上海师范大学电气信息系 chapter 3 lightwave fundamentals....

上海师范大学电气信息系

Chapter 3

Lightwave Fundamentals


Contents

Reflection at a plane boundary

Resonant cavities

Polarization

Dispersion, pulse distortion, information rate

Electromagnetic waves

Critical-angle reflection


3.1 Electromagnetic Waves

Wave Properties

•velocity

•power

•polarization

•interference

•refraction


Wave traveling in the z direction

On the figure, t1 < t2 < t3

t1 t3t2

Electric Field

Position (z)

Electric field:

At t1 < t2 < t3 , peak amplitude E0 is fixed, Ф = wt - kz

(3.1)

Electromagnetic Waves


This is a solution of the wave equation:2 2

22

2

0

0

E k E

EE

z v

(3.2)Propagation factor:

Frequency: f = v/

Radian frequency: = 2f (rad/s)

Wave peak amplitude: E0

Wave phase: = t-kz

kv



Relationships for the Propagation Factor k

k = /v = /(c/n) = n/c

In free space, n = 1, so that k = ko = /c

In general, then k = kon

o

c

f

(3.6)

o: wavelength in free space

: wavelength in the medium

/o n (3.7)

2 2k

vv

f

/ ov c n

f f n

Then,

f is fixed, f = f0



Power in a resistor: Pr = V2/R

Power is proportional to the voltage (V) squared.

In an optical beam, define Intensity: I = E2

Since P E∝ 2, P ∝ I.

Power


Irradiance S = Power Density (watts/m2)

For a plane wave, the irradiance and intensity are given

by:

NowWe conclude that the

intensity is proportional to

the irradiance I S∝ .

Power

: 材料的磁导率

: 材料的介电常数



Recall the plane wave given by

This expression represents a wave traveling with zero

loss.

(3.1)

If loss occurs, the field is represented by

sin( )zoE E t kze (3.8)

is the attenuation coefficient for E.

The frequency and phase do not vary with loss, only

the amplitude of the wave Eoe-z changes with loss.


P∝E2, for a path length L in a lossy medium, the power

diminishes by a factor of:2 Le

The corresponding P (or I) reduction in dB is:210lg LdB e

This will be a negative number for propagation through a

lossy medium.

Define: (dB/km) in terms of the attenuation coefficient

. = -8.685 proofIf L is in unit of km, then is in units of km-1.

2 is the attenuation coefficient for P



Wave Traveling in a Lossy Medium

t2

t1

Distance (z)

Ele

ctri

c F

ield



3.2 Dispersion, Pulse Distortion, Information Rate

When we write E = Eosin (t – kz), we imply a single

frequency source.

Frequency

Radio oscillators approximate single f pretty well.

Optical sources do not produce single f.


Example ： Emission Spectrum of an Optical Source

f = source bandwidth (range of frequencies emitted by

the source).

f is the central frequency.

1

0.5

f

Frequency

Normalized Power

ff1 f2

Dispersion, Pulse Distortion, Information Rate


Alternatively, we can plot the wavelength emission

spectrum as follows:

= linewidth or spectral width

1

0.5

Wavelength

Normalized Power

21



Example: If = 0.82 m, = 30 nm

so we have 3.7% bandwidth.

The conversion between wavelength and frequency is:

(3.9)

f

f



Proof:

The mean frequency is: f = c/

Define the mean wavelength as:

Then,

Now, we have

212



Spectral Widths for Typical Light Sources (table 3.1)

Source Spectral Width (nm)

LED 20-100

Laser Diode 1-5

Nd:YAG-Laser(固态钇铝石榴石 ) 0.1

He-Ne Laser 0.002



If = 0, (f = 0), the source is perfectly coherent.

It is monochromatic （单色） .

Laser diodes are more coherent than LEDs, but are

not perfectly coherent.

We will see how source bandwidth limits the

information capacity of fiber transmission lines.



3.2.1 Material Dispersion and Pulse Distortion

Recall that v = c/n.

For glass, n varies with wavelength. Thus, waves

of different wavelengths (frequencies) travel at

different speeds.

Dispersion:

Wavelength dependent propagation velocity.

Material Dispersion: Dispersion caused by the material.

Waveguide Dispersion: Dispersion caused by the

structure of the waveguide.


Consider a pulse of light emitted by a source which contains a range of wavelengths (say 1, 2, 3).

