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How to compute autocorrelation of signal defined by difference equations?
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I have no experience with difference equations and I want to learn how to compute the following, but I found no resource online. Any help would be greatly appreciated.
Find:
$$mathbbEleft[d[n]d[n + k]right]$$
Where:
$d[n]$ is a discrete time signal of the form:
$$d[n] = alpha ,d[n−1]+v[n] quad (alpha in [0;1])$$
$v[n]$ is a sequence of uncorrelated zero-mean unit-variance random variables.
discrete-signals autocorrelation
asked 2 hours ago
Julien__
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add a comment |Â
up vote
2
down vote
favorite
I have no experience with difference equations and I want to learn how to compute the following, but I found no resource online. Any help would be greatly appreciated.
Find:
$$mathbbEleft[d[n]d[n + k]right]$$
Where:
$d[n]$ is a discrete time signal of the form:
$$d[n] = alpha ,d[n−1]+v[n] quad (alpha in [0;1])$$
$v[n]$ is a sequence of uncorrelated zero-mean unit-variance random variables.
discrete-signals autocorrelation
asked 2 hours ago
Julien__
1112
New contributor
Julien__ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hint: Your definition for $d[n]$ still works if we replace $n$ by $m=n+1$ and so you know how $d[n+1]$ relates to $d[n]$. Proceed iteratively to find $d[n+2]$ in terms of $d[n+1]$ and replace $d[n+1]$ by the just discovered formula $alpha d[n]+nu[n+1]$ for $d[n+2]$. Lather, rinse, repeat, working backwards to ultimately find $d[n+k]$ in terms of $d[n]$ and $nu[n+i], 0 leq i leq k$. Then substitute into $E[d[n]d[n+k]]$ and work from there.
– Dilip Sarwate
2 hours ago
Thanks @DilipSarwate, but it doesn't help because I'm stuck with terms ind[n]v[n+k]. Even for $$mathbbE[d[n]d[n+1]] = mathbbE[alpha d[n]^2 + d[n]v[n+1]]$$ I don't know how to proceed
– Julien__
1 hour ago
What is $$mathbbE[ d[n]v[n+1] ]$$?
– Julien__
1 hour ago
Your model probably forgot to mention this explicitly or maybe you forgot to include this in your description of the problem, but it is typically assumed than the noise $nu$ is independent of the signal $d$ and so $E[d[n]nu[m]]=E[d[n]]cdot E[nu[m] = E[d[n]]cdot 0 = 0$ for all $n$ and $m$.
– Dilip Sarwate
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have no experience with difference equations and I want to learn how to compute the following, but I found no resource online. Any help would be greatly appreciated.
Find:
$$mathbbEleft[d[n]d[n + k]right]$$
Where:
$d[n]$ is a discrete time signal of the form:
$$d[n] = alpha ,d[n−1]+v[n] quad (alpha in [0;1])$$
$v[n]$ is a sequence of uncorrelated zero-mean unit-variance random variables.
discrete-signals autocorrelation
asked 2 hours ago
Julien__
1112
New contributor
Julien__ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have no experience with difference equations and I want to learn how to compute the following, but I found no resource online. Any help would be greatly appreciated.
Find:
$$mathbbEleft[d[n]d[n + k]right]$$
Where:
$d[n]$ is a discrete time signal of the form:
$$d[n] = alpha ,d[n−1]+v[n] quad (alpha in [0;1])$$
$v[n]$ is a sequence of uncorrelated zero-mean unit-variance random variables.
discrete-signals autocorrelation
discrete-signals autocorrelation
asked 2 hours ago
Julien__
1112
New contributor
Julien__ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Julien__
1112
New contributor
Julien__ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Julien__
1112
New contributor
Julien__ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Julien__
1112
asked 2 hours ago
Julien__
1112
1112
New contributor
Julien__ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Julien__ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Julien__ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hint: Your definition for $d[n]$ still works if we replace $n$ by $m=n+1$ and so you know how $d[n+1]$ relates to $d[n]$. Proceed iteratively to find $d[n+2]$ in terms of $d[n+1]$ and replace $d[n+1]$ by the just discovered formula $alpha d[n]+nu[n+1]$ for $d[n+2]$. Lather, rinse, repeat, working backwards to ultimately find $d[n+k]$ in terms of $d[n]$ and $nu[n+i], 0 leq i leq k$. Then substitute into $E[d[n]d[n+k]]$ and work from there.
