Hello again:
I believe I have an answer that's responsive to your question--the
technique was designed in 1999 which makes me worry whether it's still
state-of-the-art, but the explanation of how to maximize tag reading
under anti-collision conditions is compelling and matches up well with
what we found before.
In summary:
"When a large volume of tags must be read together in the same RF
field, the application needs an anti-collision feature that enables
the reader to receive data from the each tag on a one-by-one basis.
Microchip?s MCRF250 is an example of an RFID tag IC featuring
anti-collision technology. When multiple tags are in the same RF field
and transmit data together, the reader must communicate with the tags
to prevent collisions of data. This is accomplished by transmitting a
?gap pulse? for the MCRF250.
When the tag recognizes the gap pulse, it doesn?t transmit data until
it counts a number that?s generated by a random number counter. Each
tag will finish counting the number in a different time. Therefore
each tag will retransmit its data again in a different time slot."
http://www.elecdesign.com/Articles/Index.cfm?AD=1&ArticleID=6176
Figure 1
http://www.elecdesign.com/Files/29/6176/Figure_01.gif
Technical Specifications:
http://ww1.microchip.com/downloads/en/DeviceDoc/21267f.pdf
See especially page 11 in the .pdf file.
The timeline starts with a 60 microsec gap. The device then takes
about 5 msec (millisecs) to broadcast random number sequencing codes
to all the rfids within range. All the rfids in range will respond,
and in the event of a collision (two or more responses received
simultaneously) there's another 60 microsec gap, and the process
repeats.
I believe this gives us what we need to calculate the maximum
responses that can be logged in (say) 7.5 secs (the interval used in
the cite in my earlier clarification). It seems to me that if there
were a large number of rfids within range, there could always be a
collision, and so a 60 microsec gap to allow the retransmittal, and
then a further 5 msec = 5,000 microsec for the re-interrogation would
be needed.
By that logic, because there are 1/(.000060+.005) = 197.6 such cycles
per second, one could hope to resolve 1,482 rfid reports in 7.5
seconds.
Taking that as a theoretical maximum, I'm satisfied that the earlier
cite of 500 tags in 7.5 seconds is the answer you're seeking.
Search terms used:
http://www.rfidjournal.com/article/archive/1/
://www.google.com/search?as_q=rfid+&num=10&hl=en&rls=GGLD%2CGGLD%3A2004-08%2CGGLD%3Aen&btnG=Google+Search&as_epq=anti+collision&as_oq=&as_eq=&lr=&as_ft=i&as_filetype=&as_qdr=y&as_occt=any&as_dt=i&as_sitesearch=&safe=images
[the above is a search for rfid anti-collision within the past year]
I hope you find this useful. If you require further clarification,
please hold off on rating my answer until I have a chance to respond.
Sincerely,
Google Answers Researcher
Richard-ga |