参考书:现代通信原理(第二版) 沈保锁 侯春萍 等编著
注意:有的公式并不是所有的情况都满足,这里只给出书上还有我自己总结的公式,用之前要想一想能否适用
符号 | 含义 |
---|---|
τ \tau τ | 脉冲宽度 |
T s T_s Ts | 码元宽度 |
f 0 f_0 f0 | 中心频率(载频) |
f m f_m fm | 最高频率 |
f L f_L fL | 最低频率 |
f H f_H fH | 最高频率 |
f s f_s fs | 抽样频率 |
f c f_c fc | 载频 |
e j x = c o s ( x ) + j s i n ( x ) e^{jx}=cos(x)+jsin(x) ejx=cos(x)+jsin(x)
S N ( d B ) = 10 l o g 10 S N ( W ) \frac{S}{N} (dB)=10log_{10}\frac{S}{N} (W) NS(dB)=10log10NS(W)
H ( x ) = P ( x 1 ) l o g 2 1 P ( x 1 ) + P ( x 2 ) l o g 2 1 P ( x 2 ) + . . . + P ( x n ) l o g 2 1 P ( x n ) = − ∑ n i = 1 P ( x i ) l o g 2 1 P ( x i ) (比特 / 符号) H(x)=P(x_1)log_{2}\frac{1}{P(x_1)} +P(x_2)log_{2}\frac{1}{P(x_2)}+...+P(x_n)log_{2}\frac{1}{P(x_n)}\\=-\sum_{n}^{i=1}P(x_i)log_{2}\frac{1}{P(x_i)} (比特/符号) H(x)=P(x1)log2P(x1)1+P(x2)log2P(x2)1+...+P(xn)log2P(xn)1=−n∑i=1P(xi)log2P(xi)1(比特/符号)
R b = R B l o g 2 N = 1 T s ( b / s ) R_b=R_Blog_{2}N=\frac{1}{T_s} (b/s) Rb=RBlog2N=Ts1(b/s)
N为进制数
R B = R b l o g 2 N ( B , 波特) R_B=\frac{R_b}{log_2N}(B,波特) RB=log2NRb(B,波特)
Ts为码元持续时间
P c = 接收的错误码元数 传输总码元数 P_c=\frac{接收的错误码元数}{传输总码元数} Pc=传输总码元数接收的错误码元数
P b = 接收的错误码元比特数 传输总比特数 P_b=\frac{接收的错误码元比特数}{传输总比特数} Pb=传输总比特数接收的错误码元比特数
f = 4 k ( H Z ) f=4k(HZ) f=4k(HZ)
S Y ( ω ) = S X ( ω ) ∣ H ( ω ) ∣ 2 S_Y(\omega )=S_X(\omega )\left | H(\omega ) \right | ^2 SY(ω)=SX(ω)∣H(ω)∣2
C = B l o g 2 ( 1 + S N ) ( b / s ) C=Blog_2(1+\frac{S}{N} )~~~~(b/s) C=Blog2(1+NS) (b/s)
S A M ( t ) = [ A 0 + f ( t ) ] c o s ( ω 0 + θ 0 ) S_{AM}(t)=[A_0+f(t)]cos(\omega _0+\theta_0) SAM(t)=[A0+f(t)]cos(ω0+θ0)
β A M = A m A 0 = 信号幅度 载波幅度 \beta _{AM}=\frac{A_m}{A_0}=\frac{信号幅度}{载波幅度} βAM=A0Am=载波幅度信号幅度
( S i ) A M = 1 2 [ A 0 2 + f 2 ( t ) ‾ ] (S_i)_{AM}=\frac{1}{2}[A^2_0+\overline{f^2(t) } ] (Si)AM=21[A02+f2(t)]
中括号内是载波功率+边带功率
( N i ) A M = n 0 2 × B × 2 = n 0 B = 2 n 0 f m (N_i)_{AM}=\frac{n_0}{2}\times B \times 2=n_0B=2n_0f_m (Ni)AM=2n0×B×2=n0B=2n0fm
B = 2 f m B=2f_m B=2fm
