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CN105356953B - Frequency spectrum detection systematic function parameter preparation method after any moving period is undergone in cognitive radio networks - Google Patents

Frequency spectrum detection systematic function parameter preparation method after any moving period is undergone in cognitive radio networks Download PDF

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CN105356953B
CN105356953B CN201510706056.3A CN201510706056A CN105356953B CN 105356953 B CN105356953 B CN 105356953B CN 201510706056 A CN201510706056 A CN 201510706056A CN 105356953 B CN105356953 B CN 105356953B
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贾敏
王欣玉
郭庆
顾学迈
王雪
陈子研
朱思宇
史瑶
杨健
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Harbin Institute of Technology Shenzhen
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Abstract

认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法,涉及信息与通信技术领域。本发明是为了解决现有方法无法对认知无线电网络中移动场景下单节点和多节点协作进行感知的问题。本发明首先获得认知用户的移动速度、方向、当前周期认知用户起点与主用户的距离、终点与主用户的距离之间的函数关系。然后求得主用户与认知用户移动终点之间距离的PDF。然后获得移动用户在当前移动周期内的移动终点处检测到的主用户信号功率与当前移动周期认知用户移动终点与主用户距离之间的函数关系,得出此功率的PDF。最后求得移动场景下单节点感知,硬判决多节点协作感知和软判决协作感知系统中检测概率、漏检概率以及虚警概率的结果。

The invention discloses a method for obtaining performance parameters of a frequency spectrum detection system after an arbitrary moving period in a cognitive radio network, and relates to the field of information and communication technologies. The present invention aims to solve the problem that the existing method cannot sense the cooperation of single node and multi-node in the mobile scene in the cognitive radio network. The present invention first obtains the functional relationship between the moving speed and direction of the cognitive user, the distance between the starting point of the cognitive user and the main user in the current period, and the distance between the end point and the main user. Then obtain the PDF of the distance between the main user and the end point of the cognitive user's movement. Then obtain the functional relationship between the signal power of the primary user detected by the mobile user at the mobile end point in the current mobile cycle and the distance between the mobile end point of the cognitive user and the primary user in the current mobile cycle, and obtain the PDF of this power. Finally, the results of detection probability, missed detection probability and false alarm probability in single-node sensing, hard-decision multi-node cooperative sensing and soft-decision cooperative sensing systems in mobile scenarios are obtained.

Description

认知无线电网络中经历任意移动周期后频谱检测系统性能参 量获得方法Performance parameters of spectrum detection system after experiencing any mobile period in cognitive radio network Quantity acquisition method

技术领域technical field

本发明涉及信息与通信技术领域,具体涉及认知用户不断随机移动的移动场景下频谱感知系统检测性能指标获得方法。The invention relates to the field of information and communication technologies, in particular to a method for obtaining detection performance indicators of a spectrum sensing system in a mobile scene where cognitive users are constantly moving randomly.

背景技术Background technique

近年来,无线通信技术的快速发展导致人们对于无线频谱资源的需求量急剧增加。然而,目前固定式的频谱分配方式使得大多数频带没有得到充分的利用。为了提升频谱利用率,研究人员提出了认知无线电(CR)技术。CR作为解决无线频谱资源匮乏问题的关键技术,近年来已经引起了国际标准化组织、各国频谱资源分配管理机构、学术界和产业界的极大兴趣和研究热情。CR是在不影响主用户信号的授权频段,即固定分配给用户的频段,正常进行通信的前提下,让具有无线电环境感知功能的无线通信设备动态接入主用户授权频段,从而完成频谱资源的共享。认知无线电技术是一项复杂的技术,其研究内容包括无线频谱感知、动态频谱共享和管理等诸多技术。频谱感知是认知无线电的核心技术之一,是认知系统的基本功能,是实现频谱管理、频谱共享的前提。In recent years, the rapid development of wireless communication technology has led to a sharp increase in the demand for wireless spectrum resources. However, the current fixed spectrum allocation method makes most frequency bands not fully utilized. In order to improve spectrum utilization, researchers proposed cognitive radio (CR) technology. As a key technology to solve the problem of scarcity of wireless spectrum resources, CR has attracted great interest and research enthusiasm from the International Organization for Standardization, national spectrum resource allocation management agencies, academia and industry in recent years. CR is to allow the wireless communication equipment with radio environment awareness function to dynamically access the authorized frequency band of the main user without affecting the authorized frequency band of the main user signal, that is, the frequency band fixedly allocated to the user, and to communicate normally, so as to complete the allocation of spectrum resources. shared. Cognitive radio technology is a complex technology, and its research content includes wireless spectrum sensing, dynamic spectrum sharing and management and many other technologies. Spectrum sensing is one of the core technologies of cognitive radio, a basic function of cognitive systems, and a prerequisite for spectrum management and spectrum sharing.

频谱感知技术有两个任务,首先需要在认知用户需要传输数据时检测到频谱空隙,而且需要在认知用户传输数据过程中不间断地检测授权用户是否出现。对于各种频谱感知算法的研究已经成为了无线通信技术中的研究热点。频谱感知可分为单用户频谱检测法以及多用户协作检测法。能量检测算法是一种单节点频谱检测算法,其原理简单易于实现且不需要知道主用户的先验信息,没有对信号作任何假设。能量检测法对任何信号都适用,因此得到了广泛的使用。实际的认知系统中,单节点频谱感知算法是有较大的局限性的,例如隐藏终端问题和阴影效应等问题,这些都会大大降低单节点检测的检测性能。因此,在实际系统中,单节点检测往往是不可行的。在这样的背景下,多用户协作检测法得到了许多科研人员的关注。Spectrum sensing technology has two tasks. First, it needs to detect the spectrum gap when the cognitive user needs to transmit data, and it needs to continuously detect whether the authorized user is present during the data transmission process of the cognitive user. Research on various spectrum sensing algorithms has become a research hotspot in wireless communication technology. Spectrum sensing can be divided into single-user spectrum detection method and multi-user cooperative detection method. The energy detection algorithm is a single-node spectrum detection algorithm. Its principle is simple and easy to implement, and it does not need to know the prior information of the primary user, and does not make any assumptions about the signal. The energy detection method is applicable to any signal, so it has been widely used. In the actual cognitive system, the single-node spectrum sensing algorithm has relatively large limitations, such as hidden terminal problems and shadow effects, which will greatly reduce the detection performance of single-node detection. Therefore, single-node detection is often not feasible in practical systems. In this context, multi-user collaborative detection method has attracted the attention of many researchers.

当前绝大多数关于认知无线电频谱感知的研究都是在认知用户保持静止的假设上进行的,然而,认知用户的移动性是无线网络的固有属性。现有方法无法对认知无线电网络中移动场景下单节点和多节点协作进行感知。The vast majority of current research on spectrum sensing in cognitive radios is conducted on the assumption that cognitive users remain stationary, however, the mobility of cognitive users is an inherent property of wireless networks. Existing methods cannot perceive single-node and multi-node cooperation in mobile scenarios in cognitive radio networks.

发明内容Contents of the invention

本发明是为了解决现有方法无法对认知无线电网络中移动场景下单节点和多节点协作进行感知的问题,从而提供一种认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法。The present invention aims to solve the problem that existing methods cannot perceive single-node and multi-node cooperation in a mobile scene in a cognitive radio network, thereby providing a method for obtaining performance parameters of a spectrum detection system in a cognitive radio network after experiencing any mobile cycle.

认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法,A method for obtaining performance parameters of a spectrum detection system after experiencing any mobile period in a cognitive radio network,

该方法可以应用于单节点频谱检测系统;The method can be applied to a single-node spectrum detection system;

假设移动场景下,认知用户服从随机游走模型,认知网络中存在一个主用户和一个认知用户;Assume that in the mobile scenario, cognitive users obey the random walk model, and there is a primary user and a cognitive user in the cognitive network;

步骤一、采用公式:Step 1. Use the formula:

yn=K/(dn)2 (2)y n =K/(d n ) 2 (2)

获得认知用户检测到主用户信号的实时功率ynObtain the real-time power y n of the primary user signal detected by the cognitive user;

式中:K是常量,且:Where: K is a constant, and:

K=GrPtGtw/4π)2 K=G r P t G tw /4π) 2

式中:Gr表示认知用户的天线增益,Pt表示主用户的发射信号功率,Gt是授权用户的天线增益;λw是主用户信号的波长;In the formula: G r represents the antenna gain of the cognitive user, P t represents the transmit signal power of the primary user, G t is the antenna gain of the authorized user; λ w is the wavelength of the primary user signal;

步骤二、在每次能量检测中,认知用户采集M个点,M为正整数;且在高斯信道中,噪声的均值为0,方差为σ2,预设能量检测门限为λ;Step 2. In each energy detection, the cognitive user collects M points, where M is a positive integer; and in the Gaussian channel, the mean value of the noise is 0, the variance is σ 2 , and the preset energy detection threshold is λ;

则在第n个感知周期后,单节点频谱感知的检测概率为:Then after the nth sensing period, the detection probability of single-node spectrum sensing for:

漏检概率为:Probability of missed detection for:

虚警概率为:false alarm probability for:

式中:In the formula:

步骤三、根据公式:Step 3, according to the formula:

获得认知用户和第n个移动周期的终点之间的间距di,n;1≤i≤N;vi,n表示在第n个移动周期中,该认知用户的移动速度;θi,n表示在第n个移动周期中,该认知用户的移动方向;△t两次检测之间的时间间隔,即每个移动周期的持续的时间;Obtain the distance d i,n between the cognitive user and the end point of the nth mobile cycle; 1≤i≤N; v i,n represents the moving speed of the cognitive user in the nth mobile cycle; θ i , n represents the moving direction of the cognitive user in the nth moving cycle; Δt is the time interval between two detections, that is, the duration of each moving cycle;

步骤四、根据公式:Step 4, according to the formula:

获得间距di,n的条件概率密度函数;vmax表示该认知用户的最大移动速度;vmin表示该认知用户的最小移动速度;Obtain the conditional probability density function of the distance d i, n ; v max represents the maximum moving speed of the cognitive user; v min represents the minimum moving speed of the cognitive user;

步骤五、根据式(4)推导出:Step 5, deduce according to formula (4):

式中:In the formula:

