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EE 230: Optical Fiber Communication Lecture 15 From the movie Warriors of the Net WDM Components.

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Presentation on theme: "EE 230: Optical Fiber Communication Lecture 15 From the movie Warriors of the Net WDM Components."— Presentation transcript:

1 EE 230: Optical Fiber Communication Lecture 15 From the movie Warriors of the Net WDM Components

2 ITU Grid Wavelengths for CWDM and frequencies for DWDM defined by International Telecommunication Union, a part of the United Nations located in Geneva Central frequency is 193.1 THz, equivalent to 1552.52 nm Frequencies for 50 GHz channel spacings are thus defined as 193.1 + 0.05n THz where n is a positive or negative integer

3 Active vs. Passive Devices Passive: requires no electrical power and transfer function cannot be modified by user Active: allows user to manipulate what it does to light pulses. Requires power.

4 Platforms for WDM components Discrete optics: thin-film filters, microelectromechanical systems (MEMS), isolators, circulators All-fiber components: couplers, Mach- Zehnder interferometers Planar lightwave circuits (PLC): arrayed-waveguide gratings (AWG), couplers, MZs, etc.

5 Coupler parameters Splitting ratio: P 2 /(P 1 +P 2 ) Excess loss: 10 log (P 0 /P 1 +P 2 ) Insertion loss: 10 log (P in /P out ) Crosstalk: 10 log (P 3 /P 0 )

6 Coupling as function of length

7 Mach-Zehnder Interferometer where n eff is determined from the P core /P graphs

8 Multiplexing/demultiplexing criterion where  L is the path length difference between the two arms

9 Wavelength dependence of MZ output For wavelengths 1 entering at input port 1, and 2 entering at input port 2,

10 Wavelength adjustment (“trim”) Coarse adjustment possible with fiber MZs by heating and pulling shorter arm to increase channel spacing Fine adjustment for both fiber and PLCs done with UV irradiation to line transmission peaks up with ITU grid

11 Example To multiplex four wavelengths separated by 50 GHz (0.4 nm) How many stages needed? 2.(log 2 W). How many total MZs? 3.Two in one stage, one in the next. What is  L for each stage?

12 Example, continued If first frequency is ITU center, what are other three, and their wavelengths? 193.10, 193.15, 193.20, and 193.25 THz 1552.52, 1552.12, 1551.72, and 1551.32 nm If n eff =1.45, determine  L values

13 Example, continued First stages have 100 GHz channel spacing, one for even-numbered wavelengths and one for odd.  L equals c/2n(100x10 9 )=1.0 mm Second stages have 50 GHz channel spacing.  L =c/2n(50x10 9 )=2.1 mm As channel spacing gets smaller, it gets easier to make MZs (larger  L)!

14 General MZ expression For a multiplexer or demultiplexer with N wavelengths, you need n=log 2 N stages where the path length difference for stage i is

15 Arrayed-Waveguide Grating

16 AWG channel spacing where n s =input/output waveguide index, n c =central waveguide array index, and

17 Tuning an AWG Each input waveguide corresponds to a different center wavelength and channel spacing. Several waveguides around the center one will correspond to the correct channel spacing within the tolerance, and the peak wavelengths will vary from one waveguide to another.

18 WDM Muxes and Demuxes

19 Grating Based Demultiplexer

20 Optical Filters

21 Interference Filter Based WDM

22 Thermal drift in waveguide devices  n/  T for silica=7.5x10 -6 per degree  for silicon=2.63 ppm per degree d /dT = 12 pm per degree (red shift) 2/3 due to thermooptic effect, 1/3 to CTE

23 Effect of thermal drift Channel spacing=100 GHz=0.8 nm=800 pm DWDM device completely transparent every 800 pm, opaque between Silica-on-silicon drifts 12 pm/  Device becomes a beam stop if temperature changes by ? 33  ! “Passive” devices routinely T stabilized; customers unhappy

24 Athermalization Techniques Mechanical compensation: flex entire chip, adjust point at which signal injected into device Materials compensation: design waveguide to be inherently athermal


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