OPTICAL FIBRE COMMUNICATION

OCT/NOV 2004 (Regular/Supplementary)

OPTICAL FIBRE COMMUNICATION

OCT/NOV 2004 (Regular/Supplementary)

PART-A

1. What is the difference between Meridinal ray and Skew ray? Explain.2. Define Total Internal Reflection? What is its significance?3. What are the two important characteristics of the fibre that determine fibre

information carrying capacity? Explain them,4. What are SE LED and EE LED? Draw their schematic sketches. Explain them.5. What is a fibre pigtail? Explain its need.6. Name the different types of optical fibre connectors and splicing techniques.7. Explain the duplex operation of optical WDM.8. Enumerate the potential applications of optical fibre communication systems.9. Write notes on 1) LOS

2) OTH systems

10. Explain the need of microwave repeaters.

PART-B

11. Broadly compare the characteristics of MMSIF and MMGIF. Explain the construction and advantages of MMGIF. Or

12. Derive the equations in cylindrical coordinates for multimode step index glass fibres, starting from Maxwell’s equations.

13. Write technical notes on 1) Attenuation2) Scattering in glass fibres

Or

14. a) List the requirements of the ideal optical detector,b) With a neat sketch explain the detection principle of a PIN photo detector.

15. Explain in detail the need for optical splicing. Explain the principles of different splicing units.Or

16. Show that power coupling efficiency of source to glass fibre can be improved by using lensing schemes. Derive supporting equations.

17. a) Compare plastic fibres with glass fibres.

b) Explain the various modulation formats for optical communication.Or

18. Derive an expression for the rms pulse broadening in multimode fiber due to intermodal dispersion

19. a) Explain the basic principles of a typical microwave link.

b) Explain the significance of C band or X band microwave frequencies for microwave communication.

Or

20. a) Draw the neat block schematics of microwave terminal transmitter and receiver and explain their basic principle of operation,

b) Explain the construction of a microwave link.

Q). what is meant by fiber pigtail? What is the need for it?

A) A fiber pigtail is a single short ,usually unbuffered , optical fiber that has an optical connector on one end and a length of exposed fiber on the other end .the end of the fiber is stripped and fusion spliced to a single fiber of a multi fiber trunk to break out the multi fiber cable into its component fibers for connection to the end equipments . Pigtails can have female connectors and be mounted in a patch panel, often in pairs although single fiber solutions exist , to allow them to be connected to end ports or

other fibers runs with patch fibers .alternatively they can have male connectors and plug directly to the trans-receiver

(Q): Explain in detail the need for optical splicing? Explain the principles of different optical splicing units?

ANSWER

The continuous of an optical fiber along a communication link is determined by three factors

(1). the continuous length of fiber that can be produced by prevalent manufacturing method

(2). the length of the cable that can be produced and installed as a continuous section along the link

(3). the cable length between the repeaters

This uninterrupted length of optical fiber along a link therefore is not more than 10km.Thus for establishing long haul transmission links, optical fibers are required to be connected. The fiber to fiber connection may be achieved by two ways.

(1) Splices which are permanent joint between two fibers

(2)Connectors which are demountable joints

The major consideration in making these connections is the optical loss associated with them .The loss mechanisms associated with fiber to fiber connections may be grouped into two categories

(1)Losses due to extrinsic parameters

(2)Losses due to intrinsic parameters

FIBER SPLICES

A fiber splice is a permanent joint formed between two optical fibers .Splicing is required

(a) When the length of the system span is more than the manufactured cable length

(b) When the cable is broken and needs to be repaired

The primary objective of splicing is to establish transmission continuity in the fiber optic link. This can be done in two ways, namely through

(a) Fusion splices

(b) Mechanical splices

In order to achieve a low loss splice, it is essential for the fiber ends to be smooth, flat and perpendicular to the core axes. This is normally achieved using a cleaving tool. This technique is called scribe and break or score and break. It involves scoring the fiber under tension with a cleaving tool .This generates a crack in the fiber surface that propagates in the transverse direction and a flat fiber end is produced.

FUSION SPLICES

A good quality permanent joint may be obtained by fusion or welding the prepared fiber ends. A widely used heating source for fusion is the electric arc. The prepared fiber ends are placed in a precision alignment jig. The alignment is done with the help of an inspection microscope. After the initial setting, a short arc discharge is applied to fire polish the fiber ends. This removes any defects due to imperfect cleaving. In the final step, the two ends are pressed together and fused with a stronger arc, thus producing a fusion splice. A possible drawback of such a splicing mechanism is that the heat produced by the welding arc may weaken the fiber in the vicinity of the splice.

