Lecture 3

Example 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?

Example 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 a t least 130 photons/bit to achieve 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 1015 photons/s, what is the received power level for bit error rate at 1 Gbps? (c) Same as (b) but at 10 Gbps?

Four-port optical couplers By definition:

Four-port optical couplers Power division matrix

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.

Four-port optical couplers r = 1  C 11 = C 12 = L/2 3-dB coupler or ‘50-50’ coupler.

Example 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?

Multimode fiber Rays incident at angle still travel farther but also low average faster ( ) Length bandwidth product is lot greater than one of SI multimode fiber.  Useful for telecommunications.

Multimode fiber Rays incident at an angle to axis travel further than rays incident parallel to an axis. Low-length bandwidth product (<100 MHz-km)  not widely used in telecommunications.

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.

Modes in fibers

It begins with Maxwell’s equations to define a wave equation. In an isotropic medium:

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

Modes in fibers

For guided mode: n 2 < n eff < n 1 For radiation mode: n eff < n 2

Modes in Fibers If we rewrite a wave equation in scalar, we get

Modes in Fibers Solutions for the last equation are

Modes in Fibers

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.

Modes in Fibers

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.

Modes in Fibers Two mode fiber guide LP 01 and LP 11 modes:

Modes in Fibers ModeCutoff conditionV at m =1,2,3 LP 0m l=0J -1 (r)= 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

Example Find a core diameter for a single-mode fiber with λ=1330 nm.