B ≥ N n f m B\ge Nnf_m B≥Nnfm
N:传输信道数
n:n位二进制码
fm:信号最高频率
在理想传输情况下,上式可以写为
B = N n f m B= Nnf_m B=Nnfm
B ≥ 2 N n f m B\ge 2Nnf_m B≥2Nnfm
注意:二进制的情况下Rb才=RB, 此处容易和第八章搞混
R b = R B = n N f s R_b=R_B=nNf_s Rb=RB=nNfs
n:编码位数
N:复用信道数
fs:抽样速率
B N = 1 2 π ⋅ π T s = 1 2 T s = f s 2 = R B 2 B_N=\frac{1}{2\pi } \cdot \frac{\pi }{T_s}=\frac{1}{2T_s}=\frac{f_s}{2}=\frac{R_B}{2} BN=2π1⋅Tsπ=2Ts1=2fs=2RB
最小的基带带宽就是BN,BN是B基的一种
R B = 2 B N = 1 T s = f s R_B=2B_N=\frac{1}{T_s}=f_s RB=2BN=Ts1=fs
η = R B B N = 1 1 + α \eta =\frac{R_B}{B_N} =\frac{1}{1+\alpha} η=BNRB=1+α1
B = ( 1 + α ) B N = ( 1 + α ) R B 2 B=(1+\alpha )B_N=(1+\alpha )\frac{R_B}{2} B=(1+α)BN=(1+α)2RB
若无码间串扰,则a=0
e r f ( x ) = 2 π ∫ 0 x e − u 2 d u erf(x)=\frac{2}{\sqrt[]{\pi } } \int_{0}^{x}e^{-u^2} du erf(x)=π2∫0xe−u2du
e r f c ( x ) = 1 − e r f ( x ) erfc(x)=1-erf(x) erfc(x)=1−erf(x)
B 频 = 2 B 基 = 2 R B B_频=2B_基=2R_B B频=2B基=2RB
ASK,FSK,PSK的带宽都满足上公式
此公式在第六章是正确的,注意不要和第五章的BN搞混
f s = R B = 1 T s f_s=R_B=\frac{1}{Ts} fs=RB=Ts1
在二进制的情况下,fs与RB在频带传输中数值是相等的
S A S K ( t ) = { A c o s ω 0 t “ 1 ” 0 “ 0 ” S_{ASK}(t)=\left\{\begin{matrix} Acos\omega_0t ~~~“1” & \\ 0 ~~~~~~~~~~~~~~~“0” & \end{matrix}\right. SASK(t)={Acosω0t “1”0 “0”
B 2 A S K = 2 f s = 2 T s B_{2ASK}=2f_{s}=\frac{2}{T_{s}} B2ASK=2fs=Ts2
B 2 A S K = 2 B = ( α + 1 ) f s = α + 1 T s B_{2ASK}=2B=(\alpha +1)f_{s}=\frac{\alpha +1}{T_{s}} B2ASK=2B=(α+1)fs=Tsα+1
B为基带信号带宽
S F S K ( t ) = { A c o s ω 1 t “ 1 ” A c o s ω 2 t “ 0 ” S_{FSK}(t)=\left\{\begin{matrix} Acos\omega_1t ~~~“1” & \\ Acos\omega_2t ~~~“0” & \end{matrix}\right. SFSK(t)={Acosω1t “1”Acosω2t “0”
B F S K = ∣ f 2 − f 1 ∣ + 2 f s B_{FSK}=\left |f_{2}-f_{1} \right | +2f_{s} BFSK=∣f2−f1∣+2fs
B F S K = 4 f s B_{FSK}=4f_{s} BFSK=4fs
B F S K = 3 f s B_{FSK}=3f_{s} BFSK=3fs
B F S K = ∣ f 2 − f 1 ∣ + ( α + 1 ) f s B_{FSK}=\left |f_{2}-f_{1} \right | +(\alpha +1)f_{s} BFSK=∣f2−f1∣+(α+1)fs
