Hi, > On Wed, 4/22/09, kaleem ahmad <kaleem_...@yahoo.com> 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 _______________________________________________ Discuss-gnuradio mailing list Discuss-gnuradio@gnu.org http://lists.gnu.org/mailman/listinfo/discuss-gnuradio
