EE16.468/16.568Lecture 2 Discrete optics: Mirrors, lens, mechanical mounts bulky labor intensive...

EE16.468/16.568 Lecture 2

Discrete optics:

• Mirrors, lens, mechanical mounts• bulky• labor intensive alignment• ray optics • environment sensitive

Discrete optics v.s. Integrated optics

Integrated optics:

• Waveguide• Bendable, portable • Free-of-alignment• wave optics • robust• more functionalities

EE16.468/16.568 Lecture 2Discrete 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

Optical transceivers

OEIC, bio/sensor

EE16.468/16.568 Lecture 2Optical waveguides

2-D Optical waveguide

n2 < n1

Core, refractive index n1

Cladding, n2

Cladding, n2

y

x

z

d

0

1

90º-1

n0*sin(0) = n1*sin(1)

n1

n2

n0 = 1

Critical angle

n1*sin(90º-1)=n2*sin(90º)

Numerical aperture (NA)

cos(1)=n2/n1



),90sin(sin 01max0 cnn

,sin1

2

n

nc

,)(

sin0

2/122

21

max n

nn

,)(sin 2/122

21max0 nnnNA

max2 : total acceptance angle


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 1.3m and 1.55 m.

Solution:

,)(

sin0

2/122

21

max n

nn

,7.9)(

sin 0

0

2/122

211

max

n

nn

acceptance angle: ,4.192 0max Wavelength independent:

EE16.468/16.568 Lecture 2

Fresnel equations

Reflections at the interface

n1

n2

n2 < n1

1

zy

x

1

2

s-polarized beam (senkrecht: perpendicular)

Trans-electric beam (TE)

n1

n2

1 1

2

zy

p-polarized beam (parallel)

Trans-magnetic beam (TM)

sss EtE 12

E1s E3s

E2s

sss ErE 13

2211

2211

coscos

coscos

nn

nnrs

ss rt 1

ppp EtE 12 ppp ErE 13

2112

2112

coscos

coscos

nn

nnrp

)1(2

1pp r

n

nt

EE16.468/16.568 Lecture 2

Fresnel equations


n1

n2

n2 < n1

1

zy

x

1

2

n1

n2

1 1

2

zyE1s E3s

E2s

2

1

3

s

ss E

ER

2

1

2

1

2

s

ss E

E

n

nT

2

1

3

p

pp E

ER

2

1

2

1

2

p

pp E

E

n

nT

Poynting vector, energy flow rate

HES

HiE

HiaiknE

ˆ2

EnS

EE16.468/16.568 Lecture 2

Phase shift of reflection


n2 < n1

2211

2211

coscos

coscos

nn

nnrs

2/11

221

2222 )sin(cos nnn

2211 sinsin nn

when 122

122 sin nn

1

21sin

n

ni.e. csin

0coscos 2211 nn because 0)cos()cos( 22

21

222

211 nnnn

In this case, 0sr is a real number

The reflection is not associated with phase shift, or phase shift is 0

n1

n2

1

zy

x

1

2

E1s E3s

E2s

EE16.468/16.568 Lecture 2

Phase shift of reflection


n2 < n1

2211

2211

coscos

coscos

nn

nnrs

2/11

221

2222 )sin(cos nnn

when 122

122 sin nn

1

21sin

n

ni.e. csin

n1

n2

1zy

x

1

2

E1s E3s

E2s

2/1221

22111

2/1221

22111

)sin(cos

)sin(cos

nnin

nninrs

1

2/121

2

1

2/121

221

2

cos

)sin(sin

cos

)/(sin

2tan

cnn

0

30

60

90

120

150

180

0 10 20 30 40 50 60 70 80 90

c

1

EE16.468/16.568 Lecture 2

Evanescent wave


n2 < n1

n1

n2

1

zy

x

1

2

E1s E3s

E2s

)(222

222 )0()0( zkxkis

rkiss

zxeEeEE

Momentum conservation

zz kk 12 2211 sin2

sin2

nn

2

2/1

12

22

2122/1

122

122

2/122

222 1sin

2sin

2

in

nninnkkk zx

zikdxs

zikxss

zz eeEeeEE 222 /222 )0()0(

Attenuated wave, penetration depth: d

12/1

12

22

2121

2 1sin2

n

nnd

EE16.468/16.568 Lecture 2

Optical modes

k*n1*AC – k*n1*AB = 2m

AC = AB*cos(21) AB = d/sin(1)

