optical fibres

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Slideshow on optical fibre

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• 1.COMMUNICATION
OPTIC FIBRE TRANSMISSION

2. TOTAL INTERNAL REFLECTION
3. Total Internal Reflection
When a ray of light travels from a denser to a rarer medium such that the angle of incidence is greater than the critical angle, the ray reflects back into the medium. This is called total internal reflection.
4. Optical Fibres
Consists of a very thin glass core surrounded by a material of slightly lower refractive index called cladding
The thin fibre can be bent without breaking and a ray of light can be sent down the fibres core
Total internal reflection takes place at the boundary of the core and the cladding
5. Acceptance angle
The maximum angle of incidence that a ray can make that will result in total internal reflection is called acceptance angle
6. PROBLEM
The refractive index of the core of an optical fibre is 1.50 and that of the cladding is 1.40.Calculate the acceptance angle of the fibre. Ans:330
The refractive index of the core of an optical fibre is 1.50 and the critical angle of the core- cladding boundary is 750.Calculate the refractive index of the cladding. Ans: 1.45
7. Material Dispersion
Light of different wavelengths have different refractive index and hence come out of the fibre at different times. This is called material dispersion
8. Modal Dispersion
9. Modal dispersion
Rays that undergo many internal reflections are said to follow a high order mode paths
Rays undergoing fewer reflections follow low order mode paths
Set of rays having same wavelength reach the end at different times due to different paths taken. This is called modal dispersion
10. Monomode & Multimode fibre
In multimode fibres, the core has a diameter of about 100m and the cladding is about 20m thick
The rays passing through multimode fibres undergo material as well as modal dispersion
In monomode fibres, the core has a diameter of about 8 -10m and the cladding is about 125m thick.
Rays follow just one path eliminating modal dispersion
11. Step Index fibre
The refractive index of core is constant
The refractive index of cladding is constant
The refractive index of cladding is slightly lower than that of core
Refractive index of core decreases smoothly from the centre to the outer edge
Refractive index of cladding is constant
13. 14. ATTENUATION
Attenuation in an optic fibre is caused by the impurities of the glass core. The amount of attenuation depends on the wavelength of the light being transmitted.
Power loss in decibels is defined as
Power loss = 10log (Pfinal / Pinitial) in dB
Thus a power loss of 16 decibels means that the initial power of, say,8.0mW has been reduced to 0.2mW
15. PROBLEM
An amplifier amplifies an incoming signal of power 0.34mW to a signal of power 2.2mW.Calculate the power gain of the amplifier in decibels. Ans: 8.1dB
A signal of power 12mW is input to a cable of specific attenuation 4.0 dB/km. Calculate the power of the signal after it has travelled 6.0km in the cable. Ans: 0.048mW
16. VARIATION OF SPECIFIC ATTENUATION WITH WAVELENGTH
17. Attenuation & Wavelength
The specific attenuation ( power loss in dB per unit length ) actually depends on the wavelength of the radiation travelling along the optic fibre
The graph shows minima at 1310nm and 1550nm, which implies that these are desirable wavelengths for optimal transmission
These are infra red wavelengths
18. DETECTION
The light that enters an optic fibre travels down the length of the fibre and the arrival of light is registered by a photodiode
In the absence of any light, falling on the photodiode, the current is zero
When light of a specific wavelength falls on the photodiode, a current flows. The magnitude of the current is proportional to the intensity of light
19. A light detector circuit with a photodiode
20. NOISE
Source of noise in a cable:
Random motion of electrons which creates additional electric fields contaminating the signal. This increases with temperature.
Lightning
Charged particles emitted by the sun during intense solar activity
21. NOISE IN OPTICAL FIBRES
Main source is the dark current of the photodiode. This is the small current that flows even when the photodiode is dark
Signal to noise ratio (SNR) is defined as
SNR = 10log Psignal / Pnoise
22. PROBLEM
The minimum SNR considered acceptable for a certain signal is 30dB.If the power of the noise is 2.0mW, calculate the least acceptable signal power.
The SNR in a certain signal is 10dB.the signal passes through an amplifier of gain 6.0dB.What will be the signal to noise ratio after amplification?

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