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\title{An Adaptive Filter to Pick up a Wiener Filter from the Error using MSE with and Without Noise}
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             \author[1]{Dr. Ziad  Sobih}

             \affil[1]{  Northeastern University}

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\date{\small \em Received: 3 February 2015 Accepted: 4 March 2015 Published: 15 March 2015}

\maketitle


\begin{abstract}
        


Introduction-In this paper we explain the suboptimum channel equalization approach. This approach employ linear transversal filter that we will explain. This filter structure has computational complexity that is linear with the order.This filter is shown in figure 1. Its input is the sequence v k, its output is the estimate of the output sequence Ik. The estimate might be expressed asThe estimate I k is quantized to the nearest information symbol.Considerable research have been done to optimize the filter coefficients ck. A measure of performance for digital communication system is the average probability of error. In this system this is highly non linear function of ck. As we can see this method is computationally complex.

\end{abstract}


\keywords{employ linear transversal filter, computational complexity, linear with the order.}

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\let\tabcellsep& 	 	 		 
\section[{Introduction}]{Introduction}\par
n this paper we explain the suboptimum channel equalization approach. This approach employ linear transversal filter that we will explain. This filter structure has computational complexity that is linear with the order. This filter is shown in figure  {\ref 1}. Its input is the sequence v k, its output is the estimate of the output sequence Ik. The estimate might be expressed as \hyperref[b0]{(1)} The estimate I k is quantized to the nearest information symbol.\par
Considerable research have been done to optimize the filter coefficients ck. A measure of performance for digital communication system is the average probability of error. In this system this is highly non linear function of ck. As we can see this method is computationally complex.
\begin{quote}
Figure 1\end{quote}
\par
Two other methods are widely used. First, the peak distortion method. Second, the mean square error method. 
\section[{a) The peak Distortion criterion}]{a) The peak Distortion criterion}\par
The peak distortion criterion is simply the worst case inter symbol interference at the output of the equalizer. It is the minimization of this performance index. First we assume that the equalizer has infinite taps. Second the case that this number is finite.\par
The cascade filter have an impulse response f n. and the equalizer impulse response c n. this can be represented by a single filter with impulse response \hyperref[b1]{(2)} Author: Northeastern University Boston, MA, USA. e-mail: sobih84@gmail.com Q n is simply the convolution of c n and f n. the output of the k th sampling instant is F Î k = c j v k -j K ?\par
It is possible to have D(c) i. e. q n =0 for all n except n=0. This way we will cancel all ISI. This can be expressed as \hyperref[b4]{(5)} Figure  {\ref 2} By taking the z transform we have \hyperref[b5]{(6)} Or simply \hyperref[b6]{(7)} C(z) is the z transform of c j. note that the equalizer C(z) is simply the inverse of the linear filter F(z). This will result in complete elimination of ISI. We call this a zero forcing filter. Figure  {\ref 2} is the block diagram.\par
We use a noise whitening filter with transfer function 1/F (1/z) in cascade to the zero forcing function 1/F(z) which result in an equalizer with a transfer function \hyperref[b7]{(8)} As in figure \hyperref[fig_0]{3} the input is y k. The output is the desired symbols. The impulse response is (9) X(z) is a polynomial with 2 L roots. (p1, p2,???., p L, 1/p1, 1/p2,????.,1/p L). The closed contour of the integral can be the unit circle. Figure \hyperref[fig_0]{3} The performance can be expressed in terms of SNR at the output. We normalize the received signal energy to one. This mean that q o=1 and the expected value of I \textasciicircum 2 is one. Then the SNR is the reciprocal of the noise variance at the output.\par
The value ?\textasciicircum 2 can be determined by absorbing the input sequence of the equivalent equalizer which has a spectral density \hyperref[b9]{(10)} The noise sequence at the output of the equalizer has a power spectral density \hyperref[b10]{(11)} The variance of the noise at the output of the equalizer is \hyperref[b11]{(12)} The SNR for the zero forcing equalizer is (13)\par
The spectral characteristics of X(w) is the Fourier transform of the samples x n has a nice relationship to the filter H(w). Let us say that the equalizer has 2K+1 taps. The input is f n. the output q n is the convolution of f n and c n. If q 0 is normalized to unity the peak distortion is  {\ref (22)} The problem is to minimize D(c) with respect to c j. It has been shown that D(c) has a global minimum. Minimization can be done using the method of steepest descent. One special case is (23) Is less than unity. In this case the inter symbol interference is not severe. In this case we select q 0 =1 and q n =0 for n not equal to zero. 