Input Power

t

t

t

1

tT

2

3

Fastestwavelength

Slowest wavelength

Output Power

t

t

t

t

1

2

3

T +

Arrives last

Arrives first

Material Dispersion and Pulse Distortion


Because of dispersion, the components of the input

pulse at 1, 2, and 3 travel at different speeds and

thus arrive at the receiver at different times. The

previous slide displayed how this phenomenon spreads

pulses as they travel along a dispersive medium. The

output is widened by an amount we label as .


Dispersion also distorts an analog signal waveform.

Input Power Output Power

Pac,in Pac,out

1

2

Slower wavelength

1

2

Pac,out < Pac,in

Information is contained in the amplitude variation.

t t


DISPERSION

Refractive Index Variation for SiO2

First Derivative

o

n’

dn/d0

1.45

o

Inflection Point

0

n

Second Derivative

o

dn2/d2

0

n’’

Inflection point for SiO2 glass occurs near wavelength: 1300 nm


Find the amount of pulse spread due to material dispersion.

Let = time of travel of a pulse over path length L.

With No Dispersion

With Dispersion Present

/L

1 2

L

(/L)2

(/L)1

/L

The source linewidth is taken to be (with 2 > 1):

= 2 - 1


where 1 is the fastest and 2 is the slowest wavelength.

(3.10)

The pulse spread per unit length is then:

(/L)/ = d(/L)/d （ slope of the curve ）Pulse spread per unit length: (/L) = [d(/L)/d

(3.12)

Actual spread would be:

1 2

L

(/L)2

(/L)1

/L


(/L) = [d(/L)/d L’ (3.12)

Two distinct terms determine the pulse spread

1. the slope of the /L curve

2. the linewidth of the source.

The linewidth will be available from manufacturer's

data or must be measured.

Further analysis shows that:

The prime and double prime denote first and second

derivatives.

(3.13)


Proof:

Pulses travel at a speed called the group velocity u.

The group velocity is given by:

The pulse travel time is thus:

This is the pulse travel time per unit of path length.


( is the free space value)If n (), then (/L)’ = 0 and there is no dispersion and no pulse

spread.


Define material dispersion M :

(3.14)

Combining (3.12) and (3.13):

M (ps/nm/km) is in picoseconds of pulse spread per nanometer of source spectral width and per kilometer of fiber length.

1.3 1.55

0.82

110

-20

M(ps/(nm.km))

(m)

SiO2


1. For M > 0 (wavelengths < 1.3 m)

Wavelength 2 arrives before wavelength 1.

Energy at 2 travels faster than energy at 1. (2 > 1)

2. For M < 0 (wavelengths > 1.3 m)

So that 1 travels faster than wavelength 2.

3. At 1.3 m, M = 0 , and there is no material dispersive pulse spreading.


Example: Consider an LED at = 0.82 m, L = 10 km, and = 20 nm. Find (/L).

From the graph, at 0.82 m, M =110 ps/(nm·km).

Change the wavelength to = 1.5 m, = 50 nm. At 1.5 m, M = -15 ps/nm·km. Then


Example: = 0.82 m, = 1 nm. M = 110 ps/(nm·km)

ns

nspsnmkmnm

pskm

LL

1.1

1.11100111010

Example: = 1.5 m, = 1 nm. M = -15 ps/nm·km


Between 1200 nm and 1600 nm(near the inflection point), M is given by

Mo = -0.095 ps/(nm2•km) and o is the zero dispersion

wavelength ( 1300 nm).

Conclusion:

•The longer the path the greater the pulse spread.

•The greater the source spectral width, the greater

the pulse spread.


A soliton is a pulse that travel without spreading. The refractive index of glass depends upon the pulse intensity. This fiber nonlinearity is used to counter the effects of dispersion. The leading edge of the pulse can be slowed down, and the trailing edge speeded up to reduce spreading. Thus, the pulse must be properly shaped.

3.2.2 Solitons

The nonlinearity is such that solitons are only produced at wavelengths longer than the zero-dispersion wavelength in glass fibers. Compensation to overcome pulse broadening is only possible in the longer wavelength region range 1300 to 1600 nm.


单脉冲传输 800km 时孤子脉冲波形的演变

-10

0

100

200

400

600

8000

1

2

3

4

5

x 10-3


基态和高阶孤子沿光纤传输时的变化特点


Solitons overcome the bandwidth limitations of the

fiber, but not the attenuation. Optical amplifiers are

needed along the transmission path to maintain the

pulse energy above the minimum required for

soliton production.

