Bandwidths in an Optical System

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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

Re: Bandwidths in an Optical System
Deamon wrote:
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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.

Cheers

Phil Hobbs



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Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
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email: hobbs (atsign) electrooptical (period) net
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Re: Bandwidths in an Optical System
Bonsoir Deamon?,

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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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