1 2

A

B

C

1

1

1

d

md

knd

kn 2sin

)2cos(sin 1

111

1

m

ddkn 2

sinsin

sincos

11

12

12

1

mdkn 2

sin

1sincos

1

12

12

1

mdkn 2

sin

1sinsin1

1

12

12

1

mdkn 2sin

sin2

1

12

1

mdkn 2sin2 11

Ray optics approach

dmn mm 2

sinsin ,0,11

2/1

20

201

2/1

,12

01,101

)2

()(

)sin1(cos

dmkkn

knkn mmm

EE16.468/16.568 Lecture 2

Optical modes

1 2

A

B

C

1

1

1

d

Ray optics approach

Propagation constant

20

201

2 )2

()(d

mkknm

Effective index: effm nk0

EE16.468/16.568 Lecture 2

Optical modes, considering phase shift at reflection

k*n1*AC – k*n1*AB + 2* = 2m

AC = AB*cos(21) AB = d/sin(1)

1 2

A

B

C

1

1

1

d

md

knd

kn 22sin

)2cos(sin 1

111

1

m

ddkn 22

sinsin

sincos

11

12

12

1

mdkn 22

sin

1sincos

1

12

12

1

mdkn 22

sin

1sinsin1

1

12

12

1

mdkn 22sin

sin2

1

12

1

2tan)

2sin

2tan( 11

md

kn

Ray optics approach

2

2/122

2

11 cos

)sin(sin)

2sin

2tan(

cm

dkn

EE16.468/16.568 Lecture 2

Optical modes

1 2

A

B

C

1

1

1

d

Ray optics approach

Cut-off wavelength :

,/])(2

[ 2/122

21

nnd

m

c

V number, normalized thickness, or normalized frequency

,)(2/2 2/12

221 nn

dV

,/]2[ Vm

,2

)( cV

mdkn 22sin2 11 21

22

21 1cossinn

n

EE16.468/16.568 Lecture 2

Optical modes

Ray optics approach

Example: estimate the number of modes

• waveguide thickness 100m, free-space wavelength 1m,

,490.11 n

49 modes

,4.76)(2 2/12

221 nn

aV

,7.48/]2[ Vm

,470.12 n

,/])(2

[ 2/122

21

nnd

m

EE16.468/16.568 Lecture 2Ray optics approach

1

2/121

2

11 sin

)sin(cos)

2sin

2tan(

cm

dkn

Normalized waveguide equation:

,1

tan2

)1( 1

b

bmbV

,cos

22

21

221

221

22

21

22

2

nn

nn

nn

nnb eff

,01

tan2

)1(),,( 1

b

bmbVbmVf

b: normalized propagation constant,cos1 meff nn

,)(2/2 2/12

221 nn

dV


Discussion:

,cos

22

21

221

221

22

21

22

2

nn

nn

nn

nnb eff

,cos1 meff nn

Attenuated wave, penetration depth: D

1

2/122

21

12/1

222

2121

2

)2

1sin2

bnn

n

nnD

,cos)( 221

221

22

21 nnbnn


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 1.3m.

Solution:

,2

)(2 2/12

221

nna

V,/]2[ Vm ,0,1 m For single mode:

0.3

0.4

0.5

0.6

0.7

0.8

0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9

thickness

pi


m

cmm m

dkn

sin

)sin(cos)

2sin

2tan(

2/122

1

Normalized waveguide equation:

,1

)2

)1(tan(b

bmbV

,cos1 meff nn

-4

-2

0

2

4

0 0.2 0.4 0.6 0.8 1

m=0m=1(b/(1-b)) .̂5

b

V = 3.3


Asymmetric waveguide

),1

(tan)1

(tan2

)1( 11

b

b

b

bmbV

n1

n2

n0 = 1

n3

23

21

22

23

nn

nn

EE16.468/16.568 Lecture 2Wave optics approach

0

Dielectric materials

Maxwell equations:

t

BE

D

t

DJH

0 B

ED

0J HB

t

BE

0 D

t

DH

0 B

Maxwell equations in dielectric materials:

BjE

0 D

DjH

0 B

phasor

)()( BjE

EEE

22)(



n2 < n1


Cladding, n2

Cladding, n2

y

x

z

d

)(ˆ xEyE

TE mode:

022 EE

Helmholtz Equation:

022 EkE

22 k

)(ˆ xHyH TM mode:

Free-space solutions

ikzeEyE 0ˆ

ikzeExE 0ˆ


ziexEyE )(ˆ

020

22 EknE

n2 < n1


Cladding, n2

Cladding, n2

yz

d

I

II

III

yx

xd

0

n1

n2

n2

0)()()( 22

02

2

2

xEkndx

xEd


I

II

III

yx

xd

0

n1

n2

n2

0)()()( 22

0212

2

xEkndx

xEd

0)()()( 22

0222

2

xEkndx

xEd

0)()()( 22

0222

2

xEkndx

xEd

xknBxknAxE III22

021

220

21 sincos)(

xknAxE IIII20

22

2exp)(

dxknAxE IIIIII 20

22

2exp)(


I

II

III

yx

x

n1

n2

n2

0)()()( 22

0212

2

xEkndx

xEd

0)()()( 22

0222

2

xEkndx

xEd

0)()()( 22

0222

2

xEkndx

xEd

xknBxknAxE III22

021

220

21 sincos)(

2/exp)( 20

22

2 dxknAxE IIII

2/exp)( 20

22

2 dxknAxE IIIIII

d

0


I

II

III

yx

x

n1

n2

n2

0cos0sin 220

21

220

21

220

21

220

21

20

22

2 knknBknAknknA IIII

0|)()( xIII xEdx

dxE

dx

d

22

021

20

22

2

kn

knAB III

dxIIII xExE |)()(

dxIIII xEdx

dxE

dx

d |)()(

0|)()( xIII xExE

0|)()( xIII xEdx

dxE

dx

d

d

0


I

II

III

yx

x

n1

n2

n2

dknknBdknAknknA IIIII22

021

220

21

220

21

220

21

20

22

2 cossin

dknBdknAA IIIII22

021

220

21 sincos

dknBdknA

dknBdknA

kn

kn

II

II

220

21

220

21

220

21

220

21

20

22

2

220

21

cossin

sincos

dxIIII xExE |)()(

dxIIII xEdx

dxE

dx

d |)()(

0|)()( xIII xExE

0|)()( xIII xEdx

dxE

dx

d

d

0

22

021

20

22

2

kn

knAB III


dknBdknA

dknBdknA

kn

kn

II

II

220

21

220

21

220

21

220

21

20

22

2

220

21

cossin

sincos

22

021

20

22

2

kn

knAB II

dkn

kn

kndkn

dknkn

kndkn

kn

kn

220

2122

021

20

22

222

021

220

2122

021

20

22

222

021

20

22

2

220

21

cossin

sincos

220

21 knh

qhdh

hdqh

q

h

)tan(

)tan( 20

22

2 knq

)/1(

22)tan(

2222 hqh

q

qh

hqhd

Graphic solution

EE16.468/16.568 Lecture 2Dispersion

Dispersion

1

0|

d

d

d

dvg

v

ZieAA )()(

)(tA

))(()||)(())(()(

0000

)(

gv

ZtziZ

d

dtzitz

d

di

tzi eeeezA

Time delay )()(v

ZtAtA


• Material dispersion

,|0

d

dvg ,

gg v

L ,)()(

2

2

d

nd

cd

d

Lg

),(2

2

d

nd

cDm , LDmg

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.

,2.2 nsLDm

Solution:

,44psLDm

for the LED

for the Laser

EE16.468/16.568 Lecture 2

• Waveguide dispersion

,|0

d

dvg ,

gg v

L

,2)2(

984.1)(

22

2

2

cna

N

d

d

Lgg

, LDmg

Example --- waveguide dispersionn2 = 1.48, and delta n = 0.2 percent. Calculate Dw at 1310nm.

Solution:

,)()(

)(2

2212

dV

VbdV

c

nnn

d

d

Lg

,)()(

2

2212

dV

VbdV

c

nnnDw

,)/996.01428.1( 2Vb for V between 1.5 – 2.5.

,26.0)(

2

2

dV

VbdV

),/(9.1)()(

2

2212 kmnmps

dV

VbdV

c

nnnDw

Dispersion

EE16.468/16.568 Lecture 2

• Waveguide mode dispersion

Dispersion

n1

n2

n0 = 1

n3

Higher order mode, 2

~|0 n

c

d

dvg

Lower order mode, 1

~|0 n

c

d

dvg

)/(112 n

c

n

c

Lg


• chromatic dispersion (material plus waveduide dispersion)

,)(

wmg DD

L

• material dispersion is determined by the material composition of a fiber.

• waveguide dispersion is determined by the waveguide index profile of a fiber

EE16.468/16.568 Lecture 2

• Dispersion induced limitations

,2

1

2/1B

• For RZ bit With no intersymbol interference

,1

2/1B

• For NRZ bit With no intersymbol interference

Dispersion

EE16.468/16.568 Lecture 2

Dispersion induced limitations

,2

1

2/1B

• Optical and Electrical Bandwidth

,7.03 Bf dB

• Bandwidth length product

,25.0

D

BL

Dispersion

EE16.468/16.568 Lecture 2

Dispersion induced limitations

,16/ 12/1

pskmDL

,8.27.03 GHzBf dB

,9.3625.0 1kmGbs

DBL

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?

,5.21025.0 dBkmdBLoss

Dispersion

EE16.468/16.568Lecture 2 Discrete optics: Mirrors, lens, mechanical mounts bulky labor intensive...

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