MECHANICAL SPLICES

There are several mechanical techniques of splicing fibers. These normally use appropriate fixtures for aligning the fibers and holding them together. A popular technique, known as the snug tube splice, uses a glass or ceramic capillary with an inner diameter just large to accommodate the optical fibers. The prepared fiber ends are gently inserted into the capillary and a transparent adhesive is injected through a transverse hole. The adhesive ensures both mechanical bonding and index matching. A stable low loss splice may be obtained in this way but it posses stringent limits on the capillary diameters.

A slightly different technique uses an oversized metallic capillary of square cross section. The capillary is first filled with transparent adhesive, after which the prepared fiber ends are inserted into it. The two fiber ends are forced against one of the four inner corners of the capillary.

Other techniques of mechanical splicing normally employ v-grooves for securing optical fibers. The simplest technique uses an open v-groove, into which the prepared fiber ends are placed. The splice is accomplished with the aid of epoxy resin .it is also possible to obtain a suitable groove by placing two precision pins close to each other. The fibers may be placed in the cusp and the transparent adhesive ensures bonding as well as index matching and a flat spring on the top applies pressure ensuring that fibers remain in their positions. Such a groove is called spring groove.

There is yet another technique that utilizes the v- groove principle to realize what is known as an elastomeric splice. In this method, the prepared fiber ends are sandwiched between two elastomeric internal parts, one of which contains a v-groove. An outer sleeve holds these two parts compressed so that the fibers are held tightly in alignment. Index matching gel is employed to improve its performance. Originally this technique was developed for coupling multimode fibers, but it can also be used for single mode fibers as well as fibers with different code diameters.

Splicing with most of these techniques if properly carried out, results in splice loss of about 0.1dB for multimode fibers. Some of these can also be used for splicing single mode fibers.

MULTIPLE SPLICES

For ribbon cables containing linear arrays of fibers, the following technique has been used. In this method, the fiber ends are individually prepared, and then placed in a grooved substrate. Adhesive is then used for bonding an index matching. A cover plate retains the fibers in their position and also maintains mechanical stability.

By, Shruti Rajan Babu (#39)

Q).What are SE LED and EE LED? Draw their schematic sketches and explain them?

Barrus etched-well surface emitting LED

The Barrus etched-well LED emits light in all directions. The etched well helps to concentrate the emitted light into a small area, also domed lenses can be placed over emitting surface to direct light to smaller area .This LED is efficient than standard surface emitter and they allow more power to couple in to optical fiber but they are more difficult and expensive to manipulate.

Edge-emitting LED

These LED’s emits more directional light pattern than the surface emitting LED. The construction is similar to that of planar and barrus diodes expect that the emitting surface is a strip rather than a confined circular area. The light emitted from the active strip and from an elliptical beam pattern surface emitting LED’s are commonly used than edge-emitters because they emit more light. The typical emitting region for EE-LED is 8-10µm thick and 150µm wide. This LED is suitable to enhance optical coupling 50µm core fiber EE-LED is more expensive than S-LED.

2 A) List the requirement of an ideal optical detector?

1 .High fidelityTo reproduce the received signal waveform with fidelity, for analogy transmission the response of the photo-detector must be linear with regard to the optical signal over a wide range.

2. Large electrical response to the received optical signalThe photo-detector should produce a maximum electrical signal for a given amount of optical power; i.e. the quantum efficiency should be high.

3. Short response time to obtain a suitable bandwidthPresent systems extend into the hundreds of megahertz. However, it is apparent that future systems (single-mode fiber) will operate it in the gigahertz range, and possibly above.

4. A minimum noise introduced by the detectorDark currents, leakage currents and shunt conductance must be low. Also the gain mechanism within either the detector or associated circuitry must be of low noise.

5. Stability of performance characteristics

Ideally, the performance characteristics of the detector should be independent of changes in ambient conditions. However, the detector currently favoured (photo diodes) have characteristics (sensitivity, noise, internal gain) which vary with temperature, and therefore compensation for temperature effects is often necessary.

6. Small size

The physical size of the detector must be small for efficient coupling to the fiber and to allow easy packaging with the following electronics.

