 traverses six nodes traveling from the source to the desti- nation, that means traversing 12 AWGs, one at the source, one at the destination, and two for each intermediate node. The overall filtering characteristic experienced by the signal is (6.101) apart from the attenuation that is compensated by the amplifiers the overall bandwidth (in this case intended as the value at 1/e) now is Bo = BAWG /2/3 SO that if the design requires Bo=4R = it should be BAWG 2 13.85R. However, it would be wrong to assume that B is also the noise bandwidth. Let us imag- line, for simplicity, that the ring node is constituted by OADMs with identical amplifiers before and after (see Figure 6.39), and that the first amplifier compensates exactly the OADM loss and the second the span loss. The overall noise arriving at the receiver, if we imagine that no additional filter is present in front of the photodiode, is (see Figure 6.39 to compute how many filters the ASE generated by each amplifier traverse) (6.102) where j is the number of traversed spans and the spans are numbered SO that the num- ber one is the nearer to the drop node. In the aforementioned example, the optical bandwidth is 4R for the signal, but the equivalent optical bandwidth for the noise is Bn = 3.5B 2 14R. This is a difference of more than 5 dB that has to be taken into account when designing the system. Naturally, the problem can be faced by maintaining quite large the line filters and add- ing a noise selection filter in front of the receiver. Telecommunication Networks However, if this solution is selected, it is to take into account that in the filter bandwidth that is not occupied by the signal the ASE can circulate in the ring and the design has to carefully avoid the start of ASE lasing. 6.3.5.2 Transmission in Coarse Wavelength Division Multiplexing Metropolitan Ring In the case of CWDM we are talking about rings without amplifiers, as a matter of fact, the CWDM occupies such a large optical bandwidth (compare Chapter 3) that it is unpractical to use amplif