– Dilip Sarwate
2 hours ago
Thanks @DilipSarwate, but it doesn't help because I'm stuck with terms ind[n]v[n+k]. Even for $$mathbbE[d[n]d[n+1]] = mathbbE[alpha d[n]^2 + d[n]v[n+1]]$$ I don't know how to proceed
– Julien__
1 hour ago
What is $$mathbbE[ d[n]v[n+1] ]$$?
– Julien__
1 hour ago
Your model probably forgot to mention this explicitly or maybe you forgot to include this in your description of the problem, but it is typically assumed than the noise $nu$ is independent of the signal $d$ and so $E[d[n]nu[m]]=E[d[n]]cdot E[nu[m] = E[d[n]]cdot 0 = 0$ for all $n$ and $m$.
– Dilip Sarwate
1 hour ago
add a comment |Â
Hint: Your definition for $d[n]$ still works if we replace $n$ by $m=n+1$ and so you know how $d[n+1]$ relates to $d[n]$. Proceed iteratively to find $d[n+2]$ in terms of $d[n+1]$ and replace $d[n+1]$ by the just discovered formula $alpha d[n]+nu[n+1]$ for $d[n+2]$. Lather, rinse, repeat, working backwards to ultimately find $d[n+k]$ in terms of $d[n]$ and $nu[n+i], 0 leq i leq k$. Then substitute into $E[d[n]d[n+k]]$ and work from there.
– Dilip Sarwate
2 hours ago
Thanks @DilipSarwate, but it doesn't help because I'm stuck with terms ind[n]v[n+k]. Even for $$mathbbE[d[n]d[n+1]] = mathbbE[alpha d[n]^2 + d[n]v[n+1]]$$ I don't know how to proceed
– Julien__
1 hour ago
What is $$mathbbE[ d[n]v[n+1] ]$$?
– Julien__
1 hour ago
Your model probably forgot to mention this explicitly or maybe you forgot to include this in your description of the problem, but it is typically assumed than the noise $nu$ is independent of the signal $d$ and so $E[d[n]nu[m]]=E[d[n]]cdot E[nu[m] = E[d[n]]cdot 0 = 0$ for all $n$ and $m$.
– Dilip Sarwate
1 hour ago
Hint: Your definition for $d[n]$ still works if we replace $n$ by $m=n+1$ and so you know how $d[n+1]$ relates to $d[n]$. Proceed iteratively to find $d[n+2]$ in terms of $d[n+1]$ and replace $d[n+1]$ by the just discovered formula $alpha d[n]+nu[n+1]$ for $d[n+2]$. Lather, rinse, repeat, working backwards to ultimately find $d[n+k]$ in terms of $d[n]$ and $nu[n+i], 0 leq i leq k$. Then substitute into $E[d[n]d[n+k]]$ and work from there.
– Dilip Sarwate
2 hours ago
Hint: Your definition for $d[n]$ still works if we replace $n$ by $m=n+1$ and so you know how $d[n+1]$ relates to $d[n]$. Proceed iteratively to find $d[n+2]$ in terms of $d[n+1]$ and replace $d[n+1]$ by the just discovered formula $alpha d[n]+nu[n+1]$ for $d[n+2]$. Lather, rinse, repeat, working backwards to ultimately find $d[n+k]$ in terms of $d[n]$ and $nu[n+i], 0 leq i leq k$. Then substitute into $E[d[n]d[n+k]]$ and work from there.
– Dilip Sarwate
2 hours ago
Thanks @DilipSarwate, but it doesn't help because I'm stuck with terms in
d[n]v[n+k]. Even for $$mathbbE[d[n]d[n+1]] = mathbbE[alpha d[n]^2 + d[n]v[n+1]]$$ I don't know how to proceed– Julien__
1 hour ago
Thanks @DilipSarwate, but it doesn't help because I'm stuck with terms in
d[n]v[n+k]. Even for $$mathbbE[d[n]d[n+1]] = mathbbE[alpha d[n]^2 + d[n]v[n+1]]$$ I don't know how to proceed– Julien__
1 hour ago
What is $$mathbbE[ d[n]v[n+1] ]$$?
