PAPR论文阅读笔记2之On the distribution of the peak-to-average power ratio in OFDM signals
作者：互联网
从（A.2）到（A.5）：
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\begin{aligned} &x=\sqrt{2\sigma_x^2}r\cos\theta\\ &y=\sqrt{2\sigma_x^2}r\sin\theta\\ &\dot x=\sqrt{2\sigma_x^2}(\dot r\cos\theta-r\sin\theta \dot \theta)\\ &\dot y=\sqrt{2\sigma_x^2}(\dot r\sin\theta+r\cos\theta \dot \theta) \end{aligned}
x=2σx2
rcosθy=2σx2
rsinθx˙=2σx2
(r˙cosθ−rsinθθ˙)y˙=2σx2
(r˙sinθ+rcosθθ˙)
因此
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\begin{aligned} &dxdy={2\sigma_x^2}rdrd\theta\\ &d\dot xd\dot y={2\sigma_x^2}rd\dot rd \dot \theta+***\\ &dxdyd\dot xd\dot y=({2\sigma_x^2})^2r^2drd\theta d\dot rd \dot \theta \end{aligned}
dxdy=2σx2rdrdθdx˙dy˙=2σx2rdr˙dθ˙+∗∗∗dxdydx˙dy˙=(2σx2)2r2drdθdr˙dθ˙
这里
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***
∗∗∗ 表示前面已出现过的微分项，对最后的微分外积没有贡献。将变量代换到 （A.2）得到 （A.5）。
公式（A.5）到 （A.7）
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\int_0^{2\pi}d\theta\int_{-\infty}^{\infty} e^{-{1\over K}r^2\dot\theta^2}d\dot \theta=2\pi {\sqrt{\pi K}\over r}
∫02πdθ∫−∞∞e−K1r2θ˙2dθ˙=2πrπK
根据 （B.1）
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|R|=\left(\sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^6\right)^2
∣R∣=(σx2σx˙2σx¨2−σx˙6)2
矩阵 R 是块对角阵，块的拟
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R_1^{-1}={1\over \sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^6}\begin{pmatrix}\sigma_{\dot x}^2\sigma_{\ddot x}^2&0& \sigma_{\dot x}^4\\ 0& \sigma_{x}^2\sigma_{\ddot x}^2- \sigma_{\dot x}^4& 0 \\ \sigma_{\dot x}^4&0& \sigma_x^2\sigma_{\dot x}^2 \end{pmatrix}
R1−1=σx2σx˙2σx¨2−σx˙61⎝⎛σx˙2σx¨20σx˙40σx2σx¨2−σx˙40σx˙40σx2σx˙2⎠⎞
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\begin{aligned} &x=\sqrt{2\sigma_x^2}r\cos\theta\\ &y=\sqrt{2\sigma_x^2}r\sin\theta\\ &\dot x=\sqrt{2\sigma_x^2}(\dot r\cos\theta-r\sin\theta \dot \theta)\\ &\dot y=\sqrt{2\sigma_x^2}(\dot r\sin\theta+r\cos\theta \dot \theta)\\ &\ddot x=\sqrt{2\sigma_x^2}(\ddot r\cos\theta-\dot r\sin\theta \dot \theta- \dot r\sin\theta \dot \theta-r\cos\theta \dot \theta^2-r\sin\theta \ddot \theta)\\ &\ddot y=\sqrt{2\sigma_x^2}(\ddot r\sin\theta+\dot r\cos\theta \dot \theta+ \dot r\cos\theta \dot \theta-r\sin\theta \dot \theta^2+r\cos\theta \ddot \theta) \end{aligned}
x=2σx2
rcosθy=2σx2
rsinθx˙=2σx2
(r˙cosθ−rsinθθ˙)y˙=2σx2
(r˙sinθ+rcosθθ˙)x¨=2σx2
(r¨cosθ−r˙sinθθ˙−r˙sinθθ˙−rcosθθ˙2−rsinθθ¨)y¨=2σx2
(r¨sinθ+r˙cosθθ˙+r˙cosθθ˙−rsinθθ˙2+rcosθθ¨)
因此
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\begin{aligned} &dxdyd\dot xd\dot y=({2\sigma_x^2})^2r^2drd\theta d\dot rd \dot \theta\\ &d\ddot xd\ddot y=2\sigma_x^2 rd\ddot rd\ddot \theta+***\\ &dxdyd\dot xd\dot yd\ddot xd\ddot y=({2\sigma_x^2})^3r^3drd\theta d\dot rd \dot d\ddot rd\ddot \theta\theta\end{aligned}
dxdydx˙dy˙=(2σx2)2r2drdθdr˙dθ˙dx¨dy¨=2σx2rdr¨dθ¨+∗∗∗dxdydx˙dy˙dx¨dy¨=(2σx2)3r3drdθdr˙dd˙r¨dθ¨θ
这里
∗
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∗∗∗ 表示前面已出现过的微分项，对最后的微分外积没有贡献。
令
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X=(x,\dot x, \ddot x, y,\dot y, \ddot y)'
X=(x,x˙,x¨,y,y˙,y¨)′, 概率密度函数为
