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Formula for complementary cumulative distribution function (CCDF) using Laplace and inverse Laplace transform
Clash Royale CLAN TAG#URR8PPP
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I am not sure whether it is the right place to ask.
I am reading the paper. In this paper, the total received power at the origin from the set of active base stations (BSs) located at $x_i$ is defined as
$$P_r=asum_x_i in Phi Pcdot h_x_i cdot||x_i||^-alpha$$
where $h_x_i stackreliidsim exp(1)$ and $Phi$ is the Poisson point process and is independent with $h_x_i$ for all $i$.
The probability of having $P_r geq T$ is denoted as the wake-up probablity:
$$P_w=mathbb P(P_r geq T )$$
It is also the complementary cumulative distribution function (CCDF) of $P_r$.
$Theorem$ 1 in this paper gives the fomula for $P_w$ as
$$P_w=int^infty_0frac1pi uexp[-uT] times expleft[-frac2pi^2lambda deltaalpha tanleft(frac2pialpharight)(aP)^frac2alphau^frac2alpharight] times sin left(frac2pi^2lambdadeltaalpha(aP)^frac2alphau^frac2alpharight)du$$
I know the formula looks "ugly" and some of the letters are just constants, and I was trying to derive it. There are some hints and part of the proof in the appendix part of the paper. The idea is find the Laplace transform of $P_r$ first then do the inverse Laplace transform.
The Laplace transform (which I already checked with the result in the paper) is
$$mathscr L_P_r(s)=expleft-frac2pi^2lambda deltaalpha sinleft(frac2pialpharight)(aP)^frac2alphas^frac2alpharight$$
So since we want the CCDF of $P_r$, and the Laplace transform of $mathscr L_P_r(s)$ gives the pdf, so I guess we need to find the inverse Laplace transform of $frac1smathscr L_P_r(s)$, then $1-mathscr L^-1left[frac1smathscr L_P_r(s)right]$ is what we want? But I do not know how to get the $P_w$ showed above.
Any help will deeply appreciate!
laplace-transform
asked Aug 3 at 2:58
Nan
1,170214
add a comment |Â
up vote
0
down vote
favorite
I am not sure whether it is the right place to ask.
I am reading the paper. In this paper, the total received power at the origin from the set of active base stations (BSs) located at $x_i$ is defined as
$$P_r=asum_x_i in Phi Pcdot h_x_i cdot||x_i||^-alpha$$
where $h_x_i stackreliidsim exp(1)$ and $Phi$ is the Poisson point process and is independent with $h_x_i$ for all $i$.
The probability of having $P_r geq T$ is denoted as the wake-up probablity:
$$P_w=mathbb P(P_r geq T )$$
It is also the complementary cumulative distribution function (CCDF) of $P_r$.
$Theorem$ 1 in this paper gives the fomula for $P_w$ as
$$P_w=int^infty_0frac1pi uexp[-uT] times expleft[-frac2pi^2lambda deltaalpha tanleft(frac2pialpharight)(aP)^frac2alphau^frac2alpharight] times sin left(frac2pi^2lambdadeltaalpha(aP)^frac2alphau^frac2alpharight)du$$
I know the formula looks "ugly" and some of the letters are just constants, and I was trying to derive it. There are some hints and part of the proof in the appendix part of the paper. The idea is find the Laplace transform of $P_r$ first then do the inverse Laplace transform.
The Laplace transform (which I already checked with the result in the paper) is
$$mathscr L_P_r(s)=expleft-frac2pi^2lambda deltaalpha sinleft(frac2pialpharight)(aP)^frac2alphas^frac2alpharight$$
So since we want the CCDF of $P_r$, and the Laplace transform of $mathscr L_P_r(s)$ gives the pdf, so I guess we need to find the inverse Laplace transform of $frac1smathscr L_P_r(s)$, then $1-mathscr L^-1left[frac1smathscr L_P_r(s)right]$ is what we want? But I do not know how to get the $P_w$ showed above.
Any help will deeply appreciate!
laplace-transform
asked Aug 3 at 2:58
Nan
1,170214
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am not sure whether it is the right place to ask.
