Andreina, i replied to your private mail, but i also reply to this for
archiving purposes...
Andreina Toro wrote:
Hi everyone, I have a question regarding the calculation of
interarrival jitter and the information provided by Wireshark in the
"RTP Stream Analysis Wndow" for each call.
I can see that Wireshark gives me in the 4th Row of the RTP Stream
Analysis Wndow the Jitter for each packet of each call.
In the other hand I�ve read that:
"If Si is the RTP timestamp from packet i, and Ri is the time of arrival
in RTP timestamp units for packet i, then for two packets i and j, D may
be expressed as
D(i,j)=(Rj-Ri)-(Sj-Si)=(Rj-Sj)-(Ri-Si)
The interarrival jitter is calculated continuously as each data packet i
is received from source SSRC_n, using this difference D for that packet
and the previous packet i-1 in order of arrival (not necessarily in
sequence), according to the formula
J=J+(|D(i-1,i)|-J)/16
Whenever a reception report is issued, the current value of J is sampled."
What I don�t have clear is what this Jitter in the 4th Row represents in
the interarrival jitter calculation?
Well, it represents just that!
The value in 4th column *is* the value of J(i) according to the above
formula (ref. RFC 3550), starting with J(0):=0 and Ri:=frame.time(i) and
Si:=rtp.timestamp(i) in appropriate units (for conversion between units,
the clock sample rate is used - for details see the code in rtp_analysis.c).
Can I calculate the jitter J, defined to be the mean deviation, with
that data? I mean, can I use the values of the jitters of each
packet given in that RTP Stream Analysis in every call and calculate the
difference D??
D_m = \frac{1}{n} \sum_{i=1}^n \left| x_i -
\overline{x} \right|
What do you call "the jitter J"?
As said, the Jitter J(i) on a packet-by-packet basis is defined as above
and viewed in Wireshark RTP analysis in the 4th column.
If you want to have *one* value of J for a whole communication, feel
free the take the (arithmetic) mean over all J(i) (this is done and
shown on the RTP streams window by stream btw.) or use some other
mean/average.
I cannot tell you if one is more representative/common than another though.
best regards,
Lars Ruoff