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(1)
f(x,\dot x, \ddot x, y,\dot y, \ddot y)={1\over (2\pi)^3\sqrt{|R|}}e^{-{1\over 2}X'R^{-1}X}\tag 1
f(x,x˙,x¨,y,y˙,y¨)=(2π)3∣R∣
1e−21X′R−1X(1)
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X'R^{-1}X
X′R−1X 包含以下项：
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\begin{aligned} & {1\over \sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^6}[\sigma_{\dot x}^2\sigma_{\ddot x}^2x^2+\sigma_{\dot x}^4x\ddot x+(\sigma_{x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^4)\dot x^2+\sigma_{\dot x}^4x\ddot x+\sigma_{x}^2\sigma_{\dot x}^2\ddot x^2]\\ & {1\over \sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^6}[\sigma_{\dot x}^2\sigma_{\ddot x}^2y^2+\sigma_{\dot x}^4y\ddot y+(\sigma_{x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^4)\dot y^2+\sigma_{\dot x}^4y\ddot y+\sigma_{x}^2\sigma_{\dot x}^2\ddot y^2] \end{aligned}
σx2σx˙2σx¨2−σx˙61[σx˙2σx¨2x2+σx˙4xx¨+(σx2σx¨2−σx˙4)x˙2+σx˙4xx¨+σx2σx˙2x¨2]σx2σx˙2σx¨2−σx˙61[σx˙2σx¨2y2+σx˙4yy¨+(σx2σx¨2−σx˙4)y˙2+σx˙4yy¨+σx2σx˙2y¨2]
变量代换
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\begin{aligned} &x^2+y^2=2\sigma_x^2r^2\\ &\dot x^2+\dot y^2=2\sigma_x^2(\dot r^2+r^2\dot\theta^2)\\ &\ddot x^2+\ddot y^2=2\sigma_x^2(\ddot r^2+4\dot r^2\dot\theta^2+r^2\dot\theta^4+r^2\ddot\theta^2-r\ddot r\dot \theta^2+4r\dot r\dot\theta\ddot\theta)\\ &x\ddot x+y\ddot y=2\sigma_x^2(r\ddot r-r^2\dot\theta^2) \end{aligned}
x2+y2=2σx2r2x˙2+y˙2=2σx2(r˙2+r2θ˙2)x¨2+y¨2=2σx2(r¨2+4r˙2θ˙2+r2θ˙4+r2θ¨2−rr¨θ˙2+4rr˙θ˙θ¨)xx¨+yy¨=2σx2(rr¨−r2θ˙2)
根据定义
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M ={\sigma_{\dot x}^4\over \sigma_{x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^4}, \quad K={\sigma_{\dot x}^2\over \sigma_{x}^2}, \quad \phi={\dot\theta\over \sqrt K}
M=σx2σx¨2−σx˙4σx˙4,K=σx2σx˙2,ϕ=K
θ˙
(注：论文中的
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\phi=\sqrt K\dot\theta
ϕ=K
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得到
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\begin{aligned} & {\sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2\over \sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^6}=M+1\\ & {\sigma_x^2\sigma_{\dot x}^4\over \sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^6}={M\over K}\\ & {\sigma_x^2(\sigma_x^2\sigma_{\ddot x}^2-\sigma_{\dot x}^4)\over \sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^6}={1\over K}\\ & {\sigma_x^4\sigma_{\dot x}^2\over \sigma_x^2\sigma_{\dot x}^2\sigma_{\ddot x}^2-\sigma_{\dot x}^6}={M\over K^2} \end{aligned}
σx2σx˙2σx¨2−σx˙6σx2σx˙2σx¨2=M+1σx2σx˙2σx¨2−σx˙6σx2σx˙4=KMσx2σx˙2σx¨2−σx˙6σx2(σx2σx¨2−σx˙4)=K1σx2σx˙2σx¨2−σx˙6σx4σx˙2=K2M
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(2)
\begin{aligned} {1\over 2}X'R^{-1}X&={1\over 2\sigma_x^2}[(M+1)(x^2+y^2)+2{M\over K}(x\ddot x+y\ddot y)+{1\over K}(\dot x^2+\dot y^2)+{M\over K^2}(\ddot x^2+\ddot y^2)]\\ & =(M+1)r^2+{2M\over K}(r\ddot r-r^2\dot\theta^2)+{1\over K}(\dot r^2+r^2\dot\theta^2) \\&\quad\quad+{M\over K^2}(\ddot r^2+4\dot r^2\dot\theta^2+r^2\dot\theta^4+r^2\ddot\theta^2-r\ddot r\dot \theta^2+4r\dot r\dot\theta\ddot\theta)\\ & =[(M+1)+(1-2M)\phi^2+M\phi^4] r^2+{2M\over K}(1-\phi^2)r\ddot r+{1\over K}(\dot r^2+{M\over K}\ddot r^2) \\&\quad\quad+{M\over K^2}(4\dot r^2\dot\theta^2+r^2\ddot\theta^2+4r\dot r\dot\theta\ddot\theta)\end{aligned}\tag 2
