Optical Fibre Communication | One Liner | Unit -1

        Unit 1 - Optical fibre communication 

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One Liners :-

1     The optical source which provides the electrical – optical conversion may be either                    a  semiconductor laser or LED.

2.       Information source provides an electrical signal to a transmitter.

3.       The transmitter drives an optical source to give modulation of the light wave carrier.

 
4.       The transmission medium consists of an optical fibre cable.

5.       The receiver consists of an optical detector.

6.       Photo diodes, photo transistors and photoconductors are utilized for the detection of             the optical signal and optical conversion.

7.       Optical fibre are fabricated from glass or plastic polymer.

8.       Various elements of communication system are:

communication system diagram

       Advantage of optical fibre communication

 

1.       Enormous potential bandwidth in the range of 10¹³ to 10¹⁶ Hz.

2.       Small in size and weight.

3.       Electrical isolation.

4.       Optical fibres don’t exhibit earth loop and interface problems.

5.       They are immune to cross talk and interference.

6.       They provide high degree of signal security.

7.       They offer low transmission losses in the order of 0.15db km^-1

8.       They also offer ruggedness and flexibility this means they can withstand shock and pressure both.

9.       They are highly reliable and their maintenance is easy.

10.   Offer Low cost line communication.

    Optical fibre waveguide

1.      Refractive index n1 > n2

2.      Function of Cladding is to supports the wave guide structure.

optical fibre waveguide diagram

Ray Theory 

1. The Snell’s law is given by 

                                    N1 sin θ1 = N2 sin θ2

2. The Snell law states that in an oblique medium, the product of the refractive index and sine of incidence angle in medium 1 is same as that of medium 2. Thus it is given by N1 sin θ1 = N2 sin θ2

3. find the ratio of sine of incident angle to the sine of reflected angle when the refractive indices of medium 1 and 2 are given as 2.83 and 1.68 respectively.

                Use formula -

                                        N1 sin θ1 = N2 sin θ2

4. calculate the ratio of the refractive index of medium 1 to that of medium 2, when the incident and reflected angles in two mediums are given as 30 deg. and 45 deg. respectively.

                Again Use Formula 

                N1 sin θ1 = N2 sin θ2

5. refractive index is given by n = c √(με)

where,             c is the speed of light

&                 ε = 4π x 10-7

6. critical angle, it is defined as the angle of incidence at which the total internal reflection starts to occur

7. The critical angle for two media with permittivities of 21 and 14 respectively is (solve this numerical)

Use Formula - 

                            Since n = √ε, so sin θc = √ε2/√ε1

8. The angle of incidence always is equal to the angle of reflection for perfect reflection.

9. The angle of incidence of a wave of a wave with angle of transmission 45 degree and the refractive indices of the two media given by 2.1 and 1.6 is

                                Use Formula - 

                                                    N1 sin θi = N2 sin θr

10. Acceptance angle- it is the maximum angle at which a ray hits the fibre core which allows the incident light to be guided through it. This is given with the help of following formula.

acceptance angle formula

Where                               α_c = θa = acceptance angle

1.    11. Numerical aperture (NA)-  it is defined as the product of the refractive index of the beam from which the input light is received and the sine of the maximum ray angle against the axis, for which light can be transmitted through the system.

      

Numerical Apperture NA formula

Where, NA = Numerical Aperture

n0 = refractive index of receiving medium. For air its value is 1.

θa = maximum ray angle against the geometric axis

12. Relative Refractive Index Difference – it is represented by delta (∆) and given with the help of following formula 

Relative Refractive Index Difference

If ∆ << 1 

Then                             NA = n₁ (2∆)^1/2

Skew Rays

1. A skew ray is a ray which travels in a zig- zag path and never crosses the axis of the optical fibre (wave guide).

2. Bound ray – it is a ray in a multimode fibre which is confined by the core of the wave guide. 

3. Maxwell’s equations for electromagnetic wave propagation are given by curl equations for zero conductivity medium. These equations are – 

Maxwell’s Curl equations 

Where is called vector operator

Modes in a planner guide –  

1.       Transverse Electric Mode (TE) – When the electric field is perpendicular to the direction of propagation but a corresponding magnetic field is in the direction of the propagation, such a mode is called Transverse Electric Mode.

 

2.       Transverse Magnetic Mode (TM) – When component of electrical field is in the direction of wave propagation but the component of magnetic field is zero such a mode is called Transverse Magnetic mode.

 

3.       Transverse Electromagnetic Mode (TEM) – When our total field lies in transverse plane such a mode is called Transverse Electromagnetic Mode in this mode both electrical and magnetic field component are zero.

Modes in cylindrical fibre –

1.       The cylindrical waveguide is bounded with 2 dimensions rather than 1.

2.       For cylindrical waveguide both TE and TM modes corresponds to meridional rays.

 

Single Mode Fibre –

1.       **The advantage of propagation of single mode is that signal dispersion which is caused by delay dispersion can be avoided 

2.       In single mode only fundamental LP₀₁ exists.

3.       **The value of cut off normalized frequency (Vc) for LP₁₁ is 2.405.

4.       The relative refractive for single mode fibre is usually less than 1%.

5.       Single mode operation only occurs above the theoretical cut-off wavelength.

6.       The theoretical cut-off wavelength is given by the following formula. 

cut-off wavelength

Cut-off wavelength

The recommended cut-off wavelength value for primary coated fibre is 1.1 – 1.28.

Mode Field Diameter – 

It is the distance between the opposite 1/e = 0.37 times the near field strength and power is = 1/e2 = 0.135.

Group Delay and Mode Delay Factor – 

1. The transmit time which is also known as group delay is inverse of group velocity. It is given as –

                                            τg=1/vg 

2. Group Index of a plane wave is determined using

                                         Ng=C/vg 

3. Specific group delay for fundamental fibre mode is given as – 

                                        τg=Nge/c

where,   c is the velocity of light.

&    τ = tau

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                                                            Best Of Luck!