式中:PQ为第n个移动周期内认知用户移动的距离;In the formula: PQ is the moving distance of the cognitive user in the nth mobile period;

步骤六、用表达式替换式(5)中的则di,n的条件概率密度函数为:Step 6. Use expressions Replacement formula (5) in Then the conditional probability density function of d i,n is:

步骤七、根据式(2)获得yi,n和di,n之间的关系式:Step 7, according to formula (2), obtain the relationship between y i,n and d i,n :

yi,n=K/(di,n)2 (7)y i,n =K/(d i,n ) 2 (7)

则,yi,n的条件概率密度函数为:Then, the conditional probability density function of y i,n is:

式中:通过将中的di,n利用yi,n替换得到;In the formula: by putting d i, n in is replaced by y i,n ;

步骤八、通过公式:Step eight, pass the formula:

计算yi,n的条件数学期望;Calculate the conditional mathematical expectation of y i,n ;

步骤九、通过公式:Step nine, through the formula:

yi,n的概率密度函数;Probability density function of y i,n ;

其中:是由式(6)、式(8)和下式(11)计算得到的:in: is calculated by formula (6), formula (8) and the following formula (11):

yi,n的数学期望为:The mathematical expectation of y i,n is:

其中,是由式(6)、式(9)和下式(13)计算得到的:in, is calculated by formula (6), formula (9) and the following formula (13):

步骤十、则移动场景下单节点频谱感知的检测概率为:Step 10. The detection probability of single-node spectrum sensing in mobile scenarios for:

漏检概率为:Probability of missed detection for:

虚警概率为:false alarm probability for:

其中:是由(6)和下式(15)、(16)计算获得的:in: is calculated by (6) and the following formulas (15), (16):

将检测概率漏检概率和虚警概率作为单节点频谱感知系统的检测性能参量,完成认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法。will detect probability Probability of missed detection and false alarm probability As a detection performance parameter of a single-node spectrum sensing system, a method for obtaining the performance parameter of a spectrum detection system after an arbitrary movement period in a cognitive radio network is completed.

该方法可以应用于多节点硬判决协作频谱检测系统,This method can be applied to a multi-node hard-decision cooperative spectrum detection system,

假设移动场景下,认知用户服从随机游走模型,认知网络中存在N个移动的认知用户以及一个融合中心FC;N为正整数;Assume that in a mobile scenario, cognitive users obey the random walk model, and there are N mobile cognitive users and a fusion center FC in the cognitive network; N is a positive integer;

步骤一、采用公式:Step 1. Use the formula:

yn=K/(dn)2 (2)y n =K/(d n ) 2 (2)

获得认知用户检测到主用户信号的实时功率ynObtain the real-time power y n of the primary user signal detected by the cognitive user;

式中:K是常量,且:Where: K is a constant, and:

K=GrPtGtw/4π)2 K=G r P t G tw /4π) 2

式中:Gr表示认知用户的天线增益,Pt表示主用户的发射信号功率,Gt是授权用户的天线增益;λw是主用户信号的波长;In the formula: G r represents the antenna gain of the cognitive user, P t represents the transmit signal power of the primary user, G t is the antenna gain of the authorized user; λ w is the wavelength of the primary user signal;

步骤二、在每次能量检测中,每个认知用户采集M个点,M为正整数;且在高斯信道中,噪声的均值为0,方差为σ2,预设能量检测门限为λ;Step 2. In each energy detection, each cognitive user collects M points, where M is a positive integer; and in a Gaussian channel, the mean value of the noise is 0, the variance is σ 2 , and the preset energy detection threshold is λ;

则在第n个感知周期后,多节点硬判决协作频谱感知的检测概率为:Then after the nth sensing period, the detection probability of multi-node hard-decision cooperative spectrum sensing for:

漏检概率为:Probability of missed detection for:

虚警概率为:false alarm probability for:

式中:In the formula:

步骤三、根据公式:Step 3, according to the formula:

获得任一认知用户SUi和第n个移动周期的终点之间的间距di,n;1≤i≤N;vi,n表示在第n个移动周期中,该认知用户SUi的移动速度;θi,n表示在第n个移动周期中,该认知用户SUi的移动方向;△t两次检测之间的时间间隔,即每个移动周期的持续的时间;Obtain the distance d i,n between any cognitive user SU i and the end point of the nth mobile cycle; 1≤i≤N; v i,n means that in the nth mobile cycle, the cognitive user SU i The moving speed of ; θ i, n represents the moving direction of the cognitive user SU i in the nth moving cycle; Δt is the time interval between two detections, that is, the duration of each moving cycle;

步骤四、根据公式:Step 4, according to the formula:

获得间距di,n的条件概率密度函数;vmax表示该认知用户SUi的最大移动速度;vmin表示该认知用户SUi的最小移动速度;Obtain the conditional probability density function of distance d i, n ; v max represents the maximum moving speed of the cognitive user SU i ; v min represents the minimum moving speed of the cognitive user SU i ;

步骤五、根据式(4)推导出:Step 5, deduce according to formula (4):

式中:In the formula:

式中:PQ为第n个移动周期内认知用户移动的距离;In the formula: PQ is the moving distance of the cognitive user in the nth mobile cycle;

步骤六、用表达式替换式(5)中的则di,n的条件概率密度函数为:Step 6. Use expressions Replacement formula (5) in Then the conditional probability density function of d i,n is:

步骤七、根据式(2)获得yi,n和di,n之间的关系式:Step 7, according to formula (2), obtain the relationship between y i,n and d i,n :

yi,n=K/(di,n)2 (7)y i,n =K/(d i,n ) 2 (7)

则,yi,n的条件概率密度函数为:Then, the conditional probability density function of y i,n is:

式中:通过将中的di,n利用yi,n替换得到;In the formula: by putting d i, n in is replaced by y i,n ;

步骤八、通过公式:Step eight, pass the formula:

计算yi,n的条件数学期望;Calculate the conditional mathematical expectation of y i,n ;

步骤九、通过公式:Step nine, through the formula:

yi,n的概率密度函数;Probability density function of y i,n ;

其中:是由式(6)、式(8)和下式(11)计算得到的:in: is calculated by formula (6), formula (8) and the following formula (11):

yi,n的数学期望为:The mathematical expectation of y i,n is:

其中,是由式(6)、式(9)和下式(13)计算得到的:in, is calculated by formula (6), formula (9) and the following formula (13):

步骤十、则移动场景下多节点硬判决协作频谱检测系统频谱感知的检测概率为:Step 10: The detection probability of spectrum sensing in the multi-node hard-decision cooperative spectrum detection system in the mobile scene for:

漏检概率为:Probability of missed detection for:

虚警概率为:false alarm probability for:

其中:是通过式(6)、式(18)和式(19)计算获得的:in: is calculated by formula (6), formula (18) and formula (19):

其中:in:

并且f(y1,n,…,yN,n|di,n-1i,n)表示y1,n,y2,n,…,yN,n的联合条件概率密度函数:And f(y 1,n ,…,y N,n| d i,n-1i,n ) represents the joint conditional probability density function of y 1,n ,y 2,n ,…,y N,n :

q表示全部N个认知用户中不少于k个用户认定主用户信号存在这个事件发生的场景总数;q represents the total number of scenarios in which no less than k users among all N cognitive users believe that the main user signal exists;

将检测概率漏检概率和虚警概率作为多节点硬判决协作频谱检测系统的检测性能参量,完成认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法。will detect probability Probability of missed detection and false alarm probability As a detection performance parameter of a multi-node hard-decision cooperative spectrum detection system, a method for obtaining the performance parameter of the spectrum detection system after an arbitrary movement period in a cognitive radio network is completed.

该方法还可以应用于多节点软判决协作频谱检测系统,The method can also be applied to a multi-node soft-decision cooperative spectrum detection system,

假设移动场景下,认知用户服从随机游走模型,认知网络中存在N个移动的认知用户以及一个融合中心FC;N为正整数;Assume that in a mobile scenario, cognitive users obey the random walk model, and there are N mobile cognitive users and a fusion center FC in the cognitive network; N is a positive integer;

步骤一、采用公式:Step 1. Use the formula:

yn=K/(dn)2 (2)y n =K/(d n ) 2 (2)

获得认知用户检测到主用户信号的实时功率ynObtain the real-time power y n of the primary user signal detected by the cognitive user;

式中:K是常量,且:Where: K is a constant, and:

K=GrPtGtw/4π)2 K=G r P t G tw /4π) 2

式中:Gr表示认知用户的天线增益,Pt表示主用户的发射信号功率,Gt是授权用户的天线增益;λw是主用户信号的波长;In the formula: G r represents the antenna gain of the cognitive user, P t represents the transmit signal power of the primary user, G t is the antenna gain of the authorized user; λ w is the wavelength of the primary user signal;

步骤二、在每次能量检测中,每个认知用户采集M个点,M为正整数;且在高斯信道中,噪声的均值为0,方差为σ2,预设能量检测门限为λ;Step 2. In each energy detection, each cognitive user collects M points, where M is a positive integer; and in a Gaussian channel, the mean value of the noise is 0, the variance is σ 2 , and the preset energy detection threshold is λ;

则在第n个感知周期后,多节点软判决协作频谱的检测概率为:Then after the nth sensing period, the detection probability of the multi-node soft-decision cooperative spectrum for:

漏检概率为:Probability of missed detection for:

虚警概率为:false alarm probability for:

式中:In the formula:

步骤三、根据公式:Step 3, according to the formula:

获得任一认知用户SUi和第n个移动周期的终点之间的间距di,n;1≤i≤N;vi,n表示在第n个移动周期中,该认知用户SUi的移动速度;θi,n表示在第n个移动周期中,该认知用户SUi的移动方向;△t两次检测之间的时间间隔,即每个移动周期的持续的时间;Obtain the distance d i,n between any cognitive user SU i and the end point of the nth mobile cycle; 1≤i≤N; v i,n means that in the nth mobile cycle, the cognitive user SU i The moving speed of ; θ i, n represents the moving direction of the cognitive user SU i in the nth moving cycle; Δt is the time interval between two detections, that is, the duration of each moving cycle;

步骤四、根据公式:Step 4, according to the formula:

获得间距di,n的条件概率密度函数;vmax表示该认知用户SUi的最大移动速度;vmin表示该认知用户SUi的最小移动速度;Obtain the conditional probability density function of distance d i, n ; v max represents the maximum moving speed of the cognitive user SU i ; v min represents the minimum moving speed of the cognitive user SU i ;

步骤五、根据式(4)推导出:Step 5, deduce according to formula (4):

式中:In the formula:

式中:PQ为第n个移动周期内认知用户移动的距离;In the formula: PQ is the moving distance of the cognitive user in the nth mobile cycle;

步骤六、用表达式替换式(5)中的则di,n的条件概率密度函数为:Step 6. Use expressions Replacement formula (5) in Then the conditional probability density function of d i,n is:

步骤七、根据式(2)获得yi,n和di,n之间的关系式:Step 7, according to formula (2), obtain the relationship between y i,n and d i,n :

yi,n=K/(di,n)2 (7)y i,n =K/(d i,n ) 2 (7)

则,yi,n的条件概率密度函数为:Then, the conditional probability density function of y i,n is:

式中:通过将中的di,n利用yi,n替换得到;In the formula: by putting d i, n in is replaced by y i,n ;

步骤八、通过公式:Step eight, pass the formula:

计算yi,n的条件数学期望;Calculate the conditional mathematical expectation of y i,n ;

步骤九、通过公式:Step nine, through the formula:

yi,n的概率密度函数;Probability density function of y i,n ;

其中:是由式(6)、式(8)和下式(11)计算得到的:in: is calculated by formula (6), formula (8) and the following formula (11):

yi,n的数学期望为:The mathematical expectation of y i,n is:

其中,是由式(6)、式(9)和下式(13)计算得到的:in, is calculated by formula (6), formula (9) and the following formula (13):

步骤十、则移动场景下多节点软判决协作频谱检测系统频谱感知的检测概率为:Step 10. The detection probability of spectrum sensing in the multi-node soft-decision cooperative spectrum detection system in the mobile scene for:

漏检概率为:Probability of missed detection for:

虚警概率为:false alarm probability for:

其中:是由式(6)、式(22)和式(23)计算得到的:in: is calculated by formula (6), formula (22) and formula (23):

其中:in:

λs表示的是融合中心处的预设门限,wi,n代表认知用户SUi在第n个感知周期中的加权因子;λ s represents the preset threshold at the fusion center, w i,n represents the weighting factor of cognitive user SU i in the nth perception cycle;

将检测概率漏检概率和虚警概率作为多节点软判决协作频谱检测系统的检测性能参量,完成认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法。will detect probability Probability of missed detection and false alarm probability As a detection performance parameter of a multi-node soft-decision cooperative spectrum detection system, a method for obtaining the performance parameter of the spectrum detection system after experiencing any mobile cycle in a cognitive radio network is completed.

本发明针对认知无线电网络中任意移动周期后移动认知用户频谱检测系统进行了研究,给出了检测性能指标(检测概率、漏检概率、虚警概率)的计算方法。在本发明中,首先根据系统的网络模型以及认知用户的移动模型获得移动速度,运动方向,各个时刻认知用户与主用户的距离之间的函数关系;然后根据本移动周期内的速度,方向和起点与主用户的距离的概率密度函数求得终点与主用户的距离的概率密度函数;再根据信道模型和检测模型获得移动用户实时检测到的主用户信号功率和距离函数之间的函数关系并利用这个关系以及距离函数的概率密度函数计算得出此功率的概率密度函数。最后,利用此功率的概率密度函数以及概率论、信号检测理论的知识求得移动场景下任意时刻单节点感知,硬判决多节点协作感知和软判决协作感知系统中描述检测性能的三个指标,即检测概率,漏检概率以及虚警概率的计算方法。仿真结果显示,仿真值和利用本计算方法计算得出的理论值相吻合,说明了本计算方法的正确性。The invention studies the mobile cognitive user spectrum detection system after any mobile cycle in the cognitive radio network, and provides a calculation method for detection performance indexes (detection probability, missed detection probability, false alarm probability). In the present invention, at first according to the network model of the system and the mobile model of the cognitive user, the functional relationship between the moving speed, the direction of motion, and the distance between the cognitive user and the main user at each moment is obtained; then according to the speed in this mobile cycle, The probability density function of the distance between the direction and the starting point and the primary user is obtained to obtain the probability density function of the distance between the end point and the primary user; then according to the channel model and the detection model, the function between the signal power of the primary user detected by the mobile user in real time and the distance function is obtained relationship and use this relationship and the probability density function of the distance function to calculate the probability density function of this power. Finally, use the probability density function of this power and the knowledge of probability theory and signal detection theory to obtain three indicators describing the detection performance in single-node sensing, hard-decision multi-node cooperative sensing and soft-decision cooperative sensing systems at any time in mobile scenarios. That is, the calculation method of detection probability, missed detection probability and false alarm probability. The simulation results show that the simulated value is consistent with the theoretical value calculated by this calculation method, which shows the correctness of this calculation method.

附图说明Description of drawings

图1是单节点频谱感知系统的认知网络模型示意图;FIG. 1 is a schematic diagram of a cognitive network model of a single-node spectrum sensing system;

图2是多节点协作频谱感知系统的认知网络模型示意图;2 is a schematic diagram of a cognitive network model of a multi-node cooperative spectrum sensing system;

图3是认知无线电网络中移动场景下单节点频谱感知系统中的认知用户移动模型示意图;3 is a schematic diagram of a cognitive user mobility model in a single-node spectrum sensing system in a mobile scenario in a cognitive radio network;

图4是认知无线电网络中移动场景下多节点协作频谱感知系统中的认知用户移动模型示意图;4 is a schematic diagram of a cognitive user mobility model in a multi-node cooperative spectrum sensing system in a mobile scenario in a cognitive radio network;

图5是一个移动周期后,在不同移动速度下限时任一移动认知用户(第r个)实时接收到的主用户信号功率的仿真值和利用本发明计算得到的理论值仿真示意图;其中vmax=60km/h,a=40km,△t=1s;其中:Fig. 5 is after a mobile period, when any mobile cognitive user (the rth) real-time received simulation value of the main user signal power and the theoretical value simulation schematic diagram obtained by utilizing the calculation of the present invention when different moving speed lower limits; Wherein v max = 60km/h, a = 40km, △t = 1s; where:

曲线41是K=50(w×km2)下的仿真曲线;Curve 41 is a simulation curve under K=50(w×km 2 );

曲线42是K=50(w×km2)下的理论曲线;Curve 42 is a theoretical curve under K=50(w×km 2 );

曲线43是K=100(w×km2)下的仿真曲线;Curve 43 is a simulation curve under K=100(w×km 2 );

曲线44是K=100(w×km2)下的理论曲线;Curve 44 is a theoretical curve under K=100(w×km 2 );

曲线45是K=200(w×km2)下的仿真曲线;Curve 45 is a simulation curve under K=200(w×km 2 );

曲线46是K=200(w×km2)下的理论曲线;Curve 46 is a theoretical curve under K=200(w×km 2 );

曲线47是K=400(w×km2)下的仿真曲线;Curve 47 is a simulation curve under K=400(w×km 2 );

曲线48是K=400(w×km2)下的理论曲线;Curve 48 is a theoretical curve under K=400(w×km 2 );

图6是两个移动周期后,在不同移动速度下限时任一移动认知用户(第r个)实时接收到的主用户信号功率的仿真值和利用本发明计算得到的理论值仿真示意图;其中vmax=60km/h,a=40km,△t=1s;其中:Fig. 6 is after two mobile cycles, when any mobile cognitive user (the rth) real-time received simulation value of the main user signal power and the theoretical value simulation schematic diagram obtained by utilizing the calculation of the present invention when different moving speed lower limits; Wherein v max =60km/h, a=40km, △t=1s; where:

曲线51是K=50(w×km2)下的仿真曲线;Curve 51 is a simulation curve under K=50(w×km 2 );

曲线52是K=50(w×km2)下的理论曲线;Curve 52 is a theoretical curve under K=50(w×km 2 );

曲线53是K=100(w×km2)下的仿真曲线;Curve 53 is a simulation curve under K=100(w×km 2 );

曲线54是K=100(w×km2)下的理论曲线;Curve 54 is a theoretical curve under K=100(w×km 2 );

曲线55是K=200(w×km2)下的仿真曲线;Curve 55 is a simulation curve under K=200(w×km 2 );

曲线56是K=200(w×km2)下的理论曲线;Curve 56 is a theoretical curve under K=200(w×km 2 );

曲线57是K=400(w×km2)下的仿真曲线;Curve 57 is a simulation curve under K=400(w×km 2 );

曲线58是K=400(w×km2)下的理论曲线;Curve 58 is a theoretical curve under K=400(w×km 2 );

图7是三个移动周期后,在不同移动速度下限时任一移动认知用户(第r个)实时接收到的主用户信号功率的仿真值和利用本发明计算得到的理论值仿真示意图;其中vmax=60km/h,a=40km,△t=1s;其中:Fig. 7 is after three mobile periods, when any mobile cognitive user (the rth) real-time received simulation value of the main user signal power and the simulation schematic diagram of the theoretical value calculated by the present invention when different mobile speed lower limits; v max =60km/h, a=40km, △t=1s; where:

曲线61是K=50(w×km2)下的仿真曲线;Curve 61 is a simulation curve under K=50(w×km 2 );

曲线62是K=50(w×km2)下的理论曲线;Curve 62 is a theoretical curve under K=50(w×km 2 );

曲线63是K=100(w×km2)下的仿真曲线;Curve 63 is a simulation curve under K=100(w×km 2 );

曲线64是K=100(w×km2)下的理论曲线;Curve 64 is a theoretical curve under K=100(w×km 2 );

曲线65是K=200(w×km2)下的仿真曲线;Curve 65 is a simulation curve under K=200(w×km 2 );

曲线66是K=200(w×km2)下的理论曲线;Curve 66 is a theoretical curve under K=200(w×km 2 );

曲线67是K=400(w×km2)下的仿真曲线;Curve 67 is a simulation curve under K=400(w×km 2 );

曲线68是K=400(w×km2)下的理论曲线;Curve 68 is a theoretical curve under K=400(w×km 2 );

具体实施方式detailed description

具体实施方式一、认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法,由以下步骤实现:Specific Embodiments 1. A method for obtaining performance parameters of a spectrum detection system after experiencing any movement period in a cognitive radio network is implemented by the following steps:

首先需要指出的是为了能够具体地给出各个步骤对应的表达式且不失一般性,我们以以下具体的模型为例,但事实上本发明方法同样适用于其他的认知网络模型以及认知用户(Secondary User,SU)的移动模型。假设认知网络中存在一个主用户(Primary User,PU)(位于半径为a的圆形网络区域中心)。本发明可以计算移动场景下的单节点频谱感知算法以及多节点协作频谱感知算法的检测性能。First of all, it needs to be pointed out that in order to be able to give the expressions corresponding to each step in detail without loss of generality, we take the following specific models as examples, but in fact the method of the present invention is also applicable to other cognitive network models and cognitive User (Secondary User, SU) mobility model. Assume that there is a primary user (Primary User, PU) in the cognitive network (located in the center of a circular network area with a radius of a). The invention can calculate the detection performance of a single-node spectrum sensing algorithm and a multi-node cooperative spectrum sensing algorithm in a mobile scene.

