN sequence for N = 2 and z = 6. Steps

M

z

step 1 step 2 step 3

4 8 12

2 4 6

P2 P4 P6

 

 

P

P = P + P

P = P + P 

(0)(1)(3)(2) (0)(1)(3)(2) (0)(1)(3)(2)

(4)(5)(7)(6) (4)(5)(7)(6)

(8)(9)(11)(10)

Table 6 Selected SN sequences. M

N

z

Spectral nulls

4 6 6 8 8 10 12 16 16

2 2 3 4 2 5 6 4 8

2 3 2 2 4 2 2 4 2

0, 0, 0, 0, 0, 0, 0, 0, 0,

1 , 2 1 , 2 1 , 3 1 , 4 1 , 2 1 , 5 1 , 6 1 , 4 1 , 8

SN sequence

1 1 2 ,1 3 1 3 , , 2 4

(0)(1)(3)(2) (0)(1)(3)(2)(4)(5) (0)(1)(2)(5)(4)(3) (0)(1)(2)(3)(7)(6)(5)(4) (0)(1)(3)(2)(4)(5)(7)(6) (0)(1)(2)(3)(4)(9)(8)(6)(7)(5) (0)(1)(2)(3)(4)(5)(11)(10)(9)(8)(7)(6) (0)(1)(2)(3)(7)(6)(5)(4)(8)(9)(10)(11)(15)(14)(13)(12) (0)(1)(2)(3)(4)(5)(6)(7)(15)(14)(13)(12)(11)(10)(9)(8)

1

1 2 , 5 1 , 3 1 , 2 1 , 4

3 , 5 1 , 2 3 , 4 3 , 8

4 ,1 5 2 5 , , 3 6

1

1

3. Generalization of the spectral null sequence In this section we generalize the technique of the search or the calculation of the SN sequence. The following proposition has an exception with regards to the case of N = 2 and z = odd. Proposition 8. The SN sequence of any permutation sequence with any value of N and z, except for N = 2 and z = odd, is derived from the case of z = 2 as depicted in Table 4. Proof. For large values of M, we can build up the SN sequence from the shortest length of the corresponding SN sequence to the value N. Let us denote this sequence by P which will be used to calculate any sequence larger than P. In Table 4, P = P2 , where P2 is the SN sequence for z = 2. The question here is how do we build up or construct the final SN sequence from P? In general, if we take two permutation sequences of lengths M1 and M2 , where M1 = Nz , M2 = Nz and z = z + 2, we can see that: M1 = N(z + 2) = Nz + 2N = M2 + M3 , where M3 = 2N.   Our starting sequence in Table 4, which is P, has a length P

 

equal to M2 . For any permutation sequence Pz with a length Pz  =

  M1 , we can have Pz  = M2 + M3 .

If the value of z is incremented by two, we can calculate the required SN sequence by using the previous SN sequence in the table and concatenate it with the sequence derived from z = 2, where the sequences for any value of N and z = 2 have been previously calculated. Thus we have first to calculate P which is the same as the previous sequence but it symbols are added with the value of |P|. After this, we just concatenate this new sequence to the previous SN sequence with taking into consideration the order of the symbols. We carry on with the same principle until we complete the total length of the desired SN sequence. 

1 5 3 7 , , , , 2 8 4 8

1

Example 5. To explain the algorithm, we take as example M = 12 with z = 6. We allocate channel values to each symbol as introduced previously. If we first consider z = 4, for example, which is the same as z = 2 +2, then we can get twice the case of z = 2. This is shown in (11), where we can see that all variables yi after the first two are incremented by 4. We have chosen z = 2, which means Nz = 4 and we find ourselves at the stage of z = 2. Our case of z = 6 can be broken up into three cases of z = 2, with steps Nz, so the final SN sequence is the concatenation of the three SN sequences with steps of Nz as shown in Table 5. From the first step, which is the calculation of the SN sequences for N = 2 and z = 2, we increase the value of z by 2 and then we just add the value of the first SN sequence length, which is M = 4, to any symbol value of the concatenated sequence as depicted in Table 5. The final SN sequence steps are summarized as follows: • step 1: calculation of SN sequence for N = 2 and z = 2: 0132, • step 2: calculation of SN sequence for N = 2 and z = 4: 0132 “ +  (0 + 4)(1 + 4)(3 + 4)(2 + 4) = (0)(1)(3)(2)(4)(5)(7)(6), • step 3: calculation of the final SN sequence for N = 2 and z = 6: 01324576 “ +  (0 + 8)(1 + 8)(3 + 8)(2 + 8) = (0)(1)(3)(2)(4)(5)(7)(6) (8)(9)(11)(10). The addition sign “ +  appearing in the steps of the calculation of the final SN sequence has no arithmetic meaning but just to reflect the concatenation and the order of the added sequences. Table 6 presents selected SN sequences and the corresponding spectral nulls for different values of N and z. 4. Spectral-shaping magic squares Magic squares are known in mathematics as squares containing numbers with certain properties [21–23], like the sum for different directions in the square or certain specific locations in the square is always constant. We are going to see in the case of N = z, that we have certain mathematical properties for these symbols [24,25], which form the SN sequences.

y1 + y3 + y5 + y7 + y9 + y11

= y2 + y4 + y6 + y8 + y10 + y12 ,

y1 + y3 + y1+4 + y3+4 + y5+4 + y7+4

= y2 + y4 + y2+4 + y4+4 + y6+4 + y8+4 ,

y1 + y3 + y1+Nz + y3+Nz + y5+Nz + y7+Nz = y2 + y4 + y2+Nz + y4+Nz + y6+Nz + y8+Nz .

