Thread Subject:

need to compute this problem having problems with how to start this problem help need urgently

Let   U[0; 2theta] be a uniform random variable from the interval [0; 2theta] and let A  Exp(1) be exponentially distributed with mean 1. Assume  and A independent. Compute the mean mX(t) =E[X(t)] and autocorrelation RX(s; t) = E[X(s)X(t)] of the phase-shifted sinusoid.X(t) = A*
cos(t +theta ):
State also if X(t) is Wide Sense Stationary (WSS).
plot 10 realisations of X(t)
plotR(s-t,0)as a function of s-t

"pramod kumar" <pramod.kilu@gmail.com> wrote in message <jpapso$44b$1@newscl01ah.mathworks.com>...
> Let   U[0; 2theta] be a uniform random variable from the interval [0; 2theta] and let A  Exp(1) be exponentially distributed with mean 1. Assume  and A independent. Compute the mean mX(t) =E[X(t)] and autocorrelation RX(s; t) = E[X(s)X(t)] of the phase-shifted sinusoid.X(t) = A*
cos(t +theta ):
> State also if X(t) is Wide Sense Stationary (WSS).
> plot 10 realisations of X(t)
> plotR(s-t,0)as a function of s-t

So what have you tried? If you have not tried anything,
this is a suggestion that you were not paying attention
in class.

Make an effort.

John

"pramod kumar" <pramod.kilu@gmail.com> wrote in message <jpapso$44b$1@newscl01ah.mathworks.com>...
> Let   U[0; 2theta] be a uniform random variable from the interval [0; 2theta] and let A  Exp(1) be exponentially distributed with mean 1. Assume  and A independent. Compute the mean mX(t) =E[X(t)] and autocorrelation RX(s; t) = E[X(s)X(t)] of the phase-shifted sinusoid.X(t) = A*
cos(t +theta ):
=============

How to start it? OK, you've been asked to compute the mean of X(t) and you've been told that A is independent of all other variables in the problem. So because of this independence and because E[A]=1

E[X(t)]=E[A] *E[cos(t+theta)] = E[cos(t+theta)]

"Matt J" wrote in message <jpavb2$o7h$1@newscl01ah.mathworks.com>...
> "pramod kumar" <pramod.kilu@gmail.com> wrote in message <jpapso$44b$1@newscl01ah.mathworks.com>...
> > Let   U[0; 2theta] be a uniform random variable from the interval [0; 2theta] and let A  Exp(1) be exponentially distributed with mean 1. Assume  and A independent. Compute the mean mX(t) =E[X(t)] and autocorrelation RX(s; t) = E[X(s)X(t)] of the phase-shifted sinusoid.X(t) = A*
cos(t +theta ):
> =============
>
> How to start it? OK, you've been asked to compute the mean of X(t) and you've been told that A is independent of all other variables in the problem. So because of this independence and because E[A]=1
>
> E[X(t)]=E[A] *E[cos(t+theta)] = E[cos(t+theta)]

i have tried this code after that what should i do
clear all
close all
 clc
N=10;
% t=0:1:N-1;
t=linspace(-1,1,N-1);
A=exprnd(1);
theta=2*pi*rand(10,1);
y=A*cos(theta);
%Xt=(A*cosint(t+theta));

"pramod kumar" <pramod.kilu@gmail.com> wrote in message <jpavo9$pni$1@newscl01ah.mathworks.com>...
> "Matt J" wrote in message <jpavb2$o7h$1@newscl01ah.mathworks.com>...
> > "pramod kumar" <pramod.kilu@gmail.com> wrote in message <jpapso$44b$1@newscl01ah.mathworks.com>...
> > > Let   U[0; 2theta] be a uniform random variable from the interval [0; 2theta] and let A  Exp(1) be exponentially distributed with mean 1. Assume  and A independent. Compute the mean mX(t) =E[X(t)] and autocorrelation RX(s; t) = E[X(s)X(t)] of the phase-shifted sinusoid.X(t) = A*
cos(t +theta ):
> > =============
> >
> > How to start it? OK, you've been asked to compute the mean of X(t) and you've been told that A is independent of all other variables in the problem. So because of this independence and because E[A]=1
> >
> > E[X(t)]=E[A] *E[cos(t+theta)] = E[cos(t+theta)]
>
> i have tried this code after that what should i do
> clear all
> close all
> clc
> N=10;
> % t=0:1:N-1;
> t=linspace(-1,1,N-1);
> A=exprnd(1);
> theta=2*pi*rand(10,1);
> y=A*cos(theta);

This looks like it should be

y=A*cos(t+theta);

You should do this 10 more times to obtain the 10 realizations asked for.
You should then use the PLOT command to start making the plots requested in the exercise.

clear all, close all
realizations=10;
N=1;
%a.plot 10 realisations of X(t)
for i=1:realizations
theta=2*pi*rand(N,1);
t=0:0.0001:4*pi;
A=exprnd(1,N,1);
Xt=A*cos(t+theta);
plot(t,Xt); hold on;
end
by using this code i have calculated 10 iterations please let me know for each realization i would like to have different colur so let me know how can i plot this in different colors

"pramod kumar" <pramod.kilu@gmail.com> wrote in message <jpbakv$7mo$1@newscl01ah.mathworks.com>...
> clear all, close all
> realizations=10;
> N=1;
> %a.plot 10 realisations of X(t)
> for i=1:realizations
> theta=2*pi*rand(N,1);
> t=0:0.0001:4*pi;
> A=exprnd(1,N,1);
> Xt=A*cos(t+theta);
> plot(t,Xt); hold on;
> end
> by using this code i have calculated 10 iterations please let me know for each realization i would like to have different colur so let me know how can i plot this in different colors
===========

The PLOT command only offers 8 different colors, but you can alternate both colors and line styles using something like the following:


plotcolors={'r','b','g','m','k', 'r*','b*','g*','m*','k*'};

clear all, close all
realizations=10;
N=1;
%a.plot 10 realisations of X(t)
for i=1:realizations
theta=2*pi*rand(N,1);
t=0:0.0001:4*pi;
A=exprnd(1,N,1);
Xt=A*cos(t+theta);
plot(t,Xt,plotcolors{i}); hold on;
end

thank you
for your help
 do u have any idea about this problem
Let Y (t) be a short-term discounted average of the process X(t), i.e.
Y (t) =1\(1-e^-T)*integral(e^-(t-s)X(s)ds(intergral ranges from t-T,t)

for some xed T > 0.
(a) Find the impulse response h(tou ) of this lter and the corresponding transfer function H(f).
i have developed this code
clear all, close all
realizations=10;
N=1;
%a.plot 10 realisations of X(t)
for i=1:realizations
theta=2*pi*rand(N,1);
t=0:0.0001:4*pi;
A=exprnd(1,N,1);
Yt=(1\(1-e^-T))*expint(X(s));
plot(t,Xt); hold on;
end
but the integral and exponential values i could not find it exactly