How do I compute the power spectral density knowing the input signal autocorrelation modulated by a cosine...












0












$begingroup$


I have the following system:



$$S(t) = X(t) cdot d(t)$$



where



$R_{XX}(tau)=frac{1}{2pi}Sa(frac{tau}{2})cos(3tau)$ and $d(t)=cos(2t + theta)$ where $theta sim U[0,2pi)$ and is independent of $X(t)$



I would like to compute $R_{SS}(tau)$.



Normally, I would think to be given the autocorrelation of $d$, then I could apply the Winer-Khintchine theorem and using the fact that multiplication in time domain is convolution in frequency domain. But, I am stumped where to begin given that I am given the autocorrelation of X(t) but not d(t).



$R_{SS}(t,tau)=E[S(t)S(t+tau)]$



$=E[X(t)d(t)X(t+tau)d(t+tau)]$



$=E[X(t)X(t+tau) d(t)d(t+tau)]$



$=E[X(t)X(t+tau)]E[d(t)d(t+tau)]$ via independence



$=R_{SS}(tau)E[d(t)d(t+tau)]$



Attempting to solve for
$R_{dd}=E[d(t)d(t+tau)]$,



$E[cos(2t+theta)cos(2(t+tau)+theta)]$



Using product identities, I get,



$frac{1}{2}E[cos(4t + 2theta + 2tau)]$ which I believe is zero.






stochastic-processes fourier-transform







share|cite|improve this question











$endgroup$












  • $begingroup$
    At first what you can define is $r_{SS}(t,tau)=mathbb{E}[S(t)overline{S(t+tau)}]$. If $U,V$ are independent then $mathbb{E}[UV] = mathbb{E}[U]mathbb{E}[V]$. If $r_{SS}(t,tau)$ doesn't depend on $t$ (eg. when $S$ is stationary) then you set $R_{SS}(tau) = r_{SS}(0,tau)$
    $endgroup$
    – reuns
    Jan 6 at 17:51












  • $begingroup$
    I dont have U though, just the autocorrelation of U.
    $endgroup$
    – Avedis
    Jan 6 at 18:15










  • $begingroup$
    You have $mathbb{E}[U]$ with $U = X(t) X(t+tau)$ (assuming $mathbb{E}[X(t)] = 0$, otherwise substract it from $X(t)$). $overline{z}$ is the complex conjugate, if $X(t)$ is real it is unnecessary
    $endgroup$
    – reuns
    Jan 6 at 19:33








  • 1




    $begingroup$
    $cos A cos B =frac12 cos (A+B)+frac12cos (A-B)$. In the latter $2t+theta$ disappear so it is constant, the former is zero mean
    $endgroup$
    – reuns
    Jan 6 at 20:32








  • 1




    $begingroup$
    $E[cos(2t+theta)cos(2(t+tau)+theta)] = frac{1}{2}E[cos(2tau)]+frac{1}{2}E[cos(4t + 2theta + 2tau)]= frac{1}{2}cos(2tau)+ 0$
    $endgroup$
    – reuns
    Jan 6 at 23:22


















0












$begingroup$


I have the following system:



$$S(t) = X(t) cdot d(t)$$



where



$R_{XX}(tau)=frac{1}{2pi}Sa(frac{tau}{2})cos(3tau)$ and $d(t)=cos(2t + theta)$ where $theta sim U[0,2pi)$ and is independent of $X(t)$



I would like to compute $R_{SS}(tau)$.



Normally, I would think to be given the autocorrelation of $d$, then I could apply the Winer-Khintchine theorem and using the fact that multiplication in time domain is convolution in frequency domain. But, I am stumped where to begin given that I am given the autocorrelation of X(t) but not d(t).



$R_{SS}(t,tau)=E[S(t)S(t+tau)]$



$=E[X(t)d(t)X(t+tau)d(t+tau)]$



$=E[X(t)X(t+tau) d(t)d(t+tau)]$



$=E[X(t)X(t+tau)]E[d(t)d(t+tau)]$ via independence



$=R_{SS}(tau)E[d(t)d(t+tau)]$



Attempting to solve for
$R_{dd}=E[d(t)d(t+tau)]$,



$E[cos(2t+theta)cos(2(t+tau)+theta)]$



Using product identities, I get,



$frac{1}{2}E[cos(4t + 2theta + 2tau)]$ which I believe is zero.






stochastic-processes fourier-transform







share|cite|improve this question











$endgroup$












  • $begingroup$
    At first what you can define is $r_{SS}(t,tau)=mathbb{E}[S(t)overline{S(t+tau)}]$. If $U,V$ are independent then $mathbb{E}[UV] = mathbb{E}[U]mathbb{E}[V]$. If $r_{SS}(t,tau)$ doesn't depend on $t$ (eg. when $S$ is stationary) then you set $R_{SS}(tau) = r_{SS}(0,tau)$
    $endgroup$
    – reuns
    Jan 6 at 17:51












  • $begingroup$
    I dont have U though, just the autocorrelation of U.
    $endgroup$
    – Avedis
    Jan 6 at 18:15