7. Low bias voltagesIdeally the detector should not require excessive bias voltages or currents.

8. High reliabilityThe detector must be capable of continuous stable operation at room temperature for many years.

9. Low cost

Economic considerations often of prime importance in any large scale communication system application.

10. High Sensitivity at operating wavelengths.

B) With neat sketches explain the detection principle of a p-i-n photo-detector? In order to allow low operation at longer wave lengths where the light penetrates more deeply into the semiconductor material a wider depletion region is necessary. To achieve this n-type material is doped so lightly so that it can be considered intrinsic, and to make a low resistance contact highly doped n-type (n+) layer is added. This creates a p-i-n (PIN) structure, where absorption takes place in the depletion region.Three main factors that determine speed of response are:1) Drift time of carriers through the depletion layer- tdrift=w/vd

2) Diffusion time of carriers generated outside the depletion region-tdiff=d2/2Dc

3) Time constant incurred by the capacitance of the photo-diode with its load-Cj=ɛsA/w

By Sruthi Loy (#33)

Q). Define total internal reflection. What is its significance?

When a ray of light travels from an optically denser medium to an optically rarer

medium the angle of transmission will be greater than the angle of incidence so that the

refracted ray bends away from the normal. When the angle of incidence is increased the

angle of refraction also increases. For a particular value of angle of incidence the angle of

refraction is 90° then the transmitted ray will be parallel to the boundary. This value of

angle of incidence for which the angle of transmission is 90° is termed as the critical

angle θc. When the angle of incidence is further increased beyond θc the value of refracted

angle increases beyond 90° and hence the refracted ray comes back into the original

medium then the ray is said to be totally internally reflected.

Conditions for total internal reflection:

1. Light must travel from an optically denser medium to an optically rarer medium

2. The angle of incidence in the denser medium must be greater than critical angle of the

denser medium.

The transmission of optical energy through an optical fiber is achieved by using the principle

of total internal reflection, the condition is given by sinθc = n2/n1 where n1 = core refractive

index & n2 = cladding refractive index. When the angle of incidence is greater than critical

angle the ray undergoes total internal reflection. Hence by using total internal reflection

optical energy can be transmitted within an optical fiber without loss of information.

Q) Derive equations in cylindrical coordinates for a multimode step index glass fiber starting from Maxwell’s equations.

The dielectric cylinder which models the step index optical fiber can act as a

waveguide. The mechanism is that a total internal reflection at the cylindrical

boundary with the cladding results in a standing wave across the core and a decaying

field in the cladding; this entire field pattern propagates along the fiber. The standing

wave developed in the core is not sinusoidal but has the form of an oscillatory Bessel

function due to the circular geometry of the fiber. Because of the curvature of the

boundary transverse modes of the cylindrical fiber are mixed into a set of hybrid

modes that have components of electric and magnetic fields in the axial direction. For

the analysis of the optical waveguide we have to consider Maxwell’s equations which

give relationship between electric and magnetic field. Maxwell’s equations can be

written as:

Δ X E = - ∂B /∂t

Δ. D = 0

Δ X H = ∂D /∂t

Δ. B = 0

B = μH & D = εE

Δ x (Δ x E) = Δ X (- ∂B /∂t) = Δ x (- μ ∂H /∂t) = - μ ∂ /∂t (Δ x H)

= - μ ∂ /∂t ( ∂D /∂t ) = - μ ∂ / ∂t ( ε ∂E /∂t ) = - μ ε ∂2E / ∂t2

Δ x ( Δ x E ) = - μ ε ∂2E / ∂t2

We know that Δ x ( Δ x E ) = Δ ( Δ.E ) – Δ2 E

From Maxwell’s equations we have,

Δ x (Δ x E) = Δ (Δ.E) – Δ2 E = - Δ2 E

Δ2 E = μ ε ∂2E / ∂t2

Again we have, Δ2 H = μ ε ∂2 / ∂t2

The actual step index fiber has a core radius of a1 and a refractive index μ1 and its cladding

has a finite outer radius a2 and refractive index μ2 with another medium air or some protective

material beyond the cladding.

