1 / 37

Discrete optics:

Discrete optics v.s. Integrated optics. Discrete optics:. Integrated optics:. Mirrors, lens, mechanical mounts bulky labor intensive alignment ray optics environment sensitive. Waveguide Bendable, portable Free-of-alignment wave optics robust more functionalities.

huey
Download Presentation

Discrete optics:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Discrete optics v.s. Integrated optics Discrete optics: Integrated optics: • Mirrors, lens, mechanical mounts • bulky • labor intensive alignment • ray optics • environment sensitive • Waveguide • Bendable, portable • Free-of-alignment • wave optics • robust • more functionalities

  2. Discrete optics v.s. Integrated optics Applications of Integrated optics: • Transmitters and receivers, transceivers • All optical signal processing • Ultra-high speed communications (100Gbit/s), optical packet switching • RF spectrum analyzer • Smart sensors OEIC, bio/sensor Optical transceivers

  3. Cladding, n2 Core, refractive index n1 Cladding, n2 Optical waveguides 2-D Optical waveguide x z y n2 <n1 d 1 n2 n1 n0 = 1 0 90º-1 n0*sin(0) = n1*sin(1) Numerical aperture (NA) Critical angle n1*sin(90º-1)=n2*sin(90º) cos(1)=n2/n1

  4. Optical waveguides 2-D Optical waveguide : total acceptance angle

  5. Optical waveguides Example 2: Calculate the acceptance angle of a core layer with index of n1= 1.468, and cladding layer of n0 = 1.447 for wavelength of 1.3m and 1.55 m. Solution: acceptance angle: Wavelength independent:

  6. Reflections at the interface Fresnel equations n2 <n1 x z E1s E3s y z 1 1 y 1 1 n1 n1 2 n2 2 E2s n2 s-polarized beam (senkrecht: perpendicular) p-polarized beam (parallel) Trans-electric beam (TE) Trans-magnetic beam (TM)

  7. Reflections at the interface Fresnel equations n2 <n1 x z E1s E3s y z 1 1 y 1 1 n1 n1 2 n2 2 E2s n2 Poynting vector, energy flow rate

  8. Reflections at the interface Phase shift of reflection n2 <n1 x E1s E3s z y 1 1 n1 n2 2 E2s i.e. when because In this case, is a real number The reflection is not associated with phase shift, or phase shift is 0

  9. Reflections at the interface Phase shift of reflection n2 <n1 E1s E3s x z 1 1 y n1 n2 2 E2s i.e. when  c 1

  10. Reflections at the interface Evanescent wave n2 <n1 x E1s z E3s y 1 1 n1 Momentum conservation n2 2 E2s Attenuated wave, penetration depth: d

  11. Ray optics approach Optical modes C 1 A 1 1 2 d 1 B k*n1*AC – k*n1*AB = 2m AC = AB*cos(21) AB = d/sin(1)

  12. Ray optics approach Optical modes C 1 A 1 1 2 d 1 B Propagation constant Effective index:

  13. Ray optics approach Optical modes, considering phase shift at reflection C 1 A 1 1 2 d 1 B k*n1*AC – k*n1*AB + 2* = 2m AC = AB*cos(21) AB = d/sin(1)

  14. Ray optics approach Optical modes C 1 A 1 1 2 d 1 B V number, normalized thickness, or normalized frequency Cut-off wavelength :

  15. Ray optics approach Optical modes Example: estimate the number of modes • waveguide thickness 100m, free-space wavelength 1m, 49 modes

  16. Ray optics approach Normalized waveguide equation: b: normalized propagation constant

  17. Ray optics approach Discussion: Attenuated wave, penetration depth: D

  18. Ray optics approach Discussions: • mode numbers v.s. index difference and wavelength • effective index difference of higher and lower order modes • mode profiles dependence on index difference and wavelength Example 1: Calculate the thickness of a core layer with index of n1= 1.468, and cladding layer of n0 = 1.447 for wavelength of 1.3m. Solution: For single mode:

  19. Ray optics approach Normalized waveguide equation: V = 3.3 b

  20. Ray optics approach Asymmetric waveguide n2 n1 n0 = 1 n3

  21. Wave optics approach Maxwell equations: Dielectric materials Maxwell equations in dielectric materials: phasor

  22. Cladding, n2 Core, refractive index n1 Cladding, n2 Wave optics approach Helmholtz Equation: Free-space solutions 2-D Optical waveguide x z y n2 <n1 d TM mode: TE mode:

  23. Cladding, n2 Core, refractive index n1 Cladding, n2 Wave optics approach z y n2 <n1 d III n2 x d y x n1 I 0 II n2

  24. Wave optics approach III n2 x d y x n1 I 0 II n2

  25. Wave optics approach III n2 x d y x n1 I 0 II n2

  26. Wave optics approach III n2 x d y x n1 I 0 II n2

  27. Wave optics approach III n2 x d y x n1 I 0 II n2

  28. Wave optics approach III n2 x d y x n1 I 0 II n2

  29. Wave optics approach Graphic solution

  30. Dispersion Dispersion Time delay

  31. Dispersion • Material dispersion Example --- material dispersion Calculate the material dispersion effect for LED with line width of 100nm and a laser with a line width of 2nm for a fiber with dispersion coefficient of Dm = 22pskm-1nm-1 at 1310nm. Solution: for the LED for the Laser

  32. Dispersion • Waveguide dispersion Example --- waveguide dispersion n2 = 1.48, and delta n = 0.2 percent. Calculate Dw at 1310nm. Solution: for V between 1.5 – 2.5.

  33. Dispersion • Waveguide mode dispersion n2 n1 n0 = 1 n3 Higher order mode, Lower order mode,

  34. Dispersion • chromatic dispersion (material plus waveduide dispersion) • material dispersion is determined by the material composition of a fiber. • waveguide dispersion is determined by the waveguide index profile of a fiber

  35. Dispersion • Dispersion induced limitations • For RZ bit With no intersymbol interference • For NRZ bit With no intersymbol interference

  36. Dispersion Dispersion induced limitations • Optical and Electrical Bandwidth • Bandwidth length product

  37. Dispersion Dispersion induced limitations Example --- bit rate and bandwidth Calculate the bandwidth and length product for an optical fiber with chromatic dispersion coefficient 8pskm-1nm-1 and optical bandwidth for 10km of this kind of fiber and linewidth of 2nm. Solution: • Fiber limiting factor absorption or dispersion?

More Related