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Solutions to Chapter 4 - Communication Networks

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<strong>Communication</strong> <strong>Networks</strong> (2 nd Edition)<br />

<strong>Chapter</strong> 4 <strong>Solutions</strong><br />

a. How close do these systems come <strong>to</strong> using the available bandwidth in the 1550 nm range<br />

One STS-48 signal carries 2488.32 Mbps of traffic. Thus, the aggregate bit rate of each of these<br />

systems is:<br />

System<br />

Bit rate<br />

100 wavelengths 248.832 Gbps<br />

200 wavelengths 497.664 Gbps<br />

400 wavelengths 995.328 Gbps<br />

The available bandwidth of this window is 12.49 THz. Thus, none of these systems even comes<br />

close <strong>to</strong> using all of the available bandwidth.<br />

b. How many telephone calls can be carried by each of these systems How many MPEG2<br />

television signals<br />

The number of telephone calls (64 Kbps) and MPEG2 video signals (5 Mbps) that can be carried<br />

each system is shown below:<br />

System Telephone Calls MPEG2 Video Signals<br />

100 wavelengths 3.888 × 10 6 49 766<br />

200 wavelengths 7.776 × 10 6 99 532<br />

400 wavelengths 15.552 × 10 6 199 065<br />

Although, none of the systems comes close <strong>to</strong> using all of the available bandwidth, they can still<br />

carry huge amounts of traffic. This shows how WDM provides the potential for huge amounts of<br />

bandwidth and radically expands the bandwidth of existing optical links.<br />

21. Consider the SONET fault protection schemes described earlier in the chapter. Explain whether<br />

these schemes can be used with WDM rings.<br />

Solution:<br />

The fault protection schemes described in problem 15 are applicable <strong>to</strong> any ring network<br />

structure. To use these schemes in WDM, the exact same procedures can be carried out except<br />

that in the case of WDM, instead of switching streams, wavelengths are switched. Furthermore,<br />

the switching can be done in optical domain by all optical protection switches that detect the<br />

signals and switch <strong>to</strong> protection signal instantaneously. Optical protection switching can be faster<br />

and more efficient.<br />

22. Calculate the spacing between the WDM component signals in Figure 4.8. What is the spacing in<br />

hertz and how does it compare <strong>to</strong> the bandwidth of each component signal<br />

Solution:<br />

In Figure 4.8 there are 16 wavelengths distributed equally over the spectrum from 1545 nm <strong>to</strong><br />

1557 nm. Thus the spacing between each wavelength is:<br />

1557 −1545<br />

= 0.75nm<br />

16<br />

Leon-Garcia/Widjaja 11

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