We assume a solution of the equation in the form:

Hz = gf1 (r) f2 (θ) f3 (z) f4 (t)

Where g = a constant

The time and z dependent form is like f3 (z) f4 (t) = e j (ωt – βz)

The wave is sinusoidal in time and travels along the z – direction. So f 2 (θ) will be a

periodic function of the form f2 (θ) =

Now we get

∂2 f1 (r) / ∂r2 + 1/r ∂ / ∂r f1 (r) + ( p2 – ν2/r ) f1 = 0

The equation is differential equation for Bessel function. Similar equation can also be

obtained for Ez. Inside the core the solutions are Bessel functions of the first kind of

order n. then Ez and H z are expressed by the relations for r<a1

Ez = C Jn (ur) e jωt – βz

Hz = C Jn (ur) e jωt – βz

For the region outside the core we should imply the modified Bessel’s functions of the

second kind Kn (Wr)

Ez = A Kn (Wr) e iνθ e jωt – βz

Hz = G Kn (Wr) e iνθ e jωt – βz

These are spiralling waves along the axis but their transverse non – uniformity

gradually dies out with radial distance away from the core cladding interface. It is

evident from the definition of the modified Bessel function that Kn (Wr) tends to e –Wr

when Wr tends to infinity. Moreover Kn(Wr) must be zero for very high value of r so

W>0. This indicates that β≥K2 and it gives a cut – off condition.

The cut – off condition indicates that it is the point at which a mode is not confined to

the core region. Similarily a second condition on β can be deduced from the behaviour of

Jn(ur). The condition is K1≥β.

In above calculation we have excluded one solution. The solution is C1 Yn(ur). But

C1 is taken as zero to exclude Yn (ur). This is because Yn(ur) tends to infinity as r

tends to zero. This simply indicates that the field strength is infinite on the axis of the

fiber core.

By Shamy Alphons Tom (#31)

Q) What are the two important characteristics of fiber that determine fiber information carrying capacity? Explain them.

A) The two important characteristics of fiber that determine the information capacity of an optical waveguide is usually specified by the bandwidth-distance product in MHz.km and pulse dispersion. For a step index fiber the various distortion effects tend to limit the bandwidth-distance product to about 20MHz.km. In graded-index fibers the radial refractive-index profile can be carefully selected so that the pulse broadening is minimized at a specific operating wavelength. This had led to bandwidth-distance products as high as 2.5GHz.km. The information-carrying capacity can be determined by examining the deformation of short light pulses propagating along the fiber.

Q) Write technical notes on:-

I. AttenuationII. Scattering in glass fibers

A) I) Attenuation:- As light travels along a fiber, its power decreases exponentially with distance. It helps in determining the maximum transmission distance between a transmitter and the receiver or an in-line amplifier. The basic attenuation mechanism in a fiber are absorption, scattering and radiative losses of the optical energy. If P(0) is the optical power in a fiber at the origin, then the power P(z) at a distance z further down the fiber is

P(z) = P(0)e-αpz

αp = 1/zln[P(0)/P(z)]

is the fibre attenuation coefficient. The common procedure is to express the attenuation coefficient in units of decibels per kilometer, denoted by dB/km.

α(dB/km) = 10/zlog[P(0)/P(z)] = 4.343αp(km-1)

ii) Scattering in glass fibers:- scattering losses in glass arises from the microscopic variations in the material density, from compositional fluctuations and from structural inhomogenities or defects occurring during fiber manufacture. Glass is composed of randomly connected network of molecules in which the molecular density is either higher or lower than the average density in the glass. For single component glass the scattering loss at a wavelength λ resulting from density fluctuations can be approximated by

αscat = 8π3/3λ4n8p2kBTfβT

The defects that causes scattering may be in the form of trapped gas bubbles, unreacted starting materials, and crystallized regions in the glass. The losses of multimode fibers are generally higher than those of single mode fibers.

By, Sharon Varghese (#32)

Q.Name the different types of optical fiber connectors and splicing techniques?

Ans:a)Butt joint connectors-It employ a metal,ceramic or moulded plastic ferrule for each fiber and a precision sleeve in to which the ferrule fit.The fiber is epoxied in to a precision hole which has been drilled in to the ferrule.Two types of butt joint connectors are straight sleeve and tapered sleeve mechanisms.

Expanded beam connectors-it employs lenses on the ends of the fibers.These lenses either collimate the light emerging from the transmitting fiber ,or focus the expanded beam on to the core of the receiving fiber

Splicing technique includes fusion splice,the V groove mechanical splice , and the elastic tube splice.Fusion splices are made by thermally bonding together prepared fiber ends.The fiber ends are first prealigned and butted together.The butt joint is then heated with an electric arc or a laser pulse.