  • $begingroup$
    You have $mathbb{E}[U]$ with $U = X(t) X(t+tau)$ (assuming $mathbb{E}[X(t)] = 0$, otherwise substract it from $X(t)$). $overline{z}$ is the complex conjugate, if $X(t)$ is real it is unnecessary
    $endgroup$
    – reuns
    Jan 6 at 19:33








  • 1




    $begingroup$
    $cos A cos B =frac12 cos (A+B)+frac12cos (A-B)$. In the latter $2t+theta$ disappear so it is constant, the former is zero mean
    $endgroup$
    – reuns
    Jan 6 at 20:32








  • 1




    $begingroup$
    $E[cos(2t+theta)cos(2(t+tau)+theta)] = frac{1}{2}E[cos(2tau)]+frac{1}{2}E[cos(4t + 2theta + 2tau)]= frac{1}{2}cos(2tau)+ 0$
    $endgroup$
    – reuns
    Jan 6 at 23:22
















0












0








0





$begingroup$


I have the following system:



$$S(t) = X(t) cdot d(t)$$



where



$R_{XX}(tau)=frac{1}{2pi}Sa(frac{tau}{2})cos(3tau)$ and $d(t)=cos(2t + theta)$ where $theta sim U[0,2pi)$ and is independent of $X(t)$



I would like to compute $R_{SS}(tau)$.



Normally, I would think to be given the autocorrelation of $d$, then I could apply the Winer-Khintchine theorem and using the fact that multiplication in time domain is convolution in frequency domain. But, I am stumped where to begin given that I am given the autocorrelation of X(t) but not d(t).



$R_{SS}(t,tau)=E[S(t)S(t+tau)]$



$=E[X(t)d(t)X(t+tau)d(t+tau)]$



$=E[X(t)X(t+tau) d(t)d(t+tau)]$



$=E[X(t)X(t+tau)]E[d(t)d(t+tau)]$ via independence



$=R_{SS}(tau)E[d(t)d(t+tau)]$



Attempting to solve for
$R_{dd}=E[d(t)d(t+tau)]$,



$E[cos(2t+theta)cos(2(t+tau)+theta)]$



Using product identities, I get,



$frac{1}{2}E[cos(4t + 2theta + 2tau)]$ which I believe is zero.






stochastic-processes fourier-transform







share|cite|improve this question











$endgroup$




I have the following system:



$$S(t) = X(t) cdot d(t)$$



where



$R_{XX}(tau)=frac{1}{2pi}Sa(frac{tau}{2})cos(3tau)$ and $d(t)=cos(2t + theta)$ where $theta sim U[0,2pi)$ and is independent of $X(t)$



I would like to compute $R_{SS}(tau)$.



Normally, I would think to be given the autocorrelation of $d$, then I could apply the Winer-Khintchine theorem and using the fact that multiplication in time domain is convolution in frequency domain. But, I am stumped where to begin given that I am given the autocorrelation of X(t) but not d(t).



$R_{SS}(t,tau)=E[S(t)S(t+tau)]$



$=E[X(t)d(t)X(t+tau)d(t+tau)]$



$=E[X(t)X(t+tau) d(t)d(t+tau)]$



$=E[X(t)X(t+tau)]E[d(t)d(t+tau)]$ via independence



$=R_{SS}(tau)E[d(t)d(t+tau)]$



Attempting to solve for
$R_{dd}=E[d(t)d(t+tau)]$,



$E[cos(2t+theta)cos(2(t+tau)+theta)]$



Using product identities, I get,



$frac{1}{2}E[cos(4t + 2theta + 2tau)]$ which I believe is zero.






stochastic-processes fourier-transform




stochastic-processes fourier-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 21:23







Avedis

















asked Jan 6 at 17:39









AvedisAvedis

577




577












  • $begingroup$
    At first what you can define is $r_{SS}(t,tau)=mathbb{E}[S(t)overline{S(t+tau)}]$. If $U,V$ are independent then $mathbb{E}[UV] = mathbb{E}[U]mathbb{E}[V]$. If $r_{SS}(t,tau)$ doesn't depend on $t$ (eg. when $S$ is stationary) then you set $R_{SS}(tau) = r_{SS}(0,tau)$
    $endgroup$
    – reuns
    Jan 6 at 17:51












  • $begingroup$
    I dont have U though, just the autocorrelation of U.
    $endgroup$
    – Avedis
    Jan 6 at 18:15










  • $begingroup$
    You have $mathbb{E}[U]$ with $U = X(t) X(t+tau)$ (assuming $mathbb{E}[X(t)] = 0$, otherwise substract it from $X(t)$). $overline{z}$ is the complex conjugate, if $X(t)$ is real it is unnecessary
    $endgroup$
    – reuns
    Jan 6 at 19:33








  • 1




    $begingroup$
    $cos A cos B =frac12 cos (A+B)+frac12cos (A-B)$. In the latter $2t+theta$ disappear so it is constant, the former is zero mean
    $endgroup$
    – reuns
    Jan 6 at 20:32








