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Optical Transmission system

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Point-to-point Link

• It is the simplest optical link.• It has a transmitter at one end, a receiver on the other end.• An optical fiber connected between these two.• This places the least demand of the system.• It is the base of examining complex systems.

Figure 1. Simple Point-to-point link

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System Requirements

• The desired transmission distance.• The data rate/ bandwidth.• Acceptable bit error rate (BER).

– Based on above designers choose the component to ensure the desired performance level.

– This performance should be maintained over the expected lifetime of the system.

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System Considerations

• First the operation wavelength is selected.• Then the components in this wavelength region are taken.• In general procedure first the photodetector is taken.• Then we choose the optical source.• And see how far data can be transmitted over a particular fiber

without an amplifier to boost up the power level.

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Selecting the Photodetector

• The main feature is the minimum optical power that must fall on detector to satisfy the acceptable BER at specified data rate.

• Type of detector– PIN: Simpler, thermally stable, low bias voltage (5V or

less), less expensive, fast response but less sensitive.– APD: Highly sensitivity, complex, high bias voltage (40V

or more) and expensive.

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• Sensitivity. • Speed of response.• Responsivity• Operating wavelength and spectral selectivity

PIN or Avalanche Photodiode

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Selecting the Optical Source

• Emission wavelength• Spectral line width• Output power• Stability• Emission pattern• Effective radiating area

LED or LASER

LEDLASER

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LED & LASER Bit rate Distance Product

Wavelength LED Systems LASER Systems.

800-900 nm (Typically Multimode Fiber)

150 Mb/s.km 2500 Mb/s.km

1300 nm (Lowest dispersion)

1500 Mb/s.km 25 Gb/s.km(InGaAsP Laser)

1550 nm (Lowest Attenuation)

1200 Mb/s.km Up to 500 Gb/s.km

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Combination of Sources and Fibers For Different Link Capacity And Distance

1-10m 10m-.1km .1-1km 1-3km 3-10km 10-50km 50-100km >100km

LD 10k

SLED MM 10-100K

MM 100K-1M

LD GI 1-10M

LED 10-50M

GI LD 50-500M

LD LD SM 500M-1G

MM GI >1G

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• Lasers couple 10 to 15 dB more power into the fiber then an LED.

• Hence greater repeater less transmission is possible.• Laser diodes are expensive then LEDs.• Laser transmitter circuitry is also complex.• Lasers are used in long distance single mode operation.• LEDs are used in comparatively shorter distances and

multimode operation.

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Selecting the Optical Fiber

• Core size• Refractive index profile• Attenuation characteristics• Dispersion performance.• Numerical aperture.

Multimode or single mode

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Link Power Budget

• Optical power received at the photodetector depends on the – Amount of light coupled in to the fiber.– Losses occurring in the fiber.

• The optical link power budget in a fiber-optic communication link is the allocation of available optical power (launched into a given fiber by a given source) among various loss-producing mechanisms such as – launch coupling loss. – fiber attenuation.– splice losses.– connector losses.

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• Link power budget is derived from the sequential loss contribution of each element in the link.

• In order to ensure that adequate signal strength (optical power) is available at the receiver.

• Each of these loss elements are usually expressed in decibels (dB).

• Optical fiber communication power levels are generally expressed in dBm.

in

outdB P

PLoss 10log10

31010 101log10

1log10

P

mWPdBms

30 ss dBdBm

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Receiver sensitivities Vs bit rate

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Optical power-loss model

ystem MarginT s R c sp fP P P ml nl L S

: Total loss; : Source power; : Rx sensitivity connectors; splices

T s RP P Pm n

System margin is taken to incorporate some unexpected losses and future aspects. General value ranges in 6 to 8 dB.

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Example

• The Si pin photodiode has a receiver sensitivity of about –42 dBm (63.1 nW) at the required bandwidth of 20 Mb/s and in the wavelength range 800 – 900 nm. A GaAlAs LED is used, Pcoupled = 50 μW into a multi-mode fiber with a core diameter of 50 μm. The loss is Lc = 0.5 db per connector. Splices will be required for each kilometer of fiber with loss of 0.2 dB each. The attenuation loss for the fiber which was previously installed is αF = 3.5 dB/km. Determine the available margin for a 6 km long link.