对于单节点的系统,网络中只存在一个移动的认知用户,如图1所示。For a single-node system, there is only one mobile cognitive user in the network, as shown in Figure 1.

对于多节点协作系统,网络中存在N个移动的认知用户以及一个融合中心(FusionCenter,FC),如图2所示。For a multi-node cooperative system, there are N mobile cognitive users and a fusion center (FusionCenter, FC) in the network, as shown in FIG. 2 .

对于移动模型来说,假设认知用户服从随机游走模型。图3给出了单节点频谱感知系统中的移动认知用户的移动模型,其移动路线可以表示为A-B-…-P-Q-…。图3中的每个线段都代表一个移动周期,例如线段PQ。每一个线段的端点都代表一个感知周期,例如点P。时间△t表示两次检测之间的时间间隔,即每个移动周期的持续的时间。在第n个移动周期中,认知用户是从点P移动到点Q的,而且一旦移动到点Q,便进行一次能量检测(EnergyDetection,ED),记为第n个检测周期中的能量检测。认知用户随机地独立地以速度vn,按方向θn在第n个移动周期内进行移动。vn服从均匀分布,均匀分布在vmin至vmax的范围内。θn则均匀分布在0至2π的范围内。在下一个移动周期中,该认知用户会随机选取一个新的速度和方向,继续随机移动。起点A是在整个网络中随机选取的,d0是主用户和A之间的间隔。P表示第n个移动周期的移动起始点,Q代表终点,同时是第n+1个移动周期的起点。dn-1和dn分别代表主用户和点P和Q的距离。那么vn,θn和d0的概率密度函数分别为For the mobility model, it is assumed that cognitive users obey a random walk model. Figure 3 shows the movement model of the mobile cognitive user in the single-node spectrum sensing system, and its movement route can be expressed as AB-…-PQ-…. Each line segment in Fig. 3 represents a movement period, eg line segment PQ. The endpoint of each line segment represents a perception cycle, such as point P. The time Δt represents the time interval between two detections, that is, the duration of each movement cycle. In the nth mobile cycle, the cognitive user moves from point P to point Q, and once moving to point Q, an energy detection (EnergyDetection, ED) is performed, which is recorded as the energy detection in the nth detection cycle . Cognitive users move independently at speed v n in direction θ n during the nth movement period independently. v n obeys the uniform distribution, and the uniform distribution is in the range from v min to v max . θ n is evenly distributed in the range of 0 to 2π. In the next movement cycle, the cognitive user randomly picks a new speed and direction and continues to move randomly. The starting point A is randomly selected in the whole network, and d0 is the interval between the main user and A. P represents the moving start point of the nth movement cycle, Q represents the end point, and is also the starting point of the n+1th movement cycle. d n-1 and d n represent the distances between the primary user and points P and Q, respectively. Then the probability density functions of v n , θ n and d 0 are respectively

对于协作频谱感知系统中的认知用户来说,每个用户都服从相同的如上文所述移动模型。图4给出了其中任意一个移动认知用户SUi的移动模型,1≤i≤N。下面使用角标“i”进行区分协作频谱感知系统中的各个认知用户。那么在第n个移动周期中SUi的移动速度,移动方向分别是vi,n和θi,n。SUi和第n个移动周期的终点之间的间距为di,nFor cognitive users in a collaborative spectrum sensing system, each user obeys the same mobility model as described above. Figure 4 shows the mobility model of any mobile cognitive user SU i , 1≤i≤N. The superscript "i" is used below to distinguish each cognitive user in the cooperative spectrum sensing system. Then the moving speed and moving direction of SU i in the nth moving cycle are v i,n and θ i,n respectively. The distance between SU i and the end of the nth travel cycle is d i,n .

在单节点频谱检测系统中,由于能量检测法的简便易实现且可用于检测任意波形的信号的特性,假设认知用户采用能量检测法判定主用户是否存在。用yn表示在第n个感知周期中认知用户实时检测到的主用户信号功率。yn可以是距离、认知用户位置甚至是时间的函数。为简便且不失一般性,假设如下信道模型,其中yn只与dn(主用户和认知用户在进行第n个感知周期时位置的距离)有关。yn=GrPtGtw/4πdn)2,其中Gr表示认知用户的天线增益,Pt表示主用户的发射信号功率,Gr是授权用户的天线增益。λw是主用户信号的波长。那么认知用户检测到主用户信号的实时功率为In the single-node spectrum detection system, because the energy detection method is easy to implement and can be used to detect the characteristics of arbitrary waveform signals, it is assumed that the cognitive user uses the energy detection method to determine whether the primary user exists. Let y n denote the real-time primary user signal power detected by the cognitive user in the nth sensing period. y n can be a function of distance, perceived user location, or even time. For simplicity and without loss of generality, the following channel model is assumed, where y n is only related to d n (the distance between the primary user and the cognitive user at the nth sensing cycle). y n =G r Pt G tw /4πd n ) 2 , where G r represents the antenna gain of the cognitive user, P t represents the transmitted signal power of the primary user, and G r is the antenna gain of the authorized user. λw is the wavelength of the primary user signal. Then the cognitive user detects the real-time power of the primary user signal as

yn=K/(dn)2,K=GrPtGtw/4π)2 (25)y n =K/(d n ) 2 , K=G r P t G tw /4π) 2 (25)

因此,K是一个常量。Therefore, K is a constant.

在每次能量检测中,每个认知用户采M个点,且在高斯信道中,噪声的均值为0,方差为σ2,预设能量检测门限为λ。那么在第n个感知周期后,单节点频谱感知的检测性能参量(检测概率、漏检概率、虚警概率)的计算结果即为In each energy detection, each cognitive user takes M points, and in the Gaussian channel, the mean value of the noise is 0, the variance is σ 2 , and the preset energy detection threshold is λ. Then after the nth sensing cycle, the calculation results of the detection performance parameters (detection probability, missed detection probability, false alarm probability) of single-node spectrum sensing are

其中在多节点协作频谱感知系统中,全部N个认知用户都进行能量检测并把最终的检测结果上传到融合中心处,由融合中心做出主用户信号是否存在的最终判决结果并将此结果广播给各个认知用户。而协作频谱感知系统的检测性能参量可以通过(26)求出。需要说明的是,(26)的计算结果正确的前提是各个认知用户都是静止的。在式(26)的基础上,考虑认知用户的移动性,检测性能的参量可由如下方法计算。in In a multi-node cooperative spectrum sensing system, all N cognitive users perform energy detection and upload the final detection results to the fusion center, and the fusion center makes the final decision on whether the primary user signal exists and broadcasts the result to each cognitive user. The detection performance parameters of the cooperative spectrum sensing system can be calculated by (26). It should be noted that the correct premise of the calculation result of (26) is that each cognitive user is stationary. On the basis of Equation (26), considering the mobility of cognitive users, the parameters of detection performance can be calculated by the following method.

首先,根据图3所示几何关系,可以求得di,n的表达式。First, according to the geometric relationship shown in Figure 3, the expression of d i, n can be obtained.

那么根据(27)以及vi,n的概率密度函数,di,n的条件概率密度函数为Then according to (27) and the probability density function of v i, n , the conditional probability density function of d i, n is

其中di,n的取值范围可由图3中的几何关系求得。根据(27)可求出导数其中Among them, the value range of d i, n It can be obtained from the geometric relationship in Figure 3. According to (27), the derivative can be found in

那么都是(di,n-1,di,ni,n)的函数。用替换那么di,n的条件概率密度函数为So with Both are functions of (d i,n-1 ,d i,ni,n ). use replace Then the conditional probability density function of d i,n is

根据(25)可以获得yi,n和di,n之间的函数关系式:According to (25), the functional relationship between y i,n and d i,n can be obtained:

因此,yi,n的条件概率密度函数为Therefore, the conditional probability density function of y i,n is

其中是(di,n-1,yi,ni,n)的函数,而且可以通过将中的di,n利用yi,n替换而得到。同时,yi,n的取值范围能够通过获得。因此,yi,n的条件数学期望为in is a function of (d i,n-1 ,y i,ni,n ), and This can be done by passing d i,n in is obtained by replacing y i,n . At the same time, the value range of y i, n able to pass get. Therefore, the conditional mathematical expectation of y i,n is

那么根据(32),yi,n的概率密度函数为Then according to (32), the probability density function of y i, n is

其中是由(30),(32)和下式(35)计算得到的:in is calculated by (30), (32) and the following formula (35):

yi,n的数学期望为The mathematical expectation of y i,n is

其中,是由(30),(33)和下式(37)计算得到的:in, is calculated by (30), (33) and the following formula (37):

那么移动场景下单节点频谱感知的检测概率、漏检概率、虚警概率分别为:Then the detection probability, missed detection probability, and false alarm probability of single-node spectrum sensing in mobile scenarios are:

其中是由(30)和下式(39),(40)计算获得的:in is calculated by (30) and the following formulas (39), (40):

事实上,单节点频谱感知算法有很多的缺点,例如隐藏终端问题,信道衰落以及多经效应,这些都导致了系统检测性能的大大下降。为了解决这些问题,研究人员提出了协作频谱感知技术。协作频谱感知技术包括硬判决协作频谱感知以及软判决协作频谱感知。In fact, the single-node spectrum sensing algorithm has many disadvantages, such as hidden terminal problems, channel fading and multi-channel effects, which lead to a great decline in system detection performance. To address these issues, researchers propose collaborative spectrum sensing techniques. Cooperative spectrum sensing technologies include hard-decision cooperative spectrum sensing and soft-decision cooperative spectrum sensing.

