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Stability analysis of a Bang-Bang CDR loop (Read 351 times)
Sumilak1994
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Stability analysis of a Bang-Bang CDR loop
Mar 18th, 2019, 10:05am
 
Hi,
I am simulating the stability model of a Half Rate Bang Bang CDR, but am confused about the value for Kpd.

In most places, I found Kpd defined as a linear function of σ (sigma) as follows:

For Gaussian distribution: (1/σ)√(2/pi)

For Uniform distribution: 1/(σ2√3)

But, I am confused how this "sigma" relates to jitter - whether it's only the random jitter component or the deterministic jitter component or both.

Also, why is only the jitter component of the data and not the clock taken into consideration while determining the Kpd ?

I am in a bit of a fix.   Undecided




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Re: Stability analysis of a Bang-Bang CDR loop
Reply #1 - Dec 13th, 2019, 6:41am
 
https://ieeexplore.ieee.org/document/1327756

Random Jitter. Not deterministic.
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