– Julien__
1 hour ago
What is $$mathbbE[ d[n]v[n+1] ]$$?
– Julien__
1 hour ago
Your model probably forgot to mention this explicitly or maybe you forgot to include this in your description of the problem, but it is typically assumed than the noise $nu$ is independent of the signal $d$ and so $E[d[n]nu[m]]=E[d[n]]cdot E[nu[m] = E[d[n]]cdot 0 = 0$ for all $n$ and $m$.
– Dilip Sarwate
1 hour ago
Your model probably forgot to mention this explicitly or maybe you forgot to include this in your description of the problem, but it is typically assumed than the noise $nu$ is independent of the signal $d$ and so $E[d[n]nu[m]]=E[d[n]]cdot E[nu[m] = E[d[n]]cdot 0 = 0$ for all $n$ and $m$.
– Dilip Sarwate
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
I would approach this by trying to express $d(n)$ differently, as difference equations might be confusing. Notice that:
$$
beginalign
d(n)
&= alpha d(n-1) + v(n) \
&= alpha^2 d(n-2) +alpha v(n-1) +v(n) \
&=alpha^3 d(n-3) + alpha^2 v(n-2)+alpha v(n-1) +v(n) \
&=...
endalign
$$
You can see from the pattern that you can express $d(n)$ as a summation:
$$d(n) =alpha^nd(0)+sum_i=0^n-1alpha^iv(n-i)$$
where I assumed that the procces started at $n=0$. For simplicty, let me take $d(0)=0$ so that we don't have to carry that constant throughout the calculations.
Then we are left with:
$$mathbbE[d(n)d(n+k)] = mathbbEleft[ left(sum_i=0^n-1alpha^iv(n-i) right) left(sum_i=0^n+k-1alpha^iv(n+k-i) right)right]$$
Knowing that the noise is uncorrelated with zero-mean, this expression can be really simplified.
Can you go on from here?
answered 1 hour ago
Tendero
3,88611033
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up vote
2
down vote
There are different approaches to compute the requested auto-correlations but I would like to provide the simplest that suits to your case provided that.
- The difference equation is an LCCDE type indicating an LTI system.
- The input to the system $v[n]$ is WSS (wide sense stationary) random process.
Both of which I assume to be observed in your question. Then I would use the following well known relation between the input and output auto-correlations of an (real) LTI system driven by (real) WSS input $v[n]$:
$$ r_dd[k] = h[k] star h[-k] star r_vv[k] $$
where the auto-correlation of WSS $v[n]$ is $r_vv[k] = sigma_v^2 delta[k] = delta[k]$ for an uncorrelated, zero mean, unit variance process. And $h[n]$ is the impulse-response of the LTI system signified by the LCCDE.
Then the auto-correlation sequence of the WSS output $d[n]$ is
$$ r_dd[k] = h[k] star h[-k] = sum_m=-infty^infty h[m]h[-(k-m)]~~~,~~~text for k = 0,pm 1, pm 2,... $$
So assuming that your LCCDE :
$$d[n] - alpha d[n-k] = v[n] ~~~, ~~~text with |alpha| < 1 $$
signifies a causal LTI system, then its impulse response is:
$$ h[n] = (-alpha)^n u[n] $$
and yields the auto-correlations to be:
$$ r_dd[k] = sum_m=-infty^infty (-alpha)^m u[m] (-alpha)^m-k u[m-k]~~~,~~~text for k = 0,pm 1, pm 2,... $$
I hope you can proceed the rest.
answered 1 hour ago
Fat32
12.8k31127
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I would approach this by trying to express $d(n)$ differently, as difference equations might be confusing. Notice that:
$$
beginalign
d(n)
&= alpha d(n-1) + v(n) \
&= alpha^2 d(n-2) +alpha v(n-1) +v(n) \
&=alpha^3 d(n-3) + alpha^2 v(n-2)+alpha v(n-1) +v(n) \
&=...
endalign
$$
You can see from the pattern that you can express $d(n)$ as a summation:
$$d(n) =alpha^nd(0)+sum_i=0^n-1alpha^iv(n-i)$$
where I assumed that the procces started at $n=0$. For simplicty, let me take $d(0)=0$ so that we don't have to carry that constant throughout the calculations.
