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Re: [Discuss-gnuradio] cycle/period detection of a cyclic/periodic trans

From: Firas Abbas
Subject: Re: [Discuss-gnuradio] cycle/period detection of a cyclic/periodic transmitter
Date: Wed, 22 Apr 2009 05:05:45 -0700 (PDT)


> On Wed, 4/22/09, kaleem ahmad <address@hidden> 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 

Is that clear ?

Best Regards,


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