(11)

8

K. Ouahada, H.C. Ferreira / Int. J. Electron. Commun. (AEÜ) 67 (2013) 1–9

Table 7 SN sequence for odd values of N = z.

5. Spectral null sequence with repeating symbols

(N − 1)2 modN2 N + (N − 1)2 modN2 . . . N(N − 1) + (N − 1)2 modN2

0 NmodN2 . . . N(N − 1)modN2

... ... ..

. ...

(N − 1)3 modN2 N + (N − 1)3 modN2 . . . N(N − 1) + (N − 1)3 modN2

Table 8 Properties of the SN sequence for odd values of N = z. Shi modNz

N

Sd1

(N+1) 2

N

Sd2

(N+1) 2

N

(N+1) 2

Sv1

...

Svz

N

...

0

Table 9 Properties of the SN sequence for even values of N = z.

modNz

Sh

Sd1

Sd2

Sv1

...

Svz

N 2

0

N

N 2

...

N 2

(N + 1)

(N + 1)

4.1. N = z = odd Table 7 is formed by rows denoted by their indices j and columns denoted by their indices i. To fill the table is just starting by the symbol 0 at i = j = 1 and then increment the content of the table using the following summation (i − 1)(N − 1)2 + (j − 1)N with 1 ≤ i, j ≤ N. We module the sum to N2 and the resultant value of the table is the remainder of that operation. In the case where the result is zero, the corresponding value is M. The final sequence or the SN sequence is the concatenation of the columns from Table 7. Definition 3. The SN sequence of any permutation sequence with any equal odd values of N and z, is obtained from concatenating the values of the columns of Table 7, which are the remainder of the modulo N2 addition. We denoted by Shi the sum of each row or horizontal, by Svi the sum of each column and by Sdi the sum of the ith diagonal in Table 7. Table 8 shows some properties when applying the modulo-Nz to the sums previously introduced. Proposition 9. The sum of the corresponding channel values of the symbols of the SN sequence for the case of N = z = odd, is always a multiple of Nz. (M + Proof. Any sequence of integers of length M, has the sum of M 2 1). In our case, since M = Nz, where N = z = odd, we can say that the value M is to be an odd integer. Thus the value M + 1 is even and the value of M+1 is an integer, therefore M (M+1) is a multiple of M = Nz. 2 2  4.2. N = z = even In this section we are not looking for the calculation of the SN sequence since it was presented in Section 2.1. We present here a few properties related the sum of the channel symbols as presented in Table 1 but for the case of N = z. We denote by Shi the sum of each horizontal, by Svi the sum of each column or vertical and by Sdi the sum of each diagonal. Table 9 shows some properties when applying the modulo-Nz to the previously introduced sums. Proposition 10. The total sum of the SN sequence for the case of N = z = even, is always a result of Nz modulo M . 2 (M + Proof. Any sequence of integers of length M has the sum of M 2 1). In our case, where M = Nz with N = z = even, it is clear that the value of M is an even integer. Thus the value M is an integer. Finally, 2 we can say that the value of M×M is a multiple of M = Nz and the sum 2 is equal to M modulo Nz.  2

The generalized form of the SN sequence calculated in Section 3, will be used to generate an SN sequence with repeating symbols [26]. The idea is to repeat symbols and their corresponding channel symbols in such a way that (1) is always satisfied. In each grouping Ai we have z symbols and then we can repeat (z − 1) symbols to end up with all symbols in the SN sequence repeated. Usually when the symbols in a permutation sequence are not equal, the number of symbols in a codeword, which is M = Nz, is also the number of the channel symbols. In the case of repetitive symbols in a permutation sequences, and since we usually repeat the symbols in each grouping to satisfy the equation of spectral nulls, we can say that the final number of the channel symbols is what exists in each grouping and this depends on the number of times that we repeat symbols. As it is known, the length of the codewords is M = Nz, which is an integer and also a product of two integer numbers. This will make the values of M not equal to the values of 3, 7, 11, etc., since N and z are integers. Thus the technique of repeating symbols will solve such a problem because by repeating symbols we are able to have levels, which are not equal to the value of M and could be even equal to the value of z if all symbols in each grouping are totally repeated. We consider the example of M = 6 with N = 2, the spectral null equation is,



z=3







z=3





y1 + y3 + y5 = y2 + y4 + y6 .

(12)

If we take the case of non-repeated symbols as presented by their corresponding positions in (12), we will have six different channel symbols. In the case of repeating three symbols, presented by their positions y1 , y3 and y5 , we can write the equation differently and the resultant sequence has six symbols but only three channel symbols. The corresponding spectral null equation, where we repeat the positions to show that their corresponding symbols are the same, with y2 = y1 , y4 = y3 and y6 = y5 is presented as follows:



z=3







z=3





y1 + y3 + y5 = y1 + y3 + y5 .

(13)

The resultant SN sequence satisfying (13) is (0)(1)(2)(0)(1)(2). To generate a SN sequence with only five channel symbols, we can use the same Eq. (12) but this time we repeat only one symbol in each grouping and the corresponding SN equation for the positions of the symbols is presented as follows:



z=3







z=3





y1 + y3 + y5 = y1 + y4 + y6 .

(14)

From (14) we have six symbols but since we have repeated only one symbol, we get five channel symbols. The resultant SN sequence satisfying (14) is (0)(0)(2)(1)(3)(4). 6. Conclusion The spectral shaping sequence with permutation sequences and the search for the SN sequences play the key role in ensuring of the spectral nulls of our codes. We have presented the technique of the generation and design of all types of these sequences. The generalization of the spectral null sequences is given for different values of N and z. The generation of spectral null codes is based on the conditions made on N and z, which can introduce more constraints on the generation of this type of codes simply compared to the DC-free codes. The complexity related to the values of N and z was wisely

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