数字通信第5章 加性高斯白噪声的最佳接收机

Thus, -r1 <nm<r1, correct
1 P ( nm < r1 r1 > 0 ) = πN 0 1 = 2π
x ∫− r1 exp − N 0 dx
r1 2 2
∫
r1
Βιβλιοθήκη Baidu
N0 2 N0
− r1
x exp − dx 2 2
M 2 −1
Use Gray coding,
1 Pb ≈ PM k
4PSK and 4QAM are comparable When M>4, M-ary QAM is better than Mary PSK
The pdf of first correlation is
r− ε 1 s − 1 p (r1 | s1 ) = exp N0 π N0
数字调制方式的比较
带宽和频带利用率（归一化数据速率R/W） 给定的错误概率情况下，R/W和比特SNR 的关系曲线
带宽受限区域 功率受限区域 信道容量限
5.5 有线和无线通信系统 性能分析
本节内容
再生中继器 无线通信中链路预算分析
2 PM PM Pb = k ≈ 2 −1 2
k −1
Orthogonal signals
M-ary biorthogonal signals Constructed from M/2 orthogonal signals Demodulation: Use M/2 orthogonal cross correlation or matched filters Tx: s1(t)
Differential PSK •The received signal is compared with phase of the preceding signal •no carrier information needed
rk = ε s exp[ j (θ k − φ )] + nk
rk −1 = ε s exp[ j (θ k −1 − φ )] + nk −1
r1 = ε s + n1 r2 = n2 ... rM 2 = nM
2
Cross correlation
C (r, sm ) = r ⋅ sm = ∑ rk smk , m = 1,2,... M / 2
k =1
M 2
The sign used to determine s(t) or -s(t) was transmitted The correct decision for s1(t) •r1 > 0 •r1 > rm=nm
εb 1 Pb = exp − N 2 0
•Binary PSK and DPSK-> less than 3dB differences •Pb get smaller, differences becomes small
•BPSK & 4PSK have the same BER •4 phase DPSK worse than 2 phase DPSK
* k −1
)+ n n
k
* k −1
* x = ε s + Re(nk + nk −1 )
y = Im(nk + n
* k −1
)
x, y are the uncorrelated Gaussian random variable , N0
Θ r = tan
−1
y x
•Similar to PSK •noise is twice larger •DPSK has 3 dB poorer performance
−∞
x exp − dx 2
2
M −1
p (r1 )dr1
Symbol error rate
PM = 1 − Pc
For the bit error rate
PM PM = k M −1 2 −1
k −1 k PM 2 ∑ n 2k − 1 = k 2k − 1 PM n =1 k
The rest of correlation
2 rm 1 p (rm | s1 ) = exp − π N0 N0
(
)
2

The probability of correct decision is
Pc = ∫ P(n2 < r1 , n3 < r1 ,...nM < r1 s1 ) p(r1 | s1 )dr1
Error probability estimation of DPSK Assumption: θk-θk-1=0 θ
exp(− j (θ k − φ ))
-> compared with nk-1 and nk
* k −1
nk n
* k −1
<< ε s (nk + n
)
rr
* k k −1
= ε s + ε s (nk + n
rr
* k k −1
= ε s exp[ j (θ k − θ k −1 )] + ε s exp[ j (θ k − φ )]n
* k −1
* k −1
+ ε s exp[− j (θ k −1 − φ )]nk + nk n
No noise, -> θk-θk-1 No noise, -> rkr*k is independent of carrier phase
The probability of correct decision is
Pc = ∫
∞
0
1 2π
∫
r1
N0 2 N0
− r1
x exp − dx 2 2
2
p (r1 )dr1
PM = 1 − Pc
Biorthogonal signals
∞ −∞
For each term
P(nm < r1 s1 ) = ∫ p(rm | sm )dxm
r1 −∞
1 = 2π
∫
r1 2 N 0
−∞
x2 exp − dx 2
The correct decision
1 Pc = ∫ −∞ 2π
∞
∫
r1 2 N 0
Binary signals
Exercises 5-4, 5-8,
•M=2, equals to the antipodal signals •k increase by one, 4dB SNR is required •Use Gray coding,
1 Pb ≈ PM k
第5章 加性高斯白噪声信道的 章 最佳接收机
５.1 受加性高斯白噪声恶化信号的 最佳接收机
相关解调器
结 论
最佳AWGN接收机的实现形式 接收机的实现形式 最佳
5.2 无记忆调制的最佳接收机 性能分析
二进制调制的错误概率
1、2PAM 2、正交信号