Mixing
How do I compute the power spectral density knowing the input signal autocorrelation modulated by a cosine...
$begingroup$
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
asked Jan 6 at 17:39
AvedisAvedis
476
$endgroup$
|
show 5 more comments
$begingroup$
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
asked Jan 6 at 17:39
AvedisAvedis
476
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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
asked Jan 6 at 17:39
AvedisAvedis
476
$endgroup$
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
asked Jan 6 at 17:39
AvedisAvedis
476
asked Jan 6 at 17:39
AvedisAvedis
476
edited Jan 6 at 21:23
Avedis
asked Jan 6 at 17:39
AvedisAvedis
476
asked Jan 6 at 17:39
AvedisAvedis
476
asked Jan 6 at 17:39
AvedisAvedis
476
476
$begingroup$
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
|
show 5 more comments
$begingroup$
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
|
show 5 more comments
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$begingroup$
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