ISI in optical link
I get a question for ISI ( inter symbol interference) in optical link.
In Razavi'S book,<design of integrated circuits for optical communications> P65, Fig4.2 (b)
It says that if the bandwidth of the receiver is 1/4 of the data rate, then voltage decrease due to ISI is roughly 0.2*Vpp.
However, in Saeckinger's Book, < broadband circuits for optical fiber communication>, P59
It says that if the bandwidth of the receiver is 1/3 of the data rate, the voltage decrease due to ISI is 0.5*Vpp.
I was wondering which one is right? Or they are both right and I am wrong.
I don't have these books. But, are you sure they are talking about the same receiver ?
"the same receiver？" I think so. They all talking about optical receiver for NRZ data.
are you talking about the optical link, or just the receiver. I would assume if you tried to turn on/off a laser diode at a data rate 3 times its bandwidth, it would not look pretty
just the receiver, assuming the input is perfectly rectangular.
So, please enlighten me. Here they are saying that for 10 GBps data rate, you need 7.5 GHz 3dB bandwidth:
http://books.google.com/books?id=ynv...ate%22&f=false
Yes, usually, the bandwidth is 0.7 or 3/4 of the data rate. This is the trade off between ISI and noise.
If your bandwidth is 1/3 of the data rate, for example, 15GHz for 40Gbps. It still works.
I just want to know how does ISI do damage to the signal. this two books give a different answer.