Jitter frequency (digital data)

Altera_Forum · ‎10-28-2008

Lets assume I have digital data with a known ideal period.

Lets assume that there is some random jitter (gaussian) that variates the edge transition position from their ideal place in time.

Lets assume that *somehow* I could know exactly how much random jitter each bit contains.. 1st bit: 0.1 UI (unit intervals. 1 UI = bit ideal width). 2nd bit 0.11UI.. etc

using such information, how can I draw a graph of (power (dB) vs jitter frequency (Hz))?

To be honest I dont fully understand that graph, maybe thats why I am unable to draw it from the data I have.

Also using such information I want to draw (BER vs Eb/No) graph.

How?

Altera_Forum · ‎10-28-2008

Hi,

I don't pretend I can understand this issue but my gut-feeling is that you need to prove a gaussian graph of noise/signal power(jitter/signal voltage squared and averaged over unit time then converted to dB) Versus frequency of jitter in terms of probabilty of occurence within the total values.

Kaz

Altera_Forum · ‎10-29-2008

can u speak in more details?

Altera_Forum · ‎10-29-2008

Given only one set of data(nothing else) and then a barrage of vague questions from your kind tutor, we can only sail together in the uncertaintities. Here is my part:

1) frequency:

let us look at a section of these jitter values say all the values in the range

(.100 - .200) and assume that you have 6 values in this section from a total vector of 1000 values.

The concept of frequency can only make sense in terms of probability i.e. f = 6/1000 for the above section.

There is no physical time variable in your data and (Hz) as event/sec has no meaning whatsoever unless your tutor has redefined it.(A known period implies a fixed frequency)

2)dB:

For every period, your jitter ratio is time ratio of two signals equal in voltage but with different duration.

e.g. assuming the signal voltage is 5V then a jitter of .100 means 5V for .100 of period and 0V for .900 while the rectangular pulse will be 0V for .100 and 5V for .900

Thus the mean voltage ratio of noise to signal = your given time ratio

In short:

dB for this section = 10*log10(mean[.100 ... ... .200].^2);

This is noise/signal ratio. Normally the opposite(signal/noise ratio) is in use.

There are further equations and Q-tables that relate the so-called Q functions of a given Gaussian curve to possible error rate.

kaz