\section[{b) Mean Square Error (MSE) criterion}]{b) Mean Square Error (MSE) criterion}\par
In this method c k is adjusted to minimize the mean square value of the error (24)\par
The performance index j is We see that the deference of the two methods is the factor N0. This mean that if N0=0 the two methods give the same results. The first method is to minimize the peak distortion D(c) and the second method is to minimize the MSE performance index j.\par
J= E (e\textasciicircum 2). To minimize this we use the fact that the error e is equal to I minus the estimate of I. then we use the orthogonality condition to get  Performance in the presence of severe inter symbol interference is shown in figure  {\ref 10}. This is the probability of error measured by Monte Carlo simulation for three channels of figure 11. To compare also included the graph with no interference. Figure  {\ref 11 (a} We can say that the performance is good in channels were the spectral characteristics are well behaved and do not have spectral nulls. In general spectral nulls result in noise enhancement at the output of the linear equalizer.\par
This equalizer has a problem with sever ISI which motivated research into non linear equalizer with low computational complexity. We will talk about the decision feed back equalizer next. We have developed a model to receive information from a channel with ISI noise and additive noise. We say that the additive noise is colored. We simplified this model by using a whitening filter so the noise is AWGN. We reduced ISI corruption by using a linear transversal filter. We say this is the suboptimum solution.\par
In this part we use non linear equalizer which is also suboptimum, but its performance is better than the linear. This consist of two filters the first is feed forward and the second is feed back as shown in figure \hyperref[fig_0]{13}. This is called decision feedback equalizer. The input of the first filter is the received sequence. The feedback filter has its input the sequence of decisions on previously detected symbols. This mean that the feedback filter is used to remove ISI from the present symbol caused by previously detected symbols. Since the detector feeds hard decision to the feedback filter, the DFE is nonlinear.\par
In our treatment we focus on finite duration impulse response filters and apply the MSE criterion to optimize the coefficients. 
\section[{II.}]{II.} 
\section[{Recursive Least Squares Algorithms for Adaptive Equalization}]{Recursive Least Squares Algorithms for Adaptive Equalization}\par
The LMS algorithm that we talked about for adaptively adjusting the coefficient of the equalizer is a stochastic steepest descent algorithm in which the gradient is obtained from the data.\par
The advantage of the method of steepest decent is computational simplicity. However the convergence rate is slow specially if the autocorrelation matrix has large Lambda max / Lambda min. The only way to control this is delta. This is a fundamental limitation.\par
In order to have faster convergence we need more complex algorithm with more parameters. The matrix T has N Eigen values. The optimum selection of these parameters is in this section.\par
To get faster convergence we use the method of least squares. So we deal directly with the data and minimize the quadratic performance index. In the past we use to deal with the error and minimize the expected value of its square. This mean that the performance index is in time average not statistical average.\par
We need this in matrix form. So we need to introduce some notation and for the linear equalizer the estimate of the information symbol at time t. By changing the index the estimate is Recursive Least-Squares (kalman)  Algorithm\par
First we will do the formulation for RLS. In this method we want to minimize the time average weighted squared error \hyperref[b1]{(2)} Where the error is \hyperref[b2]{(3)} w is a weighting factor between zero and one which introduce exponential weighting to the past , this is good for time variant channel. Minimization of the error give the set of linear equations R (t) is the signal correlation matrix defined as D(t) is the cross correlation vector 
\section[{The solution is}]{The solution is}\par
The matrix R is the autocorrelation matrix. D is the cross correlation. R may be ill conditioned so we initially add delta I. With exponential weighting into the past the effect of adding delta I dissipates with time. Now suppose we have C (t-1) for time (t-1). It is impractical to solve the set of N linear equations for each new signal component that is received to find C(t).\par
To avoid this first we find R(t) may be corrupted in noise as. \hyperref[b7]{(8)} Since the inverse of R is needed we use the matrix inverse identity \hyperref[b8]{(9)} Were the inverse is computed recursively.\par