首先推导移动场景下硬判决协作频谱感知技术的检测性能参量表达式。由于“AND”、“OR”、“MAJORITY”等硬判决协作感知都是“K秩”(只有当系统中存在不少于k个认知用户认为主用户信号是存在的,融合中心才判断主用户是存在的)的特例,这里给出更为一般的“K秩”硬判决协作频谱感知系统的检测性能(检测概率、漏检概率、虚警概率)计算结果。Firstly, the expression of detection performance parameters of hard-decision cooperative spectrum sensing technology in mobile scenarios is derived. Since "AND", "OR", "MAJORITY" and other hard-decision collaborative perceptions are all "K rank" (only when there are no less than k cognitive users in the system who think that the main user signal exists, the fusion center can judge the main user's signal. user exists), here are the calculation results of the detection performance (detection probability, missed detection probability, false alarm probability) of a more general "K rank" hard decision cooperative spectrum sensing system.

其中是通过(30),(42)和(43)计算获得的。in is calculated by (30), (42) and (43).

其中, in,

并且f(y1,n,…,yN,n|di,n-1i,n)表示y1,n,y2,n,…,yN,n的联合条件概率密度函数:And f(y 1,n ,…,y N,n |d i,n-1i,n ) represents the joint conditional probability density function of y 1,n ,y 2,n ,…,y N,n :

q表示全部N个认知用户中不少于k个用户认定主用户信号存在这个事件发生的场景总数,对应的概率则为 q represents the total number of scenarios where no less than k users among all N cognitive users believe that the main user signal exists, and the corresponding probability is

在移动场景下软判决协作频谱感知的检测性能参量(检测概率、漏检概率、虚警概率)为:The detection performance parameters (detection probability, missed detection probability, and false alarm probability) of soft-decision cooperative spectrum sensing in mobile scenarios are:

其中是由(30),(46)和(47)计算得到的。in is calculated by (30), (46) and (47).

其中,in,

而且λs表示的是融合中心处的预设门限,wi,n代表SUi在第n个感知周期中的加权因子。Moreover, λ s represents the preset threshold at the fusion center, and w i,n represents the weighting factor of SU i in the nth perception cycle.

本发明针对认知无线电网络中任意移动周期后移动认知用户频谱检测系统进行了研究,给出了检测性能指标(检测概率、漏检概率、虚警概率)的计算方法。在本发明中,首先根据系统的网络模型以及认知用户的移动模型获得移动速度,运动方向,各个时刻认知用户与主用户的距离之间的函数关系。然后根据本移动周期内的速度,方向和起点与主用户的距离的概率密度函数求得终点与主用户的距离的概率密度函数。再根据信道模型和检测模型获得移动用户实时检测到的主用户信号功率和距离函数之间的函数关系并利用这个关系以及距离函数的概率密度函数计算得出此功率的概率密度函数。最后,利用此功率的概率密度函数以及概率论、信号检测理论的知识求得移动场景下任意时刻单节点感知,硬判决多节点协作感知和软判决协作感知系统中描述检测性能的三个指标,即检测概率,漏检概率以及虚警概率的获得方法。仿真结果显示,仿真值和利用本发明的方法得出的理论值相吻合,说明了本发明的正确性。The invention studies the mobile cognitive user spectrum detection system after any mobile cycle in the cognitive radio network, and provides a calculation method for detection performance indexes (detection probability, missed detection probability, false alarm probability). In the present invention, firstly, according to the network model of the system and the movement model of the cognitive user, the functional relationship between the moving speed, the direction of movement, and the distance between the cognitive user and the main user at each moment is obtained. Then obtain the probability density function of the distance between the end point and the main user according to the speed, direction and the probability density function of the distance between the starting point and the main user in this mobile cycle. According to the channel model and the detection model, the functional relationship between the real-time detected primary user signal power and the distance function is obtained, and the probability density function of this power is calculated by using this relationship and the probability density function of the distance function. Finally, use the probability density function of this power and the knowledge of probability theory and signal detection theory to obtain three indicators describing the detection performance in single-node sensing, hard-decision multi-node cooperative sensing and soft-decision cooperative sensing systems at any time in mobile scenarios. That is, the method of obtaining detection probability, missed detection probability and false alarm probability. The simulation result shows that the simulated value coincides with the theoretical value obtained by using the method of the present invention, which illustrates the correctness of the present invention.

通过理论值与实验值吻合这一事实证明了本发明方法的正确性。The correctness of the method of the present invention is proved by the fact that the theoretical value coincides with the experimental value.

本发明具有以下特点和显著进步:The present invention has following characteristics and remarkable progress:

1、本发明的方法是针对移动认知用户提出的。相对于当前绝大多数假设认知用户静止不动的方法来说要更加符合实际认知网络的情况。1. The method of the present invention is proposed for mobile cognitive users. Compared with most of the current methods that assume that cognitive users are stationary, it is more in line with the actual cognitive network situation.

2、本发明的方法可广泛用于单节点频谱感知,硬判决协作频谱感知以及软判决协作频谱感知系统。2. The method of the present invention can be widely used in single-node spectrum sensing, hard-decision cooperative spectrum sensing and soft-decision cooperative spectrum sensing systems.

3、本发明的方法可以用于任何形状的认知网络,可用于任意运动的认知用户检测系统中,适用范围广。3. The method of the present invention can be used in cognitive networks of any shape, and can be used in a cognitive user detection system for any movement, and has a wide range of applications.

4、本发明的方法可以计算在任意移动周期后的频谱感知检测性能参量给出了一个一般化的结果。4. The method of the present invention can calculate the spectrum sensing detection performance parameters after any movement period and give a generalized result.

Claims (3)

