Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 6.5 Response to White Noise
In the example we will derive of the correlation function for a process which is the result of the
application of white noise to a linear system. The correlation function for white noise is an impulse
function. We will use the convolution integrals to solve for the output correlation function.
Let us apply white noise to h(t) at t = 0 or
The correlation function is computed according to the following diagram. This diagram represents the nonstationary form of Figure 6.5-2
Notice that t1 and t2 both must be greater than zero and for this derivation we will assume that t 2 > t1.
RXY(t1,t2) is just a convolution of a system impulse response function with an impulse. The lower limit of
the integral is zero because of h(t) = 0 when t < 0.
syms t1 t2 c alp q
'Rxy='
cgt=sym('c>0');
maple('assume',cgt);
maple('int','q*exp(-c*alp)*Dirac(t2-alp-t1),alp=0..inf')
ans =
Rxy=
ans =
1/2*(signum(t2-t1)+1)*exp(-(-t2+t1)*c)*q
This expression can be rewritten using the step function as
R XY t1 , t 2   q  e ct2 t1   ut 2  t1 
Now let us plot RXY(t1,t2) as a function of t1 and t1- separately. First Plot
t1
tt2=2;
tt1=0:.05:3;
Rxyp=1/2*5*(sign(tt2-tt1)+1).*exp(-1.2*(tt2-tt1));
plot(tt1,Rxyp)
xlabel 't1 -sec'
RXY t1 ,t 2  ,as a function of
ylabel 'RXY(t1,t2)'
5
4.5
4
RXY(t1,t2)
3.5
3
2.5
2
1.5
1
0.5
0
Then plot
0
0.5
1
1.5
t1 -sec
2
2.5
3
RXY x, y  as in the convolution integral kernel
t1=1; tt2=2;
aa=-1.5:.05:3;
Rxyp=1/2*5*(sign(tt2-t1+aa)+1).*exp(-1.2*(tt2-t1+aa));
plot(aa,Rxyp)
xlabel 'aa - sec'
ylabel 'RXY(t1-aa,t2)'
5
4.5
4
RXY(t1-aa,t2)
3.5
3
2.5
2
1.5
1
0.5
0
-1.5
Now we may compute RYY
-1
t1 ,t 2  .
-0.5
0
0.5
1
aa -sec
1.5
2
2.5
The zero in the lower limit of the integral is due to the causality of
h(t1) and the t1 in the upper limit is the value where largest nonzero value of
axis.
syms t1 t2 alp
'Ryy='
cgt=sym('c>0');
maple('assume',cgt);
3
RXY t1 ,t 2  occurs on the t1
RYY=int(q*exp(-c*(t2-(t1-alp)))*exp(-c*alp),alp,0,t1);
pretty(RYY)
ans =
Ryy=
q exp(-c~ (t1 + t2))
q exp(c~ (-t2 + t1))
- 1/2 -------------------- + 1/2 -------------------c~
c~
We may factor exp(-c(t2-t1)) because examining the numerator the exponent can be written as
 ct1  t 2   2  c  t1  c  t 2  t1 
and obtain the final result
RYY t1 , t 2  
1
e  ct2 t1 
q
 1  e 2ct1
2
c


We notice that the above correlation function contains a dependence upon t 1 and is nonstationary. For this
case as we take t1   we find that the correlation functions wide sense stationary or
RYY t1 , t 2  
1 q ct2 t1 
 e
2 c
The same results can be obtained by using the convolution integral directly. The first integral for R YX()
contains the impulse function and is identical to the results above. Notice that  = t1-t2 and hence the
symbol YX for XY in the correlation function. The convolution integral from Figure 6.5-2b is

RYY    q   h  u   h u   du
0
The lower limit on the integral is taken at zero because h(t) = 0 for t < 0. The dummy variable can be
changed from u=-v to obtain

RYY    q   h  v   hv   dv
0
Substitution for h(v) and h(+v) we have after expanding the integrand
syms c tau v
tgt=sym('tau>0');
maple('assume',tgt);
RYYt=int(exp(-c*(tau+v))*exp(-c*v),v,0,inf);
pretty(RYYt)
exp(-c~ tau~)
1/2 ------------c~
Using the assume property of Matlab the limit can be set directly to and we do not have to set the limit to
n and integrate and then let n   to obtain the desired result. The RYY() can be obtained in the same
manner and compares to the stationary correlation function which was found above
RYY t1 , t 2  
1 q c
 e
2 c
