Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 6.2 Common Transformation of Processes
In this example we will compute the mean and autocorrelation function of the memoryless
transformations that are associated with the detection process. The input to the detector is a zero mean
gaussian process and the output process will be shown to be non-gaussian. We make use of the
transformation of the density function that we have derived in example 3.7. In this example the three
detection processes studied are
y (t )  x(t ) 2
1.
The square Law Detector
2.
The linear full wave detector
3.
The linear half wave detector
z (t )  x(t )
 xt  xt   0
wt   
 0 xt   0
1. Square Law detector - The output of the square law detector is not normal even though the input is
normal. Using the transformation techniques of example 3.7 we have y = x 2. We first invert the
transformation to obtain x = - sqrt(y) and sqrt (y) because there are two roots. Using Equation 3.4-10 we
have after first differentiation
y  x2
gy 
d 2
x  2 x
dx
and
1
2 y
  y   f  y 
 f
When we assume that y is a zero mean gaussian or
f x  
1
2   2
e
1x
  
2  
2
Substitution of f(x) in the g(y) equation we obtain
gy 
1
2   2  y
e
1 y 
  2 
2  
syms sig u mpi
sgt=sym('sig>0');
maple('assume',sgt);
EY=int((u/2/pi/sig^2)^(1/2)*exp(-1/2*u/sig^2),u,0,inf)
EY =
sig^2
The correlation function can be computed assuming a jointly gaussian density function directly. In order to
complete the computation of the correlation function directly using Equation 6.2-4 we would be required to
setup and evaluate the integral

 
2
2


  u
E xt    xt   E X 12 X 22 
 v 2  f u , v   du  dv
2
 
This is a difficult integral to evaluate directly. We can rewrite the expected value, as we show in Example
6.1, as

 


E y t    y t   E xt    xt   E xt    E xt   2  E xt   xt 
2
2
2
2
2
The results are expressed in term of the input correlation function
RY    R X 0  2  R X  
2
2
and
EY    2  R X 0
The variance of y becomes as
 Y2  RY 0  E Y 2
 Y2  RY 0  R X 02  2  R X 02
2. Full wave detector - The density function of the output of the full wave detector is not normal even
though the input density function is normal. Using the transformation techniques of example 3.7 we have z
= |x|. We first invert the transformation to obtain z = - y and y because there are two roots. Using Equation
3.4-10 we have after differentiation of the density function
z x
d
z 1
dx
1
g z    2  f x z 
1
The mean value can be computed from
EZ=int(2/(2*mpi*sig^2)^(1/2)*u*exp(-1/2*u^2/sig^2),u,0,inf);
simplify(EZ)
ans =
2^(1/2)*sig/mpi^(1/2)
'E[Z]='
rx0=sym('(Rx(0))^(1/2)');
pretty(subs(EZ,sig,rx0))
ans =
E[Z]=
1/2
2
Rx(0)
--------------
1/2
(mpi Rx(0))
The Correlation function can be computed directly from Equation 6.2-4 as we show in Example 6.1
E z t   z t   E  xt     xt    2  R X 0 
cos     sin  

Where
sin   
R X  
R X 0


 
2
2
In example 6.1 we derived the covariance relationships that are now needed. We may compute the
variance of z by setting  = 0 and then sin() = 1 or  = /2 and using the above equation


2

2
E z t   R X 0 and  Z2  1    R X 0 
 
3. Half Wave detector - The output of the full wave detector is not normal distributed even though the input
is normal. The mean and variance can be deduced from the values derived for the full wave detector. We
expect the mean and to be value to be 1/2 of the full wave detector or
EW  
2 
 
 2
R X 0
2 
The correlation function of the half wave detector is related to the full wave one. Let us examine the
function x + |x|
w
1
x  x   0 x  0
2
x x  0
Now we can compute
Ewt   wt  
1
 E xt     xt    xt   xt  
4
1
 Ext   xt   E  xt    xt   E xt    xt    E  xt    xt  
4
The 1/4 is given cause the half wave detector representation has a gain of 1/2 for each term. Examining the
cross terms of the above expression
E  xt    xt  


  x  y  f x, y   dx  dy
 
We find the kernel of the integral is composed of an odd function, x, times an even function, |x|, times and
even function, f(x,y), yielding an odd function. When an odd function is integrated over symmetrical limits
the result is zero. Similarly we find E[x(t+)|x(t)|] = 0. The correlation function that results is
Rw    
Substitution for the correlation function
1
 R    R X   
4 Z
RZ 0 the full wave detector we have

1  R X (0)
2
 w2   1  
 