In the V groove splicing technique,the prepared fiber ends are first butted together in a V shaped groove.They are then bonded together with an adhesive

In the elastic tube splicing method,the fiber is inserted in a tube made of elastic material.When fibre is inserted,the hole diameter expands so that the elastic material exerts a symmetrical force on the fiber.It allows an accurate and automatic alignment of the axes of two fibers to be joined.

Q.Show that the power coupling efficiency of source to glass fiber can be improved by using lensing schemes.Giving equations

Ans: The optical power launching analysis is based on centering a flat fiber end face directly over the light source as close to it as possible.If the source emitting area is larger than the fiber core area,then the resulting optical power coupled in to the fiber is the maximum that can be achieved.However if the emitting area of the source is smaller than the core area,a miniature lens is placed between the source and the fiber to improve the power coupling efficiency.

The function of microlens is to magnify the emitting area of the source to match exactly the core area of the fiber end face.If the emitting area is increased by a magnification factor M,the solid angle within which optical power is coupled to the fiber from the LED is increased by the same factor.

Several lensing schemes are rounded end fiber,a small glass sphere in contact with with both the fiber and the source,a large spherical lens used to image the source on the core area of the fiber end,a cylindrical lens generally formed from a short section of fiber,a system consisting of a spherical surfaced LED and a spherical ended fiber ,and a taper ended fiber.

For Nonimaging microsphere,Placing the LED close to the lens surface results in a magnification M of the emitting area.This is given by the ratio of the cross sectional area of the lens to that of the emitting area. 2

M = π R2 πr2 = R r

2

With the lens the optic power PL = Ps R r sin2θ

Where Ps is the total output power from the LED without lens

The theoretical coupling efficiency that can be achieved is based on the energy and radiance conservation principle.This efficiency is usually determined by the size of the fiber.For a fiber of radius a and numerical aperture NA,the maximum coupling efficiency is

By, Sibin Koshy (#35)

Q) Explain the duplex operation of critical WDM .

Ans)

There are more sophisticated systems which make a fuller utilization of the transmission capacity of an optical fiber . In standard point-to-point links a single-fiber line has one optical source at its transmitting end and one photodetector at the receiving end . Signals from different light sources require separate and uniquely assigned optical fibres . Since an optical source has a narrow spectral width , this type of transmission makes use of a very narrow portion of the transmission bandwidth capability of a fiber

We also see that many additional spectral operating regions are possible . Ideally, a dramatic increase in the information capacity of a fiber can thus be achieved by the simultaneous transmission of optical signals over the same fiber from many different light sources having properly spaced peak emission wavelengths . By operating each source at a different peak wavelength, the integrity of the independent messages from each source is maintained for subsequent conversion to electric signals at the receiving end . This is the basis of wavelength division multiplexing (WDM). Conceptually , the WDM scheme is the same as frequency division multiplexing(FDM) used in microwave radio and satellite systems . Various WDM techniques have been examined for optical fiber networks .

Two different WDM setups have been shown . In the first one , a unidirectional WDM device is used to combine different signal carrier wavelengths onto a single fibre at one end and to separate them into their corresponding detectors at the other end . A bidirectional WDM scheme is shown in the second figure. This involves sending information in one direction at a wavelength λ1 and simultaneously in the opposite direction at a wavelength λ2 .

Let us first examine the system requirements of the WDM technique. The three basic performance criteria are insertion loss , channel width , and cross talk . Insertion loss defines the amount of power loss that arises in the fiber optic line from the addition of a WDM coupling device . This includes losers occurring at the connection points of the WDM element to the fiber line and any intrinsic losses within the multiplexing element itself. In practice , designers can tolerate insertion losses of a few decibels at each end .

Channel width is the wavelength range that is allocated to a particular optical source. If laser diodes are used , channel widths of several tens of nanometers are required to ensure that no interchannel interference results from source instability(for example, drift of peak operating output wavelength with temperature changes). For LED sources, channel widths that are 10 to 20 times larger are required because of the wider spectral output width of these sources .

Cross talk refers to the amount of signal coupling from one channel to another. The tolerable interchannel cross-talk levels can vary widely depending on the application. However, in general, a -10 dB level is not sufficient whereas a -30 dB level is.