Then we are left with:
$$mathbbE[d(n)d(n+k)] = mathbbEleft[ left(sum_i=0^n-1alpha^iv(n-i) right) left(sum_i=0^n+k-1alpha^iv(n+k-i) right)right]$$
Knowing that the noise is uncorrelated with zero-mean, this expression can be really simplified.
Can you go on from here?
answered 1 hour ago
Tendero
3,88611033
add a comment |Â
up vote
2
down vote
I would approach this by trying to express $d(n)$ differently, as difference equations might be confusing. Notice that:
$$
beginalign
d(n)
&= alpha d(n-1) + v(n) \
&= alpha^2 d(n-2) +alpha v(n-1) +v(n) \
&=alpha^3 d(n-3) + alpha^2 v(n-2)+alpha v(n-1) +v(n) \
&=...
endalign
$$
You can see from the pattern that you can express $d(n)$ as a summation:
$$d(n) =alpha^nd(0)+sum_i=0^n-1alpha^iv(n-i)$$
where I assumed that the procces started at $n=0$. For simplicty, let me take $d(0)=0$ so that we don't have to carry that constant throughout the calculations.
Then we are left with:
$$mathbbE[d(n)d(n+k)] = mathbbEleft[ left(sum_i=0^n-1alpha^iv(n-i) right) left(sum_i=0^n+k-1alpha^iv(n+k-i) right)right]$$
Knowing that the noise is uncorrelated with zero-mean, this expression can be really simplified.
Can you go on from here?
answered 1 hour ago
Tendero
3,88611033
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I would approach this by trying to express $d(n)$ differently, as difference equations might be confusing. Notice that:
$$
beginalign
d(n)
&= alpha d(n-1) + v(n) \
&= alpha^2 d(n-2) +alpha v(n-1) +v(n) \
&=alpha^3 d(n-3) + alpha^2 v(n-2)+alpha v(n-1) +v(n) \
&=...
endalign
$$
You can see from the pattern that you can express $d(n)$ as a summation:
$$d(n) =alpha^nd(0)+sum_i=0^n-1alpha^iv(n-i)$$
where I assumed that the procces started at $n=0$. For simplicty, let me take $d(0)=0$ so that we don't have to carry that constant throughout the calculations.
Then we are left with:
$$mathbbE[d(n)d(n+k)] = mathbbEleft[ left(sum_i=0^n-1alpha^iv(n-i) right) left(sum_i=0^n+k-1alpha^iv(n+k-i) right)right]$$
Knowing that the noise is uncorrelated with zero-mean, this expression can be really simplified.
Can you go on from here?
answered 1 hour ago
Tendero
3,88611033
I would approach this by trying to express $d(n)$ differently, as difference equations might be confusing. Notice that:
$$
beginalign
d(n)
&= alpha d(n-1) + v(n) \
&= alpha^2 d(n-2) +alpha v(n-1) +v(n) \
&=alpha^3 d(n-3) + alpha^2 v(n-2)+alpha v(n-1) +v(n) \
&=...
endalign
$$
You can see from the pattern that you can express $d(n)$ as a summation:
$$d(n) =alpha^nd(0)+sum_i=0^n-1alpha^iv(n-i)$$
where I assumed that the procces started at $n=0$. For simplicty, let me take $d(0)=0$ so that we don't have to carry that constant throughout the calculations.
Then we are left with:
$$mathbbE[d(n)d(n+k)] = mathbbEleft[ left(sum_i=0^n-1alpha^iv(n-i) right) left(sum_i=0^n+k-1alpha^iv(n+k-i) right)right]$$
Knowing that the noise is uncorrelated with zero-mean, this expression can be really simplified.