Fiber

Amplifier


3.2.3 Information RateConsider sinusoidal modulation of the light source with modulation frequency f. Modulation period T = 1/f.

0 1 2 3 4 5 6 70

0.5

1

1.5

2

2.5

3

3.5

4

time

Optic P

ow

er

Tra

sm

itte

dSinusiodal Modulation Of The Light Source

PT

Pavg

T

Po

wer


Information Rate

consder

0 1 2 3 4 5 6 70

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Maximum Allowable Pulse Spead At The Receiver

time

Opt

ic P

ower

Rec

eive

r

PR

T

T/2

This spread reduces the total power variation to zero.

Modulation is canceled.

Blue: 1

Red: 2

TimeT/2

Information Rate


The limit on the allowable pulse spread will be

taken to be:

2

T (2)

Information Rate


so that the modulation frequency has the limits: 1

2f

The maximum modulation frequency is then:

(4)

From (2) we have the requirement that 1/T < 1/(2)(3)

This modulation frequency turns out to be the 3-dB

bandwidth. The signal is actually reduced by half (3-dB)

at this modulation frequency.

2

13 dBf (5)3-dB optical bandwidth:

Information Rate

max

1

2f


The total signal loss has two parts and can be expressed

by the equation:

(6)

La = Loss due to absorption and scattering (fixed loss).

Lf = Modulation (message) frequency dependent loss.

The modulation frequency dependent loss is given by:

(7)

Information Rate


Example: Suppose f = f3-dB. Compute the loss.

Example: Suppose f << f3-dB. Compute the loss.

The equation predicts no modulation frequency loss for modulation frequencies well below the 3-dB frequency.

Information Rate


Example: Suppose f = 0.1 f3-dB. Compute the loss.

Maximum frequency length product is calculated from (5) as follows:

3 2dB

Lf L

(8)

(3.16)

3

1 0.5

2dBf (5)

3

1

2dBf L

L

Information Rate


Find the frequency at which Lf = 1.5 dB.

Use (7) for Lf

2

3

ln 2

1.5 10lg dB

f

fe

2

3

ln 2

10lg dB

f

ffL e

(7)

1.5 30.71dB dBf f

Solving for the frequency at which the loss is 1.5

dB, we obtain

Information Rate


Now consider the photodetection circuit:

RL

P

Photodetector

Optical Power

i

P = incident optic power

i = P detector output current

= detector responsivity (A/w)

The electrical power in the load resistor RL is:

222

2

)( PRPRP

iRP

LLe

Le

Information Rate


Consider two optical power levels P1 and P2 and their corresponding electrical power levels Pe1 and Pe2.

2electrical opticaldB dB

Information Rate


Examples:

•A loss of 3 dB in optical power yields a loss of 6

dB in the corresponding electrical power.

•A loss of 1.5 dB in optical power yields a loss of 3

dB in the corresponding electrical power.

Information Rate


We found that the modulation frequency at which

the optical loss is 1.5 dB was:

3 ( )

0.35dB electricalf L

L

Electrical 3-dB bandwidth length product is:

(3.19)

(3.18)

Information Rate


t

tp

Consider a Return-to-Zero (RZ) digital signal.

Power 1 1 1 0 1

0 T 2T 3T 4T 5T 6T 7T

Each bit is allotted a time T.

tp = T/2 pulse width

R = 1/T data rate, b/s

Information Rate


Spectrum of the RZ Signal

Most of the signal power lies below 1/T Hz, so the required transmission bandwidth by a system is:

RT

BRZ 1

FrequencyptTT

121

PowerSpectralDensity(Watts/Hz)

0

Information Rate


If the system passes this band of frequencies the

pulses will be recognizable. To be conservative,

use the 3-dB electrical bandwidth.

(3.20)0.35

RZR L

L

The RZ rate length product is then:

Information Rate


We obtain the same result by allowing a pulse

spread of 70% of the initial pulse duration.

As on the preceding slide.

