EE 16 46816 568 Discrete optics v s

EE 16. 468/16. 568 Discrete optics v. s. Integrated optics Discrete optics: • Mirrors, lens, mechanical mounts • bulky • labor intensive alignment • ray optics • environment sensitive Integrated optics: • Waveguide • Bendable, portable • Free-of-alignment • wave optics • robust • more functionalities Lecture 2

EE 16. 468/16. 568 Discrete optics v. s. Integrated optics Applications of Integrated optics: • Transmitters and receivers, transceivers • All optical signal processing • Ultra-high speed communications (100 Gbit/s), optical packet switching • RF spectrum analyzer • Smart sensors OEIC, bio/sensor Optical transceivers Lecture 2

EE 16. 468/16. 568 Optical waveguides Lecture 2 2 -D Optical waveguide x z y n 2 < n 1 Cladding, n 2 d Core, refractive index n 1 Cladding, n 2 1 0 n 2 n 1 n 0 = 1 90º- 1 n 0*sin( 0)= n 1*sin( 1) Numerical aperture (NA) Critical angle n 1*sin(90º- 1)=n 2*sin(90º) cos( 1)=n 2/n 1

EE 16. 468/16. 568 Optical waveguides Lecture 2 2 -D Optical waveguide : total acceptance angle

EE 16. 468/16. 568 Optical waveguides Lecture 2 Example 2: Calculate the acceptance angle of a core layer with index of n 1= 1. 468, and cladding layer of n 0 = 1. 447 for wavelength of 1. 3 m and 1. 55 m. Solution: acceptance angle: Wavelength independent:

EE 16. 468/16. 568 Reflections at the interface Fresnel equations x y n 2 < n 1 E 1 s z Lecture 2 1 E 3 s 1 1 1 n 1 n 2 2 E 2 s s-polarized beam (senkrecht: perpendicular) Trans-electric beam (TE) z y 2 n 2 p-polarized beam (parallel) Trans-magnetic beam (TM)

EE 16. 468/16. 568 Reflections at the interface Fresnel equations x y n 2 < n 1 E 1 s z Lecture 2 1 E 3 s 1 1 1 n 1 n 2 2 z y E 2 s Poynting vector, energy flow rate 2 n 2

EE 16. 468/16. 568 Phase shift of reflection Reflections at the interface n 2 < n 1 Lecture 2 x y E 1 s z 1 E 3 s 1 n 2 when 2 i. e. because In this case, is a real number The reflection is not associated with phase shift, or phase shift is 0 E 2 s

EE 16. 468/16. 568 Reflections at the interface Phase shift of reflection n 2 < n 1 Lecture 2 E 1 s x 1 z y 1 n 1 2 n 2 when E 3 s i. e. c 1 E 2 s

EE 16. 468/16. 568 Reflections at the interface Evanescent wave Lecture 2 n 2 < n 1 x E 1 s 1 Momentum conservation z y 1 n 2 Attenuated wave, penetration depth: d E 3 s 2 E 2 s

EE 16. 468/16. 568 Lecture 2 Ray optics approach Optical modes 1 1 A 2 C B k*n 1*AC – k*n 1*AB = 2 m AC = AB*cos(2 1) d AB = d/sin( 1)

EE 16. 468/16. 568 Lecture 2 Ray optics approach Optical modes 1 1 A 2 1 1 C d B Propagation constant Effective index:

EE 16. 468/16. 568 Lecture 2 Ray optics approach Optical modes, considering phase shift at reflection 1 1 A 2 1 1 C B k*n 1*AC – k*n 1*AB + 2* = 2 m AC = AB*cos(2 1) d AB = d/sin( 1)

EE 16. 468/16. 568 Lecture 2 Ray optics approach Optical modes 1 1 A 2 C d B V number, normalized thickness, or normalized frequency Cut-off wavelength :

