Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 7.3 MSE Estimation of a Discrete Process
In the example we will make use of the results in section 7.3 to derive MS optimal processing of discrete
processes. We first derive the results for a set of discrete samples from a process whose correlation
function is discrete. For a continuous process examined in Example 7.2 we have a continuos correlation
function in contrast a discrete correlation function in this example. We begin by expressing the results of
the section in term of samples of a vector process.
A linear MS estimate, Y , of the discrete RV, Y, in terms of the previous n measurements expressed as a
vector Xi where Xt = [ x0 x1 ... xn-1] is given by
Yn  a0  x0  a1  x1    an1  xn1
where xi represent samples form the random processes Xi
The process X is modeled by
X i  Yi  N i

 
The correlation function of the error in estimate is, with X 0 a RV
P  E Yn  Y   E Yn  a0  X 0  a1  X 1    an1  X n1 
It was shown that
2
2

 
P  E Yi  Y  Yi   E Yi 2  E Yi  Y 
n 1
P  RYY  0   a m  RYX  m
m 0
The coefficients ai can be obtained by the set of linear equations
E Y  Yi X i  m   0 m  0 n  1
The results of this equation can be expressed In terms of a set of equations in terms of the correlation
coefficients.
R XX 0  a0  R XX 1  a1    R XX n   a n  RYX 0
R XX  1  a 0  R XX 0  a1    R XX n  1  a n  RYX 1

R XX  n   a 0  R XX  n  1  a1    R XX 0  a n  RYX n 
In matrix form
R  A  R0
With these expression we can find a expression for the one dimensional MS estimation of Y
X i  Yi  N i
and
R XX i   RYY i   RNN i 
We have assumed that the Noise N is uncorrelated to the signal Y. The Correlation coefficients for the
signal, Y and the noise N are respectively
RYY (i )   2  r i
R NN   N2    i 
R XX  i    2  r i   N2    i 
We must now compute RYX(i)
RYX m  EYi X i m   EYi Yi m  N i m   RYY m
since
EYi N i m   0
Solving for the coefficient a0
a0 
RYX 0 RYY 0
2

 2
R XX 0 R XX 0  N   2
and the MS error

P  RYY 0  a0  RYY 0   2  1 


  N2 
2
    2

 N2   2 
N   2 
2