Can you go on from here?
answered 1 hour ago
Tendero
3,88611033
edited 1 hour ago
answered 1 hour ago
Tendero
3,88611033
answered 1 hour ago
Tendero
3,88611033
answered 1 hour ago
Tendero
3,88611033
3,88611033
add a comment |Â
add a comment |Â
up vote
2
down vote
There are different approaches to compute the requested auto-correlations but I would like to provide the simplest that suits to your case provided that.
- The difference equation is an LCCDE type indicating an LTI system.
- The input to the system $v[n]$ is WSS (wide sense stationary) random process.
Both of which I assume to be observed in your question. Then I would use the following well known relation between the input and output auto-correlations of an (real) LTI system driven by (real) WSS input $v[n]$:
$$ r_dd[k] = h[k] star h[-k] star r_vv[k] $$
where the auto-correlation of WSS $v[n]$ is $r_vv[k] = sigma_v^2 delta[k] = delta[k]$ for an uncorrelated, zero mean, unit variance process. And $h[n]$ is the impulse-response of the LTI system signified by the LCCDE.
Then the auto-correlation sequence of the WSS output $d[n]$ is
$$ r_dd[k] = h[k] star h[-k] = sum_m=-infty^infty h[m]h[-(k-m)]~~~,~~~text for k = 0,pm 1, pm 2,... $$
So assuming that your LCCDE :
$$d[n] - alpha d[n-k] = v[n] ~~~, ~~~text with |alpha| < 1 $$
signifies a causal LTI system, then its impulse response is:
$$ h[n] = (-alpha)^n u[n] $$
and yields the auto-correlations to be:
$$ r_dd[k] = sum_m=-infty^infty (-alpha)^m u[m] (-alpha)^m-k u[m-k]~~~,~~~text for k = 0,pm 1, pm 2,... $$
I hope you can proceed the rest.
answered 1 hour ago
Fat32
12.8k31127
add a comment |Â
up vote
2
down vote
There are different approaches to compute the requested auto-correlations but I would like to provide the simplest that suits to your case provided that.
- The difference equation is an LCCDE type indicating an LTI system.
- The input to the system $v[n]$ is WSS (wide sense stationary) random process.
Both of which I assume to be observed in your question. Then I would use the following well known relation between the input and output auto-correlations of an (real) LTI system driven by (real) WSS input $v[n]$:
$$ r_dd[k] = h[k] star h[-k] star r_vv[k] $$
where the auto-correlation of WSS $v[n]$ is $r_vv[k] = sigma_v^2 delta[k] = delta[k]$ for an uncorrelated, zero mean, unit variance process. And $h[n]$ is the impulse-response of the LTI system signified by the LCCDE.
Then the auto-correlation sequence of the WSS output $d[n]$ is
$$ r_dd[k] = h[k] star h[-k] = sum_m=-infty^infty h[m]h[-(k-m)]~~~,~~~text for k = 0,pm 1, pm 2,... $$
So assuming that your LCCDE :
$$d[n] - alpha d[n-k] = v[n] ~~~, ~~~text with |alpha| < 1 $$
signifies a causal LTI system, then its impulse response is:
$$ h[n] = (-alpha)^n u[n] $$
and yields the auto-correlations to be:
$$ r_dd[k] = sum_m=-infty^infty (-alpha)^m u[m] (-alpha)^m-k u[m-k]~~~,~~~text for k = 0,pm 1, pm 2,... $$
I hope you can proceed the rest.
answered 1 hour ago
Fat32
12.8k31127
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There are different approaches to compute the requested auto-correlations but I would like to provide the simplest that suits to your case provided that.
- The difference equation is an LCCDE type indicating an LTI system.
- The input to the system $v[n]$ is WSS (wide sense stationary) random process.
Both of which I assume to be observed in your question. Then I would use the following well known relation between the input and output auto-correlations of an (real) LTI system driven by (real) WSS input $v[n]$:
$$ r_dd[k] = h[k] star h[-k] star r_vv[k] $$
where the auto-correlation of WSS $v[n]$ is $r_vv[k] = sigma_v^2 delta[k] = delta[k]$ for an uncorrelated, zero mean, unit variance process. And $h[n]$ is the impulse-response of the LTI system signified by the LCCDE.
