光纤通信基础_第四章 光无源元件（波分复用元件）

Optical fiber communication 第四节浪分复用元件 Absorption in Optical Fibers 1280mn1610nn Utilizing the Fiber Bandwidth

1-1 Copyright Wang Yan 2021/2/19 Optical fiber communications 第四节 波分复用元件 一、Absorption in Optical Fibers 二、Utilizing the Fiber Bandwidth

Optical fiber communication Utilizing the Fiber Bandwidth TDM:在时域对信息进行快速转换。信道: channels CH CH. 1 CH 2 CH 2 CHN CH N WDM:在频将不同波长的信道结合起来信道间间隔越大越容易设 计建设。 CH CH. 1 CH 2 入1, CH 2 CH N CH N

1-2 Copyright Wang Yan 2021/2/19 Optical fiber communications 二、Utilizing the Fiber Bandwidth TDM:在时域对信息进行快速转换。信道：channels。 WDM:在频域将不同波长的信道结合起来信道间间隔越大越容易设 计建设

Optical fiber communication 021/2/19 1、在实际光纤通信中色散和非线性效应限制光纤的带宽的应 用 a)SMF有材料色散和波导色散 bMMF有横式色散 c)有色散位移光纤和色散平坦光纤,非线性传播应 d)在高容量的光纤中需要对色散进行控制 2、WDM可以比TDM更充分地利用光纤带宽。 a)WDM对非线性效应(FWM&XPM)更敏感(TDM)但是可以 以应付 bWDM并设有充分地里利用带宽。技术上要求在不同的信道间要 加保护带( guard lands)

1-3 Copyright Wang Yan 2021/2/19 Optical fiber communications 二、1、在实际光纤通信中色散和非线性效应限制光纤的带宽的应 用。 a)SMF有材料色散和波导色散 b)MMF有横式色散 c)有色散位移光纤和色散平坦光纤，非线性传播应 d)在高容量的光纤中需要对色散进行控制 2、WDM可以比TDM更充分地利用光纤带宽。 a)WDM对非线性效应（FWM&XPM）更敏感（TDM）但是可以 以应付 b)WDM并设有充分地里利用带宽。技术上要求在不同的信道间要 加保护带（guard lands）

Optical fib communication 42021/2/19 EDFA wigan equalizer Modulator a中囚 Wavelength Division multiplexer 半导体 Isolator 激光二极管 Receiver dd/drop multiplexer Dispersion compensator PMD compensator WDM demultiplexer 0XC光交叉连线

1 - 4 Copyright Wang Yan 2021/2/19 Optical fiber communications

Optical fiber communication Optical filters, multiplexers and switches should have a)Low insertion loss b)Low polarization dependence (waveguide devices have high PDL) C)Low temperature dependence of center frequency and filter shape dFlat pass band e)Good cross talk fLow cost

1-5 Copyright Wang Yan 2021/2/19 Optical fiber communications a)Low insertion loss b)Low polarization dependence (waveguide devices have high PDL) c)Low temperature dependence of center frequency and filter shape d)Flat pass band e)Good cross talk f)Low cost Optical filters, multiplexers, and switches should have:

Optical fiber communication -62021/2/19 F-P( Fabry-Pery)腔型光滤波器:解复用 A:工作原理多光束干涉有频率选择物性 At,= Anrt At,=Arri t2 At= anr t At,= Att 当r1r2反射率较高时必须考虑每光束干涉

1-6 Copyright Wang Yan 2021/2/19 Optical fiber communications 一、F-P(Fabry-Pery)腔型光滤波器：解复用 A:工作原理多光束干涉有频率选择物性 2 3 1 3 4 2 At = Ar r t 2 2 1 2 3 2 At = Ar r t 2 2 1 2 At = Ar r t 1 2 1 At = At t − A = A 1 t 1 r 2 t 2 r 当r1 r2反射率较高时必须考虑每光束干涉

Optical fiber communication 72021/2/19 每一束透射波比前一束在相径上延迟Φ=k2=4Tn/A0=4Tnf/c 而振幅减小须乘以因子r1r2。令 则:b=exP(=/ At1=A12exp(一j中) At2=At t, exp(- rri exp(一少小)t2=A1h At= At,h

1-7 Copyright Wang Yan 2021/2/19 Optical fiber communications 每一束透射波比前一束在相径上延迟Φ＝k2l=4πnl/λ0=4πnlf/c 而振幅减小须乘以因子r1 r2 。令 exp( ) h = r1 r2 − j 则： exp( ) At1 = A1 t 1 t 2 − j At2 = A1 t 1 t 2 exp(− j)r2 r1 exp(− j)t 2 = At1 h =  2 At3 At1 h

Optical fiber communication 82021/2/19 透射光的复振幅为歌词透射光的叠加 At= At,+ At +at+ t t At1(1+h+h2+…)= 4(t exp(jo) 1-h1-h 着:r1=r2=rt1=t2=t(振幅 功率:R=r2,T=t2无损耗情况下R+T=1 Tt= A T exp(一j) 1-Re apCo) exp(jD)=1

1-8 Copyright Wang Yan 2021/2/19 Optical fiber communications 透射光的复振幅为歌词透射光的叠加 ( j) h At t h A t At h h At At At At − − = − = + + + = = + + + exp 1 1 ( ) (1 ) 2 1 2 1 1 2 3 若：r1=r2=r t1=t2=t (振幅) 功率：R=r2 , T=t2 无损耗情况下 R＋T＝1 1 Re ( ) exp( )   x p j AT j Tt i − − − = exp(− j) =1

Optical fiber communication 92021/2/19 输出光强: l= KAtAt Ka.AT 1-Re xp(jo) (1-R)2+4Rsm2如 2 K为常数:L=KA(入射光波的光强 B)FP腔的功率传输系数( Fabry Perot Transfer Fuction) 1+[4R/(1-R)/sm2p 1+(2F/x)2sin2(2m/c)

1-9 Copyright Wang Yan 2021/2/19 Optical fiber communications 输出光强： ( ) 2 Re (1 ) 4 sin 1 2 2 2 2   R R I T x p j KA A T I KAtAt i i i t − + = − − = =   K为常数： 为入射光波的光强 B )F—P腔的功率传输系数(Fabry Perot Transfer Fuction)  I = KAtAt i ( F ) ( nlf c) R R I I i t 1 2 / sin 2 / 1 2 1 [4 /(1 ) ]sin 1 2 2 2 2     + = + − = =

Optical fiber communication 02021/2/19 Where F=z√R/(-R)f cg an Ca 谐振频率:m-=2123,… 谐振波长: 2nl maX 传输系数的最大值: m1+(2F/r)21+[4R/1-R)

1-10 Copyright Wang Yan 2021/2/19 Optical fiber communications Where F =  R /(1− R) nl cq f 2 =  f 谐振频率： q=1,2,3,…… 谐振波长： 传输系数的最大值： nl cq f oq 2 = q nl oq 2  =  max =1 1 [4 /(1 ) ] 1 1 (2 / ) 1 min 2 2 F + R − R = + =  