In implementing a unidirectional WDM system , a multiplexer is needed at the transmitting end to combine optical signals from several light sources onto a single optical fiber .

17)a) Compare plastic fibres with glass fibres .

Ans)

Plastic fibres Glass fibres

. made of plastic . made from extremely fine fibres of glass

. used in sensing . used as a reinforcing agent for many polymer

products

. PMMA is the core and fluorinated polymers make up the cladding

. made up of glass cladding and glass core

. more tough material . less tough material

. greater optical signal attenuation in plastic fibres

. less optical signal attenuation in glass fibres

. more durability . less durable

. mechanically more flexible . mechanically less flexible

. high numerical apertures and acceptance angles can be accomodated

. there exists a limit as to the maximum numerical aperture and acceptance angle that can be accomodated

. examples of plastic fibre constructions are :

1) A polystyrene core and a methyl methacrylate cladding to give an NA of 0.60

2) A polymethyl methacrylate core and a cladding made of its copolymer to give an NA of 0.50

. examples of glass fibre constructions are:

1) Halide glass fibres – fluoride glasses are used which have extremely low transmission losses

2) Active glass fibres – rare earth elements are used which give the resulting material new optical and magnetic properties

. permits the use of inexpensive large-area light-emitting diodes which in conjunction with the less expensive plastic fibres , make an economically attractive system

. no such provision provided

. cost is low . relatively expensive

b) Explain the various modulation formats for optical communication.

Ans)

There are three fundamental ways by which information can be sent in a coherent optical transmission system. These are phase shift keying (PSK) , frequency shift keying(FSK) , or amplitude shift keying(ASK) . In binary digital systems a common ASK technique is on-off keying (OOK). To recover the information at the receiver, one can use either homodyne or heterodyne optical detection techniques together with either synchronous

or asynchronous electrical detection. The particular choice of modulation and demodulation methods determines the fundamental receiver sensitivity.

Generally one characterizes the performance of a digital communication system in terms of the bit error rate (BER). The BER depends on the signal-to-noise ratio(S/N) and the probability density function (PDF) at the receiver output (at the input to the comparator). Since for high local-oscillator powers the PDF is Gaussian for both heterodyne and homodyne techniques, the BER depends only on the signal-to-noise ratio. Thus one can describe receiver sensitivity in terms of the S/N available at the receiver output, which is directly proportional to the received optical signal power. Traditionally the receiver sensitivity for various coherent optical detection techniques has been described in terms of the average number of photons required to achieve a 10-9 BER.

Direct detection OOK :

To make a comparison between various detection techniques, let us first look at a direct- detection OOK system. Suppose we send an OOK sequence of 1 and 0 pulses which occur with equal probability. Since the OOK data stream is on only half of the time on the average, the required number of photons per bit of information is half the number required per pulse. Thus if N and 0 electron-hole pairs are created during 1 and 0 pulses ,

respectively , then the average number of photons per bit NP for unity quantum efficiency(η=1) is

NP N + (0)

or , N=2NP . The chance of making an error is

Pr (0) = e-2Np

The above equation implies that about 10 photons per bit are required to get a BER of 10-9 for a direct-detection OOK system.

In practice this fundamental quantum limit is very difficult to achieve for direct-detection receivers. The amplification electronics that follow the photo detector add both thermal noise and shot noise, so that the required power level lies between 13 and 20 dB above the quantum limit.

OOK Homodyne System :

As has already been mentioned, either homodyne or heterotype type receivers can be used with OOK modulation. Let us first look at the homodyne case. When a 0 pulse of duration T is received, the average number N0 of electron-hole pairs created is simply the number generated by the local oscillator; that is

N0 =A2LO T

For a pulse , the average number of electron-hole pairs , N1 , is

N1=(ALO+AS)2=( A2LO + 2ALOAS)T

Where the approximation arises from the condition

A2LO > A2

S

Since the local oscillator output power is much higher than the received signal level , the voltage V seen by the decoder in the receiver during a pulse is

V = N1 – N0 = 2ALOAST

And the associated rms noise n is

n = N10.5

= N00.5

Thus we have BER as

BER = (1-erf ( = (erfc (

Where ,

V = 2nAST0.5

x=20.5

erfc(x) = I – erf(x) is the complimentary error function

In the ideal case when the quantum efficiency is unity, 10-9 BER is achieved with an average received optical energy of 36 photons per pulse . again , if we assume an OOK sequence of 1 and 0 pulses which occur with equal probability , then the average number of received photons per bit of information , NP , IS 18(half the number required per pulse)

Thus for OOK homodyne detection the BER is given by

BER = erfc(ηNP)0.5

To simplify this we note that a useful approximation to erfc(x0.5) for x>=5 is

erfc(x0.5)=e-x/(πx)0.5

so that

BER = e-ηNp/(πηNP)0.5

For ηNP >= 5 in OOK homodyne detection.