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Solution

Input optical power -13 dBm

Receiver sensitivity -42 dBm

Available power 29 dB

Fiber Loss (3.5x6) 21 dB

Connector Losses (0.5x2) 1 dB

Splice losses (0.2x5) 1 dB

Total losses 23 dB

Available margin 6 dB

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Example link-loss budget

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Rise Time Budget• It is a method of determining the dispersion limitation of an

optical fiber link.• The total rise time of the link (tsys) is equal to the root-mean-

square of the rise times from each contributor ti

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1

2

N

iisys tt

• Mainly four elements limits the system bandwidth– Transmitter rise time (ttx){generally known to the designer}– Group Velocity Dispersion (tGVD)– Modal dispersion rise time (tmod)– Receiver rise time (trx)

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• The receiver rise time results from the photodetector response and the 3-dB bandwidth of the receiver front end.

• Receiver front end can be modeled by a first-order low pass filter having a step response

)()2exp(1)( tutBtg rx

• Brx is the 3-dB bandwidth of the receiver in MHz.• u(t) is the unit step function.• trx is the time taken by receiver signal to rise from 10 to 90%

(receiver front end rise time).

nsB

trx

rx350

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• The signal rise time due to the group velocity dispersion (GVD) over a length L, for a source with spectral width σλ can be given as

LDtGVD

• Fibers are seldom joint-less, normally they are series of connected fibers.

• It is experimentally observed that total BW is a function of the order in which fibers are joined.

• A verity of empirical expressions for modal dispersion have been developed.

• The most reasonable approximation BM in a link of length L is given by-

qM LB

LB 0)(

B0 is the BW of a 1-km length of link

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• Here q is a parameter ranges between 0.5 to 1.0.• Here we assume that the optical power emerging from the

fiber has the gaussian response

)2

exp(21)( 2

2

ttg

The Fourier transform of this function is

)2

exp(21)(

22

G

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• Let t1/2 required for pulse to reach its half-maximum power.

2ln221 t

• If we define the full width of pulse at its half-maximum value then

2ln222 2/1 ttFWHM

• The 3-dB optical BW is the modulation frequency at which the received power has fallen to half of its ‘0’ frequency value.

2/)0()( 2/1 gtg

2ln22FWHMt

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• From Fourier transform of optical power, power at 3-dB frequency will be

2/)0()( 3 GG dB

2exp2

)()0( 22

3

3

dB

dBGG

2ln2

3 dB

22ln2

3 dBf

FWHMdBdB t

Bf 44.033

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• tFWHM is nothing but the rise time resulting form modal dispersion.

• The B3dB is the BW of the link of length L.

0mod

44.044.0B

LB

tq

M

• If B0 is in MHz then tmod in ns is given as

0mod

440B

Ltq

ns

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• Hence the total rise time of the system is given by

21

2

0

22 350440

rx

q

txsys BBLLDtt

Total rise time of a digital link should not exceed 70% for a NRZ code bit period35% of a RZ code bit period

codeNRZfortt bsys 7.0(max)

codeRZfortt bsys 35.0(max)

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Example

A Laser diode of 1550 nm range, with its drive circuit has a rise time of 0.025 ns and spectral width of 0.1 nm . It shows an average dispersion of 2 ps/nm.km, over a 60 km link. The APD receiver has a 2.5 GHz BW. If single mode operation and NRZ code used for communication, comment about the performance of the system.

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Example

Let us take typical parameters for a link.Data rate = 1 GHz.DFB Laser spectral width = 0.1nmSM fiber dispersion at 1550nm = -20 ps/km/nm = -0.02 ns/km/nmRise time of the receiver = 0.1 nsecRise time of the transmitter = 0.1nsecFiber loss = 0.4dB/kmTransmitter power -3 dBmMin Detectable power -40 dBmNeglect splice and connector losses.

Determine the distance at which the repeater has to be installed, for SM and NRZ codes.

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Transmission Distance for 800 MHz-km Multimode Fiber at 800 nm, (BER=10-9)