Thanks Marcus. given parameters : bit rate = 100,000 samples per symbol = 2 constellation = 16
then, bits per symbol = log2(16) = 4 symbol rate = bit rate / bits per symbol = 100,000 / 4 = 25,000 sample rate = samples per symbol * symbol rate = 2 * 25,000 = 50,000 also, as I know, sample rate = bandwidth. so, bandwidth is 50KHz right ? Q1. what is difference between sample rate and samples per second ? Q2. USRP N210 maximum sample rate = 25 MS/s, CBX daughterboard analog bandwidth = 40 MHz. what is the maximum bandwidth ? Thank you again! 2016-03-02 15:51 GMT+09:00 Marcus Müller <marcus.muel...@ettus.com>: > Hi, > > > I think it is up to bit rate, but I don't know exactly what it is. > > The number of bit per second you're going to be transmitting/receiving. > It's (symbol rate) * (bits per symbol). "bits per symbol" is of course > log2(constellation points). > > > Is there unlimited bit rate option for check to modulation throughput ? > > No, a given modulation with a given sample rate and a given samples per > second must result in a fixed transmission rate. > > Best regards, > Marcus > > > On 02.03.2016 06:47, SangHyuk Kim wrote: > > Hi all. > > I'm using benchmark_tx.py and rx (/digital/example/narrowband) > > As I know, ex) QAM modulation is faster than BPSK one. > > However, when I used both of modulation schemes, I couldn't measure > throughput. > > I think it is up to bit rate, but I don't know exactly what it is. > > Is there unlimited bit rate option for check to modulation throughput ? > > Thanks. > > > _______________________________________________ > Discuss-gnuradio mailing > listDiscuss-gnuradio@gnu.orghttps://lists.gnu.org/mailman/listinfo/discuss-gnuradio > > > > _______________________________________________ > Discuss-gnuradio mailing list > Discuss-gnuradio@gnu.org > https://lists.gnu.org/mailman/listinfo/discuss-gnuradio > >
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