Then the auto-correlation sequence of the WSS output $d[n]$ is
$$ r_dd[k] = h[k] star h[-k] = sum_m=-infty^infty h[m]h[-(k-m)]~~~,~~~text for k = 0,pm 1, pm 2,... $$
So assuming that your LCCDE :
$$d[n] - alpha d[n-k] = v[n] ~~~, ~~~text with |alpha| < 1 $$
signifies a causal LTI system, then its impulse response is:
$$ h[n] = (-alpha)^n u[n] $$
and yields the auto-correlations to be:
$$ r_dd[k] = sum_m=-infty^infty (-alpha)^m u[m] (-alpha)^m-k u[m-k]~~~,~~~text for k = 0,pm 1, pm 2,... $$
I hope you can proceed the rest.
answered 1 hour ago
Fat32
12.8k31127
There are different approaches to compute the requested auto-correlations but I would like to provide the simplest that suits to your case provided that.
- The difference equation is an LCCDE type indicating an LTI system.
- The input to the system $v[n]$ is WSS (wide sense stationary) random process.
Both of which I assume to be observed in your question. Then I would use the following well known relation between the input and output auto-correlations of an (real) LTI system driven by (real) WSS input $v[n]$:
$$ r_dd[k] = h[k] star h[-k] star r_vv[k] $$
where the auto-correlation of WSS $v[n]$ is $r_vv[k] = sigma_v^2 delta[k] = delta[k]$ for an uncorrelated, zero mean, unit variance process. And $h[n]$ is the impulse-response of the LTI system signified by the LCCDE.
Then the auto-correlation sequence of the WSS output $d[n]$ is
$$ r_dd[k] = h[k] star h[-k] = sum_m=-infty^infty h[m]h[-(k-m)]~~~,~~~text for k = 0,pm 1, pm 2,... $$
So assuming that your LCCDE :
$$d[n] - alpha d[n-k] = v[n] ~~~, ~~~text with |alpha| < 1 $$
signifies a causal LTI system, then its impulse response is:
$$ h[n] = (-alpha)^n u[n] $$
and yields the auto-correlations to be:
$$ r_dd[k] = sum_m=-infty^infty (-alpha)^m u[m] (-alpha)^m-k u[m-k]~~~,~~~text for k = 0,pm 1, pm 2,... $$
I hope you can proceed the rest.
answered 1 hour ago
Fat32
12.8k31127
answered 1 hour ago
Fat32
12.8k31127
answered 1 hour ago
Fat32
12.8k31127
answered 1 hour ago
Fat32
12.8k31127
12.8k31127
add a comment |Â
add a comment |Â
Julien__ is a new contributor. Be nice, and check out our Code of Conduct.
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Hint: Your definition for $d[n]$ still works if we replace $n$ by $m=n+1$ and so you know how $d[n+1]$ relates to $d[n]$. Proceed iteratively to find $d[n+2]$ in terms of $d[n+1]$ and replace $d[n+1]$ by the just discovered formula $alpha d[n]+nu[n+1]$ for $d[n+2]$. Lather, rinse, repeat, working backwards to ultimately find $d[n+k]$ in terms of $d[n]$ and $nu[n+i], 0 leq i leq k$. Then substitute into $E[d[n]d[n+k]]$ and work from there.
– Dilip Sarwate
2 hours ago
Thanks @DilipSarwate, but it doesn't help because I'm stuck with terms in
d[n]v[n+k]. Even for $$mathbbE[d[n]d[n+1]] = mathbbE[alpha d[n]^2 + d[n]v[n+1]]$$ I don't know how to proceed– Julien__
1 hour ago
What is $$mathbbE[ d[n]v[n+1] ]$$?
– Julien__
1 hour ago
Your model probably forgot to mention this explicitly or maybe you forgot to include this in your description of the problem, but it is typically assumed than the noise $nu$ is independent of the signal $d$ and so $E[d[n]nu[m]]=E[d[n]]cdot E[nu[m] = E[d[n]]cdot 0 = 0$ for all $n$ and $m$.
– Dilip Sarwate
1 hour ago