I am reading the paper. In this paper, the total received power at the origin from the set of active base stations (BSs) located at $x_i$ is defined as
$$P_r=asum_x_i in Phi Pcdot h_x_i cdot||x_i||^-alpha$$
where $h_x_i stackreliidsim exp(1)$ and $Phi$ is the Poisson point process and is independent with $h_x_i$ for all $i$.
The probability of having $P_r geq T$ is denoted as the wake-up probablity:
$$P_w=mathbb P(P_r geq T )$$
It is also the complementary cumulative distribution function (CCDF) of $P_r$.
$Theorem$ 1 in this paper gives the fomula for $P_w$ as
$$P_w=int^infty_0frac1pi uexp[-uT] times expleft[-frac2pi^2lambda deltaalpha tanleft(frac2pialpharight)(aP)^frac2alphau^frac2alpharight] times sin left(frac2pi^2lambdadeltaalpha(aP)^frac2alphau^frac2alpharight)du$$
I know the formula looks "ugly" and some of the letters are just constants, and I was trying to derive it. There are some hints and part of the proof in the appendix part of the paper. The idea is find the Laplace transform of $P_r$ first then do the inverse Laplace transform.
The Laplace transform (which I already checked with the result in the paper) is
$$mathscr L_P_r(s)=expleft-frac2pi^2lambda deltaalpha sinleft(frac2pialpharight)(aP)^frac2alphas^frac2alpharight$$
So since we want the CCDF of $P_r$, and the Laplace transform of $mathscr L_P_r(s)$ gives the pdf, so I guess we need to find the inverse Laplace transform of $frac1smathscr L_P_r(s)$, then $1-mathscr L^-1left[frac1smathscr L_P_r(s)right]$ is what we want? But I do not know how to get the $P_w$ showed above.
Any help will deeply appreciate!
laplace-transform
asked Aug 3 at 2:58
Nan
1,170214
I am not sure whether it is the right place to ask.
I am reading the paper. In this paper, the total received power at the origin from the set of active base stations (BSs) located at $x_i$ is defined as
$$P_r=asum_x_i in Phi Pcdot h_x_i cdot||x_i||^-alpha$$
where $h_x_i stackreliidsim exp(1)$ and $Phi$ is the Poisson point process and is independent with $h_x_i$ for all $i$.
The probability of having $P_r geq T$ is denoted as the wake-up probablity:
$$P_w=mathbb P(P_r geq T )$$
It is also the complementary cumulative distribution function (CCDF) of $P_r$.
$Theorem$ 1 in this paper gives the fomula for $P_w$ as
$$P_w=int^infty_0frac1pi uexp[-uT] times expleft[-frac2pi^2lambda deltaalpha tanleft(frac2pialpharight)(aP)^frac2alphau^frac2alpharight] times sin left(frac2pi^2lambdadeltaalpha(aP)^frac2alphau^frac2alpharight)du$$
I know the formula looks "ugly" and some of the letters are just constants, and I was trying to derive it. There are some hints and part of the proof in the appendix part of the paper. The idea is find the Laplace transform of $P_r$ first then do the inverse Laplace transform.
The Laplace transform (which I already checked with the result in the paper) is
$$mathscr L_P_r(s)=expleft-frac2pi^2lambda deltaalpha sinleft(frac2pialpharight)(aP)^frac2alphas^frac2alpharight$$
So since we want the CCDF of $P_r$, and the Laplace transform of $mathscr L_P_r(s)$ gives the pdf, so I guess we need to find the inverse Laplace transform of $frac1smathscr L_P_r(s)$, then $1-mathscr L^-1left[frac1smathscr L_P_r(s)right]$ is what we want? But I do not know how to get the $P_w$ showed above.
Any help will deeply appreciate!
laplace-transform
asked Aug 3 at 2:58
Nan
1,170214
asked Aug 3 at 2:58
Nan
1,170214
asked Aug 3 at 2:58
Nan
1,170214
asked Aug 3 at 2:58
Nan
1,170214
1,170214
add a comment |Â
add a comment |Â
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