1.认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法,其特征是:1. A method for obtaining performance parameters of a spectrum detection system after experiencing any mobile period in a cognitive radio network, characterized in that: 该方法应用于单节点频谱检测系统;The method is applied to a single-node spectrum detection system; 假设移动场景下,认知用户服从随机游走模型,认知网络中存在一个主用户和一个认知用户;Assume that in the mobile scenario, cognitive users obey the random walk model, and there is a primary user and a cognitive user in the cognitive network; 步骤一、采用公式:Step 1. Use the formula: yn=K/(dn)2 (2)y n =K/(d n ) 2 (2) 式中:dn表示主用户和认知用户在进行第n个感知周期时位置的距离;In the formula: d n represents the distance between the primary user and the cognitive user during the nth sensing cycle; 获得认知用户检测到主用户信号的实时功率ynObtain the real-time power y n of the primary user signal detected by the cognitive user; 式中:K是常量,且:Where: K is a constant, and: K=GrPtGtw/4π)2 K=G r P t G tw /4π) 2 式中:Gr表示认知用户的天线增益,Pt表示主用户的发射信号功率,Gt是授权用户的天线增益;λw是主用户信号的波长;In the formula: G r represents the antenna gain of the cognitive user, P t represents the transmit signal power of the primary user, G t is the antenna gain of the authorized user; λ w is the wavelength of the primary user signal; 步骤二、在每次能量检测中,认知用户采集M个点,M为正整数;且在高斯信道中,噪声的均值为0,方差为σ2,预设能量检测门限为λ;Step 2. In each energy detection, the cognitive user collects M points, where M is a positive integer; and in the Gaussian channel, the mean value of the noise is 0, the variance is σ 2 , and the preset energy detection threshold is λ; 则在第n个感知周期后,单节点频谱感知的检测概率为:Then after the nth sensing period, the detection probability of single-node spectrum sensing for: <mrow> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> 漏检概率为:Probability of missed detection for: <mrow> <msub> <mi>P</mi> <msub> <mi>m</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>m</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> 虚警概率为:false alarm probability for: <mrow> <msub> <mi>P</mi> <msub> <mi>f</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>f</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> 式中:In the formula: <mrow> <mi>Q</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mi>x</mi> <mrow> <mo>+</mo> <mi>&amp;infin;</mi> </mrow> </msubsup> <mfrac> <mn>1</mn> <msqrt> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msqrt> </mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mfrac> <msup> <mi>t</mi> <mn>2</mn> </msup> <mn>2</mn> </mfrac> </mrow> </msup> <mi>d</mi> <mi>t</mi> </mrow> <mrow> <mi>Q</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mi>x</mi> <mrow> <mo>+</mo> <mi>&amp;infin;</mi> </mrow> </msubsup> <mfrac> <mn>1</mn> <msqrt> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msqrt> </mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mfrac> <msup> <mi>t</mi> <mn>2</mn> </msup> <mn>2</mn> </mfrac> </mrow> </msup> <mi>d</mi> <mi>t</mi> </mrow> 步骤三、根据公式:Step 3, according to the formula: <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>=</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>=</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> 获得第i个认知用户和第n个移动周期的终点之间的间距di,n;1≤i≤N,N是网络中认知用户的总数;vi,n表示在第n个移动周期中,该认知用户的移动速度;θi,n表示在第n个移动周期中,该认知用户的移动方向;Δt两次检测之间的时间间隔,即每个移动周期的持续的时间;Obtain the distance d i,n between the i-th cognitive user and the end point of the n-th mobile cycle; 1≤i≤N, N is the total number of cognitive users in the network; cycle, the moving speed of the cognitive user; θi ,n represents the moving direction of the cognitive user in the nth mobile cycle; Δt is the time interval between two detections, that is, the continuous time; 步骤四、根据公式:Step 4, according to the formula: <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mi>min</mi> </msub> </mrow> </mfrac> <mo>|</mo> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>/</mo> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mi>min</mi> </msub> </mrow> </mfrac> <mo>|</mo> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>/</mo> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> </mrow> 获得间距di,n的条件概率密度函数;vmax表示该认知用户的最大移动速度;vmin表示该认知用户的最小移动速度;Obtain the conditional probability density function of the distance d i, n ; v max represents the maximum moving speed of the cognitive user; v min represents the minimum moving speed of the cognitive user; 步骤五、根据式(4)推导出:Step 5, deduce according to formula (4): <mrow> <mo>|</mo> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mfrac> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>&amp;CenterDot;</mo> <mrow> <mo>(</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>|</mo> </mrow> <mrow> <mo>|</mo> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mfrac> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>&amp;CenterDot;</mo> <mrow> <mo>(</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>|</mo> </mrow> 式中:In the formula: <mrow> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>=</mo> <mo>-</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <mi>cos</mi> <mn>2</mn> </msup> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>=</mo> <mo>-</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <mi>cos</mi> <mn>2</mn> </msup> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> 式中:PQ为第n个移动周期内认知用户移动的距离;In the formula: PQ is the moving distance of the cognitive user in the nth mobile period; 步骤六、用表达式替换式(5)中的则di,n的条件概率密度函数为:Step 6. Use expressions Replacement formula (5) in Then the conditional probability density function of d i,n is: <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> 步骤七、根据式(2)获得yi,n和di,n之间的关系式:Step 7, according to formula (2), obtain the relationship between y i,n and d i,n : yi,n=K/(di,n)2 (7)y i,n =K/(d i,n ) 2 (7) 则,yi,n的条件概率密度函数为:Then, the conditional probability density function of y i,n is: <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mfrac> <msqrt> <mi>K</mi> </msqrt> <mrow> <mn>2</mn> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msqrt> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </msqrt> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mfrac> <msqrt> <mi>K</mi> </msqrt> <mrow> <mn>2</mn> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msqrt> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </msqrt> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> 式中:通过将中的di,n利用yi,n替换得到;In the formula: by putting d i, n in is replaced by y i,n ; 步骤八、通过公式:Step eight, pass the formula: <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </msub> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msub> <mi>dy</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </msub> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msub> <mi>dy</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> 式中:表示yi,n的取值范围;In the formula: Indicates the value range of y i, n ; 计算yi,n的条件数学期望;Calculate the conditional mathematical expectation of y i,n ; 步骤九、通过公式:Step nine, through the formula: <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> yi,n的概率密度函数;Probability density function of y i,n ; 其中:是由式(6)、式(8)和下式(11)计算得到的:in: is calculated by formula (6), formula (8) and the following formula (11): <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> 2 <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> 2 式中:表示di,n-1的取值范围;In the formula: Indicates the value range of d i,n-1 ; yi,n的数学期望为:The mathematical expectation of y i,n is: <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> 其中,是由式(6)、式(9)和下式(13)计算得到的:in, is calculated by formula (6), formula (9) and the following formula (13): <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow> 步骤十、则移动场景下单节点频谱感知的检测概率为:Step 10. The detection probability of single-node spectrum sensing in mobile scenarios for: <mrow> <msubsup> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <mi>n</mi> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mn>0</mn> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <msub> <mi>dd</mi> <mn>0</mn> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mn>1</mn> </msub> </mrow> <mrow> <msubsup> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <mi>n</mi> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mn>0</mn> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <msub> <mi>dd</mi> <mn>0</mn> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mn>1</mn> </msub> </mrow> 漏检概率为:Probability of missed detection for: <mrow> <msubsup> <mi>P</mi> <msub> <mi>m</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> </mrow> <mrow> <msubsup> <mi>P</mi> <msub> <mi>m</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> </mrow> 虚警概率为:false alarm probability for: <mrow> <msubsup> <mi>P</mi> <msub> <mi>f</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <msubsup> <mi>P</mi> <msub> <mi>f</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> </mrow> 其中:是由(6)和下式(15)、(16)计算获得的:in: is calculated by (6) and the following formulas (15), (16): <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <mi>n</mi> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <mi>n</mi> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mi>n</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>15</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <mi>n</mi> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <mi>n</mi> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mi>n</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>15</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <mi>n</mi> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <msub> <mi>dy</mi> <mi>n</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>16</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <mi>n</mi> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <msub> <mi>dy</mi> <mi>n</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>16</mn> <mo>)</mo> </mrow> </mrow> 将检测概率漏检概率和虚警概率作为单节点频谱感知系统的检测性能参量,完成认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法。will detect probability Probability of missed detection and false alarm probability As a detection performance parameter of a single-node spectrum sensing system, a method for obtaining the performance parameter of a spectrum detection system after an arbitrary movement period in a cognitive radio network is completed. 2.认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法,其特征是:2. A method for obtaining performance parameters of a spectrum detection system after experiencing any mobile cycle in a cognitive radio network, characterized in that: 该方法应用于多节点硬判决协作频谱检测系统,This method is applied to multi-node hard-decision cooperative spectrum detection system, 假设移动场景下,认知用户服从随机游走模型,认知网络中存在N个移动的认知用户以及一个融合中心FC;N为正整数;Assume that in a mobile scenario, cognitive users obey the random walk model, and there are N mobile cognitive users and a fusion center FC in the cognitive network; N is a positive integer; 步骤一、采用公式:Step 1. Use the formula: yn=K/(dn)2 (2)y n =K/(d n ) 2 (2) 式中:dn表示主用户和认知用户在进行第n个感知周期时位置的距离;In the formula: d n represents the distance between the primary user and the cognitive user during the nth sensing cycle; 获得认知用户检测到主用户信号的实时功率ynObtain the real-time power y n of the primary user signal detected by the cognitive user; 式中:K是常量,且:Where: K is a constant, and: K=GrPtGtw/4π)2 K=G r P t G tw /4π) 2 式中:Gr表示认知用户的天线增益,Pt表示主用户的发射信号功率,Gt是授权用户的天线增益;λw是主用户信号的波长;In the formula: G r represents the antenna gain of the cognitive user, P t represents the transmit signal power of the primary user, G t is the antenna gain of the authorized user; λ w is the wavelength of the primary user signal; 步骤二、在每次能量检测中,每个认知用户采集M个点,M为正整数;且在高斯信道中,噪声的均值为0,方差为σ2,预设能量检测门限为λ;Step 2. In each energy detection, each cognitive user collects M points, where M is a positive integer; and in a Gaussian channel, the mean value of the noise is 0, the variance is σ 2 , and the preset energy detection threshold is λ; 则在第n个感知周期后,多节点硬判决协作频谱感知的检测概率为:Then after the nth sensing period, the detection probability of multi-node hard-decision cooperative spectrum sensing for: <mrow> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> 漏检概率为:Probability of missed detection for: <mrow> <msub> <mi>P</mi> <msub> <mi>m</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>m</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> 虚警概率为:false alarm probability for: <mrow> <msub> <mi>P</mi> <msub> <mi>f</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>f</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> 式中:In the formula: <mrow> <mi>Q</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mi>x</mi> <mrow> <mo>+</mo> <mi>&amp;infin;</mi> </mrow> </msubsup> <mfrac> <mn>1</mn> <msqrt> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msqrt> </mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mfrac> <msup> <mi>t</mi> <mn>2</mn> </msup> <mn>2</mn> </mfrac> </mrow> </msup> <mi>d</mi> <mi>t</mi> </mrow> <mrow> <mi>Q</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mi>x</mi> <mrow> <mo>+</mo> <mi>&amp;infin;</mi> </mrow> </msubsup> <mfrac> <mn>1</mn> <msqrt> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msqrt> </mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mfrac> <msup> <mi>t</mi> <mn>2</mn> </msup> <mn>2</mn> </mfrac> </mrow> </msup> <mi>d</mi> <mi>t</mi> </mrow> 步骤三、根据公式:Step 3, according to the formula: <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>=</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>=</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> 获得任一认知用户SUi和第n个移动周期的终点之间的间距di,n;1≤i≤N;vi,n表示在第n个移动周期中,该认知用户SUi的移动速度;θi,n表示在第n个移动周期中,该认知用户SUi的移动方向;Δt两次检测之间的时间间隔,即每个移动周期的持续的时间;Obtain