Capacity problem formulation BPSK在AWGN中的性能 a Given a power constraint of the input signal ■AWGN信道模型(仅考虑实部信号) Imk [=x[+m[m]-(0a2) 当采用BPSK调制时,x[咧=±,误码率为 a What is the maximum achievable rate under the reliable communication requirement Pe =gNp/o R=log I bitslchannel use ■当采用重复N次传输时, a Whatever a error probability Parget is required ■误码率减小 there always exists a code such that P=Q(NP/o) p=(i+i <paget 口但bt传输速率也相应减小 填充球( Packing Sphere Sphere packing ■从几何上理解,重复编码其实是将码字排列 a For a given transmitted codeword x,, the 在1个方向(维度)上 distance between the received signal y and x, is ■如果为增加速率,采用MASK调制,结合重 ly-xl+ m 复编码,各星座点等间距排列在1个方向 a With the law of large numbers [] =O" P+5 Do ■ y is within the sphere 此时bt速率为 centered at x with radius√Na2 bit/channel use with probability one No2 when m→∞ 後照大手 後照k季D◼ What is the maximum achievable rate under the reliable communication requirement   2 Capacity problem formulation ◼ Given a power constraint of the input signal N 1 N m =1 x m  P R = log2 bits/channel use N ◼ Whatever a error probability ptarget is required, there always exists a code such that pe = i ˆ  iptarget 21 BPSK在AWGN中的性能 ◼ AWGN信道模型 (仅考虑实部信号) ◼ 当采用重复 N 次传输时, 但bit传输速率也相应减小 y m= xm+ wm, wm ~ (0, 2 ) ◼ 当采用BPSK调制时, xm =  P , 误码率为 p e = Q ( P  2 ) x A = ◼ 误码率减小 P 1,,1, xB = P−1,,−1 p e = Q ( NP  2 ) 22 填充球 (Packing Spheres) ◼ 从几何上理解,重复编码其实是将码字排列 在1个方向 (维度) 上. ◼ 如果为增加速率,采用M-ASK调制, 结合重 复编码, 各星座点等间距排列在1个方向 ◼ 此时bit速率为 e NP  p  Q   (M − 1)    log M bit/channel use N ◼ With the law of large numbers ◼ is within the sphere i x N 2 0 lim 1 N N → N m =1 w 2 m= 2  w 2 m N  m =1 y − x = w = Sphere packing ◼ For a given transmitted codeword xi , the distance between the received signal y and xi is centered at xi , with radius N 2 with probability one when N →  y 23 24