Information Rate


Consider the Non-Return-to-Zero (NRZ) digital signal.

t

tp

1 1 1 0 1

T 2T 3T 4T 5T 6T 7T0

Spectrum of the (NRZ) Signal

Required transmission bandwidth:22

1 R

TBNRZ

FrequencyptTT

11

2

1

PowerSpectralDensity(Watts/Hz)

0

Information Rate


The allowed data rate is:

Use the electrical 3-dB bandwidth:

0.7NRZR L

L

NRZ rate length product is:

(3.21)

Comparing the results for the RZ and NRZ data rates:

Information Rate


BANDWIDTH DATA RATE SUMMARY

Information Rate


3.3 Polarization

Linearly polarized: An electric field points in just one direction, it always

points along a single line.a. Linearly polarized in x direction and traveling in the

z direction.b. linearly polarized in y direction and traveling in the

z direction.

yz

xE

v

yz

xE

v

(a) (b)


1. The two orthogonal linear polarizations are the

plane wave modes of an unbounded media.

2. They can exist simultaneously.

3. The actual polarization is determined by the

polarization of the light source and by other

polarization sensitive components in the optical

system.

Polarization


If the direction of electric field E varies randomly (as

shown) the wave is unpolarized.

Most fibers depolarize the input light. Only special

fibers maintain the light polarization.

x

y

E

Polarization


L

镜子

镜子

A

1 、让任意一个光波从左边的镜子传向右边的镜子，如 A 图所示。绿波在右边的镜子处发生发射，因此这个波经历了一次 180 度相移。从 A 图我们可以看出，这个波在其相位上发生了中断，在这里应该是不可能的，也就是说，这个谐振器不支持这个波。

L

镜子

镜子

B

2 、在图 B 中，在右边的镜子处，这个波也发生了一个 180 度相移，然后继续传播，在左边的镜子处，同样经历了一个 180 度相移，然后继续传播。因此，图 B 所示的波有着一个稳定的模式，我们称之为驻波

3.4 Resonant Cavities


2

2

L m

mL

Design:The cavity must be an integral number of half wavelengths long to support a wave.

The wavelength is that in the medium filling the cavity.

(3.22)L

Standing-wave pattern in a cavity (m = 4)

Resonant Cavities


m

L2

The resonant wavelengths are:

Ln

cm

L

m

n

cf

n

cf

vf

22

The corresponding resonant frequencies are:

(3.23)

Resonant Cavities


Cavity Resonant frequencies

2c

cf

nL

211 mmmm ffff

This picture shows the longitudinal modes of the cavity.

Frequency

The resonant frequency spacing is:

1

( 1)

2 2 2

c m m

c

f f f

m c mc cf

Ln Ln Ln

(3.25)

Resonant Cavities


2

c c

o

o c o cc

o

f

f

f fc c

The free space wavelength spacing corresponding

to fc is c calculated from:

(3.26)

This equation refers to the free space wavelengths.

Resonant Cavities


Example: Consider an AlGaAs laser cavity.

L = 0.3 mm = 300 m; n = 3.6; o = 0.82 m.

Find the cavity resonant wavelength spacing c.

Resonant Cavities


Example: Suppose the AlGaAs, LD has a spectral width of 2 nm. Determine the number of longitudinal modes in the output.

0.82 m

2 nm

0.82 m

Gain (AlGaAs)

Cavity Resonances

c

Resonant Cavities


2 nm

0.82 m

c

Laser Output

The laser emits 6 longitudinal modes.

A laser emitting only one longitudinal mode is a

single-mode laser.

Resonant Cavities


3.5 Reflection at a Plane Boundary

Consider normal incidence of light at a boundary.

This is referred to as Fresnel Reflection.n1 n2

Incident Wave Transmitted Wave

Reflected Wave

Reflection Coefficient:

reflected electric field

incident electric field

Boundary

1 2

1 2

n n

n n

Reflection Coefficient:

Reflection at a Plane Boundary


2

2

1 2

1 2

power reflectedR

power incident

R

n nR

n n

Define Reflectance R ( 反射比 ) as:

This result is valid for normal incidence.



For air-to-glass, compute the transmitted power.

4% power reflected. 96% power transmitted.

In dB, the transmitted power is:

10 lg (0.96) = -0.177 dB

Typically we round this off to 0.2 dB (omitting the

minus sign). This is called the Fresnel loss.



Consider arbitrary incidence:

Perpendicular Polarization (s) 垂直偏振

n1 n2

r

t

i

Ei

Er

Et



Consider arbitrary incidence:

Parallel Polarization (p)平行偏振

n1 n2

r

t

i

Ei

Er

Et



Plane of Incidence( 入射平面 )

Defined by the normal to the boundary and the ray

direction of the incident beam.

z

x

The xz plane is the plane of incidence in this example.