  • 1




    $begingroup$
    $E[cos(2t+theta)cos(2(t+tau)+theta)] = frac{1}{2}E[cos(2tau)]+frac{1}{2}E[cos(4t + 2theta + 2tau)]= frac{1}{2}cos(2tau)+ 0$
    $endgroup$
    – reuns
    Jan 6 at 23:22




















  • $begingroup$
    At first what you can define is $r_{SS}(t,tau)=mathbb{E}[S(t)overline{S(t+tau)}]$. If $U,V$ are independent then $mathbb{E}[UV] = mathbb{E}[U]mathbb{E}[V]$. If $r_{SS}(t,tau)$ doesn't depend on $t$ (eg. when $S$ is stationary) then you set $R_{SS}(tau) = r_{SS}(0,tau)$
    $endgroup$
    – reuns
    Jan 6 at 17:51












  • $begingroup$
    I dont have U though, just the autocorrelation of U.
    $endgroup$
    – Avedis
    Jan 6 at 18:15










  • $begingroup$
    You have $mathbb{E}[U]$ with $U = X(t) X(t+tau)$ (assuming $mathbb{E}[X(t)] = 0$, otherwise substract it from $X(t)$). $overline{z}$ is the complex conjugate, if $X(t)$ is real it is unnecessary
    $endgroup$
    – reuns
    Jan 6 at 19:33








  • 1




    $begingroup$
    $cos A cos B =frac12 cos (A+B)+frac12cos (A-B)$. In the latter $2t+theta$ disappear so it is constant, the former is zero mean
    $endgroup$
    – reuns
    Jan 6 at 20:32








  • 1




    $begingroup$
    $E[cos(2t+theta)cos(2(t+tau)+theta)] = frac{1}{2}E[cos(2tau)]+frac{1}{2}E[cos(4t + 2theta + 2tau)]= frac{1}{2}cos(2tau)+ 0$
    $endgroup$
    – reuns
    Jan 6 at 23:22


















$begingroup$
At first what you can define is $r_{SS}(t,tau)=mathbb{E}[S(t)overline{S(t+tau)}]$. If $U,V$ are independent then $mathbb{E}[UV] = mathbb{E}[U]mathbb{E}[V]$. If $r_{SS}(t,tau)$ doesn't depend on $t$ (eg. when $S$ is stationary) then you set $R_{SS}(tau) = r_{SS}(0,tau)$
$endgroup$
– reuns
Jan 6 at 17:51






$begingroup$
At first what you can define is $r_{SS}(t,tau)=mathbb{E}[S(t)overline{S(t+tau)}]$. If $U,V$ are independent then $mathbb{E}[UV] = mathbb{E}[U]mathbb{E}[V]$. If $r_{SS}(t,tau)$ doesn't depend on $t$ (eg. when $S$ is stationary) then you set $R_{SS}(tau) = r_{SS}(0,tau)$
$endgroup$
– reuns
Jan 6 at 17:51














$begingroup$
I dont have U though, just the autocorrelation of U.
$endgroup$
– Avedis
Jan 6 at 18:15




$begingroup$
I dont have U though, just the autocorrelation of U.
$endgroup$
– Avedis
Jan 6 at 18:15












$begingroup$
You have $mathbb{E}[U]$ with $U = X(t) X(t+tau)$ (assuming $mathbb{E}[X(t)] = 0$, otherwise substract it from $X(t)$). $overline{z}$ is the complex conjugate, if $X(t)$ is real it is unnecessary
$endgroup$
– reuns
Jan 6 at 19:33






$begingroup$
You have $mathbb{E}[U]$ with $U = X(t) X(t+tau)$ (assuming $mathbb{E}[X(t)] = 0$, otherwise substract it from $X(t)$). $overline{z}$ is the complex conjugate, if $X(t)$ is real it is unnecessary
$endgroup$
– reuns
Jan 6 at 19:33






1




1




$begingroup$
$cos A cos B =frac12 cos (A+B)+frac12cos (A-B)$. In the latter $2t+theta$ disappear so it is constant, the former is zero mean
$endgroup$
– reuns
Jan 6 at 20:32






$begingroup$
$cos A cos B =frac12 cos (A+B)+frac12cos (A-B)$. In the latter $2t+theta$ disappear so it is constant, the former is zero mean
$endgroup$
– reuns
Jan 6 at 20:32






1




1




$begingroup$
$E[cos(2t+theta)cos(2(t+tau)+theta)] = frac{1}{2}E[cos(2tau)]+frac{1}{2}E[cos(4t + 2theta + 2tau)]= frac{1}{2}cos(2tau)+ 0$
$endgroup$
– reuns
Jan 6 at 23:22






$begingroup$
$E[cos(2t+theta)cos(2(t+tau)+theta)] = frac{1}{2}E[cos(2tau)]+frac{1}{2}E[cos(4t + 2theta + 2tau)]= frac{1}{2}cos(2tau)+ 0$
$endgroup$
– reuns
Jan 6 at 23:22












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