Optical Fiber Technologydrzaidi.seecs.nust.edu.pk/lectures/Lec-5 Fiber (27-03-2007).pdf · Infra-red: primary intrinsic absorption for optical communications Extrinsic: caused by

Photonic NetworkBy Dr. M H Zaidi

Optical Fiber Technology


Numerical Aperture (NA)

What is numerical aperture (NA)? Numerical aperture is the measure of the lightgathering ability of optical fiberThe higher the NA, the larger the core of lightacceptance of the fiber and the easier it is to couple the light signal into the fiberAt the same time, the higher the numerical aperture, the lower the bandwidthThe two specifications must be balanced for optimum performance


Specifying numerical aperture62.5/125 –– 0.275 +- 0.015

Numerical Aperture


Index of RefractionC= 3×108 meters per second, but it is reduced when it passes through matter. The index of refraction n:

υcn =

υspeed of light in a vacuum, 3×108 m/sspeed of light in the given material

f⋅= λυ 100 ≥==n

nλλ

υν

cf =⋅= 00 λυ

wavelength of light in a vacuumwavelength of light in the given material

c0λλ

λ0λ


Index of refraction and speed of light for various materials.

Index of Refraction Speed of Light

Free space (vacuum) 1.0 3×108 m/s

Air at sea level 1.003 2.99×108 m/s

Ice 1.31 2.29×108 m/s

Water 1.33 2.26×108 m/s

Glass (minimum) 1.45 2.07×108 m/s

Glass (maximum) 1.80 1.67×108 m/s

Diamond 2.42 1.24×108 m/s


θ1 : The incident angle (from the surface normal)θ2 : The angle of refracted light (from the surface normal)n1 : index of refraction in the incident mediumn2 : index of refraction in the refracting mediumLight that is not absorbed or refracted will be reflected. The incident ray, the reflected ray, the refracted ray, and the normal to the surface will all lie in the same plane.

2211 sinsin θθ ⋅=⋅ nn

Refraction with Snell's Law


Critical Angle

We want to find the critical case of total internal reflection at the core-cladding boundary. Using Snell’s Law with ϕ2 = 90º, we can find the critical angle ϕCR :

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛==

1

2

1

2 arcsinor,sinnn

nn

CRCR ϕϕ

φ φ

Incident ray Reflected ray

Cladding n2

Air n0

Core n1

Cladding

Unguided ray

θr

θr

φ´

φ2

θi θi

φ2 = 90º if φ = φCR


Since we can relate θr, CR to angle ϕCR by simple geometry, and we can make the approximate n0 = 1, this equation can be simplified:

The negated and shifted sine function is identical to the cosine, and we can relate this cosine to the sine by the trigonometric identity:

this sine is replaced in terms of n1 and n2 :

( ) ⎟⎠⎞

⎜⎝⎛ −⋅=⋅= CR1CR,1CR, 2

sinsinsin ϕπθθ nn ri

( ) ( ) ( )( )2CR1CR1CR1CR, sin1cos2

sinsin ϕϕϕπθ −⋅=⋅=⎟⎠⎞

⎜⎝⎛ −⋅= nnni

( ) NAnnnnni =−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−⋅= 2

221

2

1

21CR, 1sin θ

Numerical Aperture -- Mathematically


For n1 ≈ n2 , we can simplify the numerical aperture calculation:

( ) ( ) ( ) ( )( )

Λ⋅=−

⋅=

−⋅≅−⋅+=

22

2sin

11

211

2112121CR,

nn

nnn

nnnnnnniθ

1

21

nnn −

=ΛFor Δ <<1


Acceptance Angle

θa is the maximum angle to the axis at which light may enter the fiber in order to be propagated, and is often referred to as the acceptance angle for the fiber.NA can be specified in terms of acceptance angle as, NA = no sin θa = (n1

2 – n22)1/2


A silica optical fiber with a core diameter large enough to be considered by ray theory analysis has acore refractive index of 1.50 and a cladding ref. index of 1.47.

