這是一個求時間序列時延(delay time)的matlab源代碼。
上傳時間： 2016-10-13
上傳用戶：TRIFCT
Space-Time Codes and MIMO Systems，figure 2.6的程序，關于遍歷容量和終端容量
標簽： Space-Time Systems Codes MIMO
上傳時間： 2013-12-21
上傳用戶：sunjet
Space-Time Codes and MIMO Systems，figure 2.8 的程序，信道信息已知和未知條件下的遍歷容量
標簽： Space-Time Systems Codes MIMO
上傳時間： 2016-10-13
上傳用戶：lnnn30
Space-Time Codes and MIMO Systems，figure 2.9 的程序，信道信息已知和未知條件下的中斷容量
標簽： Space-Time Systems Codes MIMO
上傳時間： 2014-01-03
上傳用戶：13215175592
Space-Time Codes and MIMO Systems，figure 2.10 的程序，分集條件下的遍歷容量
標簽： Space-Time Systems Codes MIMO
上傳時間： 2013-12-15
上傳用戶：hasan2015
外國人開發的電磁時域有限差分方法工具包 Electromagnetic Finite-Difference Time-Domain (EmFDTD) is a basic two-dimensional FDTD code developed at the School of Electrical Engineering, Sharif University of Technology. This code has been written based on the standard Yee s FDTD algorithm. Applications include propagation, scattering, and diffraction of electromagnetic waves in homogeneous and non-homogeneous isotropic media for in-plane propagating waves. Negative permittivites or permeabilities as well as dispersion is not included. Zero, Periodic, and Perfectly Matched Layer boundary conditions may be selectively applied to the solution domain. The program is best suited for study of propagation and diffraction of electromagnetic waves in Photonic Crystal structures. EmFDTD is written in MATLAB language and has been tested under MATLAB 5.0 and higher versions.
標簽： Finite-Difference Electromagnetic two-dimensio Time-Domain
上傳時間： 2014-11-24
上傳用戶：watch100
bruce powel douglass的實時嵌入式領域著作real time uml 第二版
上傳時間： 2014-11-10
上傳用戶：tonyshao
a song under dos use masm6.0 to get exe to go music can change any time
上傳時間： 2016-10-22
上傳用戶：dongqiangqiang
simulation of BER of orthogonal space time block code over frequency flat Rayleigh faing channel
標簽： simulation orthogonal frequency Rayleigh
上傳時間： 2013-12-02
上傳用戶：tyler
For solving the following problem: "There is No Free Lunch" Time Limit: 1 Second Memory Limit: 32768 KB One day, CYJJ found an interesting piece of commercial from newspaper: the Cyber-restaurant was offering a kind of "Lunch Special" which was said that one could "buy one get two for free". That is, if you buy one of the dishes on their menu, denoted by di with price pi , you may get the two neighboring dishes di-1 and di+1 for free! If you pick up d1, then you may get d2 and the last one dn for free, and if you choose the last one dn, you may get dn-1 and d1 for free. However, after investigation CYJJ realized that there was no free lunch at all. The price pi of the i-th dish was actually calculated by adding up twice the cost ci of the dish and half of the costs of the two "free" dishes. Now given all the prices on the menu, you are asked to help CYJJ find the cost of each of the dishes.
標簽： Limit following solving problem
上傳時間： 2014-01-12
上傳用戶：362279997
