Optical Communications Analysis of transmission systems 2007-2008 Henrique Salgado hsalgado@fe.up.pt 1 Point-to-point system The project of a point-to-point link involves, in general, many interrelated variables: fiber, optical source and photodetector. Hence the design and analysis of the system may require several iterations before the they completed satisfactorily. System requirements: The desired (or possible) transmission distance The data rate or channel bandwidth The bit-error rate (BER) 2 2
Point-to-point link Components Optical fiber Multimode or monomode Step/graded-index Interrelation with optical source and fiber dispersion 3 3 Point-to-point link LED MM fiber Relative index difference! Higher! means more injected power but higher dispersion LD MM fiber Maximum transmission rate! distance is maximum Less injected power Design of transmitter more complex Fiber splices more critical 4 4
Link analysis Two types of analysis are usually carried out to ensure the required system performance is achieved: Link Power Budget Rise-Time Budget α f (db/km): fiber loss coefficient l c (db): connector insertion loss l sp (db): splice loss 5 5 Link power budget The transmission range of the system is obtained taking into consideration: Power margin between the coupled power at transmitter and minimum required power at the receiver Loss present in the link Loss = 10 log P out P in P s (db): coupled power into the fiber by the optical source P r (db): Sensitivity of the receiver P T (db): Total loss 6 6
Link power budget Example: P T = P s P r = 2l c + α f L + system margin Bit rate 20 Mb/s, BER = 10-9 PIN @ 850 nm, Pr = - 42 dbm LED @ -13 dbm coupled power into fiber P T = 12 + 42 = 29 db 2 conectors : 1 db/conector system margin = 6 db 29 db = 2 db + α f L + 6 db 7 7 Link power budget α f =3.5 db/km L = 6 km power budget plot 8 8
Rise-time budget The dispersion analysis in digital systems is equivalent to assessing the rise time of the link. In the power budget we neglect the dispersion effect, which is the same as consider the bandwidth of the system to be large enough to be able to transmit the required bit rate.! The dispersion reduces the available bandwidth which may limit not only the transmission rate, but also the sensitivity of the receiver and consequently the power budget due to intersymbol interference. t sys = i t 2 i = [ t 2 tx + t 2 rx + t 2 mod + t 2 ] 1/2 mat 9 9 Rise-time budget Empirical criteria NRZ: t sys < < 0.7T b, 0.7/B, T b : bit period B : Bit rate RZ: t sys < 0.35T b, =0.35/B Relation between bandwidth and rise time of Rx assume a low-pass filter of first-order rise-time measured between 10 and 90% the response to a step input, u(t), is g(t) = ( 1 e 2πB rxt ) u(t) t rx = 350 B rx, B rx in MHz and t rx in nanoseconds 10 10
Rise-time budget Material dispersion!": spectral width of the source (nm) D mat : material dispersion parameter (ps/(nm km)) L: fiber length (km) Modal dispersion t mat = σ λ D mat (λ)l empirical expression for the bandwidth BM in a link of length L B M (L) = B 0, 0.5 q 0.1 (typical q =0.7) Lq B 0 : bandwidth of 1 km length of cable 11 11 Rise-time budget Relation between fiber rise time (modal dispersion) and the 3-dB bandwidth assume the optical power emerging from the fiber has a Gaussian temporal response g(t) = 1 2πσ e t2 /2σ 2 taking the Fourier transform G(ω) = 1 2π e ω2 σ 2 /2 the time to t1/2 for the pulse to reach its half-maximum value, g(t 1/2 ) = 0.5g(0), is t 1/2 = (2 ln 2) 1/2 σ 12 12
Fiber rise time Full width of the pulse at half-maximum t FWHM 3-dB optical bandwidth t F W HM =2t 1/2 =2σ(2 ln 2) 1/2 frequency at which the received optical power has fallen to 0.5 of the zero frequency e ω2 σ 2 /2 = 1 2 f 3dB = 1 (2 ln 2) 1/2 2π σ f 3dB = 2 (ln 2) 2 = 0.44 π t F W HM t F W HM 13 13 Rise-time budget Using B M, defined previously, as the 3-dB bandwidth of the fiber and letting tfwhm be the rise time for modal dispersion we obtain t mod = 0.44 B M = 0.44Lq B 0 If t mod is expressed in nanoseconds and B M is given in Megahertz, then t mod = 440Lq t sys = [ t 2 tx + D 2 matσ 2 λl 2 + B 0 ( ) ( 440L q 350 + B 0 B rx )] 1/2 14 14
Example LED: t tx = 15 ns,! = 40 nm Dispersion Dmat (850 nm) = 0.0875 ns/(nm km) t mat = 21 ns Receiver: B rx = 25 MHz, t rx = 350/B rx t rx = 14 ns MM fiber: B 0 = 400 MHz km (q = 0.7) t mod =3.9 ns t sys = [ (15 ns) 2 + (21 ns) 2 + (3.9 ns) 2 + (14 ns) 2] 1/2 = 30 ns t sys < 0.7 = 35 ns 20 106 15 15 Example For the 20 Mb/s NRZ data stream t sys falls below the maximum allowable rise time degradation. System is not dispersion limited but rather power limited. 16 16