Covariance in and input-output filter [Stationary Stochastic Processes]
$begingroup$
The weakly stationary processes $X_t$, $;t=0$, $,pm$$1$,
$,pm$$2$,$,ldots$ and $Y_t$,$;$ $t=0$, $,pm$$1$, $,pm$$2$,$ldots$
are input and output of a linear filter according to
$Y_t;$$+;0.5Y_{t-1};$$;=;$$X_t$, $quad$ for $t=0$, $,pm$$1$,
$,pm$$2$,$ldots$
The process $X_t,$ has the covariance function $r_X(0)=1$,$,$
$r_X(pm2)=0.2$,$,$ and zero for all other values. Determine the
cross-covariance funtion
$r_{X,Y}(tau)=C[X_t,,Y_{t+tau}]$
for all values of $tau$.
I began by assigning the impulse function h(x) the following values:
$h(0) = 1qquad$
$h(1) = 0.5$
Does it makes sense?
statistics stochastic-processes covariance filters
asked Jan 7 at 17:24
LowEnergyLowEnergy
11
$endgroup$
add a comment |
$begingroup$
The weakly stationary processes $X_t$, $;t=0$, $,pm$$1$,
$,pm$$2$,$,ldots$ and $Y_t$,$;$ $t=0$, $,pm$$1$, $,pm$$2$,$ldots$
are input and output of a linear filter according to
$Y_t;$$+;0.5Y_{t-1};$$;=;$$X_t$, $quad$ for $t=0$, $,pm$$1$,
$,pm$$2$,$ldots$
The process $X_t,$ has the covariance function $r_X(0)=1$,$,$
$r_X(pm2)=0.2$,$,$ and zero for all other values. Determine the
cross-covariance funtion
$r_{X,Y}(tau)=C[X_t,,Y_{t+tau}]$
for all values of $tau$.
I began by assigning the impulse function h(x) the following values:
$h(0) = 1qquad$
$h(1) = 0.5$
Does it makes sense?
statistics stochastic-processes covariance filters
asked Jan 7 at 17:24
LowEnergyLowEnergy
11
$endgroup$
add a comment |
$begingroup$
The weakly stationary processes $X_t$, $;t=0$, $,pm$$1$,
$,pm$$2$,$,ldots$ and $Y_t$,$;$ $t=0$, $,pm$$1$, $,pm$$2$,$ldots$
are input and output of a linear filter according to
$Y_t;$$+;0.5Y_{t-1};$$;=;$$X_t$, $quad$ for $t=0$, $,pm$$1$,
$,pm$$2$,$ldots$
The process $X_t,$ has the covariance function $r_X(0)=1$,$,$
$r_X(pm2)=0.2$,$,$ and zero for all other values. Determine the
cross-covariance funtion
$r_{X,Y}(tau)=C[X_t,,Y_{t+tau}]$
for all values of $tau$.
I began by assigning the impulse function h(x) the following values:
$h(0) = 1qquad$
$h(1) = 0.5$
Does it makes sense?
statistics stochastic-processes covariance filters
asked Jan 7 at 17:24
LowEnergyLowEnergy
11
$endgroup$
The weakly stationary processes $X_t$, $;t=0$, $,pm$$1$,
$,pm$$2$,$,ldots$ and $Y_t$,$;$ $t=0$, $,pm$$1$, $,pm$$2$,$ldots$
are input and output of a linear filter according to
$Y_t;$$+;0.5Y_{t-1};$$;=;$$X_t$, $quad$ for $t=0$, $,pm$$1$,
$,pm$$2$,$ldots$
The process $X_t,$ has the covariance function $r_X(0)=1$,$,$
$r_X(pm2)=0.2$,$,$ and zero for all other values. Determine the
cross-covariance funtion
$r_{X,Y}(tau)=C[X_t,,Y_{t+tau}]$
for all values of $tau$.
I began by assigning the impulse function h(x) the following values:
$h(0) = 1qquad$
$h(1) = 0.5$
Does it makes sense?
statistics stochastic-processes covariance filters
statistics stochastic-processes covariance filters
asked Jan 7 at 17:24
LowEnergyLowEnergy
11
asked Jan 7 at 17:24
LowEnergyLowEnergy
11
asked Jan 7 at 17:24
LowEnergyLowEnergy
11
asked Jan 7 at 17:24
LowEnergyLowEnergy
11
asked Jan 7 at 17:24
LowEnergyLowEnergy
11
11
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.
