Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 3.9 Random Number Sequences with Continuous Distributions
In this example we will generate a random sequence with an exponential distribution and show as
we increase the number of samples in the sequence the mean and variance approaches the theoretical values
for the exponential distribution. We will also show that the density distribution for the sequence
approaches an exponential distribution.
The transformation used to generate the sequence is
exp l    
1

 log rand (1, N ) 
N=500;
r=rand(N,1);
The terms in the sequence are with  = 0.2
lam=0.2;
P=-1/lam*log(r);
The mean as a function of n is defined. We have used the cumsum to sum the first n of the P terms. We
plot this function with the theoretical value for the mean = 1 / for comparison.
k=1:N;
me=cumsum(P)'./k;
tme=1/lam*ones(1,N);
plot(k,me,k,tme)
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A similar function is defined for the variance and plotted with the theoretical value
m2=cumsum(P.^2)'./[1:N];
tm2=1/lam^2*ones(1,N);
plot(k,m2-me.^2,k,tm2)
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In order to generate a random sequence with an arbitrary distribution we need to have an analytical
expression for the inverse of the cumulative distribution function. We may also use Matlab to invert a
function or solve a nonlinear equation using solve. Let us define the inverse exponential function as
syms x lambda u
f=1-exp(-lambda*x);
solve(f-u)
ans =
-log(1-u)/lambda
We now generate a sequence using the inverse function and compare the analytical density function with
the histogram.
[w,v]=hist(P,40);
fp=lam*exp(-lam*v);
bar(v,w/N,0.2);hold on; plot(v,fp); hold off
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
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0.02
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