# Bandwidths in an Optical System

#### Have a question or want to start a discussion? Post it! No Registration Necessary.  Now with pictures!

•  Subject
• Author
• Posted on
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:

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

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

Re: Bandwidths in an Optical System
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