Hi, > On Wed, 4/22/09, kaleem ahmad <removed_email_address@domain.invalid> wrote: > > Thanks Firas, > > But I can simply tell that it always transmit a fix packet (with > bitrate=500kbps) of 1220 micro sec including preamble. It means that > after every T ms (T is repetition cycle time and can be a fixed > value from range: 1ms...200ms, I selected 10ms for above given data) it > transmits a 1220 microsec long signal. > If your transmitter works for 1220 microsec (1.22 msec) and it repeat the transmission (for example) every 10 msec, then your calculations is wrong. If you want to sense the time of a signal, you have to run your FFT frames with a rate at least equal to required minimum signal time. For example in your case, If the signal to be detected has a 1.22 msec then you have to collect the data at rate of 500K (USRP decimation =128). The 512 FFT length will be 1.024 msec. This means (after using appropriate spectrum threshold value) you need to count number of FFT frames that the desired signal is exist in it. So, back to our example (signal with duty 1.22msec and repetition of 10msec), if data rate is 500k, and you used 512 FFT points, then you will see this signal once in every 10 FFT frames. The resolution will be 1.024 msec. However, if you received the signal with data rate of 8MHz (USRP decimation =8), and you computed 512 FFT points then your resolution will be 64 usec, which means that the signal will be ON for 19 FFT frames and OFF for 137 FFT frames. Is that clear ? Best Regards, Firas

on 2009-04-22 16:06

on 2009-04-22 16:21

Thanks Firas, It is clear and I will implement it and will come back soon hopefully with successfull results. Best Regards Firas A. wrote: >> after every T ms (T is repetition cycle time and can be a fixed > collect the data at rate of 500K (USRP decimation =128). The 512 FFT > decimation =8), and you computed 512 FFT points then your resolution will > > > _______________________________________________ > Discuss-gnuradio mailing list > removed_email_address@domain.invalid > http://lists.gnu.org/mailman/listinfo/discuss-gnuradio > > -- View this message in context: http://www.nabble.com/cycle-period-detection-of-a-... Sent from the GnuRadio mailing list archive at Nabble.com.

on 2009-05-25 16:15

Dear All, A few weeks ago I was involved in a discussion about calculating the cycle time of an unknown system present in the area by sensing the spectrum with the help of RFX2400+USRP1 system. For details you may have a look on previous mails on this thread. As Firas suggested to choose a decimation factor with respect to duty cycle of the cyclic system. I implemented it in this way but I am facing two problems: 1- I am using following formula to calculate the cycle time: if ADC_rate = 64MHz, D=decimation rate, fft_size, count= total no. of fft_scans where signal is not present Then following would give me the value of cycle time: [1/(ADC_rate/D)]*fft_size*count*1e3 ---> where [1/(ADC_rate/D)] simply gives the time resolution or time gap between two points in fft bin. 1e3 for conversion in ms. with D=128, fft_size=512, I scanned the spectrum for more than one times and stored the results in an array, then I choose maximum occerence in that array as 'count' in above formula e.g. for 200ms cycle time my array look like [27, 200, 27, 27, 27, 27, 276, 27], here count = 27, other values are actually some errors for 100ms cycle time my array look like [13, 13, 87, 13, 13, 13, 13, 167], here count = 13, other values are actually some errors for 50ms cycle time my array look like [6, 6, 6, 6, 6, 6, 6, 6], here count = 6. From this array with multiple scans I was able to see that some cyclic behaviour is correctly detected. BUT when I put all values in above mentioned formula then I found that to get exact value of cycle time I must multiply my answer with a 'factor' ranging from 7-8 (some times 7.25, or 7.4, or 7.5 etc). But this 'factor' is not a fixed value and variation in its value depend on the cycle time. So final formula becomes [1/(ADC_rate/D)]*fft_size*count * factor *1.e3 ---> The value of factor is changing on following pattern: for 200ms cycle time factor ~ 7.25 would give me exact answer for 150ms cycle time factor ~ 7.35 would give me exact answer for 100ms cycle time factor ~ 7.5 would give me exact answer for 50ms cycle time factor ~ 8 would give me exact answer I cant understand what this factor is and where it is comming from. Is there any processing in USRP which I am forgetting? Can someone help me to figureout this factor. 2-My second problem is that, I am able to detect cycle time for any duty cycle with the help of 128, and 256 decimation rates , but with any other decimation rate my answers are quite random and I cant get a cyclic answer in my above mentioned array. Or simply you can say as it was looking from theretical side that for any specific decimation rate I should get a cyclic answer but the only thing which would change was the no. of detected and undetected scans. But its not happening in this way. e.g with any other decimation rate except 128, 256, I am getting answer like follows: for 200ms cycle time my array look like [20, 2, 187, 17, 87, 65, 13, 167], which dont provide me any clue. Similarly for any other cycle time I am getting random values. Is there any thing wrong with selection of decimation rate? can some one help me to figure out where I am doing wrong. Thanks and Best Regards Firas A. wrote: >> after every T ms (T is repetition cycle time and can be a fixed > collect the data at rate of 500K (USRP decimation =128). The 512 FFT > decimation =8), and you computed 512 FFT points then your resolution will > > > _______________________________________________ > Discuss-gnuradio mailing list > removed_email_address@domain.invalid > http://lists.gnu.org/mailman/listinfo/discuss-gnuradio > > -- View this message in context: http://www.nabble.com/cycle-period-detection-of-a-... Sent from the GnuRadio mailing list archive at Nabble.com.