the distance d i,n between any cognitive user SU i and the end point of the nth mobile cycle; 1≤i≤N; v i,n means that in the nth mobile cycle, the cognitive user SU i The moving speed of ; θ i, n represents the moving direction of the cognitive user SU i in the nth moving cycle; Δt is the time interval between two detections, that is, the duration of each moving cycle; 步骤四、根据公式:Step 4, according to the formula: <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mi>min</mi> </msub> </mrow> </mfrac> <mo>|</mo> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>/</mo> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mi>min</mi> </msub> </mrow> </mfrac> <mo>|</mo> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>/</mo> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> </mrow> 获得间距di,n的条件概率密度函数;vmax表示该认知用户SUi的最大移动速度;vmin表示该认知用户SUi的最小移动速度;Obtain the conditional probability density function of distance d i, n ; v max represents the maximum moving speed of the cognitive user SU i ; v min represents the minimum moving speed of the cognitive user SU i ; 步骤五、根据式(4)推导出:Step 5, deduce according to formula (4): <mrow> <mo>|</mo> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mfrac> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>&amp;CenterDot;</mo> <mrow> <mo>(</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>|</mo> </mrow> 4 <mrow> <mo>|</mo> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mfrac> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>&amp;CenterDot;</mo> <mrow> <mo>(</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>|</mo> </mrow> 4 式中:In the formula: <mrow> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>=</mo> <mo>-</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <mi>cos</mi> <mn>2</mn> </msup> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>=</mo> <mo>-</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <mi>cos</mi> <mn>2</mn> </msup> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> 式中:PQ为第n个移动周期内认知用户移动的距离;In the formula: PQ is the moving distance of the cognitive user in the nth mobile period; 步骤六、用表达式替换式(5)中的则di,n的条件概率密度函数为:Step 6. Use expressions Replacement formula (5) in Then the conditional probability density function of d i,n is: <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> 步骤七、根据式(2)获得yi,n和di,n之间的关系式:Step 7, according to formula (2), obtain the relationship between y i,n and d i,n : yi,n=K/(di,n)2 (7)y i,n =K/(d i,n ) 2 (7) 则,yi,n的条件概率密度函数为:Then, the conditional probability density function of y i, n is: <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mfrac> <msqrt> <mi>K</mi> </msqrt> <mrow> <mn>2</mn> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msqrt> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </msqrt> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mfrac> <msqrt> <mi>K</mi> </msqrt> <mrow> <mn>2</mn> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msqrt> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </msqrt> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> 式中:通过将中的di,n利用yi,n替换得到;In the formula: by putting d i, n in is replaced by y i,n ; 步骤八、通过公式:Step eight, pass the formula: <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </msub> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msub> <mi>dy</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </msub> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msub> <mi>dy</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> 式中:表示yi,n的取值范围;In the formula: Indicates the value range of y i, n ; 计算yi,n的条件数学期望;Calculate the conditional mathematical expectation of y i,n ; 步骤九、通过公式:Step nine, through the formula: <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> yi,n的概率密度函数;Probability density function of y i,n ; 其中:是由式(6)、式(8)和下式(11)计算得到的:in: is calculated by formula (6), formula (8) and the following formula (11): <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> 式中:表示di,n-1的取值范围;In the formula: Indicates the value range of d i,n-1 ; yi,n的数学期望为:The mathematical expectation of y i,n is: <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> 其中,是由式(6)、式(9)和下式(13)计算得到的:in, is calculated by formula (6), formula (9) and the following formula (13): <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow> 步骤十、则移动场景下多节点硬判决协作频谱检测系统频谱感知的检测概率为:Step 10: The detection probability of spectrum sensing in the multi-node hard-decision cooperative spectrum detection system in the mobile scene for: <mrow> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>H</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>H</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> 漏检概率为:Probability of missed detection for: <mrow> <mn>1</mn> <mo>-</mo> <msubsup> <mi>Q</mi> <mrow> <mi>d</mi> <mo>-</mo> <msub> <mi>H</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <msubsup> <mi>Q</mi> <mrow> <mi>d</mi> <mo>-</mo> <msub> <mi>H</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> </mrow> 虚警概率为:false alarm probability for: <mrow> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mi>k</mi> </mrow> <mi>N</mi> </munderover> <msubsup> <mi>C</mi> <mi>N</mi> <mi>i</mi> </msubsup> <msup> <mrow> <mo>(</mo> <msubsup> <mi>P</mi> <mi>f</mi> <mi>v</mi> </msubsup> <mo>)</mo> </mrow> <mi>i</mi> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>P</mi> <mi>f</mi> <mi>v</mi> </msubsup> <mo>)</mo> </mrow> <mrow> <mi>N</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> </mrow> <mrow> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mi>k</mi> </mrow> <mi>N</mi> </munderover> <msubsup> <mi>C</mi> <mi>N</mi> <mi>i</mi> </msubsup> <msup> <mrow> <mo>(</mo> <msubsup> <mi>P</mi> <mi>f</mi> <mi>v</mi> </msubsup> <mo>)</mo> </mrow> <mi>i</mi> </msup> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>P</mi> <mi>f</mi> <mi>v</mi> </msubsup> <mo>)</mo> </mrow> <mrow> <mi>N</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> </mrow> 其中:是通过式(6)、式(18)和式(19)计算获得的:in: is calculated by formula (6), formula (18) and formula (19): <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>H</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>H</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>18</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>H</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>H</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>18</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>H</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <mo>...</mo> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <msubsup> <mi>Q</mi> <mi>H</mi> <mi>v</mi> </msubsup> <mi>f</mi> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <mo>...</mo> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>dy</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>...</mo> <msub> <mi>dy</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>19</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>H</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <mo>...</mo> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <msubsup> <mi>Q</mi> <mi>H</mi> <mi>v</mi> </msubsup> <mi>f</mi> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <mo>...</mo> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>dy</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>...</mo> <msub> <mi>dy</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>19</mn> <mo>)</mo> </mrow> </mrow> 其中:in: <mrow> <msubsup> <mi>Q</mi> <mi>H</mi> <mi>v</mi> </msubsup> <mo>=</mo> <munder> <mo>&amp;Sigma;</mo> <mi>q</mi> </munder> <mrow> <mo>(</mo> <msubsup> <mi>Q</mi> <mrow> <mi>h</mi> <mi>a</mi> <mi>r</mi> <mi>d</mi> </mrow> <mi>v</mi> </msubsup> <mo>&amp;CenterDot;</mo> <munderover> <mo>&amp;Pi;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> <mrow> <mo>(</mo> <mi>j</mi> <mo>&amp;NotEqual;</mo> <mi>&amp;xi;</mi> <mo>,</mo> <mi>&amp;eta;</mi> <mo>,</mo> <mo>...</mo> <mo>,</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> </mrow> <mi>N</mi> </munderover> <mo>(</mo> <mrow> <mn>1</mn> <mo>-</mo> <msub> <mi>P</mi> <msub> <mi>d</mi> <mrow> <mi>j</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>j</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mo>)</mo> <mo>)</mo> </mrow> </mrow> <mrow> <msubsup> <mi>Q</mi> <mi>H</mi> <mi>v</mi> </msubsup> <mo>=</mo> <munder> <mo>&amp;Sigma;</mo> <mi>q</mi> </munder> <mrow> <mo>(</mo> <msubsup> <mi>Q</mi> <mrow> <mi>h</mi> <mi>a</mi> <mi>r</mi> <mi>d</mi> </mrow> <mi>v</mi> </msubsup> <mo>&amp;CenterDot;</mo> <munderover> <mo>&amp;Pi;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> <mrow> <mo>(</mo> <mi>j</mi> <mo>&amp;NotEqual;</mo> <mi>&amp;xi;</mi> <mo>,</mo> <mi>&amp;eta;</mi> <mo>,</mo> <mo>...</mo> <mo>,</mo> <mi>&amp;omega;</mi> <mo>)</mo> </mrow> </mrow> <mi>N</mi> </munderover> <mo>(</mo> <mrow> <mn>1</mn> <mo>-</mo> <msub> <mi>P</mi> <msub> <mi>d</mi> <mrow> <mi>j</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>j</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> <mo>)</mo> <mo>)</mo> </mrow> </mrow> 并且f(y1,n,…,yN,n|di,n-1i,n)表示y1,n,y2,n,…,yN,n的联合条件概率密度函数:And f(y 1,n ,…,y N,n |d i,n-1i,n ) represents the joint conditional probability density function of y 1,n ,y 2,n ,…,y N,n : <mrow> <mtable> <mtr> <mtd> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mn>...</mn> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>N</mi> </msup> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>20</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mn>...</mn> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>N</mi> </msup> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>20</mn> <mo>)</mo> </mrow> </mrow> q表示全部N个认知用户中不少于k个用户认定主用户信号存在这个事件发生的场景总数;q represents the total number of scenarios in which no less than k users among all N cognitive users believe that the main user signal exists; 将检测概率漏检概率和虚警概率作为多节点硬判决协作频谱检测系统的检测性能参量,完成认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法。will detect probability Probability of missed detection and false alarm probability As a detection performance parameter of a multi-node hard-decision cooperative spectrum detection system, a method for obtaining the performance parameter of the spectrum detection system after experiencing any mobile period in a cognitive radio network is completed. 3.认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法,其特征是:3. A method for obtaining performance parameters of a spectrum detection system after experiencing any mobile period in a cognitive radio network, characterized in that: 该方法应用于多节点软判决协作频谱检测系统,The method is applied to a multi-node soft-decision cooperative spectrum detection system, 假设移动场景下,认知用户服从随机游走模型,认知网络中存在N个移动的认知用户以及一个融合中心FC;N为正整数;Assume that in a mobile scenario, cognitive users obey the random walk model, and there are N mobile cognitive users and a fusion center FC in the cognitive network; N is a positive integer; 步骤一、采用公式:Step 1. Use the formula: yn=K/(dn)2 (2)y n =K/(d n ) 2 (2) 式中:dn表示主用户和认知用户在进行第n个感知周期时位置的距离;In the formula: d n represents the distance between the primary user and the cognitive user during the nth sensing cycle; 获得认知用户检测到主用户信号的实时功率ynObtain the real-time power y n of the primary user signal detected by the cognitive user; 式中:K是常量,且:Where: K is a constant, and: K=GrPtGtw/4π)2 K=G r P t G tw /4π) 2 式中:Gr表示认知用户的天线增益,Pt表示主用户的发射信号功率,Gt是授权用户的天线增益;λw是主用户信号的波长;In the formula: G r represents the antenna gain of the cognitive user, P t represents the transmitted signal power of the primary user, G t is the antenna gain of the authorized user; λ w is the wavelength of the primary user signal; 步骤二、在每次能量检测中,每个认知用户采集M个点,M为正整数;且在高斯信道中,噪声的均值为0,方差为σ2,预设能量检测门限为λ;Step 2. In each energy detection, each cognitive user collects M points, where M is a positive integer; and in a Gaussian channel, the mean value of the noise is 0, the variance is σ 2 , and the preset energy detection threshold is λ; 则在第n个感知周期后,多节点软判决协作频谱的检测概率为:Then after the nth sensing period, the detection probability of the multi-node soft-decision cooperative spectrum for: <mrow> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> 漏检概率为:Probability of missed detection for: <mrow> <msub> <mi>P</mi> <msub> <mi>m</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>m</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>P</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> 虚警概率为:false alarm probability for: <mrow> <msub> <mi>P</mi> <msub> <mi>f</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>P</mi> <msub> <mi>f</mi> <mi>n</mi> </msub> </msub> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>&amp;lambda;</mi> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>-</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> 式中:In the formula: <mrow> <mi>Q</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mi>x</mi> <mrow> <mo>+</mo> <mi>&amp;infin;</mi> </mrow> </msubsup> <mfrac> <mn>1</mn> <msqrt> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msqrt> </mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mfrac> <msup> <mi>t</mi> <mn>2</mn> </msup> <mn>2</mn> </mfrac> </mrow> </msup> <mi>d</mi> <mi>t</mi> </mrow> <mrow> <mi>Q</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mi>x</mi> <mrow> <mo>+</mo> <mi>&amp;infin;</mi> </mrow> </msubsup> <mfrac> <mn>1</mn> <msqrt> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msqrt> </mfrac> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mfrac> <msup> <mi>t</mi> <mn>2</mn> </msup> <mn>2</mn> </mfrac> </mrow> </msup> <mi>d</mi> <mi>t</mi> </mrow> 