Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 6.7 Correlation Function for Shot Noise
In the example we will derive of the correlation function for shot noise. This procedure is a two step
process as shown in the diagram in example 6.4. The first part of the process, the differentiation of the
Poisson process, was covered in this example 6.4. We will illustrate two forms of shot noise which depend
upon the assumed linear transfer function. The transformation by the linear process represented by h(t) that
is a ramp function will be covered in this example. When h(t) is an exponential will be covered in example
6.8. To illustrate the process ,we first will generate a shot noise waveform. This begins by generating a set
of random time points in the interval [0 - 10] and then summing impulses at these points the impulses
responses will be of the form of ramps or exponential.
Let
t=10*rand(10,1);
T=1;
The ramp input form, which is also called transit delay, is given by function
function y=f6_7(t,T)
%impulse ramp function
y=t.*stepfun(t,0)-(t-T).*stepfun(t,T)-T*stepfun(t,T);
The shot noise output is given by the summation. When an impulse is applied to a linear system with an
impulse response function h(t), the output is h(t). When we sum the output of all the inputs due the
linearity property of linear systems we obtain by the function
function y=f6_7_1(x,t,T)
%summation function of f6_7
y=zeros(1,length(x));
for i=1:length(t)
u=x-t(i);
y=y+f6_7(u,T);
end
The impulse response and the output is plotted below
tp=0:.05:2;
ts=0:.05:10;
subplot(1,2,1); plot(tp,f6_7(tp,T));xlabel 'time - sec';ylabel
'Impulse'
subplot(1,2,2);plot(ts,f6_7_1(ts,t,T));xlabel 'time - sec';ylabel
'Noise'
1
3
0.9
2.5
0.8
0.7
2
Noise
Impulse
0.6
0.5
1.5
0.4
1
0.3
0.2
0.5
0.1
0
0
0.5
1
time - sec
1.5
0
2
0
5
time - sec
10
The power density function response can be obtained by using the transform relationships. First we
transform RZZ to SZZ and we must change t2 - t1 =  in RZZ
S ZZ    2    2      
The expected value of y(t) is first computed

E  y t   E z t    ht   dt    H 0 

For the power density function for the output we have
S YY    S ZZ    H    2    2      H      H  
2
2
2
The first term can be simplified to
2    2      H    2    2      H 0
2
2
and the inverse of SYY or RYY is formed. The first term which contains and impulse becomes a constant
and the second term represents a convolution in the time domain to obtain

RYY    2 H 0     h     h d
2

The covariance expression becomes

CYY    RYY    E Y      h     h d
2

and the variance becomes

 2     h 2  d

We can now calculate the variances for the shot noise made up of the ramp functions. The details of this
process will be amplified further in Example 6.8. The impulse response given above as h(t) can be
transformed to H(term by term using Matlab. We use the Maple Heaviside function for the step
function so the symbolic processing can correctly recognize it. Using the Fourier transform we obtain
syms t x om
Hom=maple('fourier','t*Heaviside(t)-(t-x)*Heaviside(t-x)-x*Heaviside(tx),t,om')
Hom =
i*(pi*Dirac(1,om)+i/om^2)-i*(-i*x*exp(-i*x*om)*(pi*Dirac(om)i/om)+exp(-i*x*om)*(pi*Dirac(1,om)+i/om^2))
Hom=-1/om^2-i*(-x*exp(-i*x*om)/om+i*exp(-i*x*om)/om^2);
pretty(Hom)
1
/ x exp(-i x om)
i exp(-i x om)\
- --- - i |- -------------- + --------------|
2
|
om
2
|
om
\
om
/
Obtaining the magnitude function using symbolic capability of Matlab
pretty(simplify(Hom*subs(Hom,om,-om)))
2
(2 + i exp(i x om) x om - exp(i x om) - i exp(-i x om) x om + x
/
- exp(-i x om))
/
/
4
om
Rewriting using the exponential notation for simplicity.
Mhomsc= 1/om^4*(2-2*cos(om*x)+om^2*x^2-2*om*x*sin(om*x));
pretty(Mhomsc)
Substitution of sines and cosines for the exponential we obtain
2
2
2 - 2 cos(om x) + x om - 2 om x sin(om x)
------------------------------------------4
om
and
limit(Mhomsc,om,0)
ans =
1/4*x^4
2
om
H() can be found by evaluation of | H() |2 directly and taking the srqt of the result or by integration of
the impulse function
H0=int(t,t,0,x)
H0 =
1/2*x^2
and the power spectrum for y(t) becomes
S YY   

1
  2  2  cos  T 
  2    T  sin   T 
   2  T 2      4  
2
2

2
     T


and the variance is by direct evaluation
syms lam
sigyy=int( lam*t^2,t,0,x)
sigyy =
1/3*x^3*lam
Another form of the impulse response of the linear transfer function is the exponential. The computations
need to obtain the power density function can be repeated using the form of
ht   e ct
for t > 0
and for the transfer function by using Matlab (Maple) Fourier transform capability
syms c
cgt=sym('c>0');
maple('assume',cgt);
maple('fourier','exp(-c*t)*Heaviside(t),t,om')
ans =
1/(c+i*om)
hom=1/(c+i*om);
mhom=simplify(hom*subs(hom,om,-om))
mhom =
-1/(c+i*om)/(-c+i*om)
The denominator can be expanded
den=expand((c+i*om)*(-c+i*om))
den =
-c^2-om^2
pretty(1/-den)
1
--------2
2
c~ + om
By the formulas for output power density function we have
\
2    2   ( )

SYY ( )  S ZZ ( )  H ( ) 
 2
2
2
c 
c 2
2
The delta function forces the evaluation of the rest of the expression at  = 0 we have to invert
2    2   ( )

SYY ( ) 
 2
2
c
c 2
the expression can be rewritten to allow symbolic inverse transformation. Inverse transforming using
Matlab we obtain
cgt=sym('c>0');
maple('assume',cgt);
maple('invfourier','2*pi*lam^2*Dirac(om)/c^2+lam/(c^2+om^2),om,tau')
ans =
1/2*lam*(2*lam+c*exp(-c*tau)*Heaviside(tau)+c*exp(c*tau)*Heaviside(tau))/c^2
to obtain after rewriting
syms tau
pretty( lam^2/c^2+ lam/2/c*exp(-c*abs(tau)))
2
lam
lam exp(-c~ | tau |)
---- + 1/2 -------------------2
c~
c~
We can notice that E[y(t)] = | H(0)|  and then the covariance function becomes
CYY ( )  RYY ( ) 
and the variance becomes
 y2 

c c


2c
e  c 

2c
We can compare the non-impulsive portion of the power density functions for both forms of the linear
transfer function. For the ramp and for the exponential we have
omp=0.01:.05:8;
SRYM=subs(Mhomsc,x,1);
SRY=2*subs(SRYM,om,omp);
SEY=2./(2^2+omp.^2);
figure
plot(omp,SRY,omp,SEY)
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
