Hi John - 16/D bits/sample x 64e10 samples/sec x 1 B/8 bits x 1 MB / 2^20 B <= 32 MB/sec
Therefore, D >= 16*64e10 /(32*8*2^20) = 3.814 -> 4 So, if you want to transmit twice as many sample across the bus... D > = 2*16*64e10 /(32*8*2^20) = 7.63 -> 8 ...you have to halve your sampling frequency. Are you aware that there is a -8 option in some of the scripts that allows you to send 8 bits instead of 16. Hope that helps, - Lee ----- Original Message ----- From: jjw <[EMAIL PROTECTED]> Date: Wednesday, April 5, 2006 5:31 pm Subject: [Discuss-gnuradio] Bandwidth Across USB To: Discuss-gnuradio@gnu.org > > I know this is probably an extremely stupid question, but I wanted > to get > some verification. I've noticed that the minimum decimation rate > you can > use in all of the sample programs is 4 (giving you an effective > samplingrate of 16 MHz). If you are transmitting 1 real channel of > data (16 bits) > at this rate, that seems to be maxing out the USB transfer rate of > 32 MB/s. > Am I correct in my deduction that you need to use a decimation rate > of at > least 8 in order to transmit complex data (such as for FM > demodulation)successfully across the USB? Thanks. > > Thanks, > John > -- > View this message in context: http://www.nabble.com/Bandwidth- > Across-USB-t1402318.html#a3773742 > Sent from the GnuRadio forum at Nabble.com. > > > > _______________________________________________ > Discuss-gnuradio mailing list > Discuss-gnuradio@gnu.org > http://lists.gnu.org/mailman/listinfo/discuss-gnuradio > _______________________________________________ Discuss-gnuradio mailing list Discuss-gnuradio@gnu.org http://lists.gnu.org/mailman/listinfo/discuss-gnuradio
