f

#### 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 0 5/20/2012 1:00:09 PM comp.soft-sys.matlab  211264 articles. 26 followers. 7 Replies 5133 Views Similar Articles

[PageSpeed] 9

"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 0 5/20/2012 1:16:07 PM
"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)] 0 5/20/2012 2:33:07 PM
"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)); 0 5/20/2012 2:40:09 PM
"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. 0 5/20/2012 2:58:09 PM
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 0 5/20/2012 5:46:07 PM
"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 0 5/20/2012 6:02:05 PM
thank you
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 0 5/20/2012 6:27:07 PM Similar Artilces:

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