Optical Fibre Communications
Tutorial on Devices
DE2
1- Discuss briefly the four key system features of Wavelength Division Multiplexing.
2- An optical transmission system is constrained to have 500 GHz channel spacing.
How many wavelength channels can be utilized in the 1536-to-1556 nm spectral
band?
3- An N x N star coupler is constructed from n 3-dB 2x2 couplers, each of which has a
0.1 dB excess loss. Find the maximum value of n and hence the maximum size of N if
the power budget for the star coupler is 30 dB.
4- Sketch block diagrams of the three most common topologies used for optical
networks.
5- Briefly discuss the followings and comment on the suitably of them for long haul
optical communications:
(i) Optical mechnical splice
(ii) Optical fusion splice
6- Sketch a schematic diagram of a typical 2 x 2 coupler and show that it does operate
as a switch.
7- Describe the followings commenting on their applications:
(i) Optical filters and their applications,
(ii) Fibre Bragg grating
(iii) Optical Isolators
(iv) Optical circulators
8- Consider 2 x 2 coupler shown in Fig. 7, where A and B are the matrices
representing the field strengths of the input and output propagating waves,
respectively. For a given input a1, we impose the condition that there is no power
emerging from the second input port; i.e a2 = 0. Find expressions for the
transmissivity T and the reflectivity R in terms of the elements sij in the scattering
matrix S, as in the lecture note
9- For a 7x7 star coupler, the insertion loss from input port I to each output port is
shown in Table 7. Find the total excess loss through the coupler for inputs to port 1.
Exit port no.
1
2
3
4
5
6
7
9.33 7.93 7.53 9.03
9.63
8.64
9.04
Insertion loss (dB)
Table 7.
Solution
2- Channel spacing ∆λ = λ2(∆f)/c = (1546 nm)2(500 GHz)/3 x 108 = 4 nm.
The number of channels N = (1556 – 1536)/4 = 5 Not too many!
3- Excess loss 10 Log ((P0/2P1) = 0.1 dB.
Therefore, P1 = 0.5 P0/(10 0.01) = 0.977 P0/2.
Thus fractional power traversing the 3-dB coupler FP = 0.977.
The total loss LT = -10 Log[(Fp (Log2 N))/N] = -10(Log2 N Log (FP) – Log N)
= -10(1 – 3.322 Log (FP)) Log N = -10(1 – 3.322 log 0.977)Log N ≤ 30.
Therefore Log N = 30/10.33 = Log2 n = 2.904.
Thus n = 9.64. And N = 2n = 512.
8-
9-