PSK Homodyne System :

Homodyne detection of PSK modulation gives the best theoretical receiver sensitivity, but it is also the most difficult method to implement. The figure given below shows the fundamental setup for a homodyne receiver. The incoming optical signal is first combined with a strong optical wave being emitted from the local oscillator. This is done using either a fiber directional coupler and a partially reflecting plate called a beam splitter. When a beam splitter is used, transparency comes into play since the incoming signal is weaker than the local-oscillator output.

Heterodyne Detection Schemes :

The analysis for heterodyne receivers is more complicated than in the homodyne case , because the photo detector output appears at an intermediate frequency ωIF .

An attractive feature of heterodyne receivers is that they employ either synchronous or asynchronous detection. The following figure shows the general receiver configuration. Let us look at this for PSK. In synchronous PSK detection one uses a carrier-recovery circuit, which is usually a microwave phase-locked loop(PLL), to generate a local phase reference . The intermediate-frequency carrier is recovered by mixing the output of the PLL with the intermediate-frequency signal. One then uses a low pass filter to recover the baseband signal . The BER for synchronous heterodyne PSK is given by

BER = erfc (ηNP)0.5

By, Sijo Cyriac (#36)

Q. Enumerate the potential applications of optical fibre communication systems.

The small size and large information-carrying capacity of optic fibers is much high as compared to copper twisted-pair cables in telephone systems. Continuous passive links more than 100 km long have been produced. With repeaters, messages over thousands of kilometres of fiber can be sent. Because of low losses, separation between repeaters in a fiber system is greater than in a coaxial link.

Because repeaters can be spaced very far apart in fiber systems, underwater fiber links have been designed to span the oceans. The low weight of fiber cables as compared to coaxial lines gives them distinct advantages of submerged cable applications because of the relative ease of transporting ad laying fibers.

Optic fiber communications are compatible with electrified railways because they are not affected by EMI. Similarly fibers can be placed near high-voltage power lines

without adverse effects whereas wire systems would be noisy. Optic   cables  can be suspended directly from power line towers or poles if clearance space permits and if the load can be tolerated. Alternatively, the fiber can be embedded in a wire cable.

 Military applications of fiber optics include communications, command and control links on ships and aircrafts, data links for satellite earth stations and transmissions lines for tactical command post communication

Q. Derive an expression for the rms pulse broadening in multimode fiber due to intermodal dispersion

Considering a perfect step index fiber, a useful quantity regard to intermodal dispersion on an optical fiber link is the rms pulse broadening resulting from this dispersion mechanism along the fiber. When the optical output to the fiber is a pulse pi (t) of unit area as illustrated

--------------- (1)

It may be noted that pi(t) has constant amplitude of , over the range

≤ pi (t) ≤

The rms pulse broadening at the fiber output due to intermodal dispersion for the multimode step index fiber σs (i.e. standard deviation) may be given in terms of the variance σs

2.

σs2 = M2 – M1

2

Where M1 is the first temporal moment which is equivalent to the mean value of the pulse and M2 the second temporal moment is equivalent to the mean square value of the pulse. Hence

M1 = ----------- (2)

And

M2 = ----------- (3)

The mean value M1 for the unit input pulse of the figure is zero, and assuming that is maintained for the output pulse, then from eqs. (2) and (3).

σ s2 = M2 = -----------(4)

Integrating over the limits of the input pulse and substituting for P i (t) in (3) over this range gives:

σ s2 = t2dt ----------------(5)

= ( )2 ---------------------- (6)

Hence substituting for gives:

σ s =

By Sreejith.R (#37)

1. Explain the propagation of the skew rays and meridional rays through an optical fibre

What are the characteristics of MMSIF and MMGIF? Explain the construction and advantages of MMGIF?

MULTI MODE STEP INDEX FIBRE (MMSIF)

MULTIMODE GRADED INDEX FIBRES (MMGIF)

By Sanjay Mohan(#30)