Question

2. Consider the random process x(t) defined by x(t) a cos(wt + 6).where w and a are constants, and 0 is a random variable uniformly distributed in the range (-T, ) Sketch the ensemble (sample functions) representing x(t). (2.5 points). a. b. Find the mean and variance of the random variable 0. (2.5 points). Find the mean of x(t), m (t) E(x(t)). (2.5 points). c. d. Find the autocorrelation of x(t), R (t,, t) = E(x, (t)x2 (t)). (5 points). Is the random process x(t) stationary or nonstationary ? (2.5 points). e. Is the random process x(t) ergodic ? (2.5 points). f. g. What is the power of x(t)? (2.5 points).

Answer 1

Below is MAtlab Code:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

close all,
clear all,
clc,

Fs=1000;
t = 0:(1/Fs):1-(1/Fs);

A=1;
F=10;
w = 2*pi*F;

b=pi;
a=-pi;
Xt=[];
Theta=[];
for n=1:Fs
Theta(n) = (b-a)*rand + a;
Xt(n) = A*cos(w*t(n)+Theta(n));
end
subplot(2,1,1); plot(Theta); str=strcat('Theta Plot between -pi to pi'); title(str);
subplot(2,1,2); plot(Xt); str=strcat('Xt Plot at Fs = ',num2str(Fs),' Hz and Freq. of Signal = ',num2str(F),' Hz'); title(str);

MeanTheta = mean(Theta);
VarTheta = var(Theta);
MeanXt = mean(Xt);
PowerXt = sumsqr(Xt)/length(Xt);
[ACF, Lags, Bounds] = autocorr(Xt, [], 2);

disp(['Mean of Theta = ',num2str(MeanTheta)]);
disp(['Var. of Theta = ',num2str(VarTheta)]);
disp(['Mean of Xt = ',num2str(MeanXt)]);
disp(['Power of Xt = ',num2str(PowerXt)]);
disp(['AutoCorr. of Xt = ',num2str(ACF)]);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Matlab Output

Mean of Theta = 0.040587
Var. of Theta = 3.2977
Mean of Xt = 0.0047037
Power of Xt = 0.49912
AutoCorr. of Xt = 1 0.051507 -0.018127 0.008337 -0.023698 0.0062431 0.0075699 -0.037567 -0.034422 -0.022629 0.0096917 -0.0034487 0.0007964 0.010817 -0.00454 0.037404 -0.033741 -0.057873 -0.017376 0.017305 -0.021233
>>

Theta Plot between -pi to pi 1000 900 800 700 FOO 500 400 300 200 100 Xt Plot at Fs 1000 Hz and Freq, of Sianal 10 Hz 0.5 0.5 1000 800 F00 500 400 300 200