Bandwidths in an Optical System

Please i am a little confused about how the bandwidth of an optical system is calculated. The units involved are a little not standardized so it confuses me a lot . How much bandwidth does a 32nm window with a 50Ghz channel spacing have. Please I would be grateful for any explanation as to how this was calculated. Thanks

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Bandwidth is bandwidth. It's a bit confusing when everything else is in wavelength units--just write

nu = c/lambda

then differentiate to get d(nu)--i.e. the bandwidth:

d(nu) = -c d(lambda)/lambda**2.


Phil Hobbs

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Phil Hobbs

Bonsoir Deamon?,

The 50 GHz spacing gives about 0,4 nm spacing in DWDM 1550 nm. So, in the 32 nm window you can put 80 wavelengths. The global bandwidth is 80 x 50 Gbit/s =3D> 4 Tbit/s.

Best regards, Michelot

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