PAPR论文阅读笔记2之On the distribution of the peak-to-average power ratio in OFDM signals

从（A.2）到（A.5）：
x=2σx2rcos⁡θy=2σx2rsin⁡θx˙=2σx2(r˙cos⁡θ−rsin⁡θθ˙)y˙=2σx2(r˙sin⁡θ+rcos⁡θθ˙)\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=2σx2rcosθy=2σx2rsinθx˙=2σx2(r˙cosθ−rsinθθ˙)y˙=2σx2(r˙sinθ+rcosθθ˙)
因此
dxdy=2σx2rdrdθdx˙dy˙=2σx2rdr˙dθ˙+∗∗∗dxdydx˙dy˙=(2σx2)2r2drdθdr˙dθ˙\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 ∗∗∗***∗∗∗ 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）
∫02πdθ∫−∞∞e−1Kr2θ˙2dθ˙=2ππKr\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）
∣R∣=(σx2σx˙2σx¨2−σx˙6)2|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
R1−1=1σx2σx˙2σx¨2−σx˙6(σx˙2σx¨20σx˙40σx2σx¨2−σx˙40σx˙40σx2σx˙2)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⎠⎞

x=2σx2rcos⁡θy=2σx2rsin⁡θx˙=2σx2(r˙cos⁡θ−rsin⁡θθ˙)y˙=2σx2(r˙sin⁡θ+rcos⁡θθ˙)x¨=2σx2(r¨cos⁡θ−r˙sin⁡θθ˙−r˙sin⁡θθ˙−rcos⁡θθ˙2−rsin⁡θθ¨)y¨=2σx2(r¨sin⁡θ+r˙cos⁡θθ˙+r˙cos⁡θθ˙−rsin⁡θθ˙2+rcos⁡θθ¨)\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=2σx2rcosθy=2σx2rsinθx˙=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
dxdydx˙dy˙=(2σx2)2r2drdθdr˙dθ˙dx¨dy¨=2σx2rdr¨dθ¨+∗∗∗dxdydx˙dy˙dx¨dy¨=(2σx2)3r3drdθdr˙dd˙r¨dθ¨θ\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 ∗∗∗***∗∗∗ 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 X=(x,x˙,x¨,y,y˙,y¨)′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
f(x,x˙,x¨,y,y˙,y¨)=1(2π)3∣R∣e−12X′R−1X(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 1f(x,x˙,x¨,y,y˙,y¨)=(2π)3∣R∣1e−21X′R−1X(1)
X′R−1XX'R^{-1}XX′R−1X 包含以下项：
1σx2σx˙2σx¨2−σx˙6[σx˙2σx¨2x2+σx˙4xx¨+(σx2σx¨2−σx˙4)x˙2+σx˙4xx¨+σx2σx˙2x¨2]1σx2σx˙2σx¨2−σx˙6[σx˙2σx¨2y2+σx˙4yy¨+(σx2σx¨2−σx˙4)y˙2+σx˙4yy¨+σx2σx˙2y¨2]\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
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)\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M=σx˙4σx2σx¨2−σx˙4,K=σx˙2σx2,ϕ=θ˙KM ={\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=Kθ˙
(注：论文中的ϕ=Kθ˙\phi=\sqrt K\dot\thetaϕ=Kθ˙ 应该是笔误）
得到
σx2σx˙2σx¨2σx2σx˙2σx¨2−σx˙6=M+1σx2σx˙4σx2σx˙2σx¨2−σx˙6=MKσx2(σx2σx¨2−σx˙4)σx2σx˙2σx¨2−σx˙6=1Kσx4σx˙2σx2σx˙2σx¨2−σx˙6=MK2\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
12X′R−1X=12σx2[(M+1)(x2+y2)+2MK(xx¨+yy¨)+1K(x˙2+y˙2)+MK2(x¨2+y¨2)]=(M+1)r2+2MK(rr¨−r2θ˙2)+1K(r˙2+r2θ˙2)+MK2(r¨2+4r˙2θ˙2+r2θ˙4+r2θ¨2−rr¨θ˙2+4rr˙θ˙θ¨)=[(M+1)+(1−2M)ϕ2+Mϕ4]r2+2MK(1−ϕ2)rr¨+1K(r˙2+MKr¨2)+MK2(4r˙2θ˙2+r2θ¨2+4rr˙θ˙θ¨)(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 221X′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
∫−∞∞e−MK2(4r˙2θ˙2+r2θ¨2+4rr˙θ˙θ¨)dθ¨=KπrM(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¨=rMKπ(3)
将（2）（3）及积分变量雅可比代入（1）得到公式(B.2)
f(r,r˙,r¨)=∫02πdθ∫−∞∞(2σx2)3r3(2π)3∣R∣KπrMe−Tdθ˙=∫−∞∞2r2M(Kπ)3/2e−Tdϕ\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)3r3rMKπe−Tdθ˙=∫−∞∞(Kπ)3/22r2Me−Tdϕ
这里 T=[(M+1)+(1−2M)ϕ2+Mϕ4]r2+2MK(1−ϕ2)rr¨+1K(r˙2+MKr¨2)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）
∫−∞0−r¨e−2MK(1−ϕ2)ur¨−MK2r¨2dr¨=∫−∞0−r¨e−M(r¨2K+(1−ϕ2)u)2eM(1−ϕ2)2u2dr¨=eM(1−ϕ2)2u2∫−∞0−r¨e−M(r¨2K+(1−ϕ2)u)2dr¨=eM(1−ϕ2)2u2K2M∫−∞M(1−ϕ2)u[−M(r¨2K+(1−ϕ2)u)+M(1−ϕ2)u]⋅e−M(r¨2K+(1−ϕ2)u)2d[M(r¨2K+(1−ϕ2)u)]=eM(1−ϕ2)2u2K2M∫−∞M(1−ϕ2)u−xe−x2+M(1−ϕ2)ue−x2dx\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
以上两部分积分为
K2M−K2eM(ϕ2−1)2u2MπM(ϕ2−1)u⋅erfc(M(ϕ2−1)u){K^2\over M}-{K^2e^{M(\phi^2-1)^2u^2}\over M}\sqrt {\pi M}(\phi^2-1)u\cdot\mathrm{erfc} (\sqrt M(\phi^2-1)u) MK2−MK2eM(ϕ2−1)2u2πM(ϕ2−1)u⋅erfc(M(ϕ2−1)u)
得到公式（B.5）。

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.

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