We define P(t)= inverse R (t). The Kalman gain vectot \hyperref[b9]{(10)} Were µ (t) is a scalar defined as \hyperref[b10]{(11)} With this equation 11 becomes  As a result of Makhoul work we can see that in the lattice filter we can remove the last stage without effecting the coefficient of the previous stages. This mean that with taking E m as a performance index we can stop when E m is satisfactory.\par
As we can see the linear prediction problem can be implemented by transverse filter or lattice filter. The lattice filter is order recursive. This mean the number of sections can be increased or decreased without effecting the parameters of the remaining sections. On the other hand if the order of the transverse filter is changed all coefficients to minimize the MSE will change. This mean that the Kalman filter is recursive in time but not order.\par
Let us study the computational complexity. The lattice computational complexity increase linearly with the order. A graph to compare deferent algorithms is Figure  {\ref 14}. The study of the RLS lattice was done by papers by  {\ref Lee (1978)} and Alexander (1979) and books by Proakis (2002) and  {\ref Haykin (2002)}.\begin{figure}[htbp]
\noindent\textbf{3}\includegraphics[]{image-2.png}
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\caption{\label{fig_3}Figure 4 :Figure 7 :Figure 6 :Figure 9 :FFigure 8 :}\end{figure}
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\caption{\label{fig_5}}\end{figure}
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\noindent\textbf{1011}\includegraphics[]{image-8.png}
\caption{\label{fig_6}Figure 10 Figure 11 ©}\end{figure}
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\caption{\label{fig_7}}\end{figure}
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\noindent\textbf{12}\includegraphics[]{image-10.png}
\caption{\label{fig_8}( 12 )}\end{figure}
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\caption{\label{fig_9}F}\end{figure}
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\caption{\label{figure12}}\end{figure}
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\caption{\label{figure13}}\end{figure}
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 			\footnote{© 2015 Global Journals Inc. (US)} 		 		\backmatter  			 \par
The only variable parameter is the step size delta.\par
Figure  {\ref 13} show the convergence rate of the two algorithms for a channel with fixed parameters f0=.26, f1=.93, f2= .26 and a linear equalizer with 11 taps. The Eigen value ratio for this channel lambda max / lambda min =11. We start with zero coefficients. Delta for the steepest descent is .02. It is clear that the Kalman algorithm is great. This is important for time variant channel. For example the ionosphere high frequency radio channel is too fast to use steepest descent. Kalman algorithm adapt fast to track it. 
\subsection[{Figure 13}]{Figure 13}\par
The kalman algorithm has a good performance. It has two disadvantages. First its complexity second the sensitivity to round off noise that accumulate due to the recursive computation. This may cause instability.\par
The number of computation to calculate C is proportional to N\textasciicircum 2. Most of them for updating P. Updating P may introduce round off noise and to solve this problem we use (21\par
Were S is a lower triangular matrix with unity diagonal and ? is a diagonal matrix. This is called square root factorization.\par
The square root algorithm is used in control systems with kalman algorithm involved.\par
It is also possible to develop fast RLS that has linear complexity with the number of coefficient. This was described by the paper of Kailath (1991).\par
IV. 
\subsection[{Linear Prediction and the Lattice Filter}]{Linear Prediction and the Lattice Filter}\par
We will develop a relation between linear FIR and a Lattice filter. This connection can be seen by considering the problem of linear prediction.\par
The prediction problem may be stated as follows: given a set of data predict the value of the next data point. The predictor of order p is\par
The MSE is This is the Yule Walker equations\par
The matrix R is a Toeplitz matrix and the Levinson Durbin algorithm is a good way to solve the linear equations recursively. Starting from the first order and going recursively to order p. 
\subsection[{The linear prediction FIR filter of order m has a transfer function}]{The linear prediction FIR filter of order m has a transfer function}\par
The input is the data y(t). The error e is e (t) = y(t) -(estimate y (t)). This filter may be realized in lattice form as we will show.\par
We start with the use of Levinson-Durbin algorithm ti find a m k in equation 29. The coefficients of G m-1 (z) are exactly the same as A m-1 (z) but in reverse order.\par
We can see more if we look at the output if the input is y(t) and we take the z transform 			 			  				\begin{bibitemlist}{1}
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