步骤三、根据公式:Step 3, according to the formula: <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>=</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>=</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> 获得任一认知用户SUi和第n个移动周期的终点之间的间距di,n;1≤i≤N;vi,n表示在第n个移动周期中,该认知用户SUi的移动速度;θi,n表示在第n个移动周期中,该认知用户SUi的移动方向;Δt两次检测之间的时间间隔,即每个移动周期的持续的时间;Obtain the distance d i,n between any cognitive user SU i and the end point of the nth mobile cycle; 1≤i≤N; v i,n means that in the nth mobile cycle, the cognitive user SU i The moving speed of ; θ i, n represents the moving direction of the cognitive user SU i in the nth moving cycle; Δt is the time interval between two detections, that is, the duration of each moving cycle; 步骤四、根据公式:Step 4, according to the formula: <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mi>min</mi> </msub> </mrow> </mfrac> <mo>|</mo> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>/</mo> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mi>min</mi> </msub> </mrow> </mfrac> <mo>|</mo> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>/</mo> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> </mrow> 获得间距di,n的条件概率密度函数;vmax表示该认知用户SUi的最大移动速度;vmin表示该认知用户SUi的最小移动速度;Obtain the conditional probability density function of the distance d i, n ; v max represents the maximum moving speed of the cognitive user SU i ; v min represents the minimum moving speed of the cognitive user SU i ; 步骤五、根据式(4)推导出:Step 5, deduce according to formula (4): <mrow> <mo>|</mo> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mfrac> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>&amp;CenterDot;</mo> <mrow> <mo>(</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>|</mo> </mrow> <mrow> <mo>|</mo> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>v</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <mo>|</mo> <mo>=</mo> <mo>|</mo> <mfrac> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mrow> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mi>&amp;Delta;</mi> <mi>t</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>&amp;CenterDot;</mo> <mrow> <mo>(</mo> <mi>&amp;Delta;</mi> <mi>t</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>|</mo> </mrow> 式中:In the formula: <mrow> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>=</mo> <mo>-</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <mi>cos</mi> <mn>2</mn> </msup> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mover> <mrow> <mi>P</mi> <mi>Q</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>=</mo> <mo>-</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <msqrt> <mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <mi>cos</mi> <mn>2</mn> </msup> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> </mrow> </msqrt> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> 式中:PQ为第n个移动周期内认知用户移动的距离;In the formula: PQ is the moving distance of the cognitive user in the nth mobile period; 步骤六、用表达式替换式(5)中的则di,n的条件概率密度函数为:Step 6. Use expressions Replacement formula (5) in Then the conditional probability density function of d i,n is: <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>d</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> 步骤七、根据式(2)获得yi,n和di,n之间的关系式:Step 7, according to formula (2), obtain the relationship between y i,n and d i,n : yi,n=K/(di,n)2 (7)y i,n =K/(d i,n ) 2 (7) 则,yi,n的条件概率密度函数为:Then, the conditional probability density function of y i,n is: <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mi>min</mi> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mfrac> <msqrt> <mi>K</mi> </msqrt> <mrow> <mn>2</mn> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msqrt> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </msqrt> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <msub> <mi>v</mi> <mrow> <mi>m</mi> <mi>a</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>v</mi> <mi>min</mi> </msub> </mrow> </mfrac> <msub> <mi>V</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mfrac> <msqrt> <mi>K</mi> </msqrt> <mrow> <mn>2</mn> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msqrt> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </msqrt> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> 式中:通过将中的di,n利用yi,n替换得到;In the formula: by putting d i, n in is replaced by y i,n ; 步骤八、通过公式:Step eight, pass the formula: <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </msub> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msub> <mi>dy</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </msub> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <msub> <mi>dy</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> 式中:表示yi,n的取值范围;In the formula: Indicates the value range of y i, n ; 计算yi,n的条件数学期望;Calculate the conditional mathematical expectation of y i,n ; 步骤九、通过公式:Step nine, through the formula: <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> yi,n的概率密度函数;Probability density function of y i,n ; 其中:是由式(6)、式(8)和下式(11)计算得到的:in: is calculated by formula (6), formula (8) and the following formula (11): <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> 式中:表示di,n-1的取值范围;In the formula: Indicates the value range of d i,n-1 ; yi,n的数学期望为:The mathematical expectation of y i,n is: <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mo>(</mo> <mrow> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> 其中,是由式(6)、式(9)和下式(13)计算得到的:in, is calculated by formula (6), formula (9) and the following formula (13): <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>E</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>E</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>E</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow> 步骤十、则移动场景下多节点软判决协作频谱检测系统频谱感知的检测概率为:Step 10. The detection probability of spectrum sensing in the multi-node soft-decision cooperative spectrum detection system in the mobile scene for: <mrow> <msubsup> <mi>Q</mi> <mrow> <mi>d</mi> <mo>-</mo> <msub> <mi>S</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>S</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msubsup> <mi>Q</mi> <mrow> <mi>d</mi> <mo>-</mo> <msub> <mi>S</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </msubsup> <mo>&amp;lsqb;</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>a</mi> </msubsup> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>S</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <mn>2</mn> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> </mrow> <msup> <mi>a</mi> <mn>2</mn> </msup> </mfrac> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>0</mn> </mrow> </msub> <mo>&amp;rsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mn>1</mn> </mrow> </msub> </mrow> 漏检概率为:Probability of missed detection for: <mrow> <msubsup> <mi>Q</mi> <mrow> <mi>m</mi> <mo>-</mo> <msub> <mi>S</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>Q</mi> <mrow> <mi>d</mi> <mo>-</mo> <msub> <mi>S</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> </mrow> <mrow> <msubsup> <mi>Q</mi> <mrow> <mi>m</mi> <mo>-</mo> <msub> <mi>S</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> <mo>=</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>Q</mi> <mrow> <mi>d</mi> <mo>-</mo> <msub> <mi>S</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> </mrow> 虚警概率为:false alarm probability for: <mrow> <msubsup> <mi>Q</mi> <mrow> <mi>f</mi> <mo>-</mo> <msub> <mi>S</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&amp;lambda;</mi> <mi>s</mi> </msub> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>&amp;CenterDot;</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </msubsup> <msub> <mi>w</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <msubsup> <mi>Q</mi> <mrow> <mi>f</mi> <mo>-</mo> <msub> <mi>S</mi> <mi>n</mi> </msub> </mrow> <mi>v</mi> </msubsup> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&amp;lambda;</mi> <mi>s</mi> </msub> <mo>-</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>&amp;CenterDot;</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </msubsup> <msub> <mi>w</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <msqrt> <mrow> <mn>2</mn> <msup> <mi>&amp;sigma;</mi> <mn>4</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> </mrow> 其中:是由式(6)、式(22)和式(23)计算得到的:in: is calculated by formula (6), formula (22) and formula (23): <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>S</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>S</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>22</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>S</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </munderover> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> </munder> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>S</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>&amp;CenterDot;</mo> <msub> <mi>f</mi> <msub> <mi>d</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </msub> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <msub> <mi>dd</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>d&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>22</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>S</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <mn>...</mn> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <msubsup> <mi>Q</mi> <mi>S</mi> <mi>v</mi> </msubsup> <mi>f</mi> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>dy</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mn>...</mn> <msub> <mi>dy</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>23</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>Pr</mi> <mrow> <mo>(</mo> <msubsup> <mi>Pd</mi> <msub> <mi>S</mi> <mi>n</mi> </msub> <mi>v</mi> </msubsup> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <mn>...</mn> <munder> <mo>&amp;Integral;</mo> <msub> <mi>&amp;Phi;</mi> <msub> <mi>y</mi> <mi>n</mi> </msub> </msub> </munder> <msubsup> <mi>Q</mi> <mi>S</mi> <mi>v</mi> </msubsup> <mi>f</mi> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>,</mo> <mn>...</mn> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>|</mo> <msub> <mi>d</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mi>&amp;theta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>dy</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mn>...</mn> <msub> <mi>dy</mi> <mrow> <mi>N</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>23</mn> <mo>)</mo> </mrow> </mrow> 其中:in: <mrow> <msubsup> <mi>Q</mi> <mi>S</mi> <mi>v</mi> </msubsup> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&amp;lambda;</mi> <mi>s</mi> </msub> <mo>-</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </msubsup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>w</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <msqrt> <mrow> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </msubsup> <mn>2</mn> <msubsup> <mi>w</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> <mn>2</mn> </msubsup> <msup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <msubsup> <mi>Q</mi> <mi>S</mi> <mi>v</mi> </msubsup> <mo>=</mo> <mi>Q</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&amp;lambda;</mi> <mi>s</mi> </msub> <mo>-</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </msubsup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <msub> <mi>w</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> </mrow> <msqrt> <mrow> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </msubsup> <mn>2</mn> <msubsup> <mi>w</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> <mn>2</mn> </msubsup> <msup> <mrow> <mo>(</mo> <msup> <mi>&amp;sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>/</mo> <mi>M</mi> </mrow> </msqrt> </mfrac> <mo>)</mo> </mrow> </mrow> λs表示的是融合中心处的预设门限,wi,n代表认知用户SUi在第n个感知周期中的加权因子;λ s represents the preset threshold at the fusion center, w i,n represents the weighting factor of cognitive user SU i in the nth perception cycle; 将检测概率漏检概率和虚警概率作为多节点软判决协作频谱检测系统的检测性能参量,完成认知无线电网络中经历任意移动周期后频谱检测系统性能参量获得方法。will detect probability Probability of missed detection and false alarm probability As a detection performance parameter of a multi-node soft-decision cooperative spectrum detection system, a method for obtaining the performance parameter of the spectrum detection system after experiencing any mobile period in a cognitive radio network is completed.
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CN101931963A (en) * 2009-06-18 2010-12-29 新邮通信设备有限公司 Multi-user ranging detection and collision processing method and device
EP2456266A1 (en) * 2009-07-17 2012-05-23 National Institute of Information and Communication Technology Power control method in cognitive radio communication, cognitive radio communication system, and radio communication device
CN104796898A (en) * 2015-03-16 2015-07-22 哈尔滨工业大学 Time-based estimation method for spectrum sensing system user capacity in mobile scene

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CN101931963A (en) * 2009-06-18 2010-12-29 新邮通信设备有限公司 Multi-user ranging detection and collision processing method and device
EP2456266A1 (en) * 2009-07-17 2012-05-23 National Institute of Information and Communication Technology Power control method in cognitive radio communication, cognitive radio communication system, and radio communication device
CN104796898A (en) * 2015-03-16 2015-07-22 哈尔滨工业大学 Time-based estimation method for spectrum sensing system user capacity in mobile scene

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