We can then use the relation
$$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
$$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$
answered Jan 7 at 18:19
rzchrzch
1765
$endgroup$
$begingroup$
I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
$endgroup$
– LowEnergy
Jan 7 at 18:39
$begingroup$
The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
$endgroup$
– rzch
Jan 7 at 18:42
$begingroup$
I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
$endgroup$
– LowEnergy
Jan 7 at 18:52
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.
We can then use the relation
$$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
$$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$
answered Jan 7 at 18:19
rzchrzch
1765
$endgroup$
$begingroup$
I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
$endgroup$
– LowEnergy
Jan 7 at 18:39
$begingroup$
The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
$endgroup$
– rzch
Jan 7 at 18:42
$begingroup$
I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
$endgroup$
– LowEnergy
Jan 7 at 18:52
add a comment |
$begingroup$
The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.
We can then use the relation
$$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
$$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$
answered Jan 7 at 18:19
rzchrzch
1765
$endgroup$
$begingroup$
I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
$endgroup$
– LowEnergy
Jan 7 at 18:39
$begingroup$
The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
$endgroup$
– rzch
Jan 7 at 18:42
$begingroup$
I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
$endgroup$
– LowEnergy
Jan 7 at 18:52
add a comment |
$begingroup$
The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.
We can then use the relation
$$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
$$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$
answered Jan 7 at 18:19
rzchrzch
1765
$endgroup$
The impulse response $h_{t}$ is actually given by $h_{t} = -0.5^{t}u_{t}$ (where $u_{t}$ is the step function). To obtain this, you can rearrange to get $Y_{t} = X_{t} - 0.5Y_{t - 1}$ and plug in values for $X_{t}$ being the discrete delta function, or you can take the $z$-transform to get the transfer function of $Hleft(zright) = dfrac{1}{1 + 0.5z^{-1}}$ and use a table of $z$-transform pairs to find the corresponding sequence. Either way yields the sequence $h_{0} = 1, h_{1} = -0.5, h_{2} = 0.25,$ etc.
We can then use the relation
$$r_{X, Y}left(tauright) = sum_{i = -infty}^{infty}h_{i}r_{X}left(tau - iright)$$
which is found in several textbooks (eg. Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers by Yates and Goodman, Theorem 11.5) with the provided covariance function for $X_{t}$ to obtain
$$r_{X, Y}left(tauright) = h_{tau} + 0.2h_{tau + 2} + 0.2h_{tau - 2}$$
answered Jan 7 at 18:19
rzchrzch
1765
answered Jan 7 at 18:19
rzchrzch
1765
answered Jan 7 at 18:19
rzchrzch
1765
answered Jan 7 at 18:19
rzchrzch
1765
1765
$begingroup$
I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
$endgroup$
– LowEnergy
Jan 7 at 18:39
$begingroup$
The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
$endgroup$
– rzch
Jan 7 at 18:42
$begingroup$
I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
$endgroup$
– LowEnergy
Jan 7 at 18:52
add a comment |
$begingroup$
I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
$endgroup$
– LowEnergy
Jan 7 at 18:39
$begingroup$
The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
$endgroup$
– rzch
Jan 7 at 18:42
$begingroup$
I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
$endgroup$
– LowEnergy
Jan 7 at 18:52
$begingroup$
I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
$endgroup$
– LowEnergy
Jan 7 at 18:39
$begingroup$
I really wanna thank you for your time and answer. Whilst I remember the step function very well from Automatic Control, I can't really recall it ever being discussed in my Stochastic Process class. Very strange.
$endgroup$
– LowEnergy
Jan 7 at 18:39
$begingroup$
The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
$endgroup$
– rzch
Jan 7 at 18:42
$begingroup$
The step function is really only used to address one subtle technicality (often overlooked), which is that the impulse response is zero for $t < 0$. Writing the impulse response using the step function is just an elegant way to express/acknowledge this.
$endgroup$
– rzch
Jan 7 at 18:42
$begingroup$
I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
$endgroup$
– LowEnergy
Jan 7 at 18:52
$begingroup$
I see. I basically came up with the values above by using the formula: $X(t) = sum_{u = -infty}^{infty}h_{u}Y(t - u)$
$endgroup$
– LowEnergy
Jan 7 at 18:52
add a comment |
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