21X′R−1X=2σx21[(M+1)(x2+y2)+2KM(xx¨+yy¨)+K1(x˙2+y˙2)+K2M(x¨2+y¨2)]=(M+1)r2+K2M(rr¨−r2θ˙2)+K1(r˙2+r2θ˙2)+K2M(r¨2+4r˙2θ˙2+r2θ˙4+r2θ¨2−rr¨θ˙2+4rr˙θ˙θ¨)=[(M+1)+(1−2M)ϕ2+Mϕ4]r2+K2M(1−ϕ2)rr¨+K1(r˙2+KMr¨2)+K2M(4r˙2θ˙2+r2θ¨2+4rr˙θ˙θ¨)(2)
对最后一部分积分
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(3)
\int_{-\infty}^{\infty}e^{-{M\over K^2}(4\dot r^2\dot\theta^2+r^2\ddot\theta^2+4r\dot r\dot\theta\ddot\theta)}d\ddot\theta={K\sqrt \pi\over r \sqrt M} \tag 3
∫−∞∞e−K2M(4r˙2θ˙2+r2θ¨2+4rr˙θ˙θ¨)dθ¨=rM
Kπ
(3)
将（2）（3）及积分变量雅可比代入（1）得到公式(B.2)
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\begin{aligned}f(r,\dot r, \ddot r)=\int_0^{2\pi}d\theta\int_{-\infty}^\infty{(2\sigma_x^2)^3r^3\over (2\pi)^3\sqrt{|R|}}{K\sqrt \pi\over r \sqrt M}e^{-T}d\dot\theta=\int_{-\infty}^\infty{2r^2 \sqrt M\over (K\pi)^{3/2}}e^{-T}d\phi\\ \end{aligned}
f(r,r˙,r¨)=∫02πdθ∫−∞∞(2π)3∣R∣
(2σx2)3r3rM
Kπ
e−Tdθ˙=∫−∞∞(Kπ)3/22r2M
e−Tdϕ
这里
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)
T=[(M+1)+(1-2M)\phi^2+M\phi^4] r^2+{2M\over K}(1-\phi^2)r\ddot r+{1\over K}(\dot r^2+{M\over K}\ddot r^2)
T=[(M+1)+(1−2M)ϕ2+Mϕ4]r2+K2M(1−ϕ2)rr¨+K1(r˙2+KMr¨2).
从公式（B.2）到（B.5）
∫
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∞
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M
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M
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r
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∫
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M
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[
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∫
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M
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x
\begin{aligned}\int_{-\infty}^0 -\ddot re^{-{2M\over K}(1-\phi^2)u\ddot r-{M\over K^2}\ddot r^2}d\ddot r=& \int_{-\infty}^0 -\ddot re^{-M({\ddot r^2\over K}+(1-\phi^2)u)^2}e^{M(1-\phi^2)^2u^2}d\ddot r\\ =e^{M(1-\phi^2)^2u^2} &\int_{-\infty}^0 -\ddot re^{-M({\ddot r^2\over K}+(1-\phi^2)u)^2}d\ddot r\\ ={e^{M(1-\phi^2)^2u^2}K^2\over M} &\int_{-\infty}^{\sqrt M(1-\phi^2)u} \left[-\sqrt M({\ddot r^2\over K}+(1-\phi^2)u)+\sqrt M(1-\phi^2)u\right]\\&\cdot e^{-M({\ddot r^2\over K}+(1-\phi^2)u)^2}d\left[\sqrt M({\ddot r^2\over K}+(1-\phi^2)u)\right]\\ ={e^{M(1-\phi^2)^2u^2}K^2\over M} &\int_{-\infty}^{\sqrt M(1-\phi^2)u}-xe^{-x^2}+\sqrt M(1-\phi^2)ue^{-x^2}dx \\ \end{aligned}
∫−∞0−r¨e−K2M(1−ϕ2)ur¨−K2Mr¨2dr¨==eM(1−ϕ2)2u2=MeM(1−ϕ2)2u2K2=MeM(1−ϕ2)2u2K2∫−∞0−r¨e−M(Kr¨2+(1−ϕ2)u)2eM(1−ϕ2)2u2dr¨∫−∞0−r¨e−M(Kr¨2+(1−ϕ2)u)2dr¨∫−∞M
(1−ϕ2)u[−M
(Kr¨2+(1−ϕ2)u)+M
(1−ϕ2)u]⋅e−M(Kr¨2+(1−ϕ2)u)2d[M
(Kr¨2+(1−ϕ2)u)]∫−∞M
(1−ϕ2)u−xe−x2+M
(1−ϕ2)ue−x2dx
Ref.
H. Ochiai and H. Imai, “On the distribution of the peak-to-average power ratio in OFDM signals,” IEEE Trans. Commun., vol. 49, pp. 282–289, Feb. 2001.
标签：ratio,OFDM,power,over,ddot,x2,theta,sigma,dot 来源： https://blog.csdn.net/wubyatseu/article/details/123582398