S P S K ( t ) = { A c o s ω 0 t “ 1 ” A c o s ( ω 0 t + π ) “ 0 ” S_{PSK}(t)=\left\{\begin{matrix} Acos\omega_0t ~~~~~~~~~~“1” & \\ Acos(\omega_0t+\pi ) “0” & \end{matrix}\right. SPSK(t)={Acosω0t “1”Acos(ω0t+π)“0”
B P S K = 2 f s = 2 T s B_{PSK}=2f_{s}=\frac{2}{T_{s}} BPSK=2fs=Ts2
η = R B B 频 \eta =\frac{R_B}{B_频} η=B频RB
r o m a x = 2 E n 0 r_{omax}=\frac{2E}{n_0} romax=n02E
n0是单边带噪声功率谱密度
H ( ω ) = K ⋅ S ∗ ( ω ) e − j ω t 0 H(\omega )=K \cdot S^{\ast }(\omega )e^{-j\omega t_{0}} H(ω)=K⋅S∗(ω)e−jωt0
E 1 = A 2 T 2 = E E_1=\frac{A^2T}{2}=E E1=2A2T=E
E 1 = E 2 = A 2 T s 2 E_1=E_2=\frac{A^2T_s}{2} E1=E2=2A2Ts
S N = E b R b n 0 B \frac{S}{N} =\frac{E_bR_b}{n_0B} NS=n0BEbRb
B M P S K = 2 f s = 2 T s B_{MPSK}=2f_{s}=\frac{2}{T_{s}} BMPSK=2fs=Ts2
此处ASK的最佳接受误码率和FSK相关的误码率错误,我拿笔改正了
以16QAM为例
B = 2 R B = 2 R b l o g 2 16 B=2R_B=2\frac{R_b}{log_216} B=2RB=2log216Rb
S M S K ( t ) = c o s ( ω 0 + π a k 2 T s + φ k ) k T s ≤ t ≤ ( k + 1 ) T s S_{MSK}(t)=cos(\omega_{0}+\frac{\pi a_k}{2T_s}+\varphi_{k} )~~~~kT_s\le t\le(k+1)T_s SMSK(t)=cos(ω0+2Tsπak+φk) kTs≤t≤(k+1)Ts
f 2 = 1 2 π ( ω 0 + π 2 T s ) = f 0 + 1 4 T s = f 0 + 1 4 R B f_2=\frac{1}{2\pi }(\omega _0+\frac{\pi }{2T_s} ) =f_0+\frac{1}{4T_s}= f_0+\frac{1}{4}R_B f2=2π1(ω0+2Tsπ)=f0+4Ts1=f0+41RB
f 1 = 1 2 π ( ω 0 − π 2 T s ) = f 0 − 1 4 T s = f 0 − 1 4 R B f_1=\frac{1}{2\pi }(\omega _0-\frac{\pi }{2T_s} ) =f_0-\frac{1}{4T_s}= f_0-\frac{1}{4}R_B f1=2π1(ω0−2Tsπ)=f0−4Ts1=f0−41RB
Δ f = f 2 − f 1 = 1 2 T s = f s 2 \Delta f=f_2-f_1=\frac{1}{2T_s}=\frac{f_s}{2} Δf=f2−f1=2Ts1=2fs
β = Δ f T s = Δ f f s = 1 2 T s T s = 1 2 \beta =\Delta fT_s=\frac{\Delta f}{f_s}=\frac{1}{2T_s}T_s=\frac{1}{2} β=ΔfTs=fsΔf=2Ts1Ts=21
φ k = φ k − 1 + ( a k − 1 − a k ) π k 2 \varphi _k=\varphi _{k-1}+(a_{k-1}-a_k)\frac{\pi k}{2} φk=φk−1+(ak−1−ak)2πk
B M S K = 1.5 f s B_{MSK}=1.5f_s BMSK=1.5fs
R b = 2048 k b / s R_b=2048kb/s Rb=2048kb/s
R b = 1544 k b / s R_b=1544kb/s Rb=1544kb/s