Determine:

a) critical angle

b) NA

c) Acceptance angle

Numerical Aperture – Example 2.1


Solution:

a)

θc = sin-1n2/n1 = sin-1 1.47/1.5 = 78.5o

b)

NA = (n12-n2

2)1/2 = (1.52 – 1.472)1/2 = 0.30

c)

θa= sin-1NA = sin-1 0.30 = 17.4o

Numerical Aperture – Example 2.1


For instance, if n1 = 1.5 and Λ = 0.01, then the numerical aperture is 0.212 and the critical angle θ cr,

is about 12.5 degrees.

See also example 2.2 and 2.3

Numerical Aperture -- Example


Loss and Bandwidth -- Attenuation

•Attenuation ranges from 0.1 dB/km (single-mode silica fibers) to over 300 dB/km (plastic fiber)

•There are two reasons for attenuation: Scattering; Absorption A ttenuation (dB /km )

W avelength (nm )

850 nm W indow

1300 nm W indow

O H A bsorptionPeak

1550 nm W indow

2.5

2.0

1.5

1.0

0.5

0.0 800 900 1000 1100 1200 1300 1400 1500 1600 1700

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

1

210log10

PPAttenuation (dB) =


Loss and Bandwidth

Loss or attenuation is a limiting parameter in fiber optic systemsFiber optic transmission systems became competitive with electrical transmission lines only when losses were reduced to allow signal transmission over distances greater than 10 kmFiber attenuation can be described by the general relation:

Pout= Pin– α L

where α is the power attenuation coefficient per unit length


Loss and Bandwidth

Attenuation is conveniently expressed in terms of dB/km

Power is often expressed in dBm (dBm is dB from 1mW)

( )

( ) ( )

α

α

α

α

34.4

log10

log10

log10

10

10

10

=

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−

eLL

PeP

L

PP

LkmdB

in

Lin

in

out

dBmmWmWmWP 10

110log1010 10 =⎟⎟

⎠

⎞⎜⎜⎝

⎛==

mWmWdBmP 50110127 1027

=⎟⎟⎠

⎞⎜⎜⎝

⎛==


Loss and Bandwidth

Example: 10mW of power is launched into an optical fiber that has an attenuation of α=0.6 dB/km. What is the received power after traveling a distance of 100 km?

Initial power is: Pin = 10 dBmReceived power is: Pout= Pin

– α L =10 dBm – (0.6)(100)

= -50 dBm

( ) nWmWPout 10110 1050 == −


Loss and Bandwidth

Example: 8mW of power is launched into an optical fiber that has an attenuation of α=0.6 dB/km. The received power needs to be -22dBm. What is the maximum transmission distance?

Initial power is: Pin = 10log10(8) = 9 dBmReceived power is: Pout = 1mW 10-2.2 = 6.3 μWPout - Pin = 9dBm - (-22dBm) = 31dB = 0.6 LL=51.7 km


Causes of AttenuationAttenuation, or losses, in a fiber link come from a variety of sources

Bending losses Absorption

Atomic AbsorptionScattering

Rayleigh ScatteringMie-ScatteringBrillouin Scattering


Absorption

The portion of attenuation resulting from the conversion of optical power into another energy form, such as heat.Every material absorbs some light energy The amount of absorption can vary greatly with wavelength It depends very strongly on the composition of a substance


Absorption is uniform

The same amount of the same material always absorbs the same fraction of light at the same wavelength.

The total amount of material the light passes through

Material absorbs the same fraction of the light for each unit length

Absorption is cumulative


Atomic Absorption

The atoms of any material are capable of absorbing specific wavelengths of light.because of their electron orbital

structure. As light passes along an optical fibre. more and more light is absorbed by the atoms as it continues on its path


Intrinsic Absorption

is caused by basic fiber-material properties. Intrinsic absorption sets the minimal level of absorption.


