[span style='font-size:14pt;line-height:100%']2-pass encoding: why it is necessary to extract samples from an entire encoding and the original track.[/span]
wmeditor.exe is a tool provided with WMEncoder.exe. The purpose of this small application is to cut WMA files without recoding them. The tool is not extremely precise (accuracy seems to be around ½ sec, at least with VBR) but it’s enough to extract a part from a VBR encoding in order to measure the bitrate of this small part.
I encoded four samples from the latest multiformat listening test at 128 kbps with WMA VBR 2-pass, with three different method or environment. As you know, the allocated bitrate of one sample depends a lot from the content of the whole file.
• Method#1 = the sample was extracted from a full track encoded in 2-pass mode. The full track is the original one. The bitrate of the full track should correspond to the targeted one, but our extracted sample could differ from it, and will depend from the complexity of the sample as well from the complexity of the track.
• Method#2 = the sample is directly encoded in 2-pass mode. The final bitrate necessary correspond to the targeted one (here: 128 kbps), regardless of the complexity of the sample.
• Method#3 = the sample was extracted from a full track encoded in 2-pass mode. The full track is a virtual track composed with 18 samples (used in latest 128 multiformat test) merged together. The bitrate of the full track should correspond to the targeted one, but our extracted sample could differ from it, and will depend from the complexity of the sample as well from the complexity of the track.
In summary:
method#1 = 1st step = encoding (full & original track) – 2nd step: extraction
method#2 = 1st step = extraction – 2nd step: encoding
method#3 = 1st step = encoding (full but fake track) – 2nd step: extraction
BARTOK.WAV
#1 = 0:22.895 356 kb 124,39 kbps
#2 = 0:23.024 373 kb 129,60 kbps
#3 = 0:23.231 338 kb 116,40 kbps
DEBUSSY.WAV
#1 = 0:29.210 385 kb 105,44 kbps
#2 = 0:29.999 478 kb 127,47 kbps
#3 = 0:29.674 291 kb 78,45 kbps
HONGROISE.WAV
#1 = 0:29.393 490 kb 130,93 kbps
#2 = 0:30.000 478 kb 127,47 kbps
#3 = 0:29.396 326 kb 88,72 kbps
MAHLER.WAV
#1 = 0:29.767 717 kb 192,70 kbps
#2 = 0:29.999 484 kb 129,07 kbps
#3 = 0:30.058 536 kb 142,66 kbps
#1 #2 #3
bartok 124,39 129,60 116,40
debussy 105,44 127,47 78,45
hongroise 130,93 127,47 88,72
mahler 192,70 129,07 142,66
__________________________
138,37 128,40 106,56
=> method#2 and method#3 are both inaccurate. None correspond to what the listener would get by encoding his own CD with VBR 2-pass mode. The third method is in this case the worse, and would handicap WMA by 30 kbps per (classical) sample!