No, the D(i,j) to be used in the formula is the difference in relative
trasmit time (see section 6.3.1-interarrival jitter of RFC1889 (RTP),
http://www.ietf.org/rfc/rfc1889.txt),
whereas the values displayed in the "Delay" column on RTP analysis are plain
packet arrival time differences Rj-Ri (expressed in seconds).
We have D(i,j) = (Rj-Ri)-(Sj-Si) = (Rj-Sj)-(Ri-Si),
where Si is the RTP timestamp (expressed in seconds) from packet i.
now given your example:
R_108-R_107=0.006785,
S_108-S_107=0.030000 (RTP clock difference between two packets at 30ms
framing as i assumed from the sample)
with that we have
|D(i-1,i)| = 0.023215
and hence
J = J+(|D(i-1,i)|-J)/16 = 0.001689 + (0.023215-0.001689)/16 = 0.003034...
best regards,
Lars Ruoff
----- Original Message -----
From: "Szymon Fedor" <Szymon.Fedor@xxxxxxxxxxxx>
To: <ethereal-dev@xxxxxxxxxxxx>
Sent: Thursday, April 29, 2004 9:33 AM
Subject: [Ethereal-dev] Jitter validity
Hello,
I would like to analyse the RTP protocol with Ethereal, especially I am
concerned about the Jitter calculation. Below is the example of a RTP
statistic
displayed with Ehereal.
Packet Sequence Delay Jitter
106 39378 0 0
107 39379 0.002983 0.001689
108 39380 0.006785 0.003034
109 39381 0.003869 0.004478
110 39382 0.004391 0.005798
As I read on the mailing list, the Jitter is calculated with the following
formula:
J = J + ( | D(i-1, i) | - J) / 16
where J is jitter and D is delay between two frames
If I apply this formula to the sequence above I don't get wright result
(for example: 0.001689+(0.006785-0.001689)/16=0.0020075<>0.003034)
Do you know where is the problem? Do you really use the above formula to
calculate the Jitter or I just apply it in a wrong way?
Thank's for the answers
regards
Simon
--
INSA Lyon
Département Télécommunications, Services et Usages
Courriel: szymon.fedor@xxxxxxxxxxxx
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