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Simulations

于 2021-04-19 发布

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下载积分: 1 下载次数: 168

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代码说明：

说明：对NOMA下行链路的用户和信道功率分配的优化(Energy-Efficient Resource Allocation for Downlink Non-Orthogonal Multiple Access (NOMA) Network)

文件列表：

Simulations_2, 0 , 2018-11-22
Simulations_2\Capacity_power_Fig4.m, 14455 , 2018-03-08
Simulations_2\DC_Power_EnEfMax.m, 1429 , 2015-09-12
Simulations_2\DC_Power_SumRateMax.m, 1402 , 2017-10-11
Simulations_2\Energy Efficiency-Number of users per BS.fig, 18663 , 2018-05-23
Simulations_2\Energy efficient.fig, 19667 , 2018-03-08
Simulations_2\Energy-Efficient Resource Allocation for Downlink.pdf, 210211 , 2017-10-10
Simulations_2\EnergyEfficiency_Powersallcocation_Fig7.m, 22255 , 2018-03-08
Simulations_2\EnergyEfficiency_power_Fig5.m, 14671 , 2018-03-08
Simulations_2\EnergyEfficiencyfunc.m, 1275 , 2015-07-09
Simulations_2\EnergyEffiency_Pc_Fig6.m, 14658 , 2018-03-08
Simulations_2\NOMA_Capacity_Fig2.m, 16147 , 2018-11-23
Simulations_2\Noma_EnergyEfficiency_Fig3.m, 15310 , 2018-03-08
Simulations_2\SumRate_DC.m, 741 , 2015-07-08
Simulations_2\SumRate_PartRatio.m, 471 , 2015-06-22
Simulations_2\Sum_Rate_Caculation.m, 926 , 2017-10-11
Simulations_2\Sum_Rate_FTPA.m, 367 , 2015-08-31
Simulations_2\compute_distance.m, 160 , 2015-06-17
Simulations_2\compute_pathloss_BS_UE_outdoor.m, 325 , 2017-10-10
Simulations_2\myfunc.m, 330 , 2015-06-17
Simulations_2\qfunc.m, 99 , 2015-06-22

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说明： 2. 关于无线信道的多普勒谱，请回答以下问题： (1) 在某些无线信道中，多普勒频移会引起无线信道输出信号的混叠。请描述产生混叠的各种原因； (2) 无线信道中的多普勒谱有一种经典谱(classic spectrum)，请解释产生这种谱形状的机理； (3) 请用Simulink工具编写代码(S-Function方式)，产生一单径瑞利信道，其多普勒谱为经典谱，其中移动速率为120km/h。并请描述：自己用计算机产生的经典谱是正确的依据。(2. Regarding the doppler spectrum of the wireless channel, please answer the following questions: (1) in some wireless channels, doppler frequency shift will cause aliasing of the output signals of the wireless channel.Describe the various causes of aliasing; (2) there is a classic spectrum in the doppler spectrum in the wireless channel, please explain the mechanism of producing this spectrum shape; (3) please use Simulink tool to write code (s-function mode) to generate a single-path Rayleigh channel, whose doppler spectrum is the classical spectrum, in which the moving rate is 120km/h.And please describe: your own computer generated classical spectrum is the correct basis.)

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