Extrinsic Absorption. is caused by impurities introduced into the fiber material. Extrinsic absorption also occurs when hydroxyl ions (OH-) are introduced into the fiber.Water in silica glass forms (Si-OH) bond



Material Absorption

Material absorptionIntrinsic: caused by atomic resonance of the fiber material

Ultra-violetInfra-red: primary intrinsic absorption for optical communications

Extrinsic: caused by atomic absorptions of external particles in the fiber

Primarily caused by the O-H bond in water that has absorption peaks at λ=2.8, 1.4, 0.93, 0.7 μmInteraction between O-H bond and SiO2 glass at λ=1.24 μmThe most important absorption peaks are at λ=1.4 μm and 1.24 μm


Scattering


Scattering

The inhomogeneities of the refractive index of the media are responsible for this phenomena. Light traveling through the fiber interacts with the density areas.Light is then partially scattered in all directions.

The interaction of light with density fluctuations within a fiber


Types of Scattering

Rayleigh ScatteringMie-ScatteringBrillouin Scattering Raman Scattering


Rayleigh Scattering

Is the scattering of light by particles smaller than the wavelength of the light

Occurs when the size of the density fluctuation (fiber defect) is less than one-tenth of the operating wavelength of light. is more effective at short wavelengths

Therefore the light scattered down to the earth at a large anglewith respect to the direction of the sun's light is predominantly in the blue end of the spectrum.


intensity of the scattered light isinversely proportional to the fourth power of the wavelength

http://www.wikipedia.org/wiki/Inversely_proportional


Mie Scattering

If the size of the defect is greater than one-tenth of the wavelength of light, the scattering mechanism is called Mie scattering. Mie scattering, caused by these large defects in the fiber core.scatters light out of the fiber core. However, in

commercial fibers, the effects of Mie scattering are insignificant


Rayleigh and Mie Scattering


Brillouin scattering

spontaneous Brillouin scattering

simulated Brillouin scattering


Spontaneous Brillouin scattering

Scattering of light through Index variations induced by the pressure differences of an acoustic wave traveling through a transparent material.

spontaneous Brillouin scattering, can also be described using the quantum physics: a photon from a pump lightwave is transformed in a new Stokes photon of lower frequency and a new phonon adding to the acoustic wave.


Absorption and Scattering Loss


External LossesBending loss

Radiation loss at bends in the optical fiberInsignificant unless R<1mmLarger radius of curvature becomes more significant if there are accumulated bending losses over a long distance

Coupling and splicing lossMisalignment of core centersTiltAir gapsEnd face reflectionsMode mismatches


BENDING LOSSES


BENDING RADIUS The bend radius that causes loss due to light leaking from the core. When you exceed the minimum bend radius, your signal strength will drop.

Typical radius is three to five inches.


Microbends

Small microscopic bends

Microbend loss increases attenuation because low-order modes become coupled with high-order modes that are naturally lossy

Loss caused by microbending can still occur even if the fiber iscabled correctly


Macrobend losses

Radius of curvature is large compared to the fiber diameter. During installation, if fibers are bent too sharply, macrobend losses will occur


Loss on Standard Optical Fiber

Wavelength SMF28 62.5/125

850 nm 1.8 dB/km 2.72 dB/km

1380 nm 0.50 dB/km 0.92 dB/km1300 nm 0.35 dB/km 0.52 dB/km

1550 nm 0.19 dB/km 0.29 dB/km


Indoor/Outdoor cables


Dispersion

Dispersive medium: velocity of propagation depends on frequencyDispersion causes temporal pulse spreading

Pulse overlap results in indistinguishable dataInter symbol interference (ISI)

Dispersion is related to the velocity of the pulse


Material Dispersion

Since optical sources do not emit just a single frequency but a band of frequencies, then there may be propagation delay differences between the different spectral components of the transmitted signal. The delay differences may be caused by material dispersion and waveguide dispersion.For a source with rms spectral width σλ and mean wavelength λ, the rms pulse broadening due to material dispersion σm is given by

