One issue in relating RMS jitter to phase noise
时间:04-11
整理:3721RD
点击:
Longly we believe:
ΔTabs,rms=(2π/To)2 *lim∑Φ2n,j / N (1)
then in Razavi's book IC in optical circuits' P28, it will becomes:
ΔTabs,rms=(2π/To)2 lim∫Φ2n(t)dt / T (2)
then by Parceval's law: ΔTabs,rms= (2π/To)2∫SΦ(f)df ; (3)
Therefore, we accomplish the translation between phase noise and jitter;
I realized that i ignored one important effects before. From (1), we multiple ΔT (cycle) in numerator and denumerator then we get (4).
ΔTabs,rms=(2π/To)2 *lim∑Φ2n,j*ΔT / (N*ΔT ) (4)
(2) approximate (4) by integration. However, they are still very different. Because (4) approximate (2) by sample and hold Φ2n(t) by ΔT as cycle.
I believe (2) is smaller than (4), since Φ2n(t)dt will be greatest during transitions. Sample and hold them during transitions will result in bigger value than real average.
All in all, by using (3) to estimate (4) might under-estimate the jitter.
Isn't it?
ΔTabs,rms=(2π/To)2 *lim∑Φ2n,j / N (1)
then in Razavi's book IC in optical circuits' P28, it will becomes:
ΔTabs,rms=(2π/To)2 lim∫Φ2n(t)dt / T (2)
then by Parceval's law: ΔTabs,rms= (2π/To)2∫SΦ(f)df ; (3)
Therefore, we accomplish the translation between phase noise and jitter;
I realized that i ignored one important effects before. From (1), we multiple ΔT (cycle) in numerator and denumerator then we get (4).
ΔTabs,rms=(2π/To)2 *lim∑Φ2n,j*ΔT / (N*ΔT ) (4)
(2) approximate (4) by integration. However, they are still very different. Because (4) approximate (2) by sample and hold Φ2n(t) by ΔT as cycle.
I believe (2) is smaller than (4), since Φ2n(t)dt will be greatest during transitions. Sample and hold them during transitions will result in bigger value than real average.
All in all, by using (3) to estimate (4) might under-estimate the jitter.
Isn't it?
Hi
What is dimension of Φ in your relationships?
ΔT is time
(2π/To)^2 is freq^2
What is Φ indeed?
Regards
it should To/(2*pai). sorry for the typo.