Incident Boundary



Fresnel’s Law of Reflection

2 2 2 22 1 2 1

2 2 2 22 1 2 1

cos sin

cos sin

i ip

i i

n n n n

n n n n

For parallel polarization, the reflection coefficient:

Note that may be complex.

(3.29)

(3.30)

For perpendicular polarization, the reflection coefficient:


2 2 21 2 1

2 2 21 2 1

cos sin

cos sin

i is

i i

n n n

n n n


Plots of p and s for n1 = 1 (air), n2 = 1.48 (glass)

Parallel (p)

Perpendicular (s)

s

p

Angle of incidence (i)



From equation (3.29) for parallel polarization, we can get total transmission (no reflection) if

2 2 2 22 1 2 1cos sini in n n n

The angle satisfying this equation is the Brewster

angle B. The solution is:2

1

tan B

n

n

Compute B for air-to-glass and glass-to-air:

For n1 = 1, n2 = 1.5

For n1 = 1.5, n2 = 1

For perpendicular polarization there is no Brewster angle. No i s.t. Equ.3.30 = 0.



Antireflection CoatingsWe have just seen that we can transmit a beam from one material to another without reflection under Brewster angle conditions. We can also transmit with no (or very little) reflection by placing a coating between the two materials.

The thickness of the coating is a quarter wavelength. The reflectance R for this configuration is:

n1 n2 n3

/4

221 3 2

221 3 2

n n nR

n n n



Clearly, the reflectance becomes zero if: 2 1 3n n n

A coating that reduces the reflectance is called an

antireflection (AR) coating . 消反射涂覆

Example: Compute the reflectance when a quarter

wavelength of magnesium fluoride ( 氟化镁 n = 1.38) is

coated onto a piece of glass (n = 1.5).

Solution:

The reflectance is:

22

22

1.5 1.380.014

1.5 1.38R

4% → 1.4%



3.6 Critical Angle Reflection

Fresnel’s Law of Reflection

2 2 2 22 1 2 1

2 2 2 22 1 2 1

cos sin

cos sin

i ip

i i

n n n n

n n n n

For parallel polarization:

(3.29)

(3.30)

For perpendicular polarization:

2 2 21 2 1

2 2 21 2 1

cos sin

cos sin

i is

i i

n n n

n n n


From equations (3.29) and (3.30), we find that

1

2sinn

nc

Call the solution c, the critical angle.

c exists only if n1 > n2. That is, travel from a high index to a low index material. This result is valid for both polarizations.

(3.34)

The incident angle satisfying this equation is the angle whose sine is given by:

Critical Angle Reflection


is purely imaginary.

If then

Under this condition, equations (3.29) and (3.30) can

be written in the form:

where A, B, C, and D are real and j is the imaginary

term



Then:

We conclude that there is complete reflection (called

critical angle reflection) for all rays which satisfy the

condition: ci



Consider waves undergoing critical angle reflections:

n1 n2

In region n1 we have a standing wave due to the

interference of the incident and reflected waves.

i c i



In region n2 the electric field is not zero. The

boundary conditions require the electric field to be

continuous at the boundary. The field in n2 termed as

evanescent is a decaying exponentially carrying no

power.

2 2 20 1 2

00

sin

2

k n n

k

where the attenuation coefficient is given by


zE e


Consider a wave where

The decaying wave carries no power in the z-direction.

Evanescent Wave

E

z

Standing Wave

n1 n2Envelope

e-z

i c

At the critical angle,

In this case, there is no decay. The wave penetrates deeply into the second medium.


这里衰减因子和前面提到的衰减系数不同，衰减系数是指功率的实际损耗，这里衰减因子并不具有这样的含义，仅仅指电磁波回到入射区之前，场在第二种介质中要传播多远。


As i increases, increases and the decay becomes

greater. = 0, = c |E|

z

i > c

As i increases from c towards 90o, increases

and the evanescent field penetrates less and less

into the second medium.

0

i

e-z



LIGHTWAVE FUNDAMENTALS

•Pulse spread

•3-dB bandwidths

•Rate-length products

•Reflectance

•Critical angle reflections

上海师范大学电气信息系 chapter 3 lightwave fundamentals....

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