EE 16. 468/16. 568 Ray optics approach Optical modes Example: estimate the number of modes • waveguide thickness 100 m, free-space wavelength 1 m, 48 modes Lecture 2

EE 16. 468/16. 568 Ray optics approach Lecture 2 Normalized waveguide equation: b: normalized propagation constant

EE 16. 468/16. 568 Discussion: Ray optics approach Lecture 2 Attenuated wave, penetration depth: D

EE 16. 468/16. 568 Ray optics approach Lecture 2 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 thickness of a core layer with index of n 1= 1. 468, and cladding layer of n 0 = 1. 447 for wavelength of 1. 3 m. Solution: For single mode:

EE 16. 468/16. 568 Lecture 2 Ray optics approach Normalized waveguide equation: V = 3. 3 b

EE 16. 468/16. 568 Ray optics approach Lecture 2 Asymmetric waveguide n 2 n 1 n 0 = 1 n 3 For asymmetric waveguide

EE 16. 468/16. 568 Wave optics approach Maxwell equations: Dielectric materials Maxwell equations in dielectric materials: phasor Lecture 2

EE 16. 468/16. 568 Wave optics approach Lecture 2 Helmholtz Equation: Free-space solutions 2 -D Optical waveguide x z y n 2 < n 1 Cladding, n 2 d Core, refractive index n 1 Cladding, n 2 TE mode: TM mode:

EE 16. 468/16. 568 Wave optics approach Lecture 2 z y Cladding, n 2 < n 1 Core, refractive index n 1 d Cladding, n 2 III x x y I n 2 d n 1 0 II n 2

EE 16. 468/16. 568 III x x y I Wave optics approach n 2 d n 1 0 II n 2 Lecture 2

EE 16. 468/16. 568 III x x y I Wave optics approach n 2 d n 1 0 II n 2 Lecture 2

EE 16. 468/16. 568 III x x y I Wave optics approach n 2 d n 1 0 II n 2 Lecture 2

EE 16. 468/16. 568 III x x y I Wave optics approach n 2 d n 1 0 II n 2 Lecture 2

EE 16. 468/16. 568 x x y Wave optics approach III n 2 I n 1 d 0 II n 2 Lecture 2

EE 16. 468/16. 568 Graphic solution Wave optics approach Lecture 2

EE 16. 468/16. 568 Dispersion Time delay Dispersion Lecture 2

EE 16. 468/16. 568 Lecture 2 Dispersion • Material dispersion for just Example --- material dispersion Calculate the material dispersion effect for LED with line width of 100 nm and a laser with a line width of 2 nm for a fiber with dispersion coefficient of Dm = 22 pskm-1 nm-1 at 1310 nm. Solution: for the LED for the Laser

EE 16. 468/16. 568 Dispersion Lecture 2 • Waveguide dispersion Example --- waveguide dispersion n 2 = 1. 48, and delta n = 0. 2 percent. Calculate Dw at 1310 nm. Solution: for V between 1. 5 – 2. 5.

EE 16. 468/16. 568 Lecture 2

EE 16. 468/16. 568 Dispersion • Waveguide mode dispersion n 2 n 1 n 3 Higher order mode, Lower order mode, n 0 = 1 Lecture 2

EE 16. 468/16. 568 Dispersion Lecture 2 • 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

EE 16. 468/16. 568 Dispersion • Dispersion induced limitations • For RZ bit With no intersymbol interference • For NRZ bit With no intersymbol interference Lecture 2

EE 16. 468/16. 568 Dispersion induced limitations • Optical and Electrical Bandwidth • Bandwidth length product Dispersion Lecture 2

EE 16. 468/16. 568 Dispersion Lecture 2 Dispersion induced limitations Example --- bit rate and bandwidth Calculate the bandwidth and length product for an optical fiber with chromatic dispersion coefficient 8 pskm-1 nm-1 and optical bandwidth for 10 km of this kind of fiber and linewidth of 2 nm. Solution: • Fiber limiting factor absorption or dispersion?