212| |m

L d nc dλσσ λ

λ


Material Dispersion

212

1 | |md d nML d c d

τ λλ λ

= =

22 1

2| ( ) |d nd

λλ

212| |d n

dλ

The Material Dispersion for optical fibers is sometimes quoted as a value for

It may be given in terms of a material dispersion parameter M defined as:

or simply

expressed in units of ps nm -1 km -1

Where m is the pulse delay due to material dispersionτ


Example

A glass fiber exhibits material dispersion given by 2

2 12| ( ) |d n

dλ

λof 0.025. Determine the material dispersion parameter at a wavelength of 0.85 μm, and estimate the rms pulse broadening per kilometer for a good LED source with an rms spectral width of 20nm at this wavelength.


Solution

21 12 2

1| | | |dn dnMc d c dλ λ

λ λ λ= =

1 15

0.0252.998 10 850

snm kmx x

− −= 1 198.1psnm km− −=The material dispersion parameter may be obtained

The rms pulse broadening is given as2

12| |m

L d nc dλσσ λ

λ

Therefore in terms of material dispersion parameter M

m LMλσ σ

Hence, the rms pulse broadning per kilometer due to material dispersion

12 1(1 ) 20 1 98.1 10 1.96m km x x x nskmσ − −= =


Example 2

Estimate the rms pulse broadening per kilometer for the fiber in the above example when the optical source used is an injection laser with a relative spectral width σλ/λ of 0.0012 at a wavelength of 0.85 μm


Solution

The rms spectral width may be obtained from the relative spectral width byσλ= 0.0012 λ = 0.0012 x 0.85 x 10-6

= 1.02nmThe rms pulse broadening in terms of material dispersion parameter is given by

m LMλσ σσm = 1.02 x 1x 98.1 x 10-12 = 0.10 ns km-1

Hence the rms pulse broadening is reduced by a factor of 20 compared with the LED source in the previous example


Polarization mode dispersion (PMD) is another complex optical effect that can occur in single-mode optical fibers.

Single-mode fibers support two perpendicular polarizations of the original transmitted signal.

If a were perfectly round and free from all stresses, both polarization modes would propagate at exactly the same speed, resulting in zero PMD.

Polarization mode Dispersion (PMD)


However, practical fibers are not perfect, thus, the two perpendicular polarizations may travel at different speeds and, consequently, arrive at the end of the fiber at different times.

The fiber is said to have a fast axis, and a slow axis. The difference in arrival times, normalized with length, is known as PMD (ps/km0.5).

Polarization mode dispersion (PMD)


Polarization Mode Dispersion (PMD)

Polarization mode dispersion is an inherent property of all optical media. It is caused by the difference in the propagation velocities of light in the orthogonal principal polarization states of the transmission medium. The net effect is that if an optical pulse contains both polarization components, then the different polarization components will travel at different speeds and arrive at different times, smearing the received optical signal.



NUST Institute Of Information Technology



Non-flammable No fire hazard

Low power saves provider and money.


AssignmentThe material dispersion parameter for a glass fiber is 20 ps nm-1

km-1 at a wavelength of 1.5 μm. Estimate the pulse broadening due to material dispersion within the fiber when a light is launched from an injection laser source with a peak wavelength of 1.5 μm and an rms spectral width of 2nm into a 30 km length of fiber.The material distribution in an optical fiber defined |d2n1/dλ2| is 4.0 x 10-2 μm-2. Estimate the pulse broadening per kilometer due to material dispersion within the fiber when it is illuminated with an LED source with a peak wavelength of 0.9 μm and an rmsspectral width of 45 nm.Questions 2.2, 2.4, 2.5

Ramaswami

Documents

Optical Fiber Technologydrzaidi.seecs.nust.edu.pk/lectures/Lec-5 Fiber (27-03-2007).pdf · Infra-red: primary intrinsic absorption for optical communications Extrinsic: caused by