Calculate how many WDM channels can be amplified by EDFA which amplifies signals in the wavelength range between 1.530 μm and 1.570 μm if the channel spacing is 100 GHz.
Write a brief essay (typical length of 1-2 A4 pages) about:
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(b) Describe with the aid of suitable diagrams the mechanisms of the emission of light from an injection semiconductor laser. Explain the mechanism of optical feedback to provide light amplification within the laser. Explain the main difference between an injection laser and an LED and sketch the L-I characteristics of these two devices.
(a) Explain the physical principles of telecommunication semiconductor detector. Give the definitions of the two main parameters of this device and write the corresponding formulas for these parameters.
(b) With the aid of a suitable diagram describe briefly a p-i-n photodiode and an avalanche photodiode (APD). What are the advantages and drawbacks of each of these devices in telecommunication systems?
(c) Given that the GaAs bandgap energy is 1.42 eV, calculate the cut-off wavelength of the GaAs photodiode and explain its significance.
(a) What is a fractional bandwidth and what is it used for? Evaluate the range of carrier frequencies which are necessary for DSBTC modulation by a tone modulation signal x(t) = 20sin(8π104t) , where t is in seconds. Calculate maximum value of modulation index μ in order to prevent phase reversal in the DSBTC tone modulation.
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Calculate how many WDM channels can be amplified by EDFA which amplifies signals in the wavelength range between 1.530 m and 1.570 m if the channel spacing is 100 GHz.
The number of WDM channels ideally a number of different colours or number of wavelength occupied in the wavelength range of 1.530 m and 1.570 m if the channel spacing is 100 GHz
The wavelength grid is given in between 1.530 m and 1.570 m .i.e. 1.530 m- 1.570 m
in nanometer, its value is 1530–1570 nm. These values are cover in C-band.
Thus the signal bandwidth is the difference of maximum value and minimum values of wavelength=1570-1530=40nm....