Formulation

We first consider a transmission scheme where the source symbols are transmitted through a channel with impulse response _h [ (0), (1),..., ( )]_{h h h L} T

zeros padded at the beginning of each block [5]. Without loss of generality, we assume that

(0) 0

h . Also it is assumed that Q L. Denote by Zblock block

, 0,..

( .

[ )T ,0]T

k k

s s the k-th block of the transmitted symbols with zero paddings, where block

( ), ( 1),..., (1) [ k k k ]

T

k s Q s Q s

s . The length-Q vector block

k

s is

defined this way so that its first symbol is the latest transmitted one, and the last one is the one that transmitted earliest. The length of the padded zeros is L, and is equal to the

assumed maximum order of the channel. The corresponding noise vector that affects the channel outputs is Zblock

k

v . The whole block with length Q L including the paddings will affect a total of Q channel outputs. According to the model of BE in matrix form in (2.44), the observation block Zblock

k

x can be written as

Zblock Zblock Zblock

k k k

x Hs v . (3.1)

In the above model (3.1) the last L symbols of Zblock

k

s are the padded zeros and will not contribute to the observed mixture Zblock

k

x . As a result, equation (3.1) reduces equivalently to

Zblock block Zblock

T

k k k

x H s v , (3.2)

where HT is a Q Q square Toeplitz matrix with lower diagonals zero, consisting of the first Q columns of matrix H. From (3.2) it can be seen that the block of channel outputs at time k is affected only by the source symbols in the k-th block.

Since it is assumed that (0) 0h , H_{T is full rank. As a result, matrix} H_T has a unique left inverse. The recovery of block

k

s from Zblock

k

x in (3.2) is a standard BSS problem, and it can be attempted using a standard BSS algorithm.

Let W be the “separating” matrix in this zero-padded block transmission BE problem, then the goal is to find a matrix W such that C WH_T I. When the BSS adaptation converges, if the separating matrix W is a good approximation of 1

T

H , the source symbols will be well recovered. From the structure of H in (2.44), we know that H_T is a square Toeplitz matrix with lower triangular elements zero. The Toeplitz structure is maintained under inversion [6], [7], so 1

T

H should also be a square Toeplitz matrix with lower diagonals zero.

Now let us take a slight detour to consider the elements in 1

T T

W H . Let the first row of WT be wTT [ (1),wT wT(2),...,w QT( )]. Then

1

T T

T T

w H e , (3.3)

where e₁ is a length Q column vector with e₁(1) 1, e j₁( ) 0 for j 2,...,Q. Because of the Toeplitz structure of H_T, the vector on the right side of (3.3) via matrix multiplication can be equivalently obtained as the truncated convolution of w_T and h. We denote by

1:

(h w_T) _Q the column vector of the first Q terms of the convolution of w_T and h. We have 1: 1 ( ) T T T T T T Q w H h w e . (3.4)

Let us define the “inverse” of h as the column vector containing the coefficients of the inverse z-transform of h, which can be obtained by long division. From (3.4) we see that

T

w is the truncated version of the inverse of h containing only the first Q taps. Since H_T has Toeplitz structure, based on (3.3), we also have

1 zeros [0 ... 0 _T(1) _T(2) ... _T( 1)] T i T i w w w Q i H =e (3.5)

where i 1, 2,...,Q , and e_i is a length-Q vector with the i-th element one and zero elsewhere. Denoting the transpose of the vector on the left side in (3.5) by ( )i

T w (note (1) T T w w ), we have (1) (2) ( ) 1 2 ... Q T ... T T T T T Q Q w w w H = e e e I . (3.6) From the above equation we see that the inverse of H_T , where

1 (1) (2) _... ( )Q T T T

T T

H w w w , is a Q Q Toeplitz matrix containing the first Q taps of

the inverse of h, with lower triangular elements zero.

Let _wT _{be the first row of the separating matrix W. If}

T

C WH is close to the identity matrix, then W is approximately 1

T

H , and then w is approximately w_T , the truncated inverse of h. In this case, all the symbols in the block block

k

s will be recovered without any arbitrary permutation. With this model, we are able to recover the source symbols, but will not obtain the impulse response of the equalizer.