OCS 2-1 Wave Propagation

8/9/2019 OCS 2-1 Wave Propagation

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Optical Communication Systems

Chapter 2.1: Light wave propagation

Pham Quang Thai – [email protected]

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Guiding light using total reflection

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Snell’s law Willebrord van Royen Snell

1580 - 1626

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2

2 2 221 1 1 1 2

1

2 2

1 2

1sin sin sin 1 cos 1

arcsin

a c a c

a

nn n n n n

n

NA

NA n n

Condition for light to enter a fiber: acceptance angle and numerical aperture (NA)

Condition for light to travel inside a fiber: n1>n2


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Examples for chapter 2.1

• Problems 2-11:

– Calculate the numerical aperture (NA) of a step

index fiber having n1=1.48 and n2=1.46. What is

the maximum entrance angle θ0max? (nair=1)

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Content

• Geometrical optics

• Optical propagation in fiber

• Signal degradation in fiber• Types of fiber

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Guided and

Unguided ray

Meridional ray

Skewed ray


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Field distribution in

transverse plane (video)

Johann CarlFriedrich Gauss

MichaelFaraday

André-Marie Ampère

James

Clerk

Maxwell

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Wave equation in Helmholtz equation form

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22

2

22

2

0

0

0 ( )

0

0

, 0 in homogeneuos isotropic dielectric

t t

t t t t

t

B HE

D E H EH JJ E E E

B HD

D E B

EE E

The wave equation hold for each component of E and H in cylindrical coordinates


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( )

2 22 2 2 2 2 2

02 2

2 2 22 2 2

2 2 2 2

2 2 22

2 2 2 2

22 2

2 2

( , ) ( )

1 10,

1 10

1 1

j z j t j jl j t

z z

Z

Z Z Z Z z

z

z

z z z

E E r e e r e e e

E n

E E E k n E k E t c

k E r r r r z

k E

r r r r z

E E l E

r r r r

2

2 22 2

2 2

0

10

z z

z z z

E k E

E E l k E

r r r r

Wave propagation along the z-

axis: very low loss and periodic

with φ


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2 22 2

2 2

2 2

1 1 2 1

2 2

1 1 2 1

1The Bessel equation: 0, has solutions:

( ) ( ) ( ), if 0

( ) ( ) ( ), if 0

l k

r r r r

r c J hr c Y hr k

r c I hr c K hr k


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Radial distribution in the cladding and the core

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In the cladding: fields must be evanescent, only Kl is acceptable

In the core: fields must be finite when r =0 and continuous at r =a, only Jl is acceptable


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Radial distribution of light inside fiber


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Guided and unguided conditions

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2 2 22 2 2

2 0

1 0 2 02 2 2 2 2 2

1 0

00 (cladding) 0

0 (core) 0 0

n k k hn k n k

k n k q

Condition for a ray to be confined (guided) inside the core

2 2 2 2 2 2 2

1 2 0 0( )h q n n k NA k The sum is a constant

As β decrease, q increases and h decreases: the field penetrates deeper into the cladding

As β increase, h increases and q decreases: the field is mostly confined in the core

As q exceeds NA.k 0, h becomes imaginary and the wave become unguided


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Find E and H from β: need to find A, B, C, D

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00

0

0

0

t

t

H

E

EHJ

B HD

D EB


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Relations between Ez, Hz and the remained components

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Bessel solutions for components inside the core

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Bessel solutions for components in the cladding

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Use boundary conditions to find A, B, C, D

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When l =0

B=0 and D=0 Hr =0 and Hz=0

The first equation is EH mode, which will reduce to TM mode when l =0

When l =0

A=0 and C=0 E r =0 and E z=0

The second equation is HE mode, which will reduce to TE mode when l =0

By definition:TE mode: transverse electric mode, no electric field along the propagation direction

TM mode: transverse magnetic mode, no magnetic field along the propagation direction

HE mode: hybrid electromagnetic mode, but magnetic component is larger along the

propagation direction

EH mode: hybrid electromagnetic mode, but electric component is larger along the

propagation direction


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What are HE, EH, TE and TM mode?

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TE or TM mode

Skewed rayHE or EH mode


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Modes in fiber

• In practice, longitudinal modes of fiber is very weak, fiber “mode”usually refer to transverse modes

• Each value of β satisfying the determinant equation provides thevalues of E and H of one mode

• Since Bessel solution is oscillating, there are m values of β for each

value of l=0,1,2,… Each mode is named MODElm• TE and TM mode only occur when l =0. There are only TM0m and

TE0m modes

• TE and TM modes are meridional rays. When l =0,

• All remaining modes are hybrid modes

• Minus or plus sign of l in the Bessel equation of E and H leads todifferent E and H results but same radiation distribution (same β).Each mode in fiber have 2 angular momentum stages thatdegenerate

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/ 0


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Graphical solution of the determinant equation

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TE mode and TM mode

Right hand side: always negative and monotonically decrease

Left hand side: monotonically increase, diverges to ±∞ at zeros of J0


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Graphical solution of the determinant equation

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EH mode

Right hand side: always positive and monotonically increase

Left hand side: monotonically decrease, diverges to ±∞ at zeros of J0

HE mode

Right hand side: always negative and monotonically decrease

Left hand side: monotonically increase, diverges to ±∞ at zeros of J0


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The normalized frequency V

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2 2 2 2 2 2 2

1 2 0 0 0( ) 0h q n n k NA k ha NAk a V

Consider the mode functions as functions of ha. Range of ha satisfying guided mode

On the right hand side, the range of qa is

The right hand side diverges to -∞ at ha=V

If V


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• There is always HE11 mode• The normalized frequency V determine the number of guidedmodes

• The value of V for a mode to appear is its “cut-off frequency”


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Conclusion

• Light propagates in different modes (or different rays)

• Guided modes satisfy the total reflection conditions,unguided mode do not

• Each mode is a portion of the total input power

• Each mode propagates both in the core and the cladding.Higher mode have less power in the core

• Modes consists of TE/TM modes (medirional rays) andHE/EH modes (skewed rays)

• Mode HE11 always exists• Other modes exist when the normalized frequency larger

that theirs cut-off frequencies

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Examples for chapter 2.2

• Problems 2-19:

– Determine the normalized frequency (V) at 820

nm for a step index fiber having a 25-μm core

radius, n1=1.48 and n2=1.46

OCS 2-1 Wave Propagation

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