Re: Wrong calculation of jitter by wireshark

[capricorn 80 <cool_capricorn80 () hotmail com>]

Its my mistake .. i am not taking care of msec ..

 

thanks.
 


From: cool_capricorn80 () hotmail com
To: wireshark-users () wireshark org
Date: Sat, 15 May 2010 11:58:45 +0000
Subject: [Wireshark-users] Wrong calculation of jitter by wireshark




 Hi!

  I have captured the RTP packets on receiving side and check the values of jitter and than i calculated the values 
manually and it looks like wire shark is not calculating the values in correct way or may be I am doing some thing 
wrong. I am pasting all me calculation so that you people can verify it.




Frame 18
Time of Arrival: 29.490533000
Timestamp: 365000
 
Frame 20
Time of Arrival : 29.501315000
Timestamp: 365160
 
D(i,j) = (Rj - Ri) - (Sj - Si) = (Rj - Sj) - (Ri - Si)
J(i) = J(i-1) + (|D(i-1,i)| - J(i-1))/16
D = (29. 501315000- 29. 490533000) – (365160*0.000125 – 365000*0.000125)
    = -0.009218 
J = 0 + (|-0.009218| - 0)/16
  = 0.576 msec
 
Frame 22
 
Time of Arrival: 29.520124000
TimeStamp: 365320
 
(29.520124000 - 29.501315000) – (365320*0.000125 – 365160*0.000125)
=-0.001191
J= 0.576+(|-0.001191|-.576)/16
 
J= 0.54             , 
 
Wireshark calculation for frame 22 = 0.61
 
Frame 24
 
Time of Arrival: 29.539717000
Time Stamp: 365480
 
D=(29.539717000 - 29.520124000) – (365480*.000125-365320*.000125)
 
 
D= -0.000407
 
J=0.54+(|-.000407|-.0.54)/16
 
J=0.50 
 
Wireshark Calculation for frame 24 = 0.60




Really thanks for your time.




Regards,







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