ene 623/eie 696 optical communication lecture 3. example 1 a common optical component is the...

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  • Slide 1
  • ENE 623/EIE 696 Optical Communication Lecture 3
  • Slide 2
  • Example 1 A common optical component is the equal-power splitter which splits the incoming optical power evenly among M outputs. By reversing this component, we can make a combiner, which can be made to deliver to a single output the sum of the input powers if multimode fiber is used, but which splits the power incoming to each port by a factor of M if single- mode fiber is used. (a) Compare the loss in dB between the worst-case pair of nodes for the 3 topologies if the number of nodes is N = 128 and multimode is used. Assume that, for the tree, there are 32 nodes in each of the top two clusters and 64 nodes in the bottom one. (b) What would these numbers become if single-mode fiber were used? (c) How would you go about reducing the very large accumulated splitting loss for the bus?
  • Slide 3
  • Example 1 Sol n
  • Slide 4
  • Example 2 A light wave communication link, operating at a wavelength of 1500 nm and a bit rate of 1 Gbps, has a receiver consisting of a cascaded optical amplifier, narrow optical filter, and a photodetector. It ideally takes at least 130 photons/bit to achieve 10 -15 bit error rate. (a) How many photons/bit would it take to achieve the same error rate at 10 Gbps? (b) At this wavelength, 1 mW of power is carried by 7.5 x 10 15 photons/s, what is the received power level for 10 -15 bit error rate at 1 Gbps? (c) Same as (b) but at 10 Gbps?
  • Slide 5
  • Example 2 Sol n
  • Slide 6
  • Four-port optical couplers By definition:
  • Slide 7
  • Four-port optical couplers Power division matrix where C ij is called incoherent additional input amplitude.
  • Slide 8
  • Four-port optical couplers r = 0 C 11 = 0, C 12 = L All input power crossover to output 2. r = C 11 = L, C 12 = 0 All input power goes straight through.
  • Slide 9
  • Four-port optical couplers r = 1 C 11 = C 12 = L/2 3-dB coupler or 50-50 coupler.
  • Slide 10
  • Example 3 For the 4-port fiber optic directional coupler, the network below uses 8 of these couplers in a unidirectional bus. Assume that the excess loss of each coupler is 1 dB. (a) If the splitting ratio is 1 for all of the couplers, what is the worst case loss between any Tx and Rx combination in dB? (b) What is the least loss between any Tx and Rx?
  • Slide 11
  • Example 3 Sol n
  • Slide 12
  • Multimode fiber (SI fiber) Rays incident at an angle to axis travel further than rays incident parallel to an axis. Low-length bandwidth product (
  • Single-mode fiber Only one mode propagates: neglecting dispulsion all incident light arrives at fiber end at the same time. Length bandwidth product > 100 GHz-km. Much greater bandwidth than any multimode fiber. suitable for long live intercity applications.
  • Slide 15
  • Modes in fibers
  • Slide 16
  • It begins with Maxwells equations to define a wave equation. In an isotropic medium:
  • Slide 17
  • Modes in fibers We have 3 equations with solution of E i for each axis which is not generally independent. Assume that wave travels in z-direction: Substitute these into a wave equation, it yields
  • Slide 18
  • Modes in fibers
  • Slide 19
  • For guided mode: n 2 < n eff < n 1 For radiation mode: n eff < n 2
  • Slide 20
  • Modes in Fibers If we rewrite a wave equation in scalar, we get
  • Slide 21
  • Modes in Fibers Solutions for the last equation are
  • Slide 22
  • Modes in Fibers
  • Slide 23
  • It is convenient to define a useful parameter called V- number as V is dimensionless. V determines Number of modes. Strength of guiding of guided modes.
  • Slide 24
  • Modes in Fibers
  • Slide 25
  • Mode designation LP lm l = angular dependence of field amplitude e il (l = 0,1,..) m = number of zeroes in radial function u(r) Fundamental LP 01 mode: no cutoff. It can guide no matter how small r is. u(r, ) = u 01 (r) .circular symmetric maximum at r = 0.
  • Slide 26
  • Modes in Fibers Two mode fiber guide LP 01 and LP 11 modes:
  • Slide 27
  • Modes in Fibers ModeCutoff conditionV at cutoff @ m =1,2,3 LP 0m l=0J -1 (r)=003.8327.016 LP 1m l=1J 0 (r)=0 LP 2m l=2J 1 (r)=0 LP 01 HE 11 LP 11 TE 01, TM 01, HE 21 For large V, number of guide modes = V 2 /2
  • Slide 28
  • Example 4 Find a core diameter for a single-mode fiber with =1330 nm.